TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Xxxx XXXX, Pages 000–000 S 0002-9947(XX)0000-0

RELATIVE COMPLETIONS OF LINEAR GROUPS OVER ℤ[t] AND ℤ[t, t^{−1}]

KEVIN P. KNUDSON

Abstract. We compute the completion of the groups SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]) relative to the obvious homomorphisms to SL_{n}(ℚ); this is a generalization of the classical Malcev completion. We also make partial computations of the rational second cohomology of these groups.

The Malcev (or ℚ-) completion of a group Γ is a prounipotent group P defined over ℚ together with a homomorphism φ : Γ → P satisfying the following universal mapping property: If ψ : Γ → U is a map of Γ into a prounipotent group then there is a unique map Φ : P → U such that ψ = Φφ. If H_{1}(Γ, ℚ) = 0, then the group P is trivial and is therefore useless for studying Γ. In particular, the Malcev completions of the groups SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]) are trivial when n ≥ 3 (this follows from the work of Suslin [14]).

Here we consider Deligne’s notion of relative completion. Suppose ρ : Γ → S is a representation of Γ in a semisimple linear algebraic group over ℚ. Suppose that the image of ρ is Zariski dense in S. The completion of Γ relative to ρ is a proalgebraic group G over ℚ, which is an extension of S by a prounipotent group U, and homomorphism ρ̃: Γ → G which lifts ρ and has Zariski dense image. When S is the trivial group, G is simply the classical Malcev completion. The relative completion satisfies an obvious universal mapping property. The basic theory of relative completion was developed by R. Hain [5] (and independently by E. Looijenga (unpublished)), and is reviewed in Section 2 below.

In this paper we consider the completions of the groups SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]) relative to the homomorphisms to SL_{n}(ℚ) given by setting t = 0 (respectively, t = 1). There is an obvious candidate for the relative completion, namely the proalgebraic group SL_{n}(ℚ[[T]]). The map

SL_{n}(ℤ[t]) → SL_{n}(ℚ[[T]])

is the obvious inclusion and the map

SL_{n}(ℤ[t, t^{−1}]) → SL_{n}(ℚ[[T]])
\begin{matrix}c \\ ◯ \end{matrix}1998 American Mathematical Society

Received by the editors January 16, 1998.

1991 Mathematics Subject Classification. Primary 55P60, 20G35, 20H05; Secondary 20G10, 20F14.

Supported by an NSF Postdoctoral Fellowship, grant no. DMS-9627503.

is induced by the map t ↦ 1 + T.

Theorem. For all n ≥ 3, the group SL_{n}(ℚ[[T]]) is the relative completion of both SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]).

Remark. We expect that the theorem holds for an arbitrary simple group G of sufficiently large rank (large enough to guarantee the vanishing of H^{2}(G(ℤ), A) for nontrivial G(ℚ)-modules A). We have chosen to work with SL_{n} just to be concrete.

The theorem does not hold for n = 2 (see Section 5 below). Our proof breaks down in this case essentially because the ℤ-Lie algebra 𝔰l_{2}(ℤ) is not perfect.

We use this result to study the cohomology of the groups SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]). This is motivated by an attempt to find unstable analogues of the Fundamental Theorem of Algebraic K–theory. Recall that if A is a regular ring, then there are natural isomorphisms K_{•}(A[t]) ≅ K_{•}(A) and K_{•}(A[t, t^{−1}]) ≅ K_{•}(A) ⊕ K_{•−1}(A). An unstable analogue does exist for infinite fields: If k is an infinite field, then H_{•}(SL_{n}(k[t]), ℤ) ≅ H_{•}(SL_{n}(k), ℤ) for all n [9]. Since ℤ is regular, one might guess that an analogous statement holds for n sufficiently large. We note, however, that if such a result holds, we must have n ≥ 3 since H_{1}(SL_{2}(ℤ[t]), ℤ) has infinite rank [3], while H_{1}(SL_{2}(ℤ), ℤ) ≅ ℤ/12.

The basic idea is to use continuous cohomology. Following Hain [4], we define the continuous cohomology of a group π to be

H_{cts}^{•}(π, ℚ) = \underrightarrow{lim} H^{•}(π/Γ^{r}π, ℚ),

where Γ^{•}π denotes the lower central series of π. There is a natural map

H_{cts}^{•}(π, ℚ) ⟶ H^{•}(π, ℚ)

which is injective in degree 2 provided H_{1}(π, ℚ) is finite dimensional. Consider the extension

1 ⟶ K(R) ⟶ SL_{n}(R) ⟶ SL_{n}(ℤ) ⟶ 1

for R = ℤ[t], ℤ[t, t^{−1}]. This yields a spectral sequence for computing the rational cohomology of SL_{n}(R). In light of the following result it is reasonable to conjecture that H^{2}(SL_{n}(R), ℚ) = 0 for n ≥ 3.

