- Research
- Open Access
Non-residually finite extensions of arithmetic groups
- Richard M. Hill^{1}Email author
- Received: 30 July 2018
- Accepted: 20 October 2018
- Published: 2 November 2018
Abstract
Keywords
- Cohomology of arithmetic groups
- Congruence subgroup property
- Residually finite group
Mathematics Subject Classification
- 11F77
- 11F06
1 Introduction
An abstract group G is said to be residually finite if, for every non-trivial element g, there is a subgroup H of finite index in the group, which does not contain g. The content of this statement is not changed if we insist that H is a normal subgroup of G. This is equivalent to the statement that the canonical map from the group to its profinite completion is injective.
In this note, we show that a weaker version of Deligne’s result holds for a large class of arithmetic groups.
To show that this generalization is not vacuous, we remark that \(\pi _{1}(G(\mathbb {R}))\) is infinite whenever there is a Shimura variety associated to G, and the congruence subgroup property is known to hold for simple, simply connected groups of rational rank at least 2.
In this paper, we shall deal also with groups G, for which Deligne’s construction cannot be used. The most easily stated consequence of our results is the following.
Theorem 1
1.1 Residual finiteness of hyperbolic groups
It is an important open question in geometric group theory whether every Gromov-hyperbolic group is residually finite (see for example [1, 9, 11, 15, 26]). This question turns out to be related to the following conjecture of Serre [20].
Conjecture 1
Let \(G/\mathbb {Q}\) be a simple, simply connected algebraic group of real rank 1. Then the congruence kernel of G is infinite.
As a consequence of Theorem 1, we obtain the following.
Corollary 1
If every Gromov–hyperbolic group is residually finite then Conjecture 1 is true.
Proof
Let \(\varGamma \) be an arithmetic subgroup of a Lie group with real rank 1. It is known that \(\varGamma \) is Gromov–hyperbolic (see chapter 7 of [10]). Since hyperbolicity is invariant under quasi-isometry, every finite extension of \(\varGamma \) is also hyperbolic, and hence by assumption residually finite. If the congruence kernel were finite, then the groups \({\tilde{\varGamma }}\) from Theorem 1 would provide a counterexample to this.
In fact one can show as a consequence of the results proved here the following slightly more precise result.
Corollary 2
Assume that every Gromov–hyperbolic group is residually finite. If \(G/\mathbb {Q}\) is a simple, simply connected group of real rank 1 then for every positive integer n, the congruence kernel of G has a subquotient isomorphic to \(\mathbb {Z}/n\).
2 Statement of results
- 1.
G is (algebraically) simply connected;
- 2.
G has positive real rank (i.e. \(G(\mathbb {R})\) is not compact, and arithmetic subgroups of G are infinite);
- 3.
The congruence kernel of \(G/\mathbb {Q}\) is finite (and hence conjecturally the real rank of G is at least 2).
We’ll show that Theorem 1 is a consequence of the following result.
Theorem 2
Proof of Theorem 1
2.1 Some refinements of Theorem 2
Let \(G/\mathbb {Q}\) be simple, simply connected, and have real rank at least 1. Furthermore assume that the congruence kernel of G is finite (and hence, conjecturally at least, the real rank of G is at least 2). Fix an arithmetic subgroup \(\varGamma \) of G.
Theorem 3
For every positive integer n, there are infinitely many elements of order n in \({\bar{H}}^{2}(\mathbb {Z}/n)\).
Theorem 4
Let L be an S-arithmetic level in \({\widehat{G(\mathbb {Q})}}\) for some finite set of primes S. For every positive integer n, there are infinitely many elements of order n in \({\bar{H}}^{2}(\mathbb {Z}/n)^L\).
Theorem 4 will be proved in Sect. 4. The proof requires a technical result on the cohomology of finite groups of Lie type, which is proved in Sect. 5. By modifying the argument slightly, one can also prove the following result.
Theorem 5
Let n be a positive integer. Then there are infinitely many elements \(\sigma \) of order n in \({\bar{H}}^2(\mathbb {Z}/n)\) with the following property. There is a prime number p depending on \(\sigma \), such that for all primes \(q\ne p\) the element \(\sigma \) is fixed by \(\mathrm {pr}^{-1}\left( G(\mathbb {Q}_q) \right) \).
