Divisibility of Spheres with Measurable Pieces
aa r X i v : . [ m a t h . M G ] D ec Divisibility of Spheres with Measurable Pieces
Clinton T. ConleyDepartment of Mathematical SciencesCarnegie Mellon UniversityPittsburgh, PA 15213, USA Jan Greb´ıkMathematics InstituteUniversity of WarwickCoventry CV4 7AL, UKOleg PikhurkoMathematics Institute and DIMAPUniversity of WarwickCoventry CV4 7AL, UKDecember 15, 2020
Abstract
For an r -tuple ( γ , ... , γ r ) of special orthogonal d × d matrices, we say theEuclidean ( d − S d − is ( γ , ... , γ r ) -divisible if there is a subset A ⊆ S d − such that its translations by the rotations γ , ... , γ r partition the sphere.Motivated by some old open questions of Mycielski and Wagon, we investigate theversion of this notion where the set A has to be measurable with respect to thespherical measure. Our main result shows that measurable divisibility is impossiblefor a “generic” (in various meanings) r -tuple of rotations. This is in stark contrastto the recent result of Conley, Marks and Unger which implies that, for every“generic” r -tuple, divisibility is possible with parts that have the property of Baire. Let SO( d ) denote the group of special orthogonal d × d matrices, that is, real d × d matrices M such that the determinant of M is 1 and M T M = I d , where I d denotes the identity d × d matrix. The elements of this group are naturally identified with orientation-preservingisometries of the Euclidean unit sphere S d − := { x ∈ R d : k x k = 1 } , and we will often refer to them as rotations .1or an r -tuple γ = ( γ , ... , γ r ) ∈ SO( d ) r , we say that S d − is γ -divisible (or admitsa γ -division ) if there is A ⊆ S d − such that its translates γ .A, ... , γ r .A partition S d − (that is, for every x ∈ S d − there are unique y ∈ A and i ∈ [ r ] such that x = γ i . y , wherewe denote [ r ] := { , ... , r } ). Of course, a set A works for γ if and only if γ r .A works for β := ( γ γ − r , ... , γ r − γ − r , I d ). However, we do not assume that any particular rotationis the identity, mostly for the notational convenience so that all indices can be treateduniformly.We say that S d − is r -divisible if there is an r -tuple γ ∈ SO( d ) r such that S d − is γ -divisible (or, in other words, if we can partition S d − into r congruent pieces). Theinteger pairs d, r > S d − is r -divisible have been completely classified (see e.g.Theorem 6.6 in the book by Tomkowicz and Wagon [21]). Namely, the only pairs whenthe answer is in the negative are when r = 2 and d is odd. In this case, the impossibilityof any ( γ , γ )-division follows from considering a fixed point x ∈ S d − of γ − γ whichexists as the dimension d − A can work here: thetranslates γ .A and γ .A intersect if x ∈ A and do not cover γ . x if x A .) On theother hand, the case of d = 2 is trivial (e.g. one can take the r rotations of the circle S by multiples of the angle 2 π/r ) while the first published solution for S seems to beby Robinson [20, Page 254]. Furthermore, the r -divisibility for S d − easily implies the r -divisibility of S d +1 , see e.g. the proof of Theorem 6.6 in [21] or Lemma 16 here.Mycielski [16] showed that there is a subset A ⊆ S such that for every r > γ , ... , γ r with γ .A, ... , γ r .A partitioning the sphere. This should be compared withthe classical paradox of Hausdorff [12] who produced such a set A that works, apart acountable subset of S of errors, for every r >
2. (Note that we cannot take r = 2 inMycielski’s result because S is not 2-divisible.)Let µ be the spherical measure on S d − , which can be defined as the ( d − arc-length distance on the sphere (wherethe distance between x , y ∈ S d − is the angle between the vectors x and y ). We call asubset of S d − measurable if it belongs to the µ -completion of the Borel σ -algebra. Ofcourse, the paradoxical set A in the results of Hausdorff [12] and Mycielski [16] cannot bemeasurable with respect to the (rotation-invariant) measure µ on S . Mycielski [17, 18]asked if one can show that S is r -divisible without using the Axiom of Choice. Wagon [22,Question 4.15] (or Question 5.15 in [21]) asked if the 3-divisibility of S can be shown withmeasurable sets (thus the Axiom of Choice can be applied on a µ -null set). Measurabledivisibility for higher dimensional spheres is easier because of a constructive way of liftingup a division from S d − to S d +1 . It is known that S d − is r -divisible with measurablepieces for r > d > r > d > r = 2 and d is odd, the sphere S d − is r -divisible so that each piece has the propertyof Baire (that is, under one of equivalent definitions, can be represented as the symmetricdifference of a Borel set and a meager set), see Proposition 2 here.Here we propose to study the more general question of describing the set of those r -tuples γ ∈ SO( d ) r such that S d − is γ -divisible with measurable pieces.First, we consider the case when the rotations are “generic”. More precisely, let us callan r -tuple of matrices γ = ( γ , ... , γ r ) ∈ SO( d ) r generic if, for every polynomial p ∈ Q [ β ]where β ∈ R d r encodes an r -tuple of d × d matrices (that is, p is a polynomial in the d r individual matrix entries with rational coefficients), p ( γ ) = 0 implies that p ( β ) = 0 forevery β ∈ SO( d ) r . In other words, this property states that if a polynomial with rational(equivalently, integer) coefficients vanishes on γ then it necessarily vanishes everywhereon SO( d ) r .Our main result shows that no generic γ works in the measurable setting, even in arather relaxed fractional version. Theorem 1.
Let d > and r > be integers. Let ( γ , ... , γ r ) ∈ SO( d ) r be generic.Then every f ∈ L ( S d − , µ ) with P ri =1 γ i .f = 1 µ -almost everywhere is the constantfunction /r µ -almost everywhere, where γ i .f denotes the function that maps x ∈ S d − to f ( γ − i . x ) . In sharp contrast, we can derive with some extra work from the results in [4] that every generic γ works with pieces that have the property of Baire. Proposition 2.
Let r > and d > be arbitrary integers, except if d is odd then werequire that r > . Let ( γ , ... , γ r ) ∈ SO( d ) r be generic. Then there is a subset A of S d − with the property of Baire such that γ .A, ... , γ r .A partition S d − . Theorem 1 and Proposition 2 add to a growing body of results in measurable combina-torics (see e.g. the recent survey by Kechris and Marks [14]), where the requirements thatthe pieces are measurable and have the property of Baire respectively lead to differentanswers.The following lemma shows that, in various meanings, “most” elements of SO( d ) r aregeneric. Lemma 3.
Let r > , d > and N be the set of r -tuples in SO( d ) r that are not generic.Then the following statements hold.(i) The set N has measure 0 with respect to the Haar measure on the group SO( d ) r .(ii) The set N in a meager subset of SO( d ) r with respect to the topology induced by theEuclidean topology on R d r ⊇ SO( d ) r . r -tuples of rotations. In particular, the following lemma allows us to write an“explicit” generic point: just let the entries above diagonals be sufficiently small realsthat are algebraically independent over Q and extend this to an element of SO( d ) r byClaim 27 here. Lemma 4.
Let r > , d > , and γ ∈ SO( d ) r . Then γ is generic if and only if the (cid:0) d (cid:1) r -tuple of the entries in γ strictly above the diagonals is algebraically independentover Q . In the extreme opposite case, we show that, for odd d > γ -divisibility cannot beattained when γ generates a finite subgroup of SO( d ). Proposition 5.
Let d > be odd and let r > . Suppose that γ , ... , γ r ∈ SO( d ) , r > ,generate a finite subgroup Γ ⊆ SO( d ) . Then S d − is not ( γ , ... , γ r ) -divisible. Some standard general results of Borel combinatorics (e.g. Lemma 5.12 and Theo-rem 5.23 from [19]) imply that if S d − is γ -divisible and every orbit of the subgroup ofSO( d ) generated by γ , ... , γ r is finite, then there is a Borel γ -division. The followingresult gives that just one finite orbit is enough to convert a γ -division into a measurableone. Proposition 6.
