On surjectivity of word maps on \mathrm{PSL}_2
aa r X i v : . [ m a t h . G R ] J a n ON SURJECTIVITY OF WORD MAPS ON
PSL URBAN JEZERNIK AND JONATAN SÁNCHEZ
Abstract.
Let w = [[ x k , y l ] , [ x m , y n ]] be a non-trivial double commutatorword. We show that w is surjective on PSL ( K ) , where K is an algebraicallyclosed field of characteristic . Introduction
Words, word maps and their surjectivity. A word in two variables w isan element of the free group F = h x, y i . Given a group G , the word w induces a word map ˜ w on G by evalution, ˜ w : G × G → G, ( g, h ) w ( g, h ) . When the underlying group G is a connected semisimple algebraic group, say SL n ( K ) for an algebraically closed field K , every non-trivial word map is dominantby a theorem of Borel [Bor83].For certain words, one can even prove surjectivity and possibly further propertiesof the word map, say flatness. All of these can then be used to descend to the caseof finite simple groups, say PSL n ( F q ) , and deduce surjectivity or even uniformdistribution there. See [LST19, Lar19] for a recent application of this technique.Despite being dominant, not all word maps are surjective on linear algebraicgroups, for example taking powers on SL ( C ) is not always surjective. The situationis different for adjoint groups – the surjectivity problem asking whether or not allword maps are surjective is still open for the case of two variable word maps on PSL ( K ) , where K is algebraically closed of characteristic . It is this problem thatwe address in the present paper.It was shown by Bandman and Zarhin [BZ16], and later reproved by Gordeev,Kunyavski˘ı and Plotkin [GKP18], that the surjectivity problem for PSL ( K ) hasa positive solution for words w not belonging to the second derived subgroup F (2)2 of F . Their results furthermore imply that for any word w ∈ F , the image im ˜ w contains all semisimple elements of PSL ( K ) . As im ˜ w is closed for conjugation,the surjectivity problem is then reduced to finding a single non-trivial unipotent in im ˜ w for w ∈ F (2)2 . This has been shown to hold for a handful of concrete words,and recently Gnutov and Gordeev [GG20] showed that each such example w ∈ F (2)2 Date : February 1, 2021.The first author has received funding from the European Union’s Horizon 2020 research andinnovation programme under the Marie Sklodowska-Curie grant agreement No. 748129, as well asfunding from the European Research Council (ERC) under the European Union’s Horizon 2020research and innovation programme (grant agreement No. 741420). The second author is partiallysupported by Universidad Politécnica de Madrid (UPM). can be used to produce a sequence of words { w i } i ≥ with w i ∈ F ( i )2 so that ˜ w i isalso surjective. Here F ( i )2 is the i -th term of the derived series of F .1.2. Main contribution and strategy of proof.
The second derived subgroup F (2)2 is generated as a normal subgroup of F by the double commutators [[ x k , y l ] , [ x m , y n ]] with k, l, m, n ∈ Z . Here [ x, y ] stands for x − y − xy . We show that the surjectivity problem for PSL ( K ) has a positive solution for these generating words. Theorem 1.
Every non-trivial double commutator induces a surjective word mapon
PSL ( K ) , where K is an algebraically closed field of characteristic . The method of the proof works for general words and is restricted to doublecommutators only in its final stage. Our strategy is inspired by that of [GKP18],where the authors show how the existence of unipotents in im ˜ w is related to thegeometric structure of the representation variety of the -relator group F / h w i . Therequired computations to extract this geometric structure are made easier if one isable to pass to the quotient variety by the action of SL ( C ) . In our approach, wework directly inside the group SL ( K ) . Note that it suffices to show the existenceof a non-trivial unipotent in im ˜ w for this group. We focus on the Zariski opensubset of semisimple pairs in G and consider its image in the quotient by SL ( K ) .We call this procedure wiggling (see Section 3). Our ultimate aim is to show thatone can find diagonal matrices x, y ∈ SL ( K ) for which the image of the wigglemap contains a non-trivial unipotent. Example 1.