2.2 Virtual lifting to characteristic zero
Theorem 6
For comparison, we note that the construction of Deligne implies the bound \(c \ge m\); this is because \(\pi _1(G(\mathbb {R}))\) has a finite index subgroup isomorphic to \(\mathbb {Z}^m\).
As an easy consequence of the theorem, we obtain the following:
Corollary 3
Let \(G/\mathbb {Q}\) be simple and simply connected with finite congruence kernel. There is a subgroup of \({\bar{H}}^2(\mathbb {Z}/l^t)^{\widehat{G(\mathbb {Q})}}\) isomorphic to \((\mathbb {Z}/l^t)^{c}\), all of whose elements virtually lift to characteristic zero, where c is the positive integer in Theorem 6.
Theorem 6 and its corollary will be proved in section 6. The proof requires a result on the cohomology of compact symmetric spaces, which is proved in the appendix.
Remark 1
We stress that Theorem 6 implies \({\bar{H}}^2({\mathbb {Z}_l})^{{\widehat{G(\mathbb {Q})}}}\) is non-zero even in cases where \(H^2(\varGamma ,\mathbb {C})=0\) for all arithmetic subgroups \(\varGamma \) of \(G(\mathbb {Q})\). This happens when G has large real rank and the symmetric space associated to G has no complex structure, for example when \(G=\mathrm {SL}_5/\mathbb {Q}\). The extensions constructed by the method of Deligne exist only in the case \(m>0\); our result shows that \({\bar{H}}^2({\mathbb {Z}_l})^{\widehat{G(\mathbb {Q})}}\) is non-zero even in cases where \(m=0\).
Remark 2
The author suspects that \(\mathrm {rank}_{\mathbb {Z}_l}\left( \bar{H}^{2}({\mathbb {Z}_l})^{{\widehat{G(\mathbb {Q})}}}\right) =b_\mathbb {R}+b_\mathbb {C}+m\). Proving this would amount to showing that the restriction map \(H^3_\mathrm {cts}(G(\mathbb {Q}_l),\mathbb {Q}_l) \rightarrow H^3(G(\mathbb {Q}),\mathbb {Q}_l)\) is surjective. The evidence for this is very slight, but we note that \(\dim H^3_\mathrm {cts}(G(\mathbb {Q}_l),\mathbb {Q}_l)\) is at least twice as big as \(\dim H^3(G(\mathbb {Q}),\mathbb {Q}_l)\).
As an example, consider the case \(G=\mathrm {Res}^k_\mathbb {Q}(SL_{\ge 3}/k)\), where k is an imaginary quadratic field. In this case \(m=0\), \(b_\mathbb {R}=0\) and \(b_\mathbb {C}=1\), so our result implies that the rank c is either 1 or 2. In this case \(H^3_\mathrm {cts}(G(\mathbb {Q}_l),\mathbb {Q}_l)\) is 2-dimensional and \(H^3(G(\mathbb {Q}),\mathbb {Q}_l)\) is 1-dimensional (see Sect. 6.3), so the restriction map is either surjective or zero. If the restriction map is non-zero, then the rank is 1. One might expect to prove that the rank is 1 by evaluating an appropriate l-adic Borel regulator; however the author has not done this in any case.
