The automorphism group of a free-by-cyclic group in rank 2

THE AUTOMORPHISM GROUP OF A FREE-BY-CYCLIC GROUP IN RANK 2 O. BOGOPOLSKI, A. MARTINO AND E. VENTURA Abstract. Let φbe an automorphism of a free group F2of rank 2 and let Mφ=F2oφZbe the corresponding mapping torus of φ. We prove that the group Out(Mφ) is usually virtually cyclic. Moreover, we classify the cases when this group is finite depending on the conjugacy class of the image of φin GL2(Z). 1. Introduction Let Fnbe the free group of rank nfreely generated by x1, . . . , xn, and let us denote automorphisms φ∈Aut(Fn) as acting on the right, x7→ xφ. In this paper we consider extensions of finitely generated free groups by the infinite cyclic group ([f.g. free]-by-Zgroups, for short). More precisely, for any given φ∈Aut(Fn), we consider the mapping torus, Mφ=FoφZ, of φ i.e. the extension of Fnpresented by Mφ=hx1, . . . , xn, t |t−1xit=xiφi. The aim of the paper is to study the automorphism group of such groups, Aut(Mφ). We shall give partial results for arbitrary rank n, and a complete description for the cases n= 1,2. To help avoiding possible confusions, we will use greek letters (such as φor ψ) to denote automorphisms of Fn, and capital greek letters (such as Φ or Ψ) to denote automorphisms of Mφ. Accordingly, for every word w∈Fn, we shall write γwto denote the inner automorphism of Fngiven by right conjugation by w,xγw=w−1xw. And, for every element g∈Mφ, we shall write Γgto denote the inner automorphism of Mφgiven by right conjugation by g,xΓg=g−1xg. As usual, Inn(G) denotes the group of inner automorphisms of a group G, and Out(G) = Aut(G)/Inn(G). Although [f.g. free]-by-Zgroups have received a great deal of attention in recent years, there has been no real systematic study of their automorphisms. In fact, it still seems to be an open question whether or not they have finitely generated or finitely presented automorphism groups. Having 1 2 O. BOGOPOLSKI, A. MARTINO AND E. VENTURA said that, there are certain cases in which the automorphism group is understood. For instance, when Mφis word hyperbolic, it is known to have finite outer automorphism group (this can be deduced from papers [1, 3]). However, note that in the rank 2 case, the group Mφcan never be hyperbolic. In fact, by a result of Nielsen (see Proposition 5.1 in [4]), ([x1, x2])φand so t−1[x1, x2]t, must be conjugate to [x1, x2]±1in F2. Hence, Mφcontains a free abelian subgroup of rank 2 implying that Mφis not hyperbolic. It is straightforward to verify (see Lemma 2.1 below) that the isomorphism type of Mφdepends only on the conjugacy class of the outer automorphism [φ]±1∈Out(Fn) = Aut(Fn)/Inn(Fn) determined by φ±1. While this characterises the isomorphism class of Mφwhen n= 2, it seems also possible, for n > 2, that two [f.g. free]-by-Zgroups are isomorphic even if they do not arise from conjugate or conjugate-inverse outer automorphisms. In Section 2, we analyse Aut(Mφ) for arbitrary nbut under certain technical restrictions for φ(see Theorems 2.4 and 2.5). After dedicating Section 3 to recall a standard classification of 2 ×2 matrices, the main result comes in Section 4, where we analyse Aut(Mφ) in the case when the underlying free group has rank n= 2, and without conditions on φ. The rank 2 case is doubtless the easiest to deal with, but we believe that some of our methods may be of general interest. For instance, our detailed look at the parabolic case is certain to be of use in the more general UPG case (the