Normal subgroup generated by a plane polynomial automorphism
aa r X i v : . [ m a t h . G R ] A p r NORMAL SUBGROUP GENERATED BY A PLANE POLYNOMIALAUTOMORPHISM
JEAN-PHILIPPE FURTER AND STÉPHANE LAMYA
BSTRACT . We study the normal subgroup h f i N generated by an element f = id in the group G of complex plane polynomial automorphisms having Jacobian determinant 1. On one hand if f has length at most 8 relatively to the classical amalgamated product structure of G , we provethat h f i N = G . On the other hand if f is a sufficiently generic element of even length at least14, we prove that h f i N = G . I NTRODUCTION
Let Aut [ C ] denote the group of complex plane polynomial automorphisms and let G be thesubgroup of automorphisms having Jacobian determinant 1. In this paper, we deal with normalsubgroups of G generated by a single element.It is easy to check that G is equal to the commutator subgroup of Aut [ C ] and to its owncommutator subgroup as well (see Proposition 10). It is more difficult to decide whether G is a simple group or not. There does not seem to exist any natural morphism whose kernelis a proper normal subgroup of G . However, in a short note published in 1974 that seems tohave been quite forgotten, V. I. Danilov [Dan74] proves that G is not a simple group. He usesresults from P. Schupp [Sch71], namely the so-called small cancellation theory in the contextof an amalgamated product. To be precise, he shows that the normal subgroup generated by theautomorphism ( ea ) where a = ( y , − x ) and e = ( x , y + x − x ) is a strict subgroup of G . Infact, he writes ( ea ) , because he uses a slightly erroneous definition of the condition C ′ ( / ) (see subsection 3.1).We now introduce the algebraic length of an automorphism in order to state our main result.The theorem of Jung, Van der Kulk and Nagata asserts that Aut [ C ] is the amalgamated productover their intersection of the groups A and E of affine and elementary automorphisms (see 1.1).Let f be an element of Aut [ C ] . If f is not in the amalgamated part A ∩ E , its algebraic length | f | is defined as the least integer m such that f can be expressed as a composition f = g . . . g m ,where each g i is in some factor ( A or E ) of Aut [ C ] . If f is in the amalgamated part, byconvention we set | f | = f of G will be denoted by h f i N . Of course, h f i N remains unchanged when replacing f by one of its conjugates in G . So, one can assume f of minimal algebraic length in its conjugacy class (see subsection 1.4 ). If | f | 6 =
1, this amountsto saying that | f | is even (indeed, if | f | is even, it is clear that f is strictly cyclically reduced inthe sense of subsection 3.1 below). This is for example the case for the previous automorphism ( ea ) which has length 26. Date : March 2010.The second author, on leave from Institut Camille Jordan, Université Lyon 1, France, was partially supported byan IEF Marie Curie Fellowship.
Here are the two main results of our paper:
Theorem 1.
If f ∈ G satisfies | f | ≤ and f = id , then h f i N = G. Theorem 2.
If f ∈ G is a generic element of even length | f | ≥ , then the normal subgroupgenerated by f in Aut [ C ] (or a fortiori in G) is different from G. Here the genericness means that if we write f ± = a e . . . a l e l , where l ≥ a , . . . , a l ∈ A \ E and each e i = ( x + P i ( y ) , y ) , then there exists an integer D such that for any sequence d , . . . , d l of integers ≥ D , ( P , . . . , P l ) can be chosen generically (in the sense of algebraic geometry,i.e. outside a Zariski-closed hypersurface) in the affine space (cid:213) ≤ i ≤ l C [ y ] ≤ d i , where we have set C [ y ] ≤ d = { P ∈ C [ y ] ; deg P ≤ d } .Theorems 1 and 2 correspond in the text below to Theorems 32 and 45. Note that in the latterstatements we use a geometric notion of length coming from Bass-Serre theory (see subsection1.2). This geometric length allows us to obtain more natural statements. In fact, Theorem 45deals with automorphisms satisfying the special condition ( C ) (see Definition 27). The proofthat this condition is indeed generic is postponed to the annex. To convince the reader that sucha condition is necessary, we now give examples of automorphisms of arbitrary even length andgenerating normal subgroups equal to G . Example 3.
Consider the three automorphisms a = ( y , − x ) , e = ( x + P ( y ) , y ) , e = ( x + Q ( y ) , y ) , where P (resp. Q ) is an even (resp. odd) polynomial of degree ≥
2, and set f = ae ( ae ) n ,where n ≥ u = − id, we get au = ua , e u = ue and e u = ue − , so that thecommutator [ f , u ] = f u f − u − is equal to [ f , u ] = ae ( ae ) n u ( ae ) − n ( ae ) − u − = ae ue − a − u − = ae a − . Since [ f , u ] ∈ h f i N , we get e ∈ h f i N , so that h f i N = G by Theorem 1 (or by Lemma 30 below).One motivation for this work is the still open question of the simplicity of the Cremonagroup Cr , i.e. the group of birational transformations of C . For instance in [Giz94] thequestion is explicitly stated and Gizatullin gives several criterion that would prove that Cr issimple. Recently Blanc [Bla10] proved that Cr is simple as an infinite dimensional algebraicgroup. In this respect, we should mention that Shafarevich claimed that the group Aut [ C n ] ofautomorphisms of the affine space C n having Jacobian determinant 1 is simple as an infinitedimensional algebraic group for any n ≥ , view as an abstract group,could be not simple as well. Indeed, it is known since Iskovskikh [Isk85] that Cr admits apresentation as the quotient of an amalgamated product by the normal subgroup generated by asingle element. Take H = ( PGL ( ) × PGL ( )) ⋊ Z / Z the group of birational transformationsthat extend as automorphisms of P × P and take H the group of transformations that preservethe pencil of vertical lines x = cte . Take t = ( y , x ) ∈ H \ H and e = ( / x , y / x ) ∈ H \ H ; thenCr is equal to the quotient ( H ∗ H ∩ H H ) / h f i N where f = ( t e ) . To prove that Cr is not simple it would be sufficient to find an element g in the amalgamated product of H and H (that should correspond to a sufficiently general ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 3 birational transformation) such that the normal subgroup h f , g i N is proper. This is similar tothe results we obtain in this paper; but the problem seems harder in the birational setting.As a final remark on these matters, we would like to mention a nice reinterpretation ofIskovskikh’s result by Wright (see [Wri92, Th. 3.13]). Let H = PGL ( ) be the group of bira-tional transformations that extend as automorphisms of P . Then Wright proves that the groupCr is the free product of H , H , H amalgamated along their pairwise intersection in Cr .In this paper we chose to work over the field C of complex numbers, even if most of theresults could be adapted to any base field. Note that in the case of a finite field the nonsimplicityresult is almost immediate. Let F q denote the finite field of q = p n elements, where p is primeand n ≥
1. Let Aut [ F ] be the group of automorphisms of the affine plane A F q = F q and letAut [ F ] be the normal subgroup of automorphisms having Jacobian determinant 1. If X is afinite set, let Per ( X ) (resp. Per + ( X ) ) be the group of permutations (resp. even permutations)of X . Since the natural morphism f : Aut [ F ] → Per ( F ) induces a non-constant morphismAut [ F ] → Per ( F ) (consider the translations!), it is clear that Aut [ F ] is not simple. Remark . If q is odd (i.e. the characteristic p of F q is odd), one can easily check that f ( Aut [ F ]) = Per + ( F ) . Indeed, f is surjective (see [Mau01]), so that f ( Aut [ F ]) is a normalsubgroup of Per ( F q ) . However, if the cardinal of X is different from 4, it is well known that Per + ( X ) is the only nontrivial normal subgroup of Per ( X ) (see e.g. [Rot95, ex. 3.21, p. 51]).Therefore, it is enough to show that f ( Aut [ F ]) ⊆ Per + ( F ) . But on one hand Aut [ F ] isgenerated by the elementary automorphisms ( x + P ( y ) , y ) and ( x , y + Q ( x )) , where P ∈ C [ y ] , Q ∈ C [ x ] are any polynomials. On the other hand, it is straightforward to check that such auto-morphisms induce even permutations of F q .As a final remark, we would like to stress the importance of translations in getting ourresults. Let Aut [ C ] be the group of automorphisms fixing the origin and let J n be the naturalgroup-morphism associating to an element of Aut [ C ] its n -jet at the origin (for n ≥ n ≥
2, the kernel of J n is a nontrivial normal subgroup of G = G ∩ Aut [ C ] , so that this lattergroup is not simple. Of course for Aut [ C ] the morphism J n does not exist. This explains thefact that our paper strongly relies on translations (see Lemmas 7 and 16). Remark . It results from [Ani83] that the image of J n is exactly the group of n -jets of polyno-mial endomorphisms fixing the origin and whose Jacobian determinant is a non-zero constant.The precise statement can be found in [Fur07, Proposition 3.2].Our paper is organized as follows.In section 1 we gather the results from Bass-Serre theory that we need: this includes somebasic definitions and facts but also some quite intricate computations, such as in the characteri-zation of tripods (subsection 1.7). This is also the place where we define precisely the condition ( C ) that we need in Theorem 45.Section 2 is devoted to the proof of Theorem 1. This is the most elementary part of the paper.We only use Lemma 7 from section 1.In section 3 we deal with R-diagrams. This field of combinatorial group theory has beenintroduced by Lyndon and Schupp in relation with condition C ′ ( / ) from small cancellationtheory (see 3.1). A noteworthy feature of our work is that we use R-diagrams in a completelyopposite setting (positive curvature). ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 4
In section 4 we are able to give a proof of Theorem 2, using the full force of both Bass-Serreand Lyndon-Schupp theories.We briefly discuss in section 5 the cases not covered by Theorems 1 and 2, that is to saywhen the automorphism has length 10 or 12.Finally, in the annex, we prove that condition ( C ) is generic and we also give explicitexamples of automorphisms satisfying this condition.1. T HE B ASS -S ERRE TREE
Generalities.
The classical theorem of Jung, van der Kulk and Nagata states that thegroup Aut [ C ] is the amalgamated product of the affine group A = { ( a x + b y + g , d x + e y + z ) ; a , . . . , e ∈ C , ae − bd = } and the elementary group E = { ( a x + P ( y ) , b y + g ) ; a , b , g ∈ C , ab = , P ∈ C [ y ] } over their intersection (see [Jun42, vdK53, Nag72]). This is usually written in the followingway: Theorem 6.