Let γ = ( γ , ... , γ r ) ∈ SO( d ) r and let Γ be the subgroup of SO( d ) generated by γ , ... , γ r . Suppose that there is z ∈ S d − such that its Γ -orbit Γ. z is finite.Then S d − is γ -divisible if and only if S d − is γ -divisible with measurable pieces. Of course, this leaves a very wide range of unresolved cases. As an initial partial step,we completely characterise those r -tuples of rotations for which the circle S is divisiblewith measurable pieces for r r = 4 (via Claim 19).This paper is organised as follows. In Section 2 we give a quick overview of basicdefinitions and facts about spherical harmonics and use these to prove Theorem 1, whichis the main result of this paper. Proposition 5 is proved in Section 3 using Euler’scharacteristic. Propositions 6 and 2 are proved in Sections 4 and 7 respectively. InSection 5 we describe the standard construction of how an r -division of S d − can be liftedto S d +1 and observe that this gives measurable pieces (Lemma 16). In Section 6 we studyvarious versions of measurable divisibility when d = 2; in particular, we characterise r -tuples γ ∈ SO(2) r for which the circle S is γ -divisible with measurable pieces for r d ) r and use them to prove Lemma 3. In particular, we show that the variety SO( d ) r ⊆ R d r is irreducible and the entries above the diagonals form a transcendence basis for itsfunction field. While these results are fairly standard, we present their proofs since wecould not find any published statements that suffice for our purposes. In Section 8.3 weprove an auxiliary lemma from algebraic geometry and use it to derive Lemma 4.4 Spherical harmonics
For an introduction to spherical harmonics we refer to the book by Groemer [9] whosenotation we generally follow. Recall that µ denotes the spherical measure on S d − . Thusthe total measure of the sphere is σ d := µ ( S d − ) = 2 π d/ Γ ( d/ . As d is fixed, the dependence on d is usually not mentioned except for σ d (since σ d − willalso appear in some formulas). Also, the shorthand a.e. stands for µ -almost everywhere.By [9, Lemma 1.3.1], the density of the push-forward of µ under the projection to anycoordinate axis is ρ ( t ) := (cid:26) σ d − (1 − t ) ( d − / , − < t < , , otherwise. (1)A polynomial p ∈ R [ x ], x = ( x , ... , x d ), is called harmonic if ∆ p = 0, where∆ := ∂ ∂x + ... + ∂ ∂x d is the Laplace operator . A spherical harmonic is a function from S d − to the reals which isthe restriction to S d − of a harmonic polynomial on R d . Let H be the vector space of allspherical harmonics. For an integer n >
0, let H n ⊆ H be the linear subspace consistingof all functions f that are the restrictions to S d − of some harmonic polynomial p whichis homogeneous of degree n , where we regard the zero polynomial as homogeneous of anydegree. By [9, Lemma 3.1.3], the polynomial p is uniquely determined by f ∈ H n , so wemay switch between these two representations without mention. It can be derived fromthis ([9, Theorem 3.1.4]) that the dimension of H n is N n := (cid:18) d + n − n (cid:19) − (cid:18) d + n − n − (cid:19) . (2)Let h· , ·i denote the scalar product on L ( S d − , µ ) (while x · y := P di =1 x i y i denotes thescalar product of x , y ∈ R d ). It is known ([9, Theorem 3.2.1]) that h f, g i = 0 , for all f ∈ H i and g ∈ H j with i = j, (3)that is, H , H , ... are pairwise orthogonal subspaces of H ⊆ L ( S d − , µ ). Note that thegroup SO( d ) acts naturally on L ( S d − , µ ) via the shift action( γ.f )( v ) := f ( γ − . v ) , for γ ∈ SO( d ) , f ∈ L ( S d − , µ ) , v ∈ S d − . (4)Since the Laplace operator is invariant under isometries of R d , it follows that each H n isinvariant under this action ([9, Proposition 3.2.4]).5n important role is played by the Legendre (or
Gegenbauer ) polynomials ( P , P , ... )which are obtained from (1 , t, t , ... ) by the Gram-Schmidt orthonormalization processon L ([ − , , ρ ( t ) d t ), except they are normalised to assume value 1 at t = 1 (insteadof being unit vectors in the L -norm). Of course, the degree of P n is exactly n . Let uscollect some of their standard properties that we will use. Lemma 7.
For every integer n > the following holds.(i) The polynomial P n has rational coefficients.(ii) For every v ∈ S d − , the function P v n : S n − → R , defined by P v n ( x ) := P n ( v · x ) , for x ∈ S d − , (5) belongs to H n .(iii) There is a choice of v , ... , v N n ∈ S d − such that the functions P v i n , i ∈ [ N n ] , forma basis of the vector space H n .(iv) For every u , v ∈ S d − , we have h P u n , P v n i = σ d N n P n ( u · v ) .Proof. Part (i) follows from the formula of Rodrigues ([9, Proposition 3.3.7]) that providesan explicit expression for P n , or from the standard recurrence relation that writes P n +1 interms of P n and P n − for n > P − ( t ) := 0 and P ( t ) = 1.Part (ii), namely the claim that each P v n is in H n , is one of the statements of [9,Theorem 3.3.3].Part (iii) is the content of [9, Theorem 3.3.14]. Alternatively, notice that under theaction in (4) we have for every u , v ∈ S d − and γ ∈ SO( d ) that ( γ.P v n )( u ) = P n ( v · ( γ − . u )) = P n (( γ. v ) · u ), that is, γ.P v n = P γ. v n . (6)Thus the linear span of P v n , v ∈ S d − , is a non-zero SO( d )-invariant subspace of H n . By[9, Theorem 3.3.4], the only such subspace is H n itself, giving the required.Part (iv) follows from h P u n , P v n i = (cid:18)Z − ( P n ( t )) ρ ( t ) d t (cid:19) P n ( u · v ) = σ d N n P n ( u · v ) , where the first equality is a special case of the Funk-Hecke Formula ([9, Theorem 3.4.1])and the second equality (which by (1) amounts to computing the L -norm of any P u n ∈ L ( S d − , µ )) is proved in [9, Proposition 3.3.6].We need the following strengthening of Lemma 7.(iii), where we additionally requirethat the vectors v i are rational. Lemma 8.
There is a choice of v , ... , v N n ∈ S d − ∩ Q d such that the functions P v i n , i ∈ [ N n ] , form a basis of the vector space H n . roof. We pick v i in S d − ∩ Q d one by one as long as possible so that the correspondingfunctions P v i n are linearly independent as elements of H n . Let this procedure produce v , ... , v ℓ . Suppose that ℓ < N n as otherwise we are done. Let v ℓ +1 = x , with x =( x , ... , x d ) ∈ R d being viewed as a vector of unknown variables. Consider the ( ℓ + 1) × ( ℓ + 1) matrix M = M ( x ) with entries M ij := 1 σ d h P v i n , P v j n i , for i, j ∈ [ ℓ + 1] . (7)In other words, σ d M is the Gram matrix of the vectors P v n , ... , P v ℓ +1 n ∈ L ( S d − , µ ). Inparticular, the determinant det( M ) of M is 0 if and only if P v ℓ +1 n is in the span of the(linearly independent) vectors P v n , ... , P v ℓ n (by e.g. [13, Theorem 7.2.10]).By Lemma 7.(iv) we have that M ij = N n P n ( v i · v j ). Thus the determinant of M is apolynomial function of x .By Lemma 7.(iii) and ℓ < N d (and the linear independence of P v n , ... , P v ℓ n ), there issome choice of v ℓ +1 ∈ S d − with det( M ) = 0. That is, the polynomial det( M ) is notidentically zero on S d − .We need the following easy claim that can be proved, for example, by induction on d > d = 2 following from S containing all points of the form m + n ( m − n , mn ) for ( m, n ) ∈ Z \ { (0 , } . Claim 9.
For every d > , the set S d − ∩ Q d of the points on the sphere with allcoordinates rational is dense in S d − with respect to the standard topology on the sphere(i.e. the one inherited from the Euclidean space R d ⊇ S d − ). Since det( M ), as a polynomial function S d − → R , is continuous and not identicallyzero, it has to be non-zero on some point x of the dense subset S d − ∩ Q d . Thus, if we let v ℓ +1 to be such a vector x , then the functions P v n , ... , P v ℓ +1 n ∈ L ( S d − , µ ) are linearlyindependent. This contradiction proves the lemma.For an r -tuple γ = ( γ , ... , γ r ) ∈ SO( d ) r and a unit vector v ∈ S n − define G v n, γ := r X i =1 P γ − i . v n . (8)By Lemma 7.(ii), each function G v n, γ : S d − → R , as a linear combination of somespherical harmonics P γ − i . v n ∈ H n , is itself in H n . Lemma 10. If γ ∈ SO( d ) r is generic, then the linear span of { G v n, γ : v ∈ S d − } is thewhole space H n .Proof. By Lemma 8, fix some vectors v , ... , v N n ∈ S d − ∩ Q d such that P v n , ... , P v Nn n forma basis for H n . Let β = ( β , ... , β r ) be an arbitrary element of SO( d ) r (not necessarily7eneric). Consider the N n × N n matrix L = L ( β ) with entries L ij := 1 σ d h G v i n, β , P v j n i , for i, j ∈ [ N n ] . Recall that the vectors P v i n , i ∈ [ N n ], form a (not necessarily orthonormal) basis ofthe linear space H n . Write the vectors G v i n, β in this basis:( G v n, β , ... , G v Nn n, β ) T = A ( P v n , ... , P v Nn n ) T , for some N n × N n matrix A . Then L is the matrix product AM , where M is the Grammatrix of the vectors P v i n multiplied by the constant σ − d (that is, the entries of M aredefined by the formula in (7)). The matrix M is non-singular by the linear independenceof P v i n , i ∈ [ N n ]. Thus det( L ) = 0 if and only if G v n, β , ... , G v Nn n, β are linearly independentas vectors in H n .By Lemma 7.(iv), we have for every i, j ∈ [ N d ] that L ij := 1 σ d r X s =1 h P β − s . v i n , P v j n i = 1 N n r X s =1 P n (( β − s . v i ) · v j ) = 1 N n r X s =1 P n ( v i · ( β s . v j )) . Since v , ... , v N n are fixed, this writes each L ij as a polynomial in the d r entries ofthe matrices β , ... , β r . Moreover, all coefficients of this polynomial are rational sinceeach v i belongs to Q d and all coefficients of P n are rational by Lemma 7.(i). Thus thedeterminant of L is equal to p ( β ) for some polynomial p with coefficients in Q .Note that if we let each β i to be the identity matrix I d , then G v n, β becomes rP v n forevery v ∈ S d − and we have L ij = rσ d h P v i n , P v j n i for i, j ∈ [ N n ] and det( L ) = 0 (since P v n , ... , P v Nn n are linearly independent). Thus p ( I d , ... , I d ) = 0. Since γ ∈ SO( d ) r isgeneric, we have that p ( γ ) = 0, that is, the matrix L for β := γ is non-singular. Thismeans that the functions G v i n, γ , i ∈ [ N n ], are linearly independent. Since they all lie in H n and their number equals the dimension of this linear space, they span H n . The lemma isproved. Proof of Theorem 1.
