Infinite-dimensional cohomology of SL 2

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1 Infinite-dimensional cohomology of SL 2 ( Z[t, t 1 ] ) Sarah Cobb June 5, 2015 Abstract For J an integral domain and F its field of fractions, we construct a map from the 3-skeleton of the classifying space for Γ = SL 2 (J[t, t 1 ]) to a Euclidean building on which Γ acts. We then find an infinite family of independent cocycles in the building and lift them to the classifying space, thus proving that the cohomology group H (SL 2 ( 2 J[t, t 1 ] ) ) ; F is infinite-dimensional. 1 Introduction Let J be an integral domain with identity and F its field of fractions. Our goal in this paper is to prove the following theorem: ( Theorem 1. H (SL 2 2 J[t, t 1 ] ) ) ; F is infinite-dimensional. This theorem generalizes several earlier results about the finiteness properties of SL 2 ( Z[t, t 1 ] ). In [KM97], Krstić-McCool prove that SL 2 (J[t, t 1 ]), among other related groups, is not F 2. In [BW06], Bux-Wortman use geometric methods to prove that SL 2 (Z[t, t 1 ]) is also not F P 2. In particular, they use the action of SL 2 (Z[t, t 1 ]) on a product of locally infinite trees. Bux-Wortman also ask whether their proof can be extended to show the stronger result that H 2 (SL 2 (Z[t, t 1 ]); Z) is infinitely generated. Knudson proves that this is the case in [Knu08] using algebraic methods. Note that this result also follows from Theorem 1: Since Q is a field, H 2 (SL 2 (Z[t, t 1 ]); Q) Hom(H 2 (SL 2 (Z[t, t 1 ]); Q), Q) Since Hom(H 2 (SL 2 (Z[t, t 1 ]); Q), Q) is infinite dimensional, so is H 2 (SL 2 (Z[t, t 1 ]); Q), which implies that H 2 (SL 2 (Z[t, t 1 ]); Z) is not finitely generated as a Z-module. 1

2 The methods in this paper will be geometric. We will define two spaces on which SL 2 (J[t, t 1 ]) acts: one a Euclidean building as in [BW06], and the other a classifying space for SL 2 (J[t, t 1 ]). A map between these spaces will allow us to explicitly define an infinite family of independent cocycles in H 2 (SL 2 (J[t, t 1 ]); F ). The methods used are based on those of Cesa-Kelly in [CK15], where they are used to show that H ( 2 SL 3 (Z[t]); Q ) is infinite-dimensional. Wortman follows a similar outline in [Wor13]. Acknowledgments This paper is adapted from my dissertation, completed at the University of Utah. I am thankful for support from the institution and its faculty. I am deeply grateful to Kevin Wortman, my thesis advisor, for guiding me through this work. I would also like to thank Kai-Uwe Bux for pointing out some errors in an earlier draft; Brendan Kelly and Morgan Cesa, for detailed explanations of their results; and Jon Chaika and Yair Glasner for helpful conversations. 2 The Euclidean Building Throughout, let J be an integral domain with identity, F the field of fractions over J, and Γ = SL 2 (J[t, t 1 ]). We begin by recalling the structure of a Euclidean building on which Γ acts. The following construction uses the notation of Bux-Wortman in [BW06]. Let ν and ν 0 be the valuations on F (t) giving multiplicity of zeros at infinity and at zero, respectively. More precisely, ν ( p(t) q(t) ) = deg(q(t)) deg(p(t)), and ν 0 ( r(t) s(t) tn ) = n, where t does not divide the polynomials r and s. Let T and T 0 be the Bruhat-Tits trees associated to SL 2 (F (t)) with the valuations ν and ν 0, respectively. We will consider each tree as a metric space with edges having length 1. Let X = T T 0. Since F ((t 1 )) (respectively F ((t))) is the completion of F (t) with respect to ν (resp. ν 0 ), SL 2 (F ((t 1 ))) (resp. SL 2 (F (t))) acts on the tree T (resp. T 0 ). Therefore the group SL 2 (F ((t 1 ))) SL 2 (F ((t))) acts on X. Throughout this paper, we will regard Γ and SL 2 (F (t)) as diagonal subgroups of SL 2 (F ((t 1 ))) SL 2 (F ((t))), which act on X via that embedding. Let L (resp. L 0 ) be the unique geodesic line in T (resp. T 0 ) stabilized by the diagonal subgroup of SL 2 (F (t)). Let l : R L (resp. l 0 : R L 0 ) be an isometry with l (0) (resp. l 0 (0)) the unique vertex with stabilizer SL 2 (F [t 1 ]) (resp. 2

