aa r X i v : . [ m a t h . G R ] J a n SL n ( Z [ t ]) is not F P n − ∗ December 12, 2007
Abstract
We prove the result from the title using the geometry of Euclideanbuildings.
Little is known about the finiteness properties of SL n ( Z [ t ]) for arbitrary n .In 1959 Nagao proved that if k is a field then SL ( k [ t ]) is a free productwith amalgamation [Na]. It follows from his description that SL ( Z [ t ]) andits abelianization are not finitely generated.In 1977 Suslin proved that when n ≥ , SL n ( Z [ t ]) is finitely generatedby elementary matrices [Su]. It follows that H ( SL n ( Z [ t ]) , Z ) is trivial when n ≥ SL ( Z [ t ]) is not finitely pre-sented [Kr-Mc].It’s also worth pointing out that since SL n ( Z [ t ]) surjects onto SL n ( Z ),that SL n ( Z [ t ]) has finite index torsion-free subgroups.In this paper we provide a generalization of the results of Nagao andKrsti´c-McCool mentioned above for the groups SL n ( Z [ t ]). Theorem 1. If n ≥ , then SL n ( Z [ t ]) is not of type F P n − . Recall that a group Γ is of type
F P m if if there exists a projective reso-lution of Z as the trivial Z Γ module P m → P m − → · · · → P → P → Z → ∗ Supported in part by an N.S.F. Grant DMS-0604885. P i is a finitely generated Z Γ module.In particular, Theorem 1 implies that there is no K ( SL n ( Z [ t ]) ,
1) withfinite ( n − K ( G,
1) is the Eilenberg-Mac Lane space for G . The general outline of this paper is modelled on the proofs in [Bu-Wo 1] and[Bu-Wo 2], though some important modifications have to be made to carryout the proof in this setting.As in [Bu-Wo 1] and [Bu-Wo 2], our approach is to apply Brown’s fil-tration criterion [Br]. Here we will examine the action of SL n ( Z [ t ]) on thelocally infinite Euclidean building for SL n ( Q (( t − ))). In Section 2 we willshow that the infinite groups that arise as cell stabilizers for this action areof type F P m for all m , which is a technical condition that is needed for ourapplication of Brown’s criterion.In Section 3 we will demonstrate the existence of a family of diagonal ma-trices that will imply the existence of a “nice” isometrically embedded codi-mension 1 Euclidean space in the building for SL n ( Q (( t − ))). In [Bu-Wo 1]analogous families of diagonal matrices were constructed using some stan-dard results from the theory of algebraic groups over locally compact fields.Because Q (( t − )) is not locally compact, our treatment in Section 3 is quitea bit more hands on.Section 4 contains the main body of our proof. We use translates of por-tions of the codimension 1 Euclidean subspace found in Section 3 to constructspheres in the Euclidean building for SL n ( Q (( t − ))) (also of codimension 1).These spheres will lie “near” an orbit of SL n ( Z [ t ]), but will be nonzero inthe homology of cells “not as near” the same SL n ( Z [ t ]) orbit. Theorem 1will then follow from Brown’s criterion. Lemma 2. If X is the Euclidean building for SL n ( Q (( t − ))) , then the SL n ( Z [ t ]) stabilizers of cells in X are F P m for all m .Proof. Let x ∈ X be the vertex stabilized by SL n ( Q [[ t − ]]). We denote adiagonal matrix in GL n ( Q (( t − ))) with entries s , s , ..., s n ∈ Q (( t − )) × by D ( s , s , ..., s n ), and we let S ⊆ X be the sector based at x and containing2ertices of the form D ( t m , t m , ..., t m n ) x where each m i ∈ Z and m ≥ m ≥ ... ≥ m n .The sector S is a fundamental domain for the action of SL n ( Q [ t ]) on X (see [So]). In particular, for any vertex z ∈ X , there is some h ′ z ∈ SL n ( Q [ t ])and some integers m ≥ m ≥ ... ≥ m n with z = h ′ z D z ( t m , t m , ..., t m n ) x .We let h z = h ′ z D z ( t m , t m , ..., t m n ) . For any N ∈ N , let W N be the ( N + 1)-dimensional vector space W N = { p ( t ) ∈ C [ t ] | deg (cid:0) p ( t ) (cid:1) ≤ N } which is endowed with the obvious Q − structure. If N , · · · , N n in N arearbitrary then let G { N , ··· ,N