Aut [ C ] = A ∗ A ∩ E E . A geometric proof of this theorem and many references may be found in [Lam02]. Let usalso recall that elements of E are often called triangular automorphisms.The Bass-Serre theory ([Ser77]) associates a simplicial tree to any amalgamated product. Inour context, let us denote by T this tree. By definition, the vertices of T are the disjoint unionof the left cosets modulo A (vertices of type A ) and modulo E (vertices of type E ). The edgesof T are the left cosets modulo ( A ∩ E ) . Finally, if f ∈ Aut [ C ] , the edge f ( A ∩ E ) links thevertices f A and f E . Since Aut [ C ] is generated by A and E , T is connected. Thanks to theamalgamated structure, T contains no loop, so that it is indeed a tree.The group Aut [ C ] acts naturally on T by left multiplication: for any g , f ∈ Aut [ C ] , we set g . f A = ( g f ) A , g . f E = ( g f ) E and g . f ( A ∩ E ) = ( g f )( A ∩ E ) . It turns out that this action givesan embedding of Aut [ C ] into the group of simplicial isometries of T (see [Lam01, Remark3.5]). This action is transitive on the set of edges, on the set of vertices of type A and on theset of vertices of type E . The stabilizer of a vertex f A (resp. of a vertex f E , resp. of an edge f ( A ∩ E ) ) is the group f A f − (resp. f E f − , resp. f ( A ∩ E ) f − ).Following [Wri79, Lam01], one can define systems of representatives of the nontrivial leftcosets A / A ∩ E and E / A ∩ E by taking: a ( l ) = ( l x + y , − x ) ; l ∈ C e ( Q ) = ( x + Q ( y ) , y ) ; Q ( y ) ∈ y C [ y ] \ { } . Note that the minus sign in the expression of a ( l ) did not appear in [Wri79, Lam01]. We haveto introduce it in the present paper in order to get automorphisms with Jacobian determinant 1(see subsection 1.4).Then any element g ∈ Aut [ C ] may be uniquely written g = ws where w is a product offactors of the form a ( l ) or e ( Q ) , successive factors being of different forms, and s ∈ A ∩ E (seee.g. [Ser77, chap. I, 1.2, th. 1]). Similarly, any edge (resp. vertex of type A , resp. vertex oftype E ) may be uniquely written w ( A ∩ E ) (resp wA , resp. wE ) where w is as above. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 5
We call a (directed) path a sequence of consecutive edges in T . To denote a path we enu-merate its vertices separated by − . For instance the path P of two edges containing the verticesid A , id E , eA , where e ∈ E \ A will be denoted P = id A − id E − eA . If we are only interested inthe type of the vertices, we say for example that P is of type A − E − A .If two vertices of T are fixed by an automorphism of Aut [ C ] , then the path relating them isalso fixed. Therefore, the subset of T fixed by an automorphism is either empty or a subtree.Up to conjugation, this subset has been computed for any automorphism in [Lam01, proofof Proposition 3.3]. In particular, it has been computed for the translation ( x + , y ) . Thefollowing easy and technical lemma is a slight variation of this computation. As in the latterpaper, this analogous statement turns out to be very useful. The proof is given for the sake ofcompleteness. Lemma 7.
The subtree of T fixed by the translation ( x + c , y ) , c ∈ C ∗ , is exactly the union ofthe paths id E − e ( P ) A − e ( P ) a ( l ) E − e ( P ) a ( l ) e ( Q ) Awhere P ∈ y C [ y ] , l ∈ C and Q ( y ) = a y , a ∈ C ∗ .Note that we (exceptionally) allow P to be zero. In that case, the path should rather bewritten id E − id A − a ( l ) E − a ( l ) e ( Q ) A . In particular, the fixed subtree does not depend on c, has diameter and contains the closedball of radius 2 centered at id E, i.e. the union of the paths id E − e ( P ) A − e ( P ) a ( l ) E , P ∈ y C [ y ] , l ∈ C . Proof. If P , Q ∈ y C [ y ] and l ∈ C we have ( x + c , y ) ◦ e ( P ) = e ( P ) ◦ ( x + c , y ) ; ( x + c , y ) ◦ a ( l ) = a ( l ) ◦ ( x , y + c ) ; ( x , y + c ) ◦ e ( Q ) = e ( Q ) ◦ f ;where f = ( x + Q ( y ) − Q ( y + c ) , y + c ) , so that ( x + c , y ) e ( P ) a ( l ) e ( Q ) = e ( P ) a ( l ) e ( Q ) f . Therefore, the vertex e ( P ) a ( l ) e ( Q ) A is fixed by ( x + c , y ) if and only if f ∈ A , i.e. deg ( Q ( y ) − Q ( y + c )) ≤
1, i.e. deg ( Q ) ≤
2. If Q = a y , this vertex is fixed. Since the vertex id E is also(obviously) fixed, this shows that the following path is fixed:id E − e ( P ) A − e ( P ) a ( l ) E − e ( P ) a ( l ) e ( Q ) A . If Q = a y , where a = µ ∈ C , it remains to show that the vertex e ( P ) a ( l ) e ( Q ) a ( µ ) E isnot fixed. Indeed, an easy computation shows that ( x + c , y ) e ( P ) a ( l ) e ( Q ) a ( µ ) = e ( P ) a ( l ) e ( Q ) a ( µ ) g , where g = ( x − c , a cx + y + µc − a c ) / ∈ E . (cid:3) ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 6
Algebraic and geometric lengths.
We will use two notions of length on Aut [ C ] .The algebraic length has been defined in the introduction: if g ∈ Aut [ C ] is not in the amal-gamated part, | g | is defined as the least integer m such that g can be expressed as a composition g = g . . . g m where each g i is in some factor of the amalgam. If g is in the amalgamated part,we set | g | = geometric length is defined by lg ( g ) = inf v ∈ V dist ( g . v , v ) , where V is the set of verticesof T and dist ( ., . ) is the simplicial distance on T .By Lemma 8 we almost always have lg ( g ) = min {| f g f − | ; f ∈ Aut [ C ] } , the only exceptionbeing when g is conjugate to an elementary automorphism which is not conjugate to an elementin the amalgamated part.1.3. Elliptic and hyperbolic elements.
Elements g of Aut [ C ] may be sorted into two classesaccording to their action on T .If lg ( g ) = g has at least one fixed point on T ), we say that g is elliptic . This cor-responds to the case where g is conjugate to an element belonging to some factor ( A or E ) ofAut [ C ] . Since any element of A is conjugate to some element of E , this amounts to saying that g is triangularizable (i.e. conjugate to some triangular automorphism).If lg ( g ) >
0, we say that g is hyperbolic . This corresponds to the case where g is conjugateto a composition of generalized Hénon transformations h . . . h l (see [FM89]). We recall that ageneralized Hénon transformation is a map of the form h = ( y , ax + P ( y )) = ( y , x ) ◦ ( ax + P ( y ) , y ) ,where a ∈ C ∗ and P ( y ) is a polynomial of degree at least 2. Equivalently, g is conjugate to anautomorphism of the form f = a e . . . a l e l ,where each a i ∈ A \ E and each e i ∈ E \ A .The set of points v ∈ T satisfying dist ( g . v , v ) = lg ( g ) defines an infinite geodesic of T de-noted by Geo ( g ) . Furthermore, g acts on Geo ( g ) by translation of length lg ( g ) . It is notdifficult to check that lg ( g ) = lg ( f ) = | f | = l and that the geodesic of f is composed of thepath id A − a E − a e A − · · · − a e . . . a l e l A and its translated by the f k ’s ( k ∈ Z ). If g = f f f − with f ∈ Aut [ C ] , we have of course Geo ( g ) = f ( Geo ( f )) .The proof of the following easy result is left to the reader. Note that these two sets of equiv-alent conditions correspond to the notions of strictly and weakly cyclically reduced elementsgiven in subsection 3.1. Lemma 8.
Let g ∈ Aut [ C ] be a hyperbolic element.(1) The following assertions are equivalent:(i) | g | = lg ( g ) ; (ii) Geo ( g ) contains the vertices id A and id E.(2) The following assertions are equivalent:(iii) | g | ≤ lg ( g ) + ; (iv) Geo ( g ) contains the vertex id A or id E. The group G . In this subsection we prove two basic facts about G . Let us set A = A ∩ G and E = E ∩ G . Theorem 6 easily implies the following result: Proposition 9. G = A ∗ A ∩ E E .Proof. By [Ser77, chap. I, n ◦ g ∈ G is a composi-tion of affine and triangular automorphisms with Jacobian determinant 1. We know that we canwrite g as a composition of a ( l ) and e ( Q ) , with a correcting term s ∈ A ∩ E . Note that the a ( l ) and e ( Q ) are automorphisms with Jacobian determinant 1, so s is also of Jacobian determinant1 and we are done. (cid:3) ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 7
As a consequence of this proposition the whole discussion of the previous subsection stillapplies to G . In particular we can make the same choice of representatives a ( l ) and e ( Q ) towrite edges and vertices, so that there exists a natural bijection between the trees associated toAut [ C ] and to G . Proposition 10.
The group G is the commutator subgroup of the group Aut [ C ] , and is alsoequal to its own commutator subgroup.Proof. Using Proposition 9, it is sufficient to check that the commutator subgroup of G containsSL ( , C ) and all triangular automorphisms of the form ( x + P ( y ) , y ) . But on one hand it is well-known that SL ( , C ) is equal to its own commutator subgroup; on the other hand any triangularautomorphism ( x + l y n , y ) , with n ≥ l ∈ C , is the commutator of ( x + l ( − b ) − y n , y ) and (cid:0) bx , b − y (cid:1) , where b = n -th root of the unity. Finally, any translation ( x + c , y ) is thecommutator of ( − x , − y ) and ( x − c , y ) . (cid:3) The color.
We now introduce the color of a path of type A − E − A . This notion willbe used to make precise the genericness assumptions we need. Note that any path of type A − E − A can be written P = y e A − y E − y e A , where y ∈ Aut [ C ] and e , e ∈ E . Definition 11.
The color of P is the double coset ( A ∩ E ) e − e ( A ∩ E ) .One verifies easily that this definition does not depend on the choice of e , e . The color isclearly invariant under the action of Aut [ C ] . In fact, given two paths of type A − E − A onecould even show that one can send one to the other (by an element of Aut [ C ] ) if and only ifthey have the same color. However, we will not use this result. As an illustration of the notionof color, we can note that the color of the path e ( P ) A − e ( P ) a ( l ) E − e ( P ) a ( l ) e ( Q ) A appearingin Lemma 7 has color ( A ∩ E ) e ( Q )( A ∩ E ) .If P ∈ C [ y ] is such that the color of P is equal to the double coset ( A ∩ E ) e ( P )( A ∩ E ) , wesay that P represents the color of P . The following lemma implies that this notion does notdepend on the orientation of the path. Its proof is easy and left to the reader. Lemma 12.
Let P , Q ∈ C [ y ] be polynomials of degree ≥ . Then P and Q represent the samecolor if and only if there exist a , . . . , e with ab = such that Q ( y ) = a P ( b y + g ) + d y + e .Remark . Note that any path of type A − E − A can be sent by an automorphism to a path ofthe form id A − id E − e ( P ) A . It is easy to check that the vertices e ( P ) A and e ( Q ) A are equal ifand only if there exists a , b ∈ C such that Q ( y ) = P ( y ) + a y + b . Fundamental example 14.