Theorem. If n ≥ 3, then H^{0}(SL_{n}(ℤ), H_{cts}^{2}(K(R), ℚ)) = 0.

Of course, one can see that H^{2}(SL_{n}(R), ℚ) = 0 for n ≥ 5 by using van der Kallen’s stability theorem [7] and the Fundamental Theorem of Algebraic K-theory. The above result provides evidence for the vanishing of H^{2}(SL_{n}(R), ℚ) for n = 3, 4. We note, however, that H^{2}(SL_{2}(ℤ[t]), ℚ) has infinite rank (this is a consequence of results of Grunewald, et. al. [3]).

The study of the relative completion of the fundamental group of a complex algebraic variety X is related to the study of variations of mixed Hodge structure over X [5]. Moreover, relative completions were used with great success by R. Hain in his study of mapping class groups M_{g} and Torelli groups T_{g} [5],[6]. In particular, he was able to provide a presentation of the Malcev Lie algebra of T_{g} which in turn gives a partial computation of H^{2}(T_{g}, ℚ). This also yields a description of the completion G_{g} of M_{g} with respect to its representation on the first homology of the surface. However, the map M_{g} → G_{g} remains a mystery. As far as we know, the results of this paper provide the first concrete descriptions of relative completions and the associated homomorphisms aside from the obvious trivial ones (e.g., SL_{n}(ℤ) → SL_{n}(ℚ), n ≥ 3).

This paper obviously owes a great deal to the work of Dick Hain and I thank him for suggesting this problem to me.

1. Malcev Completions

Recall that the Malcev completion of a group Γ is a prounipotent group M over ℚ, together with a map Γ → M which satisfies the obvious universal mapping property. We recall the construction of M as given by Bousfield [2].

First suppose that G is a nilpotent group. The Malcev completion of G consists of a group Ĝ and a homomorphism j : G → Ĝ. It is characterized by the following three properties [12, Appendix A, Cor. 3.8]:

1. Ĝ is nilpotent and uniquely divisible.

2. The kernel of j is the torsion subgroup of G.

3. If x ∈ Ĝ, then x^{n} ∈ im j for some n ≠ 0. Quillen constructs Ĝ as the set of grouplike elements of the completed group algebra ℚ̂G (completed with respect to the augmentation ideal).

Now, if G is an arbitrary group, denote by G_{r} the nilpotent group G/Γ^{r}G, where Γ^{•}G is the lower central series of G. Following Bousfield [2], we define the Malcev completion of G to be

M = \underleftarrow{lim} Ĝ_{r}

where Ĝ_{r} is the Malcev completion of G_{r}. One checks easily that the group M satisfies the universal mapping property.

2. Relative Completions

In this section we review the theory of relative completion. All results in this section are due to R. Hain [5].

Let Γ be a group and ρ : Γ → S a Zariski dense representation of Γ in a semisimple algebraic group S over ℚ. The completion of Γ relative to ρ may be constructed as follows. Consider all commutative diagrams of the form

1 → U → E → S → 1
ρ̃ ↑ ↗ ρ

Γ

where E is a linear algebraic group over ℚ, U is a unipotent subgroup of E, and the image of ρ̃ is Zariski dense. The collection of all such diagrams forms an inverse system [5, Prop 2.1] and we define the completion of Γ relative to ρ to be

G = \underleftarrow{lim} E.

The group G satisfies the following universal mapping property. Suppose that E is a proalgebraic group over ℚ such that there is a map E → S with prounipotent kernel. If φ : Γ → E is a homomorphism whose composition with E → S is ρ, then there is a unique map τ : G → E such that the diagram

G

↗  ↘
Γ τ ↓ S
φ ↘  ↗
E

commutes.

Denote by L the image of ρ : Γ → S and by T the kernel. Let Γ → G be the completion of Γ relative to ρ and let U be the prounipotent radical of G. Consider the commutative diagram

1 → T → Γ → L → 1
↓ ↓ ↓
1 → U → G → S → 1.

Denote by T the classical Malcev completion of T. The universal mapping property of T gives a map Φ : T → U whose composition with T → T is the map T → U. Denote the kernel of Φ by K. We have the following two results.

Proposition 2.1 ([5], Prop. 4.6). Suppose that the natural action of L on H_{1}(T, ℚ) extends to a rational representation of S. If H^{1}(L, A) = 0 for all rational representations A of S, then Φ is surjective.

Proposition 2.2 ([5], Prop. 4.13). Suppose H_{1}(T, ℚ) is finite dimensional and that H^{1}(L, A) vanishes for all rational representations A of S. Suppose further that H^{2}(L, A) = 0 for all nontrivial rational representations of S. Then there is a surjective map H_{2}(L, ℚ) → K.

