CONGRUENCE SUBGROUPS AND TWISTED COHOMOLOGY OF SL_{n}(F [t]) II: FINITE FIELDS AND

NUMBER FIELDS

KEVIN P. KNUDSON

Abstract. Let K be the subgroup of SL_{n}(F [t]) consisting of matrices congruent to the identity modulo t. In [4], the author conjectured that if F is a finite field, then H_{1}(K) is the adjoint representation 𝔰𝔩_{n}(F). A proof of this conjecture is provided in this note. The argument also works in case F is a number field. Applications to the cohomology of SL_{n}(F [t]) are included as well.

The fundamental theorem of algebraic K-theory asserts that if R is a regular ring, then the inclusion R → R[t] induces an isomorphism K_{i}(R) →^{≅} K_{i}(R[t]) for i ≥ 0. In [3], the author proved an unstable analogue of this result: If F is an infinite field, then for all n the inclusion SL_{n}(F) → SL_{n}(F [t]) induces an isomorphism

H_{•}(SL_{n}(F), ℤ) ⟶ H_{•}(SL_{n}(F [t]), ℤ).

A natural question to investigate is what happens when the trivial ℤ coefficients are replaced by an irreducible rational representation of SL_{n}(F). The group SL_{n}(F [t]) acts on such a representation via the map SL_{n}(F [t]) →^{t=0} SL_{n}(F).

In [4], the author began this program and obtained the following result.

Theorem 0.1 ([4]). If F is a field of characteristic zero, then for n ≥ 3

H^{1}(SL_{n}(F [t]), V) ≅ \begin{cases}H^{1}(SL_{n}(F), V) & V ≠ Ad \\ H^{1}(SL_{n}(F), V) ⊕ F & V = Ad. \end{cases}

This was proved by examining the action of SL_{n}(F [t]) on the Bruhat–Tits building associated to the function field F (t) (with valuation v(f/g) = deg g − deg f).

An alternate approach to the computation of H^{1}(SL_{n}(F [t]), V) was also given in [4]. Consider the split extension

1 ⟶ K ⟶ SL_{n}(F [t]) ⟶^{t=0} SL_{n}(F) ⟶ 1.  (1)

The group K consists of those matrices which are congruent to the identity modulo t. If V is an SL_{n}(F)-module, then K acts trivially on V; thus, H^{1}(K, V) = Hom(H_{1}(K), V). The Hochschild–Serre spectral sequence associated to (1) then yields an exact sequence

Date: October 26, 1999.

1991 Mathematics Subject Classification. 20G10.

\begin{matrix}0 ⟶ H^{1}(SL_{n}(F), V) ⟶ H^{1}(SL_{n}(F [t]), V) ⟶ \\ ⟶ H^{0}(SL_{n}(F), H^{1}(K, V)) ⟶^{d_{2}^{0,1}} H^{2}(SL_{n}(F), V). \end{matrix}

Since the extension (1) is split, the map d_{2}^{0,1} vanishes. Moreover,

H^{0}(SL_{n}(F), H^{1}(K, V)) = Hom(H_{1}(K), V)^{SL_{n}(F)} = Hom_{SL_{n}(F)}(H_{1}(K), V).

Thus, we have an isomorphism

H^{1}(SL_{n}(F [t]), V) ≅ H^{1}(SL_{n}(F), V) ⊕ Hom_{SL_{n}(F)}(H_{1}(K), V)

(this follows from the fact that (1) is split).

This reduces the question to the computation of H_{1}(K) as an SL_{n}(F)- module. There is a surjective map

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

and in general the kernel is nonzero. Indeed, in [4], Section 7, it was shown that the kernel surjects onto the module of differentials Ω_{F}^{1} , provided char(F) ≠ 2, 3 or F is perfect.

Now if F is a finite field, Ω_{F}^{1} = 0 ([2], p. 402) and hence it is possible for ρ_{n} to be an isomorphism. In [4], the author made the following conjecture.

Conjecture 0.2. If F is a finite field, then for n ≥ 3, the map

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

This was verified for n = 3, F = 𝔽_{2}, 𝔽_{3} by examining the action of K on the same Bruhat–Tits building used to prove Theorem 0.1. The complexity involved in extending the method to larger finite fields and higher n prohibited the author from proving Conjecture 0.2 in general. This paper is devoted to proving the following result, which includes Conjecture 0.2.

Theorem 0.3. If F is a finite field or a number field, then for all n ≥ 3, the map

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

Note that this does not contradict the result in [4] since Ω_{F}^{1} = 0 for number fields ([2], p. 402).

The proof of Theorem 0.3 involves completely different techniques from those in [4]. Results of Bousfield [1] play a crucial role; the requisite facts are recalled in Section 1 below.

