SL(n,Z[t]) is not FP_{n-1}

0.17 MB·

by  Kai-Uwe Bux

SL (Z[t]) is not FP
n n−1
8
0
0 Kai-Uwe Bux, Amir Mohammadi, Kevin Wortman ∗
2

n December 12, 2007
a
J

8

] Abstract
R
G We prove the result from the title usingthe geometry of Euclidean
. buildings.
h
t
a
m
1 Introduction
[

1 Little is known about the finiteness properties of SLn(Z[t]) for arbitrary n.
v
2 In 1959 Nagao proved that if k is a field then SL2(k[t]) is a free product
3 with amalgamation [Na]. It follows from his description that SL2(Z[t]) and
3
its abelianization are not finitely generated.
1
1. In 1977 Suslin proved that when n ≥ 3, SLn(Z[t]) is finitely generated
0 by elementary matrices [Su]. It follows that H1(SLn(Z[t]),Z) is trivial when
8
n ≥ 3.
0
: More recent, Krsti´c-McCool proved that SL3(Z[t]) is not finitely pre-
v
i sented [Kr-Mc].
X
It’s also worth pointing out that since SLn(Z[t]) surjects onto SLn(Z),
r
a that SLn(Z[t]) has finite index torsion-free subgroups.
In this paper we provide a generalization of the results of Nagao and
Krsti´c-McCool mentioned above for the groups SLn(Z[t]).
Theorem 1. If n ≥ 2, then SLn(Z[t]) is not of type FPn−1.
Recall that a group Γ is of type FP if if there exists a projective reso-
m
lution of Z as the trivial ZΓ module
P → P → ··· → P → P → Z → 0
m m−1 1 0
∗Supported in part by an N.S.F. Grant DMS-0604885.
1
where each P is a finitely generated ZΓ module.
i
In particular, Theorem 1 implies that there is no K(SLn(Z[t]),1) with
finite (n−1)-skeleton, where K(G,1) is the Eilenberg-MacLane space for G.
1.1 Outline of paper
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 carry
out 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 SLn(Z[t]) on the
locally infinite Euclidean building for SLn(Q((t−1))). In Section 2 we will
show that the infinite groups that arise as cell stabilizers for this action are
of type FP for all m, which is a technical condition that is needed for our
m
application 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 SLn(Q((t−1))). 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−1)) is not locally compact, our treatment in Section 3 is quite
a bit more hands on.
Section 4 contains the main body of our proof. We use translates of por-
tionsofthecodimension 1Euclideansubspace foundinSection3toconstruct
spheres in the Euclidean building for SLn(Q((t−1))) (also of codimension 1).
These spheres will lie “near” an orbit of SLn(Z[t]), but will be nonzero in
the homology of cells “not as near” the same SLn(Z[t]) orbit. Theorem 1
will then follow from Brown’s criterion.
2 Stabilizers
Lemma 2. IfX is theEuclidean buildingforSLn(Q((t−1))), then the SLn(Z[t])
stabilizers of cells in X are FP for all m.
m
Proof. Let x0 ∈ X be the vertex stabilized by SLn(Q[[t−1]]). We denote a
diagonal matrix in GLn(Q((t−1))) with entries s1,s2,...,sn ∈ Q((t−1))× by
D(s ,s ,...,s ), and we let S ⊆ X be the sector based at x and containing
1 2 n 0
2
vertices of the form D(tm1,tm2,...,tmn)x where each m ∈ Z and m ≥ m ≥
0 i 1 2
... ≥ m .
n
The sector S is a fundamental domain for the action of SLn(Q[t]) on X
(see [So]). In particular, for any vertex z ∈ X, there is some h′z ∈ SLn(Q[t])
and some integers m ≥ m ≥ ... ≥ m with z = h′D (tm1,tm2,...,tmn)x .
1 2 n z z 0
We let h = h′D (tm1,tm2,...,tmn).
z z z
For any N ∈ N, let W be the (N +1)-dimensional vector space
N
W = {p(t) ∈ C[t] | deg p(t) ≤ N}
N
(cid:0) (cid:1)
which is endowed with the obvious Q−structure. If N1,··· ,Nn2 in N are
arbitrary then let
n2
G = {x ∈ W |det(x) = 1}
{N1,···,Nn2} Y Ni
i=1
where det(x) is a polynomial in the coordinates of x. To be more precise this
isobtainedfromtheusualdeterminant functionwhenoneconsiders theusual
n×nmatrixpresentationofx,andcalculatesthedeterminant inMat (C[t]).
n
For our choice of vertex z ∈ X above, the stabilizer of z in SLn(Q((t−1)))
equals hzSLn(Q[[t−1]])h−z1. And with our fixed choice of hz, there clearly