Let g be a hyperbolic automorphism of geometric length lg ( g ) = l . We know that g is conjugate to an automorphism of the form f = a e . . . a l e l where each a i ∈ A \ E and each e i ∈ E \ A . Then, the geodesic of g (and f ) carries the l colors ( A ∩ E ) e i ( A ∩ E ) (1 ≤ i ≤ l ) which are repeated periodically.1.6. General color.Definition 15.
A polynomial P ∈ C [ y ] of degree d ≥ general if it satisfies: ∀ a , b , g ∈ C , deg ( P ( y ) − a P ( b y + g )) ≤ d − = ⇒ a = b = g = ( A ∩ E ) e ( P )( A ∩ E ) is said to be general if P is general. Lemma 12 implies thatthis notion does not depend on the choice of a representative P . Lemma 16.
Let Q ∈ y C [ y ] be general. The stabilizer of the path P = e ( Q ) A − id E − id A isequal to { ( x + b y + g , y ) ; b , g ∈ C } . Furthermore, if b = , the automorphism ( x + b y + g , y ) does not fix any path strictly containing P . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 8
Proof.
We know that f ∈ Aut [ C ] fixes the path id E − id A if and only if f ∈ A ∩ E . In thiscase, there exists constants a , . . . , z , with ae = f = ( a x + b y + g , e y + z ) . Since f e ( Q ) = e ( Q ) g , where g = ( a x + b y + a Q ( y ) − Q ( e y + z ) , e y + z ) , the vertex e ( Q ) A is fixed by f if and only if g ∈ A , i.e. deg ( a Q ( y ) − Q ( e y + z )) ≤
1. The polynomial Q being general, thisis equivalent to a = e = z = ( x + b y + g , y ) a ( l ) E = a ( l − b ) E . Indeed, since ( x + b y + g , y ) e ( Q ) = e ( Q )( x + b y + g , y ) , we also have ( x + b y + g , y ) e ( Q ) a ( l ) E = e ( Q ) a ( l − b ) E . Therefore, the vertices a ( l ) E and e ( Q ) a ( l ) E are fixed by ( x + b y + g , y ) if and only if b = (cid:3) Remark . Lemma 16 is a kind of converse to Lemma 7. Precisely, we obtain that if f fixes ageneral path of 4 edges centered on id E , then f = ( x + c , y ) (Here by general we mean that thecolor supported by the two central edges of the path is general; see Def. 11 and below).Note also that since ( x , y + c ) = a ( ) ◦ ( x − c , y ) ◦ a ( ) − , the subset of T fixed by ( x , y + c ) is the image by a ( ) of the subset fixed by ( x − c , y ) . In particular, it contains the closed ball ofradius 2 centered at a ( ) E . Furthermore, if f fixes a general path of 4 edges centered at a ( ) E ,it can be written as f = ( x , y + c ) .We now apply the notion of a general color to prove a technical result that we need to proveTheorem 45. We consider a hyperbolic automorphism f and g = j f j − = f a conjugate of f .We want to show that if f is sufficiently general then Geo ( f ) ∩ Geo ( g ) is a path of length atmost 4. More precisely, we also describe all possibles types of such paths. Definition 18.
We say that a hyperbolic automorphism of geometric length 2 l satisfies condi-tion ( C ) if the l colors supported by its geodesic (see Example 14) are general and distinct.In the annex we show that this condition is generic in a natural sense. Proposition 19.
Let f and g = f f f − be two distinct conjugate automorphisms satisfyingcondition ( C ) . If the intersection Geo ( f ) ∩ Geo ( g ) contains at least one edge then this path isof type: A − E , E − A − E , A − E − A , or E − A − E − A − EProof.
There is no restriction to assume that P ′ = Geo ( f ) ∩ Geo ( g ) = Geo ( f ) ∩ f ( Geo ( f )) contains a path of type A − E − A , because otherwise P ′ is at most a path of type E − A − E .Let us call v the central vertex of type E of this subpath of P ′ . Since f − ( v ) ∈ Geo ( f ) , thereexists an integer k such that dist ( f k ( v ) , f − ( v )) = dist (( f f k )( v ) , v ) < lg ( f ) = l . Replacing f by f f k , we do not change g , but we now have dist ( f ( v ) , v ) < l . By condition ( C ) , thegeodesic of f carries l distinct colors which are repeated periodically. Therefore, dist ( f ( v ) , v ) ∈ l Z and finally we get f ( v ) = v , so that f is elliptic.Let us set P = f − ( P ′ ) = Geo ( f ) ∩ f − ( Geo ( f )) . Equivalently, one may define P as themaximal path such that P ⊆ Geo ( f ) and f ( P ) ⊆ Geo ( f ) .The path P contains a path of type A − E − A whose central vertex is v . Without loss ofgenerality, one can now conjugate and assume that this subpath is of the form e ( Q ) A − id E − id A . In particular v = id E .There are two subcases, depending on whether f : P → f ( P ) preserves the orientation in-duced by Geo ( f ) . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 9 If f preserves this orientation, then f fixes P point by point. We may assume that P isstrictly greater than e ( Q ) A − id E − id A , because otherwise there is nothing to show. Then, byLemmas 7 and 16, we get f = ( x + g , y ) . Since the colors of Geo ( f ) are general, Lemma 7shows us that P is of the form e ( Q ) a ( l ) E − e ( Q ) A − id E − id A − a ( µ ) E , so that it is of type E − A − E − A − E .If f does not preserve this orientation, then f fixes only the vertex v of Geo ( f ) . One canshow that f has to be an involution (see Lemma 20 below). This implies that P contains aneven number of edges and is centered on v . Since the l colors supported by Geo ( f ) are distinct, P contains only one color, so that it is of type A − E − A or E − A − E − A − E . (cid:3) Lemma 20.
Let P be a path of type A − E − A carrying a general color. If f ∈ Aut [ C ] exchanges the two ends of P then f = id .Proof. Without loss of generality, one can conjugate and assume that the path P is of theform e ( Q ) A − id E − id A (see Remark 13). Note that f = e ( Q ) ◦ ( − x , y ) is an involution thatexchanges the two vertices e ( Q ) A and id A . Thus f f fixes the path P point by point, and since Q is general by Lemma 16 we get f = f ◦ ( x + b y + g , y ) . Remark that f ◦ ( x + b y + g , y ) =( x + b y + g , y ) − ◦ f , hence f = f ◦ ( x + b y + g , y ) ◦ ( x + b y + g , y ) − ◦ f = id . (cid:3) Example 21.
Here we show that all cases allowed by Proposition 19 can be realized. In thefollowing examples we suppose that Geo ( f ) contains the path a ( ) E − id A − id E − e ( Q ) A − e ( Q ) a ( µ ) E where Q is a general polynomial and we choose f such that the path P has variousforms.(1) Examples with f fixing at least one edge: • f = ( x + P ( y ) , y ) with deg P ≥ P = id A − id E ; • f = ( a x , b y ) with ab = ( a , b ) = ( , ) , P = a ( ) E − id A − id E ; • f = ( x + by , y ) with b = P = id A − id E − e ( Q ) A ; • f = ( x + c , y ) with c = P = a ( ) E − id A − id E − e ( Q ) A − e ( Q ) a ( µ ) E .(2) Examples with f reversing the orientation: • f = ( y , x ) exchanges a ( ) E and id E , P = a ( ) E − id A − id E ; • f = ( − x + Q ( y ) , y ) exchanges id A and e ( Q ) A , P is of length 4 or 2 depending if µ = f hyperbolic: • f = e ( Q ) a ( µ ) u with u = ( − x , − y ) sends P = a ( ) E − id A − id E to f ( P ) = id E − e ( Q ) A − e ( Q ) a ( µ ) E (the reader should verify that a ( µ ) ua ( ) = ( x − µy , y ) ∈ A ∩ E ).1.7. Independent colors and tripods.Definition 22.
A family of polynomials P i ∈ C [ y ] (1 ≤ i ≤ l ) is said to be independent if givenany a k , b k , g k ∈ C with a k b k = i k ∈ { , . . . , l } , for 1 ≤ k ≤
3, we have:deg (cid:229) ≤ k ≤ a k P i k ( b k y + g k ) ≤ = ⇒ i = i = i . The family of colors ( A ∩ E ) e ( P i )( A ∩ E ) (1 ≤ i ≤ l ) is said to be independent if the family P i (1 ≤ i ≤ l ) is independent. Lemma 12 implies that this notion does not depend on the choiceof the representatives P i . Definition 23.
Three paths P , P , P of the tree T define a tripod if ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 10 • For each i = j , P i ∩ P j contains at least one edge; • The intersection P ∩ P ∩ P consists of exactly one vertex v .The three paths P i ∩ P j are called the branches of the tripod. The vertex v is called the center of the tripod.If we have a center of type E , we can consider the three colors associated with the threepaths of type A − E − A containing the center and included in the tripod. In this situation wesay that any one of these colors is a mixture of the two other ones. Lemma 24.
Let P , P , P ∈ C [ y ] be polynomials of degree ≥ . The following assertions areequivalent:(1) ( A ∩ E ) e ( P )( A ∩ E ) is a mixture of the ( A ∩ E ) e ( P i )( A ∩ E ) ’s ( ≤ i ≤ );(2) ∃ a , b , g , a , b , g , d , e ∈ C with a b a b = such thatP ( y ) = a P ( b y + g ) + a P ( b y + g ) + d y + e . Proof. (1) = ⇒ (2). Assume that there exists a tripod admitting the 3 colors ( A ∩ E ) e ( P i )( A ∩ E ) (1 ≤ i ≤ E and that one of its branch is id E − id A . Let e P , e P ∈ C [ y ] be such that the 2 other branches are id E − e ( e P ) A , and id E − e ( e P ) A and such that ( A ∩ E ) e ( P )( A ∩ E ) = ( A ∩ E ) e ( e P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) = ( A ∩ E ) e ( e P )( A ∩ E ) .By Lemma 12, for 1 ≤ i ≤
2, there exists a i , b i , g i , d i , e i with a i b i = e P i = a i P i ( b i y + g i ) + d i y + e i .We then have ( A ∩ E ) e ( P )( A ∩ E ) = ( A ∩ E ) e ( e P )( A ∩ E ) , where e P = e P − e P , so (still byLemma 12) this shows that P has the desired form.(2) = ⇒ (1). Set e P = a P ( b y + g ) , e P = − a P ( b y + g ) and e P = e P − e P = a P ( b y + g ) + a P ( b y + g ) . By Lemma 12, we have ( A ∩ E ) e ( e P i )( A ∩ E ) = ( A ∩ E ) e ( P i )( A ∩ E ) for 1 ≤ i ≤
3. Since e ( e P ) − e ( e P ) = e ( e P ) / ∈ A , the vertices e ( e P ) A and e ( e P ) A are distinct.Consider the tripod with center id E and branches id E − id A , id E − e ( e P ) A and id E − e ( e P ) A .Its three colors are ( A ∩ E ) e ( e P i )( A ∩ E ) for 1 ≤ i ≤
3. This shows that ( A ∩ E ) e ( P )( A ∩ E ) is amixture of ( A ∩ E ) e ( P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) . (cid:3) Remark . The second condition of Lemma 24 may be written under the following symmetricform:For 1 ≤ k ≤
3, there exists a k , b k , g k ∈ C with a k b k = (cid:229) ≤ k ≤ a k P k ( b k y + g k ) ≤ . Therefore, the following lemma is an easy consequence of the previous one.