n } = { x ∈ n Y i =1 W N i | det( x ) = 1 } where det( x ) is a polynomial in the coordinates of x . To be more precise thisis obtained from the usual determinant function when one considers the usual n × n matrix presentation of x , and calculates the determinant in Mat n ( C [ t ]) . For our choice of vertex z ∈ X above, the stabilizer of z in SL n ( Q (( t − )))equals h z SL n ( Q [[ t − ]]) h − z . And with our fixed choice of h z , there clearlyexist some N zi ∈ N such that the stabilizer of the vertex z in SL n ( Q [ t ]) is G { N z , ··· ,N zn } ( Q ). Furthermore, conditions on N zi force a group structure on G z = G { N z , ··· ,N zn } . Therefore, the stabilizer of z in SL n ( Q [ t ]) is the Q − pointsof the affine Q -group G z , and the stabilizer of z in SL n ( Z [ t ]) is G z ( Z ).The action of SL n ( Q [ t ]) on X is type preserving, so if σ ⊂ S is a simplexwith vertices z , z , ..., z m , then the stabilizer of σ in SL n ( Z [ t ]) is simply (cid:0) G z ∩ ... ∩ G z m (cid:1) ( Z )That is, the stabilizer of σ in SL n ( Z [ t ]) is an arithmetic group, and Borel-Serre proved that any such group is F P m for all m [Bo-Se]. This section is devoted exclusively to a proof of the following
Proposition 3.
There is a group A ≤ SL n ( Z [ t ]) such that i) A ∼ = Z n − (ii) There is some g ∈ SL n ( Q (( t − ))) such that gAg − is a groupof diagonal matrices(iii) No nontrivial element of A fixes a point in the Euclideanbuilding for SL n ( Q (( t − ))) . The proof of this proposition is modelled on a classical approach to findingdiagonalizable subgroups of SL n ( Z ). The proof will take a few steps. Z [ t ] with roots in Q (( t − )) Let { p , p , p , ... } = { , , , ... } be the sequence of prime numbers. Let q = 1. For 2 ≤ i ≤ n , let q i = p i − + 1.Let f ( x ) ∈ Z [ t ][ x ] be the polynomial given by f ( x ) = h n Y i =1 ( x + q i t ) i − f ( x ) is irreducible over Q ( t ), but we will not need to use this directly. Lemma 4.
There is some α ∈ Q (( t − )) such that f ( α ) = 0 .Proof. We want to show that there are c i ∈ Q such that if α = P ∞ i =0 c i t − in then f ( α ) = 0.To begin let c = −
1. We will define the remaining c i recursively. Define c i,k by α + q k t = P ∞ i =0 c i,k t − in . Thus, c i,k = c i when i ≥
1, each c ,k iscontained in Q , and c , = 0.That α is a root of f is equivalent to1 = n Y k =1 ( α + q k t ) = n Y k =1 (cid:0) ∞ X i =0 c i,k t − in (cid:1) = ∞ X i =0 (cid:0) X P nk =1 i k = i (cid:0) n Y k =1 c i k ,k (cid:1)(cid:1) t n (1 − i ) Our task is to find c m ’s so that the above is satisfied.Note that for the above equation to hold we must have0 · t n = X P nk =1 i k =0 (cid:0) n Y k =1 c i k ,k (cid:1) t n (1 − n Y k =1 c ,k which is an equation we know is satisfied because c , = 0. Now assume thatwe have determined c , c , ..., c m − ∈ Q . We will find c m ∈ Q .Notice that the first coefficient in our Laurent series expansion abovewhich involves c m is the coefficient for the t − nm term. This follows from thefact that each i k is nonnegative.Since X P nk =1 i k = m (cid:0) n Y k =1 c i k ,k (cid:1) is the coefficient of the t − nm term in the expansion of 1, we have0 = X P nk =1 i k = m (cid:0) n Y k =1 c i k ,k (cid:1) The above equation is linear over Q in the single variable c m and the coeffi-cient of c m is nonzero. Indeed, P nk =1 i k = m , each i k ≥
0, and c , ..., c m − ∈ Q are assumed to be known quantities. Thus, c m ∈ Q . By Lemma 4 we have that the field Q ( t )( α ) ≤ Q (( t − )) is an extension of Q ( t ) of degree d where d ≤ n . It follows that Z [ t ][ α ] is a free Z [ t ]-module ofrank d with basis { , α, α , ..., α d − } .For any y ∈ Z [ t ][ α ], the action of y on Q ( t )( α ) by multiplication is alinear transformation that stabilizes Z [ t ][ α ]. Thus, we have a representationof Z [ t ][ α ] into the ring of d × d matrices with entries in Z [ t ]. We embed thering of d × d matrices with entries in Z [ t ] into the upper left corner of thering of n × n matrices with entries in Z [ t ].By Lemma 4 n Y i =1 ( α + q i t ) = 1so each of the following matrices are invertible: α + q t, α + q t, ..., α + q n t Z [ t ][ α ] and thematrices that represent them.)For 1 ≤ i ≤ n −
