RELATIVE COMPLETIONS AND THE COHOMOLOGY OF LINEAR GROUPS OVER LOCAL RINGS

KEVIN P. KNUDSON

For a discrete group G there are two well-known completions. The first is the Malcev (or unipotent) completion. This is a prounipotent group U, defined over ℚ, together with a homomorphism ψ : G → U that is universal among maps from G into prounipotent ℚ-groups. To construct U, it suffices to consider the case where G is nilpotent; the general case is handled by taking the inverse limit of the Malcev completions of the G/Γ^{r}G, where Γ^{•}G denotes the lower central series of G. If G is abelian, then U = G ⊗ ℚ. We review this construction in Section 2.

The second completion of G is the p-completion. For a prime p, we set G^{∧p} = \underleftarrow{lim} G/Γ_{p}^{r}G, where Γ_{p}^{•}G is the p-lower central series of G. If G is a finitely generated abelian group, then G^{∧p} = ℤ_{p} ⊗ G, where ℤ_{p} is the ring of p-adic integers [3]. The group G^{∧p} is a pro-p-group and each G/Γ_{p}^{r}G is nilpotent provided H_{1}(G, 𝔽_{p}) is finite dimensional.

Both of these completions are instances of a general construction. Let k be a field. The unipotent k-completion of a group G is a prounipotent k-group U_{k} together with a homomorphism G → U_{k}. The group U_{k} is required to satisfy the obvious universal mapping property. The Malcev completion is the case k = ℚ and the p-completion is the case k = 𝔽_{p}. This construction for other fields k is probably well-known to the experts, but it does not seem to be in the literature.

One reason to study such completions is that they may be used to gain cohomological information about the groups G and U. The restriction map H^{i}(U, k) → H^{i}(G, k) is an isomorphism for i = 0, 1, and the map H_{cts}^{2}(U, k) → H^{2}(G, k) is injective (the definition of H_{cts}^{2} will be recalled below). This allows one to obtain either a lower bound for dim_{k} H^{2}(G, k) or an upper bound for dim_{k} H_{cts}^{2}(U, k).

Unfortunately, the group U_{k} may be trivial (e.g., G perfect, or more generally, if H_{1}(G, k) = 0). To circumvent this, Deligne suggested the notion of relative completion. Suppose that ρ : G → S is a representation of G in a reductive group S defined over k. Assume that the image of ρ is Zariski dense. The completion of G relative to ρ is a proalgebraic group G over k which is an extension of S by a prounipotent group U together with a map G → G. The group G should satisfy the obvious universal mapping property.

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

Partially supported by NSF grant no. DMS-0070119.

The basic theory of relative completion in characteristic zero was worked out by R. Hain[6]. Many of his results remain valid in positive characteristic. We shall review this in Section 4 below. For examples of relative completions in characteristic zero beyond those presented here, the reader is referred to Hain’s study [6] of the completion of the mapping class group Γ_{g,r} of a surface S of genus g with r marked points relative to its symplectic representation as the group of automorphisms of H_{1}(S, ℤ). Other examples, due to the author [12], are the groups SL_{n}(ℤ[t]) and SL_{n}(ℤ[t, t^{−1}]) relative to their obvious representations in SL_{n}(ℤ). Recent work by Hain and M. Matsumoto [7] tackles the absolute Galois group Gal(\overline{ℚ}/ℚ) relative to the cyclotomic character χ : Gal(\overline{ℚ}/ℚ) → ℤ_{ℓ}^{×}.

In this paper we study the completions of groups such as SL_{n}(O_{X,x}), where O_{X,x} is the local ring of a closed point x on a smooth affine curve X. We also study the completion of SL_{n}(ℤ) relative to the reduction map SL_{n}(ℤ) → SL_{n}(𝔽_{p}). We use these completions to compute the second continuous cohomology of special linear groups over complete local rings. A sample result is the following.

Theorem 5.4. If k is a finite field or a number field, then for all n ≥ 3, H_{cts}^{2}(SL_{n}(k[[T]]), k) = 0.

As far as the author knows, this is the first calculation of continuous cohomology with coefficients of the same characteristic as k.

This paper is organized as follows. In Section 1, we review Hain’s construction of the completion of ρ : G → S. In Section 2, we discuss in detail the case where S is the trivial group; this is the unipotent completion mentioned above. In Section 3, we present several examples of unipotent completions. Section 4 deals with the basic theory of relative completion and the computation of examples. Finally, in Section 5 we carry out some cohomology calculations. Acknowledgements. I would like to thank Dick Hain and Chuck Weibel for many useful discussions.

1. A Construction of the Relative Completion

The following construction is due to R. Hain [6]. Let G be a group and suppose that ρ : G → S is a Zariski dense representation in a reductive k-group S. Consider all commutative diagrams of the form

\xymatrix{1 \ar[r] & U \ar[r] & E \ar[r] & S \ar[r] & 1 \\  &  & G \ar[u]^{ρ̃} \ar@{-->}[ur]_{ρ}}

where E is a linear algebraic group over k, U is a unipotent subgroup of E, and the image of ρ̃ is Zariski dense. It is clear how to define morphisms of such diagrams. Moreover, the collection of such diagrams forms an inverse system [6], Prop. 2.1. The completion G of G relative to ρ is defined to be the inverse limit

G = \underleftarrow{lim} E

over all the above commutative diagrams.

The group G satisfies the following universal mapping property [6], Prop. 2.3. Suppose that E is a proalgebraic group over k and that E → S is a homomorphism of proalgebraic groups with prounipotent kernel. If ϕ : G → E is a map whose composition with E → S is ρ, then there is a unique map Φ : G → E such that the diagram

\xymatrix{ & G \ar[dd]_{Φ} \ar[dr] \\ G \ar@{-->}[ur] \ar[dr]_{ϕ} &  & S \\  & E \ar@{-->}[ur]}

commutes.