definition of UPG automorphisms can be found in [2]). The information obtained there is summarised in the following theorem, which is the main result of the paper. Theorem 1.1. Let F2=ha, bibe a free group of rank 2, let φ∈Aut(F2), and consider the mapping torus Mφ=F2oφZ. Let φab ∈GL2(Z)be the map induced by φon Fab 2∼ =Z2(written in row form with respect to {a, b}). i) If φab =I2, then Out(Mφ)∼ =(Z2oC2)oGL2(Z), where C2is the cyclic group of order 2 acting on Z2by sending uto −u,u∈Z2, and where GL2(Z)acts trivially on C2and naturally on Z2(thinking vectors as columns there). ii) If φab =−I2, then Out(Mφ)∼ =PGL2(Z)×C2. iii) If φab 6=−I2and does not have 1 as an eigenvalue then Out(Mφ) is finite. iv) If φab is conjugate to ¡1k 0−1¢for some integer k, then Out(Mφ)has an infinite cyclic subgroup of finite index. v) If φab is conjugate to (1k 0 1 )for some k6= 0, then Out(Mφ)has an infinite cyclic subgroup of finite index. Furthermore, for every φ∈Aut(F2),φab fits into exactly one of the above cases. THE AUTOMORPHISM GROUP OF A FREE-BY-CYCLIC GROUP IN RANK 2 3 Note that for n= 2, Mφis always the fundamental group of a 3-manifold, namely the corresponding mapping torus of a once punctured torus, say M. When φab is hyperbolic, Mis a hyperbolic manifold and it is known that, in this case, Out(π1M) = Out(Mφ) is finite. This fact is contained in the statement of (iii), which will be proven later using algebraic methods. Finally, we remark that the case of rank 1 is straightforward to analyse. Proposition 1.2. Let F1=haibe a free group of rank 1, let φ∈Aut(F1), and consider the mapping torus Mφ=F1oφZ. i) If φ=Id then Out(Mφ) = GL2(Z). ii) If φis the inversion then Out(Mφ)is finite. ¤ 2. Results for general rank Using the defining relations of Mφunder the form wt =t(wφ) and wt−1= t−1(wφ−1), w∈Fn, it is clear that in any element of Mφwe can push all the tletters to one side. That is, we have a normal form in Mφwhereby we can write elements uniquely in the form tkw, where kis an integer and w∈Fn. Our first observation is that the isomorphism type of Mφdepends only on the outer automorphism determined by φ±1, up to conjugacy in Out(Fn). Lemma 2.1. Let Fnbe a free group of rank n, let φ, ψ ∈Aut(Fn)and consider Mφ=hx1, . . . , xn, t |t−1xit=xiφi and Mψ=hx1, . . . , xn, s |s−1xis=xiψi. If the automorphisms φand ψare conjugate or conjugate-inverse to each other in Out(Fn), then Mφand Mψare isomorphic. More precisely, if χ∈Aut(Fn)is such that χ−1φχ =ψ²γw, for some ²=±1and w∈Fn, then the map Ω: Mφ→Mψ,xi7→ xiχ,t7→ s²wextends to an isomorphism. Proof. The map Ω is extended multiplicatively and one needs to show that it is well defined. In order to do this, it is enough to show that the relators in Mφare all sent to the trivial element in Mψ: t−1xit(xiφ)−17→ w−1s−²(xiχ)s²w(xiφχ)−1 =w−1(xiχψ²)w(xiφχ)−1 = (xiχψ²γw)(xiφχ)−1 = (xiφχ)(xiφχ)−1 = 1 ∈Mψ. Thus, we have a well defined homomorphism Ω: Mφ→Mψ, which is surjective by inspection. Moreover, if an element tkx∈Mφis in the kernel of Ω, 4 O. BOGOPOLSKI, A. MARTINO AND E. VENTURA we immediately deduce that k= 0 and xχ = 1. Since χis an automorphism of Fn, we have x= 1 and Ω has trivial kernel. 