3 SL 2 (F [t])). Let x 0 = (l (0), l 0 (0)) serve as a basepoint of X and Σ = L L 0 so that Σ is an apartment of X. 3 The Action of Γ on X The goal of this section is to establish certain large-scale features of the action of Γ on X. In particular, we will find a horoball containing a sequence of points far from x 0 and show that the Γ-translates of this horoball are disjoint. The techniques here are similar to those used by Bux-Wortman in [BW11]. For certain parts of the proof, it will be convenient to work with Γ F = SL 2 (F [t, t 1 ]) instead of with Γ. Note that Γ F also acts on T and on T 0 and therefore acts diagonally on X. We also define the following useful subgroups: ( ) P = a b 0 a 1 a, b F (t) ( ) 1 x U = x F (t) ( ) a 0 A = 0 a 1 a F (t) For any diagonal subgroup G of SL 2 (F ((t 1 ))) SL 2 (F ((t))), we will use the notation G Γ = G Γ and G ΓF = G Γ F Lemma 2. The double coset space Γ F \SL 2 (F (t))/p is a single point. ( ) [ ] [ ] Proof. Consider the action of SL 2 (F (t)) on P 1 a b x ax + by (F (t)) by =. c d y cx + dy [ ] 1 Under this action, P is the stabilizer of, so SL 0 2 (F (t))/p = P 1 (F (t)). Since Γ F acts transitively on P 1 (F (t)), Γ F \SL 2 (F (t))/p is a single point. 3

4 Let β be the Busemann function in T for l [0, ) and β 0 the Busemann function in T 0 for l 0 [0, ) Let ρ : [0, ) X be the geodesic ray with ρ(s) = (l (s), l 0 (s)). Let β ρ be the Busemann function for ρ with β ρ (x 0 ) = 0 such that β ρ (x, y) = β (x) + β 0 (y) for any x T and y T 0. Thus βρ 1 ([d, )) is a horoball based at ρ( ). Lemma 3. P ΓF stabilizes the horosphere β 1 ρ (r) for every r R. Proof. Since P ΓF = A ΓF U ΓF, it suffices to show that each of these groups preserves βρ 1 (r). Since β and β 0 are U ΓF -equivariant, so is β ρ. Thus U ΓF preserves βρ 1 (r). Let ( ) a 0 A F = 0 a 1 a F and D = ( ) t 0 0 t 1. Then A ΓF = A F D Z For any x T, y T 0 we have β (Dx) = β (x) + 2 and β 0 (Dy) = β 0 (y) 2. Therefore β ρ (D(x, y)) = β ρ (x, y) and D preserves βρ 1 (r). Since A F acts trivially on X, it preserves βρ 1 (r). Note that this also implies that the horoball β 1 ρ ([d, )) is stabilized by P ΓF, since it is a union of horospheres. The following lemma will be useful for describing the action of U on the boundaries T and T 0. Lemma 4. For any open sets V F ((t 1 )) and W F ((t)), we have V W F (t) Proof. Consider the sequence α i = ti, which converges to 1 in F t i +1 ((t 1 )) and to 0 in F ((t)). Let v V and w W. Then the sequence β i = (v w)α i + w converges to v in F ((t 1 )) and to w in F ((t)). Therefore β i V W F (t) for any sufficiently large i. Lemma 5. Σ is a fundamental domain for the action of U on X. 4