2. Unipotent Completion

The unipotent completion of a group is the case where S is the trivial group. The following construction of the unipotent completion is due to D. Quillen [15], although he considered only the characteristic zero case.

Let k be a field. For a group G, denote by J_{k} the augmentation ideal of the group ring kG. The map g ↦ (g − 1) + J_{k}^{2} induces an isomorphism

H_{1}(G, k) ⟶^{≅} J_{k}/J_{k}^{2}.

Denote by kG^{∧} the completed algebra

kG^{∧} = \underleftarrow{lim} kG/J_{k}^{l};

it is a complete Hopf algebra (kG is a Hopf algebra via Δ(g) = g ⊗ g). Denote the completion of J_{k} by Ĵ_{k}.

Consider the set of grouplike elements

P = {x ∈ kG^{∧} : Δ(x) = x ⊗ x, ε(x) = 1}.

This is a subgroup of the group of units of kG^{∧}. Define a filtration of P by

P^{l} = P ∩ (1 + Ĵ_{k}^{l})

and set

P_{l} = P/P^{l+1}.

Lemma 2.1. If H_{1}(G, k) is finite dimensional, then P_{l} is a linear algebraic group over k.

Proof. Since H_{1}(G, k) is finite dimensional and the graded algebra Gr• ; ĴkG^{∧} is generated by

Gr1 ; ĴkG^{∧} = Ĵ_{k}/Ĵ_{k}^{2} = H_{1}(G, k),

the quotient algebra kG^{∧}/Ĵ_{k}^{l+1} is a finite dimensional k-algebra. Left multiplication gives an embedding P → Aut(kG^{∧}) and since the elements of P^{l+1} act trivially on kG^{∧}/Ĵ_{k}^{l+1}, we have an induced embedding P_{l} → Aut(kG^{∧}/Ĵ_{k}^{l+1}).

Note that each P_{l} is actually a unipotent group: the filtration

P_{l} = P/P^{l+1} ⊇ P^{2}/P^{l+1} ⊇ ⋯ ⊇ P^{l}/P^{l+1} ⊇ 0

is a central series and hence P_{l} is nilpotent.

We have an inclusion G → kG. Composing with the completion map kG → kG^{∧} gives a map G → kG^{∧}. The image of this map lies in P and hence for each l there is a homomorphism G → P_{l}.

Lemma 2.2. If H_{1}(G, k) is finite dimensional, then the image of the map G → P_{l} is Zariski dense.

Proof. Denote by Z_{l} the Zariski closure of the image of G in P_{l}. The composite

H_{1}(G, k) ⟶ H_{1}(Z_{l}, k) ⟶ H_{1}(P_{l}, k)

is an isomorphism and hence the map H_{1}(Z_{l}, k) → H_{1}(P_{l}, k) is surjective. We claim that the map Z_{l} → P_{l} is surjective; that is, Z_{l} = P_{l}. To see this, suppose that U is a unipotent group over k and let V be a subgroup such that the induced map H_{1}(V, k) → H_{1}(U, k) is surjective. Since U = V if and only if the Lie algebras 𝔲 and 𝔳 coincide, we may work in the Lie algebra. The algebra 𝔲 is filtered by its lower central series

𝔲 = 𝔲^{1} ⊃ 𝔲^{2} ⊃ ⋯ ⊃ 𝔲^{l} ⊃ 0.

We proceed by induction on l. If l = 1, then [𝔲, 𝔲] = 0 and hence [𝔳, 𝔳] = 0. Since 𝔳 = H_{1}(𝔳) surjects onto H_{1}(𝔲) = 𝔲, we see that 𝔳 = 𝔲. Now suppose that l > 1. Our surjectivity assumption asserts that 𝔳 generates 𝔲 modulo 𝔲^{2}. An easy induction argument shows that 𝔳^{k} generates 𝔲^{k} modulo 𝔲^{k+1}. In particular, we see that 𝔳^{l} generates 𝔲^{l} modulo 𝔲^{l+1} = 0; i.e., 𝔳^{l} = 𝔲^{l}. Repeated application of the five lemma shows that 𝔳^{2} = 𝔲^{2} and also that 𝔳 = 𝔲. Thus, V = U and we are done.

Proposition 2.3. If H_{1}(G, k) is finite dimensional, then the map G → P is the unipotent k-completion.

Proof. Note that P = \underleftarrow{lim} P_{l} is a prounipotent group over k and that the image of G in P is Zariski dense. Denote by U the unipotent kcompletion of G. By the universal mapping property, there is a unique map φ : U → P making the diagram

\xymatrix{G \ar[r] \ar[dr] & U \ar[d]^{φ} \\  & P}

commute. We show that φ is an isomorphism.

Since the map G → P has Zariski dense image, so does the map U → P. It follows that φ is surjective since φ(U) is a closed subgroup of P.

To see that φ is injective, suppose that U is a unipotent group over k. Then U is a subgroup of some U_{n} ⊂ GL_{n}(k) for some n (here, U_{n} is the subgroup of upper triangular unipotent matrices). A representation p : G → U induces a ring homomorphism p̃: kG → 𝔤l_{n}(k). The image of J_{k} under p̃ lies in the Lie algebra 𝔫 of nilpotent upper triangular matrices. Thus, the kernel of p̃ contains J_{k}^{n} and hence p̃ induces a map

\overline{p} : kG/J_{k}^{n} ⟶ 𝔤l_{n}(k).

