On the vanishing of the lower K-theory of the Holomorph of a free group on two generators

Certain obstructions that appear in problems of topological rigidity of manifolds are elements of algebraic K-groups, specially lower K-groups. For this reason, the calculation of the lower K-groups has implications in geometric topology. The main modern tool for calculating lower K-groups (and other geometrically interesting obstruction groups) is the Farrell-Jones Fibered Isomorphism Conjecture. The Conjecture provides an inductive method for calculating the obstruction groups of a group from those of certain subgroups. More specifically, if a group satisfies the Fibered Isomorphism Conjecture for a specific theory, then the obstruction groups can be calculated from the obstruction groups of the virtually cyclic subgroups. The last class of subgroups consists the finite subgroups and groups that are virtually infinite cyclic. The virtually infinite cyclic groups are of two types: • Groups V that surject onto the infinite cyclic group Z with finite kernel, i.e. V = H ⋊ Z with H finite. • Groups W that surject onto the infinite dihedral group D ∞ with finite kernel, i.e. W = A * B C with B finite and [A : Obstruction groups that can be calculated this way are pseudoisotopy groups, Kgroups of group rings, K-groups of C * -algebras, L -∞ -groups. The fundamental work of Farrell-Jones ( [8]) deals with the Fibered Isomorphism Conjecture for the pseudoisotopy spectrum. It should be noted that if a group satisfies the Fibered Isomorphism Conjecture for pseudoisotopies, then it satisfies the Isomorphism Conjecture for the lower K-groups. The reason for this is that the lower homotopy groups of the pseudoisotopy spectrum and the K-theory spectrum are isomorphic. For K-groups there is a refinement of the Conjecture that was given in [6]. They showed that finite and virtually infinite cyclic subgroups of the first type suffice in detecting the K-theory of the group. Our main interest is in computing the lower K-groups of the holomorph of F 2 , the free group on two generators. For a group G, the holomorph of G is the universal split extension of G. Thus, it fits into a split exact sequence: The main result of the paper is the following: Theorem (Main Theorem). The group Hol(F 2 ) satisfies the Fibered Isomorphism Conjecture for pseudoisotopies. Furthermore, if Γ < Hol(F 2 ) then Notice that Hol(F 2 ) is equipped with a sequence of surjections: The second surjection is induced by sending an automorphism of F 2 to an automorphism of its abelianization Z 2 . Notice that the kernel of both surjections is isomorphic to F 2 . To show that Hol(F 2 ) satisfies the Fibered Isomorphism Conjecture, we use the fact that every automorphism of F 2 is geometric, i.e. it can be realized by a diffeomorphism on a surface with boundary. That allows us to show first that Aut(F 2 ) satisfies the Conjecture and using the same fact again that Hol(F 2 ) does also. The group GL 2 (Z) splits as an amalgamated free product of finite dihedral groups: GL Thus both Aut(F 2 ) and Hol(F 2 ) split as amalgamated free products. Using this fact and the properties of elements of finite order in Aut(F 2 ) given in [16,7], we determine the finite and the virtually infinite cyclic subgroups of Aut(F 2 ) and Hol(F 2 ), up to isomorphism. The list of groups is short and their lower K-theory vanishes. The Main Theorem follows from this observation. The authors would like to thank Tom Farrell for asking the question on the lower K-theory of Aut(F 2 ). Let G be a discrete group. By a class of subgroups of G we wean a collection of subgroups of G that is closed under taking subgroups and conjugates. In our application we consider the following classes of subgroups: • F in, the class of finite subgroups of G. Let C be a class of subgroups of G. The classifying space for C, E C G is a G-CWcomplex such that the isotropy groups of the actions are in C and, for each H ∈ C, the fixed point set of H is contractible (for more details [5], [13]). Remark 2.1. For Aut(F 2 ) the classifying space for finite groups is the auter space ( [10], [11]). The Fibered Isomorphism Conjecture (FIC) was stated by ). For the groups that holds, it provides