1, we let a i = α + q i +1 t . Since a i is invertible, itis an element of GL n ( Z [ t ]), and hence has determinate ±
1. By replacingeach a i with its square, we may assume that a i ∈ SL n ( Z [ t ]) for all i . Welet A = h a , ...a n − i so that A is clearly abelian as it is a representation ofmultiplication in an integral domain. This group A will satisfy Proposition 3. A is free abelian on the a i To prove part (i) of Proposition 3 we have to show that if there are m i ∈ Z with n − Y i =1 a m i i = 1then each m i = 0. But the first nonzero term in the Laurent series expansionfor α is − t , which implies that the first nonzero term in the Laurent seriesexpansion for each a i is − t + q i +1 t = p i t . Hence, the first nonzero term of n − Y i =1 a m i i = 1is n − Y i =1 ( p i t ) m i = t Thus n − Y i =1 p m i i = 1and it follows by the uniqueness of prime factorization that m i = 0 for all i as desired.Thus, part (i) of Proposition 3 is proved. A is diagonalizable Recall that α is a d × d matrix with entries in Z [ t ] where d is the degreeof the minimal polynomial of α over Q ( t ). Let that minimal polynomial be6 ( x ). Because the characteristic of Q ( t ) equals 0, q ( x ) has distinct roots in Q ( t )( α ).Let Q ( x ) be the characteristic polynomial of the matrix α . The polyno-mial Q also has degree d and leading coefficient ± Q ( α ) = 0. Therefore, q = ± Q . Hence, Q has distinct roots in Q ( t )( α ) which implies that α is di-agonalizable over Q ( t )( α ) ≤ Q (( t − )). That is to say that there is some g ∈ SL n ( Q (( t − ))) such that gαg − is diagonal.Because every element of Z [ t ][ α ] is a linear combination of powers of α ,we have that g ( Z [ t ][ α ]) g − is a set of diagonal matrices. In particular, wehave proved part (ii) of Proposition 3. A has trivial stabilizers To prove part (iii) of Proposition 3 we begin with the following
Lemma 5. If Γ ≤ SL n ( Q [ t ]) is bounded under the valuation for Q (( t − )) ,then the eigenvalues for any γ ∈ Γ lie in Q .Proof. We let X be the Euclidean building for SL n ( Q (( t − ))). By assump-tion, Γ z = z for some z ∈ X .Let x ∈ X be the vertex stabilized by SL n ( Q [[ t − ]]). We denote adiagonal matrix in GL n ( Q (( t − ))) with entries s , s , ..., s n ∈ Q (( t − )) × by D ( s , s , ..., s n ), and we let S ⊆ X be the sector based at x and containingvertices of the form D ( t m , t m , ..., t m n ) x where each m i ∈ Z and m ≥ m ≥ ... ≥ m n .The sector S is a fundamental domain for the action of SL n ( Q [ t ]) on X [So] which implies that there is some h ∈ SL n ( Q [ t ]) with hz ∈ S .Clearly we have ( h Γ h − ) hz = hz , and since eigenvalues of h Γ h − are thesame as those for Γ, we may assume that Γ fixes a vertex z ∈ S .Fix m , ..., m n ∈ Z with m ≥ ... ≥ m n ≥ z = D ( t m , ..., t m n ) x . Without loss of generality, there is a partition of n —say { k , ..., k ℓ } — such that { m , ..., m n } = { q , ..., q , q , ..., q , ..., q ℓ , ...q ℓ } where each q i occurs exactly k i times and q > q > ... > q ℓ
7e have that D ( t m , ..., t m n ) − Γ D ( t m , ..., t m n ) x = x . That gives us, D ( t m , ..., t m n ) − Γ D ( t m , ..., t m n ) ⊂ SL n ( Q [[ t − ]]). Furthermore, a trivial cal-culation of resulting valuation restrictions for the entries of D ( t m , ..., t m n ) SL n ( Q [[ t − ]]) D ( t m , ..., t m n ) − shows that Γ is contained in asubgroup of SL n ( Q (( t − ))) that is isomorphic to ℓ Y i =1 SL k i ( Q ) ⋉ U where U ≤ SL n ( Q (( t − ))) is a group of upper-triangular unipotent matrices.The lemma is proved.Our proof of Proposition 3 will conclude by proving Lemma 6.