Since p̃(J_{k}) ⊆ 𝔫, we see that P_{n−1} ⊆ I + J_{k}/J_{k}^{n} is contained in U_{n}. If the image of G is Zariski dense in U, then P_{n−1} ⊆ U. Thus, the diagram

\xymatrix{G \ar[r] \ar[dr]_{p} & P_{n−1} \ar[d] \\  & U}

commutes. Now, U = \underleftarrow{lim} U_{α}, with each U_{α} unipotent over k. Applying the above construction to the compositions G → U → U_{α}, we see that φ is injective as follows. If φ(u) = 1, then φ(u) maps to 1 in each P_{n−1}. But the diagram

\xymatrix{U \ar[r] \ar[d]_{φ} & U_{α} & u \ar[r] \ar[d] & u_{α} \\ P \ar[r] & P_{n−1} \ar[u] & φ(u) = 1 \ar[r] & 1 \ar[u]}

shows that u_{α} = 1 for each α; i.e., u = 1.

Corollary 2.4. If k ⊂ F is a field extension, then the map U_{F} → U_{k}(F) is an isomorphism.

Proof. The group U_{F} is the set of grouplike elements in F G^{∧}. On the other hand, the group U_{k}(F) is the set of grouplike elements of kG^{∧} ⊗ F = F G^{∧}.

3. Examples of Unipotent Completions

3.1. G = ℤ. Let k be a field. The group algebra kG is the ring of Laurent polynomials k[t, t^{−1}]. The augmentation is principal and is generated by t − 1. The completion kG^{∧} is the power series ring k[[T]] with augmentation T ↦ 0; the ideal Ĵ_{k} is the principal ideal (T). The canonical map kG → kG^{∧} sends t to 1+T. If k has characteristic zero, then P = {(1 + T)^{α} : α ∈ k}, where (1 + T)^{α} = exp(α log(1 + T)). Also, P ∩ (1 + Ĵ_{k}^{l}) = {1} for l ≥ 2 and hence P = P_{l} for all l. Thus, P ≅ k = k ⊗_{ℤ} G, as one would expect.

If k = 𝔽_{p}, then the situation is much different. Since (1 + T)^{p} = 1 + T^{p} ∈ 1 + Ĵ_{k}^{p}, we see that 1 + T has finite order in each P_{l} = P/P ∩ (1 + Ĵ_{k}^{l+1}). In fact, 1 + T has order p^{d} in P_{p^{d}}, P_{p^{d}+1}, …, P_{p^{d+1}−1}. Note also that the maps ℤ → P_{l} are surjective since there are no proper Zariski dense subgroups of an 𝔽_{p}-group (the Zariski topology is discrete). Each of the homomorphisms

P_{p^{d}} ⟵ P_{p^{d}+1} ⟵ ⋯ ⟵ P_{p^{d+1}−1}

is the identity ℤ/p^{d} → ℤ/p^{d}. It follows that P = \underleftarrow{lim} P_{l} ≅ ℤ_{p}. Thus, P ≅ ℤ_{p} ⊗_{ℤ} G.

Note that this must be correct. If one hoped that the answer were 𝔽_{p} = 𝔽_{p} ⊗_{ℤ} G, then the universal mapping property would imply the existence of a nontrivial homomorphism 𝔽_{p} → ℤ_{p}.

Let us make the case p = 2 explicit. The group P_{l} is a subgroup of Aut(kG^{∧}/Ĵ_{k}^{l+1}), acting as left multiplication by elements of P. We shall write down the coordinate rings of the first few P_{l}. Note that with respect to the basis 1, T, T^{2}, …, T^{l} of k[[T]]/(T^{l+1}), the element 1 + T acts via the lower triangular (l + 1) × (l + 1) matrix

( \begin{matrix}1 & 0 & 0 & ⋯ & & 0 \\ 1 & 1 & 0 & ⋯ & & 0 \\ 0 & 1 & 1 & ⋯ & & 0 \\ ... & & & ... \\ & & & & 1 & 0 \\ 0 & 0 & ⋯ & & 1 & 1 \end{matrix} ) .

For l = 1, this matrix has order 2. The coordinate ring of P_{1} is

O(P_{1}) = 𝔽_{2}[T_{11}, T_{12}, T_{21}, T_{22}]/(T_{12}, T_{11} − 1, T_{22} − 1).

For l = 2, 3, the above matrix has order 4. The coordinate ring of P_{2} is

O(P_{2}) = 𝔽_{2}[T_{ij}, 1 ≤ i, j ≤ 3]/(T_{ij}, i < j; T_{ii} − 1; T_{21} − T_{32}),

and that of P_{3} is

O(P_{3}) = 𝔽_{2}[T_{ij}, 1 ≤ i, j ≤ 4]/I,

where I is the ideal generated by (T_{ij}, i < j; T_{ii} − 1; T_{21} − T_{32}; T_{32} − T_{43}; T_{31} − T_{42}; T_{21} − T_{41}). Note that each of these are two-dimensional 𝔽_{2}-algebras. For l = 4, the matrix representing 1 + T has order 8. The coordinate ring of P_{4} is

O(P_{4}) = 𝔽_{2}[T_{ij}, 1 ≤ i, j ≤ 5]/I

where I is the ideal generated by T_{ij}, i < j; T_{ii} − 1; T_{21} − T_{41}; T_{21} − T_{32}; T_{32} − T_{43}; T_{43} − T_{54}; T_{31} − T_{42}; T_{42} − T_{53}; T_{41} − T_{52}. It is a threedimensional 𝔽_{2}-algebra. To obtain the 𝔽_{2^{d}}-unipotent completion of ℤ, we take the inverse limit of the groups Hom_{𝔽_{2}}(O(P_{l}), 𝔽_{2^{d}}).