Let x = (cid:0) λ λ − (cid:1) and y = (cid:16) µ µ − (cid:17) be a pair of diagonal matrices in SL ( K ) . The wiggle map associated to the ordinary commutator word is Wiggle x,y : SL ( K ) → SL ( K ) , g [ x, y g ] . One directly computes that, for g = (cid:0) a bc d (cid:1) , we have tr Wiggle x,y ( g ) = (cid:0) λ − λ − (cid:1) (cid:0) µ − µ − (cid:1) bc ( bc + 1) + 2 . As long as λ, µ = ± , the trace is equal to if and only if bc ∈ { , − } . On theother hand, the (1 , -entry of the image of g under the wiggle map is Wiggle x,y ( g ) = (cid:0) λ − λ − (cid:1) (cid:0) µ − µ − (cid:1) (cid:0) (cid:0) − µ (cid:1) bc (cid:1) bd. As long as λ, µ = ± , this entry is equal to if and only if bd = 0 or bc = − / (1 − µ ) .Taking λ = µ = 2 and g = (cid:0) − (cid:1) , we see that the image of the wiggle map containsa non-identity element of trace , which must be a non-trivial unipotent.We first show that the wiggle map associated to any word can be expressed in anormal form in terms of two matrix polynomials in the variable bc as in the exampleabove. We also explain the meaning of the coefficients of these polynomials. Wethen give an effective way of computing these polynomials. Furthermore, we showthat the wiggle map has a particularly symmetric form. This can be exploited (see N SURJECTIVITY OF WORD MAPS ON
PSL Section 4) to reduce the problem of finding non-trivial unipotents to the problem offinding a particular root of a single polynomial, similar to the one in the exampleabove. This polynomial is related to the trace polynomial as developed in generalby Fricke [Fri97]. For the purposes of this paper, however, additional informationregarding the structure of this polynomial in needed and we achieve this with thematrix form of the wiggle map. We conclude by applying all of the above to thecase of double commutators, where we are able to explicitly compute the relevantpolynomial.Some of the symbolic calculations were verified using
Wolfram Mathematica , therelevant notebook is available on the website of the first author. Notation.
We will write
Mat ( K ) for the matrix algebra and Diag ( K ) forthe set of all diagonal matrices in Mat ( K ) . Then Torus ( K ) = SL ( K ) ∩ Diag ( K ) is the standard maximal torus in SL ( K ) . We will use the same notation for moregeneral matrix rings Mat ( R ) over a ring R . For a matrix A ∈ Mat ( R ) , we willlet A ji denote the element of A at position ( i, j ) .2. Conjugating semisimple elements
Conjugation map.
Our objective will be to evaluate the word map ˜ w at apair of semisimple elements in a matrix group G ⊆ GL ( K ) . To achieve this, we willtransform occurrences of diagonalizable matrices with their diagonal forms underconjugation by some element g ∈ G . In order to control the situation, we thereforestudy the linear map C g : Mat ( K ) → Mat ( K ) , x x g − x. Here and throughout, we denote x g = g − xg .2.2. Conjugation map on diagonal matrices.
Set ξ g = C g (( )) . The imageof a diagonal matrix (cid:0) λ µ (cid:1) under C g can be expressed in terms of the matrix ξ g asfollows:(2.1) C g ( (cid:0) λ µ (cid:1) ) = C g (( λ
00 0 )) + C g ( (cid:0) µ (cid:1) − (cid:0) µ µ (cid:1) ) = ( λ − µ ) ξ g . This means that a conjugate of a diagonal matrix can be expressed as(2.2) (cid:0) λ µ (cid:1) g = (cid:0) λ µ (cid:1) + ( λ − µ ) ξ g . We will write this more compactly; for a diagonal matrix x = (cid:0) λ µ (cid:1) , let ¯ x = λ − µ .Then(2.3) x g = x + ¯ xξ g . https://drive.google.com/file/d/15AKylXoKmwW4hsnSPhv5GvUulRSkFZPQ/view?usp=sharing URBAN JEZERNIK AND JONATAN SÁNCHEZ
Properties of the conjugation map.
Let us collect some properties of themap C g and of the matrix ξ g . The identity matrix of GL ( K ) will be denoted by throughout. Lemma 1.
Let x, y ∈ Mat ( K ) and g, h ∈ GL ( K ) .(1) tr C g ( x ) = 0 (2) C g ( x ) = − det C g ( x ) · (3) C g h ( x h ) = C g ( x ) h (4) C g ( xy ) − C g ( x ) C g ( y ) = xC g ( y ) + C g ( x ) y Proof. (1) tr C g ( x ) = tr x g − tr x = 0 . (2) Immediate from the Cayley-Hamiltontheorem. (3) C g h ( x h ) = x hg h − x h = x gh − x h = C g ( x ) h . (4) Expand C g ( xy ) − C g ( x ) C g ( y ) = ( xy ) g − xy − ( x g − x )( y g − y ) = x g y + xy g − xy, and notice that the latter is precisely C g ( x ) y + xC g ( y ) . (cid:3) It follows from the lemma that tr ξ g = 0 and ξ g = − det ξ g · . We now collectsome more properties of the matrix ξ g that will be useful in the rest of the paper. Lemma 2.