Observe that H^{1}(SL_{n}(ℤ), A) = 0 for n ≥ 3 by Raghunathan’s theorem [13]. Moreover the second condition that H^{2}(SL_{n}(ℤ), A) = 0 for all nontrivial A holds for n ≥ 3 as well [1].^{1}

3. The Malcev Completion of K(R)

Consider the short exact sequences

1 ⟶ K(R) ⟶ SL_{n}(R) ⟶ SL_{n}(ℤ) ⟶ 1

for R = ℤ[t], ℤ[t, t^{−1}], and n ≥ 3. In this section we compute the Malcev completion of K(R).

1The result in [1] only implies vanishing for n ≥ 8. However, this is easily strengthened to n ≥ 3; compare with [6, Theorem 3.2].

Denote by 𝔪_{ℤ[t]} (resp. 𝔪_{ℤ[t,t^{−1}]}) the ideal (t) (resp. (t − 1)) of ℤ[t] (resp. ℤ[t, t^{−1}]). For each l ≥ 1, define a subgroup K^{l}(R) by

K^{l}(R) = {X ∈ K(R) : X ≡ I mod 𝔪_{R}^{l}}.

One checks easily that K^{•}(R) is a descending central series; that is,

[K^{i}, K^{j}] ⊆ K^{i+j}.

It follows that for each i, Γ^{i}K ⊆ K^{i}.

For each i ≥ 1, define homomorphisms ρ_{i}, σ_{i} as follows. If X ∈ K^{i}(ℤ[t]), write

X = I + t^{i}X_{i} + ⋯ + t^{m}X_{m}

where each X_{j} is a matrix with integer entries. Define

ρ_{i} : K^{i}(ℤ[t]) → 𝔰l_{n}(ℤ)

by ρ_{i}(X) = X_{i}. Similarly, if Y ∈ K^{i}(ℤ[t, t^{−1}]) we may write

Y = I + (t − 1)^{i}Y_{i} mod (t − 1)^{i+1}

since (t^{−1} − 1)^{i} ≡ (−1)^{i}(t − 1)^{i} mod (t − 1)^{i+1}. Now define

σ_{i} : K^{i}(ℤ[t, t^{−1}]) → 𝔰l_{n}(ℤ)

by σ_{i}(Y) = Y_{i}. These maps are well-defined since the condition det Z = 1 in K^{i}(R) forces trace Z_{i} = 0. Moreover, it is easy to see that the maps ρ_{i}, σ_{i} are surjective group homomorphisms with kernel K^{i+1}. Thus for each i ≥ 1, we have

K^{i}(R)/K^{i+1}(R) ≅ 𝔰l_{n}(ℤ).

Consider the associated graded ℤ-Lie algebra

Gr^{•}K(R) = ⨁_{i≥1} K^{i}(R)/K^{i+1}(R).

If n ≥ 3, the Lie algebra 𝔰l_{n}(ℤ) satisfies 𝔰l_{n}(ℤ) = [𝔰l_{n}(ℤ), 𝔰l_{n}(ℤ)]. It follows that the graded algebra Gr^{•}K(R) is generated by Gr^{1}K(R). The following lemma is easily proved (compare with [12, Appendix A, Prop. 3.5]).

Lemma 3.1. Let G be a group with filtration G = G^{1} ⊇ G^{2} ⊇ · · ·. Then the associated graded Lie algebra Gr^{•}G is generated by Gr^{1}G if and only if G^{r} = G^{r+1}Γ^{r} for each r ≥ 1.

Corollary 3.2. Suppose ⋂ G^{r} = {1}. If Gr^{•}G is generated by Gr^{1}G, then the completions of G with respect to the filtration G^{•} and the lower central series Γ^{•}G are isomorphic; that is,

\underleftarrow{lim} G/G^{r} ≅ \underleftarrow{lim} G/Γ^{r}G.

Proof. Consider the short exact sequence

1 ⟶ G^{r}/Γ^{r} ⟶ G/Γ^{r} ⟶ G/G^{r} ⟶ 1.

Since Gr^{•}G is generated by Gr^{1}G, we have G^{r} = G^{r+1}Γ^{r} for each r. It follows that the inverse system {G^{r}/Γ^{r}} is surjective. This, in turn, implies that the natural map

\underleftarrow{lim} G/Γ^{r} ⟶ \underleftarrow{lim} G/G^{r}

is surjective. Injectivity follows since the assumption that ⋂ G^{r} = {1} implies that \underleftarrow{lim} G^{r}/Γ^{r} = \{1\}.

We now compute the Malcev completions of the groups K(R)/K^{i}(R). We first provide the following result.

Lemma 3.3. The completion of ℤ[t, t^{−1}] with respect to the ideal (t − 1) is the power series ring ℤ[[T]]. The canonical map ℤ[t, t^{−1}] → ℤ[[T]] sends t to 1 + T.