No nontrivial element of A fixes a point in the Euclidean buildingfor SL n ( Q (( t − ))) .Proof. Suppose a ∈ A fixes a point in the building. We will show that a = 1.Let F ( x ) ∈ Z [ t ][ x ] be the characteristic polynomial for a ∈ SL n ( Z [ t ]). Then F ( x ) = ± n Y i =1 ( x − β i )where each β i ∈ Q (( t − )) is an eigenvalue of a . By the previous lemma, each β i ∈ Q . Hence, each β i ∈ Q = Q ∩ Q (( t − )). It follows that F ( x ) ∈ Z [ x ] sothat each β i is an algebraic integer contained in Q . We conclude that each β i is contained in Z .Recall, that a has determinate 1, and that the determinate of a canbe expressed as Q ni =1 β i . Hence, each β i is a unit in Z , so each eigenvalue β i = ±
1. It follows – by the diagonalizability of a – that a is a finite orderelement of A ∼ = Z n − . That is, a = 1.We have completed our proof of Proposition 3.8 Body of the proof
Let P ≤ SL n ( Q (( t − ))) be the subgroup where each of the first n − R u ( P ) ≤ P be the subgroup of elementsthat contain a ( n − × ( n −
1) copy of the identity matrix in the upper leftcorner. Thus R u ( P ) ∼ = Q (( t − )) n − with the operation of vector addition.Let L ≤ P be the copy of SL n − ( Q (( t − ))) in the upper left cornerof SL n ( Q (( t − ))). We apply Proposition 3 to L (notice that the n in theproposition is now an n −
1) to derive a subgroup A ≤ L that is isomorphicto Z n − . By the same proposition, there is a matrix g ∈ L such that gAg − is diagonal.Let b ∈ SL n ( Q (( t − ))) be the diagonal matrix given in the notation fromthe proofs of Lemmas 2 and 5 as D ( t, t, ..., t, t − ( n − ). Note that b ∈ P com-mutes with L , and therefore, with A . Thus the Zariski closure of the groupgenerated by b and A determines an apartment in X , namely g − A where A is the apartment corresponding to the diagonal subgroup of SL n ( Q (( t − ))). g − A . If x ∗ ∈ g − A , then it follows from Proposition 3 that the convex hull of theorbit of x ∗ under A is an ( n − V x ∗ . Furthermore, the orbit Ax ∗ forms a lattice in the space V x ∗ .We let g − A ( ∞ ) be the visual boundary of g − A in the Tits boundaryof X . The visual image of V x ∗ is clearly an equatorial sphere in g − A ( ∞ ).Precisely, we let P − be the transpose of P . Then P and P − are oppositevertices in g − A ( ∞ ). It follows that there is a unique sphere in g − A ( ∞ )that is realized by all points equidistant to P and P − . We call this sphere S P,P − . Lemma 7.
The visual boundary of V x ∗ equals S P,P − .Proof. Since g ∈ P ∩ P − , it suffices to prove that gV x ∗ is the sphere in theboundary of A that is determined by the vertices P and P − .Note that gV x ∗ is a finite Hausdorff distance from any orbit of a pointin A under the action of the diagonal subgroup of L . The result followsby observing that the inverse transpose map on SL n ( Q (( t − ))) stabilizesdiagonal matrices while interchanging P and P − .9e let R , R , ..., R n − be the standard root subgroups of R u ( P ). Recallthat associated to each R i there is a closed geodesic hemisphere H i ⊆ A ( ∞ )such that any nontrivial element of R i fixes H i pointwise and translates anypoint in the open hemisphere A ( ∞ ) − H i outside of A ( ∞ ). Note that ∂H i is a codimension 1 geodesic sphere in A ( ∞ ).We let M ⊆ g − A ( ∞ ) be the union of chambers in g − A ( ∞ ) that containthe vertex P . There is also an equivalent geometric description of M : Lemma 8.