Recall that γ = ( γ , ... , γ r ) ∈ SO( d ) r , r >
2, is generic and wehave to show that S d − is not “fractionally” γ -divisible.So take any f ∈ L ( S d − , µ ) such that P ri =1 γ i .f = 1 a.e. Since spherical harmonics aredense in L ( S d − , µ ) ([9, Corollary 3.2.7]) and we have the direct sum H = ⊕ ∞ n =0 H n whosecomponents are orthogonal to each other by (3), we can uniquely write f = P ∞ n =0 F n in L ( S d − , µ ) with F n ∈ H n for every n >
0. Since the action of SO( d ) preserves eachspace H n as well as the scalar product on L ( S d − , µ ), we have that γ.f = P ∞ n =0 γ.F n isthe harmonic expansion of γ.f ∈ L ( S d − , µ ).Take any integer n >
1. Recall that the sum P ri =1 γ i .f is a constant function 1 a.e.By (3), the invariance of the scalar product under SO( d ) and by (6), we have that, for8very v ∈ S d − ,0 = h P v n , i = h P v n , γ .f + ... + γ r .f r i = h P v n , γ .F n + ... + γ r .F n i = h γ − .P v n + ... + γ − r .P v n , F n i = h G v n, γ , F n i , where G v n, γ was defined by (8). Since the functions G v n, γ , v ∈ S d − , span the whole space H n by Lemma 10, we must have that F n = 0.As n > f is a constant function a.e. (whose value mustbe 1 /r ). This finishes the proof of Theorem 1. Remark . The statement of Theorem 1 remains true also when γ r = I d and ( γ , ... , γ r − )is a generic point of SO( d ) r − . One way to see this is to run the same proof except the r -th component of each encountered r -tuple of matrices is always set to be the identitymatrix I d . Proof of Proposition 5.
We have to show that an even-dimensional sphere S d − is not( γ , ... , γ r )-divisible if the subgroup Γ of SO( d ) generated by the rotations γ i is finite.Since d is odd, the 2-divisibility of S d − is impossible because of a fixed point of γ − γ .So assume that r >
3. Let V := Γ. {± e , ... , ± e d } , that is, we take all possible images of the standard basis vectors and their negationswhen moved by Γ . Clearly, the set V is a finite. Let P be the convex hull of V . Then P is a full-dimensional polytope containing in its interior (as already the convex hullof {± e , ... , ± e d } ⊆ V has these properties). Its boundary ∂P is homeomorphic to S d − by the map that sends x ∈ ∂P to x / k x k ∈ S d − .Let a hyperplane mean a ( d − R d . Identify eachoriented hyperplane H ⊆ R d with the pair ( n , a ) ∈ S d − × R so that H = { x ∈ R d : n · x = a } . Its open half-spaces are H + := { x ∈ R d : n · x > a } and H − := { x ∈ R d : n · x < a } . Call H supporting if H ∩ P = ∅ and H − ∩ P = ∅ . Call H a facet hyperplane if it is supporting and dim aff ( H ∩ P ) = d −
1, where dim aff ( X ) denotes the dimension ofthe affine subspace of R d spanned by X .The intersections of supporting hyperplanes with ∂P represent the boundary of thepolytope P as a CW-complex. Namely, for i ∈ { , ... , d − } , its i -dimensional cells areprecisely the i -dimensional faces of P , that is, the convex hulls of the sets in C i := { X ⊆ V : dim aff ( X ) = i & ∃ supporting hyperplane H with H ∩ V = X } . For a finite non-empty set X ⊆ R d , let m X := | X | P x ∈ X x be the centre of mass of X .9et us show that for every i ∈ { , ... , d − } and distinct X, Y ∈ C i we have m X = m Y .As it is well-known, see e.g. [10, Theorem 3.1.7], we can pick facet hyperplanes H , ... , H k such that V ∩ ( ∩ kj =1 H j ) = X . Since X = Y , the affine subspaces that these two sets spandiffer. Since these subspaces have the same dimension, there is y ∈ Y not in the affinespan of X . Since y ∈ V and each H j is supporting, there is j ∈ [ k ] such that y belongsto the open half-space H + j . From Y ⊆ H j ∪ H + j , it follows that m Y belongs to H + j andcannot be equal to m X ∈ H j , as claimed.Also, it holds that m X = for any X ∈ C i . Indeed, with H , ... , H k as above we havethat , which is in the interior of P , belongs to, say, the open half-space H +1 so cannotbe equal to m X ∈ H .Thus | M i | = |C i | , where M i := { m X / k m X k : X ∈ C i } ⊆ S d − denotes the set ofthe normalised centres of mass of the vertex sets of i -dimensional faces. Clearly, the setfamily C i is invariant under the natural action of Γ on finite subsets of S d − . Thus theset M i ⊆ S d − is also Γ -invariant.Since d is odd, the Euler characteristic χ ( S d − ) of the ( d − ∂P give a representation of the sphere asa CW-complex, we have (by e.g. [23, Theorem 4.2.20]) that2 = χ ( S d − ) = d − X i =0 ( − i |C i | , Thus, for at least one i ∈ { , ... , d − } , it holds that r > |C i | = | M i | . Bythe Γ -invariance of M i , there is no choice of A ∩ M i such that its translates by γ , ... , γ r partition M i . Thus S d − is not ( γ , ... , γ r )-divisible. Remark . The proof of Proposition 5 gives that if there are d linearly independentvectors on S d − such that each has a finite orbit under Γ (where some of these orbitsmay coincide) then S d − is not γ -divisible. However, this seemingly weaker assumptionis equivalent to the assumption that Γ is finite (e.g. via a version of Claim 14 below). Here we prove Proposition 6 that, in the presence of just one finite orbit, γ -divisibility isequivalent to measurable γ -divisibility. Proof of Proposition 6.
Recall that Γ is the subgroup of SO( d ) generated by γ , ... , γ r .For x ∈ S d − , let L x be the linear subspace of R d spanned by Γ. x ⊆ R d . Claim 13.
For every x ∈ S d , both L x ⊆ R d and its orthogonal complement L ⊥ x ⊆ R d are invariant under the action of Γ on R d . roof of Claim. Any γ ∈ Γ permutes the set Γ. x . Since γ is a linear map, it preservesthe linear subspace L x spanned by Γ. x . Thus L x is Γ -invariant.Since Γ consists of orthogonal matrices, its action preserves the scalar product on R d .Thus if y ∈ R d is orthogonal to L x then, for every γ ∈ Γ , we have that γ. y is orthogonalto γ.L x = L x . It follows that L ⊥ x is Γ -invariant.Recall that z ∈ S d − is a vector such that its orbit Γ. z is finite. Let Γ. z = { z , ... , z n } . Claim 14. If x ∈ L z ∩ S d − then | Γ. x | n ! .Proof of Claim. Write x ∈ L z as P ni =1 c i z i for some reals c , ... , c n . For every α ∈ Γ , wehave by linearity that α. x = P ni =1 c i ( α. z i ). Since z , ... , z n enumerate a whole orbit of Γ , the element α ∈ Γ permutes these vectors. Thus every element of Γ. x is of the form P ni =1 c i z σ ( i ) for some permutation σ of [ n ]. Thus Γ. x indeed has at most n ! elements.Now we are ready to prove the (non-trivial) forward direction of Proposition 6. Byrotating the sphere (and moving z and conjugating γ i ’s accordingly), we can assume that L z = R m × and L ⊥ z = × R d − m for some m ∈ [ d ]. By Claim 13, every matrix γ i , i ∈ [ r ], consists now of two diagonal blocks that correspond to some α i ∈ O( m ) and β i ∈ O( d − m ). (Note that these matrices may have determinant − R d as ( x , y ), we mean that x ∈ R m and y ∈ R d − m ; thus γ i . ( x , y ) = ( α i . x , β i . y ).Fix C ⊆ S d − such that γ .C, ... , γ r .C partition S d − . By the invariance of L z and L ⊥ z , the translates of the set C ∩ ( R m × ) (resp. C ∩ ( × R m − d )) by γ , ... , γ r partition S m − × (resp. × S d − m − ). By Claim 14, every orbit of the action of Γ on the invariantsubset X := S m − × has at most n ! elements. Obviously, the same holds for theaction on S m − of the subgroup Γ ′ ⊆ O( m ) generated by α , ... , α r . Fix a Borel totalorder on S m − (e.g. the restriction of the lexicographic order on R m ) and let A ′ ⊆ X be obtained by picking from every orbit Γ ′ . x ⊆ S m − the lexicographically smallestsubset such that its translates by α , ... , α r partition Γ ′ . x . Such a set always exists since { y ∈ Γ ′ . x : ( y , ) ∈ C } is one possible choice. In the terminology of [19], the set A ′ canbe computed by a local rule of radius n ! on the coloured Schreier digraph of Γ ′ y S m − (where we put a directed colour- i arc from y to α i . y for all y ∈ S m − and i ∈ [ r ]). As theaction is Borel, this is known to imply (see e.g. [19, Lemma 5.17]) that the constructedset A ′ ⊆ S m − is Borel. Define A := ∪ ρ ∈ [0 , ( p − ρ A ′ × ρ S d − m − )= ∪ ρ ∈ [0 , { ( p − ρ x , ρ y ) : x ∈ A ′ , y ∈ S d − m − } and B := C ∩ ( × R m − d ). Then γ .A, ... , γ r .A partition S d − \ ( × S d − m − ) and, aswe observed earlier, γ .B, ... , γ r .B partition × S d − m − . Thus A ∪ B witnesses the γ -divisibility of S d − . Note that the set B , which lies inside the intersection of S d − withthe linear subspace L ⊥ z of dimension less than d , has measure zero. On the other hand,11he set A can be equivalently defined as the pre-image of the Borel set A ′ × R d − m underthe natural homeomorphism between S d − \ ( × S d − m − ) and S m − × R d − m that maps( x , y ) to ( x / k x k , y / k x k ). Thus A is Borel and A ∪ B is measurable, proving theproposition. Remark . One can show via Claims 13 and 14 that if d = 3 and a subgroup Γ ⊆ SO( d )has a finite orbit of size at least 3, then Γ is finite (and thus Proposition 5 applies).However, this implication is not true in general for d > As we mentioned in the Introduction, S d − is r -divisible with Borel pieces for every r > d > r > d > Lemma 16.