exist some Niz ∈ N such that the stabilizer of the vertex z in SLn(Q[t]) is
G{N1z,···,Nnz2}(Q). Furthermore, conditions on Niz force a group structure on
Gz = G{N1z,···,Nnz2}.Therefore, thestabilizer ofz inSLn(Q[t]) istheQ−points
of the affine Q-group Gz, and the stabilizer of z in SLn(Z[t]) is Gz(Z).
The action of SLn(Q[t]) on X is type preserving, so if σ ⊂ S is a simplex
with vertices z1,z2,...,zm, then the stabilizer of σ in SLn(Z[t]) is simply
G ∩...∩G (Z)
(cid:0) z1 zm(cid:1)
That is, the stabilizer of σ in SLn(Z[t]) is an arithmetic group, and Borel-
Serre proved that any such group is FP for all m [Bo-Se].
m
3 Polynomial points of tori
This section is devoted exclusively to a proof of the following
Proposition 3. There is a group A ≤ SLn(Z[t]) such that
3
(i) A ∼= Zn−1
(ii) There is some g ∈ SLn(Q((t−1))) such that gAg−1 is a group
of diagonal matrices
(iii) No nontrivial element of A fixes a point in the Euclidean
building for SLn(Q((t−1))).
Theproofofthispropositionismodelledonaclassicalapproachtofinding
diagonalizable subgroups of SLn(Z). The proof will take a few steps.
3.1 A polynomial over Z[t] with roots in Q((t−1))
Let {p ,p ,p ,...} = {2,3,5,...} be the sequence of prime numbers. Let
1 2 3
q = 1. For 2 ≤ i ≤ n, let q = p +1.
1 i i−1
Let f(x) ∈ Z[t][x] be the polynomial given by
n
f(x) = (x+q t) −1
hY i i
i=1
It will be clear by the conclusion of our proof that f(x) is irreducible over
Q(t), but we will not need to use this directly.
Lemma 4. There is some α ∈ Q((t−1)) such that f(α) = 0.
Proof. We want to show that there are c ∈ Q such that if α = ∞ c t1−in
i Pi=0 i
then f(α) = 0.
To begin let c = −1. We will define the remaining c recursively. Define
0 i
c by α + q t = ∞ c t1−in. Thus, c = c when i ≥ 1, each c is
i,k k Pi=0 i,k i,k i 0,k
contained in Q, and c = 0.
0,1
That α is a root of f is equivalent to
n n ∞
1 = (α+q t) = c t1−in
Y k Y(cid:0)X i,k (cid:1)
k=1 k=1 i=0
∞ n
= c tn(1−i)
X(cid:0) X (cid:0)Y ik,k(cid:1)(cid:1)
i=0 Pnk=1ik=i k=1
Our task is to find c ’s so that the above is satisfied.
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Note that for the above equation to hold we must have
n
0·tn = c tn(1−0)
X (cid:0)Y ik,k(cid:1)
Pnk=1ik=0 k=1
4
That is
n
0 = c
Y 0,k
k=1
which is an equation we know is satisfied because c = 0. Now assume that
0,1
we have determined c ,c ,...,c ∈ Q. We will find c ∈ Q.
0 1 m−1 m
Notice that the first coefficient in our Laurent series expansion above
which involves c is the coefficient for the t−nm term. This follows from the
m
fact that each i is nonnegative.
k
Since
n
c
X (cid:0)Y ik,k(cid:1)
Pnk=1ik=m k=1
is the coefficient of the t−nm term in the expansion of 1, we have
n
0 = c
X (cid:0)Y ik,k(cid:1)
Pnk=1ik=m k=1
TheaboveequationislinearoverQinthesinglevariablec andthecoeffi-
m
n
cient ofc isnonzero. Indeed, i = m, eachi ≥ 0,andc ,...,c ∈ Q
m Pk=1 k k 0 m−1
are assumed to be known quantities. Thus, c ∈ Q.
m
3.2 Matrices representing ring multiplication
By Lemma 4 we have that the field Q(t)(α) ≤ Q((t−1)) is an extension of
Q(t) of degree d where d ≤ n. It follows that Z[t][α] is a free Z[t]-module of
rank d with basis {1,α,α2,...,αd−1}.
For any y ∈ Z[t][α], the action of y on Q(t)(α) by multiplication is a
linear transformation that stabilizes Z[t][α]. Thus, we have a representation
of Z[t][α] into the ring of d×d matrices with entries in Z[t]. We embed the
ring of d × d matrices with entries in Z[t] into the upper left corner of the
ring of n×n matrices with entries in Z[t].
By Lemma 4
n
(α+q t) = 1
Y i
i=1
so each of the following matrices are invertible:
α+q t, α+q t, ..., α+q t
1 2 n
5
(We will be blurring the distinction between the elements of Z[t][α] and the
matrices that represent them.)
For 1 ≤ i ≤ n − 1, we let a = α + q t. Since a is invertible, it
i i+1 i
is an element of GLn(Z[t]), and hence has determinate ±1. By replacing
each ai with its square, we may assume that ai ∈ SLn(Z[t]) for all i. We