3.2. G = ℤ/n. Here the group algebra is kG = k[t]/(t^{n} − 1); the augmentation ideal is J_{k} = (t − 1). If k has characteristic zero, then since

(t − 1)^{2} = t^{2} − 2t + 1
= t^{n} + t^{2} − 2t
= t(t^{n−2} + ⋯ + t + 2)(t − 1),

and since both t and t^{n−2} +⋯+t+2 are units in kG, we have J_{k}^{2} = J_{k}. Thus, the completion kG^{∧} is the algebra \underleftarrow{lim} kG/J_{k}^{l} = k. The only grouplike element is 1, and hence the unipotent k-completion is trivial.

If k = 𝔽_{p}, then we must consider separately two cases. First assume that p is prime to n. In this case, the above factorization shows again that kG^{∧} = k and that P = 1. If p divides n, then we may assume that n = p^{d} for some d ≥ 1 (note that the completion of G × H is the product of the completions of G and H). In this case we have J^{p^{d}} = 0 since (1 − t)^{p^{d}} = (1 − t^{p^{d}}) = (1 − t^{n}) = 0. Thus, kG^{∧} = kG and P is the collection of the t^{i}, 0 ≤ i ≤ p^{d}. This group is ℤ/p^{d}.

In the characteristic zero case, then, we see that the completion of ℤ/n is ℤ/n⊗_{ℤ}k and in the characteristic p case it is ℤ/n⊗_{ℤ}ℤ_{p}. So, if G is a finitely generated abelian group, the ℚ-completion of G is G ⊗_{ℤ} ℚ and the 𝔽_{p}-completion is G ⊗_{ℤ} ℤ_{p}.

3.3. p-completion v. 𝔽_{p}-completion. Suppose that H_{1}(G, 𝔽_{p}) is finite dimensional. Then the 𝔽_{p}-completion of G is the p-completion. Indeed, the p-completion of G is the inverse limit

G^{∧p} = \underleftarrow{lim} G/Γ_{p}^{r}G,

where Γ_{p}^{•}G is the p-lower central series of G. Each group G/Γ_{p}^{r}G is a finite p-group if H_{1}(G, 𝔽_{p}) is finite dimensional. On the other hand, each group P_{l} is a unipotent group over 𝔽_{p}. Such groups are nilpotent and since P_{l} is a closed subgroup of some U_{n}(𝔽_{p}), it is finite. Thus, we have a homomorphism

G^{∧p} ⟶ P.

But since G^{∧p} is a prounipotent group over 𝔽_{p}, we have a unique map

P ⟶ G^{∧p},

and it is clear that these maps are inverse isomorphisms. Alternatively, both G^{∧p} and P satisfy the same universal mapping property and are therefore isomorphic via a unique isomorphism.

3.4. Coordinate rings of curves. Let X be a smooth affine curve over k and let x be a closed point on X. Denote by O_{X,x} the local ring at x; it has maximal ideal 𝔪_{x}. We have a split extension

1 ⟶ K(O_{X,x}) ⟶ SL_{n}(O_{X,x}) ⟶^{mod 𝔪_{x}} SL_{n}(k) ⟶ 1.

The group K(O_{X,x}) is filtered by powers of the maximal ideal 𝔪_{x}; denote this filtration by K^{•}(O_{X,x}). By a theorem of W. Klingenberg [10], K^{•}(O_{X,x}) is the lower central series of K(O_{X,x}) provided char(k) ≠ 2 or k ≠ 𝔽_{3}. The graded quotients satisfy

K^{l}/K^{l+1} ≅ 𝔰l_{n}(k);

in particular, H_{1}(K(O_{X,x}), ℤ) ≅ 𝔰l_{n}(k).

This is a special case of the following.

Proposition 3.1. Let G be a group such that the graded quotients Γ^{l}G/Γ^{l+1}G are k-vector spaces. Let Ĝ = \underleftarrow{lim} G/Γ^{l}G. Then the map G → Ĝ is the unipotent k-completion.

Proof. Each group G/Γ^{l}G is a unipotent k-group; hence the universal mapping property provides a unique map P → Ĝ; it is clearly surjective. On the other hand, we have P = \underleftarrow{lim} P_{α}, where each P_{α} is a unipotent k-group. Such groups are nilpotent and hence the composition G → P → P_{α} factors through some G/Γ^{l}G. It follows that P → Ĝ is injective.

Thus, for the group K(O_{X,x}), char(k) ≠ 2, k ≠ 𝔽_{3}, the unipotent k-completion is the group

K(Ô_{X,x}) = ker{SL_{n}(Ô_{X,x}) ⟶ SL_{n}(k)},

where Ô_{X,x} is the 𝔪_{x}-adic completion of O_{X,x}.

3.5. G = SL_{n}(ℤ). We have the exact sequence

1 ⟶ Γ(n, p) ⟶ SL_{n}(ℤ) ⟶ SL_{n}(𝔽_{p}) ⟶ 1.

If n ≥ 3, the filtration of Γ(n, p) by powers of p is the lower central series [1]. Since the graded quotients Γ(n, p^{l})/Γ(n, p^{l+1}) are isomorphic to 𝔰l_{n}(𝔽_{p}), Proposition 3.1 implies that the 𝔽_{p}-completion of Γ(n, p) is the group

Γ̂ = \underleftarrow{lim} Γ(n, p)/Γ(n, p^{l}).

This group fits into the exact sequence

1 ⟶ Γ̂ ⟶ SL_{n}(ℤ_{p}) ⟶ SL_{n}(𝔽_{p}) ⟶ 1.

Note that the 𝔽_{p}-completion of SL_{n}(ℤ) is trivial. Also, if ℓ ≠ p the 𝔽_{ℓ}-completion of Γ(n, p) is trivial.