Lemma 26.
Consider three colors represented by P , P , P ∈ C [ y ] which are polynomials ofdegree ≥ . The following assertions are equivalent:(1) the three colors ( A ∩ E ) e ( P i )( A ∩ E ) (i = , , ) are independent;(2) For any i , i , i ∈ { , , } , if ( A ∩ E ) e ( P i )( A ∩ E ) is a mixture of ( A ∩ E ) e ( P i )( A ∩ E ) and ( A ∩ E ) e ( P i )( A ∩ E ) , then i = i = i . Definition 27.
We say that a hyperbolic automorphism of geometric length 2 l satisfies condi-tion ( C ) if the l colors supported by its geodesic (see Example 14) are general and indepen-dent.In the annex we show that this condition is generic in a natural sense. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 11
Remark . One could easily check that independent colors are necessarily distinct. Therefore,condition ( C ) is stronger than condition ( C ) .By misuse of language, we will say that three hyperbolic automorphisms g , g , g define a tripod if their geodesics Geo ( g ) , Geo ( g ) , Geo ( g ) define a tripod. Lemma 29.
A tripod associated with three conjugates of a hyperbolic automorphism f satis-fying condition ( C ) admits branches of length at most 2.Proof. If the center of the tripod is of type A , by Proposition 19 there is nothing to do. Assumenow that the center of the tripod is of type E . Without loss of generality one can conjugate andassume that the center is id E , and that Geo ( f ) contains the vertices id A and a ( ) E . We denoteby g = u f u − and h = v f v − the two conjugates of f involved in the tripod.By condition ( C ) the three colors centered on id E in the tripod must be equal. Indeed, if ( A ∩ E ) e ( P i )( A ∩ E ) , 1 ≤ i ≤ l are the l colors supported by Geo ( f ) , then there exist i , i , i ∈{ , . . . , l } such that these three colors are ( A ∩ E ) e ( P i k )( A ∩ E ) , 1 ≤ k ≤
3. By Definition 22and Lemma 24 (see also Remark 25), we get i = i = i , so that the three colors are equal.Let us prove that u can be chosen fixing the center a = id E of the tripod. Since a ∈ Geo ( f ) ∩ Geo ( g ) = Geo ( f ) ∩ u ( Geo ( f )) , we get u − ( a ) ∈ Geo ( f ) , so that there exists aninteger k such that dist ( f k ( a ) , u − ( a )) < lg ( f ) = l . Replacing u by u f k , we do not change g , but we now have dist ( u ( a ) , a ) < l . By condition ( C ) (cf. Remark 28), the geodesic of g carries l distinct colors which are repeated periodically. Therefore, dist ( u ( a ) , a ) ∈ l Z andfinally we get u ( a ) = a . We would prove in the same way that v can be chosen fixing a = id E .In other words, we have u , v ∈ E .Let us now assume that there exists a branch, say Geo ( f ) ∩ Geo ( h ) , of length strictly greaterthan 2. Then, by Proposition 19, this branch has length 4, with middle point a ( ) E (see Fig. 1).Since v fixes point by point the general path Geo ( f ) ∩ Geo ( h ) , by Remark 17, it can be writtenas v = ( x , y + c ) .Let e = e ( P ) = ( x + P ( y ) , y ) ∈ E be such that the vertex eA ∈ Geo ( f ) ∩ Geo ( g ) . SinceGeo ( h ) = v ( Geo ( f )) , the vertex veA ∈ Geo ( h ) and finally veA ∈ Geo ( g ) ∩ Geo ( h ) .We assume that the orientation induced by g on id E − eA is opposite to the one of f , theother case being symmetric. •• •• • • • • • eA ??????? veA (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) id E id A a ( ) e ( Q ) Aa ( ) e ( Q ) a ( l ) Ea ( ) E ??????? ??????? f d d h z z g (cid:2) (cid:2) F IGURE ( g ) = u ( Geo ( f )) , u sends the path id A − id E − eA to the path eA − id E − veA .On one hand, u sends id A to eA , i.e. uA = eA , i.e. e − u ∈ A , i.e. e − u ∈ A ∩ E . Since e − u ∈ A ∩ E , it can be written as s s , where s = ( a x , b y + c ) , s = ( x + b y + g , y ) ∈ A ∩ E and we have u = es s . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 12
On the other hand u sends eA to veA , i.e. ueA = veA , i.e. es s eA = veA . Since s e = es , wehave es s eA = es eA , so that es eA = veA . This last equality is still equivalent to e − v − es e ∈ A . We compute e − v − es e = ( a x + a P ( y ) + P ( b y + c ) − P ( b y + c − c ) , b y + c − c ) . We should have deg ( a P ( y )+ P ( b y + c ) − P ( b y + c − c )) ≤
1. Since a = ( P ( b y + c ) − P ( b y + c − c )) < deg P , this is impossible. (cid:3)
2. T
HE PROOF OF T HEOREM ≤
1, i.e. a triangularor affine automorphism. Note that similar results in the context of birational transformationsare proved in [Giz94] and [CD08].
Lemma 30.
If f ∈ G satisfies | f | ≤ and f = id , then h f i N = G.Proof.
Let g , h ∈ G . Note that if g or h belongs to h f i N , then so does the commutator [ g , h ] = ghg − h − . We show that G = h f i N by making the following observations: • If f ∈ SL ( , C ) and f = ± id, we obtain SL ( , C ) ⊆ h f i N . We used the fact that {± id } isthe unique nontrivial normal subgroup of SL ( , C ) . Indeed, if H is a normal subgroupof SL ( , C ) not included into {± id } , we get SL ( , C ) = H ∪ ( − H ) by simplicity ofPSL ( , C ) . Therefore, if g = ( y , − x ) , we get g ∈ H or − g ∈ H , so that − id = g =( − g ) ∈ H and finally H = SL ( , C ) .Now, if a , b ∈ C , we get [( x + a , y + b ) , ( − x , − y )] = ( x + a , y + b ) so that A ⊆ h f i N .If b = n -th root of the unity ( n ≥
2) and l ∈ C , we get [( x + l ( − b ) − y n , y ) , ( bx , b − y )] = ( x + l y n , y ) and we are done. • If f is a translation, then, conjugating by SL ( , C ) , we see that h f i N contains all trans-lations. So, it contains the commutator [( x , y + ) , ( x + y , y )] = ( x − y + , y ) and also the linear automorphism ( x − y , y ) . We conclude by the previous case. • If f is an affine automorphism which is not a translation, then there exists a translation g which does not commute with f . Therefore, the commutator [ f , g ] is a nontrivialtranslation belonging to h f i N and we conclude by the previous case. • Finally if f = ( ax + P ( y ) , a − y + c ) is a triangular non affine automorphism, then, up toreplacing f by [ f , g ] , where g is a triangular automorphism non commuting with f , wemay assume that a =
1. Still replacing f by [ f , g ] , where g is a triangular automorphismnon commuting with f , we may even assume that c =
0. Therefore, f is of the form ( x + P ( y ) , y ) . Remark then that the commutator [( x , y + ) , ( x + P ( y ) , y )] is a triangular automorphism of the form ( x + R ( y ) , y ) , with deg R = deg P −
1. Byinduction on the degree we obtain the existence of a nontrivial translation ( x + c , y ) in h f i N . This case has already been done. (cid:3) ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 13
Corollary 31.
If f ∈ G is elliptic (i.e. triangularizable) and f = id , then h f i N = G. We are now ready to prove Theorem 1. In fact, we will prove the following stronger andmore geometric version:
Theorem 32.
If f ∈ G satisfies lg ( f ) ≤ and f = id , then h f i N = G.Proof.
The crucial fact we use here is the knowledge of the subtree fixed by translations ( x + c , y ) . We know that this subtree is of diameter 6, centered in id E , and that the closed ball ofradius 2 and center id E is contained in this subtree (see Lemma 7). In consequence, given anarbitrary path of type E − A − E − A − E , there exists a conjugate y of ( x + , y ) fixing this pathpoint by point.Let us choose such a path contained in the geodesic of f and let us set g = y f y − . Then iflg ( f ) = f ◦ g − is elliptic, so we can conclude by Corollary 31. If lg ( f ) = ( f ◦ g − ) ≤ ( f ) = f by one of its conjugates, we may assume | f | = lg ( f ) =
8. We can then assume (maybereplacing f by f − ) that f = e a e a e a e a where a i ∈ A \ E , e j ∈ E \ A . Without loss of generality we can further assume that each e j isof the form e j = e ( P j ) = ( x + P j ( y ) , y ) and that deg ( e ) ≤ deg ( e j ) for j = , , ( x + c , y ) fixes the closed ball of radius 2 and center id E . Notealso that for any s ∈ A ∩ E , s ( x + , y ) s − is still a translation of the form ( x + c , y ) . In conse-quence, if we write e a under the form e a = e ( P ) a ( l ) s with s ∈ A ∩ E , the automorphism˜ e = e a ( x + , y ) a − e − = ( x + P ( y ) , y ) ◦ ( l x + y , − x ) ◦ ( x + c , y ) ◦ ( − y , l y + x ) ◦ ( x − P ( y ) , y )= ( x + l c + P ( y − c ) − P ( y ) , y − c ) fixes the closed ball of radius 2 and center e a E . Note that deg ˜ e = deg e −
1. Consider g = ˜ e f ˜ e − and h = g − f . By construction the geodesics Geo ( g ) and Geo ( f ) have at least 4 edges in common. By Lemma7 we also know that they have at most 6 edges in common. Then we can check (see Fig. 2)that h sends the vertex v = a − e − a − E to a vertex at distance at most 8 (and at least 6) of v .Explicitly, one can compute h = ˜ e a − e − a − ˜ e a e a where ˜ e = e − a − ( x − , y ) a e is a triangular automorphism with deg ( ˜ e ) = deg ( e ) − ( ˜ e ) = deg ( ˜ e ) = ( h ) =
4. This corresponds to the case when Geo ( g ) andGeo ( f ) share 6 edges. Note that a − ˜ e a and a ˜ e a − are indeed non triangular affine auto-morphisms.If deg ( ˜ e ) = ( ˜ e ) ≥ ( h ) =
6. In this case Geo ( g ) and Geo ( f ) share 5 edges: the vertices id A and ˜ e A coincide.In the two cases above we are done by the first part of the proof.Finally if deg ( ˜ e ) ≥ h admits a factorization similar to the one of the f we startedwith except that the first triangular automorphism has a strictly smaller degree. By induction,we can produce an element of length 8 in h f i N with the first triangular automorphism of degree2, and we are done by the previous argument. (cid:3) ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 14 ••• •• • • • • • •••• id A OOOOOO ˜ e A oooooo id E e A e a E = e ( P ) a ( l ) Ev = a − e − a − E OOOOOO OOOOOO f ( v ) e a e a e A oooooooooooo h ( v )= g − ( f ( v )) oooooo OOOOOOoooooo f h (cid:2) (cid:2) g - - F IGURE
2. Proof of Theorem 323. R-
DIAGRAMS
Generalities on small cancellation theory.