5 Proof. Let x T, y T 0. Then the set ( ) V = α F ((t 1 )) 1 α l passes through x is open in F ((t 1 )). Similarly, ( ) W = β F ((t)) 1 β l 0 passes through y is open in F ((t)). ( ) 1 γ By lemma 4, there is some γ V W F (t). Since (x, y) Σ, we have UΣ = X. To see that Σ contains only one point from each U-orbit, observe that a point x Σ is uniquely determined by the values β (x) and β 0 (x). Since U preserves β and β 0, Ux Σ = x. Let x n = (l (n), l 0 (n)). Lemma 6. There is some s > 0 such that for every R > 0 β 1 ρ (R) Nbhd s (Γ F x m ) where m is an integer such that R m 1 2. Proof. Let y = (l (R), l 0 (R)) = β 1 ρ (R) ρ and D = and Since we see that ( ) t 0 0 t 1. β 1 ρ (R) Σ = { (l (x), l 0 (R x)) x R } D n (l (a), l 0 (b)) = (l (a + 2n), l 0 (b 2n)) β 1 ρ (R) Σ Nbhd 2 2 (A Γ y) 5

6 Since X = UΣ and U preserves βρ 1 (R), β 1 ρ (R) Nbhd 2 2 (UA Γ y) Let m Z be such that R m 1. Since d(x 2 m, y) < 2, we have 2 Let β 1 ρ (R) Nbhd (UA Γ x m ) ( ) ( ) C = 1 a 1 b, a F [[t 1 ]], b F [[t]] 2 Then C SL 2 (F ((t 1 ))) SL 2 (F ((t))) is bounded with U CU Γ. Thus, there is some r R such that Ux Nbhd r (U Γ x) for every x X, which yields β 1 ρ (R) Nbhd r(u ΓA Γ x m ) Since U Γ A Γ < Γ F, this gives the desired inclusion β 1 ρ (R) Nbhd r(γ F x m ) Lemma 7. For every C > 0, there is some N N such that d(x n, Γ F x 0 ) > C for every n > N. Proof. The following proof is based on the proof of Lemma 2.2 by Bux-Wortman in [BW06]. Let C > 0. We will show that any subsequence of {x 2n } is unbounded in the quotient space Γ F \X. This implies that only a finite number of points of {x 2n } can be contained in any neighborhood of Γ F x 0. Since a finite number of points of {x 2n } are contained in Nbhd C+ 2 (Γ F x 0 ) and d(x 2n, x 2n+1 ) = 2, a finite number of points of {x 2n+1 } are contained in Nbhd C (Γ F x 0 ). This suffices to prove the lemma. The group SL 2 (F (t)) SL 2 (F (t)) acts componentwise on X, and has a metric induced by the valuations v and v 0. Under this metric, the stabilizer of x 0 is a bounded subgroup. Thus, to prove that a set of vertices in Γ F \X is not bounded, it suffices to prove that it has unbounded preimage under the projection Γ F \ ( SL 2 (F (t)) SL 2 (F (t)) ) Γ F \X 6

7 given by Γ F g( Γ F ) gx 0. t 0 Let D = 0 t 1 and δ = (D, D 1 ) SL 2 (F (t)) SL 2 (F (t)). Then A n x 0 = x 2n. It therefore suffices to prove that any infinite subset of {Γ F δ n } n N is unbounded in Γ F \ ( SL 2 (F (t)) SL 2 (F (t)) ). Assume that this is not the case. That is, assume that there is some infinite subset I N such that {Γ F δ i } i I is ( bounded. ) Then there is some L such that for every i I there is some M i = a i c i b i d i Γ F such that the values of v of the coefficients of M i D i are bounded from below by L and the values of v 0 of the coefficients of M i D i are also bounded from below by L. Then and L v (a n t n ) = v (a n ) + nv (t) = v (a n ) n L v 0 (a n t n ) = v 0 (a n ) nv 0 (t) = v 0 (a n ) n This gives that v (a n ) 1 and v 0 (a n ) 1 whenever n 1 L, which implies that a n = 0. The same argument shows that c n = 0, which implies that M i is not in Γ F when i 1 L. Lemma 8. There is some R > 0 such that Γ F x 0 β 1 ρ ([R, )) =. Proof. Since βρ 1 ([R, )) is a union of the horospheres βρ 1 (r) for r [R, ), Lemma 6 guarantees that βρ 1 ([R, )) {m Z m>r 1 2} Nbhd s (Γ F x m ) for some s. By Lemma 7, d(x m, Γ F x 0 ) > s for large values of R. This implies that Thus, we may conclude that d(γ F x m, Γ F x 0 ) > s Γ F x 0 βρ 1 ([R, )) = 7