2 We continue with some basic facts about the automorphism group of Mφ. Lemma 2.2. Let Fnbe a free group of rank n, and let φ∈Aut(Fn). The group Mφ=FnoφZhas non-trivial centre if and only if φk=γwfor some k6= 0 and some w∈Fn. If this equation holds and n⩾2, then wφ =w. Proof. The result is clear for n= 0,1. So, we may assume that n⩾2. A straightforward calculation shows that the element tkw−1∈Mφcommutes with every x∈Fnif and only if φk=γw. Similarly, tkw−1commutes with tif and only if wφ =w. So, tkw−1is central in Mφif and only if φk=γwand wφ =w. Hence, Mφhas non-trivial centre if and only if φk=γwand wφ =wfor some integer kand some w∈Fnsuch that (k, w)6= (0,1). Now, using the fact n⩾2, we can simplify this. Assume the equation φk=γwholds. Since, γw=φ−1γwφ=γwφ we have wφ =w. Also, note that k= 0 implies w= 1 (because Fnhas trivial centre for n⩾2). Thus, Mφhas non-trivial centre if and only if φk=γwfor some integer k6= 0 and some w∈Fn.¤ Let Ψ ∈Aut(Mφ) and suppose it leaves Fninvariant. In this situation, its restriction to Fn,ψ= Ψ|Fn, is an endomorphism of Fnthat will be seen in the next proposition to be always an automorphism. On the other hand, factorising by the normal and Ψ-invariant subgroup Fn, we get an automorphism Ψ of Mφ/Fn∼ =Z. If Ψ = Id we shall say that Ψ is a positive automorphism of Mφ. Otherwise, Ψ is the inversion of Zand we say that Ψ is negative. In any case, tΨ = t²wfor some w∈F, where ²=±1 is the signum of Ψ. Proposition 2.3. Let Fnbe a free group of rank n, let φ∈Aut(Fn)and consider Mφ=FnoφZ. Let Ψ∈Aut(Mφ)be such that FnΨ⩽Fn, and denote by ψ:Fn→Fnits restriction to Fn. Then, i) ψis an automorphism of Fn, ii) there exists w∈Fnsuch that φψ =ψφ²γw, where ²is the signum of Ψ. Furthermore, if n⩾2then the word wis unique and satisfies the equation tΨ = t²w. Proof. Since FnΨ⩽Fn, we must have that tΨ = t±1wfor some w∈Fn (otherwise, twould not be in the image of Ψ). Now, clearly, FnΨ is a normal subgroup of Mφ=MφΨ = hFnψ, t±1wi. Hence any element g∈Mφcan be written in the form g= (vψ)(t±1w)kfor some v∈Fnand k∈Z. And here g∈Fnif and only if k= 0. Thus, FnΨ = Fnand Ψ induces an automorphism on Fn. This proves (i). THE AUTOMORPHISM GROUP OF A FREE-BY-CYCLIC GROUP IN RANK 2 5 Let ²=±1 be the signum of Ψ, that is, tΨ = t²wfor some w∈Fn. Applying Ψ to both sides of the equality xφ =t−1xt we get xφψ =w−1t−²(xψ)t²w=xψφ²γw, for all x∈Fn. Hence, φψ =ψφ²γw. Furthermore, if n⩾2, this last equation can only happen for a unique w∈Fn. This proves (ii). 2 In the following result, we impose certain hypothesis on φto ensure that every automorphism of Mφleaves the free part invariant. Under these circumstances, computing Out(Mφ) is fairly straightforward. Theorem 2.4. Let Fnbe a free group of rank n, let φ∈Aut(Fn)and consider Mφ=FnoφZ. Let Mab φdenote the abelianisation of Mφ, and Fab n∼ =Znthe abelianisation of Fn(which is not in general the image of Fn⩽Mφin Mab φ). Let φab ∈GLn(Z)be the map induced by φon Fab n, and [φ]be the class of φin Out(Fn). The following are equivalent: a) Mab φis the direct sum of an infinite cyclic group and a finite abelian group, b) the matrix φab has no eigenvalue 1. Furthermore, if these conditions hold then every automorphism of Mφ leaves Fninvariant, Aut+(Mφ) = {Ψ∈Aut(Mφ)|Ψis positive} is a normal subgroup of Aut(Mφ)of index at most 2 and moreover, if n⩾2, its image Out+(Mφ)in Out(Mφ)is also normal, of index at most two, and isomorphic to C([φ])/h[φ]i, where C([φ]) denotes the centraliser of [φ]in Out(Fn). Proof. To prove the equivalence of (a) and (b), note that Mab φ∼ =ht| i ⊕ Fab n/Im(φab −Id). Then, Fab n/Im(φab −Id) is finite if and only if rankZ(Im(φab −Id)) = rankZ(Fab n) = n. And this happens if and only if rankZ(Ker(φab −Id)) = 0, which is the same as saying that φab has no eigenvalue 1. We shall now prove the remaining