an inductive method for computing obstruction groups in geometric topology (for a review see [14]). If G satisfies the FIC then the natural map Notice that the left hand side of the isomorphism can be computed from the virtually cyclic subgroups of G. In general, there are "forgetful maps" The difference between the class F in and the class VC is that the second class can be captured by the Waldhausen and Bass-Farrell Nil-groups of the infinite virtually cyclic subgroups. In [4] and [6], it was shown that the second map is an isomorphism. Essentially, the authors proved that the Waldhausen's Nil-groups that appear in the K-theory of virtually infinite cyclic subgroups can be detected by the Bass-Farrell Nil-groups that appear in the F BC class. The FIC is known to hold for certain classes of groups. One class of interest for this paper is the class of strongly poly-free groups. A group Γ is called strongly poly-free if there is a filtration: such that: (1) Γ i is normal in Γ for each i. (2) Γ i /Γ i+1 is finitely generated free for all 0 ≤ i ≤ n -1. (3) For each γ ∈ Γ there is a compact surface S and a diffeomorphism f : S → S such that the induced homomorphism f * on π 1 (S) is equal to c γ in Out(π 1 (S)), where c γ is the action of γ on Γ i /Γ i+1 by conjugation and π 1 (S) is identified with Γ i /Γ i+1 via a suitable isomorphism. In [2] and [9] it was shown that a finite extension of a strongly poly-free group satisfies the FIC. Remark 2.2. Let Γ be a group that satisfies (1) and ( 2) above. We assume that Γ i /Γ i+1 ∼ = F 2 . Then G is strongly poly-free. For this, let T 2 be the torus and where Aut(F 2 ) denotes the automorphism group of F 2 . Let c γ be an induced homomorphism as in (3) above. Then the image of c γ to Out(F 2 ) can be represented by a diffeomorphism f of T 2 that fixes p. After an isotopy starting at the identity on T 2 , we can assume that f fixes a small open disk D around p. Then f induces a diffeomorphism on the compact surface Start with an exact sequence of groups. In the Appendix of [8], it was shown the FIC holds for B if: • It holds for C. • For each virtually cyclic subgroup V of C, it holds for r -1 (V ). Using this result we show the following be an exact sequence. If the FIC holds for H, then it holds for G. Proof. Using the result in [8], it is enough to show that the FIC holds for r -1 (V ), where V is a virtually cyclic subgroup of H. If V is finite, then r -1 (V ) is a finite extension of F 2 and r -1 (V ) is a finite extension of a free group. The result follows from Remark 2.2. If V is infinite, then V contains an infinite cyclic normal subgroup W of finite index. Then r -1 (W ) is a normal subgroup of r -1 (V ) and fits into an exact sequence: Then there is a filtration r -1 (W ) > F 2 > {1}, with the first quotient being an infinite cyclic group. Obviously, every homomorphism of Z is realized by a diffeomorphism of S 1 ×[0, 1]. Using Remark 2.2, we see that r -1 (W ) is strongly poly-free. Therefore, r -1 (V ) is a finite extension of a strongly poly-free group. By [9], it satisfies the FIC, completing the proof of the proposition. Let Hol(F 2 ) denote the holomorph of F 2 , namely, the universal split extension of F 2 : Notice that there is an exact sequence that is induced by mapping the automorphisms of F 2 to the automorphisms of its abelianization. That induces an exact sequence: Proposition 2.4. The FIC holds for Aut(F 2 ) and Hol(F 2 ). Proof. The group GL 2 (Z) contains a subgroup of finite index that is isomorphic to F 2 . In fact, the following short exact sequence is known to hold, as a result of the standard action that GL 2 (Z) admits on the upper half plane: (see for example [7]). Thus the FIC holds for GL 2 (Z). Now Aut(F 2 ) fits into an exact sequence: By Proposition 2.3, the FIC holds for Aut(F 2 ). Also, Hol(F 2 ) fits into an exact sequence: By Proposition 2.3 again, the FIC holds for Hol(F 2 ). Since there is an exact sequence and F 2 = a, b is torsion free, every finite subgroup of Aut(F 2 ) maps isomorphically to a finite subgroup of GL 2 (Z). On the other hand, GL 2 (Z) admits a decomposition as an amalgamated free product of the form where D 2 , D 4 and D 6 are dihedral groups of orders 4, 8 and 12 