For any d > and r > , S d − is r -divisible with measurable pieces.Proof. Informally speaking, we will use the Borel r -divisibility of S in the last twocoordinates of S d − ⊆ R d , resorting to the r -divisibility of S d − only on the null set ofpoints where the last two coordinates are zero.Namely, choose rotations α , ... , α r ∈ SO( d −
2) and a (not necessarily measurable)subset A ⊆ S d − such that α .A, ... , α r .A partition S d − , which is possible by e.g. [21,Theorem 6.6]. Let β ∈ SO(2) be the rotation of the circle S by the angle 2 π/r . (Thusthe order of β , as an element of the group SO(2), is r .) For i ∈ [ r ], let γ i send ( x , y ) ∈ R d − × R to ( α i . x , β i . y ), where we view SO( m ) as also acting on R m . Clearly, γ i preservesboth the scalar product on R d and the orientation; thus it is an element of SO( d ).Let B := { (cos θ, sin θ ) : 0 θ < π/r } ⊆ S . Then the half-open arcs β.B, ... , β r .B partition S . Let C := A ′ ∪ B ′ , where A ′ := A × { (0 , } and B ′ := ∪ ρ ∈ [0 , ( ρ S d − × p − ρ B ) . Clearly, A ′ is a µ -null subset of S d − and B ′ is a Borel subset of S d − . Thus C ismeasurable. Also, γ .C, ... , γ r .C partition S d − . Indeed, γ i .A ′ = α i .A × { (0 , } , i ∈ [ r ],partition S d − ×{ (0 , } while γ i .B ′ = ∪ ρ ∈ [0 , ( ρ S d − × p − ρ ( β i .B )), i ∈ [ r ], partitionthe rest of S d − . d = 2 and r We parametrise S = { (cos t, sin t ) : t ∈ [0 , π ) } and use the parameter t instead ofthe Cartesian coordinates. Thus we have the interval [0 , π ) with µ being the Lebesgue12easure on it. The space H n for n > nt and sin nt (while, ofcourse, H consists of all constant functions). Here, the harmonic expansion is nothingelse as the Fourier series. We identify SO(2) with the additive group T := R / π Z ofreals taken modulo 2 π . Thus the action of γ ∈ T on [0 , π ) is to send t ∈ [0 , π ) to t + γ (mod 2 π ). We also identity [0 , π ) with T ; thus we have the natural action T y T .Let us investigate various possible versions of “measurable” divisibility, stated in termsof the action T y T . Let B r (resp. M r ) consist of those r -tuples ( t , ... , t r ) ∈ T r forwhich there is a Borel (resp. measurable) subset A ⊆ T such that t + A, ... , t r + A partition T , where we denote t + A := { t + a : a ∈ A } . Also, let M ′ r consist of those( t , ... , t r ) ∈ T r for which there is a measurable (equivalently, Borel) A ⊆ T such thatthe translates t + A, ... , t r + A are pairwise disjoint and the set of elements of T notcovered by them has measure zero. Finally, let F r consist of those ( t , ... , t r ) ∈ T r forwhich there is f ∈ L ([0 , π ) , µ ) such that t .f + ... + t r .f = 1 a.e. while f = 1 /r on a setof positive measure. As it is easy to see, the definition of F r does not change if we require t .f + ... + t r .f = 1 to hold everywhere. Trivially, it holds that B r ⊆ M r ⊆ M ′ r ⊆ F r .First, we investigate F r . Suppose that we have f ∈ L ([0 , π ) , µ ) with t .f + ... + t r .f =1 a.e. Take the Fourier series, f ( t ) = c + ∞ X n =1 ( c n cos nt + s n sin nt ) , for a.e. t ∈ [0 , π ) . Clearly, c = 1 /r . For i ∈ [ r ], by translating everything by t i we get that t i .f ( t ) = 1 r + ∞ X n =1 ( c n cos n ( t − t i ) + s n sin n ( t − t i )) , for a.e. t ∈ [0 , π ) . Summing this up for all i ∈ [ r ] and using the formula for the sine and the cosine of adifference of two angles, we get that for a.e. t ∈ [0 , π )1 = 1 + r X i =1 ∞ X n =1 c n (cos nt cos nt i + sin nt sin nt i ) + s n (sin nt cos nt i − cos nt sin nt i ) ! = 1 + ∞ X n =1 r X i =1 ( c n cos nt i − s n sin nt i ) cos nt + r X i =1 ( c n sin nt i + s n cos nt i ) sin nt ! . (Recall that P ri =1 t i .f = 1 a.e.)Let n >
1. By the uniqueness of the Fourier coefficients, we have that r X i =1 ( c n cos nt i − s n sin nt i ) = 0 r X i =1 ( c n sin nt i + s n cos nt i ) = 0 . c n , s n ) = (0 , c n and s n (resp. by − s n and c n ) and add up, we get after dividing by c n + s n that r X i =1 cos nt i = 0 and r X i =1 sin nt i = 0 , (9)that is, the vectors (cos nt i , sin nt i ) ∈ R , i ∈ [ r ], sum up to zero.If f differs from 1 /r on a set of positive measure then, for at least one integer n > c n , s n ) = (0 ,
0) and thus (9) holds. Conversely, if (9) holds for some n >
1, thenwe can take, for example, f ( t ) := (1 + cos nt ) /r for t ∈ [0 , π ). This completely describesthe set of r -tuples in SO(2) for which the circle S is “fractionally” divisible: Proposition 17. An r -tuple ( t , ... , t r ) ∈ T r belongs to F r if and only if (9) holds forat least one integer n > . Let us investigate the sets B r and M ′ r for r
4. As we will see, it holds for each r B r = M ′ r (and, in particular, this set is also equal to M r ).Let ( t , ... , t r ) ∈ T r . By replacing ( t , ... , t r ) by ( t − t r , ... , t r − t r ), which does notaffect divisibility, we can assume for convenience that t r = 0. Since M ′ r ⊆ F r , assumethat (9) holds for some n >
1. Let n > r = 2. By (9) we have nt = (2 k + 1) π for some integer k > n , it is coprime to 2 k + 1. Thus the subgroup of T generated by t = (2 k + 1) π/n is { πm/n : m ∈ { , ... , n − }} , which is the additive cyclic groupof order 2 n with t corresponding to an odd multiple of the generator π/n . Since theaddition of t swaps odd and even multiples of π/n , we have that A := (cid:8) πmn : m ∈ { , , ... , n − } (cid:9) + (cid:2) , πn (cid:1) satisfies t + A = [0 , π ) \ A and shows that ( r , ∈ B , where for B, C ⊆ T we denote B + C := { b + c : b ∈ B, c ∈ C } . Thus B = M = M ′ = F and this set can beequivalently described as consisting of precisely those ( t , t ) ∈ T such that t − t ∈ T generates a finite subgroup of even order.Suppose that r = 3. Three vectors on the unit circle sum to if and only if they forman equilateral triangle. (Indeed, the sum of any two unit vectors has norm 1 if and onlyif the angle between the vectors is 2 π/ t and t , we can assumethat nt ≡ π/ nt ≡ π/ π . Each of t , t ∈ [0 , π ) is a (non-zero) integermultiple of 2 π/ (3 n ). Let k , k ∈ [3 n −
1] satisfy t i = 2 πk i / (3 n ). By the minimality of n , the greatest common divisor gcd( k , k , n ) = 1. Furthermore, it is impossible that 3divides both k and k , for otherwise by e.g. 2 πk / (3 n ) ≡ π/ π ) we have that 3also divides n , a contradiction to gcd( k , k , n ) = 1. Therefore, the subgroup generatedby t , t ∈ T is { πk n : k ∈ { , ... , n − }} , which is the cyclic group of order 3 n . For14 = 1 ,
2, we have k i n ≡ in (mod 3 n ) and thus k i ≡ i (mod 3). Thus if we take A := (cid:8) πm n : m ∈ { , , ... , n − } (cid:9) + (cid:2) , π n (cid:1) , then t + A , t + A and t + A = A partition [0 , π ). We conclude that B = M = M ′ = F and this set can be alternatively described as consisting precisely of the triples( πk n + t, πk n + t, t ) with n > k , k ∈ [3 n −
1] and t ∈ T such that { k , k } ≡ { , } (mod 3) while the greatest common divisor of k , k and n is 1.Suppose that r = 4. We need the following geometric claim. Claim 18.