assertions under the assumption that these conditions hold. Consider the abelianisation map Mφ→Mab φ. Since Fab n/Im(φab −Id) is the torsion subgroup of Mab φ, its full pre-image Fnis characteristic in Mφ. Hence, every automorphism of Mφleaves Fninvariant. At this point, it is clear that Aut+(Mφ) is a normal subgroup of Aut(Mφ) of index at most 2, and so is Out+(Mφ) in Out(Mφ). Assuming n⩾2, it remains to prove that Out+(Mφ)∼ =C([φ])/h[φ]i. Define the map f:Aut+(Mφ)→C([φ])/h[φ]i Ψ7→ [Ψ|Fn]h[φ]i. 6 O. BOGOPOLSKI, A. MARTINO AND E. VENTURA Note that by Proposition 2.3, the image of this map lies in C([φ])/h[φ]i. Clearly, fis a homomorphism. First, we will prove that Im f =C([φ])/h[φ]i. Let ψ∈Aut(Fn) be such that [ψ]∈C([φ]). Then, φψ =ψφγwfor some w∈Fn. In this situation, it is straightforward to verify that ψextends to a well defined automorphism Ψ of Mφby just sending tto tw. Clearly, Ψ ∈Aut+(Mφ) and its f-image is [ψ]h[φ]i. Now we will prove that Ker f =Inn(Mφ). For every element g=tkw∈ Mφ, we have that Γg|Fn=φkγwand so, Γgmaps under fto the identity element of C([φ])/h[φ]i. This means that Ker f ⩾Inn(Mφ). Conversely, let Ψ ∈Ker f. Then, Ψ|Fn=φkγw= Γtkw|Fnfor some integer kand some w∈Fn. Also, applying Ψ to both sides of the equation t−1xt =xφ and using the positivity of Ψ and the fact n⩾2, it is straightforward to check that tΨ = t(wφ)−1w=tΓtkw. Thus, Ψ = Γtkw. This completes the prove that Ker f =Inn(Mφ) and so, Out+(Mφ)∼ =C([φ])/h[φ]i.2 The extreme opposite case to the one considered above is when φis the identity automorphism (or, in fact, an inner automorphism). We also calculate the automorphism group in this case. Theorem 2.5. Let Fn=hx1, . . . , xnibe a free group of rank n⩾2, and let M=MId =Fn×Z. Consider the group ZnoC2where C2is the cyclic group of order 2 which acts by sending uto −ufor all u∈Zn(think uas a column vector); also, consider the action of Aut(Fn)(and also Out(Fn)) on it given by the trivial action on C2, and the natural action after abelianisation on Zn. Then, Aut(M)∼ =(ZnoC2)oAut(Fn)and Out(M)∼ =(ZnoC2)oOut(Fn). Proof. Clearly, distinct automorphisms of Fnextend to distinct positive automorphisms of Mby sending tto t. In this sense, we shall think Aut(Fn) as a subgroup of Aut(M). On the other hand, consider ZnoC2=Znohvi so that v−1uv =−ufor all u∈Zn. It is straightforward to verify that this group acts faithfully on M=Fn×Z, whereby an element (v², u), u= (u1, . . . , un)T, sends xito tuixiand tto t1−2², where ²= 0,1. So, we shall think ZnoC2⩽Aut(M). Note that Aut(Fn) and ZnoC2have trivial intersection as subgroups of Aut(M). We shall now show that they generate Aut(M). Let Ψ ∈Aut(M). It will be sufficient to show that we can multiply Ψ by elements in Aut(Fn) and ZnoC2until we get the identity. Note that, since the centre of Mis its infinite cyclic subgroup generated by t(use n⩾2 and see Lemma 2.2), htiis characteristic in Mand so, tΨ = t±1. Thus, after possibly composing with v∈ZnoC2, we can assume that tΨ = t. Then, write xiΨ = tuiwi for i= 1, . . . , n, and u= (u1, . . . , un)T. Since Ψ is an automorphism, {w1, . . . , wn}must generate (and so form a basis of) Fn. Composing Ψ with the automorphism which fixes tand sends wiback to xi,i= 1, . . . , n, THE AUTOMORPHISM GROUP OF A FREE-BY-CYCLIC GROUP IN RANK 2 7 we obtain the element (v0, u)∈ZnoC2. This proves that Aut(Fn) together with ZnoC2generate Aut(M). Consider now elements of Aut(M), Θ ∈Aut(Fn) and (v², u)∈ZnoC2, u= (u1, . . . , un)T. We shall calculate the