Quantitative behavior of non-integrable systems (IV)
QQUANTITATIVE BEHAVIOROF NON-INTEGRABLE SYSTEMS (IV)
J. BECK, W.W.L. CHEN, AND Y. YANG
Abstract.
In this paper, there are two sections. In Section 7, we simplify theeigenvalue-based surplus shortline method for arbitrary finite polysquare surfaces.This makes it substantially simpler to determine the irregularity exponents ofsome infinite orbits, and quicker to find the escape rate to infinity of some orbitsin some infinite models. In Section 8, our primary goal is to extend the surplusshortline method, both this eigenvalue-based version as well as the eigenvalue-freeversion, for application to a large class of 2-dimensional flat dynamical systemsbeyond polysquares, including all Veech surfaces, and establish time-quantitativeequidistribution and time-quantitative superdensity of some infinite orbits in thesenew systems. More on the eigenvalue-based shortline method
The shortline method and the Edge-cutting Lemma.
Here in part (IV)we assume that the reader is more or less familiar with the earlier parts [2, 3, 4].The eigenvalue-based surplus shortline method has been developed in the specialcase of the L-surface in Sections 3–4 in [2, 3].For a 4-direction billiard flow in a general finite polysquare surface, we can apply unfolding introduced in [2] to convert the problem to one concerning a 1-directiongeodesic flow in some other finite polysquare surface. We therefore concentrate ourattention here on 1-direction geodesic flows in a finite polysquare surface.We shall show that the surplus shortcut-ancestor process, introduced earlier in[2, 3] in connection with the L-surface, can be adapted to every finite polysquaresurface with 1-direction geodesic flow. For a quick introduction, we first illustratethe method by applying it to a 1-direction geodesic flow with a particular quadraticirrational slope on the surface S given in the picture on the left in Figure 7.1.1. v v v v v v v v v v h h h h h h h h h h ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l Figure 7.1.1: the surface S with a gapWe shall compute the 2-step transition matrix, and also determine the irregularityexponent for this particular quadratic irrational slope. We shall illustrate the fact Mathematics Subject Classification.
Key words and phrases. geodesics, billiards, time-quantitative equidistribution, superdensity. a r X i v : . [ n li n . C D ] D ec BECK, CHEN, AND YANG that this 2-step transition matrix is usually “very redundant”. This will motivatethe so-called Edge-cutting Lemma which we shall formulate later.We shall consider 1-direction geodesic flow on S . Here the boundary pairingscome from simple perpendicular translations. The pair of vertical edges v andthe pair of horizontal edges h around the missing square represent a slightly lessstraightforward form of perpendicular translation.The surface S has 1 vertical street of length 1, 3 vertical streets of length 3, 1horizontal street of length 1, and 3 horizontal streets of length 3, so the street-LCMis equal to 3. The picture on the right in Figure 7.1.1 shows the streets where, forinstance, the entries ↔ (cid:108) α or α − for which the continued fraction expression of α has the special form α = [3 c ; 3 c , c , c , . . . ] = 3 c + 13 c + c + c ··· , (7.1.1)where the digits c i (cid:62) i (cid:62)
0, are integers.For illustration we consider here the simplest slope satisfying (7.1.1), namely α = [3; 3 , , , . . . ] = 3 + 13 + ··· = 3 + √ , (7.1.2)and its reciprocal α − . We study the long-term behavior of two particular geodesics V ( t ) and H ( t ) on S that start from the origin, which is some chosen vertex of oneof the square faces of S . Crucially, this guarantees that V ( t ) and H ( t ) are surplusshortlines of each other. The almost vertical geodesic V ( t ) has slope α , while thealmost horizontal geodesic H ( t ) has slope α − . Following Section 3 in [2], we brieflyelaborate on the details of the surplus shortline method in this particular case.We distinguish the 20 types of almost vertical units a i , 1 (cid:54) i (cid:54)
20, in the pictureon the left in Figure 7.1.2. a a a a a a a a a a a a a a a a a a a a a a a a a b b b b b b b b b b b b b b b b b b b b b b b b b Figure 7.1.2: the 20 types of almost vertical units and the 20 typesof almost horizontal units in the surface S We say that each of the almost vertical units a , a , a , a , a , a , a , a , a , a is of type ↑ , and starts at the bottom edge of a square face and ends at the top edgeof the same square face, while each of the almost vertical units a , a , a , a , a , a , a , a , a , a is of type − ↑ , and starts at the bottom edge of a square face and ends at the top edgeof an adjoining square face. Of particular interest is the unit a which hits the left ON-INTEGRABLE SYSTEMS (IV) 3 edge of the gap at some point and continues from the corresponding point on theidentified right edge of the gap.Similarly, we distinguish the 20 types of almost horizontal units b j , 1 (cid:54) j (cid:54) S exhibits a 45-degree reflection symmetry. But symmetryis not necessary, and indeed totally irrelevant, for the success of the shortline method.To emphasize this point, we have deliberately used labelling in the picture on theright in Figure 7.1.2 which is not a 45-degree reflection of the labelling in the pictureon the left in Figure 7.1.2.Applying a straightforward adaptation of the surplus shortline method in Section 3in [2], we can determine the surplus ancestor units of a i , 1 (cid:54) i (cid:54)
20. Similarly, wecan determine the surplus ancestor units of b j , 1 (cid:54) j (cid:54) a . It is not difficult to see from Figure 7.1.3,which shows only the top horizontal street of S , that a is the shortcut of an almosthorizontal detour crossing of the this horizontal street, made up of a fractional almosthorizontal unit b , full almost horizontal units b , b and b , and then a fractionalalmost horizontal unit b . We apply the Delete-End Rule, meaning that we keepthe initial fractional almost horizontal unit b as a full unit and discard the finalfractional almost horizontal unit b , and call b , b , b , b the ancestor units of a . a a b b b b b Figure 7.1.3: almost horizontal detour crossing for which a is the shortcutWe do likewise for a i , 2 (cid:54) i (cid:54)
20, and can summarize in the form a (cid:42) b , b , b , b ,a (cid:42) b , b , b ,a (cid:42) b , b , b , b , ... a (cid:42) b , b , b , b ,a (cid:42) b , b , b . (7.1.3)This leads to a transition matrix M ( S ), where the i -th row captures the informationin the i -th ancestor relation in (7.1.3), with the entry on the j -th column displayingthe multiplicity of b j .A similar exercise with the roles of the horizontal and vertical interchanged, againusing the Delete-End Rule, results in b (cid:42) a , a , a , a ,b (cid:42) a , a , a ,b (cid:42) a , a , a , a , ... b (cid:42) a , a , a , a ,b (cid:42) a , a , a . (7.1.4) BECK, CHEN, AND YANG
This leads to an analogous transition matrix M ( S ). Remark.
Instead of using the Delete-End Rule, we may also use the Keep-End Rule,meaning that we keep the final fractional unit as a full unit and discard the initialfractional unit. Both rules are for bookkeeping purposes only, as we do not wantto count any unit twice. Depending on which rule we use, (7.1.3) and (7.1.4), andhence also the matrices M ( S ) and M ( S ), may be a little different, but this willnot affect the subsequent argument.Composition of the two transition matrices arising from (7.1.3)–(7.1.4) leads tothe product matrix M ( S ) M ( S ). The transpose of this product matrix happensto be the 20-by-20 matrix M ( S ) = . (7.1.5)We use the transpose, because in the Edge-cutting Lemma, to be formulated later,we are interested in eigenvectors as column vectors of M ( S ). And it helps thatMATLAB, like any other linear algebra computer program, automatically computesright or column eigenvectors.We start with the 3 eigenvalues of (7.1.5) with the largest absolute values. ByMATLAB, the largest eigenvalue is λ = λ ( M ( S )) = 11 + 3 √
132 = (cid:32) √ (cid:33) = α , (7.1.6)in view of (7.1.2), with corresponding eigenvector v = v ( M ( S )) = ( v (1) , v (2) , v (3) , . . . , v (20)) T = ( c, , c, , c, , c, , c, , c, , c, , c, , c, , c, T , (7.1.7)where c = √ − . The second largest eigenvalue is λ = λ ( M ( S )) = 3 + 2 √ √ , (7.1.8) ON-INTEGRABLE SYSTEMS (IV) 5 with corresponding eigenvector v = v ( M ( S )) = ( v (1) , v (2) , v (3) , . . . , v (20)) T = ( c , c , c , c , c , c , c , c , c , c , c , c , c , c , c , c , c , c , c , T , (7.1.9)where c = 2 √ , c = − √ , c = 10 √ , c = − √ ,c = − √ , c = 5 √ − , c = 6 √ . The third largest eigenvalue is λ = λ ( M ( S )) = 3 + √ , (7.1.10)with corresponding eigenvector v = v ( M ( S )) = ( v (1) , v (2) , v (3) , . . . , v (20)) T = ( c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , c ∗ , T , (7.1.11)where c ∗ = √ , c ∗ = √ , c ∗ = − √ , c ∗ = − c ∗ ,c ∗ = − √ , c ∗ = − c ∗ , c ∗ = − √ . There are 3 other eigenvalues which are the algebraic conjugates of λ i , 1 (cid:54) i (cid:54) λ i , 1 (cid:54) i (cid:54)
3, are the only relevant eigenvalues of M ( S ), and theother 17 eigenvalues are irrelevant. We comment that the Jordan normal form ofthe matrix M ( S ) is simple, as the matrix has 20 different eigenvectors.The irregularity exponent of this special slope α given by (7.1.2) is, by definition,equal to κ ( α ) = log | λ | log | λ | = log(1 + √ √ . Before discussing the substantial redundancy and repetition of the coordinates in(7.1.7), (7.1.9) and (7.1.11), we make a short detour and state two general theorems.What we are doing here with the surface S and have done with the L-surfacein [2, 3] can be adapted for any finite polysquare surface, and we shall give somedetails in the next section. In particular, we obtain the following result. Theorem 7.1.1.
Let P be a finite polysquare surface, and let LCM( P ) denote thestreet-LCM, i.e., the least common multiple of the lengths of the horizontal andvertical streets of P . Consider a -direction geodesic flow, with a quadratic irrationalslope where the ordinary continued fraction digits are all divisible by LCM( P ) .Using the eigenvalue-based version of the surplus shortline method developed in [2, 3] , we can explicitly compute the irregularity exponent for such a slope.Combining the irregularity exponent with the method of zigzagging introduced inSection 3.3 in [2] , we can also describe, for a geodesic flow on P with such a slope,the time-quantitative behavior of the edge-cutting and face-crossing numbers, as wellas equidistribution relative to all convex sets. We remark that for the slopes in Theorem 7.1.1, Theorem 6.4.1 in [4] applies, andguarantees superdensity.
BECK, CHEN, AND YANG
In fact, Theorem 6.4.1 establishes superdensity for more slopes than those forwhich Theorem 7.1.1 provides the explicit values of the irregularity exponents. Ontop of the arithmetic condition “divisible by the street-LCM”, for superdensity inTheorem 6.4.1, we need the boundedness of the continued fraction digits given bybadly approximable slopes. On the other hand, we need in Theorem 7.1.1 thestronger condition of periodicity of the tail of the sequence of continued fractiondigits, so that the slope is a quadratic irrational.The irregularity exponent can be computed from the two eigenvalues with thelargest absolute values of an appropriate 2 d × d matrix, where d is the number ofsquare faces of the polysquare surface P .Since for every non-integrable polysquare surface the street-LCM is at least 2,the surplus shortline method does not have a chance of determining the irregularityexponent for every quadratic irrational slope.The reason why in the special case of the L-surface we are able to determinethe irregularity exponent for every quadratic irrational slope is two-fold. First, thestreet-LCM is equal to 2. More importantly, we combine the surplus and deficitversions of the shortline method according to the ± -even type continued fractionexpansion of the slope.If the street-LCM of a polysquare region/surface is equal to 2, then we call it a 2 -polysquare region/surface. One such region with the simplest boundary identificationvia perpendicular translation is called a 2 -polysquare snake . The special class of 2-polysquare snakes is surprisingly large, and Figure 7.1.4 gives an example. b bb b b bb b b b b bb b b Figure 7.1.4: 2-polysquare snake with capitals and its strictly alternating pathWe can easily characterize every 2-polysquare snake by making use of the simplegraph-theoretic term of “path”.Given a 2-polysquare snake, we can put a point at the center of every square face ofthe underlying polysquare, and call it the “capital of the square face”. Two capitalsare joined by a dashed line-segment if and only if the corresponding square facesshare a common edge. Then we obtain a non self-intersecting strictly alternatingh-v-path (or v-h-path) of the capitals, where “h” stands for a horizontal edge ofunit length and “v” stands for a vertical edge of unit length. The term “strictlyalternating” means that h is always preceded and followed by v, and v is alwayspreceded and followed by h, unless the path stops.We also have the converse, that every non self-intersecting strictly alternatingh-v-path (or v-h-path) corresponds to a 2-polysquare snake. Indeed, we cover everyvertex of the path with a unit size square such that the vertex is the capital of thesquare, and we may place some extra walls, denoted by boldface line-segments, ifnecessary.It can be shown that the special treatment of the L-surface in [2, 3] can be adaptedto every ± -even type continued fraction expansion of the slope. ON-INTEGRABLE SYSTEMS (IV) 7
Theorem 7.1.2.
Let P be a finite -polysquare surface. Then the LCM-divisibilitycondition in Theorem 7.1.1 may be dropped, so that Theorem 7.1.1 can be extendedto every quadratic irrational slope. To explain the substantial redundancy or repetition of the coordinates in theeigenvectors (7.1.7), (7.1.9) and (7.1.11), the trick is to visualize and represent thesurface S in Figure 7.1.1 as the period of a doubly periodic infinite polysquaresurface S ( ∞ ) in Figure 7.1.5 below. v v v v v v h h h h Figure 7.1.5: the doubly periodic polysquare surface S ( ∞ )The infinite polysquare surface S ( ∞ ) is essentially a square lattice of copiesof S with a noticeable difference – observe carefully the edge pairings in S ( ∞ ).For convenience, we say that S is the period surface of S ( ∞ ).Following the surplus shortline method, we study two special geodesics on theinfinite polysquare surface S ( ∞ ). The first geodesic V ∗ ( t ), t (cid:62)
0, starts from theorigin, namely a vertex of one of the square faces, and has slope α defined in (7.1.2).The second geodesic H ∗ ( t ), t (cid:62)
0, also starts from the origin and has reciprocalslope α − . Note that the projections of V ∗ ( t ) and H ∗ ( t ) on the period surface S give back the geodesics V ( t ) and H ( t ) defined earlier.A particle traveling on the almost vertical geodesic V ∗ ( t ), t (cid:62)
0, of the doublyperiodic infinite surface S ( ∞ ) moves from one copy of the period S to anothercopy of S , back and forth, up and down, and generally wanders around.Note that in S ( ∞ ), each copy of S has 4 immediate neighbors: we shall refer tothem as the North, South, East and West neighbors. Traveling along the geodesic V ∗ ( t ), t (cid:62)
0, is somewhat similar to a symmetric random walk on the 2-dimensionaldoubly periodic lattice Z . Escaping from a copy of S to the North means followinga unit of type a or a in Figure 7.1.2, while escaping to the South means a followinga unit of type a or a . Similarly, escaping to the East means following a unit oftype a , while escaping to the West means following a unit of type a .To describe an arbitrary finite segment of the geodesic V ∗ ( t ), t (cid:62)
0, of slope α on the infinite polysquare surface S ( ∞ ), it suffices to consider its projection on theperiod surface S . The difference between number of North-escapes and the numberof South-escapes gives the vertical change on S ( ∞ ), while the difference betweenthe number of East-escapes and the number of West-escapes gives the horizontalchange on S ( ∞ ). This is how we can determine the escape rate to infinity of thespecial geodesic V ∗ ( t ), t (cid:62)
0, on S ( ∞ ) from the behavior of its projection V ( t ), t (cid:62)
0, on the period surface S .Since the two special geodesics V ( t ) and H ( t ) are surplus shortlines of each otheron the period surface S , the time evolution of V ( t ), t (cid:62)
0, of slope α and startingfrom the origin is described by powers of the 2-step transition matrix M ( S ) given by BECK, CHEN, AND YANG (7.1.5). Since M ( S ) is diagonalizable, it is particularly easy to express the numberof types of units by using the relevant eigenvalues and their eigenvectors. Since thenumber of types of units is expressed as a linear combination of the powers of theeigenvalues, in case of large powers the contribution of the irrelevant eigenvalues isclearly negligible.Recall that escaping to the North means following a unit of type a or a inFigure 7.1.2, and escaping to the South means following a unit of type a or a .It follows that for the vertical change, we need to know the 3-rd, 4-th, 15-th and16-th coordinates of the relevant eigenvectors v i , i = 1 , ,
3, in (7.1.7), (7.1.9) and(7.1.11). Escaping to the East means following a unit of type a , and escaping to theWest means following a unit of type a . It follows that for the horizontal change,we need to know the 7-th and 11-th coordinates of the relevant eigenvectors.Recall that v i ( j ) denotes the j -th coordinate of v i , i = 1 , , j = 1 , , . . . , v i (3) + v i (4) = v i (15) + v i (16) , i = 1 , , , (7.1.12)and v i (7) = v i (11) , i = 1 , , . (7.1.13)The reader may perhaps find (7.1.12)–(7.1.13) a surprising coincidence. However,there is a simple explanation why these equalities must hold. The underlying reasonis the fact that the geodesic V ∗ ( t ), t (cid:62)
0, of the 3-square-maze S ( ∞ ) satisfiesthe condition of Theorem 6.5.1 and exhibits super-slow logarithmic escape rate toinfinity; see the Remark after the proof of Theorem 6.5.1 in Section 6.5 of [4].Meanwhile, a violation of (7.1.12)–(7.1.13) would imply a power-size escape rate toinfinity, exponentially larger than logarithmic escape rate to infinity. Remark.
The equalities (7.1.12)–(7.1.13) remain true even if in obtaining (7.1.3)–(7.1.4) we replace the Delete-End Rule by the Keep-End Rule, or mix the two rulesarbitrarily. It follows from the fact that the relevant eigenvalues and eigenvectorsremain the same under these changes.The Edge-cutting Lemma, which we shall formulate later, is simply a far-reachinggeneralization of the equalities (7.1.12)–(7.1.13).It can be seen from (7.1.7), (7.1.9) and (7.1.11) that (7.1.13) can be extended toinclude the 9-th coordinate, so that v i (7) = v i (9) = v i (11) , i = 1 , , . (7.1.14)Note from Figure 7.1.2 that the entries in (7.1.14) represent almost vertical unitsof type − ↑ on the second horizontal street of S , of length 3. For the first horizontalstreet and the third horizontal street, both also of length 3, we have respectively theanalogs v i (1) = v i (3) = v i (5) , i = 1 , , , (7.1.15)and v i (13) = v i (15) = v i (17) , i = 1 , , , (7.1.16)both of which hold, as can be seen from (7.1.7), (7.1.9) and (7.1.11).It can also be seen from (7.1.7), (7.1.9) and (7.1.11) that (7.1.12) can be extendedto v i (3) + v i (4) = v i (15) + v i (16) = v i (19) + v i (20) , i = 1 , , . (7.1.17)Note from Figure 7.1.2 that the entries in (7.1.17) represent almost vertical unitsof type ↑ or − ↑ on the third vertical street of S , of length 3. Note that we havechosen almost vertical units of type − ↑ that intersect the left edge of each squareface. In view of (7.1.15) and (7.1.16), it would have made no difference if we hadchosen instead any that intersects the right edge of the same square face. For the ON-INTEGRABLE SYSTEMS (IV) 9 second vertical street and the fourth vertical street, both also of length 3, we haverespectively the analogs v i (1) + v i (2) = v i (9) + v i (10) = v i (13) + v i (14) , i = 1 , , , (7.1.18)and v i (5) + v i (6) = v i (11) + v i (12) = v i (17) + v i (18) , i = 1 , , , (7.1.19)both of which hold, as can be seen from (7.1.7), (7.1.9) and (7.1.11).The Edge-cutting Lemma basically says that for any finite polysquare surface with1-direction geodesic flow, analogs of the equalities (7.1.14)–(7.1.19) hold. Section 7.2contains a proof of this fact. In fact, we prove more, see Theorem 7.2.2.We conclude this section by giving a simple proof of the Edge-cutting Lemma inthe simplest special case of the L-surface. We include it, because this simple proofalready illustrates quite well the method we use in Section 7.2.We go back to Section 3 in [2], to the L-surface with a 1-direction geodesic flowwith slope α = 1 + √
2. Figure 7.1.6 below shows the edges of the L-surface as wellas the almost vertical units. v v v v v h h h h h h h h h h h h h h h h h ∗ Figure 7.1.6: the L-surface and almost vertical unitsCorresponding to the matrix M ( S ) for the polysquare surface S , we have the2-step transition matrix M = M T M T = . (7.1.20)In (7.1.20), we have used the lexicographic order for the almost vertical units h h , h h , h h , h h , h h , h h ∗ . (7.1.21)The L-surface has precisely one vertical and one horizontal street of length greaterthan 1. If v = ( v , v , v , v , v , v ) is a relevant eigenvector of M , then the coor-dinates correspond to the 6 almost vertical units in (7.1.21). It follows that theL-surface analog of the equalities (7.1.14)–(7.1.16) is v = v , (7.1.22)and the L-surface analog of the inequalities (7.1.17)–(7.1.19) is v + v = v + v . (7.1.23)These two equations together give an invariant subspace of the matrix M . To seethis, suppose that M v T = w T = ( w , w , w , w , w , w ) T . Then w = 2 v + v + v + v ,w = v + v + 2 v + 3 v ,w = v + v + 2 v + 3 v ,w = v + v + v + 2 v ,w = 2 v + v + v + v + v + v ,w = 2 v + 3 v . It is easy to see that w = w follows from v = v . On the other hand, we have w + w = 3 v + 2 v + v + v + 2 v + 3 v ,w + w = 2 v + v + v + v + 3 v + 4 v , so that ( w + w ) − ( w + w ) = ( v + v ) − ( v + v ) , and so w + w = w + w follows from v + v = v + v .The 2-step transition matrix M has 6 eigenvalues and 6 independent eigenvectorsin R . The eigenvalues, in order of decreasing absolute value, are λ = 3 + 2 √ , λ = 3 + √ , λ = 1 ,λ = 1 , λ = 3 − √ , λ = 3 − √ . (7.1.24)Suppose that the corresponding eigenvectors of M are respectively v i , 1 (cid:54) i (cid:54) λ → v = (1 , √ , √ , , √ , T ,λ → v = (cid:32) − , √ , − √ , − , √ − , (cid:33) T ,λ → v = (0 , − , , , , T ,λ → v = ( − , , , , − , T ,λ → v = (cid:32) − , − √ − , √ , − , − √ , (cid:33) T ,λ → v = (1 , −√ , −√ , , −√ , T . Note now that the 4-dimensional invariant subspace of the matrix M given by theequations (7.1.22) and (7.1.23) contains the eigenvectors v and v correspondingto the two relevant eigenvalues λ and λ . It also contains the eigenvectors v and v of the two conjugate eigenvalues λ and λ .This simple example illustrates what we call an invariant subspace argument. Inthe next section we use a similar but more sophisticated version of this to prove theEdge-cutting Lemma and more.Finally observe that the Edge-cutting Lemma has the following vague intuitivemeaning. If the surplus shortline method works for a 1-direction geodesic flow of afinite polysquare surface with some fixed slope, then the edge-cutting numbers ofthe edges in the same street are “nearly the same”. We shall clarify this later. ON-INTEGRABLE SYSTEMS (IV) 11
Reducing the 2-step transition matrix: the street-spreading matrix.
We see from the examples in Section 7.1 that the 2-step transition matrix of theshortline method is quite redundant. Here we elaborate on this. What we areinterested in is the long-term behavior of the geodesics, and when the shortlinemethod works, it suffices to consider the relevant eigenvalues, those with absolutevalue greater than 1, and the corresponding eigenvectors. For example, while the2-step transition matrix M ( S ) in (7.1.5) has 20 eigenvalues, only 3 of them arerelevant. Moreover, each of the corresponding 20-dimensional eigenvectors (7.1.7),(7.1.9) and (7.1.11) has at most 8 distinct coordinates, with some multiplicities.Accordingly, we introduce a general reduction process that, roughly speaking,eliminates the irrelevant part of the 2-step transition matrix M . We shall give arecipe of how we can read out the relevant eigenvalues and their eigenvectors froma smaller matrix called the street-spreading matrix .We consider a vector space W with basis made up of all the distinct types ofalmost vertical units of P , of dimension equal to twice the number of distinct squarefaces of P . We shall show that the equations in the Edge-cutting Lemma define asubspace of W , and we want to show that this subspace is M -invariant and containsall the relevant eigenvectors. Thus the basic idea is to find such a relevant M -invariant subspace, and a convenient basis of it, leading to the usually substantiallysmaller street-spreading matrix. A proof of the Edge-cutting Lemma will comeas a byproduct of the reduction process. Theorem 7.2.2, the main result, will beformulated towards the end of this section. The Shortcut-Ancestor Process.
Let P be a given finite polysquare surface, with d square faces. Let m be any fixed integer multiple of the lengths of the horizontalstreets of P , and let n be any fixed integer multiple of the lengths of the verticalstreets of P .We adopt the following convention.A typical square face of P , on the i -th horizontal street and j -th vertical street, isdenoted by S i,j . For any fixed i , there is a finite set J i of indices j such that j ∈ J i if and only if S i,j ⊂ P , so that (cid:91) j ∈ J i S i,j is the i -th horizontal street of P . For any fixed j , there is a finite set I j of indices i such that i ∈ I j if and only if S i,j ⊂ P , so that (cid:91) i ∈ I j S i,j is the j -th vertical street of P . Remark.
Our convention here is an over-simplification of the general situation, asit is in fact possible for two or more distinct square faces of a polysquare surface tolie on the same horizontal street and the same vertical street simultaneously. Wehave chosen to adopt the present simplification and convention here as the notationis somewhat simpler and rather convenient, and enough to enable us to study manyof the polysquare surfaces of interest. In general, the notation S i,j does not allow usto distinguish between two distinct square faces that may lie on the i -th horizontalstreet and the j -th vertical street, and we shall have to introduce an extra parameterto enable us to identify individual square faces of the polysquare surface in question.Indeed, the first occasion in this paper where our notation becomes inadequate isat the end of Section 8.2 when we discuss the regular octagon surface, and we shalldeal with the problem on a case by case basis when it arises. We consider two types of almost vertical units in S i,j . Both types start from thebottom edge of S i,j . However, type ↑ i,j ends on the top edge of S i,j , whereas type − ↑ i,j exits S i,j through the right edge. We also consider two types of almost horizontalunits in S i,j . Both types end on the right edge of S i,j . However, type → i,j startsfrom the left edge of S i,j , whereas type + → i,j enters S i,j through the bottom edge.Figure 7.2.1: the square face S i,j , with units of types ↑ i,j , − ↑ i,j , → i,j , + → i,j We now consider a 1-direction almost vertical geodesic V on P , starting at avertex of P and with slope α = [ n ; m, n, m, . . . ] = n + 1 m + n + m + ··· , (7.2.1)so that n (cid:54) α < n + 1. The geodesic V is made up of almost vertical units of type ↑ i,j and − ↑ i,j , all with slope α . Thus we may consider a vector space W , with basis W = {↑ i,j , − ↑ i,j : S i,j ⊂ P} . (7.2.2)The shortline of V is a 1-direction almost horizontal geodesic H on P , starting atthe same vertex of P and with slope α − , where α = [ m ; n, m, . . . ] = m + 1 n + m + ··· , so that m (cid:54) α < m + 1. The geodesic H is made up of almost horizontal units oftype → i,j and + → i,j , all with slope α − . Thus we may consider a vector space W (cid:48) ,with basis W (cid:48) = {→ i,j , + → i,j : S i,j ⊂ P} . The shortline of H is the original 1-direction almost vertical geodesic V on P , withslope α . Thus we return to the vector space W , with basis W . Remark.
Note that V and H are mutual shortlines. Note the special property of α ,that its continued fraction expansion has period 2.Our task is to start with W and identify the 2-step ancestors in W of each of thealmost vertical units. Let ANC denote one step of this ancestor process. Step 1 . We consider each of the almost vertical units in W , and determine itsancestors in W (cid:48) using the Delete-End Rule, in the same spirit as in (7.1.3). Thisleads to a transition matrix M , where each row captures the information concerningthe ancestors of a particular almost vertical unit, with the columns displaying themultiplicities of the individual ancestor almost horizontal units. Indeed, M T is thematrix for this ancestor process W → W (cid:48) , where the coefficient vectors are takenas column vectors. Note that M T is a 2 d × d matrix.Thus for every i , j such that S i ,j ⊂ P , using the Delete-End Rule, we have theancestor relationshipsANC( {↑ i ,j } ) = { + → i ,j } ∪ {→ i ,j : j ∈ J ∗ i } \ {→ i ,j } , (7.2.3)ANC( { − ↑ i ,j } ) = { + → i ,j } ∪ {→ i ,j : j ∈ J ∗ i } , (7.2.4)where the ∗ in J ∗ i denotes that each edge of J i is counted with multiplicity m | J i | − ,where | J i | denotes the number of distinct elements of the set J i . We also adopt theconvention that unions and complements are taken with appropriate multiplicities. ON-INTEGRABLE SYSTEMS (IV) 13
Step 2 . We consider each of the almost horizontal units in W (cid:48) , and determineits ancestors in W using the Keep-End Rule, so not quite in the same spirit asin (7.1.4). This leads to a transition matrix M , where each row captures theinformation concerning the ancestors of a particular almost horizontal unit, withthe columns displaying the multiplicities of the individual ancestor almost verticalunits. Indeed, M T is the matrix for this ancestor process W (cid:48) → W , where thecoefficient vectors are taken as column vectors. Note that M T is a 2 d × d matrix.Thus for every i , j such that S i ,j ⊂ P , using the Keep-End Rule, we have theancestor relationshipsANC( {→ i ,j } ) = { − ↑ i ,j } ∪ {↑ i,j : i ∈ I ∗ j } \ {↑ i ,j } , (7.2.5)ANC( { + → i ,j } ) = { − ↑ i ,j } ∪ {↑ i,j : i ∈ I ∗ j } , (7.2.6)where the ∗ in I ∗ j denotes that each edge of I j is counted with multiplicity n | I j | − ,where | I j | denotes the number of distinct elements of the set I j . We also adopt theconvention that unions and complements are taken with appropriate multiplicities. Step 3 . Finally, we combine the two steps and end up with a 2-step transitionmatrix M = ( M M ) T , of size 2 d × d , in the same spirit as in (7.1.5). Thuscombining the ancestor relationships (7.2.3)–(7.2.6), we deduce thatANC(ANC( {↑ i ,j } ))= (cid:0) { − ↑ i ,j } ∪ {↑ i,j : i ∈ I ∗ j } (cid:1) ∪ (cid:0) { − ↑ i ,j : j ∈ J ∗ i } ∪ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } \ {↑ i ,j : j ∈ J ∗ i } (cid:1) \ (cid:0) { − ↑ i ,j } ∪ {↑ i,j : i ∈ I ∗ j } \ {↑ i ,j } (cid:1) = {↑ i ,j } ∪ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } ∪ { − ↑ i ,j : j ∈ J ∗ i } \ {↑ i ,j : j ∈ J ∗ i } , (7.2.7)ANC(ANC( { − ↑ i ,j } ))= (cid:0) { − ↑ i ,j } ∪ {↑ i,j : i ∈ I ∗ j } (cid:1) ∪ (cid:0) { − ↑ i ,j : j ∈ J ∗ i } ∪ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } \ {↑ i ,j : j ∈ J ∗ i } (cid:1) = { − ↑ i ,j } ∪ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j }∪ {↑ i,j : i ∈ I ∗ j } ∪ { − ↑ i ,j : j ∈ J ∗ i } \ {↑ i ,j : j ∈ J ∗ i } . (7.2.8) Remark.