Values of the rank of \(\bar{H}^2(\mathbb {Z}_l)^{\widehat{G(\mathbb {Q})}}\)
G | m | \(b_\mathbb {R}\) | \(c = \mathrm {rank}_{{\mathbb {Z}_l}}\left( {\bar{H}}^{2}({\mathbb {Z}_l})^{{\widehat{G(\mathbb {Q})}}}\right) \) | |
---|---|---|---|---|
\(\mathrm {SL}_{n}/\mathbb {Q}\) | (\(n \ge 3\)) | 0 | 1 | 1 |
\(\mathrm {Sp}_{2n}/\mathbb {Q}\) | (\(n \ge 2\)) | 1 | 1 | 2 |
\(\mathrm {Spin}(r,s)\) | (\(r \ge s \ge 3\)) | 0 | 1 | 1 |
\(\mathrm {Spin}(r,2)\) | (\(r \ge 3\)) | 1 | 1 | 2 |
\(\mathrm {Spin}(2,2)\) | 2 | 2 | 4 | |
\(\mathrm {Res}^{k}_{\mathbb {Q}}(\mathrm {SL}_{n}/k)\) | (\(n \ge 3\), k totally real) | 0 | \([k:\mathbb {Q}]\) | \([k:\mathbb {Q}]\) |
\(\mathrm {Res}^{k}_{\mathbb {Q}}(\mathrm {SL}_{2}/k)\) | (k totally real, \(k \ne \mathbb {Q}\)) | \([k:\mathbb {Q}]\) | \([k:\mathbb {Q}]\) | \(2[k:\mathbb {Q}]\) |
\(\mathrm {Res}^{k}_{\mathbb {Q}}(\mathrm {Sp}_{2n}/k)\) | (k totally real) | \([k:\mathbb {Q}]\) | \([k:\mathbb {Q}]\) | \(2[k:\mathbb {Q}]\) |
The case \(\mathrm {SL}_{2}/\mathbb {Q}\) and its forms of rank 0 are not included in the table. This is because these groups have infinite congruence kernel, and indeed for these groups we have \({\bar{H}}^{2}(\mathbb {Z}/n)=0\) and \({\bar{H}}^2({\mathbb {Z}_l})=0\).
3 Background material
3.1 Continuous cohomology
We shall make use of the continuous cohomology groups \(H^\bullet _\mathrm {cts}(G,A)\), where G is a topological group and A is an abelian topological group, which is a G-module via a continuous action \(G \times A \rightarrow A\).
In all cases under consideration here, the group G will be metrizable, locally compact, totally disconnected, separable and \(\sigma \)-compact. The coefficient group A will always be Polonais (a topological group is Polonais if its topology admits a separable complete metric; see page 3 of [18]). Under these restriction, the continuous cohomology groups defined in [7] (based on continuous cocycles) are the same as those defined in [16–18] based on Borel measurable cocycles. This is proved in Theorem 1 of [25].
If A is a continuous H-module for some closed subgroup H of G, then we shall write \(\mathrm {ind}_H^G(A)\) for the induced module, consisting of all continuous functions \(f:G \rightarrow A\) satisfying \(f(hg)=h\cdot f(g)\) for all \(g\in G\) and \(h\in H\). This agrees with the notation of [7] but not [16–18]. The following version of Shapiro’s lemma holds for these induced representations.
Theorem 7
Proof
This follows Propositions 3 and 4 of [7] in view of the remark following Proposition 4. \(\square \)
We shall also make frequent use of the following.
Theorem 8
Proof
This follows from Theorem 9 of [18] in all cases under consideration. \(\square \)
3.2 The derived functor of projective limit
The projective system \((A_{t})\) is said to satisfy the Mittag–Leffler property if for every \(t\in \mathbb {N}\), there is a \(j\in \mathbb {N}\) with the property that for all \(k>j\) the image of \(A_{k}\) in \(A_{t}\) is equal to the image of \(A_{j}\) in \(A_{t}\). For example, if the Abelian groups \(A_{t}\) are all finite then the projective system has the Mittag–Leffler property. Similarly, if the groups \(A_t\) are all finite dimensional vector spaces connected by linear maps, then the projective system satisfies the Mittag–Leffler condition.
Proposition 1
(Proposition 3.5.7 of [24]) If the projective system \((A_{t})\) satisfies the Mittag–Leffler condition then \( \left( \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array}\right) ^{1} A_{t} = 0 \).
Theorem 9
Proposition 2
For any field \(\mathbb {F}\), we have \( H^\bullet (G(\mathbb {Q}), \mathbb {F}) = \begin{array}{c} \lim \\ {\mathop {\scriptstyle S}\limits ^{\textstyle \leftarrow }} \end{array} H^\bullet (\varGamma ^S,\mathbb {F}). \) In the case \(\mathbb {F}=\mathbb {C}\) we have \( H^\bullet (G(\mathbb {Q}),\mathbb {C}) = H^\bullet (\mathfrak {g},\mathfrak {k},\mathbb {C}), \) where \(H^\bullet (\mathfrak {g},\mathfrak {k},\mathbb {C})\) are the relative Lie algebra cohomology groups studied in [5].