Proof. This follows easily once one notes that in ℤ[t, t^{−1}]/(t − 1)^{m}, we have t^{−1} = 1+(t−1)+⋯+(t−1)^{m−1}, so that any polynomial in ℤ[t, t^{−1}]/(t−1)^{m} may be written as a polynomial in nonnegative powers of (t − 1).

Consider the short exact sequence

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

Corollary 3.4. The group \overline{K} is the completion of K(R) with respect to the filtration K^{1}(R) ⊃ K^{2}(R) ⊃ ⋯ and with respect to the lower central series of K(R).

Proof. The first assertion follows from Lemma 3.3 and the second from Corollary 3.2.

Observe that the group \overline{K} has a filtration given by powers of T (exactly as K(ℤ[t]) does) and that the successive graded quotients are isomorphic to 𝔰l_{n}(ℤ). Denote the filtration by \overline{K}^{•}.

We have an analogous sequence over ℚ:

1 ⟶ U ⟶ SL_{n}(ℚ[[T]]) ⟶^{T =0} SL_{n}(ℚ) ⟶ 1,

and the corresponding T -adic filtration U^{•} in U. In this case, the successive graded quotients are isomorphic to 𝔰l_{n}(ℚ). We can assemble our exact sequences into a commutative diagram

1 → K(R) → SL_{n}(R)  → SL_{n}(ℤ) → 1
↓ ↓ ||
1 → \overline{K} → SL_{n}(ℤ[[T]]) →^{T =0} SL_{n}(ℤ) → 1
↓ ↓ ↓
1 → U → SL_{n}(ℚ[[T]]) →^{T =0} SL_{n}(ℚ) → 1.

Proposition 3.5. The map K(R)/K^{r}(R) →^{j} U/U^{r} is the Malcev completion.

Proof. According to Quillen’s criterion (see Section 1) we must check three things. First, the group U/U^{r} is nilpotent and uniquely divisible. Nilpotency is obvious, so suppose

Y = I + T Y_{1} + ⋯ + T^{r−1}Y_{r−1}

is an element of U/U^{r}. For each n > 0, we must find a unique X ∈ U/U^{r} with X^{n} = Y. For an arbitrary X ∈ U/U^{r}, write X = I + T X_{1} + ⋯ + T^{r−1}X_{r−1}, and consider the equation

Y = X^{n}
= I + T nX_{1} + T^{2}(nX_{2} + \binom{n}{2}X_{1}^{2}) + ⋯ + T^{r−1}(nX_{r−1} + p(\{X_{i}\}_{i=1}^{r−2}))

where p(X_{1}, …, X_{r−2}) is a polynomial in the X_{i}, i ≤ r − 2. Clearly, we can solve this equation inductively for the X_{i} and find a unique X.

Second, we must show that the kernel of j is the torsion subgroup of K(R)/K^{r}(R). This is clear since K(R)/K^{r}(R) is torsion-free (i.e., if some power of X ∈ K lies in K^{r}, then X ∈ K^{r} already) and the map is injective.

Finally, we must show that if X ∈ U/U^{r}, then X^{m} ∈ im j for some m ≠ 0. We prove this by induction on r, beginning at r = 2. Let X = I + T X_{1} be an element of U/U^{2}. Then there is an m > 0 such that mX_{1} consists of integer entries. Then X^{m} = I + T mX_{1} lies in the image of j. Now suppose the result holds for r − 1 and consider the commutative diagram

0 → K^{r−1}/K^{r} → K/K^{r} → K/K^{r−1} → 1
↓ ↓ ↓
0 → U^{r−1}/U^{r} → U/U^{r} → U/U^{r−1} → 1.

Suppose X ∈ U/U^{r}. Denote its image in U/U^{r−1} by \overline{X}. By the inductive hypothesis, there is an integer m ≠ 0 with \overline{X}^{m} = \overline{Y} for some \overline{Y} ∈ K/K^{r−1}. Choose a lift Y of \overline{Y} in K/K^{r}. Then Y maps to \overline{X}^{m} in U/U^{r−1}. But X^{m} also maps to \overline{X}^{m} so that X^{m}Y^{−1} = Z for some Z ∈ U^{r−1}/U^{r}. Now, there exists some W ∈ K^{r−1}/K^{r}(≅ 𝔰l_{n}(ℤ)) with Z^{p} = W for some p ≠ 0. Since Y = Z^{−1}X^{m}, we have

Y^{p} = (Z^{−1}X^{m})^{p}
= Z^{−p}X^{mp}  (since U^{r−1}/U^{r} is central)
= W^{−1}X^{mp}.

Thus, X^{mp} = W Y^{p} belongs to K/K^{r} and the induction is complete.