Four vectors ( x i , y i ) ∈ S , i ∈ [4] , have sum if and only if they can be splitinto two pairs of opposite vectors.Proof of Claim. The non-trivial direction of the claim can be derived by observing that,up to a permutation of indices, we can assume that v := ( x , y ) + ( x , y ) is a non-zerovector while, in general, there is at most one way to write ± v ∈ R \{ } as the unorderedsum of two unit vectors.Recall that n > x i :=cos( nt i ) and y i := sin( nt i ) for i ∈ [4] gives that, up to a permutation of indices, ( x , y ) = − ( x , y ) and ( x , y ) = − ( x , y ). Thus, by Proposition 17, the set F consists preciselyof those ( t , ... , t ) such that, for some integer n > n ( t − t ) ≡ n ( t − t ) ≡ π (mod 2 π ) . (10)Let us show that if t /π is irrational then ( t , ... , t )
6∈ M ′ . (Recall that we assumethat t = 0.) By (10), we can assume that t = t + k π/n and t = k π/n for some oddintegers k and k . Suppose for a sake of contradiction that for some measurable subset A ⊆ T we have that P i =1 t i . A = 1 a.e. Take the Fourier expansion A ( t ) = 14 + ∞ X m =1 ( c m cos mt + s m sin mt ) . By the argument leading to (9) and Claim 18, we see that ( c m , s m ) can be non-zero onlyif we can split ( mt , ... , mt ) ∈ T into two pairs, each pair having difference π . Since t /π is irrational, these pairs must be ( t , t ) and ( t , t ). Thus mk i π/n ≡ π (mod 2 π )for i = 2 ,
3. Clearly, the validity of these two equations is determined by the residue of m modulo n . Since n is minimal, these equations cannot both hold for any m ∈ [ n − m is a multiple of n . This means that the only non-zeroFourier terms of A have period 2 π/n . It follows that A = (2 πk/n ) + A a.e. for every15nteger k and A = n P n − k =0 (2 πk/n ) . A . Thus t . A + A = 1 n n − X k =0 (( t + 2 πk/n ) . A + (2 πk/n ) . A )= 12 n n − X k =0 (2 πk/n ) . ( t . A + t . A + t . A + t . A ) = 12 a.e.,where we used that t . A + t . A + t . A + t . A = 1 a.e. by the choice of A . We concludethat the function 2 A demonstrates that ( t , ∈ F . By the case r = 2 that was solvedearlier, this contradicts the irrationality of t /π .This gives that M ′ is strictly smaller than F : for example, ( a, a + π, π,
0) belongs to F \ M if a/π is irrational.Now, suppose that t /π is rational. Let Γ be the subgroup of T that is generated by t , t and t . (There is no need to add t as it is 0.) By (10), the group Γ is finite. Ofcourse, if 4 does not divide its order | Γ | then there is no t -division even if a null set can beremoved. So suppose that | Γ | = 4 m for some integer m , i.e. that Γ is the cyclic group oforder 4 m . For i ∈ [4], let k i ∈ { , ... , m − } satisfy that t i = πk i m . Let k := ( k , ... , k ).Let us say that the cyclic group Z m , that consists of integer residues modulo 4 m , is k -divisible if there is a subset A ⊆ Z m such that the sets k i + A , i ∈ [4], partition Z m .Of course, such a set A must have exactly m elements.The following claim implies in particular that B = M = M ′ . Claim 19. If Z m is k -divisible then t ∈ B ; otherwise, t
6∈ M ′ .Proof of Claim. Suppose first that a subset A ⊆ Z m witnesses the ( k , ... , k )-divisibilityof Z m . It corresponds to an m -subset B ⊆ [0 , π ) such that its translates by t , ... , t partition the subgroup Γ ⊆ T . Now the Borel set C := B + [0 , π m ) exhibits the t -divisibility of T .Conversely, suppose that Z m is not k -divisible. Take any measurable set C ⊆ [0 , π )such that its translates by t , ... , t are pairwise disjoint. Take any coset X := t + Γ ⊆ T of Γ . Define A to consist of those k ∈ Z m such that t + πk m ∈ C (that is, A encodes theintersection of C with the Γ -coset X ). The translates of A by k , ... , k in Z m (whichcorrespond to the intersections ( t i + C ) ∩ X , i ∈ [4]) are pairwise disjoint and, by ourassumption, omit at least one element of Z m . Thus every coset of Γ in T containsat least one element of B := T \ ( { t , ... , t } + C ). It follows that B has measure atleast 2 π/ (4 m ) (as its translates by πk m for k ∈ { , ... , m − } cover T ). This gives that t
6∈ M ′ .Unfortunately, an explicit characterization of the set B = M = M ′ for general n seems to be rather messy, although it reduces to a finite case analysis for any given t ∈ T
16y Claim 19. So we will restrict ourselves to the special cases n = 1 and n = 2, justto illustrate that the measurable t -divisibility is not determined by the order 4 m of thegroup Γ alone (which happens already for n = 2).First, assume that n = 1. By (10), we have up to a permutation that ( t , t , t ) ≡ ( a, a + π, π ) (mod 2 π ) with a
6∈ { , π } . Thus, working inside Z m (that is, modulo4 m ), we have that k = k + 2 m and k = 2 m . Since k , k + 2 m, m generate Z m ,we have that k and 2 m are coprime; in particular k is odd. As it is easy to see A := { i : i ∈ { , ... , m − }} witnesses the k -divisibility of Z m . Thus t ∈ B byClaim 19.Now, assume that n = 2. By (10), we have that each of the differences k − k and k − k modulo 4 m is either m or 3 m . By negating all k i ’s and swapping k and k ifnecessary, we can assume that, k = k + m and k = m . (Note that these operations donot affect the k -divisibility of Z m and thus the conclusion of Claim 19 is also unaffected.)Let k := k . Thus k = ( k, k + m, m, . First, let us show that if m = 2 s is even then Z m is k -divisible (and thus t ∈ B byClaim 19). It is enough to find an s -set S ⊆ { , ... , m − } such that, modulo m , the sets S and k + S partition Z m (because then A := S ∪ (2 m + S ) as a subset of Z m witnessesthe k -divisibility of Z m ). Note that S := { ik : i ∈ { , ... , s − }} works. (Indeed, bygcd( k, m ) = 1 each residue modulo m appears exactly once as ik with i ∈ { , ... , m − } and we have included every second multiple of k into the set S .)Finally, suppose that m is odd. Recall that gcd( k, m ) = 1. We claim that Z m is k -divisible if and only if k ≡ m -set A ⊆ Z m witnessesthe k -divisibility. Since m is odd, some residue i modulo m appears an odd numberof times in A . This multiplicity cannot be larger than 2 since otherwise the translates k + A, ... , k + A would cover the four points i, i + m, i + 2 m, i + 3 m ∈ Z m at least sixtimes. Thus the multiplicity of i in A modulo m is exactly 1. By the commutativity of Z m , we can replace A by any its translate. Thus assume that A contains 0 but noneof m , 2 m and 3 m . The set ( k + A ) ∪ ( k + A ) covers 0 and m but not 2 m nor 3 m .Since 2 m A , the only way to consistently cover 2 m and 3 m is that 2 m − k ∈ A . Now, { k , ... , k } + { , m − k } contains 2 m − k and 3 m − k but not − k nor m − k . Note ofthe last two elements cannot be covered by k + A or k + A (as then A modulo m wouldcontain − k (mod m ) at least twice but then the four elements 0 , m, m, m would becovered at least six times, with the extra multiplicity coming from 0 and m being coveredby 0 ∈ A when translated by k and k ). Thus the only way to consistently cover − k and m − k is that − k ∈ A . One can continue to argue in this manner, showing that for each i ∈ { , , ... } we have − ik ∈ A and − (2 i +1) k +2 m ∈ A . As the first m of these elementsof A are pairwise distinct (in fact, they have pairwise distinct residues modulo m ) and m is odd, it must hold that − mk + 2 m ≡ m ). This equation has m solutions,17amely, all k ∈ Z m with k ≡ k ≡ A consisting of elements − ik and − (2 i + 1) k + 2 m for i ∈ { , ... , ( m − / } shows the k -divisibility of Z m . Indeed, note that | A | = m (as its elements have differentresidues modulo m by gcd( k, m ) = 1) and that if we keep increasing the index i beyond( m − / A since − mk + 2 m ≡ m ). By“reverse engineering” the proof of the forward implication, we see that the translates of A by k , ... , k are pairwise disjoint and thus partition Z m , as required.We make the following conjecture which, if true, implies that B r = M r = M ′ r forevery r and reduces the question if any given t ∈ T r belongs to this set to some finitecase analysis. Conjecture 20. If ( t , . . . , t r ) ∈ M ′ r then ( t i − t j ) /π is rational for every i, j ∈ [ r ] . In order to prove Proposition 2, we need some auxiliary results first.
Lemma 21.
The kernels of real n i × n matrices A i , i ∈ [ k ] , contain a common non-zerovector x ∈ R n \{ } if and only if the n × n matrix M := P ki =1 A Ti A i has zero determinant.Proof. If some non-zero x ∈ R n satisfies A i x = for every i ∈ [ k ], then M x = P ki =1 A Ti ( A i x ) = , so the determinant of M is zero.Conversely, suppose that M is singular. Choose a non-zero vector x ∈ R n with M x = . Then0 = x · M x = k X i =1 x · ( A Ti A i x ) = k X i =1 ( A i x ) · ( A i x ) = k X i =1 k A i x k and each A i x must be the zero vector, giving the required.The results of Dekker [6], Deligne and Sullivan [8], and Borel [3] (see Theorem 6.4in [21] and the historical discussion preceding it) give the following. Lemma 22.