We choose this particular convention regarding the Delete-End Rule andKeep-End Rule, as it makes the matrix reduction particularly simple. For instance,if we use this particular convention for the surface S , then in the 2-step transitionmatrix M , the 14 eigenvalues of absolute value 1 are all the same and equal to 1.Indeed, we shall give a good reason later, when we discuss a simpler approach tothe process, why they are all equal to 1.Suppose that λ , . . . , λ s are the eigenvalues of the 2-step transition matrix M ,with multiplicities d , . . . , d s respectively, and where | λ | (cid:62) . . . (cid:62) | λ s | . Then clearly d + . . . + d s = 2 d . Furthermore, the space C d can be decomposed into a direct sum C d = W ⊕ . . . ⊕ W s , where each W i , i = 1 , . . . , s , is an M -invariant subspace of C d and also contains aneigenvector Ψ i corresponding to the eigenvalue λ i . If d i = 1, then Ψ i generates W i and gives rise to a basis of W i . If d i >
1, then we can find a basis Ψ i,j , j = 1 , . . . , d i ,of W i , with Ψ i = Ψ i, . Thus the collectionΨ i,j , i = 1 , . . . , s, j = 1 , . . . , d i , gives rise to a basis of C d .The almost vertical geodesic V starts at a vertex of P , and it starts with a finitesuccession of almost vertical units. Let w denote the column coefficient vector of this finite collection of units with respect to the basis W . Then we can findcoefficients c i,j ∈ C , i = 1 , . . . , s , j = 1 , . . . , d i , such that w = s (cid:88) i =1 d i (cid:88) j =1 c i,j Ψ i,j . (7.2.9)Suppose that w r = M r w . Then w r = s (cid:88) i =1 d i (cid:88) j =1 c i,j M r Ψ i,j . (7.2.10)We shall consider the special case where for every eigenvalue λ i of M with | λ i | > M -invariant subspace W i has a basis consisting entirely of eigenvectors of M witheigenvalue λ i . Let λ i , i = 1 , . . . , s , be these eigenvalues. ThenΨ i,j , i = 1 , . . . , s , j = 1 , . . . , d i , are all eigenvectors of M . Furthermore, | λ i | (cid:54) i = s + 1 , . . . , s .In this case, we have w r = s (cid:88) i =1 d i (cid:88) j =1 c i,j λ ri Ψ i,j + O ( r D − ) , (7.2.11)where D = max { d , . . . , d s } , and the error term O ( r D − ) gives an upper bound onthe absolute value of the coordinates of the missing vectors. To see this, note thatthe matrix M is similar to a matrix of the form J ( τ , e ) . . . J ( τ u , e u ) , where a typical Jordan block J = J ( τ, e ) is an e × e matrix of the form J ( τ, e ) = τ
1. . . . . .. . . 1 τ , where every entry on the diagonal is equal to an eigenvalue τ of M , every entry onthe superdiagonal is equal to 1, and every other entry is equal to 0.The contribution to the error term in (7.2.11) comes from those eigenvalues of M with absolute value at most 1. These are related to Jordan blocks J = J ( τ, e ) with | τ | (cid:54) e (cid:54) D . The matrix J r is e × e and upper-triangular, and the entry onrow i and column j , with j (cid:62) i , is given by (cid:18) rj − i (cid:19) τ r − ( j − i ) , where the binomial coefficient is equal to 0 if j − i > r . The bound O ( r D − ) followson observing that (cid:18) rj − i (cid:19) (cid:54) r e − (cid:54) r D − and | τ r − ( j − i ) | (cid:54) . Important Remark.
It is crucial to bring those eigenvalues with the largest absolutevalues into play. The coefficient in (7.2.11) of the eigenvector corresponding to thelargest eigenvalue is non-zero, since the left hand side of (7.2.11) contains termsof order of magnitude λ r . For the coefficient of the eigenvector corresponding tothe second largest eigenvalue, this is less obvious. The trick here is to start with ON-INTEGRABLE SYSTEMS (IV) 15 an almost vertical unit from the chosen vertex. If the coefficient corresponding tothe second eigenvector is non-zero, then we have a good initial vector w . If thecoefficient is zero, then we add the next unit, and keep on doing so but stop assoon as we get a non-zero coefficient. Take this as our starting succession of almostvertical units, and use the corresponding good w . This process must stop afterbounded time, depending only on P and the slope α , as soon every basis element in W will come into play. Of the 2 d types of almost vertical units, it is clear that atleast one of them gives rise to a non-zero coefficient. Finding a Relevant M -invariant Subspace and the Edge-cutting Lemma. The above method in rather laborious. Finding the eigenvalues and eigenvectors ofthe 2-step transition matrix M , of size 2 d × d , is not a very pleasant task.It is clear that the expression (7.2.11) is dominated by the terms arising fromthose eigenvalues λ i , i = 1 , . . . , s , with absolute values exceeding 1. We call thesethe relevant eigenvalues, and the remaining eigenvalues make little to no contribu-tion. We next seek a simpler way of finding these relevant eigenvalues and theircorresponding eigenvectors.For any set S of almost vertical units in P , counted with multiplicity, let [ S ]denote the column coefficient vector of S with respect to the basis W . Note that[ S ] ∈ C d , where d is the number of square faces of P .As the matrix M is the transition matrix of the 2-step ancestor process withrespect to the basis W , (7.2.7) and (7.2.8) can be rewritten in the form( M − I )[ {↑ i ,j } ] = [ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } ]+ [ { − ↑ i ,j : j ∈ J ∗ i } ] − [ {↑ i ,j : j ∈ J ∗ i } ] , (7.2.12)( M − I )[ { − ↑ i ,j } ] = [ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } ] + [ {↑ i,j : i ∈ I ∗ j } ]+ [ { − ↑ i ,j : j ∈ J ∗ i } ] − [ {↑ i ,j : j ∈ J ∗ i } ] . (7.2.13)Write u i = [ {↑ i,j : j ∈ J ∗ i , i ∈ I ∗ j } ] , (7.2.14) v i = [ { − ↑ i ,j : j ∈ J ∗ i } ] − [ {↑ i ,j : j ∈ J ∗ i } ] , (7.2.15) z j = [ {↑ i,j : i ∈ I ∗ j } ] . (7.2.16)Then u i , v i , z j ∈ C d , and (7.2.12) and (7.2.13) can be expressed in the form( M − I )[ {↑ i ,j } ] = u i + v i , (7.2.17)( M − I )[ { − ↑ i ,j } ] = u i + v i + z j . (7.2.18)Motivated by the equations (7.2.17) and (7.2.18), we consider Im( M − I ), theimage of the matrix M − I in C d . Note that this is an M -invariant subspace of C d .Indeed, for every vector w ∈ C d , we have M (( M − I ) w ) = ( M − M ) w = ( M − I )( M w ) = ( M − I ) w (cid:48) where w (cid:48) ∈ C d satisfies w (cid:48) = M w . It is clear from the equations (7.2.17) and(7.2.18) that the subspace Im( M − I ) is generated by the elements of the form u i , v i , z j ∈ Im( M − I ) ⊂ C d . (7.2.19) Lemma 7.2.1.
All eigenvectors corresponding to relevant eigenvalues of M belongto the subspace Im( M − I ) ⊂ C d .Proof. Suppose that w ∈ C d satisfies M w = λ w with | λ | >
1. Then( λ − w = ( M − I ) w ∈ Im( M − I ) , so that w ∈ Im( M − I ). (cid:3) Theorem 7.2.1 (“edge-cutting lemma”) . Let P be a finite polysquare surface. Let m be any fixed integer multiple of the lengths of the horizontal streets of P , and let n be any fixed integer multiple of the lengths of the vertical streets of P . Considera -direction almost vertical geodesic V on P , with slope α given by (7.2.1) . Let M denote the transition matrix of the -step ancestor process with respect to thebasis W , where W is given by (7.2.2) . Then the following hold: (i) For any two distinct almost vertical units − ↑ i ,j in a square face S i ,j and − ↑ i ,j in a square face S i ,j on the same horizontal street of P and every eigenvec-tor corresponding to a relevant eigenvalue of M , the two entries in the eigenvectorcorresponding to − ↑ i ,j and − ↑ i ,j are equal. (ii) For any two distinct pairs of almost vertical units ↑ i ,j , − ↑ i ,j in a square face S i ,j and ↑ i ,j , − ↑ i ,j in a square face S i ,j on the same vertical street of P and everyeigenvector corresponding to a relevant eigenvalue of M , the sum of the two entriesin the eigenvector corresponding to ↑ i ,j , − ↑ i ,j in the square face S i ,j is equal to thesum of the two entries in the eigenvector corresponding to ↑ i ,j , − ↑ i ,j in the squareface S i ,j .Proof. In view of Lemma 7.2.1, it is enough to concentrate our attention on thesubspace Im( M − I ). Since this subspace is generated by the vectors in (7.2.19), itsuffices to check that these generating vectors all satisfy (i) and (ii). Our argumentis reduced to a simple case study.Note that u i does not involve units of type − ↑ i,j , so (i) holds by default. On theother hand, u i satisfies (ii), due to cyclic vertical symmetry. A similar argumentapplies to z j .On the other hand, v i satisfies (i), due to cyclic horizontal symmetry. Clearly italso satisfies (ii), as the count for each − ↑ i ,j cancels the count for ↑ i ,j .This completes the proof of the Edge-cutting Lemma. (cid:3) An Important Subspace.
Let h = h ( P ) and v = v ( P ) denote respectively thenumber of horizontal and vertical streets in P .Let V be the subspace of C d generated by u i , v i , 1 (cid:54) i (cid:54) h , defined in (7.2.14)and (7.2.15), so that the dimension of V is at most 2 h . Lemma 7.2.2.
We have
Im(( M − I ) ) ⊂ V .Proof. Since Im(( M − I ) ) = ( M − I )(Im( M − I )), and the subspace Im( M − I ) isgenerated by the vectors in (7.2.19), it suffices to show that( M − I ) u i ∈ V , ( M − I ) v i ∈ V , ( M − I ) z j ∈ V . (7.2.20)From (7.2.14) and (7.2.16), it is clear that both u i and z j involve only the almostvertical units of various types ↑ i,j , so it follows from (7.2.17) that the first and thirdstatements in (7.2.20) hold. On the other hand, it follows from (7.2.14)–(7.2.18)that ( M − I ) v i = (cid:88) j ∈ J ∗ i z j = [ {↑ i,j : i ∈ I ∗ j , j ∈ J ∗ i } ] = u i , (7.2.21)and the second statement in (7.2.20) follows immediately. (cid:3) By definition V is a subspace of Im( M − I ). Thus the following lemma is anextension of Lemma 7.2.1. Lemma 7.2.3.
The vector space V is an M -invariant subspace of C d , and containsthe eigenvectors corresponding to each of the eigenvalues λ (cid:54) = 1 of M .Proof. To prove that V is M -invariant, it suffices to check that M u i ∈ V and M v i ∈ V . ON-INTEGRABLE SYSTEMS (IV) 17
This is trivial, since we know from (7.2.20) that ( M − I ) u i ∈ V and ( M − I ) v i ∈ V .Next suppose that w ∈ C d satisfies M w = λ w for some λ (cid:54) = 1. Then( λ − w = ( M − I ) w ∈ Im(( M − I ) ) , so that w ∈ Im(( M − I ) ) ⊂ V , where in the last step we have used Lemma 7.2.2. (cid:3) Next we study how the matrix M acts on V . First of all, the equation (7.2.21)implies M v i = u i + v i . (7.2.22)We already know that M u i ∈ V , so that M u i is a linear combination of the vectors u i , v i , 1 (cid:54) i (cid:54) h . Next we shall find these coefficients explicitly, but this processis much more complicated than (7.2.22), as the coefficients depend heavily on thecombinatorial arrangement of the square faces on the given polysquare surface P .In fact, they depend on the given triple ( P ; m, n ). This will eventually lead us to thedesired h × h street-spreading matrix S = S ( P ; m, n ) which contains all the “relevantinformation”.Combining (7.2.14) and (7.2.17), one can determine the desired coefficients byusing a particular algorithm. The abstract/formal definition of this algorithm inthe general case is somewhat complicated. As with most algorithms, the best wayto learn it is to study a few concrete examples with figures. We strongly recommendthe reader to study the next two examples, after which the abstract/formal definitionbecomes almost self-explanatory. Algorithm to Find the Street-spreading Matrix.
The algorithm can easily besummarized in a nutshell: spread the horizontal streets vertically . We illustrate partof the recipe by two examples.
Example . Consider again the polysquare surface S , given in Figure 7.1.1, andconsider a 1-direction geodesic starting from some vertex of S with slope α givenby (7.2.1) with m = n = 3.We shall number the horizontal streets from top to bottom, and the vertical streetsfrom left to right.Consider the first horizontal street at the top, corresponding to u . We now use(7.2.14), and since J ∗ = { , , } , it follows that u concerns almost vertical units oftype ↑ on the vertical streets 2 , ,
4. In other words, we start with the first horizontalstreet, and spread along every vertical street that intersects this horizontal street.The picture on the left in Figure 7.2.2 shows a listing of all the almost vertical unitsof type ↑ along these vertical streets. Applying (7.2.17), we have( M − I ) u = ( M − I )[ {↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , } ]= 3( u + v ) + 2( u + v ) + 3( u + v ) + ( u + v ) . (7.2.23) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , u u Figure 7.2.2: S with m = n = 3: almost vertical units of type ↑ in u , u For the second horizontal street, corresponding to u , highlighted in the pictureon the right in Figure 7.2.2, we have J ∗ = { , , } , and a similar argument gives( M − I ) u = ( M − I )[ {↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , , ↑ , , ↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , } ]= 2( u + v ) + 5( u + v ) + 2( u + v ) . (7.2.24)The multiple 3 for ↑ , in the square face S , comes from the fact that I ∗ = { , , } ,since n = 3 but the vertical street has length 1.For the third horizontal street, corresponding to u , highlighted in the picture onthe left in Figure 7.2.3, we have J ∗ = { , , } , and a similar argument gives( M − I ) u = ( M − I )[ {↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , } ]= 3( u + v ) + 2( u + v ) + 3( u + v ) + ( u + v ) . (7.2.25) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , u u Figure 7.2.3: S with m = n = 3: almost vertical units of type ↑ in u , u For the fourth horizontal street, corresponding to u , highlighted in the pictureon the right in Figure 7.2.3, we have J ∗ = { , , } , and a similar argument gives( M − I ) u = ( M − I )[ {↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , , ↑ , } ]= 3( u + v ) + 3( u + v ) + 3( u + v ) . (7.2.26)We now define the street-spreading matrix by S = , (7.2.27)where the j -th column comes from the coefficients of ( u i + v i ), i = 1 , , ,
4, in theexpression for ( M − I ) u j , j = 1 , , ,
4, as given by (7.2.23)–(7.2.26).Combining (7.2.22) and (7.2.23)–(7.2.26), we obtain M u = 3( u + v ) + 2( u + v ) + 3( u + v ) + ( u + v ) + u ,M u = 2( u + v ) + 5( u + v ) + 2( u + v ) + u ,M u = 3( u + v ) + 2( u + v ) + 3( u + v ) + ( u + v ) + u ,M u = 3( u + v ) + 3( u + v ) + 3( u + v ) + u ,M v = u + v ,M v = u + v ,M v = u + v ,M v = u + v . ON-INTEGRABLE SYSTEMS (IV) 19
For any w ∈ V , if we write w = a u + a u + a u + a u + b v + b v + b v + b v , and M w = c u + c u + c u + c u + d v + d v + d v + d v , then c c c c d d d d = a a a a b b b b , so the 8 × M | V = (cid:18) S + I I S I (cid:19) . We shall return to this example later.
Example . Consider the polysquare surface S , given in Figure 7.2.4. v v v v v v v v v v v v v v v v h h h h h h h h h h h h h h h h h h h h h h h h h h Figure 7.2.4: S with edge pairingsThis has 6 horizontal streets of lengths 3 , , , , ,
6, and 9 vertical streets oflengths 1 , , , , , , , ,
3, as illustrated in Figure 7.2.5 where, for instance, ↔ (cid:108) ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l l l l l l l l l l l l l l Figure 7.2.5: the horizontal and vertical streets of S Consider a 1-direction geodesic starting from some vertex of the polysquare surface S with slope α given by (7.2.1) with m = n = 12.The polysquare surface S has 25 square faces, so any 2-step transition matrix M has size 50 ×
50, making it an onerous task to find any eigenvalue, even withthe help of MATLAB. It is clearly much simpler to find the 6 × First of all, we define the matrices u i , v i , i = 1 , , , , ,
6, according to (7.2.14)and (7.2.15).Consider the horizontal street corresponding to u , as highlighted in the left halfof Figure 7.2.6. We see that J = { , , } and J ∗ = { , , , , , , , , , , , } , since m = 12, and u concerns almost vertical units on the vertical streets 2 , , ↑ along these vertical streets.Applying (7.2.17), we have( M − I ) u = ( M − I )[ { ↑ , , ↑ , , ↑ , } ]+ ( M − I )[ { ↑ , } ] + ( M − I )[ { ↑ , } ]+ ( M − I )[ { ↑ , , ↑ , , ↑ , } ] + ( M − I )[ { ↑ , , ↑ , } ]= 44( u + v ) + 16( u + v ) + 12( u + v )+ 44( u + v ) + 28( u + v ) . (7.2.28)We need to explain the multiples in Figure 7.2.6. Observe the horizontal streetcorresponding to u consists of 3 square faces. As m = 12, the correct multiplicityis 12 / S , consists of 4 square faces. As n = 12, the correct multiplicity is 12 / × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × × × × × u u Figure 7.2.6: S with m = n = 12: almost vertical units of type ↑ in u , u Remark.
We always have | I ∗ j | = n for every j = 1 , . . . , v , and | J ∗ i | = m for every i = 1 , . . . , h . The first of these corresponds to the fact that every almost verticaldetour crossing of V travels a vertical distance between n and n + 1. The secondof these corresponds to the fact that every almost horizontal detour crossing of H travels a horizontal distance between m and m + 1.With J ∗ = { , , , , , , , , , , , } , for the horizontal street correspondingto u , as highlighted in the right half of Figure 7.2.6, a similar argument gives( M − I ) u = 44( u + v ) + 16( u + v ) + 12( u + v )+ 44( u + v ) + 28( u + v ) . (7.2.29)With J ∗ = { , , , , , , , , , , , } , for the horizontal street correspondingto u , as highlighted in Figure 7.2.7, a similar argument gives( M − I ) u = 16( u + v ) + 16( u + v ) + 80( u + v ) + 32( u + v ) . (7.2.30) ON-INTEGRABLE SYSTEMS (IV) 21 ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × u Figure 7.2.7: S with m = n = 12: almost vertical units of type ↑ in u With J ∗ = { , , , , , , , , , , , } , for the horizontal street correspondingto u , as highlighted in Figure 7.2.8, a similar argument gives( M − I ) u = 9( u + v ) + 9( u + v ) + 54( u + v )+ 18( u + v ) + 54( u + v ) . (7.2.31) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × u Figure 7.2.8: S with m = n = 12: almost vertical units of type ↑ in u With J ∗ = { , , , , , , , , , , , } , for the horizontal street correspondingto u , as highlighted in Figure 7.2.9, a similar argument gives( M − I ) u = 22( u + v ) + 22( u + v ) + 16( u + v )+ 12( u + v ) + 44( u + v ) + 28( u + v ) . (7.2.32) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × × × × × u Figure 7.2.9: S with m = n = 12: almost vertical units of type ↑ in u With J ∗ = { , , , , , , , , , , , } , for the horizontal street correspondingto u , as highlighted in Figure 7.2.10, a similar argument gives( M − I ) u = 14( u + v ) + 14( u + v ) + 36( u + v )+ 28( u + v ) + 52( u + v ) . (7.2.33) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × × × u Figure 7.2.10: S with m = n = 12: almost vertical units of type ↑ in u We now define the street-spreading matrix by S =
44 0 16 9 22 140 44 16 9 22 1416 16 80 0 16 012 12 0 54 12 3644 44 32 18 44 2828 28 0 54 28 52 , where the j -th column comes from the coefficients of ( u i + v i ), i = 1 , , , , ,
6, inthe expression for ( M − I ) u j , j = 1 , , , , ,
6, as given by (7.2.28)–(7.2.33).
Theorem 7.2.2.
Let P be a finite polysquare surface with d square faces. Assumethat h (cid:54) v , where h is the number of horizontal streets in P and v is the numberof vertical streets in P . Let m be any fixed integer multiple of the lengths of thehorizontal streets of P , and let n be any fixed integer multiple of the lengths of thevertical streets of P .Consider a -direction almost vertical geodesic V on P , with slope α given by (7.2.1) . Let M denote the transition matrix of the -step ancestor process with respectto the basis W , where W is given by (7.2.2) . Then all the relevant eigenvalues of M and their corresponding eigenvectors can be determined as follows: (i) Let V denote the subspace of C d generated by u i , v i , i = 1 , . . . , h , given by (7.2.14) and (7.2.15) . Then V in an M -invariant subspace of C d , and contains theeigenvectors corresponding to all the relevant eigenvalues of M . Furthermore, thereexist non-negative integers s i,j , i, j = 1 , . . . , h , such that for every j = 1 , . . . , h , ( M − I ) u j = h (cid:88) i =1 s i,j ( u i + v i ) , (7.2.34) and M v j = u j + v j . (7.2.35)(ii) Consider the street-spreading matrix S = s , · · · s ,h ... ... s h, · · · s h,h , (7.2.36) as well as the matrix M | V = (cid:18) S + I I S I (cid:19) . (7.2.37) Suppose that τ is an eigenvalue of S with eigenvector ψ , so that S ψ = τ ψ . Then λ ( τ ; ± ) = 1 + τ ± √ τ + 4 τ ON-INTEGRABLE SYSTEMS (IV) 23 are two eigenvalues of M | V with product equal to , with corresponding eigenvectors Ψ( τ ; ± ) = (cid:32) ψ, − τ ± √ τ + 4 τ ψ (cid:33) T ∈ C h . (7.2.39) In particular, the h eigenvalues of S give rise to h eigenvalues of M | V . (iii) If Ψ( τ ; ± ) = ( a , a , a , a , b , b , b , b ) T is an eigenvector corresponding toan eigenvalue λ ( τ ; ± ) of M | V , then λ ( τ ; ± ) is an eigenvalue of M with eigenvector w = h (cid:88) i =1 a i u i + h (cid:88) i =1 b i v i . (7.2.40) Remark.
The assumption that the number h = h ( P ) of horizontal streets in P isless than or equal to the number v = v ( P ) of vertical streets, i.e. , h (cid:54) v , is one forconvenience, as the street-spreading matrix is smaller. For any polysquare surface P where h (cid:62) v , we can always interchange the horizontal and vertical directions. Proof of Theorem 7.2.2. (i) In view of Lemma 7.2.3, it remains to establish (7.2.34)and (7.2.35). For (7.2.34), note from (7.2.14) that each u j involves only units oftype ↑ , and so (7.2.34) follows immediately from (7.2.17), on observing that u i , v i , i = 1 , . . . , h , generate V . On the other hand, (7.2.35) follows immediately from(7.2.21).(ii) Suppose that Ψ is a column eigenvector corresponding to an eigenvalue λ of M | V . From (7.2.37), it is clear that M | V can be reduced to the matrix (cid:18) I S I (cid:19) by elementary row operations. The determinant of this matrix is clearly equal to 1.It follows that det( M | V ) = 1, and so λ (cid:54) = 0. We can writeΨ = ( ψ, ψ ∗ ) T , (7.2.41)where ψ and ψ ∗ are both column vectors with h entries. Then M | V Ψ = (cid:18) S + I I S I (cid:19) (cid:18) ψψ ∗ (cid:19) = λ (cid:18) ψψ ∗ (cid:19) , and so S ψ + ψ + ψ ∗ = λψ, S ψ + ψ ∗ = λψ ∗ . (7.2.42)Subtracting the second equation in (7.2.42) from the first, we obtain ψ = λ ( ψ − ψ ∗ ),and so ψ ∗ = λ − λ ψ. (7.2.43)Substituting this into the first equation in (7.2.42), we obtain S ψ = (cid:18) λ − λ − λ (cid:19) ψ, so that τ = λ − λ − λ is an eigenvalue of S with eigenvector ψ . This is equivalent to λ = 1 + τ ± √ τ + 4 τ . (7.2.44) This proves that any eigenvalue λ of M | V is obtained from an eigenvalue τ of S by(7.2.44), and establishes (7.2.38). Finally, note that λ − λ = 1 − λ = 1 − (cid:32) τ ∓ √ τ + 4 τ (cid:33) = − τ ± √ τ + 4 τ . In view of (7.2.41) and (7.2.43), this establishes (7.2.39).(iii) Suppose that ( a , . . . , a h , b , . . . , b h ) T is an eigenvector corresponding to aneigenvalue λ of M | V . Let the vector w be given by (7.2.40). Then M w = h (cid:88) i =1 c i u i + h (cid:88) i =1 d i v i , (7.2.45)where the coefficients are related by( c , . . . , c h , d , . . . , d h ) T = M | V ( a , . . . , a h , b , . . . , b h ) T . It is clear that ( c , . . . , c h , d , . . . , d h ) T = λ ( a , . . . , a h , b , . . . , b h ) T , and so it follows from (7.2.40) and (7.2.45) that M w = λ w , so that λ is an eigenvalueof M with eigenvector w . (cid:3) Continuation of Example . Recall from (7.1.6), (7.1.8) and (7.1.10) that therelevant eigenvalues of M ( S ) are λ = 11 + 3 √ , λ = 3 + 2 √ , λ = 3 + √ , obtained by tedious calculation using MATLAB. Let us instead retrieve the sameinformation using the street-spreading matrix S given by (7.2.27).The eigenvalues of S are τ = 9, τ = 4, τ = 1 and τ = 0. Using (7.2.38), thecorresponding eigenvalues of M are λ (9; ± ) = 11 ± √ , λ (4; ± ) = 3 ± √ , λ (1; ± ) = 3 ± √ , λ (0; ± ) = 1 . Note that in view of Lemma 7.2.3, λ (9; ± ), λ (4; ± ) and λ (1; ± ) are the onlyeigenvalues of M that are not equal to 1. Thus the multiplicity of the eigenvalue 1is 14, as we have claimed earlier in the Remark following (7.2.8). Example . Consider again the L-shape, given in Figure 7.1.6, and consider a1-direction geodesic starting from some vertex of the L-shape with slope α given by(7.2.1) with m = n = 2. It is easy to see from Figure 7.2.11 that( M − I ) u = 2( u + v ) + 2( u + v ) , (7.2.46)( M − I ) u = ( u + v ) + 3( u + v ) . (7.2.47) ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × u u Figure 7.2.11: L-shape with m = n = 2: almost vertical unitsof type ↑ in u , u It follows from (7.2.46)–(7.2.47) that the street-spreading matrix is given by S = (cid:18) (cid:19) , ON-INTEGRABLE SYSTEMS (IV) 25 with eigenvalues τ = 4 and τ = 1. Using (7.2.38) leads to the eigenvalues λ (4; ± ) = 3 ± √ λ (1; ± ) = 3 ± √ , including those given in (7.1.24) with absolute values greater than 1. Example . It is not difficult to show that geodesic flow on the surface of theunit cube is equivalent to 1-direction geodesic flow on the polysquare surface P asshown in Figure 7.2.12 that represents the 4-copy version of the cube surface. a a g f b d b g c e d f c e a a e c e c f d g b f g d b c e a a b g d f c e d b f g a a e c e c f d g b d b f g Figure 7.2.12: 4-copy version of the surface of the unit cubeNote that P has 6 horizontal streets and 6 vertical streets, all of length 4, asshown in Figure 7.2.13, where, for instance, ↔ (cid:108) ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l l l l l l l l l l l l l Figure 7.2.13: 4-copy version of the surface of the unit cube
Figures 7.2.12 and 7.2.13 are not very convenient for us to visualize the horizontaland vertical streets clearly. We can re-draw these two figures, and attempt to arrangethe square faces in an array where each row represents a horizontal street and eachcolumn represents a vertical street. In Figure 7.2.14, we do precisely this, wherefor i, j = 1 , . . . ,
6, the i -th row represents the i -th horizontal street and the j -thcolumn represents the j -th vertical street as shown in Figure 7.2.13. Note that thisexercise is somewhat cumbersome and requires very careful edge identifications asillustrated. We include the details here for the sake of completeness, and commentthat the edge identifications are not required in the process for the determinationof the street-spreading matrix. h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l Figure 7.2.14: 4-copy version of the surface of the unit cubewith square faces arranged in an arrayConsider now a 1-direction geodesic on the polysquare surface P starting fromsome vertex of P with slope α given by (7.2.1) with m = n = 4.The polysquare surface P has 24 square faces, so any 2-step transition matrix M has size 48 ×
48. It is much simpler to find the 6 × u i , v i , i = 1 , . . . ,
6, according to (7.2.14) and(7.2.15).Consider the horizontal street corresponding to u , and also the horizontal streetcorresponding to u , as highlighted in the picture on the left in Figure 7.2.15. Wesee that J ∗ = J ∗ = { , , , } , and both u and u concern almost vertical units onthe vertical streets 2 , , ,
6. Applying (7.2.17), we have( M − I ) u = ( M − I ) u = ( M − I )[ {↑ , , ↑ , , ↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , , ↑ , , ↑ , } ]+ ( M − I )[ {↑ , , ↑ , } ] + ( M − I )[ {↑ , , ↑ , } ]= 4( u + v ) + 2( u + v ) + 2( u + v )+ 4( u + v ) + 2( u + v ) + 2( u + v ) . (7.2.48) ON-INTEGRABLE SYSTEMS (IV) 27 ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × u u ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × u u Figure 7.2.15: P with m = n = 4: almost vertical units of type ↑ in u u , u , u With J ∗ = J ∗ = { , , , } , for the horizontal street corresponding to u , andalso the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 7.2.15, a similar argument gives( M − I ) u = ( M − I ) u = 2( u + v ) + 4( u + v ) + 2( u + v )+ 2( u + v ) + 4( u + v ) + 2( u + v ) . (7.2.49)With J ∗ = J ∗ = { , , , } , for the horizontal street corresponding to u , and alsothe horizontal street corresponding to u , as highlighted in Figure 7.2.16, a similarargument gives( M − I ) u = ( M − I ) u = 2( u + v ) + 2( u + v ) + 4( u + v )+ 2( u + v ) + 2( u + v ) + 4( u + v ) . (7.2.50) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × u u Figure 7.2.16: P with m = n = 4: almost vertical units of type ↑ in u , u It follows from (7.2.48)–(7.2.50) that the street-spreading matrix is given by S = , with eigenvalues τ = 16, τ = τ = 4 and τ = τ = τ = 0. Using (7.2.38), thecorresponding eigenvalues of M are λ (16; ± ) = 9 ± √ , λ (4; ± ) = 3 ± √ , λ (0; ± ) = 1 , with the appropriate multiplicities. The two largest eigenvalues of M are therefore λ = 9 + 4 √ √ and λ = 3 + 2 √ √ . Furthermore, the multiplicity of the eigenvalue 1 is 42.Note that 2 + √ , , , . . . ] and 1 + √ , , , . . . ]exhibit “digit-halving” in their continued fraction expansions, and are also the twolargest eigenvalues of the 1-step transition matrix of a 1-direction geodesic of slope α = 2 + √ α is defined by (7.2.1) with m = n = 4 k for any positive integers k . In this general case, the largest eigenvalueis the square of [4 k ; 4 k, k, k, . . . ], and the second largest eigenvalue is the squareof [2 k ; 2 k, k, k, . . . ].We complete this section by considering two consequences of Theorem 7.2.2. Thefirst one is fairly straightforward. Corollary 7.2.1.