The union of chambers M ⊆ g − A ( ∞ ) can be realized as an ( n − -simplex. Furthermore, M = n − \ i =1 g − H i and, when M is realized as a single simplex, each of the n − faces of M iscontained in a unique equatorial sphere g − ∂H i = ∂g − H i .Proof. Let M ′ ⊆ A ( ∞ ) be the union of chambers in A ( ∞ ) containing thevertex P . Since M = g − M ′ , it suffices to prove that M ′ is an ( n − M ′ = ∩ n − i =1 H i and with each face of M ′ contained in a unique ∂H i .For any nonempty, proper subset I ⊆ { , , ..., n } , we let V I be the | I | -dimensional vector subspace of Q (( t − )) n spanned by the coordinates givenby I , and we let P I be the stabilizer of V I in SL n ( Q (( t − ))). For example, P = P { , ,...,n − } .Recall that the vertices of A ( ∞ ) are given by the parabolic groups P I ,that edges connect P I and P I ′ exactly when I ⊆ I ′ or I ′ ⊆ I , and thatthe remaining simplicial description of A ( ∞ ) is given by the condition that A ( ∞ ) is a flag complex.We let V be the set of vertices in A ( ∞ ) of the form P J where ∅ 6 = J ⊆{ , , ..., n − } . Note that M ′ is exactly the set of vertices V together with thesimplices described by the incidence relations inherited from A ( ∞ ). Thus, M ′ is easily seen to be isomorphic to a barycentric subdivision of an ab-stract ( n − M ′ is the abstract simplex on vertices P { } , P { } , ..., P { n − } , then a simplex of dimension k in M ′ corresponds to aunique P J ∈ V with | J | = k + 1. So we have that M ′ can be topologicallyrealized as an ( n − F i be a face of the simplex M ′ . Then there is some 1 ≤ i ≤ n − F i is exactly { P { } , P { } , ..., P { n − } } − P { i } .10ote that R i V I = V I exactly when n ∈ I implies i ∈ I . It follows that R i fixes M ′ pointwise, and thus M ′ ⊆ H i for all 1 ≤ i ≤ n −
1. Furthermore, if P I ∈ H i for all 1 ≤ i ≤ n −
1, then R i P I = P I for all i so that n ∈ I implies i ∈ I for all 1 ≤ i ≤ n −
1. As I must be a proper subset of { , , ..., n } , wehave P I ∈ V , so that M ′ = ∩ n − i =1 H i .All that remains to be verified for this lemma is that F i ⊆ ∂H i . For thisfact, recall that F i is comprised of ( n − A ( ∞ ) whose verticesare given by P J where J ⊆ { , , ..., n − } − { i } . Hence, if σ ⊆ A ( ∞ ) is an( n −
3) simplex of A ( ∞ ) with σ ⊆ F i , then σ is a face of exactly 2 chambersin A ( ∞ ): C P and C P J ′ where C P contains P and thus C P ⊆ M ′ , and C P J ′ contains P J ′ where J ′ = { , , ..., n } − { i } and thus C P J ′ * M ′ . Furthermore, σ = C P ∩ C P J ′ .Since R i V J ′ = V J ′ , it follows that C P J ′ is not fixed by R i . Since C P J isfixed by R i we have that σ = C P ∩ C P J ′ ⊆ ∂H i . Therefore, F i ⊆ ∂H i .For any vertex y ∈ X , we let C y ⊆ X be the union of sectors basedat y and limiting to a chamber in M . Thus, C y is a cone. Note also thatbecause any chamber in g − A ( ∞ ) has diameter less than π/
2, it follows that M ∩ S P,P − = ∅ . Therefore, if we choose x ∗ , y ∈ g − A such that x ∗ is closerto P than y , then C y ⊆ g − A and V x ∗ ∩ C y is a simplex of dimension n − y before x ∗ , and we will choose y to satisfythe below Lemma 9.