If k is any field of characteristic p, then the unipotent k-completion of Γ(n, p) is obtained by taking the k-form of Γ̂. This is constructed as follows. The first group in the inverse system is U_{2} = 𝔰l_{n}(k). The next is a nonsplit extension

0 ⟶ 𝔰l_{n}(k) ⟶ U_{3} ⟶ 𝔰l_{n}(k) ⟶ 0.

The group U_{3} is the unipotent group defined by the diagram

\xymatrix{0 \ar[r] & 𝔰l_{n}(𝔽_{p}) \ar[d] \ar[r] & Γ(n, p)/Γ(n, p^{3}) \ar[d] \ar[r] & 𝔰l_{n}(𝔽_{p}) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & 𝔰l_{n}(k) \ar[r] & U_{3} \ar[r] & 𝔰l_{n}(k) \ar[r] & 0.}

In general, we have a diagram of central extensions

\xymatrix{0 \ar[r] & 𝔰l_{n}(𝔽_{p}) \ar[d] \ar[r] & Γ(n, p)/Γ(n, p^{r}) \ar[d] \ar[r] & Γ(n, p)/Γ(n, p^{r−1}) \ar[d] \ar[r] & 1 \\ 0 \ar[r] & 𝔰l_{n}(k) \ar[r] & U_{r} \ar[r] & U_{r−1} \ar[r] & 1}

and the k-completion of Γ(n, p) is U_{k} = \underleftarrow{lim} U_{r}.

4. Relative Completions

Let S be a reductive algebraic group over k and let ρ : Γ → S be a representation with Zariski dense image. Denote by G the completion of Γ with respect to ρ; it is an extension

1 ⟶ U ⟶ G ⟶ S ⟶ 1.

In this section we construct some examples of relative completions. We first prove a few basic facts.

First, let A be a rational representation of S and suppose that we have an extension in the category of algebraic groups over k:

0 ⟶ A ⟶ G ⟶ S ⟶ 1.

Then this extension splits [2], p. 158. This extends to unipotent kernels; a proof of the following may be found in [6], Prop. 4.3.

Proposition 4.1. If S is a reductive algebraic group over k and if

1 ⟶ U ⟶ G ⟶ S ⟶ 1

is an extension in the category of algebraic groups over k, where U is unipotent, then the extension splits.

From this, one easily deduces that the relative completion

1 ⟶ U ⟶ G ⟶ S ⟶ 1

is a split extension.

Denote the kernel of ρ by T and the unipotent k-completion of T by T . Then the map T → U induces a unique map Φ : T → U. Denote by K the kernel of Φ and by L the image of ρ.

Proposition 4.2. Suppose that H_{1}(T, k) is finite dimensional. If the action of L on H_{1}(T, k) extends to a rational representation of S (e.g., L = S), then K is central in T .

Proof. [6] The group Γ acts on the completion kT^{∧} by conjugation. This action preserves the filtration by powers of Ĵ, so Γ acts on the associated graded algebra

Gr• ; ĴkT^{∧} = ⨁_{m≥0} Ĵ^{m}/Ĵ^{m+1}.

If H_{1}(T, k) is finite dimensional, then each algebra kT^{∧}/Ĵ^{l} is finite dimensional. Thus, each of the groups Aut(kT/J^{l}) is an algebraic group. Since Gr• ; ĴkT^{∧} is generated by J/J^{2} = H_{1}(T, k), it follows that Aut(kT^{∧}), the group of augmentation preserving algebra automorphisms of kT^{∧}, is a proalgebraic group which is an extension of a subgroup of AutH_{1}(T, k) by a prounipotent group.

If the action of Γ on H_{1}(T, k) factors through a rational representation S → AutH_{1}(T, k), then we can form a proalgebraic group extension

1 ⟶ J^{−1}Aut(kT^{∧}) ⟶ E ⟶ S ⟶ 1

of S by the prounipotent radical of Aut(kT^{∧}) by pulling back the extension

1 ⟶ J^{−1}Aut(kT^{∧}) ⟶ Aut(kT^{∧}) ⟶ AutH_{1}(T, k)

along S → AutH_{1}(T, k). The map Γ → Aut(kT^{∧}) lifts to a map Γ → E whose composition with E → S is ρ : Γ → S. This induces a homomorphism G → E. Since the composite

T ⟶ G ⟶ E ⟶ Aut(kT^{∧})

is the action of T ⊂ kT^{∧} on kT^{∧} by inner automorphisms, we see that the kernel of this map is the center of T . It follows that the kernel of Φ is central in T .

In [6], Hain states the result (due to Deligne) that if H^{1}(L, A) = 0 for all rational representations A of S, then Φ : T → U is surjective. This works well in characteristic zero thanks to the vanishing results for arithmetic groups due to Ragunathan [16]. In positive characteristic, however, this criterion is not very useful. We present here a simplified version which suits our purposes.

Proposition 4.3. If ρ : Γ → S is surjective, then Φ : T → U is surjective.

Proof. Note that Φ is surjective if and only if the induced map

H_{1}(T, k) → H_{1}(U, k)

is surjective. Let A be the cokernel of this map; it is a rational representation of S. Pushing out the extension

1 ⟶ U ⟶ G ⟶ S ⟶ 1

along the map U → H_{1}(U, k) → A, we obtain an extension of algebraic groups

0 ⟶ A ⟶ G ⟶ S ⟶ 1.

This splits, say s : S → G is a splitting. Since the image of Γ in G is Zariski dense, the image of Γ in G is Zariski dense. But since the diagram

\xymatrix{Γ \ar[r] \ar[dr]_{ρ} & G \\  & S \ar[u]_{s}}

commutes, we see that the image of s is Zariski dense in G. This forces A to be trivial.