The determinant of the -step transition matrix M is equal to .Proof. We have shown earlier that det( M | V ) = 1.Let p ( x ) be the characteristic polynomial of M | V . In view of Theorem 7.2.2, it isclear that the characteristic polynomial of M is equal to p ( x ) q ( x ), where q ( x ) hasdegree 2 d − h and its roots are all the eigenvalues of M that are not eigenvaluesof M | V . In view of Lemma 7.2.3, we have q ( x ) = ( x − d − h . On the other hand,it follows from (7.2.38) that p ( x ) = h (cid:89) i =1 ( x − λ ( τ i ; +))( x − λ ( τ i ; − )) , where τ , . . . , τ h are the eigenvalues of S , and λ ( τ i ; +) λ ( τ i ; − ) = 1, i = 1 , . . . , h .Clearly the determinant of M is equal to p (0) q (0) = 1. (cid:3) To motivate the second consequence, we start with the L-surface, and obtain anew polysquare surface with 6 smaller square faces, each of area 1 /
2. The trick isto draw two diagonals on each of the 3 square faces of the L-surface. We use theboundary pairing in the picture on the right in Figure 7.2.17, which gives the so-called diagonal subdivision surface of the L-surface, or DS-L-surface. Of course thegenus remains the same; namely, 2. Rotating this picture on the right anticlockwiseby 45 degrees and resizing, the DS-L-surface becomes a polysquare surface in thestandard horizontal-vertical position. It is easy to see that the DS-L-surface has onlyone horizontal street, which has length 6 and is defined by the cycle 1 , , , , , , , , , ,
4. So the streetsize multiple of the DS-L-surface is equal to 6.
ON-INTEGRABLE SYSTEMS (IV) 29 h h h h v v v v a a a a a a a a a a a a a a Figure 7.2.17: L-surface and DS-L-surfaceNow Theorem 5.5.1 in [3] says that, applying the surplus shortline method tothis particular one-street surface , we obtain infinitely many explicit slopes with theproperty that a 1-direction geodesic with any such slope is superuniform .Theorem 5.5.1 is obtained by using Theorem 5.3.1 in [3], the main result for theL-surface in that paper, with an exceptionally long proof. So the mysterious con-nection between a concrete one-street surface and superuniformity in Theorem 5.5.1may seem at first sight a very deep, perhaps accidental result. However, nothingis further from the truth. There is a general connection between one-street sur-faces and superuniformity, and it is an easy consequence of Theorem 7.2.2. Indeed,we simply need the condition that the polysquare surface has only one horizontalstreet (or one vertical street), but no restriction about the perpendicular direction.Then the street-spreading matrix S is a 1 × M . It follows that the second largesteigenvalue is irrelevant, which implies that the surplus shortline method providesexplicit quadratic irrational slopes for which the geodesic is superuniform.If a polysquare surface has only one horizontal street, or only one vertical street,then we call it a one-street polysquare surface . Thus we just proved the followingresult. Corollary 7.2.2.
For every one-street polysquare surface, there exist infinitely manyexplicit quadratic irrational slopes such that every -direction geodesic with any suchslope is superuniform. We complete this section by showing that one-street polysquare surfaces are notrare. The idea is based on a a far-reaching generalization of the DS-L-surface.Recall the 2-polysquare snake surface in Figure 7.1.4. Snake surfaces are notrestricted to 2-polysquare surfaces, as illustrated in Figure 7.2.18. b b b b bb b b b b bb b b b
Figure 7.2.18: a snake surfaceMore precisely, let P be a finite polysquare surface. We put a point at the centerof every square face of P , and call it the “capital of the square face”. Consider the neighbor graph of P , where the vertices are the capitals, and where two capitals arejoined by an edge of the graph if the corresponding square faces share a commonedge. If the neighbor graph of P is a “path”, with no branching, then P is calleda snake surface, as shown in Figure 7.2.18. If the neighbor graph of P is a “tree”,then P is called a polysquare-tree surface, as shown in Figure 7.2.18 and 7.2.19. b b b b bb b b b b bb b b b b bb b b Figure 7.2.19: a polysquare-tree surfaceGiven any graph, the number of edges at a vertex is called the degree of the vertex.In a path, the degree of every vertex is 2, except at the two end-vertices where thedegree is 1. In a tree we may have higher degrees. For example, the neighbor treeof the polysquare-tree surface in Figure 7.2.19 has two vertices of degree 3.One-street polysquare surfaces are not rare, because the diagonal subdivision ofevery polysquare-tree surface has only one street.The proof goes by induction on the number of square faces in the polysquare.The inductive step is based on the elementary result in graph theory that every treehas a vertex of degree 1. Indeed, it is easy to see that a tree with n vertices hasprecisely n − n − Figure 7.2.20: removing an end square faceNote that we carefully start the labelling from a common edge shared with theremoved end square face, and keep moving in the north-east direction. Addingback the removed end square face, Figure 7.2.21 shows an explicit one-street in theoriginal polysquare-street surface, where the consecutive squares along the street arelabelled with the consecutive integers from 1 to 40.
ON-INTEGRABLE SYSTEMS (IV) 31
Figure 7.2.21: extension of the labellingNote that the labelling in Figure 7.2.21 is obtained as a trivial extension of thelabelling in Figure 7.2.20 by including 39 and 40. This completes the inductive step.We cannot entirely drop the tree condition for the neighbor graph, as cycles mayspoil the argument. Consider the two polysquare cycles in Figure 7.2.22. b b bb bb b b b bb b
Figure 7.2.22: two polysquare-cyclesNeither of these polysquare-cycles leads to a one-street diagonal decompositionsurface, as can be shown in Figure 7.2.23.
12 23 4 55 6 77 88 1 2 33 44
Figure 7.2.23: diagonal decomposition of two polysquare-cyclesIn the picture on the left, we have a tilted street of length 8, covering only half thetilted square faces of the polysquare. In the picture on the right, we have a tiltedstreet of length 4, again covering only half the tilted square faces of the polysquare.7.3.
Application of Theorem 7.2.2: super-fast escape to infinity.
Our goalin this section is to find explicit quadratic irrational slopes which give rise to orbitsthat exhibit super-fast escape rate to infinity. More precisely, given any ε >
0, weshall find explicit quadratic irrational slopes such that for any orbit with any suchslope, there exists an infinite sequence T n , n (cid:62)
1, such that T n → ∞ as n → ∞ , andthe diameter of the initial segment of length T n of the orbit exceeds c T − εn , where c is an absolute constant depending only on the given special slope.Orbits in integrable polysquare systems behave in the same way for all quadraticirrational slopes. In complete contrast, orbits in infinite non-integrable polysquaresystems may behave in completely different ways for different quadratic irrationalslopes. We may have the full spectrum from super-slow to super-fast escape rate toinfinity. Example . We return to the problem of billiard in the ∞ -L-strip region shownin Figure 7.3.1. In Theorem 6.7.2 in [4], it is shown that there are infinitely manyquadratic irrational slopes such that a billiard orbit starting from a corner pointwith such a slope is dense on the ∞ -L-strip region and also exhibits super-slow logarithmic escape rate to infinity. In particular, any initial segment of such anorbit of length T , T (cid:62)
2, has distance at most c log T from the starting point,where c is an absolute constant depending only on the given special slope.Figure 7.3.1: billiard in the ∞ -L-strip regionAs illustrated in Figure 7.3.2, the L-shape is the building block of the ∞ -L-stripregion. L i − L i L i +1 Figure 7.3.2: the ∞ -L-strip region with its building blocksThe trick of “unfolding” reduces the problem of billiard on the ∞ -L-strip region,a 4-direction flow, to the problem of a 1-direction geodesic flow on an appropriateinfinite polysquare surface. We call this surface the 1 -direction billiard surface ofthe ∞ -L-strip , and denote it by Bil( ∞ ; 1).To describe Bil( ∞ ; 1), we first “unfold” each of the L-shapes L i by reflecting ithorizontally and vertically to obtain 4 reflected copies and then put them together toform a 4-copy- L i . We then obtain Bil( ∞ ; 1) by gluing together the infinitely manycopies of these 4-copy- L i .Clearly there is billiard flow from each L-shape L i to its two immediate neighbours L i − and L i +1 . We therefore need to identify corresponding edges of these 4-copyversions very carefully. A simple examination will convince the reader that the edgeidentifications can be as illustrated in Figure 7.3.3. v ( i )1 v ( i )1 v ( i )4 v ( i )4 h ( i )1 h ( i )1 h ( i )3 h ( i )3 h ( i )2 h ( i )2 v ( i )2 v ( i )3 v ( i − v ( i − v ( i +1)1 v ( i +1)1 v ( i +1)4 v ( i +1)4 h ( i +1)3 h ( i +1)3 h ( i +1)1 h ( i +1)1 h ( i +1)2 h ( i +1)2 v ( i +1)2 v ( i +1)3 v ( i )2 v ( i )3 L i L i +1 Figure 7.3.3: 4-copy- L i and 4-copy- L i +1 together with edge identificationsNote that a 1-direction geodesic on Bil( ∞ ; 1) that goes from the vertical edge v ( i +1)2 to the vertical edge v ( i )2 corresponds to a billiard path going from L i +1 to L i , ON-INTEGRABLE SYSTEMS (IV) 33 whereas a 1-direction geodesic on Bil( ∞ ; 1) that goes from the vertical edge v ( i − to the vertical edge v ( i )3 corresponds to a billiard path going from L i to L i +1 .Thus the motion of the infinite L-strip billiard can be described in terms of1-direction geodesic flow on a finite polysquare surface P , the period-surface ofBil( ∞ ; 1), illustrated in Figure 7.3.4, where we have rotated by 90 degrees in theanticlockwise direction. Clearly P has 12 square faces, 4 horizontal streets and 6vertical streets, labelled according to the picture on the right where, for instance,the information ↔ (cid:108) v v v v v v v v h h h h h h ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l Figure 7.3.4: the period surface P of Bil( ∞ ; 1)The boundary identification in P is clearly the simplest perpendicular translation.Consider a 1-direction geodesic in P with slope given by the arrow in the pictureon the left in Figure 7.3.4. If such a geodesic hits the edge v at the top, thenit jumps down vertically to the identified edge v in the middle. In terms of the ∞ -L-strip billiard this means precisely that the billiard moves from L i to L i +1 , i.e. ,“moving right”. If such a geodesic hits the edge v in the middle, then it jumpsdown vertically to the identified edge v at the bottom. In terms of the ∞ -L-stripbilliard this means precisely that the billiard moves from L i to L i − , i.e. , “movingleft”. Thus by counting the number of edge cuttings for v and for v , we know thetotal number of steps moving to the left and the total number of steps moving tothe right. Taking the difference of these two numbers, we then know precisely whichparticular constituent L-shape happens to contain the billiard at a given moment.Now the edge cutting numbers are expressed in terms of powers of the relevanteigenvalues of the 2-step transition matrix corresponding to the given quadraticirrational slope. It makes it plausible to guess that the irregularity exponent of theperiod-surface P gives the escape rate to infinity for the infinite L-strip billiard,as long as the shortline method works. Next we expand on this and justify thisintuition.We need to choose the parameters m and n , which have to be integer multiplesof the lengths of the horizontal and vertical streets respectively. For this periodsurface, the horizontal streets have length 2 or 4, while the vertical streets all havelength 2. Thus we let m = 4 k and n = 2 k , where k (cid:62) α k = [2 k ; 4 k, k, k, . . . ] = k + √ k + 22 . (7.3.1)Since the period-surface P has 12 squares, the original shortline method leads to a24 ×
24 2-step transition matrix M . To find the relevant eigenvalues and eigenvectorsof such a large matrix directly is clearly rather inconvenient.Instead we apply Theorem 7.2.2, which leads to a much smaller 4 × S defined by (7.2.36). We follow the notation in Section 7.2. Weconsider the M -invariant subspace V generated by the 8 vectors u i , v i , i = 1 , , , defined by (7.2.14) and (7.2.15). The relevant eigenvalues of M are then eigenvaluesof the matrix M | V , defined by (7.2.37).Consider the horizontal street corresponding to u , as highlighted in the pictureon the left in Figure 7.3.5. Clearly J = { , } . Using (7.2.17), we have( M − I ) u = ( M − I )[ { k ↑ , , k ↑ , } ] + ( M − I )[ { k ↑ , , k ↑ , } ]= 4 k ( u + v ) + 4 k ( u + v ) . (7.3.2) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , k × k k × kk × k k × k k × k k × k k × k k × kk × k k × kk × k k × k u u Figure 7.3.5: almost vertical units of type ↑ in u and u With J = { , , , } , for the horizontal street corresponding to u , as highlightedin the picture on the right in Figure 7.3.5, a similar argument gives( M − I ) u = 2 k ( u + v ) + 4 k ( u + v ) + 2 k ( u + v ) . (7.3.3)With J = { , , , } , for the horizontal street corresponding to u , as highlightedin the picture on the left in Figure 7.3.6, a similar argument gives( M − I ) u = 2 k ( u + v ) + 4 k ( u + v ) + 2 k ( u + v ) . (7.3.4) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , k × k k × k k × k k × kk × k k × kk × k k × k k × k k × kk × k k × k u u Figure 7.3.6: almost vertical units of type ↑ in u and u With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture on the right in Figure 7.3.6, a similar argument gives( M − I ) u = 4 k ( u + v ) + 4 k ( u + v ) . (7.3.5)It follows from (7.3.2)–(7.3.5) that the street-spreading matrix is given by S = k k k k k
00 2 k k k k k = k . This has eigenvalues and corresponding eigenvectors τ = 8 k , ψ = (1 , , , T ,τ = 6 k , ψ = (1 , , − , − T ,τ = 2 k , ψ = (1 , − , − , T ,τ = 0 , ψ = (1 , − , , − T . ON-INTEGRABLE SYSTEMS (IV) 35
Using the formula (7.2.38), we obtain corresponding eigenvalues of M | V given by λ ( τ ; ± ) = 1 + 8 k ± √ k + 32 k k ± k √ k + 2 ,λ ( τ ; ± ) = 1 + 6 k ± √ k + 24 k k ± k √ k + 6 ,λ ( τ ; ± ) = 1 + 2 k ± √ k + 8 k k ± k √ k + 2 ,λ ( τ ; ± ) = 1 . The three relevant eigenvalues of M | V and corresponding eigenvectors are therefore λ = 1 + 4 k + 2 k √ k + 2 , Ψ = (1 , , , , b , b , b , b ) T , (7.3.6) λ = 1 + 3 k + k √ k + 6 , Ψ = (1 , , − , − , b , b , b , b ) T , (7.3.7) λ = 1 + k + k √ k + 2 , Ψ = (1 , − , − , , b , b , b , b ) T , where the values of b ij , i = 1 , , j = 1 , , ,
4, can be calculated using (7.2.39).Recall that by counting the number of edge cuttings for v and for v , we know thetotal number of steps of the billiard orbit moving to the left and the total numberof steps of the billiard orbit moving to the right. The difference between these twonumbers then gives us information on the movement of the billiard orbit.We now proceed to find the edge-cutting numbers of v and v .The eigenvector of M corresponding to the eigenvector Ψ of M | V is given by u + 2 u + 2 u + u + b v + b v + b v + b v . We first identity the number of almost vertical units counted here that cut theedge v ; see Figure 7.3.4. Concerning the contribution from those units countedby u , it is clear from the picture on the left in Figure 7.3.5 that we have 2 k units of type ↑ , and 2 k units of type ↑ , , making a total of 4 k . Concerning thecontribution from those units counted by u , it is clear from the picture on the rightin Figure 7.3.5 that we have k units of type ↑ , and k units of type ↑ , , making atotal of 2 k . Concerning the contribution from those units counted by u and u , itis clear from Figure 7.3.6 that there is none. To study the contribution from thoseunits counted by v , . . . , v , we appeal to (7.2.15) and Figure 7.2.1. There is clearlyno contribution from v , v , v , while for v , we have positive contributions from − ↑ , and − ↑ , , and negative contributions from ↑ , and ↑ , , each counted with themultiplicity of J ∗ , and so there is perfect cancellation. Thus for the edge v , we havea total count of 4 k + 2 k = 6 k .We next identity the number of almost vertical units counted here that cut theedge v ; see Figure 7.3.4. Concerning the contribution from those units countedby u and u , it is clear from Figure 7.3.5 that there is none. Concerning thecontribution from those units counted by u , it is clear from the picture on the leftin Figure 7.3.6 that we have k units of type ↑ , and k units of type ↑ , , makinga total of 2 k . Concerning the contribution from those units counted by u , it isclear from the picture on the right in Figure 7.3.6 that we have 2 k units of type ↑ , and 2 k units of type ↑ , , making a total of 4 k . There is clearly no contributionfrom v , v , v , while for v , we have positive contributions from − ↑ , and − ↑ , , andnegative contributions from ↑ , and ↑ , , each counted with the multiplicity of J ∗ ,and so there is perfect cancellation. Thus for the edge v , we have a total count of2 k + 4 k = 6 k .Since the two counts are the same, it follows that the eigenvalue λ does notcontribute to the difference between the edge-cutting numbers of v and v . Remark.
The perfect cancellation is not surprising at all. Indeed, the high powerscorresponding to the largest eigenvalue represents the main term of the problem.Any lack of cancellation here would violate the uniformity of the geodesic. With aquadratic irrational slope, this would specifically violate the Gutkin–Veech theorem[9, 16, 18] that guarantees uniformity of any 1-direction geodesic in any polysquaresurface with irrational slope. See also [5, 6].The eigenvector of M corresponding to the eigenvector Ψ of M | V is given by u + u − u − u + b v + b v + b v + b v . We first identity the number of almost vertical units counted here that cut theedge v . Recall that the contribution from u and u are 4 k and 2 k respec-tively, while there is no contribution from u and u , and also no contribution from v , . . . , v . Thus for the edge v , we have a total count of 4 k + 2 k = 6 k . Wenext identity the number of almost vertical units counted here that cut the edge v .Recall that the contribution from u and u are 2 k and 4 k respectively, while thereis no contribution from u and u , and also no contribution from v , . . . , v . Thusfor the edge v , we have a total count of − k − k = − k .The difference, in absolute value, is therefore 12 k . Let c = c , in (7.2.11).Then it follows that the second eigenvalue λ contributes 12 c λ r k to the differencebetween the edge-cuttings numbers of v and v .So the deviation from the starting point comes from the second largest eigenvalue,and this has the order of magnitude λ r compared to the order of magnitude λ r ofthe main term. Choosing T = λ r , we have λ r (cid:16) T κ , where κ = κ ( k ) = log λ log λ = 2 log k + log 62 log k + log 8 + o (1) , (7.3.8)in view of (7.3.6)–(7.3.7), is the irregularity exponent of a 1-direction geodesic ofslope (7.3.1) on the period-surface in Figure 7.3.4.Clearly κ = κ ( k ) → k → ∞ . So we have just established T κ = T − ε size super-fast escape rate to infinity for the infinite L-strip billiard with the explicit classof quadratic irrational slopes in (7.3.1) where the parameter k (cid:62) κ = κ ( k ) in (7.3.8) is precisely the escape rate to infinity of this infinite billiard. Indeed, the escape rate to infinitycannot be larger than the expression (7.3.8) coming from the two largest eigenvalues. Example . Consider the Ehrenfest wind-tree model with a = b = 1 /
2, rescaledso that we have square faces of unit area. Consider a 4-direction billiard trajectory inthe infinite region in Figure 7.3.7. Here the building block of this infinite polysquaresurface is an L-shape.Figure 7.3.7: Ehrenfest wind-tree billiard modelAs usual, this infinite billiard model is equivalent to a 1-direction geodesic flowon an infinite polysquare surface that we denote by Bil( ∞ ; 2), with the index 2here indicating that this is double periodic. To construct Bil( ∞ ; 2), we take one ON-INTEGRABLE SYSTEMS (IV) 37 of the L-shape building blocks, and unfold the 4-direction billiard flow on it to a1-direction geodesic flow on a 4-copy version of it, obtained by reflecting horizontallyand vertically. Note that each 4-copy L-shape has a right-neighbor, a left-neighbor,a down-neighbor and an up-neighbor in Bil( ∞ ; 2), and we need appropriate edgeidentification which amount to to a more complicated version of Figure 7.3.3. The period-surface Bil(2) of Bil( ∞ ; 2) is shown in the picture on the left in Figure 7.3.8. h h h h h h h h h h h h v v v v v v v v v v v v ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l Figure 7.3.8: the period surface Bil(2) of Bil( ∞ ; 2)The horizontal and vertical streets of Bil(2) are indicated in the picture on theright in Figure 7.3.8 where, for instance, the entries ↔ (cid:108) h or h atthe top, then it jumps down vertically to the identified edge h or h in the middle.In terms of the Ehrenfest wind-tree billiard this means precisely that the billiardmoves from one L-shape to the next above, i.e. , “moving up”. If such a geodesichits the edge h or h in the middle, then it jumps down vertically to the identi-fied edge h or h at the bottom. In terms of the Ehrenfest wind-tree billiard thismeans precisely that the billiard moves from one L-shape to the next below, i.e. ,“moving down”. Thus by counting the number of edge cuttings for h , h and for h , h , we know the total number of steps moving up and the total number of stepsmoving down. Taking the difference of these two numbers, we then know preciselythe vertical location of the billiard at a given moment.We need to choose the parameters m and n , which have to be integer multiplesof the lengths of the horizontal and vertical streets respectively. For this periodsurface, the horizontal and vertical streets all have length 2. Thus we let m = 2 k and n = 2 k , where k (cid:62) α k = [2 k ; 2 k, k, k, . . . ] = k + √ k + 1 . (7.3.9)We apply Theorem 7.2.2, leading to a 6 × S definedby (7.2.36). We follow the notation in Section 7.2. We consider the M -invariantsubspace V generated by the 12 vectors u i , v i , i = 1 , . . . ,
6, defined by (7.2.14) and(7.2.15). The relevant eigenvalues of M are then eigenvalues of the matrix M | V ,defined by (7.2.37).Consider the horizontal street corresponding to u , as highlighted in the pictureon the left in Figure 7.3.9. Clearly J = { , } . Using (7.2.17), we have( M − I ) u = ( M − I )[ { k ↑ , , k ↑ , } ] + ( M − I )[ { k ↑ , , k ↑ , } ]= 2 k ( u + v ) + k ( u + v ) + k ( u + v ) . (7.3.10) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , k × k k × kk × k k × k k × k k × k k × k k × kk × k k × kk × k k × k u u u Figure 7.3.9: almost vertical units of type ↑ in u , u , u With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture in the middle in Figure 7.3.9, a similar argument gives( M − I ) u = k ( u + v ) + 2 k ( u + v ) + k ( u + v ) . (7.3.11)With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture on the right in Figure 7.3.9, a similar argument gives( M − I ) u = k ( u + v ) + 2 k ( u + v ) + k ( u + v ) . (7.3.12)With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture on the left in Figure 7.3.10, a similar argument gives( M − I ) u = k ( u + v ) + 2 k ( u + v ) + k ( u + v ) . (7.3.13) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , k × k k × k k × k k × kk × k k × kk × k k × k k × k k × kk × k k × k u u u Figure 7.3.10: almost vertical units of type ↑ in u , u , u With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture in the middle in Figure 7.3.10, a similar argument gives( M − I ) u = k ( u + v ) + 2 k ( u + v ) + k ( u + v ) . (7.3.14)With J = { , } , for the horizontal street corresponding to u , as highlighted inthe picture in the middle in Figure 7.3.10, a similar argument gives( M − I ) u = k ( u + v ) + k ( u + v ) + 2 k ( u + v ) . (7.3.15)It follows from (7.3.10)–(7.3.15) that the street-spreading matrix is given by S = k k k k k k k k k k k k k k k k k k = k . ON-INTEGRABLE SYSTEMS (IV) 39
This has non-zero eigenvalues and corresponding eigenvectors τ = 4 k , ψ = (1 , , , , , T ,τ = 3 k , ψ , = ( − , , − , , , T ,ψ , = (0 , − , , − , , T ,τ = k , ψ , = (1 , , − , − , , T ,ψ , = (0 , , − , − , , T ,τ = 0 , ψ = ( − , , , − , − , T . Using the formula (7.2.38), we obtain corresponding eigenvalues of M | V given by λ ( τ ; ± ) = 1 + 4 k ± √ k + 16 k k ± k √ k + 1 ,λ ( τ ; ± ) = 1 + 3 k ± √ k + 12 k k ± k √ k + 122 ,λ ( τ ; ± ) = 1 + k ± √ k + 4 k k ± k √ k + 42 .λ ( τ ; ± ) = 1 . The three relevant eigenvalues of M | V and corresponding eigenvectors are therefore λ = 1 + 2 k + 2 k √ k + 1 , Ψ = (1 , , , , , , b , . . . , b ) T , (7.3.16) λ = 1 + 3 k k √ k + 122 , Ψ , = ( − , , − , , , , b (1)21 , . . . , b (1)26 ) T , (7.3.17)Ψ , = (0 , − , , − , , , b (2)21 , . . . , b (2)26 ) T ,λ = 1 + k k √ k + 42 , Ψ , = (1 , , − , − , , , b (1)31 , . . . , b (1)36 ) T , Ψ , = (0 , , − , − , , , b (2)31 , . . . , b (2)36 ) T , where the values of b j , b (1)2 j , b (2)2 j , b (1)3 j , b (2)3 j , j = 1 , . . . ,
6, can all be calculated using(7.2.39).We now proceed to find the edge-cutting numbers of h , h and h , h .As in Example 7.3.1, the eigenvalue λ does not contribute to their difference, asany lack of cancellation would violate the Gutkin–Veech theorem that guaranteesuniformity.The eigenvector of M corresponding to the eigenvector Ψ , of M | V is given by − u − u + u + u − b (1)21 v − b (1)23 v + b (1)24 v + b (1)26 v . (7.3.18)The eigenvector of M corresponding to the eigenvector Ψ , of M | V is given by − u + u − u + u − b (2)22 v + b (2)23 v − b (2)24 v + b (2)25 v . (7.3.19)We first identity the number of almost vertical units counted here that cut theedges h and h ; see Figure 7.3.8. The count from each of u i , i = 1 , . . . , u (cid:55)→ , u (cid:55)→ , u (cid:55)→ , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k . On the other hand, it is easy to show that the total contribution from v i , i = 1 , . . . , , and (7.3.18), the total count is 3 k .Corresponding to the eigenvector Ψ , and (7.3.19), the total count is 0. We next identity the number of almost vertical units counted here that cut theedges h and h ; see Figure 7.3.8. The count from each of u i , i = 1 , . . . , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ , u (cid:55)→ , u (cid:55)→ . Like before, the total contribution from v i , i = 1 , . . . , , and (7.3.18), the total count is − k . Corresponding to theeigenvector Ψ , and (7.3.19), the total count is 0.It follows that the second eigenvalue λ contributes 6 c , λ r k to the differencebetween the edge-cuttings numbers of h , h and h , h .So the deviation from the starting point comes from the second largest eigenvalue,and this has the order of magnitude λ r compared to the order of magnitude λ r ofthe main term. Choosing T = λ r , we have λ r (cid:16) T κ , where κ = κ ( k ) = log λ log λ = 2 log k + log 32 log k + log 4 + o (1) , (7.3.20)in view of (7.3.16)–(7.3.17), is the irregularity exponent of a 1-direction geodesic ofslope (7.3.9) on the period-surface in Figure 7.3.8.Clearly κ = κ ( k ) → k → ∞ . So we have just established T κ = T − ε size super-fast escape rate to infinity for the infinite L-strip billiard with the explicit classof quadratic irrational slopes in (7.3.9) where the parameter k (cid:62) κ = κ ( k ) in (7.3.20) is precisely the escape rate to infinity of this infinite billiard. Indeed, the escape rate to in-finity cannot be larger than the expression (7.3.20) coming from the two largesteigenvalues.This example also highlights that different quadratic irrational slopes can leadto 1-direction geodesics that exhibit vastly different escape rates to infinity. Toexplain this, we shall consider an infinite polysquare surface P where there is anabsolute bound (cid:96) such that the length of any horizontal or vertical street of P is atmost (cid:96) . This is called an (cid:96) -square-maze, and 1-direction geodesics on square-mazesare studied in Theorem 6.5.1 in [4].The problem connected to the Ehrenfest wind-tree model billiard with a = b = 1 / ∞ ; 2). Now Bil( ∞ ; 2) has infinite horizontal and vertical streets, so is not asquare-maze. However, if we consider “streets” in Bil( ∞ ; 2) at 45 degrees to thehorizontal and vertical axes, then the situation becomes very different. First ofall, observe such a street in the Ehrenfest wind-tree model billiard as shown inFigure 7.3.11. Figure 7.3.11: a diagonal street in the Ehrenfest wind-tree model billiardThe street indicated is contained in 4 constituent L-shapes. If we consider theanalogous problem of 1-direction geodesic flow in Bil( ∞ ; 2), then the image of thisstreet is shown in Figure 7.3.12, made up of the 4-copy versions of these 4 constituent ON-INTEGRABLE SYSTEMS (IV) 41
L-shapes. We have in fact shown that Bil( ∞ ; 2) rotated by 45 degrees becomes asquare-maze where every horizontal and vertical street has length 12. Figure 7.3.12: a 45-degree street of length 12 in Bil( ∞ ; 2)Theorem 6.5.1 in [4] asserts that for an (cid:96) -square-maze, there are infinitely manynumbers α of the form α = [ a ; a, a, a, . . . ] = a + 1 a + a + a + ··· , where a (cid:62) (cid:96) ! is divisible by (cid:96) !, such that there exist 1-direction geodesics with slope α that exhibit time-quantitative density and super-slow logarithmic escape rate toinfinity. For a square-maze where every street has length precisely (cid:96) , the conditionon a can be relaxed to include all integers a (cid:62) (cid:96) that are divisible by (cid:96) .It follows that for the rotated Bil( ∞ ; 2), there exist infinitely many numbers α k of the form α k = [12 k ; 12 k, k, k, . . . ] = 12 k + 112 k + k + k + ··· , (7.3.21)where k (cid:62) α k that exhibit super-slow logarithmic escape rate to infinity.Now note that slopes α k of the form (7.3.21) make up a subset of those slopes α k of the form (7.3.9). It follows that slopes α k of the form (7.3.21) give rise to1-direction geodesics on Bil( ∞ ; 2) that exhibit super-fast escape rate to infinity.It is easy to show that a line with slope α k , after a clockwise rotation of 45 degrees,now has slope equal to α ∗ k = α k − α k + 1 . (7.3.22)Thus we have shown that there are infinitely many numbers α k of the form (7.3.21)such that there exist 1-direction geodesics of slope α k in Bil( ∞ ; 2) that exhibit super-fast escape rate to infinity, and also 1-direction geodesics of slope α ∗ k in Bil( ∞ ; 2)that exhibit super-slow logarithmic escape rate to infinity. Note that in view of(7.3.21) and (7.3.22), both α k and α ∗ k are quadratic irrationals. More on the escape rate to infinity.
In this section, we consider some morecomplicated infinite billiards.