There is some y ∈ g − A such that the Q [[ t − ]] -points of R u ( P ) fix C y pointwise.Proof. Let x be the point in X stabilized by SL n ( Q [[ t − ]]). Recall that R u ( P ) M = M so that the Q [[ t − ]]-points of R u ( P ) fix C x pointwise.Because M ⊆ g − A ( ∞ ), there is a y ∈ C x ∩ g − A . Any such y satisfiesthe lemma.Choose e such that with x ∗ = e as above and with y as in Lemma 9, thereexists a fundamental domain D e for the action of A on V e that is containedin C y . The choice of e can be made by travelling arbitrarily far from y alonga geodesic ray in g − A that limits to P .By the choice of D e we have that AD e = V e and that the Q ([[ t − ]])-points of R u ( P ) fix D e .11 .2 The filtration We let X = SL n ( Z [ t ]) D e and for any i ∈ N we choose an SL n ( Z [ t ])-invariant and cocompact space X i ⊆ X somewhat arbitrarily to satisfy the inclusions X ⊆ X ⊆ X ⊆ ... ⊆ ∪ ∞ i =1 X i = X In our present context, Brown’s criterion takes on the following form [Br]
Brown’s Filtration Criterion 10.
By Lemma 2, the group SL n ( Z [ t ]) isnot of type F P n − if for any i ∈ N , there exists some class in the homologygroup e H n − ( X , Z ) which is nonzero in e H n − ( X i , Z ) . P moves away from filtration sets The following is essentially Mahler’s compactness criterion.
Lemma 11.
Given any i ∈ N , there is some k ∈ N such that b k e / ∈ X i .Proof. The lemma follows from showing that the sequence { SL n ( Z [ t ]) b k e } k ⊆ SL n ( Z [ t ]) \ X is unbounded.Since stabilizers of points in X are bounded subgroups of SL n ( Q (( t − ))),the claim above follows from showing that the sequence { SL n ( Z [ t ]) b k } k ⊆ SL n ( Z [ t ]) \ SL n ( Q (( t − )))is unbounded.But bounded sets in SL n ( Z [ t ]) \ SL n ( Q (( t − ))) do not contain sequences ofelements { SL n ( Z [ t ]) g ℓ } ℓ such that 1 ∈ g − ℓ ( SL n ( Z [ t ]) − { } ) g ℓ . And clearly b k ’s contract some root groups to 1. Thus none of the sequences above isbounded. 12 .4 Applying Brown’s criterion As is described by Brown’s criterion, we will prove Theorem 1 by fixing X i and finding an ( n − X that is nontrivial in the homology of X i .Recall that we denote the standard root subgroups of R u ( P ) by R , ..., R n − .Each group g − R j g determines a family of parallel walls in g − A . By Lemma 8,each face of the cone C y is contained in a wall of one of these families.Choose r j ∈ g − R j g for all j such that b k e is contained in the wall de-termined by r j where k is determined by i as in Lemma 11. In particular, r j b k e = b k e .The intersection of the fixed point sets in g − A of the elements r , ..., r n − determine a cone that we name Z . Note that Z is contained in – and is afinite Hausdorff distance from – the cone C y .Let Z − ⊆ g − A be the closure of the set of points in g − A that are fixedby none of the r j . The set Z − is a cone based at b k e , containing y , andasymptotically containing the vertex P − .As the walls of Z − are parallel to those of Z – and hence of C y , we havethat Z − ∩ V e is an ( n − σ . The component of Z − − V e that contains b k e is an ( n − σ as a face. Call this ( n −
1) simplex Y .For any ℓ ∈ N , there are exactly 2 n − possible subsets of the set { r ℓ , ..., r ℓn − } .For each such subset S ℓ , we let Y S ℓ = ( Y g ∈ S ℓ g ) Y and σ S ℓ = ( Y g ∈ S ℓ g ) σ Notice that the product of group elements in the equations above are well-defined regardless of the order of the multiplication since R u ( P ) is abelian.In the degenerate cases, Q g ∈∅ g = 1, so Y ∅ = Y and σ ∅ = σ .For any ℓ ∈ N , we let Y ℓ = ∪ S ℓ Y S ℓ . Because the wall in g − A determinedby r ℓj is the same as the wall determined by r j , the space Y ℓ is a closedball containing b k e whose boundary sphere is ∪ S ℓ σ S ℓ . Indeed the simplicialdecomposition of Y ℓ described above is isomorphic to the simplicial decompo-sition of the unit ball in R n − that is given by the n − ω ℓ = ∪ S ℓ σ S ℓ . Thus ω ℓ = ∂Y ℓ . Furthermore, the building X is ( n − n − ω ℓ must contain Y ℓ and thus b k e . That is for all ℓ ∈ N [ ω ℓ ] = 0 ∈ e H n − ( X − b k e , Z )If we can show that ω ℓ ⊆ X for some choice of ℓ , then we will have provedour main theorem by application of Brown’s criterion since we would have[ ω ℓ ] = 0 ∈ e H n − ( X i , Z )by Lemma 11. Lemma 12.