In this subsection we consider H = H ∗ H ∩ H H a general amalgamated product of two factors. Of course our motivation is to apply the the-ory to the group Aut [ C ] of plane automorphisms.The following definitions are taken from [LS01], chap. V, §11 (p. 285). If u is an elementof H , not in the amalgamated part H ∩ H , a normal form of u is any sequence x · · · x m suchthat u = x · · · x m , each x i is in a factor of H , successive x i come from different factors of H , andno x i is in the amalgamated part. The length of u is defined by | u | = m . This definition doesnot depend on the chosen normal form, but only on u . If u is in the amalgamated part of H , byconvention we set | u | = word an element u ∈ H given with a factorization u = u · · · u k , where u i ∈ H for i = , · · · , k . A word u = u · · · u k is said to have reduced form if | u · · · u k | = | u | + · · · + | u k | .Suppose u and v are elements of H with normal forms u = x · · · x m and v = y · · · y n . If x m y is in the amalgamated part, we say that there is cancellation between u and v in forming theproduct uv . Equivalently, this means that | uv | ≤ | u | + | v | −
2. If x m and y are in the samefactor of H and x m y is not in the amalgamated part, we say that x m and y are consolidated informing a normal form of uv . Equivalently, this means that | uv | = | u | + | v | − semi-reduced form u · · · u k if there is no cancellation in this product.Consolidation is expressly allowed.A word u = x · · · x m in normal form is strictly (resp. weakly ) cyclically reduced if m ≤ x m and x are in different factors of H (resp. the product x m x is not in the amalgamatedpart). These two notions correspond to the two sets of equivalent conditions given in Lemma 8A subset R of H is symmetrized if all elements of R are weakly cyclically reduced and foreach r ∈ R , all weakly cyclically reduced conjugates of both r and r − belong to R .If f is strictly cyclically reduced, R ( f ) denotes the symmetrized set generated by f , i.e. thesmallest symmetrized set containing f . It is clear that R ( f ) is equal to the set of conjugates of f ± of length ≤ | f | + C ′ ( l ) (mostly used with l = / ) . We do not need thisnotion in our construction, but this was the original setting where the notion of R-diagram (seenext subsection) was introduced. Let R be a symmetrized subset of H . A word b is said to be a piece (relative to R ) if there exists distinct elements r , r of R such that r = bc and r = bc in semi-reduced form. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 15
Lemma 33. If < l < and ∀ r ∈ R, | r | > / l , the following assertions are equivalent:(1) If r ∈ R admits a semi-reduced form r = bc, where b is a piece of R, then | b | < l | r | ;(2) ∀ r , r ∈ R such that r r = , | r r | > | r | + | r | − l min {| r | , | r |} + .Proof. The equivalence is easily obtained from the following claim.Let r = bc and r = bc be semi-reduced expressions with b = r = r .Claim. There exists b ′ , c ′ , c ′ such that:a) the equalities r = b ′ c ′ and r = b ′ c ′ hold;b) these expressions are semi-reduced;c) exactly one of these expressions is reduced;d) the expression ( c ′ ) − c ′ is reduced;e) | b ′ | ≥ | b | . (cid:3) Definition 34.
When the equivalent assertions of Lemma 33 are satisfied, we say that R satisfiescondition C ′ ( l ) .The first assertion is the one used by Lyndon and Schupp. The second one is used by Danilov,except that he forgets the + Theorem 35.
Let R be a symmetrized subset of the amalgamated group H. Suppose thatR satisfies condition C ′ ( l ) with l ≤ / , then the normal subgroup generated by R in H isdifferent from H. Construction of an R-diagram.
The idea of associating diagrams in the Euclidean planeto some products in amalgamated groups appears in [VK33].In 1966, Lyndon independently arrived at the same idea and Weinbaum rediscovered vanKampen’s paper (see [Lyn66, Wei66] and [LS01], p. 236). For the basic definition of a dia-gram , we refer to [LS01], chap. V, §1, p. 235. Here follows a quick review of this notion.A diagram is a plane graph (or more generally a graph on an orientable surface, we willconsider spherical diagrams in Lemma 42). Vertices are divided into two types, primary and secondary . Any edge joining two vertices gives rise to two directed edges (according to thechosen directions) which we call half-segments . If e denotes one of these half-segments, e − will refer to the other one (obtained by reversing the direction of e ). The notation ’edge’ will beused later on to refer to some special unions of half-segments (see the remark on terminologybelow). A half-segment will always join vertices of different types. By definition, segments will denote some special successions of two half-segments that we now describe. If e , . . . , e r are the half-segments arriving at some secondary vertex v and taken counterclockwise, then,by definition, the segments passing through v are the successive half-segments e i , e − i + and theirinverses e i + , e − i , for 1 ≤ i ≤ r , where i and i + r . If two successive half-segments e , e ′ define a segment, the latter will be noted ee ′ . Note that the initial and terminalvertices of a segment have to be primary. By convention, each segment (resp. half-segment)has length 1 (resp. 1 / e will be labeled by an element f ( e ) be-longing to a factor of Aut [ C ] , with the labels on successive half-segments at a secondary vertexbelonging to the same factor. The identity f ( e − ) = f ( e ) − is required. This labelling gives alabelling on segments, by taking f ( ee ′ ) = f ( e ) f ( e ′ ) . The label on an individual half-segmentmay be in the amalgamated part, but if e , e ′ are the two half-segments of a segment, we will ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 16 usually insist that f ( ee ′ ) is not in the amalgamated part (in fact, there will be only one excep-tion to this rule, see step 4 in the proof of Theorem 36). We call region a bounded connectedcomponent of the complement of the graph in the surface. A boundary cycle of a region D isa collection of half-segments that run along the entire boundary of D (say counterclockwise inthe case of the plane, or in a way compatible with the orientation in general) with initial vertexof primary type. Similarly, a boundary cycle of the diagram is a collection of half-segmentsthat run along the boundary of the diagram. Let us note that a segment necessarily belongs tothe boundary of some region and/or to the boundary of the diagram.Now let f be an element of Aut [ C ] and consider R ( f ) the associated symmetrized set. Wesay that a diagram is a R ( f ) -diagram if for any region D and any boundary cycle e . . . e s of D ,we have f ( e ) . . . f ( e s ) ∈ R ( f ) . Terminology.
Note that we use two kinds of graph in this paper: the Bass-Serre tree and thediagrams of Lyndon-Schupp. In the context of the Bass-Serre tree we have already used theterm edge , and we have called a path the union of several edges. In the context of Lyndon-Schupp diagrams, we have segments and half-segments . We call edge in this context a con-nected component of the intersection of the boundary of two regions, which is a collection ofhalf-segments.The following result will be the key ingredient for the proof of Theorem 2. Its proof willoccupy the rest of this subsection.
Theorem 36.
Let f ∈ G be a strictly cyclically reduced element of G of (even) algebraic length | f | ≥ . Assume that the normal subgroup generated by f in Aut [ C ] is equal to G. Then thereexists a planar R ( f ) -diagram M such that:(1) M is connected and simply connected;(2) The boundary of M has length or ;(3) If e e ′ . . . e t e ′ t is a boundary cycle of some region of M, then t = | f | and f ( e e ′ ) . . . f ( e t e ′ t ) is a reduced form of a strictly cyclically reduced conjugate of f .Proof. We start by choosing an element g = id with lg ( g ) =
0. By assumption we can write g = ( f f ± f − ) · · · ( f n f ± f − n ) . with f i ∈ Aut [ C ] .We assume that we have chosen g such that n is minimal. By Lemma 37, we may assume thateach f i f ± f − i is expressed under reduced form y i r i y − i (i.e. | y i r i y − i | = | y i | + | r i | + | y − i | )where r i ∈ R ( f ) . There is no restriction to assume that | y i | = y i = id. Notealso that the four following assertions are equivalent:a) r i is strictly cyclically reduced; b) | r i | = | f | ; c) | r i | is even; d) | y i r i y − i | is even.If any one of these assertions is satisfied, we necessarily have y i = id (since the expression y i r i y − i is reduced).Let us now explain the construction of M , that we perform in several steps:Step 1. We associate a diagram to each y i r i y − i .Our construction will involve a base point O which will be considered as a primary vertex.Let r i = x . . . x m be a normal form of r i . • Assume that r i is strictly cyclically reduced, i.e. m = | f | . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 17
The diagram for y i r i y − i = r i is the loop at the base point O consisting of 2 m half-segments d , d ′ , . . . , d m , d ′ m such that f ( d j d ′ j ) = x j for each j . • Assume that r i is not strictly cyclically reduced, i.e. m = | f | +
1. Note that in this case ( x m x ) x · · · x m − is strictly cyclically reduced.The diagram for y i r i y − i is a loop at a vertex v joined to the base point O by a path.Let y i = z . . . z k be a normal form of y i .The path Ov consists of 2 k half-segments e , e ′ , . . . , e k , e ′ k such that f ( e j e ′ j ) = z j for each j and an additional final half-segment e .The loop at v consists of 2 m − b , d , d ′ , . . . , d m − , d ′ m − , c such that f ( d j d ′ j ) = x j for each j .The three half-segments e , b , c which meet at the secondary vertex v are labeled to satisfythe necessary (and compatible) conditions f ( eb ) = x , f ( ce − ) = x m and f ( cb ) = x m x . Forinstance we can take f ( b ) = x , f ( c ) = x m and f ( e ) = id.Step 2. The initial diagram for the composition g = ( y r y − ) · · · ( y n r n y − n ) consists of the initial diagrams for each y i r i y − i arranged, in counterclockwise order, aroundthe base point O . This initial diagram has the desired properties (1) and (3).Step 3. We will now proceed to the identification of some half-segments of M until theboundary length of M is ≤ • We shall always identify primary vertices with primary vertices and secondary verticeswith secondary vertices, preserving this distinction; • The label of a segment will never be in the amalgamated part; • The number n of regions of M will not change and (1) and (3) will be satisfied at eachstage; • If a is a boundary cycle of M , then f ( a ) is conjugate to g .For grounds of brevity, the tiresome and easy verification of the second point (on label ofsegments) has been omitted in the two cases below.If the boundary length of M is ≥
3, there necessarily exists successive segments ee ′ and f f ′ in ¶ M such that the labels f ( ee ′ ) and f ( f f ′ ) are in the same factor of Aut [ C ] . Indeed,otherwise, any boundary cycle a = e e ′ . . . e i e ′ i of M would have even length i ≥ f ( a ) = f ( e e ′ ) . . . f ( e i e ′ i ) would be a strictly cyclically reduced conjugate of g : A contradiction.So we consider the element s = f ( ee ′ ) f ( f f ′ ) which lies in a factor of Aut [ C ] . Case 1:
Assume that s is not in the amalgamated part.Change the label on the half-segment e ′ to 1, readjusting the labels on the other half-segments at the secondary vertex separating e and e ′ . In other words this amounts, foreach half-segment g ending at this secondary vertex, to replace its label f ( g ) by f ( ge ′ ) .In the same way, change the label on the half-segment f to 1, readjusting the labelson the other half-segments at the secondary vertex separating f and f ′ .Then we identify the (oriented) half-segments e ′ and f − (which now have the samelabels) (see Fig. 3, where the • are primary vertices and the ◦ are secondary vertices). ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 18 • • •◦ ◦• • / / f ( e ) / / f ( e ′ ) / / f ( f ) / / f ( f ′ ) G G (cid:15)(cid:15)(cid:15)(cid:15) f ( g ) (cid:15)(cid:15)(cid:15)(cid:15) (cid:23) (cid:23) //// f ( h ) //// • • •◦ ◦• • / / f ( ee ′ ) / / / / / / f ( f f ′ ) G G (cid:15)(cid:15)(cid:15)(cid:15) f ( ge ′ ) (cid:15)(cid:15)(cid:15)(cid:15) (cid:23) (cid:23) //// f ( f h ) //// • • •◦• • / / f ( ee ′ ) / / f ( f f ′ ) C C (cid:6)(cid:6)(cid:6)(cid:6) f ( ge ′ ) (cid:6)(cid:6)(cid:6)(cid:6) (cid:27) (cid:27) f ( f h ) relabel / / identify (cid:10) (cid:10) F IGURE