Let x ∈ Diag ( K ) .(1) tr ( xξ g ) = ¯ x det ξ g (2) If x ∈ Torus ( K ) , then ξ g x = x − ξ g + ¯ x det ξ g · .Proof. (1) When ¯ x = 0 , we have x = ± and the claim holds. Assume now that ¯ x = 0 . Use Lemma 1 (4) with y = x to obtain C g ( x ) = − xC g ( x ) − C g ( x ) x + C g ( x ) .It follows from Lemma 1 (2) together with (2.3) that C g ( x ) = − ¯ x det ξ g · . Wenow apply the trace map to obtain − x det ξ g = tr C g ( x ) = − tr xC g ( x ) − tr C ( g ) x + tr C g ( x ) = − xC g ( x ) . Note that xC g ( x ) = ¯ xxξ g by (2.3). This implies tr xξ g = ¯ x det ξ g .(2) Use Lemma 1 (4) with y = x − to obtain C g ( x ) C g ( x − ) = − xC g ( x − ) − C g ( x ) x − . It follows from (2.3) that C g ( x ) = ¯ xξ g and C g ( x − ) = − ¯ xξ g . Wetherefore have that − ¯ x ξ g = ¯ xxξ g − ¯ xξ g x − . It follows from Lemma 1 (2) that ξ g x − = xξ g − ¯ x det ξ g · . The claim follows after replacing x by x − . (cid:3) Lemma 3.
Set t = det ξ g . Then ξ g = (cid:0) t pq − t (cid:1) for some p, q ∈ K satisfying pq = − t ( t + 1) .Proof. Use Lemma 2 (1) with x = ( ) to conclude that ( ξ g ) = t . The sameargument with x = ( ) gives ( ξ g ) = − t . The equality pq = − t ( t + 1) is nothingbut det ξ g = t . (cid:3) Remark . Writing g = (cid:0) a bc d (cid:1) ∈ SL ( K ) , we have ξ g = (cid:0) bc bd − ac − bc (cid:1) and det ξ g = bc . Example 2.
We show how the properties of ξ g determined in this section canbe used to quickly show that the commutator word w = [ x a , y b ] for a, b non-zerointegers contains non-trivial unipotents in its image. To this end, take a semisimple N SURJECTIVITY OF WORD MAPS ON
PSL element y ∈ Torus ( K ) with y b non-trivial, and let x ∈ SL ( K ) be a non-diagonalupper-triangular matrix. Thus ξ x a = ( ⋆ ) with ⋆ = 0 by Remark 1. We have ˜ w ( x, y ) = ( y − b ) x a y b = ( y − b + y − b ξ x a ) y b = + y − b ξ x a y b , hence ˜ w ( x, y ) is a non-trivial unipotent.3. Wiggling the word map on semisimple pairs
In this section, we deal with finding a description of the values of the word map ˜ w on pairs of semisimple elements in terms of certain polynomials.3.1. Reducing the word.
The word w can be written as(3.1) w = n Y i =1 x a i y b i for some integers a i , b i and a positive integer n . Since our objective is to showthat the image of ˜ w contains a non-trivial unipotent element in SL ( K ) , it sufficesto show this after replacing w by any of its conjugates xwx − or y − wy . We canrepeat this procedure of reducing w . If we end with a power of x or y , then theimage of the word map clearly contains a non-trivial unipotent. We will thereforefocus on the case when the reduction process end with a word that is cyclicallyreduced and of the form (3.1) with a i , b i = 0 for all ≤ i ≤ n . In this situation, wesay that the length of the word w is ℓ ( w ) = n . Example 3.
The double commutator word [[ x k , y l ] , [ x m , y n ]] reduces to w k,l,m,n = x m y − l + n x − k y l x k y − n x − m y n x − k + m y − l x k y l x − m y − n . As long as l = n and k = m , this word is cyclically reduced and of length .3.2. Wiggling one parameter.
Let R = k [ λ, λ − , µ, µ − ] be the commutativering of Laurent polynomials on the variables λ, µ . Consider a pair of generic semisim-ple elements x = (cid:18) λ λ − (cid:19) and y = (cid:18) µ µ − (cid:19) . For a fixed choice of λ, µ ∈ K ∗ , we have the wiggle map Wiggle x,yw : SL ( K ) → SL ( K ) , g ˜ w ( x, y g ) that conjugates the second parameter y and returns the ˜ w value. Lemma 4. im ˜ w | SL ( K ) ss × SL ( K ) ss = [ x,y ∈ Torus ( K ) [ h ∈ SL ( K ) (im Wiggle x,yw ) h Proof.