respectively. Hence, any finite subgroup of GL 2 (Z) is a subgroup of a conjugate of either D 2 or D 4 or D 6 and hence, so is every finite subgroup of Aut(F 2 ). Now a presentation for Aut(F 2 ) is given by a τ b where τ a , τ b are the inner automorphism of F 2 corresponding to a, b respectively (see for example [15]). Moreover, a presentation for GL 2 (Z) is given by As shown in [16], if g is an element of finite order in Aut(F 2 ), then g is conjugate in Aut(F 2 ), to one of the following elements p, px, pxτ a , x 2 , y 2 τ -1 b or x with orders 2, 2, 2, 2, 3 or 4 respectively. This fact implies that Aut(F 2 ) cannot contain finite subgroups isomorphic to D 6 . Moreover, any element of Aut(F 2 ) can be written uniquely in the form p r u(x, y)x 2s w(τ a , τ b ) where r, s ∈ {0, 1}, w(τ a , τ b ) is a reduced word in Inn(F 2 ) and u(x, y) is a reduced word where x, y, y -1 are the only powers of x, y appearing (see [16,15]). Also, due to the decomposition (1) of GL 2 (Z), Aut(F 2 ) is also an amalgamated free product of the form where B, C and D fit into the following short exact sequences Moreover, since every one of B, C and D are free-by-finite groups they admit an action on a tree with finite quotient graph and finite vertex and edge stabilizers (as a corollary of the Almost Stability Theorem of Dicks and Dunwoody [7]). In fact, they are also amalgamated free products of the form Once again, the elements of finite order are p, px, x 2 τ b , px 3 τ b , x 2 , px 2 , y 2 τ -1 b τ a and x. To be in accordance with Meskin, we see that By definition, G = Hol(F 2 ) is the universal split extension of Aut(F 2 ) and thus it fits to the split exact sequence So Hol(F 2 ) = F 2 ⋊ Aut(F 2 ). Hence, the above presentation for Aut(F 2 ) provides us with a presentation for Hol(F 2 ). Namely, Moreover, the decomposition (2) of Aut(F 2 ) provides an amalgamated free product decomposition for Hol(F 2 ): and based on (3) we have Based again on the Almost Stability Theorem, we see that every vertex group in the above graphs of groups is a free-by-finite group so, it also admits a decomposition as a graph of groups with finite vertex groups. An analysis, based on the presentations and also on the fact that the action of x, y, p on a, b is the same as that on τ a , τ b , would give us the following: In F 2 ⋊ B, In the above, S * t denotes the HNN-extension with base group S and stable letter t. For example, F 2 ⋊ px, x 2 τ b has a presentation of the form where ξ 1 = px and ξ 2 = x 2 τ b . By setting ζ 2 = bξ 2 and eliminating b, we get . Now by setting ξ 3 = ξ 1 a and eliminating a we get which is the desired decomposition. So now we can prove the following result which generalizes the result in [16] on the elements of finite order in Aut(F 2 ). Lemma 3.1. An element of finite order in Hol(F 2 ) is conjugate to exactly one of p, px, pxa, pxτ a , pxτ a a, x 2 , x 2 b, y 2 τ -1 b , y 2 τ -1 b a and x with orders 2, 2, 2, 2, 2, 2, 2, 3, 3 and 4 respectively. Proof. Given the above decomposition, every element of finite order is a conjugate of an element of a vertex group. So it suffices to observe the following: b τ a has no equivalent for a and b due to the semidirect product structure of G. From the fact that Hol(F 2 ) = a, b ⋊ Aut(F 2 ) we have that every element W of Hol(F 2 ) can be written uniquely in the form where V ∈ Aut(F 2 ) and z(a, b) is a word in the free group a, b . So, the normal form for the elements of Aut(F 2 ) implies the existence of a normal form for the elements of Hol(F 2 ): where w(t a , t b ) is a reduced word in the free group t a , t b , u(x, y) is a reduced word where x, y, y -1 are the only powers of x, y appearing and r, s ∈ {0, 1}. Moreover, every vertex group in the decomposition (4) has also a normal form. More specifically, every element in F 2 ⋊ B can be written uniquely in the form p r x n x 2s w(τ a , τ b )z(a, b) where r, n, s ∈ {0, 1} and every element in although such a sequence does not split. We are searching for subgroups of G = Hol(F 2 ) which are isomorphic to A ⋊ Z where A is a finite subgroup of G. In our argument we shall make extensive use of the following well known result of Bass-Serre theory [17]. Let M be a group that acts on its