3. Relabellings and identifications in case 1.
Case 2:
Assume that s is in the amalgamated part.Note first that the diagram has no loop of length ≤ [ C ] . Indeed, such a loop a would be a boundary cycle of some strictlysmaller subdomain, so that, by Lemma 38 below, f ( a ) would be the product of strictlyless that n conjugates of f . This would contradict the minimality of n .Therefore, if u is the initial vertex of ee ′ , v its terminal vertex (as well as the initialvertex of f f ′ ) and w the terminal vertex of f f ′ , then the vertices u , v , w are distinct.Recall that f ( f ) f ( f ′ ) = f ( e ′ ) − f ( e ) − s . We change the labels in the following way(see Fig. 4): • we change the label of f to f ( e ′ ) − , readjusting the labels on the other half-segments at the secondary vertex separating f and f ′ ; • we change the label of f ′ to f ( e ) − ; • for each half-segment g having w as initial vertex, we replace its label f ( g ) by s f ( g ) .Then we identify the (oriented) segments e , e ′ and f ′− , f − (which now have the samelabels).Note that after performing the identification in case 1 (resp. in case 2) the boundary lengthdrops by 1 (resp. by 2). Note also that if two regions D and D share at least one half-segment,and if r , r are two boundary cycles of these regions with respect to a common starting point,then we can not have r = r − . Indeed, if it was the case, removing the two regions from the di-agram and applying Lemma 38 we would obtain a new element in R ( f ) that would contradictsthe minimality of n . In fact, by Lemma 39, two regions in the diagram never share an edge oflength greater than 4.Step 4. By induction, the previous step gave us a diagram with a boundary length ≤
2. Wenow perform one last more identification to obtain that the boundary length of M is at most 1.If the last identification falls under case 1 there is no particular problem. However, if we arein case 2, then we can no longer assume that the vertices u and w are disjoint. So we slightlymodify the procedure: we keep the label of f ′ to be f ( e ) − s and we only identify the half-segments e ′ and f . It may happen that after this identification the label of the segment e f ′ on ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 19 • • •◦ ◦ ◦• u / / f ( e ) / / f ( e ′ ) v / / f ( f ) / / f ( f ′ ) w / / f ( g ) (cid:23) (cid:23) //// f ( h ) //// • • •◦ ◦ ◦• u / / f ( e ) / / f ( e ′ ) v / / f ( e ′ ) − / / f ( e ) − w / / s f ( g ) (cid:23) (cid:23) //// f ( e ′ f h ) //// ••◦ ◦• u = w (cid:15) (cid:15) f ( e ) v (cid:15) (cid:15) f ( e ′ ) / / s f ( g ) / / f ( e ′ f h ) relabel identify (cid:12) (cid:12) F IGURE
4. Relabellings and identifications in case 2.the boundary of M is in the amalgamated part: apart from being slightly non aesthetic, this willnot be a problem in the proof of Theorem 45. (cid:3) Lemma 37.
Any conjugate of f (notation as in Theorem 36) can be written under reducedform y r y − , where r is a weakly cyclically reduced conjugate of f .Proof. Recall that a hyperbolic element of Aut [ C ] is strictly (resp. weakly) cyclically reducedif and only if its geodesic contains (resp. intersects) the edge e = id ( A ∩ E ) in the Bass-Serretree (see Lemma 8). Let now g be a conjugate of f . If the geodesic of g intersects e , we canjust set y = id, r = g . Therefore, let us assume that this geodesic does not intersect e .Let dist be the natural distance on the Bass-Serre tree and I be the middle of the edge e . Forany element h of G , we have | h | = dist ( I , h ( I )) .Let p ∈ Geo ( g ) be the unique vertex such that dist ( Geo ( g ) , e ) = dist ( p , e ) . Since dist ( p , e ) ≥
1, there exists a unique point I ′ on the geodesic [ p , I ] such that dist ( p , I ′ ) = . The group G acting transitively on the middles of the edges of the Bass-Serre tree, there exists an element y of G such that y ( I ) = I ′ . Let us set r = y − g y . We have Geo ( r ) = y − ( Geo ( g )) and d ( Geo ( g ) , I ′ ) = , so that dist ( Geo ( r ) , I ) = and Geo ( r ) meets e , i.e. r is weakly cyclicallyreduced. Finally, we have | g | = dist ( I , g ( I )) = lg ( g ) + ( I , Geo ( g )) = | f | + ( I , I ′ ) + | y | = dist ( I , I ′ ) and | r | = | f | +
1, so that | g | = | y | + | r | + | y − | . (cid:3) The following result can be proven similarly as Lemma 1.2 in [LS01, p. 239] (that is, byinduction on the number m of regions). Lemma 38.
Let M be an oriented connected and simply connected diagram with m regionsD , . . . , D m . Let a be a boundary cycle of M (beginning at some vertex of ¶ M) and let b i be aboundary cycle of D i (beginning at some vertex of ¶ D i ), for ≤ i ≤ m. Then f ( a ) belongs tothe normal subgroup generated by the f ( b i ) , ≤ i ≤ m. More precisely, there exists u , . . . , u m in Aut [ C ] such that f ( a ) = ( u f ( b ) u − ) . . . ( u m f ( b m ) u − m ) . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 20
A dictionary between Bass-Serre and Lyndon-Schupp theories.
Let a be a boundarycycle of some region of M (as in Theorem 36) beginning at some vertex v . If v is primary (resp.secondary), f ( a ) is a reduced form of a strictly cyclically reduced (resp. non strictly cyclicallyreduced) element of R ( f ) . Lemma 39.
If D , D are two distinct regions of a diagram M having a common edge, thereexists a primary vertex v of ¶ D ∩ ¶ D such that the labels g , g of the boundary cycles ofD , D beginning at v satisfy | Geo ( g ) ∩ Geo ( g ) | ≥ | ¶ D ∩ ¶ D | .Proof. If k is the largest integer such that k < | ¶ D ∩ ¶ D | , there exists a path of k segments s , . . . , s k included into ¶ D ∩ ¶ D . We can just take for v the initial or terminal vertex of thispath (if k =
0, these two vertices coincide). Indeed, we may assume that g has the normalform g = f ( s ) . . . f ( s k ) x . . . x m (where each x i is in some factor of G ). Therefore, g − has thenormal form g − = f ( s ) . . . f ( s k ) y . . . y m (where each y i is in some factor of G ).The geodesics Geo ( g ) and Geo ( g − ) = Geo ( g ) contain the k + ( A ∩ E ) , f ( s ) ( A ∩ E ) , . . . , f ( s ) . . . f ( s k ) ( A ∩ E ) . (cid:3) Example 40.
Assume that M contains the two regions depicted in Fig. 5 (the • are primaryvertices, the secondary vertices are denoted by ◦ only when they have valence ≥ • •◦ ◦• •• • D D (cid:31) (cid:31) ???? e ???? o o a ? ? (cid:127)(cid:127)(cid:127)(cid:127) e (cid:127)(cid:127)(cid:127)(cid:127) (cid:127) (cid:127) (cid:127)(cid:127)(cid:127)(cid:127) e (cid:127)(cid:127)(cid:127)(cid:127) / / a _ _ ???? e ???? id id o o a v F IGURE g = a e a e , g = e a e a − and Fig. 6 gives the picture in the Bass-Serre tree.Note that here for simplicity we took D and D with boundary length 4, but in the context ofTheorem 36 any region has boundary length at least 10. •• • • • •• e − A ??????? e A (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) id E id A a e A (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127) a e − A ??????? a Eg : : g z z F IGURE Lemma 41.
If v is a vertex of valence of M with regions D , D , D meeting at v andif g , g , g are the labels of the boundary cycles of these regions beginning at v, then thegeodesics of the g i ’s form a tripod in the Bass-Serre tree and for all i , j’s: | Geo ( g i ) ∩ Geo ( g j )) | ≥ | ¶ D i ∩ ¶ D j | . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 21
Proof.