The left hand side is closed under conjugation by elements of SL ( K ) and itcontains im Wiggle x,yw . This proves that LHS ⊇ RHS. As for the converse, consider apair of elements x ′ , y ′ ∈ SL ( K ) ss . Since every semisimple element is in a maximal URBAN JEZERNIK AND JONATAN SÁNCHEZ torus and all maximal tori are conjugate, there exist h, g ∈ SL ( K ) with x :=( x ′ ) h − ∈ Torus ( K ) and y := (( y ′ ) h − ) g − ∈ Torus ( K ) . Then ˜ w ( x ′ , y ′ ) = ˜ w (cid:16) x, ( y ′ ) h − (cid:17) h = Wiggle x,yw ( g ) h ∈ RHS . (cid:3) Our strategy is to find a non-trivial unipotent as the image of a pair of semisimpleelements under ˜ w . It follows from Lemma 4 that it is enough to search for thisunipotent in the image of the map Wiggle x,yw for some x, y ∈ Torus ( K ) .3.3. A normal form of the wiggle.
Write the word w as in (3.1). We can use(2.3) to rewrite Wiggle x,yw ( g ) in terms of ξ g as follows:(3.2) Wiggle x,yw ( g ) = n Y i =1 x a i ( y g ) b i = n Y i =1 x a i ( y b i ) g = n Y i =1 (cid:16) x a i y b i + x a i y b i · ξ g (cid:17) . We will expand the last product using the following notation. For a subset I ⊆{ , , . . . , n } , let X I ( i ) = ( x a i y b i · ξ g if i ∈ Ix a i y b i if i I and let W I = Q ni =1 X I ( i ) . Then(3.3) Wiggle x,yw ( g ) = X I ⊆{ , ,...,n } W I . Each summand W I can be written as(3.4) W I = δ · D ξ g D ξ g D · · · D r − ξ g D r , where r = | I | , δ ∈ R and D i ∈ Diag ( R ) ∩ SL ( R ) for ≤ i ≤ r . We now showhow this expression can be further simplified. Lemma 5.
Each W I can be written as E + E ξ g with E , E ∈ Diag ( R [det ξ g ]) .Proof. Write W I as in (3.4), where the factor δ can clearly be ignored. We useinduction on r to prove the claim. If r = 0 , there is nothing to prove. If r = 1 , wecan use Lemma 2 (2) to express W I in the desired form: D ξ g D = ¯ D det ξ g · D + D D − ξ g . Now let r ≥ be a positive integer and assume by induction that the expression (3.4)can be written in the desired form for all r ′ < r . Apply Lemma 2 (2) on the term ξ g D to obtain D ξ g D ξ g D · · · D r − ξ g D r = D (cid:0) D − ξ g D + D det ξ g · ξ g D (cid:1) ξ g D · · · ξ g D r , which is the same as det ξ g · D D − D ξ g · · · ξ g D r + D det ξ g · D ξ g D · · · ξ g D r by Lemma 1. By induction, both summands can be written in the desired form.The same is therefore true for their sum and the proof is complete. (cid:3) Let R = R [ t ] be the ring of polynomials with coefficients in R . We will identifythe ring Mat ( R ) with Mat ( R )[ t ] . In this language, the wiggle can be expressedin the following unique way. N SURJECTIVITY OF WORD MAPS ON
PSL Theorem 2.
There exist unique matrix polynomials A x,yw , B x,yw ∈ Diag ( R )[ t ] sothat ∀ g ∈ SL ( K ) . Wiggle x,yw ( g ) = A x,yw ( t ) + B x,yw ( t ) · ξ g at t = det ξ g . Proof.