Example . Consider “C-wall” obstacles, as shown in Figure 7.4.1.Figure 7.4.1: infinite billiard with double periodic C-shaped wallsMore precisely, the building block of this infinite polysquare surface is a 3 × ∞ ; 2; C). Here the index 2indicates that this is double periodic, and the letter C refers to the common shapeof the obstacles. To construct Bil( ∞ ; 2; C), we take one of the building blocks,and unfold the 4-direction billiard flow on it to a 1-direction geodesic flow on a4-copy version of it, obtained by reflecting horizontally and vertically. Note thateach 4-copy version has a right-neighbor, a left-neighbor, a down-neighbor and anup-neighbor in Bil( ∞ ; 2; C), and we need appropriate edge identification for gluingthem together.The period-surface Bil(2; C) of Bil( ∞ ; 2; C) is shown in the picture on the left inFigure 7.4.2. The horizontal and vertical streets of Bil(2; C) are indicated in thepicture on the right in Figure 7.4.2 where, for instance, the entries ↔ (cid:108) h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l Figure 7.4.2: the period surface Bil(2; C) of Bil( ∞ ; 2; C)We shall consider an almost vertical geodesic V in Bil(2; C) with slope α givenby (7.2.1) with m = n = 12. As Bil(2; C) has 36 square faces, the 2-step transition ON-INTEGRABLE SYSTEMS (IV) 43 matrix M is 72 ×
72 which is mildly inconvenient. Instead, we shall determine thestreet-spreading matrix S which, at size 10 ×
10, is considerably smaller.We follow the notation in Section 7.2. We consider the M -invariant subspace V generated by the 20 vectors u i , v i , i = 1 , . . . ,
10, defined by (7.2.14) and (7.2.15).The relevant eigenvalues of M are then eigenvalues of the matrix M | V , defined by(7.2.37).For u , u , u , u , Figure 7.4.3 describes the situation. With J = J = { , , } and J = J = { , , } , for the horizontal streets corresponding to u , u , u , u ,we have, using (7.2.17),( M − I ) u = ( M − I ) u = ( M − I )[ { ↑ , , ↑ , , ↑ , } ] + ( M − I )[ { ↑ , , ↑ , } ]+ ( M − I )[ { ↑ , , ↑ , , ↑ , } ] + ( M − I )[ { ↑ , } ]+ ( M − I )[ { ↑ , } ]= 44( u + v ) + 32( u + v ) + 44( u + v )+ 12( u + v ) + 12( u + v ) , (7.4.1)( M − I ) u = ( M − I ) u = 44( u + v ) + 32( u + v ) + 44( u + v )+ 12( u + v ) + 12( u + v ) . (7.4.2) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × u u u u Figure 7.4.3: almost vertical of type ↑ in u , u , u , u For u , u , u , u , Figure 7.4.4 describes the situation. With J = J = { , , } and J = J = { , , } , for the horizontal streets corresponding to u , u , u , u ,we have ( M − I ) u = ( M − I ) u = 12( u + v ) + 12( u + v ) + 44( u + v )+ 32( u + v ) + 44( u + v ) , (7.4.3)( M − I ) u = ( M − I ) u = 12( u + v ) + 12( u + v ) + 44( u + v )+ 32( u + v ) + 44( u + v ) . (7.4.4) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × u u u u Figure 7.4.4: almost vertical of type ↑ in u , u , u , u For u , u , Figure 7.4.5 describes the situation. With J = { , , , , , } and J = { , , , , , } , for the horizontal streets corresponding to u , u , we have( M − I ) u = 16( u + v ) + 16( u + v ) + 56( u + v )+ 16( u + v ) + 16( u + v ) + 24( u + v ) , (7.4.5)( M − I ) u = 24( u + v ) + 16( u + v ) + 16( u + v )+ 56( u + v ) + 16( u + v ) + 16( u + v ) . (7.4.6) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × u ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × × × × × u Figure 7.4.5: almost vertical of type ↑ in u , u It follows from (7.4.1)–(7.4.6) that the street-spreading matrix is given by S =
44 0 16 44 0 12 0 0 12 00 44 16 0 44 0 12 0 0 1232 32 56 32 32 0 0 24 0 044 0 16 44 0 12 0 0 12 00 44 16 0 44 0 12 0 0 1212 0 0 12 0 44 0 16 44 00 12 0 0 12 0 44 16 0 440 0 24 0 0 32 32 56 32 3212 0 0 12 0 44 0 16 44 00 12 0 0 12 0 44 16 0 44 . (7.4.7) ON-INTEGRABLE SYSTEMS (IV) 45
This has non-zero eigenvalues and corresponding eigenvectors given by τ = 144 , ψ = (1 , , , , , , , , , T ,τ = 112 , ψ = ( − , , , − , , − , , , − , T ,τ = 96 , ψ = ( − , − , − , − , − , , , , , T ,τ = 64 , ψ = (1 , − , , , − , − , , , − , T ,τ = 48 , ψ = (1 , , − , , , , , − , , T . Note that the street-spreading matrix S given by (7.4.7) corresponds to the choiceof parameters m = n = 12. Switching to m = n = 12 k , where k (cid:62) S ( k ) = k S simply by multiplying thematrix S by k . Then of course the eigenvalues are also multiplied by k , but theeigenvectors remain the same. Naturally the 2-step transition matrix M is modifiedto M ( k ).We next determine some of the eigenvalues of M ( k ) using (7.2.38). The largesteigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 12 k √ k + 1 , (7.4.8)while the second largest eigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 4 k √ k + 7 , (7.4.9)with eigenvector of the formΨ = ( − , , , − , , − , , , − , , − τ ∗ , τ ∗ , , − τ ∗ , τ ∗ , − τ ∗ , τ ∗ , , − τ ∗ , τ ∗ ) T , where τ ∗ = − τ k + (cid:112) τ k + 4 τ k k √ k + 7 − k . Let us return to Figures 7.4.1 and 7.4.2. It is clear that the billiard moves to theup-neighbour if the 1-direction geodesic hits any of the edges h , . . . , h , and movesto the down-neighbor if the 1-direction geodesic hits any of the edges h , . . . , h . Itis also clear that the billiard moves to the right-neighbor if the 1-direction geodesichits any of the edges v , . . . , v , and moves to the left-neighbor if the 1-directiongeodesic hits any of the edges v , . . . , v .We now proceed to find the edge-cutting numbers of h , . . . , h and h , . . . , h .As in the example in Section 7.3, the eigenvalue λ does not contribute to theirdifference, as any lack of cancellation would violate the Gutkin–Veech theorem thatguarantees uniformity.The eigenvector of M ( k ) corresponding to the eigenvector Ψ of M | V is given by − u + u − u + u − u + u − u + u − τ ∗ v + τ ∗ v − τ ∗ v + τ ∗ v − τ ∗ v + τ ∗ v − τ ∗ v + τ ∗ v . (7.4.10)We first identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.2. The counts from u i , i = 1 , . . . ,
10, are u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ , u (cid:55)→ k , u (cid:55)→ k . (7.4.11)On the other hand, it is not difficult to show that each of v , . . . v contributesprecisely zero. Thus for the edges h , . . . , h , we have a total count of − k + 44 k − k + 44 k − k + 12 k − k + 12 k = 0 . We next identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.2. The counts from u i , i = 1 , . . . ,
10, are u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k . (7.4.12)Again, it is not difficult to show that each of v , . . . v contributes precisely zero.Thus for the edges h , . . . , h , we have a total count of − k + 12 k − k + 12 k − k + 44 k − k + 44 k = 0 . Thus there is perfect cancellation in the vertical direction.We next proceed to find the edge-cutting numbers of v , . . . , v and v , . . . , v ;see Figure 7.4.2.Again, the eigenvalue λ does not contribute to their difference, as any lack ofcancellation would violate the Gutkin–Veech theorem that guarantees uniformity.So we concentrate our attention on the contribution from the eigenvector Ψ of M | V . Here, note that none of u , . . . , u makes any non-zero contribution, as onlyunits of type − ↑ can contribute to the count. For the contributions from v , . . . , v ,we use (7.2.15).We first identity the number of almost vertical units counted in (7.4.10) that cutthe edges v , . . . , v ; see Figure 7.4.2. Since v and v do not feature in (7.4.10), itis not necessary to any counting for them. Clearly we have a count of 4 k for eachof v , v , v , v , and none for the rest, making a total of − τ ∗ k .We next identity the number of almost vertical units counted in (7.4.10) that cutthe edges v , . . . , v ; see Figure 7.4.2. Clearly we have a count of 4 k for each of v , v , v , v , and none for the rest, making a total of 16 τ ∗ k .The difference, in absolute value, is therefore 32 τ ∗ k . Let c = c , in (7.2.9) and(7.2.10). Then it follows that the second eigenvalue λ ( k ) contributes 32 c τ ∗ λ r ( k ) k to the difference between the edge-cuttings numbers of v , . . . , v and v , . . . , v .So the horizontal deviation from the starting point comes from the second largesteigenvalue, with order of magnitude λ r compared to the order of magnitude λ r ofthe main term. Choosing T = λ r , we have λ r (cid:16) T κ , where κ = κ ( k ) = log λ log λ = 2 log k + log 1122 log k + log 144 + o (1) , (7.4.13)in view of (7.4.8)–(7.4.9), is the irregularity exponent of a 1-direction geodesic ofslope α k = [12 k ; 12 k, k, k, . . . ] = 6 k + √ k + 1 = (cid:112) λ ( k ) (7.4.14)on the period-surface Bil(2; C) in Figure 7.4.2.Clearly κ = κ ( k ) → k → ∞ . So we have just established T κ = T − ε size super-fast escape rate to infinity for this infinite billiard with the explicit class ofquadratic irrational slopes in (7.4.14) where the parameter k (cid:62) κ = κ ( k ) in (7.4.13) is precisely the escape rate to infinity of this infinite billiard. The escape rate to infinity cannotbe larger than the expression (7.4.13) coming from the two largest eigenvalues.Note that this super-fast escape rate to infinity comes from horizontal deviation.As observed earlier, we have perfect cancellation in the vertical direction for the twolargest eigenvalues. So let us investigate what the third eigenvalue gives.The third largest eigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 4 k √ k + 6 , ON-INTEGRABLE SYSTEMS (IV) 47 with eigenvector of the formΨ = ( − , − , − , − , − , , , , , , − τ ∗ , − τ ∗ , − τ ∗ , − τ ∗ , − τ ∗ , τ ∗ , τ ∗ , τ ∗ , τ ∗ , τ ∗ ) T , where τ ∗ = − τ k + (cid:112) τ k + 4 τ k k √ k + 6 − k . The eigenvector of M ( k ) corresponding to the eigenvector Ψ of M | V is given by − u − u − u − u − u + u + u + 2 u + u + u − τ ∗ v − τ ∗ v − τ ∗ v − τ ∗ v − τ ∗ v + τ ∗ v + τ ∗ v + 2 τ ∗ v + τ ∗ v + τ ∗ v . We first identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.2. The counts from u i , i = 1 , . . . ,
10, are givenby (7.4.11). Again, it is not difficult to show that each of v , . . . v contributesprecisely zero. Thus for the edges h , . . . , h , we have a total count of − k − k − k − k − k + 12 k + 12 k + 12 k + 12 k = − k . We next identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.2. The counts from u i , i = 1 , . . . ,
10, are givenby (7.4.12). Again, it is not difficult to show that each of v , . . . v contributesprecisely zero. Thus for the edges h , . . . , h , we have a total count of − k − k − k − k + 44 k + 44 k + 64 k + 44 k + 44 k = 196 k . The difference, in absolute value, is therefore 392 τ ∗ k . Let c = c , in (7.2.9) and(7.2.10). Then it follows that the third eigenvalue λ ( k ) contributes 392 c λ r ( k ) k to the difference between the edge-cuttings numbers of h , . . . , h and h , . . . , h .So the vertical deviation from the starting point comes from the third largesteigenvalue, with order of magnitude λ r compared to the order of magnitude λ r ofthe main term. Example . Consider the Ehrenfest wind-tree model with a = b = 1 /
3, rescaledso that we have square faces of unit area. Consider a 4-direction billiard trajectory inthe infinite region in Figure 7.4.6. Here the building block of this infinite polysquaresurface is a 3 × ∞ ; 2; W). Here the index2 indicates that this is double periodic, and the letter W refers to the wind-treemodel. To construct Bil( ∞ ; 2; W), we take one of the building blocks, and unfoldthe 4-direction billiard flow on it to a 1-direction geodesic flow on a 4-copy versionof it, obtained by reflecting horizontally and vertically. Note that each copy ofthe period surface has a right-neighbor, a left-neighbor, a down-neighbor and anup-neighbor in Bil( ∞ ; 2; W), and we need appropriate edge identification for gluingthem together. The period-surface
Bil(2; W) of Bil( ∞ ; 2; W) is shown in the picture on the leftin Figure 7.4.7. The horizontal and vertical streets of Bil(2; W) are indicated in thepicture on the right in Figure 7.4.7 where, for instance, the entries ↔ (cid:108) h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v v ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l Figure 7.4.7: the period surface Bil(2; W) of Bil( ∞ ; 2; W)Consider now a 1-direction geodesic starting from some vertex of Bil(2; W ) withslope α given by (7.2.1) with m = n = 12.For u , u , u , u , Figure 7.4.8 describes the situation. With J = J = { , , } and J = J = { , , } , for the horizontal streets corresponding to u , u , u , u ,we have ( M − I ) u = ( M − I ) u = 44( u + v ) + 32( u + v ) + 44( u + v )+ 12( u + v ) + 12( u + v ) , (7.4.15)( M − I ) u = ( M − I ) u = 44( u + v ) + 32( u + v ) + 44( u + v )+ 12( u + v ) + 12( u + v ) . (7.4.16) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × u u u u Figure 7.4.8: almost vertical of type ↑ in u , u , u , u ON-INTEGRABLE SYSTEMS (IV) 49
For u , u , u , u , Figure 7.4.9 describes the situation. With J = J = { , , } and J = J = { , , } , for the horizontal streets corresponding to u , u , u , u ,we have ( M − I ) u = ( M − I ) u = 12( u + v ) + 12( u + v ) + 44( u + v )+ 32( u + v ) + 44( u + v ) , (7.4.17)( M − I ) u = ( M − I ) u = 12( u + v ) + 12( u + v ) + 44( u + v )+ 32( u + v ) + 44( u + v ) . (7.4.18) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × u u u u Figure 7.4.9: almost vertical of type ↑ in u , u , u , u For u , u , Figure 7.4.10 describes the situation. Noting that J = { , , , } and J = { , , , } , for the horizontal streets corresponding to u , u , we have( M − I ) u = 24( u + v ) + 24( u + v ) + 48( u + v )+ 24( u + v ) + 24( u + v ) , (7.4.19)( M − I ) u = 24( u + v ) + 24( u + v ) + 48( u + v )+ 24( u + v ) + 24( u + v ) . (7.4.20) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × u ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × u Figure 7.4.10: almost vertical of type ↑ in u It follows from (7.4.15)–(7.4.20) that the street-spreading matrix is given by S =
44 0 24 44 0 12 0 0 12 00 44 24 0 44 0 12 0 0 1232 32 48 32 32 0 0 0 0 044 0 24 44 0 12 0 0 12 00 44 24 0 44 0 12 0 0 1212 0 0 12 0 44 0 24 44 00 12 0 0 12 0 44 24 0 440 0 0 0 0 32 32 48 32 3212 0 0 12 0 44 0 24 44 00 12 0 0 12 0 44 24 0 44 . (7.4.21)This has non-zero eigenvalues and corresponding eigenvectors given by τ = 144 , ψ = (3 , , , , , , , , , T ,τ = 112 , ψ , = (0 , − , − , , − , , , , , T ,ψ , = ( − , , − , − , , , , , , T ,τ = 64 , ψ = (1 , − , , , − , − , , , − , T ,τ = 16 , ψ = (1 , , − , , , , , − , , T . Note that the street-spreading matrix S given by (7.4.21) corresponds to the choiceof parameters m = n = 12. Switching to m = n = 12 k , where k (cid:62) S ( k ) = k S simply by multiplying thematrix S by k . Then of course the eigenvalues are also multiplied by k , but theeigenvectors remain the same. Naturally the 2-step transition matrix M is modifiedto M ( k ).We next determine some of the eigenvalues of M ( k ) using (7.2.38). The largesteigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 12 k √ k + 1 , (7.4.22)while the second largest eigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 4 k √ k + 7 , (7.4.23)with eigenvectors of the formΨ , = (0 , − , − , , − , , , , , , , − τ ∗ , − τ ∗ , , − τ ∗ , τ ∗ , , τ ∗ , τ ∗ , T , Ψ , = ( − , , − , − , , , , , , , − τ ∗ , , − τ ∗ , − τ ∗ , , , τ ∗ , τ ∗ , , τ ∗ ) T , where τ ∗ = − τ k + (cid:112) τ k + 4 τ k k √ k + 7 − k . Let us return to Figures 7.4.6 and 7.4.7. It is clear that the billiard moves to theup-neighbour if the 1-direction geodesic hits any of the edges h , . . . , h , and movesto the down-neighbor if the 1-direction geodesic hits any of the edges h , . . . , h . Itis also clear that the billiard moves to the right-neighbor if the 1-direction geodesichits any of the edges v , . . . , v , and moves to the left-neighbor if the 1-directiongeodesic hits any of the edges v , . . . , v .We now proceed to find the edge-cutting numbers of h , . . . , h and h , . . . , h .As before, it is easily shown that each of v , . . . v contributes precisely zero. ON-INTEGRABLE SYSTEMS (IV) 51
As in Examples 7.3.1 and 7.3.2, the eigenvalue λ does not contribute to theirdifference, as any lack of cancellation would violate the Gutkin–Veech theorem thatguarantees uniformity.The eigenvector of M ( k ) corresponding to the eigenvector Ψ , of M | V is given by − u − u − u + u + u + u − τ ∗ v − τ ∗ v − τ ∗ v + τ ∗ v + τ ∗ v + τ ∗ v . (7.4.24)The eigenvector of M ( k ) corresponding to the eigenvector Ψ , of M | V is given by − u − u − u + u + u + u − τ ∗ v − τ ∗ v − τ ∗ v + τ ∗ v + τ ∗ v + τ ∗ v . (7.4.25)We first identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.7. The counts from u i , i = 1 , . . . ,
10, are u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ , u (cid:55)→ k , u (cid:55)→ k . Thus corresponding to the eigenvector Ψ , and (7.4.24), the total count is − k .Corresponding to the eigenvector Ψ , and (7.4.25), the total count is also − k .We next identity the number of almost vertical units counted here that cut theedges h , . . . , h ; see Figure 7.4.7. The counts from u i , i = 1 , . . . ,
10, are u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k , u (cid:55)→ k . Thus corresponding to the eigenvector Ψ , and (7.4.24), the total count is 112 k .Corresponding to the eigenvector Ψ , and (7.4.25), the total count is also 112 k .The difference, in absolute value, is therefore 224 k in each case. It follows that thesecond eigenvalue λ ( k ) contributes 224( c , + c , ) λ r ( k ) k to the difference betweenthe edge-cuttings numbers of h , . . . , h and h , . . . , h .Naturally, we can choose a starting vector w for the geodesic in question suchthat c , + c , (cid:54) = 0. This is clearly possible, since c , + c , = 0 always would violatethe linear independence of the eigenvectors Ψ , and Ψ , .So the deviation from the starting point comes from the second largest eigenvalue,with order of magnitude λ r compared to the order of magnitude λ r of the main term.Choosing T = λ r , we have λ r (cid:16) T κ , where κ = κ ( k ) = log λ log λ = 2 log k + log 1122 log k + log 144 + o (1) , (7.4.26)in view of (7.4.22)–(7.4.23), is the irregularity exponent of a 1-direction geodesic ofslope α k = [12 k ; 12 k, k, k, . . . ] = 6 k + √ k + 1 = (cid:112) λ ( k ) (7.4.27)on the period-surface Bil(2; W) in Figure 7.4.8.Again κ = κ ( k ) → k → ∞ . So we have just established T κ = T − ε size super-fast escape rate to infinity for this infinite billiard with the explicit class ofquadratic irrational slopes in (7.4.27) where the parameter k (cid:62) κ = κ ( k ) in (7.4.26) is precisely the escape rate to infinity of this infinite billiard. The escape rate to infinity cannotbe larger than the expression (7.4.26) coming from the two largest eigenvalues.In view of the examples in Sections 7.3–7.4 so far, it seems plausible to conjecturethat every periodic infinite polysquare billiard with at least one infinite street ex-hibits super-fast escape rate to infinity for some concrete infinite class of quadraticirrational slopes. It would be nice to prove this conjecture in general. We complete this section with a hybrid example.
Example . Our final example is modified from the Ehrenfest wind-tree model.Consider a 2-direction billiard trajectory in the region in Figure 7.4.11. Here thebuilding block of this infinite polysquare surface is a 3 × ∞ ; 2; H). Here the index 2indicates that this is double periodic, and the letter H refers to the hybrid model.To construct Bil( ∞ ; 2; H), we take one of the building blocks, and unfold the 2-direction billiard flow on it to a 1-direction geodesic flow on a 2-copy version of it,obtained by reflecting vertically. Note that each 2-copy version has a right-neighbor,a left-neighbor, a down-neighbor and an up-neighbor in Bil( ∞ ; 2; H), and we needappropriate edge identification for gluing them together.The period-surface Bil(2; H) of Bil( ∞ ; 2; H) is shown in the picture on the left inFigure 7.4.12. h h h h h h h h h h h h h h h h v v v v v v v v v v v v v v v v ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ ↔ l l l l l l l l l l l l l l l l Figure 7.4.12: the period surface Bil(2; H) of Bil( ∞ ; 2; H)The horizontal and vertical streets of Bil(2; H) are indicated in the picture on theright in Figure 7.4.12 where, for instance, the entries ↔ (cid:108) H ) withslope α given by (7.2.1) with m = n = 12.For u , u , u , u , Figure 7.4.13 describes the situation. With J = J = { , , } and J = J = { , , } , for the horizontal streets corresponding to u , u , u , u , wehave ( M − I ) u = ( M − I ) u = 56( u + v ) + 32( u + v ) + 56( u + v ) (7.4.28)( M − I ) u = ( M − I ) u = 56( u + v ) + 32( u + v ) + 56( u + v ) . (7.4.29) ON-INTEGRABLE SYSTEMS (IV) 53 ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × u u u u Figure 7.4.13: almost vertical units of type ↑ in u , u , u , u For u , Figure 7.4.14 describes the situation. With J = { , , , } , for thehorizontal street corresponding to u , we have( M − I ) u = 24( u + v ) + 24( u + v ) + 48( u + v )+ 24( u + v ) + 24( u + v ) . (7.4.30) ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , ↑ , × × × × × × × × × × × × u Figure 7.4.14: almost vertical units of type ↑ in u It follows from (7.4.28)–(7.4.30) that the street-spreading matrix is given by S =
56 0 24 56 00 56 24 0 5632 32 48 32 3256 0 24 56 00 56 24 0 56 . (7.4.31)This has non-zero eigenvalues and corresponding eigenvectors given by τ = 144 , ψ = (3 , , , , T ,τ = 112 , ψ = ( − , , , − , T ,τ = 16 , ψ = (1 , , − , , T . Note that the street-spreading matrix S given by (7.4.31) corresponds to the choiceof parameters m = n = 12. Switching to m = n = 12 k , where k (cid:62) S ( k ) = k S simply by multiplying thematrix S by k . Then of course the eigenvalues are also multiplied by k , but theeigenvectors remain the same. Naturally the 2-step transition matrix M is modifiedto M ( k ).We next determine some of the eigenvalues of M ( k ) using (7.2.38). The largesteigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 12 k √ k + 1 , while the second largest eigenvalue of M ( k ) is λ ( k ) = 1 + τ k + (cid:112) τ k + 4 τ k k + 4 k √ k + 7 , with eigenvector of the formΨ = ( − , , , − , , − τ ∗ , τ ∗ , , − τ ∗ , τ ∗ ) T , where τ ∗ = − τ k + (cid:112) τ k + 4 τ k k √ k + 7 − k . As observed earlier, there is no billiard in the vertical direction, so we now proceedto study its horizontal behaviour. We thus proceed to find the edge-cutting numbersof v , v , v and v , v , v ; see Figure 7.4.12.It can be shown that the eigenvalue λ does not contribute to their difference,although the Gutkin–Veech theorem does not apply here. So we concentrate ourattention on the contribution from the eigenvector Ψ of M | V .The eigenvector of M ( k ) corresponding to the eigenvector Ψ of M | V is given by − u + u − u + u − τ ∗ v + τ ∗ v − τ ∗ v + τ ∗ v . (7.4.32)Here, note that none of u , . . . , u makes any non-zero contribution, as only unitsof type − ↑ can contribute to the count. For the contributions from v , . . . , v , we use(7.2.15).We first identity the number of almost vertical units counted in (7.4.32) that cutthe edges v , v , v ; see Figure 7.4.12. Since v does not feature in (7.4.32), it is notnecessary to any counting for it. Clearly we have a count of 4 k for each of v , v ,and none for v , v , making a total of − τ ∗ k .We next identity the number of almost vertical units counted in (7.4.32) that cutthe edges v , v , v ; see Figure 7.4.12. Clearly we have a count of 4 k for each of v , v , and none for v , v , making a total of 8 τ ∗ k .The difference, in absolute value, is therefore 16 τ ∗ k . Thus the second eigenvalue λ ( k ) contributes 16 c τ ∗ λ r ( k ) k to the difference between the edge-cuttings numbersof v , v , v and v , v , v .Again the horizontal deviation from the starting point comes from the secondlargest eigenvalue, with order of magnitude λ r compared to the order of magnitude λ r of the main term.7.5. Geodesics with arbitrary starting points.
In Sections 7.3 and 7.4, we haveexhibited examples of 1-direction geodesics in infinite polysquare surfaces that startfrom a vertex and which exhibit super-fast escape rate to infinity. In Example 7.3.2,we have also shown that there are 1-direction geodesics in square-mazes that startfrom a vertex and which exhibit super-slow escape rate to infinity.We now describe a method by which we can relax the restriction that the startingpoint of the geodesic has to be a vertex of a polysquare, so that only the angle α ofthe geodesic matters, and can still establish results on the escape rate to infinity asbefore.Note, first of all, that a 1-direction geodesic modulo one is equivalent to a torusline in the unit square. To help us visualize the situation even more clearly, we canreplace the unit square by R and the torus line by a line on R .Next, note that we are considering slopes α that are badly approximable numbers.Such numbers satisfy the following two properties. Property A.
Suppose that α is badly approximable. Then there exists a constant C ∗ = C ∗ ( α ) > such that for every real number (cid:96) (cid:62) , every segment of length (cid:96) ofa straight line of slope α has distance at most C ∗ /(cid:96) from the nearest integer latticepoint in Z . ON-INTEGRABLE SYSTEMS (IV) 55 b nearest lattice point ℓ ≤ C ∗ ℓ b origin no lattice point ℓ C ∗∗ ℓ Figure 7.5.1: illustrating Property A and Property B
Property B.
Suppose that α is badly approximable. Then there exists a constant C ∗∗ = C ∗∗ ( α ) > such that for any real number (cid:96) (cid:62) , any tilted rectangle with oneside starting from the origin, with slope α and length (cid:96) , and with the perpendicularside of length C ∗∗ /(cid:96) , does not contain any integer lattice point in Z except theorigin. The idea is to consider geodesics that are parallel to our given geodesic and whichshare some of its characteristics.Consider the initial segment L ( S ; t ), 0 (cid:54) t (cid:54) T , with T (cid:62)
2, of a 1-directiongeodesic L of slope α , starting at a point S , not necessarily a vertex of a polysquaresurface P , as shown in the top half of Figure 7.5.2. Let Q = Q ( T ) = L ( S ; t )denote the point on this segment which is closest to a vertex of P , and let V be thisclosest vertex. Using Property A, we see that the distance between Q and V is atmost C ∗ /T . b b b b b b T ′ S L ( S ; t ) = QV L ( S ; t + T ′ ) = Q V L ( S ; T )no lattice point b b b b bb b b T ′′ L ( S ; t + T ′ + T ′′ ) = Q L ( S ; t + T ′ ) = Q V V L ( S ; T ) S Q V no lattice point
Figure 7.5.2: using parallel geodesicsStarting from the vertex V , we draw a new geodesic L (cid:48) of slope α , in the sameforward direction as L . Consider the longest vertex-free rectangle between L and L (cid:48) .Using Property B, we see that the side of this rectangle parallel to L has length T (cid:48) (cid:62) C ∗∗ C ∗ T. (7.5.1)Suppose that T (cid:54) t + T (cid:48) . Then the segment L ( S ; t ), t (cid:54) t (cid:54) T , of L remainsclose to the segment L (cid:48) ( V ; t ), 0 (cid:54) t (cid:54) T − t , of the new geodesic L (cid:48) , since therectangle between them does not contain any vertex of P apart from V and anothervertex at the far end if T = t + T (cid:48) .Suppose that T > t + T (cid:48) . Then the segment L ( S ; t ), t (cid:54) t (cid:54) t + T (cid:48) , of L remainsclose to the segment L (cid:48) ( V ; t ), 0 (cid:54) t (cid:54) T (cid:48) , of the new geodesic L (cid:48) , since the rectangle between them does not contain any vertex of P apart from at the two ends. Sincethis rectangle is the longest vertex-free rectangle that we can draw, clearly there isanother vertex V of P at the far end. We now repeat the argument.Starting from the vertex V , we draw a new geodesic L (cid:48)(cid:48) of slope α , in the sameforward direction as L . Consider the longest vertex-free rectangle between L and L (cid:48)(cid:48) ,as shown in the bottom half of Figure 7.5.2. Using Property B, we see that the sideof this rectangle parallel to L has length T (cid:48)(cid:48) (cid:62) C ∗∗ C ∗ T. (7.5.2)Suppose that t + T (cid:48) < T (cid:54) t + T (cid:48) + T (cid:48)(cid:48) . Then the segment L ( S ; t ), t + T (cid:48) (cid:54) t (cid:54) T ,of L remains close to the segment L (cid:48)(cid:48) ( V ; t ), 0 (cid:54) t (cid:54) T − t − T (cid:48) , of the newgeodesic L (cid:48)(cid:48) , since the rectangle between them does not contain any vertex of P apart from V and another vertex at the far end if T = t + T (cid:48) + T (cid:48)(cid:48) .Suppose that T > t + T (cid:48) + T (cid:48)(cid:48) . Then the segment L ( S ; t ), t + T (cid:48) (cid:54) t (cid:54) t + T (cid:48) + T (cid:48)(cid:48) ,of L remains close to the segment L (cid:48)(cid:48) ( V ; t ), 0 (cid:54) t (cid:54) T (cid:48)(cid:48) , of the new geodesic L (cid:48)(cid:48) ,since the rectangle between them does not contain any vertex of P apart from atthe two ends. Since this rectangle is the longest vertex-free rectangle that we candraw, clearly there is another vertex V of P at the far end. We now repeat theargument again.Clearly the argument must stop after a finite number of steps, in view of estimatessuch as (7.5.1) and (7.5.2) and their analogs. In particular it must stop after at most C ∗ /C ∗∗ steps.Note that we have moved forward from the point Q to the point L ( S ; T ). Clearlya similar argument applies when we move backward from the point Q to the startingpoint S = L ( S ; 0).What we have shown is that the finite geodesic L ( S ; t ), 0 (cid:54) t (cid:54) T , can be brokenup into a finite number of parts, each of which remains close to a parallel finitegeodesic of the same length that starts from a vertex of the polysquare surface P .We can now use this observation to study escape rates to infinity.Suppose that we have established that for a polysquare surface P , the escaperate to infinity for a 1-direction geodesic L of slope α that starts at a vertex of P is O ( f ( T )) as a function of time T , where f is an increasing function satisfying f ( t ) → ∞ as t → ∞ . This means that the diameter of the finite geodesic L ( t ),0 (cid:54) t (cid:54) T , is O ( f ( T )). Consider now a 1-direction geodesic L ( S ; t ), 0 (cid:54) t (cid:54) T , ofslope α and starting point S that is not necessarily a vertex of P . We now break L ( S ; t ), 0 (cid:54) t (cid:54) T , into a finite number of parts as described above, and note thateach part remains close to a parallel finite geodesic that starts at a vertex of P and has length at most T . The diameter of each of these parallel finite geodesics is O ( f ( T )). It follows that the diameter of L ( S ; t ), 0 (cid:54) t (cid:54) T , is also O ( f ( T )).For super-slow logarithmic escape rate to infinity, we have f ( T ) = log T . Forsuper-fast escape rate to infinity as in the examples in Sections 7.3 and 7.4, we have f ( T ) = T κ .To exhibit fluctuations of the required order of magnitude in the escape rate toinfinity, we have a similar approach but with a slightly different first step.Let n (cid:62) L ( S ; t ), t (cid:62)
0, on a polysquare surface P that starts from an arbitrary pointwith slope α . Suppose that t is the first value of t (cid:62) L ( S ; t ) hasperpendicular distance at most 1 /n from a vertex of P , as shown in Figure 7.5.3. ON-INTEGRABLE SYSTEMS (IV) 57 b b b b b n S L ( S ; t ) W L ( S ; t + n )no lattice point Figure 7.5.3: using parallel geodesicsIn view of Property A, we know that 0 (cid:54) t (cid:54) C ∗ n .Suppose that the vertex of P in question is W . Starting from this vertex W , wedraw a new geodesic L (cid:48) of slope α , in the same forward direction as L . Consider thelongest vertex-free rectangle between L and L (cid:48) . Using Property B, we see that theside of this rectangle parallel to L has length n (cid:62) C ∗∗ n. (7.5.3)Furthermore, the segment L ( S ; t ), t (cid:54) t (cid:54) t + n , of L remains close to the segment L (cid:48) ( W ; t ), 0 (cid:54) t (cid:54) n , of the new geodesic L (cid:48) , since the rectangle between them doesnot contain any vertex of P apart from at the two ends.Suppose now that fluctuations of size C T κ are exhibited for geodesics of slope α and length T that start from a vertex of P . Then L (cid:48) ( W ; t ), 0 (cid:54) t (cid:54) n , and hencealso L ( S ; t ), t (cid:54) t (cid:54) t + n , exhibits fluctuations of size C n κ (cid:62) C n κ , in view of (7.5.3). If the distance between S and L ( S ; t ) exceeds12 C n κ , (7.5.4)then L ( S ; t ), 0 (cid:54) t (cid:54) t , has diameter at least equal to (7.5.4). Otherwise L ( S ; t ),0 (cid:54) t (cid:54) t + n , has diameter at least equal to C n κ − C n κ = 12 C n κ . Beyond polysquare surfaces
Street-rational polyrectangle surfaces.