For every d > and r > there is a choice of rotations β , ... , β r ∈ SO( d ) that generate the free rank- r group F r such that its action on S d − is free for even d and locally commutative for odd d (that is, every two elements of F r that have a commonfixed element on S d − commute). Note that the above result is usually stated in the special case r = 2 as the generalcase easily follows by taking any subgroup of F isomorphic to F r .18 emma 23. If γ = ( γ , ... , γ r ) ∈ SO( d ) r is generic, then the rotations γ , ... , γ r generatethe free rank- r group F r and the corresponding action of F r on S d − is free for even d and locally commutative for odd d .Proof. For a non-trivial reduced word w in F r and β = ( β , ... , β r ) ∈ SO( d ) r , the relation w ( β ) = I d amounts to d polynomial equations, with p ij ( β ) = 0 stating that the ( i, j )-thentry of the corresponding product of the matrices of β i ’s and their inverses (which areequal to their transposes) is i = j , where i = j is 1 if i = j and 0 otherwise. Each of thesepolynomials p ij has rational coefficients. Moreover, the r -tuple of matrices β returnedby Lemma 22 (which, in particular, generates the free subgroup) gives a point where atleast one of these polynomials is non-zero, say p ij ( β ) = 0. The polynomial p ij has tobe non-zero also at the generic point γ ∈ SO( d ) r and so w ( γ ) = I d . Since w was anarbitrary non-trivial word, the rotations γ , ... , γ r indeed generate the free group.Let us show the second part in the case of odd d (with the case of even d beingsimilar). Suppose on the contrary that we have two reduced non-commuting words w and w in F r such that the corresponding elements w ( γ ) and w ( γ ) have a commonfixed point x ∈ S d − . Thus the matrices A := w ( γ ) − I d and A := w ( γ ) − I d have x = as a common zero eigenvector. By Lemma 21, this property is equivalent todet( A T A + A T A ) = 0, which is a polynomial equation in γ with rational coefficients.For the special r -tuple of matrices β returned by Lemma 22, the matrices B := w ( β ) − I d and B := w ( β ) − I d cannot have a common zero eigenvector as it would give acommon fixed point for the non-commuting elements w ( β ) and w ( β ). Thus, we haveby Lemma 21 that det( B T B + B T B ) = 0. We have found a polynomial equality withrational coefficients that holds for γ but not for β ∈ SO( d ) r . This contradicts ourassumptions that γ ∈ SO( d ) r is generic.Also, we will need the following result of Conley, Marks and Unger that directly follows(as a rather special case) from Lemmas 3.4 and 3.6 in [4]. Theorem 24 (Conley, Marks and Unger [4]) . Let F r be the free group of rank r withgenerators γ , ... , γ r and let a : F r y X be a free Borel action on a Polish space X . Thenthere is a Borel subset A ⊆ X such that γ .A, ... , γ r .A are disjoint and X \ ∪ ri =1 γ i .A ismeager.Proof of Proposition 2. We have to show that if r, d > r -tuple γ = ( γ , ... , γ r ) ∈ SO( d ) r is generic then there is a γ -division of S d − with pieces that have the property ofBaire.By Lemma 23, the elements γ , ... , γ r ∈ SO( d ) generate a free (resp. locally commu-tative) action a of the free group F r on the sphere S d − when d is even (resp. odd). Themore general Corollary 5.12 in [21] (which is attributed in [21] to Dekker [6, 7]) directlygives that S d − is γ -divisible, that is, there is a subset B ⊆ S d − with γ .B, ... , γ r .B partitioning the sphere. 19or every γ ∈ SO( d ) \ { I d } , the set of its fixed points on S d − is closed (as the preimageof under the continuous map that sends x ∈ S d − to γ. x − x ∈ R d ) and has emptyrelative interior (for otherwise one can choose d linearly independent vectors fixed by γ ,contradicting γ = I d ). In particular, this set is meager. Since the group F r is countable,the free part X of the action a (which consists of x ∈ S d − such that w. x = x for eachnon-trivial w ∈ F r ) is co-meager. Also, it is easy to show that the free part X is a Borelsubset of the sphere (see e.g. [19, Lemma 4.4]).Theorem 24, when applied to the free action of F r on X , gives a Borel set A ⊆ X with its translates γ .A, ... , γ r .A being disjoint and Z := S d − \ ∪ ri =1 γ i .A being meager.We can additionally assume that Z is a -invariant: its saturation [ Z ] := ∪ w ∈ F r w.Z is stillmeager (since the countable group F r acts by homeomorphisms) so we can replace A by A \ [ Z ] without violating the conclusion of Theorem 24.Now, we can combine the Borel γ -division of S d − \ Z given by Conley, Marks andUnger [4] with the γ -division of Dekker [6, 7] restricted to Z . Formally, take C := A ∪ ( B ∩ Z ). The set C , as the union of a Borel set and a meager set, has the propertyof Baire while its translates γ .C, ... , γ r .C partition S d − by the invariance of Z . This section is dedicated to proving Lemmas 3 and 4. Their proofs are rather technical;this is why we postponed them until the very end.
In this section we present some definitions and results from algebraic geometry that weneed. We will follow the notation from the book by Hassett [11] to which we refer formissing details (and for a nice concrete introduction to most results needed here).A field extension
K ֒ → L is called algebraic if every x ∈ L is algebraic over K , thatis, satisfies a non-trivial polynomial equation with coefficients in K . Some easy but veryuseful facts ([11, Proposition A.16]) are that, for an arbitrary field extension K ֒ → L ,the elements of L that are algebraic over K form a field (11)and, for another field extension L ֒ → M ,if K ֒ → L and L ֒ → M are both algebraic then K ֒ → M is algebraic. (12)Let us fix a field K . 20y a variety we mean a subset X of some affine space K n which is closed in the Zariskitopology , that is, X is equal to V K ( F ) := { x ∈ K n : ∀ f ∈ F f ( x ) = 0 } for some family F ⊆ K [ x ] of polynomials where x := ( x , ... , x n ). Then then coordinatering of X is K [ X ] := K [ x ] /I ( X ), where I ( X ) := { f ∈ K [ x ] : ∀ x ∈ X f ( x ) = 0 } denotes the ideal of the variety X ⊆ K n .We call a variety X ⊆ K n irreducible if we cannot write X = X ∪ X for some varieties X , X ( X . This is equivalent to the statement that the ideal I ( X ) ⊆ K [ x ] is prime([11, Theorem 6.5]). Then K [ X ] is a domain so we can define its fraction field, which iscalled the function field of X and is denoted by K ( X ). Elements of K [ X ] (resp. K ( X ))can be viewed as the restrictions of polynomial (resp. rational) functions to X moduloidentifying functions that coincide on X .The dimension dim X of an irreducible variety X is the cardinality of a transcendencebasis for the field extension K ֒ → K ( X ), which is a collection of algebraically independent(over K ) elements z , ... , z k ∈ K ( X ) such that K ( X ) is algebraic over K ( z , ... , z k ),the smallest subfield of K ( X ) containing K ∪ { z , ... , z k } . By [11, Proposition 7.15], atranscendence basis exists and every two transcendence bases have the same cardinality.Every variety X can be written as a finite union X ∪ ... ∪ X m of irreducible varieties([11, Theorem 6.4]). (In fact, this decomposition, if irredundant, is unique up to apermutation of indices.) Then the dimension of X is defined as dim X := max { dim X i : i ∈ [ m ] } . By [5, Corollary 2.68], one can equivalently definedim X := max { k : ∃ irreducible varieties Y , ... Y k with ∅ ( Y ( ... ( Y k ⊆ X } . (13)We will also need the following easy result. Lemma 25. If X , ... , X n are infinite subsets of a field K and a polynomial f ∈ K [ x , ... , x n ] vanishes on each element of X × ... × X n , then f is the zero polynomial.Proof. We use induction on n . The base case n = 1 can be proved by induction on thedegree of the univariate polynomial f ( x ) by factoring out a linear factor correspondingto a root of f .Let n >
2. Expand f ( x , ... , x n ) = P mi =0 c i x in , with c i ∈ K [ x , ... , x n − ] and c m = 0.By induction, there is ( a , ... , a n − ) in X × ... × X n − with c m ( a , ... , a n − ) = 0. Thus f ( a , ... , a n − , x n ) is a non-zero polynomial of x n so it cannot vanish on X n by the basecase n = 1. 21 .2 Variety SO( d ; K ) r In this section we show in particular that SO( d ) r , as a variety in R d r , is irreducible andthat the set of entries above the diagonals forms a transcendence basis; in particular,the dimension of SO( d ) r is (cid:0) d (cid:1) r . In fact, we will need an extension of this result, wherethe underlying field can be different from R , for the proof of Lemma 4 (even though thestatement of Lemma 4 deals only with the real case).Let d > K be a field. Consider the affine space K d × d of all d × d matrices with entries in K , writing its elements as γ = ( γ i,j ) i,j ∈ [ d ] . Let the specialorthogonal variety over K be the variety SO( d ; K ) := V K ( I SO ) ⊆ K d × d defined by theideal I SO := (cid:10) ( u i ) i ∈ [ d ] , ( f ij ) i
1. The r -th power SO( d ; K ) r = SO( d ; K ) × ... × SO( d ; K ) is avariety in K d r since a product of Zariski closed sets is Zariski closed (or since one canwrite the explicit equations defining SO( d ; K ) r ).For ( γ , ... , γ r ) ∈ SO( d ; K ) r , let γ U := (( γ s ) i,j : s ∈ [ r ] , i < j d ) , be the sequence of the (cid:0) d (cid:1) r entries strictly above the diagonals. We call these entries upper . For notational convenience, we fix an ordering of the coordinates of K d r so thatall non-upper entries (that is, those on or below the diagonals) come before all upperones; thus when we write a vector of length d r as ( x , y ) then we mean that y is theupper part. Lemma 26.