Existence follows from (3.3) and Lemma 5. As for uniqueness, suppose onehas A x,yw ( t ) + B x,yw ( t ) ξ g = A ′ x,yw ( t ) + B ′ x,yw ( t ) ξ g at t = det ξ g for all g ∈ SL ( K ) . Itwill suffice to consider only those g ∈ SL ( K ) with det ξ g / ∈ { , − } . By Lemma3, we can write ξ g = (cid:0) t pq − t (cid:1) with pq = t ( t + 1) . Then ( B x,yw ( t )) p = ( B ′ x,yw ( t )) p .This implies that the polynomials ( B x,yw ( t )) and ( B ′ x,yw ( t )) have the same valueswhenever t / ∈ { , − } , hence they are the same. A similar argument gives that ( B x,yw ( t )) = ( B ′ x,yw ( t )) . Therefore B x,yw = B ′ x,yw and hence also A x,yw = A ′ x,yw . (cid:3) By means of Theorem 2, one can compute
Wiggle x,yw ( g ) by first computing thetwo associated matrix polynomials A x,yw ( t ) , B x,yw ( t ) ∈ Diag ( R )[ t ] and evaluatingthem at t = det ξ g .3.4. Computing the associated polynomials.
We now give an effective way ofcomputing the polynomials A x,yw ( t ) , B x,yw ( t ) . This will rely on a recursive formulabased on shortening the length of the word w .In the case of the trivial word w = 1 ∈ F , we clearly have Wiggle x,y ( g ) = ,hence A x,y = and B x,y = 0 .Suppose now that ℓ ( w ) ≥ . We can thus write w = w ′ · x a y b for some w ′ ∈ F with ℓ ( w ′ ) < ℓ ( w ) . This induces Wiggle x,yw ( g ) = Wiggle x,yw ′ ( g ) · Wiggle x,yx a y b ( g ) , which is the same as A x,yw ( t ) + B x,yw ( t ) ξ g = ( A x,yw ′ ( t ) + B x,yw ′ ( t ) ξ g ) (cid:16) x a y b + y b · x a ξ g (cid:17) at t = det ξ g by Theorem 2. Expand the product on the right side and use Lemma 2(2) to write the obtained in a normal form as in Theorem 2. Uniqueness of normalforms then gives the following recursive formulae: ( A x,yw ( t ) = A x,yw ′ ( t ) x a y b + B x,yw ′ ( t ) x a y b t − B x,yw ′ ( t ) y b x − a t,B x,yw ( t ) = A x,yw ′ ( t ) y b x a + B x,yw ′ ( t ) x − a y − b + B x,yw ′ ( t ) x a · y b t. The first equality can be simplified using the fact that x a y b · = x a y b + y b x − a . Wethen arrive to the following recursion: A x,yw ( t ) = A x,yw ′ ( t ) · x a y b + B x,yw ′ ( t ) · x a y b t,B x,yw ( t ) = A x,yw ′ ( t ) · y b x a + B x,yw ′ ( t ) · (cid:16) x − a y − b + x a · y b t (cid:17) . The formulae can be thought of as linear in the recursive variables A x,yw ′ ( t ) , B x,yw ′ ( t ) over the ring Diag ( R ) . We can therefore express them in terms of matrices. Notethat the two particular cases ( a, b ) ∈ { (1 , , (0 , } of the recursion are A x,yw · x ( t ) = A x,yw ( t ) · x + B x,yw ( t ) · xt, B x,yw · x ( t ) = B x,yw ( t ) · x − ,A x,yw · y ( t ) = A x,yw ( t ) · y, B x,yw · y ( t ) = A x,yw ( t ) · y + B x,yw ( t ) · y − . URBAN JEZERNIK AND JONATAN SÁNCHEZ
In order to write these in matrix form, let X = (cid:18) x xt · x − (cid:19) and Y = (cid:18) y y · y − (cid:19) be matrices in SL (Diag ( R )) . We then have A x,yw ( t ) B x,yw ( t ) ! = Y b X a A x,yw ′ ( t ) B x,yw ′ ( t ) ! . Extending this step by step to the whole word w gives the following way of quicklycomputing the associated polynomials of the wiggle map. Proposition 1.
Let w = Q ni =1 x a i y b i and ←− w = Q i = n y b i x a i . Then A x,yw ( t ) B x,yw ( t ) ! = ←− w ( X, Y ) · ! . We record an immediate corollary.
Corollary 1.
The polynomials A x,yw ( t ) , B x,yw ( t ) ∈ Diag ( R )[ t ] are coprime.Proof. It follows from Proposition 1 that the first column of the matrix ←− w ( X, Y ) is ( A x,yw ( t ) , B x,yw ( t )) T . It now follows from det ←− w ( X, Y ) = that A x,yw ( t ) · ←− w ( X, Y ) − B x,yw ( t ) · ←− w ( X, Y ) = . (cid:3) Symmetry and the shape of the wiggle.