We are now able to compute some examples of relative completions.

4.1. Let X be a smooth affine curve over k and let O_{X,x} be the local ring at a closed point x with maximal ideal 𝔪_{x}. In the previous section we showed that the unipotent k-completion of K(O_{X,x}) = {A ∈ SL_{n}(O_{X,x}) : A ≡ I mod 𝔪_{x}} is the group K(Ô_{X,x}), provided that char(k) ≠ 2 or k ≠ 𝔽_{3}. Let ρ : SL_{n}(O_{X,x}) → SL_{n}(k) be reduction modulo 𝔪_{x} and consider the completion relative to ρ

1 ⟶ U ⟶ G ⟶ SL_{n}(k) ⟶ 1.

There is a surjective map Φ : K(Ô_{X,x}) → U with central kernel. An easy calculation shows that the center of K(Ô_{X,x}) is trivial. Since the above extension splits, we see that the group G is the group SL_{n}(Ô_{X,x}).

4.2. Consider the curve X = 𝔸_{k}^{1}. The coordinate ring is the polynomial algebra k[t]. If k is a finite field or a number field, then the filtration of

K(k[t]) = ker{SL_{n}(k[t]) ⟶^{ρ} SL_{n}(k)}

by powers of t is the lower central series, provided that n ≥ 3 [13]. Thus, the unipotent k-completion of K(k[t]) is the group K(k[[T]]). Since ρ : SL_{n}(k[t]) → SL_{n}(k) is surjective, and since K(k[[T]]) has trivial center, we see that the relative completion is SL_{n}(k[[T]]).

4.3. Let k be any field and consider the group SL_{n}(k[t]/t^{l}). Reduction modulo t gives a representation ρ : SL_{n}(k[t]/t^{l}) → SL_{n}(k). The kernel K is filtered by powers of t; this filtration is finite in length and is the lower central series. The graded quotients are all isomorphic to 𝔰l_{n}(k). It follows that the unipotent k-completion of K is

\underleftarrow{lim} K/K^{l} ≅ K

(this holds for all n). Thus, the relative completion is simply the group SL_{n}(k[t]/t^{l}).

4.4. Consider the extension

1 ⟶ Γ(n, p) ⟶ SL_{n}(ℤ) ⟶^{ρ} SL_{n}(𝔽_{p}) ⟶ 1

for n ≥ 3. The 𝔽_{p}-completion of Γ(n, p) is the group Γ̂:

1 ⟶ Γ̂ ⟶ SL_{n}(ℤ_{p}) ⟶ SL_{n}(𝔽_{p}) ⟶ 1.

Denote the completion of ρ : SL_{n}(ℤ) → SL_{n}(𝔽_{p}) by G and let U be its prounipotent radical. The map Φ : Γ̂ → U is surjective since ρ is and its kernel is central in Γ̂. Clearly, though, the center of Γ̂ is trivial and hence Φ is an isomorphism. Thus, G is the semidirect product

G ≅ Γ̂ · SL_{n}(𝔽_{p}).

(Recall that the relative completion is a split extension.) This group is not isomorphic to SL_{n}(ℤ_{p}) since the projection map SL_{n}(ℤ_{p}) → SL_{n}(𝔽_{p}) does not split. Note also that SL_{n}(ℤ_{p}) is not proalgebraic over 𝔽_{p} since the rings ℤ/p^{d} are not 𝔽_{p}-algebras.

Thus, the group G is a bit mysterious in the sense that we do not have an explicit description of it in terms of matrices, nor do we have a formula for the homomorphism SL_{n}(ℤ) → G.

4.5. Let Γ be the absolute Galois group of the finite field 𝔽_{p}. The group Γ is isomorphic to ℤ̂, the profinite completion of ℤ. Let ℓ be a prime different from p and consider the action of Γ on the ℓth roots of unity in \overline{𝔽}_{p}. This defines a homomorphism ρ : Γ → 𝔽_{ℓ}^{×} as follows. Let ζ_{ℓ} be a primitive ℓth root of unity and let σ be an element of Γ. Write

σ(ζ_{ℓ}) = ζ_{ℓ}^{ρ(σ)}, ρ(σ) ∈ {1, 2, …, ℓ − 1}.

(Note that ρ(σ) ≠ 0 since σ(1) = 1.) This defines the map ρ. Since ρ maps the Frobenius automorphism to a generator of 𝔽_{ℓ}^{×}, we see that ρ is surjective. The kernel of ρ consists of those automorphisms that fix 𝔽_{p}(ζ_{ℓ}); that is, kerρ = Gal(\overline{𝔽}_{p}/𝔽_{p}(ζ_{ℓ})). This group is also isomorphic to ℤ̂ and the inclusion kerρ → Γ is simply multiplication by ℓ − 1. Note that ℤ̂ ≅ ∏_{q} ℤ_{q}. Thus, we have an extension

0 ⟶ ℤ̂ ⟶^{ℓ−1} ℤ̂ ⟶^{ρ} 𝔽_{ℓ}^{×} ⟶ 1.

The 𝔽_{ℓ}-completion of ℤ̂ is clearly ℤ_{ℓ} and since ρ is surjective we have a surjection Φ : ℤ_{ℓ} → U, where U is the prounipotent radical of the completion of Γ relative to ρ. The kernel of Φ is central in ℤ_{ℓ}, but that is not a useful piece of information in this case. We claim that Φ is injective so that the relative completion is the split extension

0 ⟶ ℤ_{ℓ} ⟶ G ⟶ 𝔽_{ℓ}^{×} ⟶ 1,

where 𝔽_{ℓ}^{×} acts via multiplication on ℤ_{ℓ}.