8 Fix R as in Lemma 8. Let d be the maximum distance from a point in the horosphere βρ 1 (R) to the orbit Γ F x 0. Lemma 9. d is finite. Proof. By Lemma 6, βρ 1 (R) Nbhd s (Γ F x n ) for some n > 0, s > 0. Thus for every x βρ 1 (R) there is some γ Γ F such that d(x, γ x n ) < s. Since d(γ x n, γ x 0 ) = d(x n, x 0 ) = n 2 we have d(x, γ x 0 ) < s + n 2. Since this bound is independent of x, it follows that d < s + n 2. Let H = βρ 1 ([R + d, )). Lemma 10. Let γ Γ F. If γh H, then γh = H and γ P ΓF. Proof. Let C be the chamber of X containing the point ρ( ) and S X an apartment whose boundary contains C and γc. The endpoints of the arc C are l ( ) and l 0 ( ). Thus, any chamber adjacent to C contains one of these two points. Because Stab ΓF (l ( )) = Stab ΓF (l 0 ( )) < P, any element of Γ F either stabilizes both l ( ) and l 0 ( ) or neither. Therefore, C and γc cannot be adjacent. Since each apartment of X contains exactly four chambers, the two chambers are either equal or opposite. If C and γc are opposite, then S H γh is contained in a neighborhood of a hyperplane in S. Suppose x S γh βρ 1 (R). It follows from the choice of d that there is some y Γ F x 0 such that d(x, y) d. Since β γρ (x) R + d, it is clear that β γρ (y) R. Therefore β ρ (γ 1 y) R, which contradicts Lemma 8, since γ 1 y Γ F x 0. Thus γc = C, which implies that γ P ΓF. It follows that H and γh are horoballs based at the same boundary point. Since γ preserves distance from Γ F x 0, γh = H. 4 The Classifying Space We will now construct a second space on which Γ acts freely and properly discontinuously by isometries. A map between the two spaces will allow us to construct cohomology classes of Γ in the familiar context of a product of trees. Let X 0 be a discrete collection of points {x γ γ Γ}. Γ acts freely on this set by γ x γ = x γ γ. Define a Γ-equivariant map ψ 0 : X 0 X by ψ 0 (x γ ) = γ x 0. By Lemma 8, ψ 0 (X 0 ) H =. Note also that ψ 0 (X 0 ) γh = for every γ Γ. 8

9 Construct X 1 from X 0 by attaching a 1-cell E xγ,xγ of length 1 between every pair of points x γ, x γ X 0. For ζ Γ, let ζ : E xγ,xγ E xζγ,x ζγ be the unique distance-preserving map with ζ(x γ ) = x ζγ and ζ(x γ ) = x ζγ. This defines a Γ-action on X 1 that extends the Γ-action on X 0. We wish to define a Γ-equivariant map ψ 1 : X 1 X extending ψ 0 and with ψ 1 (X 1 ) X γ Γ γh For every nonidentity element γ Γ, choose a path c γ from x 0 to γx 0. Since (γh) is connected for all γ Γ, we can choose all such paths to lie outside of γh. We may then define ψ 1 (E xγ,xγ ) to lie along the path γc (γ 1 γ ). γ Γ We will build a family of spaces X n inductively. Beginning with an n-dimensional, (n 1)-connected cell complex X n on which Γ acts freely and a Γ-equivariant map ψ n : X n X, we may construct an n + 1-dimensional, n-connected cell complex X n+1 on which Γ acts freely and a Γ-equivariant map ψ n+1 : X n+1 X. For every map s : S n X n, attach an (n+1)-cell E s along γs for every γ Γ. This results in an n-connected space X n+1. Define a Γ-action on X n+1 by γe s = E γs. For each s as above, define ψ n+1,s : E s X to extend ψ n ( E s ) with ψ n+1,γs = γψ n+1,s. Because π n+1 (X) is trivial for all n, this is always possible. There is a unique Γ-equivariant map ψ n+1 extending these maps. Since the focus of Theorem 1 is second-dimensional cohomology, we will use the space X 3 extensively. Therefore we will let ψ = ψ 3. Since Γ\X 3 is the 3-skeleton of a K(Γ, 1), H 2 (Γ; F ) = H 2 (X 3 ; F ). 5 Local Cohomology Before defining cocyles on Γ \ X 3, we will define cocycles in the relative homology groups of several subspaces of X. Recall that x n = (l (n), l 0 (n)). We will assume from now on that n > R + d so that x n H. Recall also and U Γ = U Γ. ( U = 1 x ) x F (t) 9