The vertex v is necessarily secondary. Let e (resp. e , resp. e ) be the (oriented)half-segment having v as initial vertex and included into ¶ D ∩ ¶ D (resp. ¶ D ∩ ¶ D , resp. ¶ D ∩ ¶ D ). The f ( e i ) ’s are in the same factor of G and if i = j , f ( e i ) f ( e j ) − is not in theamalgamated part. As in Lemma 39, let k be the largest integer such that k < | ¶ D ∩ ¶ D | andlet s , . . . , s k be segments such that the path e , s , . . . , s k is included into ¶ D ∩ ¶ D . We mayassume that g has the normal form g = f ( e ) f ( s ) . . . f ( s k ) x . . . x m f ( e ) − ,where each x i is in some factor of G . Therefore, g ′ = f ( e ) − g f ( e ) is strictly cyclicallyreduced and has the normal form g ′ = f ( s ) . . . f ( s k ) x . . . x m + ,where x m + = f ( e ) − f ( e ) . Since the geodesic of g ′ contains the consecutive edgesid ( A ∩ E ) , f ( s ) ( A ∩ E ) , . . . , f ( s ) . . . f ( s k ) ( A ∩ E ) ,it is clear that the geodesic of g contains the consecutive edges f ( e ) ( A ∩ E ) , f ( e ) f ( s ) ( A ∩ E ) , . . . , f ( e ) f ( s ) . . . f ( s k ) ( A ∩ E ) .One would show in the same way that these edges are also contained in the geodesic of g ,so that we get | Geo ( g ) ∩ Geo ( g ) | ≥ | ¶ D ∩ ¶ D | . The other inequalities are proven similarly.We finish the proof by noting that Geo ( g ) ∩ Geo ( g ) contains the edge f ( e ) ( A ∩ E ) and thatGeo ( g ) ∩ Geo ( g ) contains the edge f ( e ) ( A ∩ E ) . If the f ( e i ) ’s are in the factor A (resp. E ),it is clear that the three edges f ( e i ) ( A ∩ E ) intersect at the vertex id A (resp. id E ). (cid:3)
4. T
HE PROOF OF T HEOREM
A result about curvature.
Let us recall some notations from [LS01]. If v is a vertex of adiagram M , the degree d ( v ) (or valence) of v will denote the number of oriented edges having v as initial vertex (thus, if an edge has both endpoints at v , we count it twice). If D is a region,the degree d ( D ) of D will denote the number of edges of D . The following formula defines acurvature contribution for each region: d ( D ) = − d ( D ) + (cid:229) v ∈ D d ( v ) . Lemma 42.
For any diagram on the 2-sphere, we have = (cid:229) D d ( D ) . Proof.
Let V , E , F be the numbers of vertices, edges and faces of the diagram. The formula is adirect consequence of Euler’s formula on the sphere 2 = V − E + F and of the obvious relations2 E = (cid:229) ( v , D ) V = (cid:229) ( v , D ) d ( v ) and F = (cid:229) ( v , D ) d ( D ) :4 = V + F − E = (cid:229) ( v , D ) (cid:18) d ( v ) + d ( D ) − (cid:19) = (cid:229) D d ( D ) where the first sum runs over the couples ( v , D ) with v a vertex and D a face such that v ∈ D . (cid:3) Corollary 43.
For any planar diagram homeomorphic to the disk, we have ≤ (cid:229) D d ( D ) . ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 22
Proof.
Let K be this diagram. Let L be the spherical diagram obtained by sticking along theirboundaries two copies K and K of K . Since L is homeomorphic to the sphere, we have4 = (cid:229) D ∈ L d ( D ) , i.e. 4 = (cid:229) D ∈ K d ( D ) + (cid:229) D ∈ K d ( D ) = (cid:229) D ∈ K d ( D ) ≤ (cid:229) D ∈ K d ( D ) . The last inequality comes from the fact that for each boundary region D in K the contributioncurvature d ( D ) computed in the disk diagram is bigger than the contribution computed in thespherical diagram. (cid:3) Remark . Here is a (non complete) list of faces D having negative or zero curvature: • D with d ( D ) ≥ • D with d ( D ) = D are tripods; • D with d ( D ) = D has valence at least 4; • D with d ( D ) = D admits a tripod, two vertices of valence at least 4 and a fourthvertex of valence at least 6; • D with d ( D ) = D has valence at least 6.4.2. The end of the proof.
We are now in position to prove Theorem 2. As in Theorem 1, wewill prove a stronger and more geometric version:
Theorem 45.
If f ∈ G is a hyperbolic element of geometric length lg ( f ) ≥ satisfying con-dition ( C ) , then the normal subgroup generated by f in Aut [ C ] is different from G.Proof. We can assume that f is a strictly cyclically reduced element of length lg ( f ) = | f | = l ≥
14. If the normal subgroup generated by f in Aut [ C ] was equal to G then, by Theorem36, there would exist an Aut [ C ] -labeled oriented diagram M such that:(1) M is connected and simply connected;(2) The perimeter of M is ≤ e e ′ . . . e t e ′ t is a boundary cycle of some region of M , then t = | f | and f ( e e ′ ) . . . f ( e t e ′ t ) is a reduced form of a strictly cyclically reduced conjugate of f .Let D , D be two distinct regions of M having a common edge. By Proposition 19 andLemma 39, we have | ¶ D ∩ ¶ D | ≤
4. Since | ¶ D | ≥
14, we conclude that any interior regionhas at least 4 edges.Furthermore, if D , D , D are three distinct regions of M having a common vertex of va-lence 3, by Lemmas 29 and 41, we know that each edge ¶ D i ∩ ¶ D j is at most of length 2. Inconsequence, if an interior region has at least 1 interior vertex of valence 3, then this region hasat least 5 edges. Similarly, if an interior region has at least 3 interior vertices of valence 3, thenthis region has at least 6 edges.By the previous observations, and using Remark 44, we conclude that the curvature contri-bution d ( D ) of any interior region D is non positive. Let us examine now the contribution ofthe boundary regions. Since the perimeter is at most 1 (i.e. at most two half-segments), thereare at most 2 boundary regions.Suppose there are exactly 2 boundary regions. Since the boundary edge of such a region D is an half-segment, it is easy to check that D has at least 5 edges, and that if at least one interiorvertex is of valence 3 then D has at least 6 edges. Thus d ( D ) ≤ D . Then the only boundary vertex of D (which has to be counted twice) has valence at least 4. So D has at least 5 edges, and if D has exactly 5 edges then the 3 interior vertices can not be of valence 3, and again we obtain d ( D ) ≤ ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 23
In conclusion we have (cid:229) d ( D ) ≤
0, which is contradictory with Lemma 43. We concludethat the normal subgroup generated by f in Aut [ C ] can not be equal to G . (cid:3)
5. T
HE REMAINING CASES : LENGTH AND
Length 12.
The main problem to adapt our strategy to the case of f with lg ( f ) =
12 isthat we have to deal with regions in a R ( f ) -diagram that are triangles with 3 edges of length 4.Then we would have to study not only tripods coming from 3 conjugates of f , but their gener-alization, which we call n -pods, coming from n conjugates f i (0 ≤ i ≤ n −
1) of f . It is the casewhere the geodesics Geo ( f i ) have a common vertex and where each pair Geo ( f i ) , Geo ( f i + ) has at least one edge in common (where i = , ..., n − n ).Precisely to be sure that the curvature of such a triangle is non positive it would be sufficient tohave the following Lemma/conjecture 46.
If n conjugates of f form a n-pod in the Bass-Serre tree, with twoconsecutive branches of length 4, then n ≥ . We believe that this result is true, but the verification seems to have to involve a very longlist of cases: that is why we do not think reasonable to try to present a proof. However it isinteresting to note that there exists 6-pods with branches of length 4. • • • • ••••• • • • ••••• • • • • ••••
RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR llllllllllllllllllllllllllllllllllllllllllllllllllllllllll a ( ) E RRRRRRR ea ( ) E lllllll id A id E RRRRRRR f eA RRRRRRR eA lllllll f A = f A f a ( ) Ef f f + + f & & f p p f n n F IGURE
7. A 6-pod with all branches of length 4 (example 47)
Example 47 (6-pod with all branches of length 4) . Let us consider the following automorphism f of length 2 l ≥ f = e ae a · · · e l a where a = a ( ) = ( y , − x ) . We suppose that e = ( x + P ( y ) , y ) , and we set e = e . We are goingto construct f , · · · , f five conjugates of f such that their geodesics form a 6-pod (see Figure7). ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 24
For i = , · · · , c i = t i = ( x , y + c i ) . We take f i = f i f f − i where f = et e − f = et e − t f = et e − t et e − f = et e − t et e − t f = et e − t et e − t et e − are all elements of E .We claim that for each i = , · · · ,
4, the geodesics of f i and f i + share a path of 4 edges withid E as an extremity.Consider the case i =
0. We have Geo ( f ) = f ( Geo ( f )) . Recall that t fixes the ball ofradius 2 centered on a ( ) E (Remark 17), so f fixes the ball of radius 2 centered on ea ( ) E ,hence the claim.Now take i =
1. Note that f = f t f t − f − = f t f − f f t − f − , and f t f − fixes theball of radius 2 centered at f a ( ) E . Thus the geodesic of f and f share 4 edges. We canmake a similar computation for i = , , c i satisfy: c + c + c = c + c + c = c + c + c = ( c , c , c , c , c ) = ( , , − , , ) .A straightforward computation shows that f = et e − t et e − t et e − =( x + P ( y + c + c + c + c + c ) − P ( y + c + c + c + c ) + P ( y + c + c + c ) − P ( y + c + c ) + P ( y + c ) − P ( y ) , y + c + c + c + c + c ) = ( x , y − c ) . Since ( x , y − c ) fixes the ball of radius 2 centered at a ( ) E , this implies that the geodesics of f and f share 4 edges, as shown on Fig. 7.5.2. Length 10.
The case of f of length 10 seems even more doubtful. For instance one couldhave pentagonal regions with all edges of length 2 and all vertices of valence 3. It is probablyeasy to rule out this case, but there are some harder ones. One could have triangular regionswith edges of length 4 , ,
2. The example 47 allows us to glue 6 such triangles along their edgeof length 4, to obtain a R ( f ) -diagram with boundary length 12. One can wonder if it possibleto glue two such diagrams to obtain a R ( f ) -diagram on a sphere (in this case our strategywould fail). One would need to have 4-pods with branches 4 , , ,
2. We do not know if thisis possible, but the following example shows again that we would have to rely on very carefulcomputations to exclude this case (note also that the assumption ’consecutive’ was crucial inthe statement of Lemma 46)
ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 25
Example 48 (4-pod with branches of length 4, 1, 4, 1) . Similarly to the previous example wetake f i = f i f f − i where f = et e − f = et e − t f = et e − t et e − with t = t = ( x + c , y ) and t = ( − x , y − c ) . Then one can verify that f = ( − x , y + c ) and thegeodesics of the f i form a 4-pod as on Figure 8. ••• • • • • •••• et e − A id A id Eet e − t eA eA ea ( ) Ef f . . f p p f n n F IGURE
8. A 4-pod with branches 4,1,4,1 (example 48)A
NNEX : GENERICNESS OF CONDITION ( C ) We begin with a reformulation of Theorem 45:
Theorem 49.