Corollary 0.4. Let F be a finite field or a number field. Then for n ≥ 3,

H^{1}(SL_{n}(F [t]), V) ≅ H^{1}(SL_{n}(F), V) ⊕ Hom_{SL_{n}(F)}(𝔰𝔩_{n}(F), V).

TWISTED COHOMOLOGY OF SL_{n}(F [t]) II 3

In the number field case, this is exactly what Theorem 0.1 gives. If F is finite field, however, the adjoint representation may not be irreducible and hence there may be nontrivial equivariant maps 𝔰𝔩_{n}(F) → V.

This paper is organized as follows. In Section 1 we provide the results of Bousfield which will be used in the sequel. In Section 2 we prove Theorem 0.3 for finite fields. Section 3 contains the proof of Theorem 0.3for number fields. Finally, in Section 4 we consider the analogous question for the group SL_{n}(F [t, t^{−1}]). Remark. This paper deals only with the cases n ≥ 3. When n = 2, complete results valid for all fields were obtained in [4].

1. Central Series and Completions

Let G be a group and let R be a subring of ℚ (i.e., R = ℤ[J^{−1}] for a set J of primes). An R-central series in G is a descending series

G = F^{1} ⊃ F^{2} ⊃ ⋯ ⊃ F^{n} ⊃ ⋯

satisfying 1. F^{i}/F^{i+1} is central in G/F^{i+1} for all i ≥ 1;

2. [F^{r}, F^{s}] ⊆ F^{r+s} for all r, s;

3. F^{i}/F^{i+1} has trivial p-torsion for all p ∈ J. The lower R-central series {Γ_{R}^{s}G}_{s≥1} is defined by

Γ_{R}^{1}G = G
Γ_{R}^{s+1}G = ker{Γ_{R}^{s}G ⟶^{1⊗π_{s}} R ⊗_{ℤ} (Γ_{R}^{s}G/[G, Γ_{R}^{s}G])},

where π_{s} : Γ_{R}^{s}G → Γ_{R}^{s}G/[G, Γ_{R}^{s}G] is the canonical projection. This is evidently the most rapidly descending R-central series in G. If R = ℤ, then Γ_{ℤ}^{•}G is just the usual lower central series of G. An alternate description of Γ_{R}^{s+1}G is that it is the subgroup of all x ∈ Γ_{R}^{s}G such that x^{j} ∈ [G, Γ_{R}^{s}G] for some integer j ≥ 1 whose prime factors lie in J.

If {F^{s}} is an R-central series in G, then each graded quotient F^{i}/F^{i+1} is an R-module. We can form the associated graded algebra

Gr^{•}G = ⨁_{i≥1} F^{i}/F^{i+1}.

This is a Lie algebra over R with bracket induced by the commutator in G.

Given a set J of primes, we say that G is (uniquely) J-divisible if for each x ∈ G and p ∈ J there exists a (unique) w ∈ G with w^{p} = x.

Lemma 1.1 ([1],13.6). Let {F^{s}}_{s≥1} be an R-central series in G such that 1. The natural map G ⟶ \underleftarrow{lim} G/F^{s} is an isomorphism;

2. F^{s}/F^{s+1} is uniquely J-divisible for s ≥ 1;

3. The Lie product

[ , ] : G/F^{2} ⊗_{ℤ} F^{s}/F^{s+1} ⟶ F^{s+1}/F^{s+2}

is surjective for s ≥ 1;

4. G/F^{2} is a finitely generated R-module. Then Γ_{ℤ}^{s}G = Γ_{R}^{s}G = F^{s} for s ≥ 1.

Proof. Since F^{s}/F^{s+1} has trivial p-torsion for p ∈ J, we must have

Γ_{ℤ}^{s}G ⊆ Γ_{R}^{s}G ⊆ F^{s}

for s ≥ 1. We must show that F^{s} ⊆ Γ_{ℤ}^{s}G and for this it suffices to show that F^{s+1} ⊆ [G, F^{s}]. Indeed, if the latter holds, then

F^{s+1} ⊆ [G, F^{s}] = [F^{1}, F^{s}] ⊆ F^{s+1},

i.e., F^{s+1} = [G, F^{s}]. But this is the defining property of Γ_{ℤ}^{s}G.