let A = ha ,...a i so that A is clearly abelian as it is a representation of
1 n−1
multiplication in anintegral domain. This groupA will satisfy Proposition 3.
3.3 A is free abelian on the a
i
To prove part (i) of Proposition 3 we have to show that if there are m ∈ Z
i
with
n−1
ami = 1
Y i
i=1
then each m = 0. But the first nonzero term in the Laurent series expansion
i
for α is −t, which implies that the first nonzero term in the Laurent series
expansion for each a is −t+q t = p t. Hence, the first nonzero term of
i i+1 i
n−1
ami = 1
Y i
i=1
is
n−1
(p t)mi = t0
Y i
i=1
Thus
n−1
pmi = 1
Y i
i=1
and it follows by the uniqueness of prime factorization that m = 0 for all i
i
as desired.
Thus, part (i) of Proposition 3 is proved.
3.4 A is diagonalizable
Recall that α is a d × d matrix with entries in Z[t] where d is the degree
of the minimal polynomial of α over Q(t). Let that minimal polynomial be
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q(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-
mialQalsohasdegreedandleadingcoefficient±1withQ(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−1)). That is to say that there is some
g ∈ SLn(Q((t−1))) such that gαg−1 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−1 is a set of diagonal matrices. In particular, we
have proved part (ii) of Proposition 3.
3.5 A has trivial stabilizers
To prove part (iii) of Proposition 3 we begin with the following
Lemma 5. If Γ ≤ SLn(Q[t]) is bounded under the valuation for Q((t−1)),
then the eigenvalues for any γ ∈ Γ lie in Q.
Proof. We let X be the Euclidean building for SLn(Q((t−1))). By assump-
tion, Γz = z for some z ∈ X.
Let x0 ∈ X be the vertex stabilized by SLn(Q[[t−1]]). We denote a
diagonal matrix in GLn(Q((t−1))) with entries s1,s2,...,sn ∈ Q((t−1))× by
D(s ,s ,...,s ), and we let S ⊆ X be the sector based at x and containing
1 2 n 0
vertices of the form D(tm1,tm2,...,tmn)x where each m ∈ Z and m ≥ m ≥
0 i 1 2
... ≥ m .
n
The sector S is a fundamental domain for the action of SLn(Q[t]) on X
[So] which implies that there is some h ∈ SLn(Q[t]) with hz ∈ S.
Clearly we have (hΓh−1)hz = hz, and since eigenvalues of hΓh−1 are the
same as those for Γ, we may assume that Γ fixes a vertex z ∈ S.
Fix m ,...,m ∈ Z with m ≥ ... ≥ m ≥ 0 and such that z =
1 n 1 n
D(tm1,...,tmn)x . Without loss of generality, there is a partition of n —
0
say {k ,...,k } — such that
1 ℓ
{m ,...,m } = {q ,...,q , q ,...,q , ..., q ,...q }
1 n 1 1 2 2 ℓ ℓ
where each q occurs exactly k times and
i i
q > q > ... > q
1 2 ℓ
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We have that D(tm1,...,tmn)−1ΓD(tm1,...,tmn)x = x . That gives us,
0 0
D(tm1,...,tmn)−1ΓD(tm1,...,tmn) ⊂ SLn(Q[[t−1]]). Furthermore, a trivialcal-
culation of resulting valuation restrictions for the entries of
D(tm1,...,tmn)SLn(Q[[t−1]])D(tm1,...,tmn)−1 shows that Γ is contained in a
subgroup of SLn(Q((t−1))) that is isomorphic to
ℓ
SLk (Q)⋉U
Y i
i=1
where U ≤ SLn(Q((t−1))) 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 building
for SLn(Q((t−1))).
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 ∈ SLn(Z[t]). Then
n
F(x) = ± (x−β )
Y i
i=1
where each β ∈ Q((t−1)) is an eigenvalue of a. By the previous lemma, each
i
β ∈ Q. Hence, each β ∈ Q = Q∩Q((t−1)). It follows that F(x) ∈ Z[x] so
i i
that each β is an algebraic integer contained in Q. We conclude that each
i
β is contained in Z.
i
Recall, that a has determinate 1, and that the determinate of a can
n
be expressed as β . Hence, each β is a unit in Z, so each eigenvalue
Qi=1 i i
β = ±1. It follows – by the diagonalizability of a – that a is a finite order
i
element of A ∼= Zn−1. That is, a = 1.
We have completed our proof of Proposition 3.
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4 Body of the proof
Let P ≤ SLn(Q((t−1))) be the subgroup where each of the first n−1 entries
along the bottom row equal 0. Let R (P) ≤ P be the subgroup of elements
u