There exists some ℓ ∈ N such that ω ℓ ⊆ X .Proof. For any u ∈ R u ( P ) there is a decomposition u = u ′ u ′′ where theentries of u ′ ∈ R u ( P ) are contained in Q [ t ] and the entries of u ′′ ∈ R u ( P ) arecontained in Q [[ t − ]].For any a ∈ A and u ∈ R u ( P ) there is a power ℓ ( a, u ) ∈ N such that( a − u ℓ ( a,u ) a ) ′ = (( a − ua ) ′ ) ℓ ( a,u ) ∈ SL n ( Z [ t ])(For the above equality recall that A ≤ L normalizes R u ( P ) and the groupoperation on R u ( P ) is vector addition.)There are only finitely many a ∈ A such that aD e ∩ σ = ∅ (or equivalently,such that aD e ∩ Z − = ∅ ). Call this finite set D ⊆ A .At this point we fix ℓ = Y a ∈D n − Y i =1 ℓ ( a, r i )Thus, [ a − ( Y g ∈ S ℓ g ) a ] ′ ∈ SL n ( Z [ t ])for any a ∈ D and any S ℓ ⊆ { r ℓi } n − i =1 .Because ω ℓ = ∪ S ℓ σ S ℓ and σ S ℓ = ( Q g ∈ S ℓ g ) σ = ( Q g ∈ S ℓ g )( AD e ∩ Z − ), wecan finish our proof of this lemma by showing (cid:0) Y g ∈ S ℓ g (cid:1) aD e ⊆ X a ∈ D ⊆ A ≤ SL n ( Z [ t ]) and each S ℓ ⊆ { r ℓi } n − i =1 . For this, recall thatthe Q [[ t − ]]-points of R u ( P ) fix D e and thus (cid:0) Y g ∈ S ℓ g (cid:1) aD e = a [ a − (cid:0) Y g ∈ S ℓ g (cid:1) a ] D e = a [ a − (cid:0) Y g ∈ S ℓ g (cid:1) a ] ′ [ a − (cid:0) Y g ∈ S ℓ g (cid:1) a ] ′′ D e = a [ a − (cid:0) Y g ∈ S ℓ g (cid:1) a ] ′ D e ⊆ SL n ( Z [ t ]) D e = X References [Bo-Se] A. Borel, J-P. Serre,
Corners and arithmetic groups.
CommentariiMathematici Helvetici (1973) p. 436 491.[Br] Brown, K., Finiteness properties of groups.
J. Pure Appl. Algebra (1987), 45-75.[Bu-Wo 1] Bux, K.-U., and Wortman, K., Finiteness properties of arithmeticgroups over function fields.
Invent. Math. (2007), 355-378.[Bu-Wo 2] Bux, K.-U., and Wortman, K.,
Geometric proof that SL ( Z [ t, t − ]) is not finitely presented. Algebr. Geom. Topol. (2006), 839-852.[Kr-Mc] Krsti´c, S., and McCool, J., Presenting GL n ( k h T i ) . J. Pure Appl.Algebra (1999), 175-183.[Na] Nagao, H., On GL(2 , K [ X ]). J. Inst. Polytech. Osaka City Univ.Ser. A (1959), 117-121.[Su] Suslin, A. A. The structure of the special linear group over rings ofpolynomials. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. (1977),no. 2, 235–252, 477. 15So] Soul´ e , C., Chevalley groups over polynomial rings.