For every subfield K ⊆ C , the variety X := (SO( d ; K )) r ⊆ K rn is irre-ducible, has dimension (cid:0) d (cid:1) r and the set of upper coordinates forms a transcendence basisof the function field K ( X ) over K .Proof. First, let us show that X is irreducible The proof of this in the case r = 1 (for anarbitrary field with 2 = 0) can be found in [2, Proposition 5-2.3]. We adopt the argument22rom [2] to work for any r >
1. (Note that products need not preserve the irreducibilitywhen the underlying field is not algebraically closed.)For x ∈ K d with x · x := P di =1 x i non-zero, the map ρ x : K d → K d that is defined by ρ x ( y ) := y − y · xx · x x , for y ∈ K d , can be thought of as the reflection of K d around the hyperplane orthogonal to x , so wecall ρ x a reflection . Each γ ∈ SO( d ; K ) can be written as a product of an even number ofreflections, see [2, Proposition 1-9.4] (and, conversely, every such product is in SO( d ; K )).In fact, the proof in [2], which proceeds by induction on d , shows that at most m := 2 d reflections are needed. By inserting the trivial composition ρ x ρ x = I d for some x ∈ K d with x · x = 0 we can write each γ ∈ SO( d ; K ) as the product of exactly m reflections.Let U := { z ∈ K d : z · z = 0 } and define f : U m → SO( d ; K ) by f ( z , ... , z m ) := ρ z ... ρ z m ∈ SO( d ; K ) , for ( z , ... , z m ) ∈ U m . Consider the product map f r : ( U m ) r → SO( d ; K ) r that applies f in each of the r coordinates. As the complement V := K dm \ U m is Zariski closed (as the finite union over i ∈ [ m ] of the sets of ( z , ... , z m ) ∈ K dm satisfying the polynomial equation z i · z i = 0),the complement W := K dmr \ U mr is also closed as the finite union over i ∈ [ r ] of theclosed sets K dm ( i − × V × K dm ( r − i ) . Clearly, f r is a rational map defined everywhereon U mr and thus continuous in the Zariski topology on U mr ⊆ K dmr . Also, the imageof f r is exactly X = SO( d ; K ) r with the surjectivity following from the choice of m . Itfollows from [2, Lemma 5-2.1] that X is irreducible. (In brief, if X can be written as aunion of two proper closed subsets X ∪ X , then K dmr is a union of two proper closedsets f − ( X ) ∪ W and f − ( X ) ∪ W , contradicting the irreducibility of K dmr since itsideal I ( K dmr ), which is { } by e.g. Lemma 25, is trivially prime.) Thus X is indeedirreducible.It remains to show that the set of upper coordinates γ U (that is, all entries above thediagonals) is a transcendence basis for the function field K ( X ) over K . This claim ismade of the following two parts.First, let us show that the field extension K ( γ U ) ֒ → K ( X ) is algebraic. By (11)and (12), it is enough to represent this field extension as a composition of field extensionswhere, at each step, every added non-upper coordinate is algebraic over the previouslyadded coordinates and the upper coordinates in the same matrix. Thus we consider justone matrix in SO( d ; K ), which we denote as γ = ( γ i,j ) i,j ∈ [ d ] . We add the non-uppercoordinates by whole rows in the natural order (with Row 1 added first, then Row 2, andso on). Take any Row m and a non-upper pair ( m, j ) (i.e. with j m ). The followingargument works for every index j ∈ [ m ] so we pick j = m for notational convenience.Thus we have to show that z := γ m,m , as an element of K ( X ), is algebraic over K ( { γ i,j : i ∈ [ m − , j ∈ [ d ] } ∪ { γ m,j : j ∈ { m + 1 , ... , d } } ) . x := ( γ m, , ... , γ m,m − ) consist of the other non-upper entries of Row m andlet M := ( γ i,j ) i,j ∈ [ m − be the square submatrix of γ which lies above x . The orthogonalityof Row m to the previous rows gives a system of m − M x T = f T , where f := ( f , ... , f m − ) with f i := − γ i,m z − P dj = m +1 γ i,j γ m,j for i ∈ [ m − M ) x T = Ad( M ) f T , where Ad( M ) denotes the adjoint matrix of M (whose ( i, j )-th entry is ( − i + j times the determinant of M with Row j and Column i removed). Take the unit “norm” relation P di =1 γ m,i = 1 for Row m , multiply it by(det( M )) and replace each (det( M )) x i by its value from Cramer’s rule. We get apolynomial equation having no x , namely,(det( M )) z + m − X i =1 m − X j =1 Ad( M ) ij f j ! + (det( M )) d X i = m +1 γ m,i = (det( M )) . (16)Let us show that the coefficient at z in this equation is non-zero. This coefficient is somepolynomial in the upper entries and the previous entries. If we take the identity matrix I d for γ , then the column above z is all zero and the matrix M is invertible (namely,it is the ( m − × ( m −
1) identity matrix I m − ). Then f does not depend on z atall and the coefficient at z is (det( M )) = 1, which is non-zero. So the coefficient at z in (16) is a non-zero polynomial, that is, z is algebraic over all previous entries, asdesired. We conclude (by (11) and (12)) that all entries on or below the diagonals arealgebraic over K ( γ U ) and thus the field extension K ( γ U ) ֒ → K ( X ) is indeed algebraic.Thus in order to show that the coordinates γ U form a transcendence basis, it remains toprove that these (cid:0) d (cid:1) r coordinates, as elements of the function field K ( X ), are algebraicallyindependent over K . It is enough to prove this for K = C . Indeed, we assumed that K ⊆ C . A non-trivial algebraic relation over K between the upper coordinates meansthat the ideal that defines SO( d ; K ) r (which, in the case r = 1, is the ideal I SO in (14))contains a non-zero polynomial g that does not depend on non-upper coordinates. Thesame polynomial g , when viewed as a polynomial in C [ γ ], then witnesses that the uppercoordinates are algebraically dependent over C .Thus let us assume that K = C . We need an easy auxiliary claim first from which wewill derive that every choice of sufficiently small in absolute value upper entries can beextended to a matrix in SO( d ; C ). For m ∈ [ d ] and an m × d matrix γ = ( γ i,j ), let theproperty P m state that for all i ∈ [ m ] we have P dj =1 γ i,j γ m,j = i = m . (Recall that i = m is 1 if i = m and 0 otherwise.) In other words, P m states that Row m has unit “norm”and is orthogonal to all previous rows. Claim 27.
For every m ∈ [ d ] and δ > there is ε = ε m ( δ ) > such that the followingholds. Take any complex numbers ( γ i,j ) ( i,j ) ∈ S , where S := ([ m − × [ d ]) ∪ { ( m, j ) : m < j d } , uch that P , ... , P m − hold and | γ i,j − i = j | ε for any ( i, j ) ∈ S . Then there is achoice of γ m, , ... , γ m,m ∈ C such that | γ m,j − m = j | δ for each j ∈ [ m ] and P m holds.Moreover, if γ i,j for each ( i, j ) ∈ S is real then γ m, , ... , γ m,m can additionally be chosento be real.Proof of Claim. Suppose that the claim fails for for some m ∈ [ d ] and δ >
0. Let real ε tend to 0 from above and let γ ∈ C S be a partial assignment violating the claim. Let ususe the notation that was introduced around (16). By our choice of γ , we have that eachentry of M is within additive ε = o (1) from the corresponding entry of the identity matrixand thus det( M ) = 1 + o (1) is non-zero. Of the two roots of the quadratic equation (16),which now reads z − o (1 + | z | ), choose z = 1 + o (1). In fact, (16) gives not only theentry z = γ m,m but the consistent remainder of Row m by x T := (det( M )) − Ad( M ) f T ,satisfying P m . By the continuity of the all involved functions (and det( M ) = 1 + o (1)),we have k x k ∞ = o (1), a contradiction to δ > γ i,j ’s are reals. In the above notation, the quadraticequation (16) has all real coefficients and, as before, states that z − o (1 + | z | ). Itsleft-hand side as a function of z ∈ R changes sign at z = 1 with its derivative 2 z beingbounded away from 0 around z = 1. Hence we can choose a real root z = 1 + o (1). Then M is a real matrix and the rest of Row m , namely x T := (det( M )) − Ad( M ) f T is alsoreal.Consider the projection π : SO( d ; C ) r → C m on the m := (cid:0) d (cid:1) r upper coordinates,which maps ( x , y ) to y . In particular, the r -tuple of the identity matrices projects tothe zero vector ∈ C m . The image of π contains some Euclidean open ballBall ε ( ) := { z ∈ C m : k z k < ε } of radius ε > ε := ε d ( ε d − ( ... ε (1 / (2 d d !)) ... )) > , (17)where ε , ... , ε d are the functions returned by Claim 27. Indeed, by the choice of theconstants we know that for every y ∈ Ball ε ( ), we can construct a d × d matrix γ row byrow so that γ projects to y and satisfies all properties P , ... , P d while it also holds that k γ − I d k ∞ < / (2 d d !). The last inequality gives, rather roughly, that | det( γ ) − | < γ ) = 1 because det( γ ) is either − π (SO( d ; C ))contains π ( γ ) = y .Now, suppose on the contrary that there is a non-trivial polynomial relation betweenthe upper coordinates. Thus there is a non-zero polynomial g which does not dependon the non-upper coordinates and belongs to the ideal generated by the polynomialsthat define SO( d ; C ) r (with those for r = 1 being listed in (14)). The polynomial g ,25s a function of the m upper coordinates, vanishes on π ( X ) ⊆ C m . This contradictsLemma 25 as π ( X ) contains a non-empty open set, namely the open ball of radius ε around the origin, and thus π ( X ) contains a product of m infinite sets.Now we are ready to show that the set N of non-generic points in SO( d ) r is “small”. Proof of Lemma 3.
When we identify an r -tuple of d × d matrices over a field K withan element of K d r , let us order the d r coordinates so that the m := (cid:0) d (cid:1) r upper entries(i.e. those above the diagonals) come at the end. Thus if we write an element of K d r as( x , y ) then y corresponds to the m upper entries. Also, we use the standard topology on S d − (the one which is inherited from the Euclidean space R d ).There are countably many polynomials in Q [ x , y ] so enumerate those that are non-zeroon at least one element of SO( d ) r as f , f , ... . By definition, if a point ( a , b ) ∈ SO( d ; R ) r is not Q -generic then some f i vanishes on ( a , b ). Thus N is a subset of the countableunion ∪ ∞ i =1 Z i , where Z i := { ( a , b ) ∈ SO( d ) r : f i ( a , b ) = 0 } . (18)Since each polynomial f i is continuous as a function R d r → R , each set Z i is closed.Let us turn to Part (i) where we have to show that the Haar measure ν assigns measure0 to N . By the countable additivity, it is enough to show that each set Z i , defined by (18),has ν -measure zero.First, let us recall how the Haar measure can be constructed for the group Γ := SO( n ) r (and, in fact, for any real Lie group), following the presentation in [15, Sections VIII.1–2]. Namely, choose some linear basis for the Lie algebra ( so ( d )) r viewed as the tangentspace T ( I d ,... ,I d ) at the identity ( I d , ... , I d ) ∈ SO( d ) r and, using the translations of thesevectors, turn them into left-invariant vector fields X , ... , X m . (Note the the Lie algebra( so ( d )) r , that consists of all r -tuples of skew-symmetric matrices, has dimension m = (cid:0) d (cid:1) r as a vector space.) For each γ ∈ Γ , let e ( γ ) , ... , e m ( γ ) ∈ T ∗ γ be the dual basis to( X ( γ ) , ... , X m ( γ )). Then ω = e ∧ ... ∧ e m (the skew-symmetric product) is a smooth m form on Γ , which is positive and left-invariant and thus defines a Borel left-invariantnon-zero measure on Γ ([15, Theorem 8.21]). By the uniqueness, this has to be a multipleof the Haar measure ν . In particular, any smooth submanifold of Γ of dimension (as amanifold) less than m has zero Haar measure ([15, Equation (8.25)]).The set Z i ( SO( d ) r , as an algebraic variety, has dimension smaller than m whichfollows from the definition of the dimension via nested chains of irreducible varieties (thatis, by (13)) and from the irreducibility of the variety SO( d ) r (that is, by Lemma 26). Somestandard results in the theory of (semi-)algebraic sets give that every bounded variety insome R n admits a triangulation into simplices each of which is a smooth submanifold of R n , see e.g. [1, Theorem 5.43]. Apply this result to every irreducible component Z ⊆ Z i .The dimension k of each obtained simplex S (as a manifold) is at most dim Z . Indeed, picka point s ∈ S and the projection from S on some k coordinates which is a homeomorphism26round s . Observe that these k coordinates are algebraically independent in the functionfield R ( Z ) because no non-zero polynomial on R k can vanish on a non-empty open setby Lemma 25.Thus we covered Z i by finitely many manifolds of dimension less than m , each havingzero Haar measure as it was observed earlier (by [15, Equation (8.25)]). We concludethat the Haar measure of Z i is indeed zero.Let us show Part (ii). It is enough to show that the relative interior of each Z i ⊆ SO( d ) r is empty (with Z i defined in (18)). Suppose on the contrary that the relative interior U of Z i is non-empty. Since the compact group SO( d ) r acts transitively on itself byhomeomorphisms, finitely many translates of U cover the whole group. As the Haarmeasure is ν is invariant under this action, we have that ν ( U ) >