For polysquare surfaces, we havedeveloped two very different versions of the shortline method. Using the eigenvalue-based version of the method, we can prove time-quantitative uniformity in termsof the irregularity exponent, as in [2, 3] and Section 7 of the present paper. Usingthe eigenvalue-free version of the method, we can prove time-quantitative density,including superdensity in some cases, as in [4].The purpose of this section is to show that both versions of the shortline methodcan be extended to the class of street-rational polyrectangle surfaces , a class whichincludes all polysquare surfaces but goes far beyond. The first example of this largerclass comes from right triangle billiard with angle π/
8. This is perhaps the simplestnon-integrable right triangle billiard. The standard trick of “unfolding” implies thatthis billiard can be described in terms of a 1-direction geodesic flow on the regularoctagon surface , as shown in Figure 8.1.1. A A A A A A A A BC Figure 8.1.1: right triangle billiard with angle π/
8, and geodesic flowon the regular octagon surface via unfoldingThe definition of the regular octagon surface is rather straightforward. We identifythe opposite parallel boundary edges by translation. Thus the four identified pairsare ( A A , A A ) , ( A A , A A ) , ( A A , A A ) , ( A A , A A ) . Thus a 1-direction geodesic flow on the regular octagon surface, a compact orientablesurface, is a 16-fold covering of the right triangle billiard with angle π/
8, in muchthe same way as a torus line flow on a 2 × C represents the center of the octagon. The right triangle A BC has angle π/ C . Reflecting the A BC billiard across the side CB is thefirst step in the unfolding process, and leads to the triangle A A C . There are threemore steps. Reflecting A A C across the side CA leads to a polygon A A A C .Reflecting A A A C across the side CA leads to a polygon A A A A A C . Finally,reflecting A A A A A C across the side CA leads to the whole octagon.Non-integrability is clear from the unfolding, since the vertices of the octagon aresplit-singularities of the geodesic flow on the surface.Of course the same elegant construction works for any right triangle billiard withangle π/k where k (cid:62) even . Unfolding will then convert the billiard orbit to a1-direction geodesic flow on the regular k -gon surface , defined by identifying parallelboundary edges of the k -gon by translation. These are non-integrable systems forevery even integer k (cid:62)
8, so the regular octagon is the simplest such system.
Remark.
Consider the regular k -gon surface with even k (cid:62)
4. If k is divisible by 4,then the boundary identification gives 1 vertex, k/ − g = χ = V − E + R = 1 − ( k/
2) + 1gives g = k/
4. If k is not divisible by 4, then the boundary identification gives 2vertices, k/ − g = χ = V − E + R = 2 − ( k/
2) + 1gives g = ( k − /
4. Thus the regular k -gon surface with even k has genus 1 when k = 4 ,
6. This is consistent with the well known fact that we can tile the plane withsquares or regular hexagons, but not with any other regular polygons with an evennumber of sides.Let us return to the regular octagon surface. While it looks completely differentfrom a polysquare surface, there is a hidden similarity. The regular octagon surfaceis in fact equivalent to a polyrectangle surface ; see Figure 8.1.2.
ON-INTEGRABLE SYSTEMS (IV) 59 A A A A A A A A a ab bccd de ef f − − Figure 8.1.2: the regular octagon surface seen as a polyrectangle surfaceThe edge A A is identified with the edge A A . This allows us to replace thetriangle labelled 7 − by the triangle labelled 7+, with the two horizontal edges b identified and the two vertical edges e identified. Likewise, the edge A A is identifiedwith the edge A A . This allows us to replace the triangle labelled 9 − by the trianglelabelled 9+, with the two horizontal edges a identified and the two vertical edges d identified. Thus the regular octagon surface becomes a polyrectangle surfaceconsisting of 7 rectangles, labelled(1 , , , (3 , , , , , . With the edge identification, this polyrectangle surface has 2 horizontal streets(1 , , , (3 , , , , , as well as 2 vertical streets(1 , , , (3 , , , , . The two identified edges e also allows us to replace the rectangle 8 at the bottomin Figure 8.1.2, indicated by 8 − in Figure 8.1.3, by a rectangle on the top rightindicated by 8+. A A A A A A A A D D D D D − − − Figure 8.1.3: street decomposition into similar rectanglesFigures 8.1.2 and 8.1.3 justify the claim that the regular octagon surface is infact a polyrectangle surface . Furthermore, we have a second crucial property of theregular octagon surface, that the three rectangles D A D A , A D D D , A A A A are similar. To see this, assume, without loss of generality, that length( A D ) = 1,so that each edge of the regular octagon has length √
2. Thenlength( A D )length( A D ) = length( D D )length( D D ) = 1 + √ , (8.1.1) and length( A A )length( A A ) = 2 + √ √ √ . (8.1.2)We also have length( A D )length( D D ) = 2(1 + √ . (8.1.3)Note that (8.1.3) shows that the cotangent of the diagonal A D of the horizontalstreet D A D D is equal to 2(1 + √ A A of the horizontal street A A A A is equal to 1 + √
2. We candraw similar conclusions for the vertical streets.
Remark.
The equality of the ratios in (8.1.1) and (8.1.2) can also be establishedby using the well known geometric property of the circle very often known as the“chord-angle relation” and which states that if we fix any chord with endpoints
P, Q on a circle, then the angle
P RQ remains the same for any point R on the samecircular arc P Q . Since A A and A A are two chords of the same length, the angles A A A and A A A are equal.The ratio of the cotangents of the diagonals of the horizontal streets, and similarly,the ratio of the cotangents of the diagonals of the vertical streets, is rational . Thuswe refer to the regular octagon surface as a street-rational polyrectangle surface . Remark.
Consider a horizontal street on a finite polyrectangle surface P . The widthof this street, i.e. the perpendicular distance between its top and bottom horizontaledges, may not be equal to 1. Let us expand or contract this street to obtain asimilar copy where the width is now equal to 1. Then we call the length of thisexpanded or contracted horizontal street the normalized length of the horizontalstreet. Some authors also use the terms modulus and cylinder in place of the terms normalized length and street . We can also think of this as the cotangent of the (anglethat the) diagonals (make with the direction) of the horizontal street. For a finitestreet-rational polyrectangle surface P , there clearly exists a smallest real number h ∗ which is an integer multiple of the normalized length of every horizontal street.We call h ∗ the normalized horizontal street-LCM of P . Analogous to this, we denoteby v ∗ the normalized vertical street-LCM of P .To have a suitable version of the surplus shortline method on a street-rationalpolyrectangle surface P , we must consider slopes of the form α = v ∗ a + 1 h ∗ a + v ∗ a + h ∗ a ··· , (8.1.4)where a , a , a , a , . . . are positive integers.Note that h ∗ plays the analogous role of an integer which is the least integermultiple of the lengths of the horizontal streets in a polysquare surface, and v ∗ playsthe analogous role of an integer which is the least integer multiple of the lengths ofthe vertical streets in a polysquare surface. We can view the expression (8.1.4) asan extension of the concept of continued fraction expansion. For the success of thesurplus shortline method in a street-rational polyrectangle surface, we require these“digits” to be integer multiples of h ∗ and v ∗ as relevant.Note that it is not necessary that h ∗ and v ∗ are rational multiples of each other,so there may not be a quantity that corresponds to the street-LCM of a finitepolysquare surface.Indeed, we have the two key ingredients needed for the success of the shortlinemethod. Working with a polyrectangle automatically gives rise to the horizontal ON-INTEGRABLE SYSTEMS (IV) 61 and vertical directions, vital for the shortline method which is an alternating pro-cess between two directions, while street-rationality guarantees that the concept of shortline is well defined.Thus the eigenvalue-free version of the shortline method developed in Sections 6.2and 6.4 in [4] also works for street-rational polyrectangle surfaces, and establishessuperdensity of the right triangle billiard with angle π/ √
2) and 1 + √
2, so that h ∗ = v ∗ = 2(1 + √ α = 2(1 + √ a + 12(1 + √ a + √ a + ··· , (8.1.5)and their reciprocals α − , where a i (cid:62) i (cid:62)
0, is an infinite sequence of positiveintegers bounded from above .The choice (8.1.5) represents the analog of those badly approximable slopes forwhich the eigenvalue-free version of the shortline method works in the case ofpolysquare surfaces.Figure 8.1.4 illustrates some almost horizontal detour crossings and their almostvertical shortcuts.Figure 8.1.4: almost horizontal detour crossingsand their almost vertical shortcutsBy a straightforward adaptation of the proof of Theorems 6.1.1 and 6.4.1 tothe street-rational polyrectangle surface in Figures 8.1.2–8.1.4 that represents theregular octagon surface, we obtain the following result concerning the right trianglebilliard with angle π/ Theorem 8.1.1. (i)
Consider the right triangle with angle π/ . Let α > be a realnumber of the form (8.1.5) . Then any half-infinite billiard orbit in the right trianglewith angle π/ with initial slope α exhibits superdensity. Figures 8.1.2–8.1.4 are based on the parallel decomposition of the regular octagoninto 3 parts as shown in the picture on the left in Figure 8.1.5. The picture on theright shows a different decomposition into 4 parts.Figure 8.1.5: parallel decompositions of the regular octagonFigure 8.1.6 illustrates the representation of the regular octagon surface as astreet-rational polyrectangle surface, based on this second decomposition. The 4rectangles are similar. Since A A and A A are two chords of the same length, theangles A A A and A A A are equal. A A A A A A A A − − − − Figure 8.1.6: another street decomposition into similar rectanglesIt is not difficult to see that this polyrectangle surface, suitably rotated, has 2horizontal streets and 2 vertical streets. For instance, in one of these two directions,one street comprises the two smaller rectangles, and the other street comprises thetwo larger rectangles. Furthermore, it can be checked that the normalized lengths ofthe streets are equal to 2(1 + √ new explicitslopes such that any half-infinite orbit with such an initial slope in the right trianglebilliard with angle π/ π/k , where k (cid:62) k -gon are all on a circle, and street-rationality comes fromthe “chord-angle relation” on this circle. We leave the details to the reader. Theorem 8.1.1. (ii)
Let k (cid:62) be an even integer, and consider the right-trianglewith angle π/k . There exist infinitely many slopes, depending on k , such that anyhalf-infinite billiard orbit in the right triangle with angle π/k with such an initialslope exhibits superdensity. (iii) In general, consider an arbitrary finite street-rational polyrectangle surface P .There exist infinitely many slopes, depending on P , such that any half-infinite -direction geodesic on P having such a slope exhibits superdensity. (iv) As in Theorem 7.1.1, for many of these slopes we can explicitly computethe irregularity exponent. Combining the irregularity exponent with the method ofzigzagging introduced in Section 3.3 in [2] , we can also describe, for a geodesic flowon P with such a slope, the time-quantitative behavior of the edge-cutting and face-crossing numbers, as well as equidistribution relative to all convex sets. In the course on the next few sections, we shall emphasize on parts (iii) and (iv)by giving a few examples; see Theorems 8.3.1 and Theorems 8.4.1–8.4.3.
Remark.
This paper is not about ergodic theory. Instead, we focus on the time-quantitative evolution of individual orbits instead of geodesic flow as a whole. Nev-ertheless, it is very interesting to point out a substantial difference between geodesicflow on a polysquare surface and geodesic flow on a regular k -gon surface with even k (cid:62)
8. From the viewpoint of ergodic theory these two flows are very different. Thelatter is weakly mixing in almost every direction, while the former is not weaklymixing in any direction; see [1].Our main point is that, despite this big difference between the two flows, theshortline method works equally well for individual orbits in either case.However, whereas for a polysquare surface, a geodesic modulo one is equivalent toa torus line in the unit square, there is no corresponding analog for the street-rationalpolyrectangle surface corresponding to the regular octagon surface.
ON-INTEGRABLE SYSTEMS (IV) 63
In the next section, we illustrate the proof of part (iv) in a special case thatreflects the whole difficulty of the general case.8.2.
Computing the irregularity exponent for the octagon surface.
We nowapply the eigenvalue-based version of the shortline method on the regular octagonsurface, and compute the irregularity exponent of 1-direction geodesics with certainslopes. In particular, we consider special slopes of the form α = 2(1 + √ a + 12(1 + √ a + √ a + ··· , (8.2.1)where a i (cid:62) i (cid:62)
0, form a sequence of integers which is eventually periodic. Thislast requirement distinguishes (8.2.1) from (8.1.5).Since the common ratio of the vertical and horizontal sides of the rectangles inFigure 8.1.4 is 1 + √
2, the shortline of a geodesic on the regular octagon surfacewith slope α given by (8.2.1) has slope α − , where α = 2(1 + √ a + 12(1 + √ a + √ a + ··· , and the shortline of this shortline has slope α = 2(1 + √ a + 12(1 + √ a + √ a + ··· , and so on. Note that for α, α , α , . . . , the usual shift of digits a , a , a , . . . applies. Theorem 8.2.1.
Consider -direction geodesic flow on the regular octagon surfacewith slope α given by (8.2.1) . If the sequence a , a , a , . . . has period k , k , . . . , k r eventually, then the irregularity exponent of the geodesic with slope α is equal to log | λ | log | Λ | , where λ is the eigenvalue with the larger absolute value of the product matrix r (cid:89) i =1 (cid:18) − √ − k i
11 0 (cid:19) , and Λ is the eigenvalue with the larger absolute value of the product matrix r (cid:89) i =1 (cid:18) √ k i
11 0 (cid:19) . Alternatively, we have the product formula
Λ = r (cid:89) i =1 β i , where for every i = 1 , . . . , r , β i = 2(1 + √ k i + 12(1 + √ k i +1 + √ k i +2 + ··· , corresponding to the periodic sequence k i , k i +1 , k i +2 , . . . , k r , k , k , . . . , k r , . . . . Remark.
If the sequence a , a , a , . . . has period k , k eventually, of length 2, then itis possible to use the street-spreading matrix in the same spirit as in Theorem 7.2.2,noting that the method there works even for street-rational polyrectangle surfaces.But here the period can be arbitrarily long. Proof of Theorem 8.2.1.
We apply an adaptation of the original eigenvalue-basedversion of the surplus shortline method developed in Sections 3 and 4 in [2, 3].First of all, we note that the regular octagon region, viewed as a street-rationalpolyrectangle surface as in Figures 8.1.2–8.1.4, has 7 horizontal and 7 vertical edges,some with identified pairs, as shown in Figure 8.2.1. h h h h h h h h h h h h v v v v v v v v v v v v Figure 8.2.1: horizontal and vertical edges of the regular octagon surfaceviewed as a street-rational polyrectangle surfaceWe distinguish the 14 types of almost vertical units h h , h h , h h , h h , h h , h h , h h , h h , h h , h h , h h , h h , h h , h h , as shown in Figure 8.2.2. h h h h h h h h h h h h h h h h h h h h h h h h h h h h Figure 8.2.2: almost vertical units of the regular octagon surfaceWe distinguish the 14 types of almost horizontal units v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , as shown in Figure 8.2.3. v v v v v v v v v v v v v v v v v v v v v v v v v v v v Figure 8.2.3: almost horizontal units of the regular octagon surfaceConsider first the almost vertical unit h h . It is not difficult to see from Figures8.2.2 and 8.2.3 that h h is the shortcut of an almost horizontal detour crossing of ON-INTEGRABLE SYSTEMS (IV) 65 a horizontal street, made up of a fractional almost horizontal unit v v , full almosthorizontal units v v , v v , v v , then ( k −
1) copies of full almost horizontal units v v , v v , v v , v v , and finally a fractional almost horizontal unit v v . This canbe summarized by h h → v v , v v , v v , v v , ( v v , v v , v v , v v ) k − , v v , where k (cid:62) v v as a fullunit, discard the final fractional almost horizontal unit v v , and call these units theancestor units of h h . Repeating the same exercise on the other 13 almost verticalunits, we can summarize this ancestor process by h h (cid:42) v v , k × v v , k × v v , k × v v , ( k − × v v , (8.2.2) h h (cid:42) v v , k × v v , k × v v , k × v v , k × v v , (8.2.3) h h (cid:42) v v , k × v v , k × v v , (2 k − × v v , (8.2.4) h h (cid:42) v v , k × v v , k × v v , k × v v , (8.2.5) h h (cid:42) v v , k × v v , k × v v , (2 k − × v v , (8.2.6) h h (cid:42) v v , k × v v , k × v v , k × v v , (8.2.7) h h (cid:42) v v , k × v v , k × v v , k × v v , (8.2.8) h h (cid:42) v v , k × v v , k × v v , (2 k − × v v , (8.2.9) h h (cid:42) v v , k × v v , k × v v , k × v v , k × v v , (8.2.10) h h (cid:42) v v , k × v v , k × v v , k × v v , ( k − × v v , (8.2.11) h h (cid:42) v v , k × v v , k × v v , k × v v , ( k − × v v , (8.2.12) h h (cid:42) v v , k × v v , k × v v , k × v v , k × v v , (8.2.13) h h (cid:42) v v , k × v v , k × v v , k × v v , ( k − × v v , (8.2.14) h h (cid:42) v v , k × v v , k × v v , k × v v , k × v v , (8.2.15)These lead to a 14 ×
14 transition matrix M ( k ) = (cid:18) M , ( k ) M , ( k ) M , ( k ) M , ( k ) (cid:19) , where any particular row captures the information in the ancestor relation of thealmost vertical unit in question by displaying the multiplicities of each of its ancestoralmost horizontal units. We have M , ( k ) = v v v v v v v v v v v v v v h h k − h h k h h k − k h h k k h h k k − h h k k h h k k , M , ( k ) = v v v v v v v v v v v v v v h h k k k h h k k k h h k h h k h h k h h k h h k ,M , ( k ) = v v v v v v v v v v v v v v h h k k h h k h h k h h k h h k h h k h h k , and M , ( k ) = v v v v v v v v v v v v v v h h k − h h k k k h h k k − k h h k − k k h h k k k h h k k k − h h k k k . Similarly, we can study the ancestor relation of each of the almost horizontalunits, again using the Delete-End Rule, and obtain the analogs of (8.2.2)–(8.2.15).These will lead to another 14 ×
14 transition matrix. Since we have listed the almostvertical units and almost horizontal units in lexicographical order, these two 14 × M ( k ), there are 10 irrelevant ones of the form ± ± iand ( − ± √ /
2. The more interesting ones are(1 + √ k ± (cid:16) (1 + √ k + 1 (cid:17) / , (1 − √ k ± (cid:16) (1 − √ k + 1 (cid:17) / . Clearly the eigenvalue with the largest absolute value isΛ = (1 + √ k + (cid:16) (1 + √ k + 1 (cid:17) / , (8.2.16)and the eigenvalue with the second largest absolute value is λ = − ( √ − k − (cid:16) ( √ − k + 1 (cid:17) / . (8.2.17)The other two eigenvalues among these four are also irrelevant.We shall show that the transition matrix M ( k ) has a conjugate with the form P − M ( k ) P = (cid:18) T ?0 A ( k ) (cid:19) , (8.2.18)where T is a 10 ×
10 triangular matrix with main diagonal entries1 , − , − , − , − i , i , − , , − − √ , − √ , (8.2.19) ON-INTEGRABLE SYSTEMS (IV) 67 in this order, the entries of T below the main diagonal are all zero, and A ( k ) = √ k − √ − k
10 0 1 0 . (8.2.20)The description of the matrix M ( k ) by (8.2.18)–(8.2.20) is extremely convenient. Itreduces the necessary eigenvalue computation of arbitrary products r (cid:89) i =1 M ( k i )of 14 ×
14 matrices with different values of the branching parameter k i to the muchsimpler eigenvalue computation of products r (cid:89) i =1 (cid:18) √ k i
11 0 (cid:19) and r (cid:89) i =1 (cid:18) − √ − k i
11 0 (cid:19) of 2 × ± ± i and ( − ±√ / M ( k ) has 6 eigenvectors that are independentof the branching parameter k . Together with the eigenvalues, they are λ = 1 , v = ( − , , , , , , , − , , , , , , T ,λ = − , v = ( − , , − , , , , , , , , , , , T ,λ = − , v = ( − , , − , , , , , , , , , , , T ,λ = − , v = ( − , , − , , , , , , , , , , , T ,λ = − i , v = (1 , , i , , − , − , , − , , , − i , − , − i , T ,λ = i , v = (1 , , − i , − , − − i , − , , − , , , i , , , T . This observation allows us to apply a “partial diagonalization” trick first discussedin Lemma 4.1.1 in [3]. Let Q be a 14 ×
14 invertible matrix such that the first 6columns are v , . . . , v . Then for every 1 (cid:54) i (cid:54)
6, the i -th column of the conjugate Q − M ( k ) Q has the special form that its i -th element is λ i , and the remainingelements are all zero. A concrete choice of Q gives Q − M ( k ) Q = (cid:18) D ?0 M ( k ) (cid:19) , where D is a 6 × λ , . . . λ , and M ( k ) = − − √ − √ − √ − √ k i6 14 − √ k i6 + √ − + √ √ + √ k i6 14 + √ k i6 − k − √ − k − − √ k − k − k − k − √ − k − − √ k + k + k k − √ k − − √ − − k − − k − − k − k − k k k k k − √ k − − √ − k − − k − − k − k k k k . We next make use of the fact that M ( k ) has 4 eigenvectors that are independentof the branching parameter k . Together with the eigenvalues, they are τ = − , w = (0 , , , , , − , , T ,τ = 1 , w = (cid:32) √ , − √ , − , − , , , , (cid:33) T ,τ = − √ , w = (cid:18) , − , , − , , , , (cid:19) T ,τ = − − √ , w = (cid:18) − , , , − , , , , (cid:19) T . We apply the same “partial diagonalization” trick. Let R be an 8 × w , . . . , w . Then for every 1 (cid:54) i (cid:54)
4, the i -th column of the conjugate R − M ( k ) R has the special form that its i -th elementis τ i , and the remaining elements are all zero. A concrete choice of R gives R − M ( k ) R = (cid:18) D ?0 M ( k ) (cid:19) , where D is a 4 × τ , . . . , τ , and M ( k ) = k − − k k k − − k k k − − k k − k − − k k . Finally, one more routine conjugation turns M ( k ) into A ( k ) in (8.2.20). Thiscompletes the deduction of (8.2.18)–(8.2.20).One can derive Theorem 8.2.1 from (8.2.18)–(8.2.20) in the usual way; we leaveit to the reader. (cid:3) To complete this section, we shall compute the eigenvalues of the 2-step transitionmatrix M of the regular octagon surface.With Figure 8.1.2 in mind, we view the regular octagon surface as a polyrectanglesurface with 7 rectangle faces, as shown in Figure 8.2.4. Here the horizontal streetsare 1 , , , , ,
6, while the vertical streets are 1 , , , , ,
8. Note herethat two distinct rectangle faces can fall into the same horizontal and the samevertical street simultaneously, for instance, faces 2 , ,
6, so we shall adaptour notation from that used in earlier street-spreading matrix determination. ↑ ↑ ↑ ↑ ↑ ↑ ↑ Figure 8.2.4: the regular octagon surfaces and almost vertical units of type ↑ We also show the almost vertical units of type ↑ in the polyrectangle surface. Wehave not shown the almost vertical units of type − ↑ here.Let J = { , , , } and J = { , , } ON-INTEGRABLE SYSTEMS (IV) 69 denote the horizontal streets, and let I = I = I = I = { , , , } and I = I = I = { , , } denote the vertical streets.For simplicity, we consider only the special case with branching parameter k = 1,so that we are considering a slope of the form α = 2(1 + √
2) + 12(1 + √
2) + √ ··· . Corresponding to (7.2.14), we define the column matrices u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] and u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] . (8.2.21)Here J ∗ , J ∗ and I ∗ j denote that the edges are counted with multiplicity. × × × × ↑ ↑ ↑ ↑ × × × × ↑ ↑ ↑ ↑ × × × × ↑ ↑ ↑ ↑ × × ↑ ↑ u × × × ↑ ↑ ↑ × × ↑ ↑ × × × ↑ ↑ ↑ × × × ↑ ↑ ↑ u Figure 8.2.5: almost vertical units of type ↑ in u and u The horizontal street J has length 2(1+ √
2) and width 1, so has normalized length2(1 + √ √ α , this means that an almost horizontal detour crossing of slope α − travels alongthis street essentially once, giving rise to a single count and J ∗ = { , , , } . On theother hand, the horizontal street J has smaller length 2 + √ √ √
2. Thus an almost horizontal detour crossingof slope α − travels along the horizontal street J essentially twice, giving rise to adouble count and J ∗ = { , , , , , } .A similar argument now gives I ∗ = I ∗ = I ∗ = I ∗ = { , , , } and I ∗ = I ∗ = I ∗ = { , , , , , } . We also define the column matrices v and v analogous to (7.2.15), but theirdetails are not important. Also, analogous to (7.2.17), we have( M − I )[ {↑ s } ] = (cid:26) u + v , if s ∈ J , u + v , if s ∈ J . (8.2.22)We now combine (8.2.21) and (8.2.22). For the horizontal street corresponding to u ,as highlighted in the picture on the left in Figure 8.2.5, we have( M − I ) u = ( M − I )[ { ↑ , ↑ , ↑ , ↑ } ]+ ( M − I )[ { ↑ , ↑ , ↑ } ]= 12( u + v ) + 8( u + v ) . (8.2.23) For the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 8.2.5, we have( M − I ) u = ( M − I )[ { ↑ , ↑ , ↑ , ↑ } ]+ ( M − I )[ { ↑ , ↑ , ↑ } ]= 16( u + v ) + 12( u + v ) . (8.2.24)It follows from (8.2.23)–(8.2.24) that the street-spreading matrix is given by S = (cid:18)
12 168 12 (cid:19) , with eigenvalues τ = 12+8 √ τ = 12 − √
2. Using (7.2.38), the correspondingeigenvalues of M are λ (12 + 8 √ ± ) = 7 + 4 √ ± (cid:16)
20 + 14 √ (cid:17) / and λ (12 − √ ± ) = 7 − √ ± (cid:16) − √ (cid:17) / . The two largest eigenvalues are therefore λ = 7 + 4 √ (cid:16)
20 + 14 √ (cid:17) / and λ = 7 − √ (cid:16) − √ (cid:17) / . Recall (8.2.16) and (8.2.17) that, for the branching parameter k = 1, the two largesteigenvalues for the 1-step transition matrix areΛ = (1 + √
2) + (cid:16) (1 + √ + 1 (cid:17) / and λ = − ( √ − − (cid:16) ( √ − + 1 (cid:17) / . Note that λ = Λ and λ = λ .8.3. Regular octagon billiard.
We now switch to billiard flow on the regularoctagon. This also leads to a street-rational polyrectangle surface with 1-directiongeodesic flow, but the details are more complicated than in Section 8.1.Clearly the method earlier of iterated reflection on a side, like in Figure 8.1.1,does not apply here, since we start with the regular octagon. In this case, we need asomewhat different, but ultimately equivalent, approach based on the constructionof the “reflected net”. The first step is illustrated by Figure 8.3.1 which shows a“reflected double-octagon net” of the regular octagon billiard surface, where theidentified boundary edges are marked with the same letter. Using Euler’s formula,it is easy to see that the Euler characteristic χ of the compact surface in Figure 8.3.1is χ = 8 − g is g = 1 − ( χ/
2) = 0. Hence this double-octagonsurface is homeomorphic to the sphere. a cg e dbfhac ged bf
Figure 8.3.1: double-octagon net of the regular octagon billiard surfaceNote the reflection symmetry of the boundary labelling, where the labelling onthe right octagon is obtained from the labelling on the left octagon by a reflectionacross the vertical dotted line that we may call a “mirror”.
ON-INTEGRABLE SYSTEMS (IV) 71
Regular octagon billiard is a complicated flow. While the “double-octagon” netin Figure 8.3.1 is quite simple, the corresponding surface still exhibits a rathercomplicated geodesic flow. We illustrate how we construct this geodesic flow inFigure 8.3.2. We start with a short billiard orbit in the left octagon, labelled , thathits the boundary edge a . Following the usual billiard rule, bounces back as shownby the dashed arrow. We reflect this dashed arrow across the “mirror” in the middle,and obtain . If is considered an initial segment of a geodesic on the compactsurface, then is the continuation of the same geodesic on this compact surface. Inthis way, the process can be viewed as a partial unfolding of the 8-direction billiardflow in the left octagon into a 4-direction flow in the double-octagon, noting that abilliard orbit which starts horizontal or vertical or with slope ± a cg e dbfhac ged bf Figure 8.3.2: partial unfolding of the billiard in the left regular octagonNext, we join up octagons in such a way that neighboring octagons are reflectionsof each other. We end up with a ring of 8 octagons, as shown in Figure 8.3.3. Thisis sometimes known as the translation surface for regular octagon billiard. a a a a a a a a b b b b b b b b c c c c c c c c d d d d d d d d e e e e f f f f f f f f g g g g g g g g h h h h Figure 8.3.3: translation surface for regular octagon billiard with edge labellings
Remark.
In general, for any even k (cid:62)
4, we can glue k copies of a regular k -gontogether in a perfect ring formation analogous to that in Figure 8.3.3 such that themidpoints of the common edges between neighboring k -gons all lie on a circle. Forodd k (cid:62)
3, we can do likewise with 2 k copies of a regular k -gon.The edge labellings in Figure 8.3.3 look very cumbersome at first sight. However,we can follow a simple convention. Start with one copy of the octagon, and label thedirected edges of this octagon by a, b, c, d, e, f, g, h , initially without subscripts. Theadjacent octagon has reflected labelling, and the next one in the ring has labellingthat is a reflection of the labelling of the second one, and so on. When we completethis process, we have labellings a, b, c, d, e, f, g, h , still without subscripts, on each octagon. The second step of the process is to look at all the edges labelled a , andthey occur as 4 directed parallel pairs. We now identify such directed parallel pairsby labelling them a , a , a , a . In Figure 8.3.3, we have labelled them in increasingorder of the angles they make with the positive horizontal direction. We then repeatthis step with the other edges. Note that the edges e and h do not appear to comein pairs, but they actually do, but the identified edges overlap.We clearly need more, as we need a corresponding net that exhibits a 1-directiongeodesic flow and which shows the streets in a more transparent fashion. Theimpatient reader may jump ahead to the net in Figures 8.3.6 and 8.3.7 from whichwe can easily obtain the corresponding street-rational polyrectangle surface, adaptthe shortline method in the proof of Theorems 6.1.1 and 6.4.1, and exhibit explicitslopes for which we can establish superdensity. Remark.