Recall that in the general situation we denote by L the image of ρ : Γ → S. Assume that H^{1}(L, A) vanishes for all rational representations A of S and that H^{2}(L, A) vanishes for all nontrivial A. Consider the extension

0 ⟶ K ⟶ G ⟶ L ⟶ 1 (1)

where K is the kernel of Φ : T → U. Since H_{1}(L, k) = 0, there is a universal central extension with kernel a k-vector space

0 ⟶ H_{2}(L, k) ⟶ L̃ ⟶ L ⟶ 1.

Its cocycle is the identity map in

Hom(H_{2}(L, k), H_{2}(L, k)) ≅ H^{2}(L, H_{2}(L, k)).

The extension (1) is classified by a linear map ψ : H_{2}(L, k) → K. Under the above stated conditions on H^{1}(L, A) and H^{2}(L, A), the map ψ is surjective [6], Prop. 4.13.

In the example under consideration, we have L = S = 𝔽_{ℓ}^{×}. If A is a rational S-module, then H^{1}(S, A) = 0 = H^{2}(S, A) since |S| = ℓ − 1 is invertible in A (A is an 𝔽_{ℓ}-vector space). Since H_{2}(𝔽_{ℓ}^{×}, 𝔽_{ℓ}) = 0, we see that K = 0.

5. Cohomology

Suppose that π is a projective limit of groups, π = \underleftarrow{lim} π_{α}, and let k be a field. We define the continuous cohomology of π to be

H_{cts}^{i}(π, k) = \underrightarrow{lim} H^{i}(π_{α}, k).

For example if π is the Galois group of a field extension L/F, then H_{cts}^{i}(π, k) is simply the usual Galois cohomology. Note that there is a natural map H_{cts}^{i}(π, k) → H^{i}(π, k). It is obvious that H_{cts}^{0}(π, k) = H^{0}(π, k) and it is easy to see that H_{cts}^{1}(π, k) = H^{1}(π, k).

Lemma 5.1. Let G be a group and let U be its unipotent k-completion. Then the map

H_{cts}^{2}(U, k) ⟶ H^{2}(G, k)

is injective.

Proof. Let α ∈ H_{cts}^{2}(U, k). This class corresponds to a central extension

0 ⟶ k ⟶ U_{α} ⟶ U ⟶ 1

in the category of proalgebraic k-groups. Observe that U_{α} is prounipotent. Suppose that α restricts to 0 in H^{2}(G, k). Then we have a commutative diagram of extensions

\xymatrix{0 \ar[r] & k \ar[r] & U_{α} \ar[r] & U \ar[r] & 1 \\ 0 \ar[r] & k \ar[u]_{=} \ar[r] & G_{α} \ar[u] \ar[r] & G \ar[u] \ar[r] & 1.}

Since α maps to 0 in H^{2}(G, k), the bottom extension splits. We then have the composite map G → G_{α} → U_{α}. By the universal mapping property of U, we get a unique map U → U_{α} making the diagram

\xymatrix{G \ar[r] \ar[dr] & U \ar[d] \\  & U_{α}}

commute; i.e., the top extension splits. Thus, α = 0.

Remark. The above argument does not work to show that H^{2}(U, k) injects into H^{2}(G, k). An element α corresponds to an extension of U by k, but this may not be an extension of proalgebraic groups. A class in H_{cts}^{2}(U, k) actually corresponds to a compatible sequence of extensions of algebraic groups; this allows us to use the universal mapping property of U.

Corollary 5.2. The map H_{cts}^{2}(U, k) → H^{2}(U, k) is injective.

Corollary 5.3. Let ρ : Γ → S be a split surjective representation and let G be the completion relative to ρ. Assume that H_{1}(T, k) is finite dimensional (T = kerρ) and that Φ : T → U is an isomorphism. Then the restriction map

H_{cts}^{2}(G, k) ⟶ H^{2}(Γ, k)

is injective.

Proof. There is a Hochschild–Serre spectral sequence for continuous cohomology [5]

E_{2}^{i,j}(G) = H^{i}(S, H_{cts}^{j}(U, k)) ⟹ H_{cts}^{i+j}(G, k).

Since Φ : T → U is an isomorphism and since the extension

1 ⟶ T ⟶ Γ ⟶ S ⟶ 1

splits, we have

E_{2}^{i,j}(Γ) = E_{2}^{i,j}(G), i = 0, 1.

Moreover, the differential d^{2} : E_{2}^{i,1} → E_{2}^{i+2,0} vanishes for both spectral sequences since the extensions are split. By Lemma 5.1, there is an inclusion of S-modules H_{cts}^{2}(U, k) → H^{2}(T, k) and hence E_{2}^{0,2}(G) injects into E_{2}^{0,2}(Γ). These facts together imply the following:

1. E_{∞}^{2,0}(G) = E_{∞}^{2,0}(Γ);

2. E_{∞}^{1,1}(G) = E_{∞}^{1,1}(Γ);

3. E_{∞}^{0,2}(G) ↪ E_{∞}^{0,2}(Γ). It follows that H_{cts}^{2}(G, k) injects into H^{2}(Γ, k).

5.1. Let X be a smooth affine curve over k and let O_{X,x} be the local ring at x ∈ X. In the previous section, we showed that the completion of SL_{n}(O_{X,x}) → SL_{n}(k) is the extension

1 ⟶ K(Ô_{X,x}) ⟶ SL_{n}(Ô_{X,x}) ⟶ SL_{n}(k) ⟶ 1,

provided that char(k) ≠ 2 or k ≠ 𝔽_{3}. By Corollary 5.3, we see that the map

H_{cts}^{2}(SL_{n}(Ô_{X,x}), k) ⟶ H^{2}(SL_{n}(O_{X,x}), k)

is injective for all n.