10 Let and ( ) 1 x U n = v (x) n, v 0 (x) n ( U n 1 ) l = i=k a it i k, l Z, a i = 0 for n i n Then U n U n = Id and U Γ = U n U n. Let S n be the star of x n in X (that is, the collection of cells having x n as a vertex) and C the cell in S n containing x n 1. Let Sn = U n C. This is also the union of 2-cells in S n with β and β 0 bounded above by n. We will define cocycles ϕ n H 2 (Sn, Sn S n ; F ). Summing these over cosets of U Γ in Γ will give us cocycles in H 2 (Γ; F ), which we will use to prove Theorem 1. Lemma 11. cells in S n. ( 1 at n + bt n ) a, b F < U n acts transitively on the set of 2- Proof. We begin by describing subsets of U n that act transitively on the sets E 0 = { l (n) e e T 0 an edge, l 0 (n) e, l 0 (n + 1) e } and and Let E = { e l 0 (n) e T an edge, l (n) e, l (n + 1) e } U E 0 n = U E n = ( ( 1 at n 1 bt n ) a F ) b F 10

11 U E 0 n acts transitively on the set of edges in T 0 incident to l 0 (n) and not to l 0 (n+1). Since U E 0 n stabilizes x n, it acts transitively ( on E 0. Denote ) by e 0 the edge in E 0 1 at between x n and (l (n), l 0 (n 1)). Let e a = n e 0 for each a F. Similarly, Un E acts transitively on the set of edges in T incident to l (n) and not to l (n + 1). Since Un E stabilizes x n, it acts transitively on ( E. ) Denote by f 0 1 bt the edge in E between x n and (l (n 1), l 0 (n)). Let f b = f 0 for each b F. Since any 2-cell in Sn contains a unique pair of edges e a and f b in its boundary, U E 0 n U E n = ( 1 at n + bt n ) a, b F acts transitively on the set of 2-cells in Sn as well as on E 0 and E. ( ) Let C0,0 n be the 2-cell in Sn containing x n 1, and let Ca,b n = 1 at n + bt n C 0,0. n Orient each edge e a and f b with initial vertex x n. We may then choose an orientation for each Ca,b n such that we have Cn a,b = e a f b + D for some chain D Sn S n. Define a cochain ϕ n by ϕ n (Ca,b n ) = ab. Since X is 2-dimensional, any 2-cochain is a cocycle. We will define cocycles in H 2 (Γ\X 3 ; F ) as sums of the ϕ n s. The following lemma will be used in proving that these cocycles are well-defined. Lemma 12. ϕ n is U n -invarant. Proof. Let a basic cycle in Sn be one of the form Cx,y n Cx n,y Cn x,y + Cn x,y. We will show first that ϕ n is U n -invariant on basic cycles, then that any cycle is a sum of basic cycles. Let ( 1 ) n u = i= n a it i U n 11