Let σ = ( ) ∈ SL ( K ) . We willexploit the action of σ on SL ( K ) by conjugation to obtain a certain symmetricalrelation between the polynomials defining the associated matrix polynomials. Notefirst that we have x σ = x − and y σ = y − . Now Wiggle x,yw ( g σ ) σ = w ( x, y g σ ) σ = w ( x σ , y g σ σ ) = w ( x σ , ( y σ ) g ) = Wiggle x − ,y − w ( g ) . It follows from Theorem 2 that we can therefore write
Wiggle x − ,y − w ( g ) = ( A x,yw (det ξ g σ ) + B x,yw (det ξ g σ ) · ξ g σ ) σ . By Lemma 1 (3), we have ξ g σ = − ξ σg . This gives A x − ,y − w ( t ) + B x − ,y − w ( t ) · ξ g = Wiggle x − ,y − w ( g ) = A x,yw ( t ) σ − B x,yw ( t ) σ · ξ g at t = det ξ g . Using the uniqueness part of Theorem 2, we now conclude thefollowing. Proposition 2 (Symmetry of associated polynomials) . A x,yw ( t ) σ = A x − ,y − w ( t ) , B x,yw ( t ) σ = − B x − ,y − w ( t ) . This symmetry allows us to reduce associated matrix polynomials to ordinarypolynomials. We will express this in terms of an extended action of σ . First of all, σ acts by conjugation on Mat ( R ) . By letting t σ = t , we have an induced actionof σ on Mat ( R ) . This action induces the standard action on matrix functions.For example, for a polynomial with matrix parameters δ x,y ( t ) ∈ R [ t ] , we have δ x,y ( t ) σ = δ x σ ,y σ ( t ) . N SURJECTIVITY OF WORD MAPS ON
PSL Corollary 2.
There exist polynomials α x,yw , β x,yw ∈ R [ t ] so that A x,yw ( t ) = α x,yw ( t ) 00 α x,yw ( t ) σ ! , B x,yw ( t ) = β x,yw ( t ) 00 − β x,yw ( t ) σ ! . Example 4.
Using Proposition 1 and extracting only the upper-left entries, wecan compute the polynomials α and β associated to the double commutator word w k,l,m,n . Both are, in the general situation, of degree . We have α x,yw k,l,m,n ( t ) = ( λ k − µ l − λ m − µ n − λ k + m ) µ l + n ) · t ( t + 1) · α ( t ) + 1 ,β x,yw k,l,m,n ( t ) = ( λ k − µ l − λ m − µ n − λ k + m ) µ l + n ) · t ( t + 1) · β ( t ) , where α ( t ) = P i =0 a i t i and β ( t ) = P i =0 b i t i . The coefficients of α are a = λ k +2 m ) µ l +2 n ) − λ k + m ) µ l + n ) ,a = − λ m ( µ l + n ) ( µ l − µ n − λ k − µ l ( µ l − µ l − µ n + µ n ) λ k + µ l ( µ l − µ l − µ n ) λ k − µ n ( µ l − λ m − µ n ( µ l − λ m + ( µ n + µ l + n ) + µ l + n ) − µ l +2 n ) ) λ k + m ) + µ l + n ) ( − µ l + µ n + 1) λ k + m ) + µ n ( µ l − λ k +2 m ) ) ,a = λ m ( λ k − µ l − − ( µ n − µ l + µ n − µ l + n ) ) λ k + ( µ l − µ l − µ n + 3 µ l + n ) + µ l + n ) − µ l +2 n ) ) λ k − µ n ( µ l − − µ l + 3 µ n − λ m + µ n ( µ l − µ n + µ l + n ) + 1) λ m − µ l ( µ l − µ n − λ k + m ) + ( µ l − µ l − µ n ) λ k + m ) − ( µ l − µ n )( µ n − λ k +2 m ) ) ,a = − λ m ( λ k − µ l − − ( µ n − µ l − µ l − µ n + 3 µ l + n ) ) λ k − (2 µ l − µ l − µ n + µ n + 2 µ l + n ) + 2 µ l + n ) − µ l +2 n ) ) λ k +( − µ l + µ l + 2 µ n − µ n − µ l + n ) + 3 µ l +2 n ) ) λ m + ( µ l − µ n + 3 µ n − µ l + n ) + µ l + n ) − µ l +2 n ) ) λ m + µ l ( µ l − µ n − λ k + m ) + ( µ l − µ l − µ n )( µ n − λ k + m ) + ( µ l − µ n − µ n − µ l ) λ k +2 m ) ) ,a = λ m ( λ k − ( λ m − λ m − λ k )( µ l − ( µ l − µ n )( µ n − , and expressions of the same form can be computed for β . We remark that α (0) , α ( − and β (0) , β ( − are all non-zero for general k, l, m, n and generic λ, µ .Using the symmetry under σ , we can now express the wiggle matrix as follows. Corollary 3.