Theorem 3.6. The inclusion K(R) → U is the Malcev completion.

Proof. Since U = \underleftarrow{lim} U/U^{r} and since U/U^{r} is the Malcev completion of K(R)/K^{r}(R), the theorem will follow immediately if we can show that K^{r}(R) = Γ^{r}K(R) for each r. This follows from the next two lemmas.

Lemma 3.7 ([2], 13.6). Let F^{•} be a central series in a group G such that 1. The natural map G → \underleftarrow{lim} G/F^{s} is an isomorphism.

2. F^{s}/F^{s+1} is torsion-free for s ≥ 1.

3. The Lie product G/F^{2}⊗F^{s}/F^{s+1} → F^{s+1}/F^{s+2} is surjective for s ≥ 1.

4. G/F^{2} is finitely generated. Then Γ^{s}G = F^{s} for s ≥ 1.

Lemma 3.8 ([2], 13.4). Let G be a group and denote by \overline{G} the completion \overline{G} = \underleftarrow{lim} G/Γ^{r}G. Then the following statements are equivalent.

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

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

Completion of the proof of Theorem 3.6. Consider the group

\overline{K} = ker(SL_{n}(ℤ[[T]]) ⟶^{T =0} SL_{n}(ℤ))

with its T -adic filtration \overline{K}^{•}. Note that Lemma 3.7 shows that \overline{K}^{r} = Γ^{r}\overline{K} for each r: the first two conditions are clear, as is the fourth; the third condition follows since the Lie algebra 𝔰l_{n}(ℤ) is perfect (it is here that we must exclude the case n = 2). Since

\overline{K} = \underleftarrow{lim} K(R)/K^{r}(R) = \underleftarrow{lim} K(R)/Γ^{r}K(R)

(the last equality is Corollary 3.4), and since

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

Lemma 3.8 implies that K(R)/Γ^{r}K(R) ≅ \overline{K}/Γ^{r}\overline{K} for all r. Consider the commutative diagram

K(R)/Γ^{r}K(R) →^{≅} \overline{K}/Γ^{r}\overline{K}
↓  ↓≅
K(R)/K^{r}(R) →^{≅} \overline{K}/\overline{K}^{r}.

It follows that K^{r}(R) = Γ^{r}K(R) and hence U is the Malcev completion of K(R).

Remark 3.9. Even if Lemma 3.8 were not available, we could still prove the result as follows. Denote by M_{r} the Malcev completion of K(R)/Γ^{r}K(R), and by M = \underleftarrow{lim} M_{r} the Malcev completion of K(R). Then the map K(R) → M factors through \overline{K}. Moreover, by the universal property of M, we get a unique map M → U which is easily seen to be an isomorphism since it has an inverse given by the universal property of the Malcev completion \overline{K} → U.

Corollary 3.10. If n ≥ 3, then H_{1}(K(R), ℤ) ≅ H_{1}(\overline{K}, ℤ) ≅ 𝔰l_{n}(ℤ).

Proof. The first isomorphism follows from Lemma 3.8 and the second isomorphism from Lemma 3.7.

4. The Relative Completion of SL_{n}(R)

We are now ready to prove the main result.

Theorem 4.1. If n ≥ 3, then the map SL_{n}(ℤ[t]) ⟶^{t↦T} SL_{n}(ℚ[[T]]) (resp. SL_{n}(ℤ[t, t^{−1}]) ⟶^{t↦1+T} SL_{n}(ℚ[[T]])) is the completion with respect to the map

SL_{n}(ℤ[t]) ⟶^{t=0} SL_{n}(ℚ) (resp. SL_{n}(ℤ[t, t^{−1}]) ⟶^{t=1} SL_{n}(ℚ)).

Proof. The relative completion is a proalgebraic group which is an extension

1 ⟶ P ⟶ G ⟶ SL_{n}(ℚ) ⟶ 1 (4.1)

where P is prounipotent. By the universal property of U, we have a unique map Φ : U → P induced by the map K(R) → P. Since H^{1}(SL_{n}(ℤ), A) = 0 for all rational SL_{n}(ℚ)-modules A [13], we see that Φ is surjective (Proposition 2.1). On the other hand, since H_{1}(K(R), ℚ) ≅ 𝔰l_{n}(ℚ) is finite dimensional by Corollary 3.10, since the action of SL_{n}(ℤ) on H_{1}(K(R), ℚ) extends to a rational representation of SL_{n}(ℚ), and since H^{2}(SL_{n}(ℤ), A) = 0 for all nontrivial rational representations A of SL_{n}(ℚ) [1], Proposition 2.2 shows that the kernel of Φ is no larger than H_{2}(SL_{n}(ℤ), ℚ) = 0. Thus, U ≅ P, and since the extension (4.1) is split ([5, Prop. 4.4]), it follows that G ≅ SL_{n}(ℚ[[T]]).