Choose g_{1}, g_{2}, …, g_{k} ∈ G such that g_{1}F^{2}, g_{2}F^{2}, …, g_{k}F^{2} generate the Rmodule G/F^{2} and let h ∈ F^{s+1}. Hypotheses 2 and 3 imply that h can be written

h = [g_{1}, x_{1,s}] ⋯ [g_{k}, x_{k,s}]ε_{s+2}

where x_{1,s}, …, x_{k,s} ∈ F^{s} and ε_{s+2} ∈ F^{s+2}. Similarly,

ε_{s+2} = [g_{1}, x_{1,s+1}] ⋯ [g_{k}, x_{k,s+1}]ε_{s+3}

where x_{i,s+1}, …, x_{k,s+1} ∈ F^{s+1} and ε_{s+3} ∈ F^{s+3}. Some elementary properties of commutators (see [1], 11.4) then show that

h = [g_{1}, x_{1,s+1}x_{1,s}] ⋯ [g_{k}, x_{k,s+1}x_{k,s}]ε_{s+3}.

Proceeding in this way we obtain elements ε_{t+2} ∈ F^{t+2} and x_{i,t} ∈ F^{t} for t ≥ s and 1 ≤ i ≤ k such that

h = [g_{1}, p_{1,t}] ⋯ [g_{k}, p_{k,t}]ε_{t+2}

where p_{i,t} = x_{i,t} ⋯ x_{i,s+1}x_{i,s} ∈ F^{s}. Now let p_{i} ∈ F^{s} be the unique element with

p_{i}F^{t+1} = p_{i,t}F^{t+1} ∈ G/F^{t+1}

for t ≥ s (this element exists because of hypothesis 1). Then

h = [g_{1}, p_{1}] ⋯ [g_{k}, p_{k}] ∈ [G, F^{s}].

Thus, F^{s+1} ⊆ [G, F^{s}], as required.

We shall also need the following result.

Lemma 1.2. Let G be a group and let \overline{G} = \underleftarrow{lim} G/Γ_{ℤ}^{s}G be its usual lower central series completion. Then the following statements are equivalent.

1. The map \overline{G} ⟶ \overline{\overline{G}} is an isomorphism.

2. The map G/Γ^{s}G ⟶ \overline{G}/Γ^{s}\overline{G} is an isomorphism for each s ≥ 1.

Proof. Fix s ≥ 1 and consider the commutative diagram

\xymatrix{\underleftarrow{lim} G/Γ^{r}G \ar[r] \ar[d] & \underleftarrow{lim} \overline{G}/Γ^{r}\overline{G} \ar[d] \\ G/Γ^{s}G \ar[r] & \overline{G}/Γ^{s}\overline{G}}

The vertical maps are surjective since each inverse system is surjective. Clearly, then, if the top arrow is an isomorphism, so is the bottom one. Conversely, if the bottom arrow is an isomorphism for each s ≥ 1, then the inverse limits are isomorphic as well.

2. Finite Fields

Recall that the group K is the subgroup of SL_{n}(F [t]) consisting of matrices which are congruent to I modulo t. Define a filtration {K^{r}}_{r≥1} of K by

K^{r} = {X ∈ K : X ≡ I mod t^{r}}.

If X ∈ K^{r}, we may write X = I + t^{r}X_{r} + ⋯ + t^{n}X_{n}. Since X^{−1} = I − t^{r}X_{r} + ⋯, we see that

[K^{r}, K^{s}] ⊆ K^{r+s}.

Moreover, since det X = 1, we see that X_{r} has trace zero. Define a map f_{r} : K^{r} → 𝔰𝔩_{n}(F) by

f_{r}(I + t^{r}X_{r} + ⋯) = X_{r}.

One checks easily that f_{r} is a surjective group homomorphism with kernel K^{r+1}. Thus, for each r, the graded quotient K^{r}/K^{r+1} is isomorphic as an SL_{n}(F)-module to the adjoint representation 𝔰𝔩_{n}(F). In particular, we have a short exact sequence

0 ⟶ K^{2}/Γ^{2}K ⟶ H_{1}(K) ⟶^{ρ_{n}} 𝔰𝔩_{n}(F) ⟶ 0.

Consider the extension

1 ⟶ \overline{K} ⟶ SL_{n}(F [[t]]) ⟶^{t=0} SL_{n}(F) ⟶ 1.

Since the power series ring F [[t]] is the t-adic completion of F [t], we see that

SL_{n}(F [[t]]) = \underleftarrow{lim} SL_{n}(F [t])/K^{r}

and that

\overline{K} = \underleftarrow{lim} K/K^{r}.

The group \overline{K} has a t-adic filtration, just as K does.

Now assume that F is a finite field, say F = 𝔽_{q} with q = p^{d}.