Let γ x,yw ( t ) = α x,yw ( t ) + β x,yw ( t ) · t ∈ R [ t ] . Then Wiggle x,yw ( g ) = γ x,yw ( t ) β x,yw ( t ) · p − β x,yw ( t ) σ · q γ x,yw ( t ) σ ! at t = det ξ g , where ξ g = (cid:0) t pq − t (cid:1) as in Lemma 3.Proof. This is immediate by joining Theorem 2 and Corollary 2. (cid:3)
Example 5.
For the double commutator word w k,l,m,n , we compute γ x,yw k,l,m,n ( t ) = ( λ k − µ l − λ m − µ n − λ k + m ) µ l + n ) · t ( t + 1) · γ ( t ) + 1 , tr Wiggle x,yw k,l,m,n ( g ) = (cid:18) ( λ k − µ l − λ m − µ n − λ k + m ) µ l + n ) (cid:19) · t ( t + 1) · τ ( t ) + 2 , where γ ( t ) = P i =0 c i t i and τ ( t ) = P i =0 t i t i . The coefficients of τ are t = ( λ k µ l − λ m µ n ) ,t = µ l ( µ l − µ n )( µ n − λ k − µ l ( − µ l + µ n + µ l + n ) + 1) λ k + µ n ( µ l − µ n − µ l ) λ m − µ n ( µ l − µ n + µ l + n ) + 1) λ m + ( µ l − µ n − µ l + µ n ) λ k + m ) − ( µ l − µ l − µ n ) λ k + m ) + ( µ l − µ n )( µ n − λ k +2 m ) ,t = (2 µ l − µ l − µ n )( µ n − λ k + ( − µ l + 3 µ l + µ n − µ l + n ) − µ l + n ) + µ l +2 n ) ) λ k + ( µ l − µ n − µ l )(2 µ n − λ m +( µ l − µ n + 3 µ n − µ l + n ) + µ l + n ) − µ l +2 n ) ) λ m + ( µ l − µ n − µ l + µ n ) λ k + m ) + ( µ l − µ l − µ n )( µ n − λ k + m ) + ( µ l − µ n − µ n − µ l ) λ k +2 m ) ,t = ( λ k − λ k − λ m )( λ m − µ l − µ l − µ n )( µ n − . In the next section, we will make use of the following relationship between thepolynomials γ x,yw and β x,yw . Corollary 4. γ x,yw ( t ) · γ x,yw ( t ) σ − β x,yw ( t ) · β x,yw ( t ) σ · t ( t + 1) = 1 . Proof.
Use det
Wiggle x,yw ( g ) = 1 together with Corollary 3 and Lemma 3. (cid:3) Finding non-trivial unipotents
Reducing to one parameter.
Set
Uni x,yw = { g ∈ SL ( K ) | Wiggle x,yw ( g ) is a non-trivial unipotent } . Our objective is to show that this is a non-empty set. Since the wiggle can beexpressed almost exclusively in terms of the parameter t = det ξ g by Corollary 3,we will rather work with a slight modification of the set Uni x,yw defined as follows:
Uni x,yw = ( t ∈ K − { , − } (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) γ x,yw ( t ) + γ x,yw ( t ) σ = 2 and ( β x,yw ( t ) = 0 or β x,yw ( t ) σ = 0) ) . We now explain the connection between these two sets in terms of the map ξ : SL ( K ) → Mat ( K ) , g ξ g . Lemma 6. (det ◦ ξ ) − ( Uni x,yw ) ⊆ Uni x,yw . Proof.
Recall first that an element v ∈ SL ( K ) is a non-trivial unipotent if and onlyif tr v = 2 and v = . Given t ∈ Uni x,yw , let g ∈ SL ( K ) be any element with det ξ g = t . Note that after writing ξ g = (cid:0) t pq − t (cid:1) as in Lemma 3, we have pq = − t ( t + 1) = 0 . N SURJECTIVITY OF WORD MAPS ON
PSL It then follows from Corollary 3 that tr Wiggle x,yw ( g ) = γ x,yw ( t ) + γ x,yw ( t ) σ = 2 andat least one of Wiggle x,yw ( g ) , Wiggle x,yw ( g ) is non-zero. Therefore Wiggle x,yw ( g ) is anon-trivial unipotent in SL ( K ) , and so g ∈ Uni x,yw . (cid:3) It follows from Remark 1 that the composition map det ◦ ξ : SL ( K ) → k issurjective. In order to find a non-trivial unipotent, it will therefore suffice to showthat the set Uni x,yw is non-empty for some x, y ∈ Torus ( K ) .4.2. Reducing to one polynomial.