12 and D be a basic cycle. ϕ n (ud) = ϕ n (u(c n x,y C n x,y C n x,y + Cn x,y )) = ϕ n (uc n x,y uc n x,y uc n x,y + ucn x,y ) = ϕ n (C n x+a n,y+a n C n x +a n,y+a n C n x+a n,y +a n + C n x +a n,y +a n ) = (x + a n )(y + a n ) (x + a n )(y + a n ) (x + a n )(y + a n ) = xy x y xy + x y = ϕ n (D) + (x + a n )(y + a n ) Thus ϕ n is U n -invariant on basic cycles. It remains to show that all cycles are sums of basic cycles. Define the length l of a chain in H 2 (Sn, Sn S n ) by l α i,j C n i,j = α i,j i,j J i,j J Suppose there are cycles which are not sums of basic cycles. Let B = i,j J α i,jc n i,j be such a cycle with the property that l(b) l(b ), for B any other such cycle. Let C n x,y be a 2-cell such that α x,y > 0. Since B is a cycle, B = 0. Therefore, there are some x, y J such that α x,y, α x,y < 0. Then l(b (Cx,y n Cx n,y Cx,y n + Cn x,y )) l(b) 2 Since this cycle differs from B by a basic cycle, it cannot be written as a sum of basic cycles, contradicting the assumption that B is the shortest cycle with that property. 6 Cohomology We now wish to define cocycles in H 2 (Γ\X 3 ; F ) by summing ϕ n over cosets of U Γ in Γ. Let Y n = Stab U (x n ) Σ Lemma 13. Y n is a fundamental domain for the action of U n on X. 12

13 Proof. We will show first that Y n intersects every U n -orbit in X. That is, X = U n Stab U (x n )Σ Since UΣ = X, it suffices to show that Ux n = U n x n. Let u U. Since u preserves β ρ, there is some m > n such that ux m = x m. By lemma 11, for every n < i m, the group ( V i = 1 a i t i + a i t i ) a i, a i F acts transitively on the set of 2-cells in S i. In particular, it acts transitively on βρ 1 (i 1) B 2 (x i) where B r (y) denotes the sphere of radius r centered at y. It follows that V m V m 1... V n+1 acts transitively on βρ 1 (n) B 2(m n) (x m). Since U acts by β ρ -preserving isometries, ux n βρ 1 (n) B 2(m n) (x m). Thus, there is some u V m V m 1... V n+1 with u x n = ux n. It remains to show that Y n contains ( ) exactly one point of each U n -orbit. Suppose there is some u n 1 α = U n and x Y n such that u n x Y n. We will first consider the case that x Σ, so that x = (l (a), l 0 (b)) for some a, b R. If u n Id, then either v (α) (n + 1) or v 0 (α) (n + 1). Therefore either or u n L π (Y n ) l ([n + 1, )) u n L 0 π 0 (Y n ) l 0 ([n + 1, )) where π and π 0 are orthogonal projections to T and T 0 respectively. Therefore if u n x Y n, we have a, b n + 1 and u n x = x. Now consider general x Y n. By the definition of Y n, x = vy for some v Stab U (x n ) and y Σ. Since U is abelian, u n x = vu n y. Since vu n y Y n, so is u n y. As above, this implies that y = u n y and thus that x = u n x. We may identify Y n with the quotient U n \X. Let θ n : X Y n be the quotient map. Note that θ n is U n -equivariant since U is abelian. We may now define Φ n H 2 (Γ\X 3 ; F ) by Φ n (ΓD) = ϕ n (θ n (ψ(gd)) Sn) U Γ g U Γ \Γ 13

14 for any 2-cell D in X 3. The next several lemmas show that Φ n is well-defined as a formal sum, always contains a finite number of nonzero terms, is a cocycle, and is U n -invariant. Lemma 14. Φ n is well-defined. In particular, it is independent of the choice of coset representatives in ΓD and U Γ g. Proof. This proof closely follows the proof of a lemma from [CK15]. Let u U Γ, u = u n u n for u n U n and u n U n. Then ϕ n (θ n (ψ(ugd)) S n) = ϕ n (θ n (uψ(gd)) S n) = ϕ n (θ n (u n u n ψ(gd)) S n) = ϕ n (θ n (u n ψ(gd)) S n) = ϕ n (u n θ n (ψ(gd)) S n) = ϕ n (θ n (ψ(gd)) S n) So Φ n is independent of coset representative in U Γ g. Let γ Γ. Then Φ n (Γ(γD)) = ϕ n (θ n (ψ(gγd) Sn)) = U Γ g U Γ \Γ U Γ (gγ) U Γ \Γ = U Γ g U Γ \Γ = Φ n (ΓD) ϕ n (θ n (ψ(gγd) S n)) ϕ n (θ n (ψ(gd) S n)) Therefore Φ n is also independent of coset representative in ΓD. Lemma 15. For any 2-cell D in X 3, the formal sum Φ n (ΓD) = ϕ n (θ n (ψ(gd)) Sn) U Γ g U Γ \Γ has a finite number of nonzero terms. Proof. Since D is compact, so is ψ(d). Therefore ψ(d) must intersect a finite number of Γ-translates of H. We may therefore write k D = (D ψ 1 (α i H)) D i=1 14