conjugate Θ(v², u)Θ−1as an element of Aut(M). Let θ= Θ|Fn∈Aut(Fn) and let θab = (bi,j)∈GLn(Z) be its abelianisation (an n×nintegral matrix whose i-th row describes the total exponent sums of xiθ). Bearing this in mind, Θ(v², u)Θ−1acts as xi7→ xiΘ7→ tci(xiΘ) 7→ tcixi t7→ t7→ t1−2²7→ t1−2², where ci=Pn j=1 bi,jujis the i-th entry of the column vector θabu. In other words, Θ(v², u)Θ−1= (v², θ abu) for every u= (u1, . . . , un)T∈Zn. This immediately shows that ZnoC2is normal in Aut(M). Hence, Aut(Fn×Z)∼ =(ZnoC2)oAut(Fn), where the action of Aut(Fn) in this last semi-direct product is the trivial one over the C2part, and the natural one after abelianisation over the Znpart. Lastly, to prove the final statement note that, since htiis central, inner automorphisms of Fn×Zare just inner automorphisms by elements of Fn, and all of them fix t. Thus, Inn(M) = Inn(Fn)⩽Aut(Fn) and so Out(M)∼ =(ZnoC2)oOut(Fn), where the actions are just like before but factorised by Inn(Fn). 2 3. Aut(F2)and GL2(Z) For the rest of the paper we will be considering only the case n= 2. Hence, we shall avoid unnecessary subscripts by using the letters {a, b}as free generators of F2=ha, bi. In this section we will briefly review some well known facts about Aut(F2), Out(F2) and GL2(Z). The abelianisation map F2→Fab 2induces naturally a surjective map Aut(F2)³GL2(Z) for which, abusing notation, we shall write φ7→ φab. More precisely, φab is the 2 ×2 integral matrix whose first (second) row counts the total a- and b-exponent sums of aφ (of bφ). Clearly, this is well defined for any rank, but in rank 2 it has some special properties which make this case easier to study. The main specificity of the rank 2 case is the following well known result. Theorem 3.1 (Nielsen, Prop. 4.5 in [4]).The kernel of the map from Aut(F2)to GL2(Z)consists of precisely the inner automorphisms of F2. That is, Out(F2)∼ =GL2(Z). This means that, for every automorphism φ∈Aut(F2), the 2×2 integral matrix φab is enough to recover the automorphism φup to conjugation. 8 O. BOGOPOLSKI, A. MARTINO AND E. VENTURA Since the isomorphism type of Mφ(and so that of Aut(Mφ)) only depends on φup to conjugation, φab contains all the algebraic information we may want about Mφand Aut(Mφ). Matrices in GL2(Z) can be classified according to their eigenvalues and dynamics, often leading to useful information for Aut(F2). This is often done in the following way. Definition 3.2. Let Abe a 2 ×2 integral invertible matrix, A∈GL2(Z). If A2is the identity matrix, A2=I2, we say that Ais elliptic. Otherwise, Ais called hyperbolic if |trace(A2)|>2, parabolic if |trace(A2)|= 2, and elliptic if |trace(A2)|<2. Suppose A∈GL2(Z) is a hyperbolic matrix. Then, A26=I2has two real eigenvalues, α, 1/α, such that |α|>1. Since Apreserves the one dimensional eigenspaces of A2,Amust also have two real eigenvalues, β, ±1/β, such that |β|>1. In particular, Adoes not have 1 as an eigenvalue. Suppose A∈GL2(Z) is a parabolic matrix. Then, A26=I2has characteristic polynomial equal to (x±1)2. This implies that A2is conjugate, in GL2(Z), to ±¡1k0 0 1 ¢for some 0 6=k0∈Z(take a rational eigenvector of eigenvalue ±1, multiply it by a scalar to obtain an integral vector vwith coprime entries, and then extend to a basis {u, v}of Z2). A simple calculation shows now that Amust then be conjugate to one of the following matrices µ1k 0 1 ¶ µ −1k 0−1¶, for some integer k6= 0. These are infinite order matrices and the first of these will turn out to be the most challenging case to consider. Finally, suppose A∈GL2(Z) is an elliptic matrix. Then, either A2=I2 or the characteristic polynomial of A2is equal to x2+ 1, x2+x+ 1 or x2−x+ 1. In the first case, Ais either ±I2or is conjugate, in GL2(Z), to ¡1k 0−1¢for some k∈Z(by a similar reasoning as above). Otherwise, A have complex conjugate roots and, in particular, it does not have 1 as an eigenvalue. 