We will in fact show that the following subsetof
Uni x,yw is non-empty.
Lemma 7. { t ∈ K − { , − } | γ x,yw ( t ) + γ x,yw ( t ) σ = 2 and γ x,yw ( t ) = 1 } ⊆ Uni x,yw . Proof.
Let t ∈ LHS. It follows from Corollary 4 that γ x,yw ( t ) · ( γ x,yw ( t ) + γ x,yw ( t ) σ −
2) = ( γ x,yw ( t ) − + β x,yw ( t ) · β x,yw ( t ) σ · t ( t + 1) . Hence β x,yw ( t ) · β x,yw ( t ) σ = 0 , which implies that t ∈ Uni x,yw . (cid:3) Finding suitable roots.
We now show that in the case of the double com-mutator word, the subset appearing in Lemma 7 is indeed non-empty (apart froman exceptional case that we translate to a non-exceptional case).
Proof of Theorem 1.
It now suffices to prove that there exists a root of the polyno-mial γ x,yw k,l,m,n ( t ) + γ x,yw k,l,m,n ( t ) σ − that is distinct from , − and is not a root of γ x,yw k,l,m,n ( t ) . To this end, we use the computations from Example 5. It then sufficesto show that τ ( t ) and γ ( t ) have no common roots. We verify this by computingtheir resultant: Res( τ ( t ) , γ ( t )) = − λ k +8 m µ l +6 n ( λ k − ( µ l − ( λ m − µ n − ( λ k − λ m ) ( µ l − µ n ) ( µ l λ m − λ k µ n )( λ m µ n − λ k µ l )(( µ l − µ n )( λ k +2 m −
1) + ( λ m − λ k )( µ l +2 n − . Apart from exceptional cases for the values of k, l, m, n , this resultant is non-zerofor generic λ, µ , thus proving our claim. The exceptional cases are the cases whenthe double commutator word [[ x k , y l ] , [ x m , y n ]] is not trivial, yet the resultant iszero. By inspecting the factors of the resultant, we see that the exceptional casesare:(1) m = k and n = l . In this case, the resultant vanishes because the degrees ofthe polynomials τ and γ both drop by . More precisely, in this case we have τ ( t ) = − λ k ( µ l − µ n ) (cid:0) t ( λ k − − (cid:1) (cid:0) t ( λ k −
1) + λ k (cid:1) , and evaluating γ at the two roots of τ gives the generically non-zero values ( λ k (cid:0) λ k + 1 (cid:1) µ l + n ) (cid:0) µ n − µ l (cid:1) λ k − , − λ k (cid:0) λ k + 1 (cid:1) µ n (cid:0) µ n − µ l (cid:1) λ k − ) . (2) n = l and m = k . In this case, the extraordinary event that τ divides γ occurs.We deal with this case in the following manner. Using the fact that [ x, y − ] =[ y, x ] x − , we deduce that the double commutator words [[ x k , y l ] , [ x m , y n ]] and [[ y l , x k ] , [ y n , x m ]] are conjugate in F . Hence this case follows from havingalready proved that im w k,l,k,n contains a non-trivial unipotent.(3) m = − k and n = − l . In this case, the polynomial τ is equal to ( λ k − µ l − λ k µ l (cid:0) t ( λ k − µ l −
1) + λ k µ l + 1 (cid:1) (cid:18) t + ( λ k µ l − ( λ k − µ l − (cid:19) . The double root of τ is also a root of γ . Evaluating γ at the third root of τ gives, however, the generically non-zero value − (cid:0) λ k + 1 (cid:1) (cid:0) λ k − µ l (cid:1) (cid:0) λ k µ l − (cid:1) (cid:0) λ k + 4 λ k µ l + λ k µ l + λ k µ l + µ l (cid:1) λ k ( λ k −
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Urban Jezernik, Alfréd Rényi Institute of Mathematics, Hungarian Academy ofSciences, Reáltanoda utca 13-15, H-1053, Budapest, Hungary
Email address : [email protected] Jonatan Sánchez, Department of Applied Mathematics (DMATIC), ETSI Inge-nieros Informáticos, Universidad Politécnica de Madrid, Campus de Montegancedo,Avenida de Montepríncipe, 28660, Boadilla del Monte, Spain
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