5. The Case n = 2

The proof of Theorem 4.1 breaks down in the case n = 2 for a variety of reasons.

1. The Lie algebra 𝔰l_{2}(ℤ) is not perfect.

2. Raghunathan’s theorem on the vanishing of H^{1}(SL_{n}(ℤ), A) does not apply for n = 2.

3. Borel’s result for the vanishing of H^{2}(SL_{n}(ℤ), A) cannot be strengthened to include n = 2.

However, one can make the following observations. Denote by G(ℤ) the completion of SL_{2}(ℤ) relative to its canonical inclusion in SL_{2}(ℚ), and by G(R) the completion of SL_{2}(R) (R = ℤ[t], ℤ[t, t^{−1}]) relative to the map SL_{2}(R) ⟶^{mod 𝔪_{R}} SL_{2}(ℚ). The group G(ℤ) is not isomorphic to SL_{2}(ℚ); in fact, it is an extension of SL_{2}(ℚ) by a free prounipotent group with infinite dimensional H_{1} (see [6, Rmk. 3.9]).

We have a commutative diagram

1 ⟶ K(R) ⟶ SL_{2}(R) ⟶^{⟵} SL_{2}(ℤ) ⟶ 1
↓ ↓ ↓
1 ⟶ W ⟶ G(R) ⟶^{⟵} G(ℤ) ⟶ 1.

The map Φ : G(R) → G(ℤ) is induced by the composition SL_{2}(R) → SL_{2}(ℤ) → G(ℤ) and the map Ψ : G(ℤ) → G(R) is induced by the composition SL_{2}(ℤ) → SL_{2}(R) → G(R). Since the composition SL_{2}(ℤ) → SL_{2}(R) → SL_{2}(ℤ) is the identity, we see that Φ ∘ Ψ = id_{G(ℤ)}.

If n ≥ 3, then the map SL_{n}(ℤ) → SL_{n}(ℚ) is the relative completion so that the completion of SL_{n}(R) is an extension of the completion of SL_{n}(ℤ) by the Malcev completion of K(R). This leads us to make the following conjecture.

Conjecture 5.1. The map K(R) ⟶ W is the Malcev completion.

Note that there is at least some hope for this since W is properly contained in the kernel of the map G(R) → SL_{2}(ℚ) so that W is prounipotent.

6. Cohomology

In this section we provide evidence for the following conjecture.

Conjecture 6.1. If n ≥ 3, then H^{2}(SL_{n}(R), ℚ) = 0.

Note that this conjecture is true for n ≥ 5 for the following reason. If n ≥ 5, then by van der Kallen’s stability theorem [7], we have

H_{2}(SL_{n}(R), ℤ) ≅ H_{2}(SL(R), ℤ) ≅ K_{2}(R).

It follows that if n ≥ 5, then H_{2}(SL_{n}(R), ℚ) ≅ K_{2}(R)⊗ℚ. Since K_{2}(ℤ[t]) ≅ K_{2}(ℤ) and K_{2}(ℤ[t, t^{−1}]) ≅ K_{2}(ℤ) ⊕ K_{1}(ℤ), we see that K_{2}(R) ⊗ ℚ = 0.

The tool that we will use is continuous cohomology. We define the continuous cohomology of a group π by

H_{cts}^{•}(π, ℚ) = \underrightarrow{lim} H^{•}(π/Γ^{r}π, ℚ).

The basic properties of continuous cohomology were established by Hain [4]. We note the following facts.

Proposition 6.2 ([4], Thm. 5.1). The natural map H_{cts}^{k}(π, ℚ) → H^{k}(π, ℚ) is an isomorphism for k = 0, 1 and is injective for k = 2.

The map on H^{2} need not be surjective in general. A group π is called pseudo-nilpotent if the natural map H_{cts}^{•}(π, ℚ) → H^{•}(π, ℚ) is an isomorphism. Examples of pseudo-nilpotent groups include the pure braid groups, free groups and the fundamental groups of affine curves over ℂ.

Proposition 6.3 ([4], Thm. 3.7). Let π be a group with H_{1}(π, ℚ) finite dimensional. Let P be the Malcev completion of π and denote by 𝔭 the Lie algebra of P. Then the natural map

H_{cts}^{•}(π, ℚ) ⟶ H_{cts}^{•}(𝔭, ℚ)

is an isomorphism.

Thus, if H_{1}(π, ℚ) is finite dimensional, we can use Lie algebra cohomology to obtain a lower bound on the dimension of H^{2}(π, ℚ). We will not compute H_{cts}^{2}(K(R), ℚ) explicitly. However, we note the following result.

Proposition 6.4. If n ≥ 3, then dim H_{cts}^{2}(K(R), ℚ) ≥ (n^{2} − 1)^{2}/4.