The remarkable result of Veech [18] on the uniform-periodic dichotomy inflat systems, particularly that every infinite billiard orbit in a regular polygon iseither periodic or exhibits uniformity. His method is ergodic in nature, however,and does not give time-quantitative results. It is the purpose of our work here tomake some quantitative statements.The plan is quite straightforward, and the details are not too cumbersome. Weshall proceed in steps and illustrate our ideas with pictures. Indeed, regular octagonbilliard represents the whole difficulty. Once we fully understand the special caseof regular octagon billiard, it is easy to visualize the general case of regular k -gonbilliard, where k (cid:62) ′ ′ ′ ′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′ ′ ′ ′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ Figure 8.3.4: first set of tilted streets on the translation surfaceof regular octagon billiardThe first big street in this direction consists of 8 trapezoids labelled (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) . Note the edge labellings in Figure 8.3.3. The right edge of the trapezoid (cid:48) is b , thesame as the left edge of the trapezoid (cid:48) . The right edge of the trapezoid (cid:48) is g , thesame as the left edge of the trapezoid (cid:48) . The right edge of the trapezoid (cid:48) is d ,the same as the left edge of the trapezoid (cid:48) . The right edge of the trapezoid (cid:48) is a , the same as the left edge of the trapezoid (cid:48) . The right edge of the trapezoid ON-INTEGRABLE SYSTEMS (IV) 73 (cid:48) is f , the same as the left edge of the trapezoid (cid:48) . The right edge of the trapezoid (cid:48) is c , the same as the left edge of the trapezoid (cid:48) . This completes the street.The second big street in this direction consists of 8 trapezoids labelled (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) . There are also two small streets in this direction, consisting of 8 triangles1 (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) , (cid:48) and 1 (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) , (cid:48)(cid:48) each.To use the shortline method, we need to consider streets in a second direction.Figure 8.3.5 is an analog of Figure 8.3.4 in this new direction. Again, we see thatthere are two big streets, each consisting of 8 trapezoids, and two small streets, eachconsisting of 8 triangles. ′ ′ ′ ′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′ ′ ′ ′ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ Figure 8.3.5: second set of tilted streets on the translation surfaceof regular octagon billiardHaving determined the streets in two perpendicular directions, we now attemptto visualize the translation surface of regular octagon billiard as a polyrectanglesurface P . To do so, we must be able to visualize the intersection of any twoperpendicular streets in the translation surface of regular octagon billiard as one ofthe rectangle faces in P . Figure 8.3.6 is a suitable modification of Figure 8.3.4. A B C D E F G HI J K L M NO P Q RS T U V W XA B C D E FG H I J K L M NO P Q R S TU V W X
Figure 8.3.6: a modification of the streets in Figure 8.3.4
Note that in Figure 8.3.4, each big tilted street consists of 8 trapezoids, eachthe union of 2 squares and 2 ( π/ π/ π/ π/ π/ π/
8, whereas a ( π/ π/ A and B are the squares inthe trapezoid (cid:48) , while the ( π/ C is made up of the ( π/ (cid:48) and the ( π/ (cid:48) . On the other hand, the ( π/ X is madeup of the ( π/ (cid:48) and the( π/ (cid:48) .Likewise, Figure 8.3.7 is a suitable modification of Figure 8.3.5. AB CDEF GH IJKLMNOPQRSTUVWX ABCDEF GH IJKL MNOPQRSTUVWX
Figure 8.3.7: a modification of the streets in Figure 8.3.5Figures 8.3.6 and 8.3.7 together clearly demonstrate that we can now visualizethe translation surface of regular octagon billiard as a polyrectangle surface P .Figures 8.3.4–8.3.7 correspond to the decomposition of the regular octagon into 4parts as shown in the picture on the right in Figure 8.1.5. If we repeat our argumentin this section for the decomposition of the regular octagon into 3 parts as shownin the picture on the left in Figure 8.1.5, we obtain a different 8-octagon net whichnevertheless has the same 1-direction geodesic flow.Adapting the shortline method proof of Theorems 6.1.1 and 6.4.1 to any suchconcrete street-rational polyrectangle surface that arise, we then obtain the specialcase k = 8 of Theorem 8.3.1 below.It is easy to see that the same method works for any regular k -gon, where k (cid:62) k -gon with oriented boundary edges, and reflect it on a side. This gives a2-copy version of the regular k -gon with boundary identification, which is a net ofthe corresponding billiard surface. To adapt the shortline method of Theorems 6.1.1and 6.4.1, we need a flat surface with 1-direction geodesic flow. To construct sucha surface we repeat the arguments illustrated by Figures 8.3.1–8.3.3. At the end wehave the translation surface of regular k -gon billiard with boundary identification.We have street-rationality , courtesy of the elementary geometric fact that thevertices of a regular polygon lie on a circle, so that we can use the “chord-angle ON-INTEGRABLE SYSTEMS (IV) 75 relation” on this circle. The last simple step is to rearrange this flat polygonal surfaceto a street-rational polyrectangle surface via translation of some corresponding parts,like converting Figures 8.3.4–8.3.5 to Figures 8.3.6–8.3.7.To get a polyrectangle, we need two perpendicular decompositions into parallelstrips like in Figure 8.1.5 for the regular octagon. And of course it can be done forevery regular polygon of k sides, where k (cid:62) k = 6. In the picture on the left, the horizontal line divides the hexagon into 2parts. In the picture on the right, the 2 vertical lines divide it into 3 parts.Figure 8.3.8: two perpendicular decompositions of the regular hexagonThus we obtain the following general result. Theorem 8.3.1.
Let k (cid:62) be an even integer, and consider billiard in a regularpolygon of k sides. There exist infinitely many explicit slopes, depending on k , suchthat any half-infinite billiard orbit having such an initial slope exhibits superdensityin the polygon.For infinitely many of these initial slopes that give rise to superdensity, we canexplicitly compute the corresponding irregularity exponent.Combining the irregularity exponent with the method of zigzagging introduced inSection 3.3 in [2] , we can also describe, for billiard orbits having these initial slopes,the time-quantitative behavior of the edge-cutting and face-crossing numbers, as wellas equidistribution relative to all convex sets. More superdensity results.
Next we switch to the regular polygons of k sides, or k -gons for short, where k (cid:62) π/ regular double-pentagon surface , as shown inFigure 8.4.1. Since we identify parallel edges, this surface has 1-direction geodesicflow. d ca bab dc Figure 8.4.1: right triangle billiard with angle π/
5, and geodesic flowon the regular double-pentagon surface via unfoldingHere the horizontal and vertical are no longer natural directions. However, viewedin the appropriate way, 1-direction geodesic flow on the regular double-pentagonsurface can be shown to be equivalent to 1-direction geodesic flow on a street-rational polyparallelogram surface. By moving the two triangles on the bottom leftto the top right, we end up with two rhombi and two other parallelograms, as shownin the picture on the left in Figure 8.4.2. Here street-rationality comes from the geometric fact that the vertices of a regular polygon all lie on the same circle, andso we can use the “chord-angle relation” on this circle. Using this, we see that theangles 716 and 534 are the same. Hence these two other parallelograms are similar.Note from the picture on the right that we have two northwest-to-southeast streets,namely
A, B, C and
D, E . We also have two southwest-to-northeast streets, namely
A, C, E and
B, D . We shall return to this surface in the next section. d ca bab dc
A B CD E
Figure 8.4.2: the double-pentagon surface as a polyparallelogram surfaceSimilar constructions show that 1-direction geodesic flow on every regular double- k -gon surface with odd k (cid:62) a bd cab dc e Figure 8.4.3: double-pentagon net of the regular pentagon billiard surfaceAs in the case for regular octagon billiard in Section 8.3, we join up pentagonsin such a way that neighboring pentagons are reflections of each other. We end upwith a ring of 10 pentagons, as shown in Figure 8.4.4. This is sometimes known asthe translation surface of regular pentagon billiard. Note that the edge labellings inFigure 8.4.4 are obtained by following the convention discussed in the Remark afterFigure 8.3.3. a a a a a a a a a a b b b b b b b b b b c c c c c d d d d d d d d d d e e e e e Figure 8.4.4: translation surface for regular pentagon billiard with edge labellingsWe clearly need more, as we need a corresponding net that exhibits a 1-directiongeodesic flow and which shows the streets in a more transparent fashion. Theimpatient reader may jump ahead to the nets in Figure 8.4.6 from which we can
ON-INTEGRABLE SYSTEMS (IV) 77 easily obtain the corresponding street-rational polyparallelogram surface, adapt theshortline method in the proof of Theorems 6.1.1 and 6.4.1, and exhibit explicit slopesfor which we can establish superdensity.The plan is quite straightforward, and again the details are not too cumbersome.We shall proceed in steps and illustrate our ideas with pictures. Indeed, regularpentagon billiard represents the whole difficulty. Once we fully understand thespecial case of regular pentagon billiard, it is easy to visualize the general case ofregular k -gon billiard, where k (cid:62)
167 8 4 52 3 9 10
Figure 8.4.5: tilted streets on the translation surfaceof regular pentagon billiardThe big street in this direction consists of 10 trapezoids labelled , , , , , , , , , . There is also a small street in this direction, consisting of 10 triangles1 , , , , , , , , , . To use the shortline method, we need to consider streets in a second direction,shown in the picture on the right in Figure 8.4.5. Again, we see that there are a bigstreet, consisting of 10 trapezoids, and a small street, consisting of 10 triangles.Having determined the streets in two different directions, we now attempt tovisualize the translation surface of regular pentagon billiard as a polyparallelogramsurface P . To do so, we must be able to visualize the intersection of any two streetsin different directions in the translation surface of regular pentagon billiard as one ofthe paralleogram faces in P . Figure 8.4.6 is a suitable modification of Figure 8.4.5. A B C D EF G HI J K LM N O
12 3 4 567 8 9 10
ABCDEF GHIJKLMNO
Figure 8.4.6: a modification of the streets in Figure 8.4.5of regular pentagon billiard
Note that in Figure 8.4.5, each big tilted street consists of 10 trapezoids, each theunion of a rhombus and a triangle. If we compare the pictures on the left in Figures8.4.5 and 8.4.6, we see that A and C are the rhombi in the trapezoids and ,while the parallelogram B is made up of the triangle on the right hand end of thetrapezoid and the triangle on the left hand end of the trapezoid . On the otherhand, the parallelogram E is made up of the triangle on the right hand end of thetrapezoid and the triangle on the left hand end of the trapezoid .Each small tilted street consists of 10 triangles, and each can be split into unequalparts by a segment parallel to the other direction and intersecting one of its vertices.If we compare the pictures on the left in Figures 8.4.5 and 8.4.6, we see that theparallelogram 1 is the union of the bigger half of the triangle 1 and the bigger halfof the triangle 2, while the rhombus 2 is the union of the smaller half of the triangle2 and the triangle 3.Thus the resulting Figure 8.4.6 shows that we have a polyparallelogram surface.We have already studied the superdensity of certain geodesic flow on the regulartetrahedron surface and on the cube surface. The next member of the famous listof the five platonic solids is the regular dodecahedron which has regular pentagonfaces. Thus it is not surprising that a similar decomposition works for geodesic flowon the dodecahedron surface. The standard net of the dodecahedron surface consistsof 12 pentagons. To construct the corresponding polygonal surface with 1-directiongeodesic flow we have to glue together 10 copies of the 12-pentagon standard net.This means 120 copies of the pentagon, far too complicated to be explicitly includedhere. For illustration, Figure 8.4.7 shows one large street on the dodecahedronsurface. We leave the complicated details of the explicit construction of the whole120-copy version of the pentagon to the interested reader.Figure 8.4.7: one large street on the dodecahedron surfaceNote that the parity condition that k (cid:62) k (cid:62) k (cid:62) Theorem 8.4.1.
Let k (cid:62) be an odd integer. (i) Consider the right-triangle with angle π/k . There exist infinitely many slopes,depending on k , such that any half-infinite billiard orbit with such an initial slopeexhibits superdensity in the right-triangle with angle π/k . (ii) Consider geodesic flow on the regular double-polygon of k sides. There existinfinitely many explicit slopes, depending on k , such that any half-infinite geodesichaving such a slope exhibits superdensity in the double-polygon. (iii) Consider billiard in a regular polygon of k sides. There exist infinitely manyexplicit slopes, depending on k , such that any half-infinite billiard orbit having suchan initial slope exhibits superdensity in the polygon. (iv) Consider geodesic flow on the dodecahedron surface. There exist infinitelymany explicit slopes such that any half-infinite geodesic having such a slope exhibitssuperdensity on the surface. (v)
For infinitely many of these slopes in parts (i) , (ii) , (iii) and (iv) that give riseto superdensity, we can explicitly compute the corresponding irregularity exponent. ON-INTEGRABLE SYSTEMS (IV) 79
Combining the irregularity exponent with the method of zigzagging introduced inSection 3.3 in [2] , we can also describe, for trajectories having these initial slopes,the time-quantitative behavior of the edge-cutting and face-crossing numbers, as wellas equidistribution relative to all convex sets.
At this point, it is perhaps appropriate to mention the study of the remarkable uniform-periodic dichotomy in flat systems. This study essentially originates fromthe work of Veech [18], with the result that every infinite billiard orbit in a regularpolygon is either periodic or exhibits uniformity.As we have shown earlier, surfaces arising from regular polygons of k sides canbe visualized as street-rational polyrectangle surfaces for even k (cid:62) k (cid:62) golden L-surface . As can be seenin the picture on the left in Figure 8.4.8, this surface exhibits 45-degree reflectionsymmetry, and it becomes a flat surface if we carry out the boundary identificationindicated in the picture on the right in Figure 8.4.8. √
52 1+ √ v v v v h h h h Figure 8.4.8: the golden L-surfaceThe street-rationality of the golden L-surface comes from a simple arithmeticproperty of the golden ratio, that1 + √
52 = 1 √ − . Of course the two special cases of square billiard and equilateral triangle billiardare well known classical results dating back to the 1920s. These classical resultsrepresent the “easy case” of integrable systems.Billiard in a regular polygon of k sides, with k (cid:62)
5, on the other hand, is a “non-integrable system”, where the vertices represent split-singularities of the orbits.A flat system exhibiting the uniform-periodic dichotomy is called optimal .For a long time regular polygon billiard remains the only known infinite family of primitive optimal systems. Here primitive means that an elementary building blocklike an “atom” cannot be obtained from a simpler system by some covering spaceconstruction. We illustrate this concept on a familiar example.Consider geodesic flow (or billiard) on the class of polysquare surfaces. By theGutkin–Veech theorem, each member of this class is an optimal system, and togetherthey exhibit every high genus. But there is only one primitive member, namely,geodesic flow (or billiard) on the unit torus [0 , . In other words, the fractionalpart of geodesic flow on a polysquare surface, or modulo one , is the torus line flowon the flat torus. The inverse map of modulo one gives back the polysquare as a(branching) covering surface of the flat torus.Similarly, we can glue together an arbitrary number of, for instance, congruent regular octagons in a natural horizontal or vertical way. The class of such polyoctagon surfaces exhibits arbitrarily high genera. But there is only one primitive member,namely, the regular octagon surface itself.Let us return to regular polygon billiard, the first infinite family of primitiveoptimal systems, discovered in the 1980s. A completely different infinite family ofprimitive optimal systems, discovered in the early 2000s independently by Calta [7]and McMullen [11], using different methods, consists of L-shaped tables for whichthe billiard is optimal. We shall refer to this as the Calta–McMullen family. Suchbilliard surfaces have genus 2, and the simplest member of this family is the goldenL-surface. The L-shape has the special property that it is the same to considergeodesic flow or billiard flow, as they are equivalent systems.The general member of the Calta–McMullen family is an L-shaped billiard table,with boundary identification given as in the picture on the right in Figure 8.4.9. v v v v h h h h b −
11 1 a − Figure 8.4.9: L-shaped billiard tableThe bottom horizontal side has length a , the left-most vertical side has length b ,the right-most vertical side has length 1, and the top-most horizontal side haslength 1. The numbers a = r √ d + r and b = r √ d + 1 − r are such that r , r are rational numbers and d (cid:62) i.e. , it representsnormalization.It is easy to check that this is a street-rational polyrectangle surface. For the twohorizontal streets, the width-height ratios of the two rectangle are a b − , and their ratio is a ( b −
1) = ( r √ d + r )( r √ d − r ) = r d − r , which is rational.For the two vertical streets, the width-height ratios of the two rectangle are a −
11 and 1 b , and their ratio is ( a − b = ( r √ d + r − r √ d + 1 − r ) = r d − ( r − , whichis also rational.The shortline method gives the following result. Theorem 8.4.2.
Consider any surface in the Calta–McMullen family. (i)
There exist infinitely many explicit slopes, depending on the surface, such thatany half-infinite geodesic on the surface having such a slope exhibits superdensity. (ii)
There exist infinitely many explicit slopes, depending on the surface, such thatany billiard orbit on the surface having such an initial slope exhibits superdensity. (iii)
For infinitely many of these slopes in parts (i) and (ii) that give rise tosuperdensity, we can explicitly compute the corresponding irregularity exponents.Combining the irregularity exponent with the method of zigzagging introduced in
ON-INTEGRABLE SYSTEMS (IV) 81
Section 3.3 in [2] , we can also describe, for trajectories having these initial slopes,the time-quantitative behavior of the edge-cutting and face-crossing numbers, as wellas equidistribution relative to all convex sets.
We shall justify part (iii) for the golden L-surface in Section 8.5.Optimality and superdensity of some orbits are two different aspects concerning adynamical system, both exhibiting “perfect” behavior. Optimality for these systemsis established by Calta and McMullen via ergodicity, making use of Birkhoff’s ergodictheorem. Since Birkhoff’s ergodic theorem does not give any error term, these resultsdo not say anything quantitative about the speed of convergence to uniformity, oreven about time-quantitative density. Our superdensity results, on the other hand,establish a best possible form of time-quantitative density, at least for some slopes.And this is the best that we can hope for, since superdensity fails for almost everyslope anyway.Note next that every member of the Calta–McMullen family is what we may calla 1 -step L-staircase , and the corresponding surface has the same genus 2. It is a2-parameter family, since a = √ q + r and b = √ q + 1 − r depend only on two rational parameters q = r d and r .We next describe a far-reaching extension of the Calta–McMullen family, leadingto surfaces with arbitrarily large genus. For any integer k (cid:62)
1, we construct aninfinite family of street-rational polysquares that are k -step L-staircases , and eachsurface has genus k + 1.The picture on the left in Figure 8.4.10 illustrates a typical street-rational 2-stepL-staircase. α α r α α α α r α Figure 8.4.10: street-rational 2-step and 3-step L-staircasesThe right-most vertical rectangle is a unit square, representing normalization toexclude equivalent systems. The left-most vertical rectangle has size α × r α , whilethe second vertical rectangle has size α × r α , where r α > r α > r , r are positive rational numbers.The top horizontal rectangle now has size α × ( r α − r α ), the middle horizontalrectangle has size ( α + α ) × ( r α − α + α + 1) ×
1. Thus the 2-step L-staircase is a street-rational polyrectanglesurface precisely if we have the relations α r α − r α = r α + α r α − r ( α + α + 1) (8.4.1)for the width-height ratios of the 3 horizontal rectangles, where r , r are somepositive rational numbers.The analogous requirement for the 3 vertical rectangles is guaranteed by definition.We wish to express the two positive real variables α , α in (8.4.1) in terms of thegiven positive rational parameters r , r , r , r . Routine calculation shows that both α , α are algebraic numbers. Indeed, using the second equality in (8.4.1), we obtain α = r r α + ( r r − r − r ) α − r r + r − r r α . (8.4.2)Substituting this into the first equality in (8.4.1) eliminates the variable α , and givesrise to a polynomial equation of degree 4 in the variable α with rational coefficients.Thus α is algebraic. In view of (8.4.2), α is also algebraic. Crucially, for the typicalchoice of the rational parameters r , r , r , r , this polynomial equation of degree 4has a positive root which is not rational.This completes the construction of the class of 2-step street-rational L-staircases.It is a 4-parameter family, depending on the rational parameters r , r , r , r , andeach member of this family has genus 3.The picture on the right in Figure 8.4.10 illustrates a typical street-rational 3-stepL-staircase.The right-most vertical rectangle is a unit square, representing normalization toexclude equivalent systems. The left-most vertical rectangle has size α × r α , thesecond vertical rectangle has size α × r α , while the third vertical rectangle hassize α × r α , where r α > r α > r α > r , r , r are positive rationalnumbers.The top horizontal rectangle now has size α × ( r α − r α ), the second horizontalrectangle has size ( α + α ) × ( r α − r α ), the third horizontal rectangle hassize ( α + α + α ) × ( r α − α + α + α + 1) ×
1. Thus the 3-step L-staircase is a street-rational polyrectanglesurface precisely if we have the relations α r α − r α = r α + α r α − r α = r α + α + α r α − r ( α + α + α + 1) (8.4.3)for the width-height ratios of the 4 horizontal rectangles, where r , r , r are somepositive rational numbers.The analogous requirement for the 4 vertical rectangles is guaranteed by definition.We wish to express the three positive real variables α , α , α in (8.4.3) in terms ofthe given positive rational parameters r , r , r , r , r , r . Routine calculation showsthat α , α , α are all algebraic numbers. We leave the details to the reader.This completes the construction of the class of 3-step street-rational L-staircases.It is a 6-parameter family, depending on the rational parameters r , r , r , r , r , r ,and each member of this family has genus 4.The construction of street-rational k -step L-staircases, where k (cid:62)
1, goes in asimilar way. This gives rise to a 2 k -parameter family, and each member of thefamily has genus k + 1.The construction gives a street-rational polyrectangle with a 1-direction geodesicflow, and the corresponding billiard surface is also street-rational. Thus the shortlinemethod works, and gives the following generalization of Theorem 8.4.2. Theorem 8.4.3.
Let k (cid:62) be an integer, and consider a k -step street-rationalL-staircase surface. (i) There exist infinitely many explicit slopes, depending on the surface, such thatany half-infinite geodesic on the surface having such a slope exhibits superdensity. (ii)
There exist infinitely many explicit slopes, depending on the surface, such thatany billiard orbit on the surface having such an initial slope exhibits superdensity. (iii)
For infinitely many of these slopes in parts (i) and (ii) that give rise tosuperdensity, we can explicitly compute the corresponding irregularity exponents.Combining the irregularity exponent with the method of zigzagging introduced inSection 3.3 in [2] , we can also describe, for trajectories having these initial slopes,
ON-INTEGRABLE SYSTEMS (IV) 83 the time-quantitative behavior of the edge-cutting and face-crossing numbers, as wellas equidistribution relative to all convex sets.Remark.
We are not aware whether these street-rational k -step L-staircases, where k (cid:62)
2, in general are optimal. What we can prove is that they all have explicitsuperdense orbits. More precisely, our construction gives primitive street-rationalL-staircase surfaces in every positive genus and which exhibit explicit superdenseorbits. On the other hand, we are not aware of any infinite family of primitiveoptimal billiard systems in a fixed high genus.To obtain new street-rational polyrectangle surfaces, there is a technique whichis algebraic in nature. We shall describe this technique by first looking at a simpleexample.Consider the polysquare surface which is a 3 × v v v v v v v v h h h h h h h h Figure 8.4.11: a polysquare surfaceWe can obtain a polyrectangle surface by varying the lengths of the edges of thesquare faces to change them to rectangle faces. To ensure that we obtain affine-different polyrectangles, we fix a horizontal side and a vertical side to have length 1.For simplicity, we have ensured that the bottom left square face does not get altered.As shown in Figure 8.4.12, we allow the other faces to have different lengths. d d e e Figure 8.4.12: a polyrectangle surface correspondingto the polysquare surface in Figure 8.4.11The original polysquare surface has 4 horizontal streets and 4 vertical streets, cor-responding to the edge pairings v , v , v , v and h , h , h , h . The 4 correspondinghorizontal streets in the polyrectangle surface have normalized lengths, i.e. streetlengths normalized by division by street widths, equal to1 + d + d e , e , d e , d + d . For horizontal street-rationality, we require positive rational numbers r , r , r suchthat e = r , d = r , e (1 + d + d ) = r . (8.4.4) On the other hand, the 4 corresponding verticalal streets in the polyrectangle surfacehave normalized lengths equal to1 + e + e , e d , d , e + e d . For vertical street-rationality, again we require e and d to be rational, alreadytaken care of in (8.4.4), as well as another positive rational numbers r such that d (1 + e + e ) = r . (8.4.5)Combining (8.4.4) and (8.4.5), we obtain e ( d + 1 + r ) = r and d ( e + 1 + r ) = r , leading to a linear equation in e and d with rational coefficients. On combiningthis linear equation with (8.4.5) and eliminating the variable e , say, we obtaina quadratic equation with rational coefficients in the variable d . One can checkthat for infinitely many choices of the rational numbers r , r , r , r , this quadraticequation has a positive irrational root d with corresponding positive e , leading toa street-rational polyrectangle surface that is not a polysquare surface.A similar argument seems to work if we start with a polysquare surface withperpendicular boundary identification. For illustration, we consider the slightlymore complicated polysquare surafce shown in Figure 8.4.13. v v v v v v v v h h h h h h h h h h h h h h Figure 8.4.13: another polysquare surfaceWe can obtain a polyrectangle surface by varying the lengths of the edges of thesquare faces to change them to rectangle faces. To ensure that we obtain affine-different polyrectangles, we fix a horizontal side and a vertical side to have length 1.For simplicity, we have ensured that the left most square face does not get altered.As shown in Figure 8.4.14, we allow the other faces to have different lengths. e e e d d d d d d Figure 8.4.14: a polyrectangle surface correspondingto the polysquare surface in Figure 8.4.13
ON-INTEGRABLE SYSTEMS (IV) 85
The original polysquare surface has 4 horizontal streets and 7 vertical streets.The 4 corresponding horizontal streets in the polyrectangle surface have normalizedlengths equal to d + . . . + d e , d + . . . + d , d + . . . + d e , d + d e , and horizontal street-rationality leads to 3 equations that involve 3 positive rationalnumbers r , r , r , say. On the other hand, the 7 corresponding vertical streets inthe polyrectangle surface have normalized lengths equal to11 , e d , e + e d , e + e d , e + e + e d , e + e d , e d , and vertical street-rationality leads to 6 equations that involve 6 positive rationalnumbers r , . . . , r , say. In other words, we have 9 equations in the 9 variables d , d , d , d , d , d , e , e , e that involve 9 positive rational parameters r , . . . , r . And it is reasonable to expectthat there are infinitely many choices of r , . . . , r that lead to positive solutions forall the variables and irrational solutions for some.Indeed, we may generalize this to any arbitrary finite polysquare surface with m horizontal streets and n vertical streets. Street-rationality will give rise to m + n − m + n − d i and e j , and these equations involve m + n − h h h h h h v v v v d d e Figure 8.4.16: an infinite family of non-optimalstreet-rational polyrectangle surfaces
The polysquare surface has 2 horizontal streets and 3 vertical streets. Let usconsider the corresponding polyrectangle surfaces shown in the picture on the rightin Figure 8.4.16. The 3 vertical streets have normalized lengths equal to e d , e d , , and for vertical street-rationality, we need rational numbers r , r > r e d = r e d = 1 , so we have the two conditions d = r e and d = r (1 + e ) . (8.4.6)On the other hand, the 2 horizontal streets have normalized lengths equal to1 + d , d + d e , and for horizontal street-rationality, we need a rational number r > r d d + d e , so we have the extra condition d + d = r e (1 + d ) . (8.4.7)The equations (8.4.6) and (8.4.7) have quadratic irrational solutions in d , d , e that can be written in a simple explicit form, in sharp contrast to L-staircaseswhere the corresponding requirements (8.4.1)–(8.4.3) lead to unpleasant higher de-gree equations, so we work out the details here.Substituting (8.4.6) into (8.4.7), we obtain r e + r (1 + e ) = r e (1 + r (1 + e )) . (8.4.8)This quadratic equation in the variable e can be rewritten in the form r r e + ( r r + r − r − r ) e − r = 0 . Since r , r >
0, such a typical quadratic equation has complex conjugate roots e = R ± R √ D , where R , R are rational numbers and D (cid:62) R D − R = − ( R + R √ D )( R − R √ D ) = 1 r > . (8.4.9)For simplicity, let us assume that e = R √ D + R > . (8.4.10)Using (8.4.10), the left hand side of (8.4.8) becomes r ( R √ D + R ) + r ( R √ D + R + 1) , and the right hand side of (8.4.8) becomes r ( R √ D + R )(1 + r ( R √ D + R + 1)) . The coefficient for √ D in (8.4.8) is equal to( r + r ) R = r (1 + r ) R + 2 r r R R , (8.4.11)while the rational term in (8.4.8) is equal to( r + r ) R + r = r (1 + r ) R + r r R + r r R D. (8.4.12) ON-INTEGRABLE SYSTEMS (IV) 87
Combining (8.4.11) and (8.4.12) and eliminating r , we deduce that( r + r ) R (1 + r ) R + 2 r R R = ( r + r ) R + r (1 + r ) R + r R + r R D , which can be simplified to r + r r + 2 r R = ( r + r ) R + r (1 + r ) R + r R + r R D .
For any given R , R , D and r , this is a linear equation in r , with solution r = 1 + r (2 R + 1) R D − R − r . (8.4.13)We require r >
0, so we have to assume the positivity condition r (2 R + 1) R D − R − r > . In view of (8.4.9), this is equivalent to 1 + r (2 R + 1) − ( R D − R ) r > i.e. ,( R D − ( R + 1) ) r < . (8.4.14)Note that R D − ( R + 1) is never zero, so there are only two possibilities:(i) If R D − ( R + 1) >
0, then r is restricted to the range0 < r < R D − ( R + 1) . (ii) If R D − ( R + 1) <
0, then any r > e = R √ D + R , d = r ( R √ D + R ) , d = r ( R √ D + R + 1) . (8.4.15)Note that (8.4.13) and (8.4.14) become particularly simple if R D − R = 1 and R >
0. Then case (ii) applies. If we write r = r , then using (8.4.13), we canreduce (8.4.15) to e = R √ D + R , d = (2 rR + 1)( R √ D + R ) , d = r ( R √ D + R + 1) . It is a lucky coincidence that we know all the integral solutions of the Pell equation R D − R = 1 and R >
0. The simplest such choice is D = 2 and R = R = 1,leading to the solution e = √ , d = (2 r + 1)( √ , d = r ( √ , where r > double-rational gluing . For a simple illustration of the ideas,we consider Figure 8.4.17, where we glue together two regular octagons with edgelengths that are rational multiples of each other. r r r r aab bcc dd eef fgg h hii b Figure 8.4.17: gluing together two regular octagonsSince the edge lengths of the two regular octagons are rational multiples of eachother, we may assume, without loss of generality, that both edge lengths are rationalnumbers, equal to r and r as shown in the picture on the left in Figure 8.4.17.We glue the two octagons together in such a way that the vertex, indicated in thepicture by the dot, of the octagon on the right lies on a point on the edge of theoctagon on the left which has rational distances r and r from the two nearestvertices. With edge identification shown in the picture on the right in Figure 8.4.17,the union of the two regular octagons becomes a surface.It is clear that the edge identification give rise to 4 vertical streets and 5 horizontalstreets, as shown in Figure 8.4.18. To check for street-rationality, we need to look atthe normalized lengths, i.e. street lengths normalized by division by street widths,of these streets. V V V V V V H H H H H H H H Figure 8.4.18: vertical and horizontal streetsThe normalized lengths of the vertical streets V , V , V , V are respectively2 + √ / √ , √ , √ / √ , √ , and these are all rational multiples of 1 + √
2. On the other hand, the normalizedlengths of the horizontal streets H , H , H , H , H are respectively2 + √ / √ , r (1 + √ r − r , ( r + r )(1 + √ r , r (1 + √ r , √ / √ , and these are also all rational multiples of 1 + √
2. Thus the surface comprisingthe two octagons is a street-rational polyrectangle surface which is not a polysquaresurface.Note that a bigger regular octagon is not a covering surface of a smaller regularoctagon. Thus this street-rational polyrectangle surface is primitive , i.e. it cannotbe obtained from a simpler surface via covering construction. ON-INTEGRABLE SYSTEMS (IV) 89
We need not stop at two regular octagons. In general, we can glue together, acrosshorizontal or vertical edges, arbitrarily many regular octagons with edge lengthsthat are rational multiples of each other in a similar way. With a typical choice ofrational edge length parameters, the resulting street-rational polyrectangle surfaceis primitive.Of course, we can replace the regular octagon by any regular polygon of k sides,where k (cid:62) √ / r r r r b Figure 8.4.19: gluing together two golden crossesIn Figure 8.4.20, we have a street-rational polyrectangle surface made up of ninegolden crosses, with six copies of the same size, two copies with double edge length,and one copy of triple edge length. Indeed, if we consider this figure as a billiardtable, the the corresponding billiard surface is also a street-rational polyrectanglesurface. Figure 8.4.20: a street-rational polyrectangle surfaceconsisting of nine golden crossesA similar consideration arises if we start with a cross made up of 4 reflected copiesof any member of the Calta–McMullen family, or a surface made up of 4 reflectedcopies of an street-rational k -step L-staircase. Note that there is no analogous results for optimal systems. We are not awareof any construction using double-rational gluing that builds optimal systems fromoptimal components.We conclude this section by considering surfaces of some rectangular boxes. Thisis a generalization of Example 7.2.4 concerning the surface of the unit cube.One such example is what we may call the “golden brick”, or a brick with goldenratio, as shown in Figure 8.4.21. h Figure 8.4.21: a brick with the golden ratioThis has two opposite faces that are unit squares, and the four parallel edgesjoining them have length h = ( √ − /
2. It is easy to see that this gives rise to astreet-rational polyrectangle surface. Indeed, it has 3 streets, each of length 4. Oneof these streets has size h ×
4, while the other two have common size 1 × (2 + 2 h ).Street-rationality is now a consequence of the observation that12 + 2 h = 2 · h , due to h + h = 1. We shall return to this example in the next section where weshall determine some irregularity exponents.Clearly, this example can be generalized. Starting with two opposite unit squarefaces, we now let z denote the length of the four parallel edges joining them. Oneof these streets has size z ×
4, while the other two have common size 1 × (2 + 2 z ).Thus we have street-rationality if we can find coprime integers a, b (cid:62) z = ab · z , precisely when az + az − b = 0. Thus z = √ a + 8 ab − a a . Note that the choice a = 2 and b = 1 gives z = h .Thus Theorem 8.4.3 can be extended to all of these special box-surfaces.We can further generalize by starting with two opposite rectangle faces. We leavethe details to the reader. Unfortunately the analogous problem for an arbitrary box-surface remains wide open.8.5. Computing more irregularity exponents.