15 where k is some finite number, α i Γ and D is disjoint from the interior of ΓH. Let D i = D ψ 1 (α i H) Then we have k Φ n (ΓD) = Φ n (ΓD) + Φ n (ΓD i ) Since θ n (ψ(gd) Sn = for every g Γ, we have Φ n (ΓD) = 0. Therefore, it suffices to show that Φ n (ΓD i ) is a finite sum for each i. Let { } C i = γ Γ γψ(d i ) Sn i=1 and C i = { U Γ γ U Γ \Γ } ξ Ci for some ξ U Γ γ We will show that C i is finite for each i. Suppose first that α i H = H. Then C i P Γ = U Γ A Γ. Suppose ua C, with u U Γ and a A Γ. Since either us n S n = or u fixes S n pointwise, we may conclude that a C. Since ψ(d i ) is compact and as n S n = for every a A, C i contains finitely many elements of A. Thus C i is finite. Now consider the general α i with nonempty C i and let γ C i. Since Φ n (ΓγD i ) = Φ n (ΓD i ) and γd i H, the result follows from the argument above. Lemma 16. Φ n is a cocycle. Proof. Let ΓD be a 3-cell in Γ\X 3 corresponding to the 3-cell D in X 3. Then D is a 2-sphere. Since X contains no nontrivial 2-spheres, ψ( D) is trivial. Thus, for any g Γ and Φ n (ΓD) = 0. ϕ n (θ n (ψ(g D)) S n) = 0 Lemma 17. Φ n is U n -invariant. Proof. It suffices to show that θ n ψ(d) S n is supported on Sn for every disk D in X 3 since ϕ n is U n -invariant. Let H n = βρ 1 ([β ρ (x n ), )). Since H n H, the definition of ψ implies that ψ(x 1 ) H n =. Thus, ψ(d) is entirely outside of H n. Since U n preserves H n, we may also conclude that θ n (ψ(d)) is outside θ n (H n ). 15