that contain a (n−1)×(n−1) copy of the identity matrix in the upper left
corner. Thus R (P) ∼= Q((t−1))n−1 with the operation of vector addition.
u
Let L ≤ P be the copy of SLn−1(Q((t−1))) in the upper left corner
of SLn(Q((t−1))). We apply Proposition 3 to L (notice that the n in the
proposition is now an n−1) to derive a subgroup A ≤ L that is isomorphic
to Zn−2. By the same proposition, there is a matrix g ∈ L such that gAg−1
is diagonal.
Let b ∈ SLn(Q((t−1))) be the diagonal matrix given in the notation from
the proofs of Lemmas 2 and 5 as D(t,t,...,t,t−(n−1)). Note that b ∈ P com-
mutes with L, and therefore, with A. Thus the Zariski closure of the group
generated by b and A determines an apartment in X, namely g−1A where A
is the apartment corresponding to the diagonal subgroup of SLn(Q((t−1))).
4.1 Actions on g−1A.
If x ∈ g−1A, then it follows from Proposition 3 that the convex hull of the
∗
orbit of x under A is an (n−2)-dimensional affine space that we will name
∗
V . Furthermore, the orbit Ax forms a lattice in the space V .
x∗ ∗ x∗
We let g−1A(∞) be the visual boundary of g−1A in the Tits boundary
of X. The visual image of V is clearly an equatorial sphere in g−1A(∞).
x∗
Precisely, we let P− be the transpose of P. Then P and P− are opposite
vertices in g−1A(∞). It follows that there is a unique sphere in g−1A(∞)
that is realized by all points equidistant to P and P−. We call this sphere
SP,P−.
Lemma 7. The visual boundary of Vx∗ equals SP,P−.
Proof. Since g ∈ P ∩P−, it suffices to prove that gV is the sphere in the
x∗
boundary of A that is determined by the vertices P and P−.
Note that gV is a finite Hausdorff distance from any orbit of a point
x∗
in A under the action of the diagonal subgroup of L. The result follows
by observing that the inverse transpose map on SLn(Q((t−1))) stabilizes
diagonal matrices while interchanging P and P−.
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We let R ,R ,...,R be the standard root subgroups of R (P). Recall
1 2 n−1 u
that associated to each R there is a closed geodesic hemisphere H ⊆ A(∞)
i i
such that any nontrivial element of R fixes H pointwise and translates any
i i
point in the open hemisphere A(∞)−H outside of A(∞). Note that ∂H
i i
is a codimension 1 geodesic sphere in A(∞).
We let M ⊆ g−1A(∞) be the union of chambers in g−1A(∞) that contain
the vertex P. There is also an equivalent geometric description of M:
Lemma 8. The union of chambers M ⊆ g−1A(∞) can be realized as an
(n−2)-simplex. Furthermore,
n−1
M = g−1H
\ i
i=1
and, when M is realized as a single simplex, each of the n−1 faces of M is
contained in a unique equatorial sphere g−1∂H = ∂g−1H .
i i
Proof. Let M′ ⊆ A(∞) be the union of chambers in A(∞) containing the
vertex P. Since M = g−1M′, it suffices to prove that M′ is an(n−2)-simplex
with M′ = ∩n−1H and with each face of M′ contained in a unique ∂H .
i=1 i i
For any nonempty, proper subset I ⊆ {1,2,...,n}, we let V be the |I|-
I
dimensional vector subspace of Q((t−1))n spanned by the coordinates given
by I, and we let PI be the stabilizer of VI in SLn(Q((t−1))). For example,
P = P .
{1,2,...,n−1}
Recall that the vertices of A(∞) are given by the parabolic groups P ,
I
that edges connect PI and PI′ exactly when I ⊆ I′ or I′ ⊆ I, and that
the 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 where ∅ 6= J ⊆
J
{1,2,...,n−1}. NotethatM′ isexactlythesetofvertices V together withthe
simplices 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 − 2)-simplex. Indeed, if M′ is the abstract simplex on vertices
P ,P ,...,P , then a simplex of dimension k in M′ corresponds to a
{1} {2} {n−1}
unique P ∈ V with |J| = k + 1. So we have that M′ can be topologically
J
realized as an (n−2)-simplex.
Let F be a face of the simplex M′. Then there is some 1 ≤ i ≤ n − 1
i
such that the set of vertices of F is exactly {P ,P ,...,P }−P .
i {1} {2} {n−1} {i}
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