In this section, we first returnto the double-pentagon surface as shown in Figure 8.4.2. Our purpose is to find theeigenvalues of its 2-step transition matrix M .This surface can be viewed as a polyparallelogram surface, and we shall use ananalogy with a corresponding polyrectangle surface as shown in Figure 8.5.1. ON-INTEGRABLE SYSTEMS (IV) 91
A B CD E ↑ ↑ ↑ ↑ ↑ Figure 8.5.1: analogy of the double-pentagon surfacewith a polyrectangle surfaceThe rhombi
A, C, D in the picture on the left become the squares 3 , , B, E become the rectangles 4 , , , ,
5, while the vertical streets are 1 , , , √ /
2. Note here that the two distinct rectangle faces 3 , ↑ in the polyrectangle surface. We have not shown that almost vertical unitsof type − ↑ here.Let J = { , } and J = { , , } denote the horizontal streets, and let I = I = { , } and I = I = I = { , , } denote the vertical streets.For the polyrectangle surface, we thus consider slopes of the form α k = 3 + √ k + 1 √ k + √ k + ··· . For simplicity, however, we consider only the special case with branching parameter k = 1.Corresponding to (7.2.14), we define the column matrices u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] and u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] . (8.5.1)Here J ∗ , J ∗ and I ∗ j denote that the edges are counted with multiplicity. × × × × × ↑ ↑ ↑ ↑ ↑ u × × × × × × × × ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ u Figure 8.5.2: almost vertical units of type ↑ in u and u We also define the column matrices v and v analogous to (7.2.15), but theirdetails are not important. Also, analogous to (7.2.17), we have( M − I )[ {↑ s } ] = (cid:26) u + v , if s ∈ J , u + v , if s ∈ J . (8.5.2) We now combine (8.5.1) and (8.5.2). For the horizontal street corresponding to u ,as highlighted in the picture on the left in Figure 8.5.2, we have( M − I ) u = ( M − I )[ {↑ , ↑ } ] + ( M − I )[ {↑ , ↑ , ↑ } ]= 2( u + v ) + 3( u + v ) . (8.5.3)For the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 8.5.2, we have( M − I ) u = ( M − I )[ {↑ , ↑ } ] + ( M − I )[ { ↑ , ↑ , ↑ } ]= 3( u + v ) + 5( u + v ) . (8.5.4)It follows from (8.5.3)–(8.5.4) that the street-spreading matrix is given by S = (cid:18) (cid:19) , with eigenvalues τ = 7 + 3 √
52 and τ = 7 − √ . Using (7.2.38), the corresponding eigenvalues of M are λ (cid:32) √
52 ; ± (cid:33) = 11 + 3 √ ± (cid:16)
150 + 66 √ (cid:17) / and λ (cid:32) − √
52 ; ± (cid:33) = 11 − √ ± (cid:16) − √ (cid:17) / . The two largest eigenvalues are therefore λ = 11 + 3 √
54 + 14 (cid:16)
150 + 66 √ (cid:17) / and λ = 11 − √
54 + 14 (cid:16) − √ (cid:17) / . Next, as promised earlier, we return to Theorem 8.4.2(iii) concerning the goldenL-surface. √
52 1+ √ v v v v v h h h h h Figure 8.5.3: the golden L-surface, detour crossing and shortcutAs in Section 8.2, we shall apply the eigenvalue-based shortline method to computethe irregularity exponent for geodesic flow on this surface.What makes the golden L-surface particularly simple is that its two horizontalstreets are similar rectangles, and its two vertical streets are also similar rectangles;see Figure 8.5.3.
ON-INTEGRABLE SYSTEMS (IV) 93
To apply the surplus shortline method, we consider slopes of the special form α = 1 + √ a + 1 √ a + √ a + ··· , (8.5.5)where a i (cid:62) i (cid:62)
0, are integers. Furthermore, to determine the irregularityexponent explicitly, we need eventual periodicity of the sequence a , a , a , . . . . The golden L-surface has 3 faces.The almost vertical units h h , h h , h h , h h , h h , h h ∗ can be defined in thesame way as shown in Figure 7.1.6.The almost horizontal units v v , v v , v v , v v , v v , v v ∗ can be defined in asimilar way.The picture on the left in Figure 8.5.3 illustrates an almost horizontal detourcrossing of a horizontal street and its almost vertical shortcut h h in the specialcase when the branching parameter is equal to 2. Using the Delete-End Rule, for ageneral branching parameter k (cid:62)
1, the ancestor process can be summarized by h h (cid:42) v v , k × v v , k × v v , (8.5.6) h h (cid:42) v v , k × v v , ( k − × v v , (8.5.7) h h (cid:42) v v ∗ , k × v v , ( k − × v v , (8.5.8) h h (cid:42) v v ∗ , k × v v , k × v v , (8.5.9) h h (cid:42) v v , ( k − × v v , (8.5.10) h h ∗ (cid:42) v v , k × v v . (8.5.11)These lead to a 6 × M ( k ) = v v v v v v v v v v v v ∗ h h k k h h k − k h h k k − h h k k h h k − h h ∗ k . Similarly, we can study the ancestor relation of each of the almost horizontal units,again using the Delete-End Rule, and obtain the analogs of (8.5.6)–(8.5.11). Thesewill lead to another 6 transition matrix. Since we have listed the almost verticalunits and almost horizontal units in lexicographical order, these two 6 × M ( k ) are two pairs of algebraic conjugates1 + √ k ± (cid:32) √ k (cid:33) + 1 / , − √ k ± (cid:32) − √ k (cid:33) + 1 / , and ( − ± √ / √ k + (cid:32) √ k (cid:33) + 1 / = 1 + √ k + 1 √ k + √ k + ··· , i.e. , Λ is equal to the slope α in (8.5.5) in the special case a i = k for all i (cid:62)
0. Theeigenvalue with the second largest absolute value is λ = 1 − √ k − (cid:32) − √ k (cid:33) + 1 / . The remaining 4 eigenvalues are irrelevant.We shall show that the transition matrix M ( k ) has a conjugate with the form P − M ( k ) P = (cid:18) T ?0 A ( k ) (cid:19) , (8.5.12)where T = (cid:32) − −√ ?0 − √ (cid:33) , (8.5.13)and A ( k ) = (1+ √ k (1 −√ k
10 0 1 0 . (8.5.14)The description of the matrix M ( k ) by (8.5.12)–(8.5.14) is extremely convenient. Itreduces the necessary eigenvalue computation of arbitrary products r (cid:89) i =1 M ( k i )of 6 × k i to the muchsimpler eigenvalue computation of products r (cid:89) i =1 (cid:32) (1+ √ k i
11 0 (cid:33) and r (cid:89) i =1 (cid:32) (1 −√ k i
11 0 (cid:33) of 2 × − ± √ / M ( k ) has 2 eigenvectors that are independentof the branching parameter k . Together with the eigenvalues, they are λ = − − √ , v = (cid:32) − − √ , − , , − √ , , (cid:33) T ,λ = − √ , v = (cid:32) − √ , − , , − − √ , , (cid:33) T . This allows us to use a “partial diagonalization” trick first discussed in Lemma 4.1.1in [3]. Let Q be a 6 × v , v ,and where the remaining columns are v = (1 , , − , − , − , T , v = (0 , , , − , , T , v = (0 , , , , , − T , v = ( − , , , , , T . ON-INTEGRABLE SYSTEMS (IV) 95
Then Q − M ( k ) Q = (cid:18) T ?0 M ( k ) (cid:19) , (8.5.15)where M ( k ) = k − − k − k − k k − k − k − k − k k − k − − k . (8.5.16)Let R be a 4 × w = (cid:32) √ , √ − , , − √ (cid:33) , w = (cid:32) , , √ , (cid:33) . Routine calculation shows that if R is invertible, then independently of the choice ofits third and fourth columns, the conjugate R − M ( k ) R has the simpler triangularform R − M ( k ) R = (cid:18) A ( k ) ?0 A ( k ) (cid:19) , (8.5.17)where A ( k ) = (cid:32) (1+ √ k
11 0 (cid:33) , (8.5.18)and A ( k ) is a 2 × R .Again with some routine calculation we can find suitable third and fourth columnsof R which give A ( k ) = (cid:32) (1 −√ k
11 0 (cid:33) . (8.5.19)Clearly (8.5.12)–(8.5.14) follow on combining (8.5.15)–(8.5.19).Consider now a geodesic on the golden L-surface with slope α of the form (8.5.5),where the sequence a i (cid:62) i (cid:62)
0, of integers is eventually periodic. Then theshortline of this geodesic has slope α − , where α = 1 + √ a + 1 √ a + √ a + ··· , and the shortline of this shortline has slope α = 1 + √ a + 1 √ a + √ a + ··· , and so on. Theorem 8.5.1. (i)
Consider geodesic flow on the golden L-surface with slope α given by (8.5.5) . If the sequence a , a , a , . . . has period k , k , . . . , k r eventually,then the irregularity exponent of the geodesic with slope α is equal to log | λ | log | Λ | , where λ is the eigenvalue with the larger absolute value of the product matrix r (cid:89) i =1 (cid:32) (1 −√ k i
11 0 (cid:33) , (8.5.20) and Λ is the eigenvalue with the larger absolute value of the product matrix r (cid:89) i =1 (cid:32) (1+ √ k i
11 0 (cid:33) . (8.5.21) Alternatively, we have the product formula
Λ = r (cid:89) i =1 β i , where for every i = 1 , . . . , r , β i = 1 + √ k i + 1 √ k i +1 + √ k i +2 + ··· , corresponding to the periodic sequence k i , k i +1 , k i +2 , . . . , k r , k , k , . . . , k r , . . . . (ii) Similarly, for every street-rational polyrectangle surface there are infinitelymany explicit slopes for which any geodesic with such a slope exhibits superdensity,and we can also compute the corresponding irregularity exponent. (iii)
Combining the irregularity exponent with the method of zigzagging introducedin Section 3.3 in [2] , we can also describe, for any geodesic and special slopes inparts (i) and (ii) , the time-quantitative behavior of the edge-cutting and face-crossingnumbers, as well as equidistribution relative to all convex sets.
For the golden L-surface, let us now calculate the eigenvalues of 2-step transitionmatrix M by first finding the street-spreading matrix for the slope α given by (8.5.5)with a i = 1 for every i (cid:62)
0. We shall follow the notation in Section 7.2, and thedetails are shown in Figure 8.5.4. ↔ l ↔ l ↔ l × ↑ , × ↑ , u × ↑ , × ↑ , × ↑ , u Figure 8.5.4: almost vertical units of type ↑ in u , u Consider the horizontal street corresponding to u , as highlighted in the picturein the middle in Figure 8.5.4. Using (7.2.17), we have( M − I ) u = ( M − I )[ {↑ , } ] + ( M − I )[ {↑ , } ]= ( u + v ) + ( u + v ) . (8.5.22)For the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 8.5.4, a similar argument gives( M − I ) u = ( M − I )[ {↑ , } ] + ( M − I )[ {↑ , , ↑ , } ]= ( u + v ) + 2( u + v ) . (8.5.23)It follows from (8.5.22)–(8.5.23) that the street-spreading matrix is given by S = (cid:18) (cid:19) , with eigenvalues τ = 3 + √
52 and τ = 3 − √ . ON-INTEGRABLE SYSTEMS (IV) 97
Using (7.2.38), the corresponding eigenvalues of M are λ (cid:32) √
52 ; ± (cid:33) = 7 + √ ± (cid:32) √ (cid:33) + 4 (cid:32) √ (cid:33) / and λ (cid:32) − √
52 ; ± (cid:33) = 7 − √ ± (cid:32) − √ (cid:33) + 4 (cid:32) − √ (cid:33) / . The two largest eigenvalues are therefore λ = 7 + √
54 + 12 (cid:32) √ (cid:33) + 4 (cid:32) √ (cid:33) / and λ = 7 − √
54 + 12 (cid:32) − √ (cid:33) + 4 (cid:32) − √ (cid:33) / . For the corresponding 1-step transition matrix, we shall determine the eigenvaluesΛ and λ by using (8.5.20) and (8.5.21) with r = 1 and k = 1. The eigenvalue withthe larger absolute value of the matrices (cid:32) √
11 0 (cid:33) and (cid:32) −√
11 0 (cid:33) are respectively Λ = 1 + √
54 + 12 (cid:32) √ (cid:33) + 4 / and λ = 1 − √
54 + 12 (cid:32) − √ (cid:33) + 4 / . Note that λ = Λ and λ = λ .We complete this section by calculating the eigenvalues of the 2-step transitionmatrix M for the “golden brick”, first introduced at the end of Section 8.4, for ageodesic with a suitable slope.For the 4-copy surface of the golden brick, note that the streets of width 1 haslength 2 + 2 h , and so normalized length 2 + 2 h , while the streets of width h haslength 4, and so normalized length4 h = 2(2 + 2 h ) , so that h ∗ = v ∗ = 2(2 + 2 h ).To visualize the 4-copy version of the golden brick, we refer the reader to Figures7.2.12–7.2.14 which illustrate the 4-copy version of the surface of the unit cube. Weobtain the 4-copy version of the golden brick if we shorten the edges a , a , a , a , andthose in between, in Figure 7.2.12 from length 1 to length h . Then the horizontalstreets 3 , , h instead of 1, and we can obtain an analog of Figure 7.2.14 with thinnerrows 3 , , For simplicity, we consider a geodesic with slope α = 2(2 + 2 h ) + 12(2 + 2 h ) + h )+ h )+ ··· . As before, we define the matrices u i , v i , i = 1 , . . . ,
6, according to (7.2.14) and(7.2.15).Figures 8.5.5–8.5.7 below are the analogs of Figures 7.2.15–7.2.16 for the surfaceof the cube.For the horizontal streets corresponding to u , u , as highlighted in Figure 8.5.5,we have, applying (7.2.17),( M − I ) u = ( M − I ) u = ( M − I )[ { ↑ , , ↑ , , ↑ , , ↑ , } ] + ( M − I )[ { ↑ , , ↑ , } ]+ ( M − I )[ { ↑ , , ↑ , } ] + ( M − I )[ { ↑ , , ↑ , , ↑ , , ↑ , } ]+ ( M − I )[ { ↑ , , ↑ , } ] + ( M − I )[ { ↑ , , ↑ , } ]= 12( u + v ) + 4( u + v ) + 8( u + v )+ 12( u + v ) + 4( u + v ) + 8( u + v ) . (8.5.24) u u ↑ , ↑ , ↑ , ↑ , × × × × ↑ , ↑ , × × ↑ , ↑ , × × ↑ , ↑ , ↑ , ↑ , × × × × ↑ , ↑ , × × ↑ , ↑ , × × Figure 8.5.5: almost vertical units of type ↑ in u , u For the horizontal streets corresponding to u , u , as highlighted in Figure 8.5.6,a similar argument gives( M − I ) u = ( M − I ) u = 4( u + v ) + 12( u + v ) + 8( u + v )+ 4( u + v ) + 12( u + v ) + 8( u + v ) . (8.5.25) ON-INTEGRABLE SYSTEMS (IV) 99 u u ↑ , ↑ , × × ↑ , ↑ , ↑ , ↑ , × × × × ↑ , ↑ , × × ↑ , ↑ , × × ↑ , ↑ , ↑ , ↑ , × × × × ↑ , ↑ , × × Figure 8.5.6: almost vertical units of type ↑ in u , u For the horizontal streets corresponding to u , u , as highlighted in Figure 8.5.7,a similar argument gives( M − I ) u = ( M − I ) u = 4( u + v ) + 4( u + v ) + 8( u + v )+ 4( u + v ) + 4( u + v ) + 8( u + v ) . (8.5.26) u u ↑ , ↑ , × × ↑ , ↑ , × × ↑ , ↑ , ↑ , ↑ , × × × × ↑ , ↑ , × × ↑ , ↑ , × × ↑ , ↑ , ↑ , ↑ , × × × × Figure 8.5.7: almost vertical units of type ↑ in u , u It follows from (8.5.24)–(8.5.26) that the street-spreading matrix is given by S =
12 4 4 12 4 44 12 4 4 12 48 8 8 8 8 812 4 4 12 4 44 12 4 4 12 48 8 8 8 8 8 ,
00 BECK, CHEN, AND YANG with non-zero eigenvalues τ = 24 + 8 √ τ = 16 and τ = 24 − √
5. Using (7.2.38),the corresponding eigenvalues of M are λ (24 + 8 √ ± ) = 13 + 4 √ ± (cid:16)
248 + 104 √ (cid:17) / ,λ (16; ± ) = 9 ± √ ,λ (24 − √ ± ) = 13 − √ ± (cid:16) − √ (cid:17) / . Surfaces tiled with congruent equilateral triangles.
We wish to completeour study of geodesic flow on the surface of each of the 5 platonic solids.We have already discussed in [2] the superdensity of geodesic flow on the regulartetrahedron surface which is integrable. Earlier in this paper, we have also discussedthe superdensity of geodesic flow on the cube surface and the regular dodacahedronsurface which are non-integrable.The last two examples of the 5 platonic solids are the regular octahedron and theregular icosahedron. Both of these surfaces belong to the large class of flat surfaceswhere the faces are congruent equilateral triangles. We refer to such a surface asa polytriangle surface . Every polytriangle surface has the crucial property thatthe union of any two equilateral triangle faces sharing an edge forms a 60-degreerhombus, which turns out to be a perfect substitute for squares. Thus polytrianglesurfaces are perfect analogs of polysquare surfaces, and both versions of the shortlinemethod work for this class.We give here a detailed discussion of geodesic flow on the regular octahedronsurface, as shown in Figure 8.6.1. Geodesic flow on the icosahedron surface goes ina similar way, and we omit it.
AFB DCE AE DB C
FB CE D
56 7 8
Figure 8.6.1: the regular octahedron, top view and bottom viewFor our convenience, we shall label the faces as in Figure 8.6.1. Thus we have faceand vertex pairings given by( , ABC ) , ( , ACD ) , ( , ADE ) , ( , AEB ) , ( , F ED ) , ( , F BE ) , ( , F CB ) , ( , F DC ) , where the three vertices of each triangle face are given in anticlockwise order. Notethat we have four pairs of vertex-disjoint and parallel faces, given by( , ) , ( , ) , ( , ) , ( , ) . Between any vertex-disjoint pair of faces we have a street made up of 6 trianglefaces. In Figure 8.6.2, we show 4 different nets of the regular octahedron, togetherwith a surplus detour crossing of a geodesic in a given direction.
ON-INTEGRABLE SYSTEMS (IV) 101
ABA D E AB C F B
ADA B C AD E F D
EAE F B EA D C A
CAC F D CA B E A
Figure 8.6.2: nets of the regular octagon surface,detour crossings and shortcutsThe picture on the top left of Figure 8.6.2 is a net of the regular octahedronsurface. The long parallelogram in the middle is a street between faces and . Moreprecisely, the cycle of triangle faces , , , , , is the street, with the left and rightedges of the parallelogram identified. This street also has a natural decompositioninto 60-degree rhombi formed from the union of the pairs ( , ), ( , ) and ( , ).The detour crossings in all 4 different nets can be thought of as all going in thesame direction. The copies on the top left and bottom right have parallel edges AB ,while the copies on the top right and bottom left have parallel edges AD . On theother hand, the two copies on the left have parallel edges CA , while the two copieson the right have parallel edges EA .Note that each shaded triangle in Figure 8.6.2 has 2 more copies of itself rotatedby 120 degrees and 240 degrees. We therefore conclude that a geodesic crossesevery triangle face in 3 different directions, at 120 degrees to each other. Note alsothat every triangle face is complemented by 3 other triangle faces to form 60-degreerhombi. For instance, for the triangle face , we have the pair ( , ) in the top left,the pair ( , ) in the bottom left, and the pair ( , ) in the bottom right.Since a 60-degree rhombus is just a tilted analog of a square, and can be viewedas one, it is straightforward to adapt the shortline method. To obtain some specialslopes, we consider the analogy illustrated in Figure 8.6.3. ABA D E AB C F BY BA BAY
Figure 8.6.3: a particular geodesic and mutual shortlines
02 BECK, CHEN, AND YANG
For the polysquare model in the picture on the right, it is clear that we consideran almost horizontal geodesic of slope α − k , where α k = [3 k ; 3 k, k, k, . . . ] = √ k + 4 + 3 k . (8.6.1)Then the shortline has slope α k . Returning to the original polytriangle surface, asuitable angle corresponding to α k can be determined. In particular, if the lengthof the edge DE is equal to 1, then the length of the segment AY should be equalto α − k . Elementary calculation shows that the detour crossing in the picture on theleft in Figure 8.6.3 must have slope β − k = √ α k + 1 , (8.6.2)which is √ a and b having different directions. This is sometimes known as the translation surfacefor 60-degree rhombus billiard. Note that the edge labellings in Figure 8.6.4 areobtained by following the convention discussed in the Remark after Figure 8.3.3.Identifying the boundary edges a , b , b, c , c , c , d , d , d , we obtain a polytrianglesurface, which is intuitively a “double hexagon”. a b d c c d c d a b e e e b a c d d c d c b a e e e Figure 8.6.4: the translation surface of 60-degree rhombus billiardwith edge labellingsFurthermore, we label the 6 rhombi as in Figure 8.6.5. T , T , T , T , T , T , T , T , T , T , T , T , Figure 8.6.5: the faces of the translation surface of 60-degree rhombus billiardIt is not difficult to see that 60-degree rhombus billiard is equivalent to 1-directiongeodesic flow on the polytriangle surface defined by the “double hexagon” shown inFigures 8.6.4 and 8.6.5. We refer to this polytriangle surface as the RB-surface, orthe rhombus billiard surface.We can list the streets of the RB-surface explicitly.
ON-INTEGRABLE SYSTEMS (IV) 103
The long parallelogram in the middle of Figure 8.6.6 is a street of the RB-surface,comprising the cycle T , , T , , T , , T , , T , , T , (8.6.3)of 6 equilateral triangles. The key fact is that this street has a natural decompositioninto 60-degree rhombi T , ∪ T , , T , ∪ T , , T , ∪ T , . T , T , T , T , T , T , T , T , T , T , T , T , d d c c d d e e e e b b e e Figure 8.6.6: a net and a street with detour crossing of the RB-surfaceBeside the street (8.6.3), there are 5 more streets of the RB-surface, made up ofthe cycles T , , T , , T , , T , , T , , T , , (8.6.4) T , , T , , T , , T , , T , , T , , (8.6.5) T , , T , , T , , T , , T , , T , , (8.6.6) T , , T , , T , , T , , T , , T , , (8.6.7) T , , T , , T , , T , , T , , T , . (8.6.8)However, we only need some of these streets.Figure 8.6.7 gives another description of the RB-surface as a polyrhombus surface. T , T , T , T , T , T , T , T , T , T , T , T , d d a a T , T , T , T , T , T , T , T , T , T , T , T , e e d d Figure 8.6.7: the RB-surface as a polyrhombus surfaceIn the picture on the left, we see 2 horizontal streets. The top horizontal streetis (8.6.3), while the bottom horizontal street is (8.6.7). In the picture on the right,we see 2 tilted streets that are the analogs of vertical streets in polysquare surfaces.The left tilted street is (8.6.8), while the right tilted street is (8.6.5).Noting the similarity between Figure 8.6.3 and Figure 8.6.6, we conclude that forthe shortline method to work, the detour crossing in Figure 8.6.6 must have slope β − k , given by (8.6.1) and (8.6.2).We have the following analog of Theorem 8.5.1(ii). Theorem 8.6.1. (i)
For any arbitrary polytriangle surface, there exist infinitelymany explicit slopes, depending on the surface, such that any half-infinite geodesicwith such a slope exhibits superdensity.
04 BECK, CHEN, AND YANG (ii)
For any arbitrary table of polytriangle shape, there exist infinitely many explicitslopes, depending on the shape of the table, such that any half-infinite billiard orbithaving such an initial slope exhibits superdensity. (iii)
For infinitely many of these slopes in parts (i) and (ii) that give rise to su-perdensity, we can explicitly compute the corresponding irregularity exponents whichdescribe the time-quantitative aspects of equidistribution.
For the RB-surface, we now attempt to find the eigenvalues of its 2-step transitionmatrix M . This surface can be viewed as a polyrhombus surface, and we shall usean analogy with a corresponding polysquare surface as shown in Figure 8.6.8. ↑ ↑ ↑ ↑ ↑ ↑ Figure 8.6.8: analogy of the double-pentagon surfacewith a polyrectangle surfaceFor simplicity of notation, the rhombi are renumbered 1 , , , , ,
6, as shown inthe picture on the left in Figure 8.6.8, and we consider the corresponding polysquaresurface in the picture on the right. Here the horizontal streets are 1 , , , , , , , ,
3. Both horizontal streets and bothvertical streets have lengths 3. Note here that distinct square faces can fall into thesame horizontal street and the same vertical street, so we shall adapt our notationfrom Section 7.2 for the determination of the street-spreading matrix.We also show in the picture on the right in Figure 8.6.8 the almost vertical unitsof type ↑ in the polysquare surface. We have not shown that almost vertical unitsof type − ↑ here.Let J = { , , } and J = { , , } denote the horizontal streets, and let I = I = I = { , , } and I = I = I = { , , } denote the vertical streets.For the polysquare surface, we thus consider slopes of the form α k = 3 k + 13 k + k + ··· . For simplicity, however, we consider only the special case with branching parameter k = 1.Corresponding to (7.2.14), we define the column matrices u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] and u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] . (8.6.9)Here J ∗ , J ∗ and I ∗ j denotes that the edges are counted with multiplicity. × × × × × × × × × ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ u × × × × × × × × × ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ u Figure 8.6.9: almost vertical units of type ↑ in u and u ON-INTEGRABLE SYSTEMS (IV) 105
We also define the column matrices v and v analogous to (7.2.15), but theirdetails are not important. Also, analogous to (7.2.17), we have( M − I )[ {↑ s } ] = (cid:26) u + v , if s ∈ J , u + v , if s ∈ J . (8.6.10)We now combine (8.6.9) and (8.6.10). For the horizontal street corresponding to u ,as highlighted in the picture on the left in Figure 8.6.9, we have( M − I ) u = ( M − I )[ {↑ , ↑ , ↑ } ] + ( M − I )[ {↑ , ↑ , ↑ } ]= 5( u + v ) + 4( u + v ) . (8.6.11)For the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 8.6.9, we have( M − I ) u = ( M − I )[ { ↑ , ↑ , ↑ } ] + ( M − I )[ { ↑ , ↑ , ↑ } ]= 4( u + v ) + 5( u + v ) . (8.6.12)It follows from (8.6.11)-(8.6.12) that the street-spreading matrix is given by S = (cid:18) (cid:19) . The eigenvalues of S are τ = 9 and τ = 1. Using (7.2.38), the correspondingeigenvalues of M are λ (9; ± ) = 112 ± √
117 and λ (1; ± ) = 32 ± √ . The two largest eigenvalues are therefore λ = 112 + 12 √
117 and λ = 32 + 12 √ . We have studied geodesics and billiards on many flat surfaces, for which bothversions of the shortline method work, and established many explicit slopes thatyield superdensity and for which we can compute the irregularity exponents. Suchsurfaces include polysquares, street-rational polyrectangles and polyparallelogramswhich include regular polygons and L-staircases, and polytriangles. We may referto them under the general name of street-rational surfaces .8.7.
The shortline method works for all Veech surfaces, and beyond.
Thetitle of this section summarizes Sections 8.1–8.6, that the surplus shortline methodworks for all Veech surfaces, providing time-quantitative results like superdensityand uniformity with explicit values of the irregularity exponents.Note that we have not defined here the concept of
Veech surfaces .As motivation, consider an arbitrary polysquare surface P and 1-direction geodesicflow on it. We now consider the following two statements:(1) The surface P has a street-rational decomposition in any arbitrary directionwith rational slope.(2) The surface P is optimal, i.e. , 1-direction geodesic flow on P exhibits uniform-periodic dichotomy.Perhaps the best way to describe Veech surfaces is to say that they are translationsurfaces that exhibit properties like (1) and (2). Unfortunately, the precise definitionis a rather technical algebraic one, and we do not really need to know that here.The interested reader is referred to the introductory article [10].Instead, our wish is to have a list of known examples of Veech surfaces, and someof the key properties of such surfaces. In particular, the following two propertiesestablished by Veech [17] are analogous to (1) and (2) respectively.
06 BECK, CHEN, AND YANG
Theorem A.
Suppose that S is a Veech surface, and that v is an arbitrary directionsuch that there exists a non-empty finite geodesic segment on S that goes betweentwo not necessarily distinct singularities. Then S has a street-rational decompositionin this direction. Theorem B.