16 Suppose there is some cell C 0 Supp(θ n (ψ(d)) with C 0 S n and C 0 S n. We will find an infinite family of cells which must also be contained in Supp(θ n (ψ(d)), contradicting the compactness of θ n (ψ(d)). Since S n U n Σ and S n is U n -invariant, we may assume without loss of generality that C 0 Σ. Then C 0 is one of l ([n 1, n]) l 0 ([n, n + 1]), l ([n, n + 1]) l 0 ([n 1, n]), or l ([n, n + 1]) l 0 ([n, n + 1]). We will examine in detail the case where C 0 = l ([n 1, n]) l 0 ([n, n + 1]). The others are parallel. Let C i = l ([n 1, n]) l 0 ([n + i, n + i + 1]) and e i = l ([n 1, n]) l 0 (n + i) so that C i, e i Σ with C i 1 and C i adjacent along e i. We will show by induction that all C i are in Supp(θ n (ψ(d)). Since e i H n, we have e i θ n (ψ(d)). Thus, if Supp(θ n (ψ(d)) contains C i 1, it must also contain some other cell with e i in its boundary. Such a cell must be of the form l ([n 1, n]) e for some edge e T 0 with l 0 (n + i) one endpoint. Since the action of U n on T 0 fixes all edges incident to l 0 (n + i), the only cells of this form in U n Σ are the ones in Σ itself: C i 1 and C i. Thus C i Supp(θ n (ψ(d)). This shows that Supp(θ n (ψ(d)) contains infinitely many cells and is not compact. Now that we have defined a family of cocyles, it remains to show that they are independent. We will do this by exhibiting a family of disks B 2n X 3 such that Φ 2n (Γ B 2n ) = 1 and Φ k (Γ B 2n ) = 0 for k > 2n. Let Z 2n be the triangle in Σ with vertices x n, (l (n), l 0 ( n)), and (l ( n), l 0 (n)). Let ( ) ( ) ( ) 1 t B 2n = Z 2n 2n 1 t Z 2n 2n 1 t Z 2n + 2n + t 2n Z 2n Then B 2n is a square with B 2n S 2n = C 2n 0,0 C 2n 1,0 C 2n 0,1 + C 2n 1,1. Lemma 18. For every n N, there is some disc B 2n X 3 such that ψ( B 2n ) H = B 2n. ( ) ( ) 1 t Proof. Letting u = 2n + t 2n t and a = n 0 0 t n, the vertices of B 2n are ax 0, a 1 x 0, uax 0, and ua 1 x 0. In particular, these four vertices are in Γx 0 and therefore in ψ(x 0 ). Let v i and v j be two vertices on an edge of the square, with ṽ i ψ 1 (v i ) and ṽ j ψ 1 (v j ). Since X 3 is connected, there is a path c i,j between ṽ i and ṽ j. Thus c i,j = ψ( c i,j ) is a path connecting v i and v j. By the definition of ψ, c i,j is disjoint from ΓH. Repeating this for all four edges gives a path c in X which bounds a disk B 2n such that B 2n H = B 2n H and is the image of a path c in X 3. Since X 3 is 2-connected, there is a disk B 2n that fills c, and ψ( B 2n ) = B 2n. 16

17 Lemma 19. Φ 2n (Γ B 2n ) = 1 and Φ k (Γ B 2n ) = 0 for k > 2n. Proof. We will prove that Φ k (Γ B 2n ) = 0 by showing that γb 2n S k = for every γ Γ. Since B 2n H βρ 1 ([R + d, 2n]) and P Γ preserves β ρ, we have pb 2n S k = for every p P Γ. For γ Γ P Γ, γh H =. Since B 2n ΓH H, this implies that γb 2n S k =. By the definitions of B 2n and ϕ 2n, we have ϕ 2n (B 2n ) = 1. It therefore suffices to show that γb 2n S 2n = for every γ Γ U Γ. Let p P Γ U Γ. Since P Γ preserves β ρ and B 2n βρ 1 ([2n 1, 2n + 1]) S 2n, it follows that pb 2n βρ 1 ([2n 1, 2n + 1]) ps 2n. Let p = au for some a A Γ and u U Γ with a Id. Since us 2n = S 2n and as 2n S 2n =, we have pb 2n S 2n =. For γ Γ P Γ, γh H =. Since B 2n ΓH H, this implies that γb 2n S 2n =. Thus, for any k and any even integer m R + d, the set of cocycles {Φ m+2, Φ m+4,... Φ m+2k } is independent, which suffices to prove Theorem 1. References [BW06] K.-U. Bux and K. Wortman, A Geometric proof that SL 2 (Z[t, t 1 ]) is not finitely presented, Algebr. Geom. Topol (2006), [BW11] [CK15], Connectivity properties of horospheres in euclidean buildings and applications to finiteness properties of discrete groups, Inventiones mathematicae 185 (2011), M. Cesa and B. Kelly, H 2 (SL(3, Z[t]), Q) is infinite dimensional, preprint (2015). [KM97] S. Krstić and J. McCool, The non-finite presentability of IA(F 3 ) and GL 2 (Z[t, t 1 ])., Inventiones Mathematicae 129 (1997), [Knu08] K. Knudson, Homology and finiteness properties of SL 2 (Z[t, t 1 ]), Algebr. Geom. Topol. 8 (2008), [Wor13] K. Wortman, An infinitely generated virtual cohomology group for noncocompact arithmetic groups over function fields, preprint (2013). 17

18 S. Cobb, Department of Mathematics, Midwestern State University 3410 Taft Boulevard, Wichita Falls, TX address, 18