5.2. Let k be a finite field or a number field and consider the completion of SL_{n}(k[t]) → SL_{n}(k) for n ≥ 3. This is the split extension

1 ⟶ K(k[[T]]) ⟶ SL_{n}(k[[T]]) ⟶ SL_{n}(k) ⟶ 1.

Theorem 5.4. Let k be a finite field or a number field. Then for all n ≥ 3, H_{cts}^{2}(SL_{n}(k[[T]]), k) = 0.

Proof. We have an injection

H_{cts}^{2}(SL_{n}(k[[T]]), k) ⟶ H^{2}(SL_{n}(k[t]), k).

If k is a number field then by [11], the group H^{2}(SL_{n}(k[t]), k) coincides with H^{2}(SL_{n}(k), k). We then have the chain of equalities

H^{2}(SL_{n}(k[t]), k) = H^{2}(SL_{n}(k), k)
= H_{2}(SL_{n}(k), k)
= H_{2}(SL_{n}(k), ℤ) ⊗ k
= K_{2}(k) ⊗ k.

Since K_{2}(k) is a torsion group when k is a number field, we see that H^{2}(SL_{n}(k[t]), k) = 0.

If k is a finite field, then we use the following chain of equalities

H^{2}(SL_{n}(k[t]), k) = H_{2}(SL_{n}(k[t]), k)
= H_{2}(SL_{n}(k[t]), ℤ) ⊗ k
= K_{2}(k[t]) ⊗ k
= K_{2}(k) ⊗ k,

where the last equality is the fundamental theorem of algebraic Ktheory. Since K_{2}(k) = 0 for finite fields, we are done.

Remark. It is easy to see that if E is a field with char(E) ≠ char(k), then H_{cts}^{•}(SL_{n}(k[[T]]), E) = H^{•}(SL_{n}(k), E).

In [5], Hain calls a prounipotent group P pseudonilpotent if

H_{cts}^{•}(P, k) ⟶^{≅} H^{•}(P, k)

(his definition is actually more general). Let k be a number field and consider the group K(k[[T]]). We know that H_{cts}^{2}(K(k[[T]]), k) injects into H^{2}(K(k[[T]]), k). Suppose that this map is an isomorphism. Then we would have an isomorphism

H^{2}(SL_{n}(k[[T]]), k) ⟶^{≅} H^{2}(SL_{n}(k[t]), k) ⟶^{≅} H^{2}(SL_{n}(k), k).

Dualizing and taking the direct limit over n, we would have an isomorphism

K_{2}(k[[T]]) ⊗ k ⟶^{≅} K_{2}(k) ⊗ k.

Gabber’s rigidity theorem [4] asserts that K_{2}(k[[T]], ℤ/n) ≅ K_{2}(k, ℤ/n). We would therefore have an isomorphism K_{2}(k[[T]]) ≅ K_{2}(k). This is not the case, however, since the kernel of the map K_{2}(k[[T]]) → K_{2}(k) is the relative group K_{2}(k[[T]], (T)) and this group is nontrivial [9] (indeed, it is uniquely divisible). It follows that the group K(k[[T]]) is not pseudonilpotent.

5.3. Consider the commutative diagram of extensions

\xymatrix{1 \ar[r] & Γ(n, p) \ar[d] \ar[r] & SL_{n}(ℤ) \ar[d] \ar[r] & SL_{n}(𝔽_{p}) \ar[d] \ar[r] & 1 \\ 1 \ar[r] & Γ̂ \ar[r] & SL_{n}(ℤ_{p}) \ar[r] & SL_{n}(𝔽_{p}) \ar[r] & 1.}

We showed (3.5) that Γ̂ is the 𝔽_{p}-completion of Γ(n, p) if n ≥ 3. While Corollary 5.3 does not apply in this case, we can still use the argument in the proof to deduce that the map

H_{cts}^{2}(SL_{n}(ℤ_{p}), 𝔽_{p}) ⟶ H^{2}(SL_{n}(ℤ), 𝔽_{p})

is injective for n ≥ 3. Note that H^{2}(SL_{n}(ℤ), 𝔽_{p}) ≅ H_{2}(SL_{n}(ℤ), 𝔽_{p}). The group H_{2}(SL_{n}(ℤ), ℤ) is 2-torsion for all n ≥ 3; it is equal to ℤ/2 ⊕ ℤ/2 for n = 3, 4 [8], and to ℤ/2 for n ≥ 5 [14]. It follows that if p ≥ 3, then H_{2}(SL_{n}(ℤ), 𝔽_{p}) = 0.

Theorem 5.5. Let n ≥ 3. If p ≥ 3, then H_{cts}^{2}(SL_{n}(ℤ_{p}), 𝔽_{p}) = 0. When p = 2 we have

dim_{𝔽_{2}} H_{cts}^{2}(SL_{n}(𝔽_{2}), 𝔽_{2}) ≤ \begin{cases}2 & n = 3, 4 \\ 1 & n ≥ 5. \end{cases}

References

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[13] K. Knudson, ‘Congruence subgroups and twisted cohomology of SL_{n}(F [t]) II: finite fields and number fields’, preprint, 2000, http://www.math.wayne.edu/∼knudson/pubs/twist2.dvi.

[14] J. Milnor, Introduction to algebraic K-theory (Princeton University Press, Princeton, 1971).

[15] D. Quillen, ‘Rational homotopy theory’, Ann. of Math. (2) 90 (1969) 205– 295.

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Department of Mathematics, 1150 F/AB, Wayne State University, Detroit, Michigan 48202

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