Proof. By a result of Lubotzky and Magid [10], if G is a nilpotent group with b_{1} = dim H_{1}(G, ℚ) finite, then the second Betti number b_{2} satisfies b_{2} ≥ b_{1}^{2}/4. In the case of K(R)/K^{r}(R), since H_{1}(K/K^{r}, ℚ) ≅ 𝔰l_{n}(ℚ), we see that b_{2}(K/K^{r}) ≥ (n^{2} − 1)^{2}/4 for each r.

To show that H^{2}(SL_{n}(R), ℚ) vanishes, it would suffice to show the following three things.

1. H^{2}(SL_{n}(ℤ), ℚ) = 0.

2. H^{1}(SL_{n}(ℤ), H^{1}(K(R), ℚ)) = 0.

3. H^{0}(SL_{n}(ℤ), H^{2}(K(R), ℚ)) = 0. The first statement is clear. The second follows from [13] since H^{1}(K(R), ℚ) is the adjoint representation 𝔰l_{n}(ℚ). The third statement is true for n ≥ 5.

Proposition 6.5. If n ≥ 5, then H^{0}(SL_{n}(ℤ), H^{2}(K(R), ℚ)) = 0.

Proof. Consider the Hochschild–Serre spectral sequence

E_{2}^{p,q} = H^{p}(SL_{n}(ℤ), H^{q}(K(R), ℚ)) ⟹ H^{p+q}(SL_{n}(R), ℚ).

We know that H^{2}(SL_{n}(R), ℚ) = 0 for n ≥ 5 (see the remarks following Conjecture 5.1). Note also that H^{2}(SL_{n}(ℤ), H^{1}(K(R), ℚ)) = 0 and H^{3}(SL_{n}(ℤ), ℚ) = 0. It follows that d_{2} : E_{2}^{0,2} → E_{2}^{2,1} and d_{3} : E_{3}^{0,2} → E_{3}^{3,0} are both the zero map and hence E_{∞}^{0,2} = H^{0}(SL_{n}(ℤ), H^{2}(K(R), ℚ)). But this group must vanish since E_{∞}^{1,1} and E_{∞}^{2,0} do.

The next result provides some evidence for the vanishing of the group H^{0}(SL_{n}(ℤ), H^{2}(K(R), ℚ)) when n = 3, 4. We first state the following lemma, which can be proved via direct computation.

Lemma 6.6. Let Γ_{a_{1},...,a_{n−1}} be the irreducible SL_{n}(ℚ)-module with highest weight (a_{1}+⋯ a_{n−1})L_{1}+⋯+a_{n−1}L_{n−1}, where L_{1}, …, L_{n−1} are the weights of the fundamental representation. Then we have the following isomorphisms of SL_{n}(ℚ)-modules:

1. 𝔰l_{3}(ℚ) ⊗ 𝔰l_{3}(ℚ) ≅ Γ_{2,2} ⊕ Γ_{3,0} ⊕ Γ_{0,3} ⊕ Γ_{1,1} ⊕ Γ_{1,1} ⊕ Γ_{0,0}

2. ⋀^{2} 𝔰l_{3}(ℚ) ≅ Γ_{3,0} ⊕ Γ_{0,3} ⊕ Γ_{1,1}

3. ⋀^{3} 𝔰l_{3}(ℚ) ≅ Γ_{2,2} ⊕ Γ_{3,0} ⊕ Γ_{0,3} ⊕ Γ_{1,1} ⊕ Γ_{0,0}

4. 𝔰l_{4}(ℚ) ⊗ 𝔰l_{4}(ℚ) ≅ Γ_{2,0,2} ⊕ Γ_{2,1,0} ⊕ Γ_{0,1,2} ⊕ Γ_{0,2,0} ⊕ Γ_{1,0,1} ⊕ Γ_{1,0,1} ⊕ Γ_{0,0,0} 5. ⋀^{2} 𝔰l_{4}(ℚ) ≅ Γ_{2,1,0} ⊕ Γ_{0,1,2} ⊕ Γ_{1,0,1}

6. ⋀^{3} 𝔰l_{4}(ℚ) ≅ Γ_{4,0,0} ⊕ Γ_{0,0,4} ⊕ Γ_{1,2,1} ⊕ Γ_{2,0,2} ⊕ Γ_{2,1,0} ⊕ Γ_{0,1,2} ⊕ Γ_{0,2,0} ⊕ Γ_{1,0,1} ⊕ Γ_{0,0,0}

Theorem 6.7. If n ≥ 3, then H^{0}(SL_{n}(ℤ), H_{cts}^{2}(K(R), ℚ)) = 0.