standard tree T and m ∈ M such that m stabilizes two distinct vertices of T . Then m stabilizes the (unique reduced) path that connects the two vertices. In particular, m is an element of every edge stabilizer of every edge that constitute the path that connects the two vertices. In fact we shall show the following: b a. Assume that there is a subgroup of G isomorphic to Z/3Z ⋊ Z. Then, conjugating if necessary, we may assume that there is an element of infinite order in G, say g, such that g -1 (y 2 τ -1 b a s )g = (y 2 τ -1 b a s ) ±1 with s ∈ {0, 1}. Based on the decomposition (4) we see that the above relation implies that y 2 τ -1 b a s stabilizes both vertices F 2 ⋊ C and g -1 (F 2 ⋊ C) and hence the path that connects them. So it belongs to the edge stabilizer F 2 ⋊ D, unless g ∈ F 2 ⋊ C. But if g ∈ F 2 ⋊ D we have a contradiction since by decomposition (5), F 2 ⋊D cannot contain elements of order 3. Now if g ∈ F 2 ⋊ C, then based again on the decomposition (5) of F 2 ⋊ C, we have that g stabilizes both F 2 ⋊ D 3 and g -1 (F 2 ⋊ D 3 ) and so it belongs to F 2 ⋊ Z/2Z, a further contradiction, unless again g ∈ F 2 ⋊ D 3 . Finally, by the decomposition of F 2 ⋊ D 3 = D 3 * Z/2Z D 3 we have again that g has to be an element of either of the two D 3 vertices and hence an element of finite order. We shall show now that G cannot contain subgroups isomorphic to Z/4Z ⋊ Z. Assume that G contains such a subgroup, say A. Then, A is generated by a conjugate of x, since the conjugacy class of x is the only class of elements of order 4, and by an element g of G. Using conjugation if necessary, we may assume the element of order 4 in A is x. Now let g ∈ G such that x, g ∼ = Z/4Z ⋊ Z. Then g -1 xg = x ±1 . Let G act to the tree that corresponds to the decomposition (4). Then, due to the above relation, x stabilizes both F 2 ⋊ B and g -1 (F 2 ⋊ B) and so it stabilizes the path between the two vertices. Hence, x ∈ F 2 ⋊ D a contradiction, unless g ∈ F 2 ⋊ B. Moreover, using the decomposition (5) we see that g can only be an element of F 2 ⋊ D 4 and using the fact that F 2 ⋊ D 4 = D 4 * Z/2Z D 2 it can only be an element of D 4 and so is of finite order. This completes the proof of claim 1. Claim 2. There are no subgroups of G of the form D 2 ⋊ Z or of the form D 3 ⋊ Z. Up to conjugacy, the possible D 2 in G are x 2 , p , px, x 2 , px, x 2 b , px, x 2 τ b , px, x 2 τ b b -1 and pxa, x 2 τ b b -1 . Now notice that all last five, px, x 2 , px, x 2 b , px, x 2 τ b , px, x 2 τ b b -1 and pxa, x 2 τ b b -1 appear only once in the graph of groups decomposition (5) of Hol(F 2 ), as vertex groups. Moreover, in all five, none of the generators is conjugate to the other, i.e. there are no g ∈ G such that gpxg -1 = x 2 or gpxg -1 = x 2 b or gpxg -1 = x 2 τ b b -1 or gpxg -1 = x 2 b or gpxag -1 = x 2 τ b b -1 by Lemma 3.1. Hence, a relation of the form gD 2 g -1 = D 2 implies (repeating again the argument of claim 1) that g is an element of finite order. So the only possibility for a semidirect product D 2 ⋊ Z lies with x 2 , p . So assume that there is an element g ∈ G such that g, p, x 2 = D 2 ⋊ Z. Then, since p and x 2 are not conjugates and px 2 is conjugate to p, the action of g is either gpg -1 = p and gx 2 g -1 = x 2 , or gpg -1 = px 2 and gx 2 g -1 = x 2 . Let us concentrate to the relation gpg -1 = p. Given the normal form of g = p r u(x, y)x 2s w(τ a , τ b )z(a, b) we have that p r u(x, y)x 2s w(τ a , τ b )z(a, b)pz -1 (a, b)w -1 (τ a , τ b )x -2s u -1 (x, y)p -r = p. The above relation implies the existence of the following relation in GL 2 (Z): P r U (X, Y )X 2s P X 2s U -1 (X, Y )P r = P Proof. We will show the theorem for G = Hol(F 2 ). The proof for Aut(F 2 ) is similar. By Proposition 2.4, G satisfies the FIC. Let Γ < Hol(F 2 ). Then by [8], Γ also satisfies the FIC. Thus the maps H G q (E F BC Γ; KZ -∞ ) → Wh q (ZΓ), q ≤ 1, are isomorphisms. There is a spectral sequence that computes the left hand side of such an isomorphism: , where V is in F BC. Now, by the decomposition of Hol(F 2 ) and Proposition 3.2: (1) If V is finite, V will isomorphic to one of the following groups: Z/2Z, Z/3Z, Z/4Z, D 2 , D 4 . But in this case from the lists in [1] and [12] Wh q (V ) = 0, for q ≤ 1.