4. The rank 2 case: proof of Theorem 1.1 Given φ∈Aut(F2), we shall analyse Aut(Mφ) and prove Theorem 1.1 by following the classification of matrices in the previous section for φab. First, we state the following two lemmas for later use. Lemma 4.1. Every non-central element in GL2(Z)generates a finite index subgroup of its own centraliser. The centre consists precisely of the matrices ±I2. THE AUTOMORPHISM GROUP OF A FREE-BY-CYCLIC GROUP IN RANK 2 9 Proof. The first statement of the lemma follows easily from the presentation of GL2(Z) as an amalgamated product, GL2(Z)∼ =D4∗D2D6, where Dnis the dihedral group of order 2n. The statement about central elements is elementary. 2 Lemma 4.2. Consider the automorphism φof F2=ha, bigiven by aφ =abk and bφ =b, where k6= 0. Then, for every integer r6= 0 and every w∈F2, i) Fix φ =Fix φr=haba−1, bi, ii) if wφris conjugate to w, then wis conjugate to an element fixed by φ. Proof. Given an arbitrary reduced word w∈F2, let us split it into pieces each of the form bm,abm,bma−1or abma−1, where mis some integer. There can be many variants of such a splitting, but we shall use the special one defined by putting breaking points precisely before each occurrence of a and after each occurrence of a−1. One can easily see that this splitting is invariant under the action of φ, and that the pieces do not interact under iterates of φ. Hence, if wφr=wthen the corresponding pieces in the splitting of wmust also be fixed by φr, which rules out the possibilities abm and bma−1(because rk 6= 0). This proves that Fix φr⩽haba−1, bi. The inclusions haba−1, bi⩽Fix φ ⩽Fix φrare obvious. This proves (i). In order to prove (ii), note that we can assume wis cyclically reduced. If wis a power of bthere is nothing to prove. So, we may also assume that w contains a±1. Moreover, by inverting and cyclically permuting if necessary, we may assume that wbegins with a. So, in this particular situation, assume that wφris conjugate to w. The splitting of wmust begin with a piece of the form abmor abma−1, and must end with a piece of the form bmor abm. But this splitting is stable under iterates of φhence, the first and last pieces in wφrwill be of the corresponding same types. In particular, wφris still cyclically reduced. Thus, wφrmust be a cyclic permutation of w. Then, for a suitable s,wφrs =wwhich, by (i), means that wis fixed by φ. This completes the proof. ¤ Proof of Theorem 1.1. All along the prove, let us fix the following notation. Let F2=ha, bibe a free group of rank n= 2, let φ∈Aut(F2) and let Mφ=F2oφZbe the mapping torus of φ. Let φab ∈GL2(Z) be the map induced by φon Fab 2∼ =Z2, i.e. the 2 ×2 integral matrix whose rows count the total a- and b-exponent sums of {aφ, bφ}. First of all, note that the discussions in the previous section show that a generic matrix φab ∈GL2(Z) fits into one of the cases distinguished in Theorem 1.1. Namely, if Ais hyperbolic then it satisfies (iii), if it is parabolic it satisfies either (iii) or (v), and if it is elliptic then it fits into either (i), (ii) or (iv). Uniqueness is a straightforward exercise in linear algebra.