0. However, thiscontradicts Part (i) that we have already proved.This finishes the proof of Lemma 3.
Our proof of the reverse (harder) implication of Lemma 4 needs Lemma 28 below. Sincewe could not find this rather natural statement anywhere in the literature we present aproof whose main idea (to use dimension) was suggested to us by Miles Reid. In fact,Miles Reid came up with a full proof of some initial version of the lemma. Since his proofrelies on the so-called universal domain of K while we would like to have this paper aselementary as possible, we present a proof that avoids universal domains.Given a field extension K ֒ → L and a variety X ⊆ L n (over the field L ), we say thatan element a ∈ X is K -generic for X if every polynomial p ∈ K [ x , ... , x n ] with p ( a ) = 0vanishes on every element of X . (Here as well as in the rest of this paper, each evaluationmixing elements of some two fields K ֒ → L is done in the larger field L .) In the specialcase when K := Q , L := R , X := SO( d ) r we get exactly the definition of a generic r -tuple of rotations from the Introduction. Lemma 28.
Let
K ֒ → L be a field extension, with L being algebraically closed. Let P ⊆ K [ x , y ] be some family of polynomials over K , where we abbreviate x := ( x , ... , x m ) and y := ( y , ... , y n ) . Suppose that X := { ( x , y ) ∈ L m + n : ∀ f ∈ P f ( x , y ) = 0 } , (19) as a variety over L , is irreducible and has dimension n with y , ... , y n forming a tran-scendence basis for the function field L ( X ) over L .Then every p = ( a , b ) ∈ X with the n -tuple b ∈ L n being algebraically independentover K is a K -generic point of X . roof. Let the ideal I p ⊆ K [ x , y ] consist of those polynomials over K that vanish on p .Let Z := V L ( I p ) = { ( x , y ) ∈ L m + n : ∀ f ∈ I p f ( x , y ) = 0 } . As P ⊆ I p , we trivially have that Z ⊆ X . We have to show that Z = X , which by thedefinition of Z = V L ( I p ) will give the required result (namely, that every f ∈ I p vanisheson X ).Let Z = Z ∪ ... ∪ Z t be a decomposition of Z into irreducible varieties ([11, Theorem6.4]).Suppose first that there is i ∈ [ t ] such that the n -tuple y , with each y j viewed as anelement of the function field L ( Z i ), is algebraically independent over L . This means thatthe dimension of the irreducible variety Z i ⊆ L m + n is at least n . Recall that Z i ⊆ Z ⊆ X .By the definition of the dimension via nested chains of irreducible subvarieties (that is,by (13)), we cannot have Z i ( X for otherwise any chain for Z i extends to a strictly largerchain for X which gives that dim X − > dim Z i > n , contradicting our assumption.Thus Z i = Z = X , as desired.Thus we can assume that for every i ∈ [ t ] there is a non-zero g i ∈ L [ y ] ∩ I ( Z i ). Since Z = ∪ ti =1 Z i , we have by [11, Proposition 3.12] that I ( Z ) = ∩ ti =1 I ( Z i ). (Recall that, forexample, the ideal I ( Z ) of Z ⊆ L m + n consists of those p ∈ L [ x , y ] that vanish on Z .)Thus the product g ... g t ∈ L [ y ], which trivially belongs to each I ( Z i ), also belongsto I ( Z ).Let I Lp be the ideal in L [ x , y ] generated by I p ⊆ K [ x , y ] ⊆ L [ x , y ]. In other words, I Lp := ( m X i =1 h i ( x , y ) f i ( x , y ) : m > , h , ... , h m ∈ L [ x , y ] , f , ... , f m ∈ I p ) , from which it easily follows that V L ( I Lp ) = V L ( I p ) = Z . Since L is algebraically closed,we have by Hilbert’s Nullstellensatz ([11, Theorem 7.3]) that I ( Z ) is equal to q I Lp := { f ∈ L [ x , y ] : ∃ N f N ∈ I Lp } , the radical of I Lp . Thus there is some integer N > g := ( g ... g t ) N belongs to I Lp .In other words, we have shown that I Lp contains a non-zero polynomial g that doesnot depend on x , that is, I Lp ∩ L [ y ] = { } . (20)We claim that, in fact, I p ∩ K [ y ] = { } . In order to show this, we analyse how a knownalgorithm for eliminating variables works, arguing that we can run two instances of thealgorithm, one for I Lp ∩ L [ y ] and the other for I p ∩ K [ y ], to produce the same generatingset of polynomials in each case.Since all following steps are fairly standard, we will be rather brief, referring the readerto [11] for a detailed exposition. First, by the Hilbert Basis Theorem ([11, Corollary 2.22]),28here is a finite set F ⊆ K [ x , y ] that generates I p . Of course, the same set F , as a subsetof L [ x , y ], generates I Lp . We fix any monomial order ≺ for ( x , y ) which is an eliminationorder for x ([11, Definition 4.6]) and apply Buchberger’s algorithm ([11, Corollary 2.29])to find a ≺ -Gr¨obner basis G for I Lp using F as its input. At a very low level, each stepof the algorithm is to pick some two previous non-zero polynomials h and h , take thecoefficients c and c at their ≺ -highest monomials and add h − ( c /c ) hh for somemonomial h to the current pool of polynomials. Thus all encountered polynomials havecoefficients in K ; in particular, the obtained Gr¨obner basis G is a subset of K [ x , y ]. Bythe Elimination Theorem ([11, Theorem 4.8]) and our choice of the monomial order ≺ ,the ideal I Lp ∩ L [ y ] is generated by G ∩ L [ y ], that is, by those polynomials in G that do notdepend on x . Moreover, if we apply Buchberger’s algorithm to find the intersection of I p = hF i ⊆ K [ x , y ] and K [ y ], we obtain the very same generating set G ∩ K [ y ] (becausethe choice of h , h and h at each low-level step of the algorithm depends only on the ≺ -highest monomials of the previous polynomials).However, we know that I p ∩ K [ y ] = { } because no non-zero polynomial in K [ y ]can vanish on p by our assumption that y is algebraically independent over K . Thus G ∩ L [ y ] = G ∩ K [ y ] can contain only the zero polynomial. This means that I Lp ∩ L [ y ] = { } , contradicting (20) and proving the lemma.Now we are ready to prove Lemma 4 that gives an alternative characterisation of Q -generic points of SO( d ) r . Proof of Lemma 4.
As before, the m := (cid:0) d (cid:1) r upper entries of SO( d ) r ⊆ R d r come atthe end and if we write an element of K d r as ( x , y ) then y corresponds to the m upperentries.The forward implication of the lemma is easy. Take any ( a , b ) ∈ SO ( d ; R ) r such that f ( b ) = 0 for some non-zero polynomial f with rational coefficients. Take any vector b ′ ∈ R m whose L ∞ -norm is at most the expression in (17) with entries algebraicallyindependent over Q . By Claim 27, there is a choice of a real vector a ′ with ( a ′ , b ′ ) ∈ SO( d ) r , that is, we can extent the vector b ′ of upper entries to an r -tuple of real specialorthogonal matrices. Since the polynomial f with rational coefficients cannot vanish on b ′ , the polynomial map ( x , y ) f ( y ) shows that ( a , b ) is not a generic point.Let us show the converse implication. Let ( a , b ) ∈ SO ( d ; R ) r be any point with the m -tuple b ∈ R m of reals being algebraically independent over Q .By Lemma 26, the complex variety X := SO( d ; C ) r ⊆ C d r is irreducible and the uppercoordinates y form a transcendence basis for the function field C ( X ). Now, Lemma 28(which requires that the field L is algebraically closed) applies with K := Q , L := C and P ⊆ Q [ x , y ] consisting of the polynomials that define the variety SO( d ; R ) r (with theones in (14) corresponding to the case r = 1). The lemma gives that ( a , b ) ∈ SO( d ; R ) r ⊆ SO( d ; C ) r is a Q -generic point of SO( d ; C ) r . Of course, this trivially implies that ( a , b )29s a Q -generic point also of SO( d ; R ) r (as every polynomial p ∈ Q [ x , y ] that vanishes on( a , b ) has to vanish on SO( d ; C ) r ⊇ SO( d ; R ) r ), as desired.This finishes the proof of Lemma 4. Acknowledgements
Clinton T. Conley was was supported by NSF Grant DMS-1855579. Jan Greb´ık andOleg Pikhurko were supported by Leverhulme Research Project Grant RPG-2018-424.The authors would like to thank Miles Reid for the very useful discussions and forsharing his proof of some initial version of Lemma 28.
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