Theorem 2.1. If n ≥ 3, then the map

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

Proof. Consider the group \overline{K}. Since \overline{K}/\overline{K}^{r} = K/K^{r} for each r, we see that the map \overline{K} → \underleftarrow{lim} \overline{K}/\overline{K}^{r} is an isomorphism. Since n ≥ 3, the Lie algebra 𝔰𝔩_{n}(F) is perfect; it follows that the Lie product

[ , ] : \overline{K}/\overline{K}^{2} ⊗_{ℤ} \overline{K}^{s}/\overline{K}^{s+1} ⟶ \overline{K}^{s+1}/\overline{K}^{s+2}

is surjective for s ≥ 1. Moreover, the group \overline{K}/\overline{K}^{2} ≅ 𝔰𝔩_{n}(F) is a finitely generated abelian group. Thus we may apply Lemma 1.1 with J = ∅ to conclude that Γ^{s}\overline{K} = \overline{K}^{s} (note that condition 2 in Lemma 1.1 is trivially satisfied if J = ∅).

Now we need the following fact, a proof of which may be found in [4].

Lemma 2.2. Let G^{•} be a central series in G. Then the graded algebra Gr^{•}G is generated by Gr^{1}G if and only if G^{k} = G^{k+1}Γ^{k}G for each k ≥ 1.

Since the graded algebra Gr^{•}K is generated by Gr^{1}K ≅ 𝔰𝔩_{n}(F), we see that K^{s} = K^{s+1}Γ^{s}K for each s ≥ 1. It follows then that

\overline{K} = \underleftarrow{lim} K/K^{r} ≅ \underleftarrow{lim} K/Γ^{r}K.

Consequently, the map \overline{K} → \overline{\overline{K}} is an isomorphism. By Lemma 1.2, the map K/Γ^{r}K → \overline{K}/Γ^{r}\overline{K} is an isomorphism for each r. Consider the commutative diagram

\xymatrix{K/Γ^{r}K \ar[d] \ar[r]^{≅} & \overline{K}/Γ^{r}\overline{K} \ar[d]^{≅} \\ K/K^{r} \ar[r]^{≅} & \overline{K}/\overline{K}^{r}}

The diagram shows that K/Γ^{r}K ≅ K/K^{r} for each r ≥ 1; that is, Γ^{r}K = K^{r}. Thus,

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

3. Number fields

If F is a number field, then the proof of Theorem 2.1 can be applied word for word with one minor modification. To see that Γ^{s}\overline{K} = \overline{K}^{s} we have to verify the hypotheses of Lemma 1.1. Statements 1 and 3 are clear. To check properties 2 and 4, let J be the set of all primes; that is, let R = ℚ. Then since \overline{K}^{s}/\overline{K}^{s+1} ≅ 𝔰𝔩_{n}(F) is uniquely p-divisible for all primes p and 𝔰𝔩_{n}(F) is a finite-dimensional ℚ-vector space, we see that

Γ^{s}\overline{K} = Γ_{ℚ}^{s}\overline{K} = \overline{K}^{s}.

Thus, we have proved the following result.

Theorem 3.1. If n ≥ 3, then the map

ρ_{n} : H_{1}(K) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

4. Laurent Polynomials

Consider the group SL_{n}(F [t, t^{−1}]). We have a split short exact sequence

1 ⟶ G ⟶ SL_{n}(F [t, t^{−1}]) ⟶^{t=1} SL_{n}(F) ⟶ 1.

The group G consists of matrices congruent to the identity modulo t − 1. We can filter G by powers of t − 1 to obtain a central series G^{•} with

G^{r}/G^{r+1} ≅ 𝔰𝔩_{n}(F).

Thus we have a homomorphism

σ_{n} : H_{1}(G) ⟶ 𝔰𝔩_{n}(F).

The proof of Theorem 2.1 may be applied to this situation to yield the following result.

Theorem 4.1. Let F be a finite field or a number field. Then if n ≥ 3, the map

σ_{n} : H_{1}(G) ⟶ 𝔰𝔩_{n}(F)

is an isomorphism.

Now if V is an irreducible SL_{n}(F)-module, we have an isomorphism

H^{1}(SL_{n}(F [t, t^{−1}]), V) ≅ H^{1}(SL_{n}(F), V) ⊕ Hom_{SL_{n}(F)}(𝔰𝔩_{n}(F), V).

References

[1] A. Bousfield, Homological localization towers for groups and π-modules, Memoirs Amer. Math. Soc. 10, no. 186, (1977).

[2] D. Eisenbud, Commutative algebra with a view toward algebraic geometry, Springer Graduate Texts in Mathematics 150, Springer–Verlag, New York, 1994.

[3] K. Knudson, The homology of special linear groups over polynomial rings, Ann. Sci. École Norm. Sup. (4) 30, no. 3 (1997), 385–416.

[4] K. Knudson, Congruence subgroups and twisted cohomology of SL_{n}(F [t]), J. Algebra 207 (1998), 695–721.

Department of Mathematics, Wayne State University, Detroit, MI 48202

  E-mail address: knudson@math.wayne.edu