Let l ≥ be an integer. Assume that the polynomials P , . . . , P l ∈ C [ y ] are generaland independent. If the element f of G can be written f = a e . . . a l e l where e i = e ( P i ) anda i ∈ A \ E for each i, then the normal subgroup generated by f in Aut [ C ] is different from G. In this section, we will show that if P , . . . , P l are generic (in some sense), then they aregeneral and independent. We will also finish by giving explicit examples. A. Genericness of condition ( C ) . The aim of this subsection is to show that condition ( C ) is generic (see Corollary 57 and Remark 58). For technical purposes we introduce a variationof the notion of general polynomial (see Def. 15). Lemma 50.
Let Q ∈ C [ y ] be a polynomial. The following assertions are equivalent:(1) ∀ a , b , g ∈ C , Q ( y ) = a Q ( b y + g ) = ⇒ a = b = and g = (2) ∀ a , b , g ∈ C , Q ( y ) = a Q ( b y + g ) = ⇒ b = . Proof. (1) ⇒ (2) is obvious. Let us prove (2) ⇒ (1). If Q satisfies (2), note that Q can not beconstant. If Q ( y ) = a Q ( y + g ) , it is enough to show that g =
0. Let z be a root of Q . Since z + n g is also a root of Q for any integer n , we must have g = (cid:3) Definition 51.
We say that Q is weakly general if it satisfies the equivalent assertions ofLemma 50. Remark . Clearly if Q ′ is weakly general then Q is also weakly general. Furthermore, Q ( k ) is weakly general if and only if the following equivalent assertions are satisfied:(1) ∀ a , b , g ∈ C , deg ( Q ( y ) − a Q ( b y + g )) < k = ⇒ a = b = g = ∀ a , b , g ∈ C , deg ( Q ( y ) − a Q ( b y + g )) < k = ⇒ b = Q of degree d ≥ Q ( d − ) is weaklygeneral. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 26
Lemma 53.
The following assertions are equivalent:(1) Q is not weakly general;(2) there exists c ∈ C , R ∈ C [ y ] , k ≥ , n ≥ such that Q ( y + c ) = y k R ( y n ) .Proof. (1) = ⇒ (2). If Q is not weakly general, there exists a , b , g with b = Q ( y ) = a Q ( b y + g ) . If we set c = g − b , then the polynomial P ( y ) = Q ( y + c ) satisfies P ( y ) = a P ( b y ) .Writing P = (cid:229) i p i y i , the last equation is equivalent to ∀ i , ( − ab i ) p i =
0. If b is not a root ofunity, this implies that there exists k ≥ P = p k y k . Assume now that b is a primitive n th root of the unity. If P =
0, there exists k ≥ p k = a = b − k . Since p i = i ≡ k ( mod n ) , we get P = y k R ( y n ) , where R ( y ) = (cid:229) i p k + ni y i .(2) = ⇒ (1). This is a consequence of the previous computation. (cid:3) Proposition 54. (1) If d ≥ , the generic element of C [ y ] ≤ d is weakly general;(2) If d ≥ , the generic element of C [ y ] ≤ d is general.Proof. If u ∈ R , we denote its integer part by [ u ] .(1) If Q ∈ C [ y ] ≤ d is not weakly general, by Lemma 53 we can write Q ( y ) = ( y − c ) k R (( y − c ) n ) where 0 ≤ k ≤ d , 2 ≤ n ≤ d , c ∈ C , e = [ d / n ] and R ∈ C [ y ] ≤ e . Therefore, Q belongs to theimage of the following morphism j k , n : C × C [ y ] ≤ e → C [ y ] , ( c , R ( y )) ( y − c ) k R (( y − c ) n ) . Howeverdim Im j k , n ≤ dim ( C × C [ y ] ≤ e ) = e + ≤ dn + ≤ d + < d + = dim C [ y ] ≤ d . (2) is a direct consequence of (1), by considering the map Q Q ( d − ) , and using Remark52. (cid:3) Proposition 55.
If d , d ≥ and ( P , P ) is a generic element of C [ y ] ≤ d × C [ y ] ≤ d , then P ,P represent different colors.Proof. By Lemma 12, if P , P represent the same color, then ( P , P ) belongs to the image ofthe following morphism: j : C [ y ] ≤ d × C → C [ y ] × C [ y ] , ( P , ( a , b , g , d , e )) ( P , a P ( b y + g ) + d y + e ) . However,dim Im j ≤ d + < dim C [ y ] ≤ d × C [ y ] ≤ d . (cid:3) (cid:3) Remark . If d = d , Proposition 55 is still more obvious. Indeed, the generic element P i of C [ y ] ≤ d i has degree d i . Therefore, if ( P , P ) is a generic element of C [ y ] ≤ d × C [ y ] ≤ d , thendeg P = deg P , which clearly implies that P , P represent different colors.Propositions 54 and 55 give us the following result. Corollary 57.
Fix a sequence of integers d , . . . , d l ≥ . If ( P , · · · , P l ) is a generic element of (cid:213) ≤ i ≤ l C [ y ] ≤ d i , then the polynomials P i are general and represent distinct colors.Remark . In other words, if a i ∈ A \ E and e i = e ( P i ) for 1 ≤ i ≤ l , then the automorphism a e . . . a l e l satisfies condition ( C ) . B. Genericness of condition ( C ) . The aim of this subsection is to show that condition ( C ) is generic (see Corollary 61 and Remark 62). Proposition 59.
If d , d , d ≥ and ( P , P , P ) is generic in (cid:213) ≤ i ≤ C [ y ] ≤ d i , then the poly-nomials P , P , P are independent. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 27
Proof.
By permutations, it is enough to show the following two points:1) If ( P , P ) is generic in C [ y ] ≤ d × C [ y ] ≤ d , then ( A ∩ E ) e ( P )( A ∩ E ) is not a mixture of ( A ∩ E ) e ( P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) .2) If ( P , P , P ) is generic in C [ y ] ≤ d × C [ y ] ≤ d × C [ y ] ≤ d , then ( A ∩ E ) e ( P )( A ∩ E ) is not amixture of ( A ∩ E ) e ( P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) .Proof of 1. Define f : C [ y ] ≤ d × C → C [ y ] ≤ d × C [ y ] , ( P , ( a , . . . , q )) ( P , a P ( b y + g ) + d P ( e y + z ) + h y + q ) . We have dim Im f ≤ d + + < dim C [ y ] ≤ d × C [ y ] ≤ d . If ( P , P ) ∈ ( C [ y ] ≤ d × C [ y ] ≤ d ) \ Im f ,it is clear that ( A ∩ E ) e ( P )( A ∩ E ) is not a mixture of ( A ∩ E ) e ( P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) .Proof of 2. Define y : C [ y ] ≤ d × C [ y ] ≤ d × C → C [ y ] ≤ d × C [ y ] ≤ d × C [ y ] , ( P , P , ( a , . . . , q )) ( P , P , a P ( b y + g ) + d P ( e y + z ) + h y + q ) . We have dim Im f ≤ ( d + ) + ( d + ) + < dim C [ y ] ≤ d × C [ y ] ≤ d × C [ y ] ≤ d . If ( P , P , P ) ∈ ( C [ y ] ≤ d × C [ y ] ≤ d × C [ y ] ≤ d ) \ Im y , it is clear that ( A ∩ E ) e ( P )( A ∩ E ) is not a mixture of ( A ∩ E ) e ( P )( A ∩ E ) and ( A ∩ E ) e ( P )( A ∩ E ) . (cid:3) Corollary 60.
Fix a sequence of integers d , . . . , d l ≥ . The generic element ( P , · · · , P l ) of (cid:213) ≤ i ≤ l C [ y ] ≤ d i is an independent sequence. Combining Corollaries 57 and 60, we get:
Corollary 61.
Fix a sequence of integers d , . . . , d l ≥ . The generic element ( P , · · · , P l ) of (cid:213) ≤ i ≤ l C [ y ] ≤ d i defines a sequence of general and independent polynomials.Remark . In other words, if a i ∈ A \ E and e i = e ( P i ) for 1 ≤ i ≤ l , then the automorphism a e . . . a l e l satisfies condition ( C ) . C. Explicit examples.
Lemmas 63 and 66 below will allow us to give explicit examples ofpolynomials P , . . . , P l ∈ C [ y ] which are general an independent (see Example 67). Lemma 63.
Let P ∈ C [ y ] be a polynomial of degree d ≥ and let M = − p d − d p d be the arithmeticmean of its roots. If there exists two consecutive integers k ≥ such that P ( k ) ( M ) = , then Pis weakly general.Proof. If P ( y ) = a P ( b y + g ) , then the automorphism f of the affine line given by f ( y ) = b y + g permutes the roots of P . Since f is affine, we must have f ( M ) = M . By substituting M for y inthe equality P ( k ) ( y ) = ab k P ( k ) ( f ( y )) , we get ( − ab k ) P ( k ) ( M ) =
0. Whence the result. (cid:3)
Remark . We always have P ( d − ) ( M ) =
0. Therefore, if P has degree 2, it is not possible tofind two consecutive integers k such that P ( k ) ( M ) =
0. As a consequence, it is not possible toshow that P is weakly general by using an analogous version of Lemma 63. In fact, it is easyto check that no polynomial of degree 2 is weakly general! Example 65.
Let P = (cid:229) i p i y i be a polynomial of degree d ≥ p d − = p d − p d − =
0, then P is general;(2) If p d − = p d − = p d − =
0, then P is general. Lemma 66.
A family ( P i ) i of general polynomials satisfying | deg P i − deg P j | > for any i = jis independent. ORMAL SUBGROUP GENERATED BY AN AUTOMORPHISM 28
Proof.
Let us assume (by contradiction) that deg (cid:229) ≤ k ≤ a k P i k ( b k y + g k ) ≤ i = i = i .First case. i , i , i are distinct.By the assumption, deg P i , deg P i , deg P i are distinct, this is impossible.Second case. i , i , i are not distinct.We may assume that i = i = i .Since P i is general, for any a , b , g , the polynomial P i ( y ) − a P i ( b y + g ) either has degree ≥ deg P i − Q ( y ) = (cid:229) ≤ k ≤ a k P i ( b k y + g k ) .But | deg P i − deg P i | > Q = deg P i . Therefore, we cannothave deg ( Q + a P i ( b y + g )) ≤ (cid:3) Example 67.
By Example 65, the polynomial y d + y d − is general for d ≥
5. Therefore, if weset P d = y d + + y d , the polynomials P , . . . , P l are general and independent (for any l ). As aconsequence, if a i ∈ A \ E and e i = e ( P i ) for 1 ≤ i ≤ l , then f = a e . . . a l e l satisfies condition ( C ) . If we assume furthermore that f ∈ G and l ≥
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