Every Veech surface S is optimal, i.e., -direction geodesic flow on S exhibits uniform-periodic dichotomy.Remark. Note that Veech and other authors use the term cylinder instead of theterm street .Classifying the class of Veech surfaces is a very difficult problem. Early examplesare the translation surfaces of regular polygon billiards discovered by Veech, togetherwith a few related examples. Progress has been slow, despite the effort of many. Thesubject is on the borderline of many fields, including ergodic theory, topology andalgebraic geometry. The goal is to understand optimal dynamics , a time-qualitativeproperty of the long-term behavior of orbits, and the existing methods do not seemto say anything about the time-quantitative aspects. Our goal here is to complementthe time-qualitative work by making some time-quantitative statements on the long-term behavior of orbits.Indeed, our surplus shortline method works for all Veech surfaces, and the reasonis very simple. For the success of our method, we need a translation surface withstreet-rational decomposition in only two different directions, and Theorem A givesfar more than that.The class of Veech surfaces include the following:(i) polysquare surfaces including the flat torus, and polytriangle surfaces;(ii) translation surfaces of regular polygon billiards and some related systems suchas infinitely many special triangle billiards, including the regular decagon surfaceand the Ward system;(iii) Calta–McMullen L-staircases;(iv) the class of cathedral surfaces ;(v) the Bouw–M¨oller family;(vi) isolated triangle billiards with angles ( π/ , π/ , π/ π/ , π/ , π/
9) and( π/ , π/ , π/ primitive Veech surfaces.Here we do not say anything about the Ward system or the Bouw–M¨oller family.However, for illustration, we elaborate a little on the class of cathedral surfaces , firstintroduced by McMullen, Mukamel and Wright [15] in 2017.Figure 8.7.1 shows two identical copies of the cathedral polygon CP( a, b ). Thepolygon is symmetric across the horizontal axis, and each boundary edge either ishorizontal or vertical, or has slope ±
1. The identification of boundary horizontal andvertical edges comes from perpendicular translation, while the identification of edgesof slope ± H , . . . , H and the decomposition into vertical streets V , . . . , V . Thus the cathedral surface iswell defined. It has 5 horizontal streets and 5 vertical streets. ON-INTEGRABLE SYSTEMS (IV) 107 H H H H H H H V V V V V V V a aa ab bb b a a Figure 8.7.1: horizontal and vertical streets of CP( a, b )Let a = r √ d + r > b = 3 r √ d − r − > , (8.7.1)where d (cid:62) r , r are rational numbers. Remarkably, aslong as (8.7.1) is satisfied, geodesic flow on the cathedral surface CP( a, b ) is optimal,as shown by McMullen, Mukamel and Wright.We now verify that under the condition (8.7.1), the cathedral surface CP( a, b ) isa street-rational polyrectangle surface.Consider first the horizontal streets H , . . . , H as indicated in the picture on theleft in Figure 8.7.1. Here the ratio of the (horizontal) lengths and (vertical) widthsof these streets are respectively1 + 2 a / , b , a , b , a / , so these streets have only two different shapes, with ratio2 b (1 + 2 a ) = 3(2 r √ d − r − r √ d + 2 r + 1) = 3(4 r d − (2 r + 1) ) , which is clearly rational.Consider next the vertical streets V , . . . , V as indicated in the picture on theright in Figure 8.7.1. Here the ratio of the (vertical) lengths and (horizontal) widthsof these streets are respectively1 a , b / , b / , a , a , so these streets have only two different shapes, with ratio2 a (3 + 2 b ) = 12( r √ d + r )( r √ d − r ) = 12( r d − r ) , which is also clearly rational.Thus the surplus shortline method works for any cathedral surface CP( a, b ) thatsatisfies (8.7.1). We therefore conclude that geodesic flow on such surfaces exhibits time-qualitative optimality, and also exhibits time-quantitative behavior in the sensethat there are infinitely many explicit slopes such that geodesics with such slopesare superdense and we can also compute the irregularity exponents.It is easy to see that the class of cathedral surfaces CP( a, b ) depends only on tworational parameters q = r d and r , so this is a 2-parameter class.While we remain very far from the complete classification of all primitive Veechsurfaces, we now know all the primitive Veech surfaces that have genus 2, throughthe breakthrough of McMullen [12, 13].
08 BECK, CHEN, AND YANG
Theorem C.
The affine-different primitive Veech surfaces in genus are preciselythe following: (i) the infinite class of Calta–McMullen L-staircases; and (ii) the regular decagon surface with parallel edge identification. The following partial converse to Theorem B is also due to McMullen [12].
Theorem D.
Suppose that a translation surface S in genus is not a Veech surface.Then there exist geodesics on S which are neither dense nor periodic. A different sort of partial converse to Theorem B is due to Cheung and Masur [8],building on the work of McMullen.
Theorem E.
Suppose that a translation surface S in genus is not a Veech surface.Then there exist geodesics on S which are dense but not uniformly distributed. Recall that the surplus shortline method works for all Veech surfaces, as we needonly street-rational decomposition in two different directions, while Veech surfacesexhibit street-rational decomposition in infinitely many different directions. Thismakes it plausible to expect that there are many surfaces that are not Veech surfaces,or not optimal, but nevertheless have sufficient street-rationality for the surplusshortline method to work, giving rise to time-quantitative results. Thus the surplusshortline method goes beyond the class of optimal systems.We next give an example of a 3-parameter family of non-optimal systems that havesufficient street-rationality for the surplus shortline method to work. Figure 8.7.2shows a family of decagons which are affine-different, due to the fixed height 2 andfixed width 2. These decagons are also symmetric across the horizontal axis andthe vertical axis as shown. Identifying the parallel edges of every such decagon, weobtain a translation surface S ( a, b, c ) in genus 2. V V V V V H H H H (1 , − ,
0) ( a,
1) ( b, c )( a,
1) ( b, c ) (0 , ,
1) (0 , − , − Figure 8.7.2: S ( a, b, c ) and its horizontal and vertical streetsNote first that S ( a, b, c ) has 2 horizontal streets H , H , as shown in the pictureon the left in Figure 8.7.2. Here the ratio of the (horizontal) lengths and (vertical)widths of these streets are respectively2 a + 2 b − c , bc , so we have street rationality if there exists a positive rational number r such that a + b − c = r bc . (8.7.2)Note next that S ( a, b, c ) has 3 vertical streets V , V , V , as shown in the pictureon the right in Figure 8.7.2. Here the ratio of the (vertical) lengths and (horizontal) ON-INTEGRABLE SYSTEMS (IV) 109 widths of these streets are respectively2 c − b , cb − a , a , so we have street rationality if there exist positive rational numbers r , r such that a = r − bc and a = r b − a c . (8.7.3)Given positive rational numbers r , r , r , we have 3 equations (8.7.2)–(8.7.3) inthe 3 variables a, b, c . It follows that there exist infinitely many affine-different non-regular decagons S ( a, b, c ) that have street-rational decompositions in the horizon-tal and vertical directions. Every one of these non-regular decagons is primitive andhas genus 2, differs from the Veech surfaces listed in Theorem C, and is therefore notoptimal. It then follows from Theorem D that each has geodesics which are neitherdense nor periodic. It also follows from Theorem E that each has geodesics whichare dense but not uniformly distributed. However, it also follows from the surplusshortline method that each has geodesics that exhibit superdensity and quantitativeequidistribution with explicit irregularity exponents. These vastly different types oforbits that can arise demonstrate the intriguing fact that we have a wide spectrumof possibilities.Unfortunately, we currently do not have analogs of Theorems C and D whenthe genus exceeds 2, and this prevents us from proving plausible conjectures aboutmore infinite families of surfaces that are not optimal but have sufficient street-rationality for the surplus shortline method to work. Nevertheless, we are of theopinion that the class of surfaces that have sufficient street-rationality for the surplusshortline method to work is much , much larger than the class of surfaces with optimaldynamics.8.8. Density in generalized mazes.
In this section, we switch to infinite flatsurfaces, and begin by generalizing the concept of square-maze first introduced inSection 6.5 in [4].It is natural to start the discussion here with arguably the simplest example ofgeneralized square-mazes, the so-called infinite halving staircase surface S ( ∞ ; 1 / infinite halving staircase region R ( ∞ ; 1 /
2) shown in Figure 8.8.1. (0 ,
0) (1 ,
0) (3 , − b b bb b b b L ( − L ( − L (0)3 x + y = 72 x + 3 y = 0 Figure 8.8.1: three halving L-shapes L (0) , L ( − , L ( −
2) of R ( ∞ ; 1 /
10 BECK, CHEN, AND YANG
The building blocks of R ( ∞ ; 1 /
2) are L-shapes L ( n ), n ∈ Z , that consist of threesquares of the same size 2 n × n . These L-shapes of different sizes are glued togetheras shown. Thus R ( ∞ ; 1 /
2) = (cid:91) −∞ 2) from the infinite region R ( ∞ ; 1 / R ( ∞ ; 1 / L ( i ). The vertical boundary edge identification is simple, and comprises apair of identified edges v (cid:48) i on the top square of L ( i ) and a pair of identified edges v (cid:48)(cid:48) i on the two bottom squares of L ( i ). The horizontal edge identification is a littleless straightforward. First of all, there is a pair of identified edges h (cid:48) i on the left halfof the right square of L ( i ). Then there are two further horizontal boundary edges.The edge h (cid:48)(cid:48) i on the top of the right half of the right square of L ( i ) is identified witha horizontal edge of the L-shape L ( i − 1) below, while the edge h (cid:48)(cid:48) i +1 on the bottomof the bottom left square of L ( i ) is identified with a horizontal edge of the L-shape L ( i + 1) above. b L ( i ) v ′ i v ′ i v ′′ i v ′′ i h ′ i h ′ i h ′′ i +1 h ′′ i h ′′ i Figure 8.8.2: identifying pairs of boundary edges of R ( ∞ ; 1 / R ( ∞ ; 1 / i.e. , every square haszero curvature, then it becomes the desired flat infinite closed surface S ( ∞ ; 1 / S ( ∞ ; 1 / S ( ∞ ; 1 / 2) as a generalized maze – more about this later.The infinite surface S ( ∞ ; 1 / 2) can be interpreted as a limit of a sequence ofcompact surfaces S ( n ; 1 / n = 0 , , , , . . . . For an arbitrary integer n (cid:62) 0, thesurface S ( n ; 1 / 2) is defined by “closing” the open surface (cid:91) − n (cid:54) i (cid:54) n L ( i ) , ON-INTEGRABLE SYSTEMS (IV) 111 which is a union of 2 n + 1 L-shapes. Here “closing” means setting up an appropriatepattern for identifying pairs of parallel boundary edges.We illustrate the pattern in the special cases n = 0 and n = 1 in Figure 8.8.3.The general case goes in a similar way. h h h h h h h h h h v v v v Figure 8.8.3: S (0; 1 / 2) and S (1; 1 / 2) (not to scale)In the picture of S (0; 1 / 2) on the left, the edges v , v and h correspond re-spectively to v (cid:48) , v (cid:48)(cid:48) and h (cid:48) in Figure 8.8.2. The only novelty comes from the twocopies of h and the two copies of h , as these identified edges do not have the samelength. The edge h on the left has twice the length of the edge h on the right,and the edge h on the left has twice the length of the edge h on the right. Thusthe identification comes from a translation combined with a dilation or contractionwith factor 2.In the picture of S (1; 1 / 2) on the right, we have only indicated the edge pairings h and h . The remaining edge pairings follow the pattern set in Figure 8.8.2. Again,the pairs of edges h and h do not have the same length. The edge h on the lefthas 8 times the length of the edge h on the right, and the edge h on the left has8 times the length of the edge h on the right. Thus the identification comes froma translation combined with a dilation or contraction with factor 8.It is easy to see that for an arbitrary integer n (cid:62) 0, the identification of the specialedges h and h comes from a translation combined with a dilation or contractionwith factor 2 n +1 .In view of this dilation or contraction, the bounded surfaces S ( n ; 1 / n (cid:62) 0, falloutside of the class of polysquare surfaces. We say that S ( n ; 1 / n (cid:62) 0, belong tothe class of d-c-polysquare surfaces , where d-c refers to the dilation and contraction of some boundary edges of the polysquare.It is easy to see that S ( ∞ ; 1 / 2) has infinite genus which, intuitively, represents“infinite complexity of the geodesic flow”. Actually, all square-mazes have infinitegenus. We have already mentioned this simple fact several times without proof. Forthe sake of completeness, we include here a sketch of the proof in the special case of S ( ∞ ; 1 / − g i = V i − E i + R i , where g i denotes the genus of S ( i ; 1 / V i , E i , R i denote respectively thenumbers of vertices, edges and regions of S ( i ; 1 / 2) after boundary edge identification. 12 BECK, CHEN, AND YANG Using Figure 8.8.3, we see that2 − g = V − E + R = 2 − − , − g = V − E + R = 2 − 24 + 12 = − . Hence the genus of S (0; 1 / 2) is 2, and the genus of S (1; 1 / 2) is 6. Similarly, one canshow that for an arbitrary integer n (cid:62) 0, the genus of S ( n ; 1 / 2) is 4 n + 2. Since S ( ∞ ; 1 / 2) is a limit of S ( n ; 1 / 2) as n → ∞ , we conclude that S ( ∞ ; 1 / 2) has infinitegenus.Since S ( ∞ ; 1 / 2) is a flat surface, its geodesics consist of parallel straight linesegments, and we have a 1-direction flow. We may call S (0; 1 / period-surface of S ( ∞ ; 1 / S ( ∞ ; 1 / (0 , 0) (2 , , , 0) (2 , , Figure 8.8.4: the horizontal and vertical streets of S ( ∞ ; 1 / S ( ∞ ; 1 / V ∗ ( t ) and H ∗ ( t ) on S ( ∞ ; 1 / V ∗ ( t ) H ∗ ( t )(0 , 0) (2 , , PQ Figure 8.8.5: two particular geodesics of S ( ∞ ; 1 / ON-INTEGRABLE SYSTEMS (IV) 113 The almost vertical geodesic V ∗ ( t ), t (cid:62) 0, starts from the origin and has specialslope α = 2 + √ ··· , (8.8.1)where the continued fraction digits are all equal to 4. The almost horizontal geodesic H ∗ ( t ), t (cid:62) 0, starts from the origin and has the reciprocal slope α − = √ − 2. Notethat V ∗ (0) = H ∗ (0) is the origin and, as usual, the parameter t denotes time; inother words, for both V ∗ ( t ) and H ∗ ( t ), we use the arc-length parametrization, whichrepresents unit-speed motion of a particle. The key fact is that V ∗ ( t ) and H ∗ ( t ) areshortlines of each other. Indeed, the initial segment of V ∗ ( t ), t (cid:62) 0, between theorigin and the point P is a detour crossing of a vertical street of 4 building blocksquares, and the initial segment of H ∗ ( t ), t (cid:62) 0, between the origin and the point P is the shortcut. Similarly, the initial segment of H ∗ ( t ), t (cid:62) 0, between the origin andthe point Q is a detour crossing of a horizontal street of 2 building block squares,and the initial segment of V ∗ ( t ), t (cid:62) 0, between the origin and the point Q is theshortcut.The “mutual shortcut” property of the special geodesics V ∗ ( t ), t > 0, and H ∗ ( t ), t > 0, means that for every vertical street of S ( ∞ ; 1 / V ∗ ( t ) coincides with the shortcut crossing points of H ∗ ( t ). Similarly, for everyhorizontal street of S ( ∞ ; 1 / H ∗ ( t ) coincides withthe shortcut crossing points of V ∗ ( t ).As a consequence of an intrinsic symmetry within S ( ∞ ; 1 / S ( ∞ ; 1 / 2) is infinite;this symmetry makes it sufficient to distinguish only 8 different types of almostvertical shortcuts or units of slope α , where the prototypes are a i , 1 (cid:54) i (cid:54) 8, inthe picture on the left in Figure 8.8.6. Similarly, it is sufficient to distinguish only8 different types of almost horizontal shortcuts of slope α − , where the prototypesare b j , 1 (cid:54) j (cid:54) 8, in the picture on the right in Figure 8.8.6. (0 , , a a a a a a a a x (0 , , b b b b b b b b y Figure 8.8.6: almost vertical and almost horizontal units of S ( ∞ ; 1 / R ( ∞ ; 1 / 2) is built from similarL-shapes L ( i ), i ∈ Z , where L ( i + 1) and L ( i − 1) are respectively obtained from L ( i )by doubling and halving. This gives rise to a transformation U , which represents“up and doubling”, as well as an inverse transformation D , which represents “downand halving”.Consider a prototype almost vertical unit a and another almost vertical unit x in the picture on the left in Figure 8.8.6, where x is the same type of unit as a .The transformation U maps x to a , which we denote formally by a = U x . Thetransformation D maps a to x , which we denote formally by x = D a . 14 BECK, CHEN, AND YANG Consider a prototype almost horizontal unit b and another almost horizontal unit y in the picture on the right in Figure 8.8.6, where y is the same type of unit as b .The transformation U maps y to b , which we denote formally by b = U y . Thetransformation D maps b to y , which we denote formally by y = D b .There are analogs for the remaining prototypes a i , 2 (cid:54) i (cid:54) 8, and b j , 1 (cid:54) j (cid:54) U , and the symmetry groupof S ( ∞ ; 1 / 2) contains the infinite cyclic group { U k : k ∈ Z } .It is straightforward to determine the ancestor units of the almost vertical units a i , 1 (cid:54) i (cid:54) 8. These include a fractional unit at the beginning and a fractional unitat the end, with whole units in the middle. We have a (cid:44) → { b } , b , { U b } , (8.8.2) a (cid:44) → { b } , b , { U b } , (8.8.3) a (cid:44) → { U b } , b , b , b , { b } , (8.8.4) a (cid:44) → { U b } , b , b , b , { b } , (8.8.5) a (cid:44) → { b } , b , b , b , { b } , (8.8.6) a (cid:44) → { b } , b , b , b , { b } , (8.8.7) a (cid:44) → { b } , b , b , b , { b } , (8.8.8) a (cid:44) → { b } , b , b , b , { b } , (8.8.9)where fractional units are indicated by { . . . } , and the whole units in the middle arelisted lexicographically.We also determine the ancestor units of the almost horizontal units b j , 1 (cid:54) j (cid:54) b (cid:44) → { D a } , D a , D a , a , { a } , (8.8.10) b (cid:44) → { a } , D a , D a , { a } , (8.8.11) b (cid:44) → { a } , D a , D a , a , { D a } , (8.8.12) b (cid:44) → { a } , a , { a } , (8.8.13) b (cid:44) → { a } , a , { a } , (8.8.14) b (cid:44) → { a } , a , U a , { a } , (8.8.15) b (cid:44) → { a } , a , a , U a , { a } , (8.8.16) b (cid:44) → { a } , a , U a , { a } . (8.8.17)The symmetry group of S ( ∞ ; 1 / 2) contains the infinite cyclic group { U k : k ∈ Z } .It follows that by applying arbitrary integer powers of the transformation U to(8.8.2)–(8.8.17), we obtain their analogs over the whole infinite surface.The infinite halving staircase surface S ( ∞ ; 1 / 2) is not a square-maze. However,as Figure 8.8.6, the relations (8.8.2)–(8.8.17) and their analogs illustrate, the keyconcept of shortline is well defined for some explicit slopes including the particularslope given by (8.8.1). We can therefore easily adapt the proof of Theorem 6.5.1in [4] and obtain density of some explicit geodesics on this infinite surface.We may say, intuitively speaking, that S ( ∞ ; 1 / 2) is an infinite polysquare surface“with bounded-ratio streets formed from squares with side lengths equal to integerpowers of 2”. We now make this intuition precise with the following definition,which includes the infinite halving staircase surface as a special case. Definition of the 2-power square-maze. An infinite closed flat surface S iscalled a 2-power square-maze if it satisfies the following four requirements:(1) The building blocks of S are axis-parallel squares with possibly different sidelengths 2 k , where k ∈ Z . ON-INTEGRABLE SYSTEMS (IV) 115 (2) Any two building block squares that have a common edge have a commonvertex, and either they have the same side length, or the side length of one is halfthe side length of the other.(3) S does not contain an infinite horizontal or vertical line. Furthermore, thereis an integer r such that any horizontal or vertical line segment fully inside S canbe covered by at most r congruent squares fully inside S .(4) We apply the simplest boundary identification via perpendicular translation,giving rise to 1-direction geodesic flow in S .The requirement (3) expresses the analogous “maze property” that the surface S has “bounded-ratio streets”, where the ratio of the long side and the short side ofany street is less than an absolute constant.If the minimum value of r in (3) is r , then we call this a 2 -power r -square-maze . The infinite halving staircase surface S ( ∞ ; 1 / 2) is a an example of a 2-power4-square-maze.In a similar way, we can define the class of k -power r -square-mazes for every fixedintegers k (cid:62) r (cid:62) S ( ∞ ; 1 / 2) provided by U and D is not really important. What is importantis that the concept of shortline is well defined for any k -power r -square-maze atleast for some explicit slopes. Moreover, the requirement (3) on “bounded-ratiostreets” guarantees that, despite the different sizes of the squares, the magnificationprocess in the shortline method still works. Thus we can adapt the proof of Theo-rem 6.5.1 in [4] and obtain density of some explicit geodesics on these generalizedmaze surfaces.Next we return to the ordinary square-maze of congruent unit square buildingblocks. We can obtain a different kind of generalization by replacing the “unitsquare” in the definition of the square-maze with, say, any other fixed regular n -gon, where n (cid:62) n -gon, we can form a horizontal-vertical Z -like grid.Making infinitely many appropriate “holes” in the grid, we can easily enforce the“maze property” that the lengths of the horizontal and vertical streets are uniformlybounded. The “maze property” then guarantees that we can adapt the proof ofTheorem 6.5.1 in [4] and obtain density of some explicit geodesics on these n -gon-maze surfaces .8.9. When the flow does not preserve the area. We now return to the infinitehalving staircase surface S ( ∞ ; 1 / S (0; 1 / 2) shown in thepicture on the left in Figure 8.8.3, and recall that S (0; 1 / 2) is not a polysquaresurface, but belongs to the class of d-c-polysquare surfaces, where there is boundaryedge identification involving dilation or contraction.We study 1 -direction line-flow on the period-surface S (0; 1 / projection of 1-direction geodesic flow on the infinite halving staircase surface S ( ∞ ; 1 / S (0; 1 / 2) particularly interesting is that it is our first exampleof a flat system with a natural 1-direction line-flow that is not area-preserving, aconsequence of the dilation or contraction at the boundary. Thus we cannot expectuniform distribution of the orbits. Our purpose in this section is therefore to examinewhat we can still say about the distribution of an orbit. 16 BECK, CHEN, AND YANG Let us return to the picture on the left in Figure 8.8.6. Since every almost verticalunit a i , 1 (cid:54) i (cid:54) 8, has the same length, it is fairly easy to compute asymptoticallythe relative time a 1-direction line-flow with slope α spends in the top square of S (0; 1 / M of the shortcut-ancestor process, assuming of course that the shortlinemethod works for the slope α .More precisely, the relative time a 1-direction line-flow with slope α spends in thetop square of S (0; 1 / 2) is asymptotically the ratio v (1) + v (2) v (1) + . . . + v (8) , (8.9.1)where ( v (1) , . . . , v (8)) T is an eigenvector corresponding to the largest eigenvalue Λof M where, for j = 1 , . . . , v ( j ) denotes the coefficient of the almost verticalunit a j . Here we make the usual assumption that the 1-direction line-flow moveswith constant speed.We can generalize the slope α given by (8.8.1) to any slope of the form α = [ n ; m, n, m, . . . ] = n + 1 m + n + m + ··· = n (cid:32) (cid:114) mn (cid:33) , (8.9.2)where both n, m (cid:62) S ( ∞ ; 1 / 2) and the period-surface S (0; 1 / S (0; 1 / 2) has 2horizontal streets, where the top street consists of 1 square, and the bottom streetconsists of 2 squares. It also has 2 vertical streets. One of these is the white rectanglewith top and bottom edges h . The more complicated one is shaded, with the leftpart consisting of 2 squares with side length 1, and the right part consisting of 2squares with side length 1 / S (0; 1 / 2) is not a polysquare surface,we can still apply Theorem 7.2.2. The pessimistic reader may verify that this resultcan indeed be extended to d-c-polysquare surfaces.Using this, we now attempt to find the eigenvalues of the 2-step transition ma-trix M . For simplicity of notation, we number various parts of S (0; 1 / 2) as inFigure 8.9.1. ↑ ↑ ↑ ↑ 12 3 4 Figure 8.9.1: S (0; 1 / 2) and almost vertical units of type ↑ Here the horizontal streets are 1 and 2 , , 4, while the vertical streets are 1 , , ↑ , so that ↑ , ↑ , ↑ , ↑ are ON-INTEGRABLE SYSTEMS (IV) 117 respectively a , a , a , a in Figure 8.8.6. We have not shown the almost verticalunits of type − ↑ , but − ↑ , − ↑ , − ↑ , − ↑ are respectively a , a , a , a .Let J = { } and J = { , , } denote the horizontal streets, and let I = I = I = { , , } and I = { } denote the vertical streets.We consider slopes of the form (8.9.2).Corresponding to (7.2.14), we define the column matrices u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] and u = [ {↑ s : j ∈ J ∗ , s ∈ I ∗ j } ] . (8.9.3)Here J ∗ , J ∗ and I ∗ j denote that the edges are counted with multiplicity. ↑ n × m ↑ n × m ↑ n × m u ↑ n × m ↑ n × m ↑ n × m ↑ n × m ↑ n × m ↑ n × m ↑ n × m u Figure 8.9.2: almost vertical units of type ↑ in u and u Corresponding to (7.2.15), we also define the column matrices v = [ { − ↑ j : j ∈ J ∗ } ] − [ {↑ j : j ∈ J ∗ } ] , (8.9.4) v = [ { − ↑ j : j ∈ J ∗ } ] − [ {↑ j : j ∈ J ∗ } ] . (8.9.5)Also, analogous to (7.2.17), we have( M − I )[ {↑ s } ] = (cid:26) u + v , if s ∈ J , u + v , if s ∈ J . (8.9.6)We now combine (8.9.3) and (8.9.6). For the horizontal street corresponding to u ,as highlighted in the picture on the left in Figure 8.9.2, we have( M − I ) u = ( M − I ) (cid:104)(cid:110) mn ↑ (cid:111)(cid:105) + ( M − I ) (cid:104)(cid:110) mn ↑ , mn ↑ (cid:111)(cid:105) = mn u + v ) + mn u + v ) . (8.9.7)For the horizontal street corresponding to u , as highlighted in the picture on theright in Figure 8.9.2, we have( M − I ) u = ( M − I ) (cid:104)(cid:110) mn ↑ (cid:111)(cid:105) + ( M − I ) (cid:104)(cid:110) mn ↑ , mn ↑ , mn ↑ (cid:111)(cid:105) = mn u + v ) + 3 mn u + v ) . (8.9.8)It follows from (8.9.7)–(8.9.8) that the street-spreading matrix is given by S = mn (cid:18) (cid:19) , 18 BECK, CHEN, AND YANG with eigenvalues τ = (2 + √ mn τ = (2 − √ mn , (8.9.9)and eigenvector ψ = (cid:32) √ − , (cid:33) T corresponding to τ . By (7.2.38) and (7.2.39), the largest eigenvalue of M | V isΛ = 1 + τ + (cid:112) τ + 4 τ , with eigenvector Ψ = ( y, , xy, x ) T , where y = √ − 12 and x = − τ + (cid:112) τ + 4 τ . (8.9.10)We know that Λ is the largest eigenvalue of M .Suppose that a corresponding eigenvector is given by V = ( v (1) , . . . , v (8)) T , (8.9.11)where v ( i ), i = 1 , . . . , 8, denote respectively the coefficients of a i , i = 1 , . . . , 8, alsorepresented in alternative form respectively by ↑ , − ↑ , ↑ , − ↑ , ↑ , − ↑ , ↑ , − ↑ . Then V = y u + u + xy v + x v . (8.9.12)From (8.9.3)–(8.9.5), we have u = (cid:104)(cid:110) mn ↑ , mn ↑ , mn ↑ (cid:111)(cid:105) , u = (cid:104)(cid:110) mn ↑ , mn ↑ , mn ↑ , mn ↑ (cid:111)(cid:105) , v = [ { m − ↑ } ] − [ { m ↑ } ] , v = (cid:104)(cid:110) m − ↑ , m − ↑ , m − ↑ (cid:111)(cid:105) − (cid:104)(cid:110) m ↑ , m ↑ , m ↑ (cid:111)(cid:105) , so that u = (cid:16) mn , , mn , , , , mn , (cid:17) T , (8.9.13) u = (cid:16) mn , , mn , , mn , , mn , (cid:17) T , (8.9.14) v = ( − m, m, , , , , , T , (8.9.15) v = (cid:16) , , − m , m , − m , m , − m , m (cid:17) T . (8.9.16)Combining (8.9.11)–(8.9.16), we conclude that v (1) = ( y + 1) mn − xym, v (2) = xym, (8.9.17) v (3) = v (7) = ( y + 1) mn − xm , (8.9.18) v (5) = mn − xm , v (4) = v (6) = v (8) = xm . (8.9.19) ON-INTEGRABLE SYSTEMS (IV) 119 It follows from (8.9.17)–(8.9.19) that v (1) + v (2) = ( y + 1) mn v (1) + . . . + v (8) = (3 y + 4) mn , (8.9.20)so that the ratio (8.9.1) is equal to v (1) + v (2) v (1) + . . . + v (8) = y + 13 y + 4 = 2 − √ , (8.9.21)in view of (8.9.10).Recall that the ratio (8.9.21) is asymptotically the relative time a 1-direction line-flow in S (0; 1 / 2) with slope α given by (8.9.2) spends in the top square. Its value of2 − √ ≈ . 268 is independent of the choice of the integer parameters m and n , andcrucially it is less than 1 / 3. Thus any such 1-direction line-flow cannot be uniformin S (0; 1 / under-visited .Consider next the left half-square of the right square of S (0; 1 / S (0; 1 / 2) with slope α given by (8.9.2) spends in this half-square.Note first of all from the edge identification given in the picture on the left inFigure 8.8.3 that every time a 1-direction line-flow enters this half-square, it spendsa constant time t in it before leaving. On the other hand, it is clear from the pictureon the left in Figure 8.8.6 that the only way that such 1-direction line-flow can enterthis half-square is along an almost vertical unit of type a . Thus the total time theline-flow spends in this half-square is equal to t v (4).To evaluate t , we note that from the edge identification given in the picture onthe left in Figure 8.8.3 that every time a 1-direction line-flow enters the top square,it spends a constant time t in it before leaving. Furthermore, t is equal to thelength of an almost vertical unit. Simple geometric consideration now shows that t /t = α/ t ( v (1) + . . . + v (8)),which is the product of the total number of almost vertical units and the length ofsuch a unit. Thus the relative time a 1-direction line-flow in S (0; 1 / 2) with slope α given by (8.9.2) spends in the half-square under consideration is equal to t v (4) t ( v (1) + . . . + v (8)) = αx (3 y + 4) n = (3 √ − 5) 1 + (cid:113) mn (cid:113) √ mn , (8.9.22)in view of (8.9.2), (8.9.9), (8.9.10), (8.9.19) and (8.9.20).Taking the limit m, n → ∞ in (8.9.22), the relative time a 1-direction line-flow in S (0; 1 / 2) with slope α given by (8.9.2) spends in the half-square under considerationconverges to 3 √ − ≈ . . 195 for every choice m, n (cid:62) / 6. Thus this half-square is over-visited .We summarize our observations in the form of a theorem. Theorem 8.9.1. Let m, n (cid:62) be arbitrary integers such that both are divisibleby , and let α = α ( m, n ) be the slope given by (8.9.2) . Let L α ( t ) , t (cid:62) , be anyhalf-infinite line-flow with slope α on the d-c-polysquare surface S (0; 1 / . (i) The relative visiting time of L α to the top square of S (0; 1 / is asymptoticallyequal to − √ ≈ . , which is independent of the choice of the integer parameters m and n . In particular, the top square is under-visited. Furthermore, in an arbitrarytime interval (cid:54) t (cid:54) T with T (cid:62) , the actual visiting time of the line-flow to the 20 BECK, CHEN, AND YANG top square is equal to (2 − √ T + O ( T κ ) , where κ = log | Λ | log | Λ | . Here Λ and Λ are respectively the eigenvalues of the -step transition matrix ofthe line-flow with the largest and second largest absolute value. (ii) The relative visiting time of L α to the left half of the right square of S (0; 1 / isasymptotically equal to (8.9.22) which depends on the choice of the integer parameters m and n . In particular, this part of S (0; 1 / is over-visited. 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Boshernitzan’s criterion for unique ergodicity of an interval exchange transfor-mation. Ergodic Theory Dynam. Systems (1987), 149–153.[17] W.A. Veech. Teichm¨uller curves in moduli space, Eisenstein series and an application totriangle billiards. Invent. Math. (1989), 553–583.[18] W.A. Veech. The billiard in a regular polygon. Geom. Funct. Anal. (1992), 341–379. Department of Mathematics, Rutgers University, Hill Center for the Mathe-matical Sciences, Piscataway NJ 08854, USA Email address : [email protected] Department of Mathematics and Statistics, Macquarie University, Sydney NSW2109, Australia Email address : [email protected] Department of Mathematics, Rutgers University, Hill Center for the Mathe-matical Sciences, Piscataway NJ 08854, USA Email address ::