Proof. We need only consider the cases n = 3, 4. It suffices to show that H^{0}(SL_{n}(ℤ), H^{2}(K/K^{l}, ℚ)) = 0 for each l. We use Lie algebra cohomology. Denote by 𝔲 the Lie algebra of U and consider the T -adic filtration 𝔲^{•}. The Malcev Lie algebra of K/K^{l} is the Lie algebra 𝔲_{l} = 𝔲/𝔲^{l}. Observe that for each l, the quotient 𝔲^{l−1}/𝔲^{l} is isomorphic as an SL_{n}(ℚ)-module to the adjoint representation 𝔰l_{n}(ℚ), but as a Lie algebra it is abelian (i.e., to compute the bracket in 𝔲^{l−1}/𝔲^{l}, we lift elements to 𝔲^{l−1}, apply [ , ], and project back; but the commutator of any two elements in 𝔲^{l−1} lies in 𝔲^{l} and so projects to 0).

We proceed by induction on l, beginning at l = 2. The Lie algebra 𝔲_{2} is abelian of dimension n^{2} − 1; as an SL_{n}(ℚ)-module it is the adjoint representation 𝔰l_{n}(ℚ). Thus H^{2}(𝔲_{2}, ℚ) ≅ ⋀^{2} 𝔰l_{n}(ℚ) as an SL_{n}(ℚ)-module. By Lemma 6.6, parts 2 and 5, we see that H^{0}(SL_{n}(ℤ), H^{2}(𝔲_{2}, ℚ)) = 0. Now, suppose that l > 2 and that H^{0}(SL_{n}(ℤ), H^{2}(𝔲_{l−1}, ℚ)) = 0. Consider the short exact sequence

0 ⟶ 𝔲^{l−1}/𝔲^{l} ⟶ 𝔲_{l} ⟶ 𝔲_{l−1} ⟶ 0.

The kernel is central in 𝔲_{l}. Consider the Hochschild–Serre spectral sequence

E_{2}^{p,q} = H^{p}(𝔲_{l−1}, H^{q}(𝔲^{l−1}/𝔲^{l}, ℚ)) ⟹ H^{p+q}(𝔲_{l}, ℚ).

We have isomorphisms of SL_{n}(ℚ)-modules:

1. H^{2}(𝔲_{l−1}, H^{0}(𝔲^{l−1}/𝔲^{l}, ℚ)) = H^{2}(𝔲_{l−1}, ℚ)

2. H^{1}(𝔲_{l−1}, H^{1}(𝔲^{l−1}/𝔲^{l}, ℚ)) ≅ 𝔰l_{n}(ℚ) ⊗ 𝔰l_{n}(ℚ)

3. H^{0}(𝔲_{l−1}, H^{2}(𝔲^{l−1}/𝔲^{l}, ℚ)) ≅ Hom_{ℚ}(⋀^{2} 𝔰l_{n}(ℚ), ℚ). By induction, the SL_{n}(ℤ) invariants of the first module are trivial and by Lemma 6.6, parts 2 and 5, so are the invariants of the last group. It follows that H^{0}(SL_{n}(ℤ), E_{∞}^{0,2}) = 0. Also, since

H^{1}(SL_{n}(ℤ), E_{∞}^{0,1}) = H^{1}(SL_{n}(ℤ), 𝔰l_{n}(ℚ)) = 0,

the long exact cohomology sequence associated to the extension

0 ⟶ E_{∞}^{0,1} ⟶^{d_{2}} H^{2}(𝔲_{l−1}, ℚ) ⟶ E_{∞}^{2,0} ⟶ 0

shows that H^{0}(SL_{n}(ℤ), E_{∞}^{2,0}) = 0. It remains to show that the group H^{0}(SL_{n}(ℤ), E_{∞}^{1,1}) vanishes.

Note that E_{2}^{1,1} contains a copy of the trivial representation (parts 1 and 4 of Lemma 6.6). However, the differential

d_{2} : E_{2}^{1,1} ⟶ H^{3}(𝔲_{l−1}, ℚ)

is easily seen to map the trivial representation onto a copy of the trivial representation in the image (this copy arises from the map in cohomology induced by the map 𝔲_{l−1} → 𝔰l_{n}(ℚ); use parts 3 and 6 of Lemma 6.6). It follows that E_{∞}^{1,1} contains no copies of the trivial representation and hence H^{0}(SL_{n}(ℤ), E_{∞}^{1,1}) = 0. Thus

H^{0}(SL_{n}(ℤ), H^{2}(𝔲_{l}, ℚ)) = 0

and the induction is complete.

One might conjecture that K(R) is pseudo-nilpotent (we do not know if this is the case). If so, it would follow that H^{0}(SL_{n}(ℤ), H^{2}(K(R), ℚ)) = 0 and hence H^{2}(SL_{n}(R), ℚ) = 0 for n ≥ 3.

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Department of Mathematics, Northwestern University, Evanston, IL 60208

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