Proof
In the case \(\mathbb {F}=\mathbb {C}\), the theorem of [2] implies that \(H^r(\varGamma ^S,\mathbb {C})=H^r(\mathfrak {g},\mathfrak {k},\mathbb {C})\) whenever S contains more than r primes. Hence the projective limit (over S) is in this case \(H^r(\mathfrak {g},\mathfrak {k},\mathbb {C})\). \(\square \)
3.3 The congruence kernel
4 Proof of Theorem 4
In this section, we assume that the group \(G/\mathbb {Q}\) is a simple, simply connected algebraic group with positive real rank. We shall also assume that the congruence kernel \(\mathrm {Cong}(G)\) is finite. Hence, conjecturally that the real rank of G is at least 2.
4.1 The groups \(\mathcal {C}(L,\mathbb {Z}/n)\)
Let L be an open subgroup of the arithmetic completion \({\widehat{G(\mathbb {Q})}}\). We shall write \({\varGamma (L)}\) for the group \(G(\mathbb {Q})\cap L\). Since \(G(\mathbb {Q})\) is dense in \({\widehat{G(\mathbb {Q})}}\), it follows that \({\varGamma (L)}\) is dense in L. If L is compact and open then \({\varGamma (L)}\) is an arithmetic group and L is its profinite completion. If L is an S-arithmetic level, then \({\varGamma (L)}\) is an S-arithmetic group.
We shall write \(\mathcal {C}(L,\mathbb {Z}/n)\) for the group of continuous functions \(f:L \rightarrow \mathbb {Z}/n\). We regard \(\mathcal {C}(L,\mathbb {Z}/n)\) as a \({\varGamma (L)}\times L\)-module, in which (for the sake of argument) \({\varGamma (L)}\) acts by left-translation and L acts by right-translation. We regard \({\varGamma (L)}\) as a discrete topological group, and L as a topological group with the subspace topology from \({\widehat{G(\mathbb {Q})}}\). We do not assume that elements of \(\mathcal {C}(L,\mathbb {Z}/n)\) are uniformly continuous, and so the action of L is not smooth unless L is compact. The action is continuous, where \(\mathcal {C}(L,\mathbb {Z}/n)\) is equipped with the compact–open topology.
Proposition 3
Proof
Lemma 1
We have \( H^{0}_{\mathrm {cts}}(L, \mathcal {C}(L,\mathbb {Z}/n)) = \mathbb {Z}/n\) and \(H^s_\mathrm {cts}(L,\mathcal {C}(L,\mathbb {Z}/n))=0\) for \(s>0\). In particular the groups \(H^{s}_{\mathrm {cts}}(L, \mathcal {C}(L,\mathbb {Z}/n))\) are Hausdorff.
Proof
As a continuous L-module, we have \( \mathcal {C}(L,\mathbb {Z}/n)= \mathrm {ind}_{1}^{L} (\mathbb {Z}/n). \) The result follows from this using Shapiro’s Lemma. \(\square \)
Proposition 4
Let L be any open subgroup of \({\widehat{G(\mathbb {Q})}}\). Then there is a first quadrant spectral sequence with \(E_2^{r,s} = H^{r}_{\mathrm {cts}}(L, \bar{H}^{s}(\mathbb {Z}/n))\) which converges to \(H^{r+s}({\varGamma (L)}, \mathbb {Z}/n)\).
Proof
4.2 Low degree terms
We shall now describe some of the low degree terms of the spectral sequence of Proposition 4.
Lemma 2
Proof
Theorem
Let S be a finite set of prime numbers and let L be an S-arithmetic level. Then the group \({\bar{H}}^2(\mathbb {Z}/n)^L\) contains infinitely many elements of order n.
Proof
- 1.
p is not in the finite set S;
- 2.
p is not a factor of \(|\mathrm {Cong}(G)|\);
- 3.
p is not a factor of n;
- 4.
G is unramified over \(\mathbb {Q}_p\).
- 5.
The group \(K_p\) is a maximal hyperspecial compact open subgroup of \(G(\mathbb {Q}_p)\) (see [23]). This implies that if we let \(K_p^0\) be the maximal pro-p normal subgroup of \(K_p\), then the quotient \(G(\mathbb {F}_p)=K_p/K_p^0\) is a product of some of the simply connected finite Lie groups described in [22].
- 6.
\(H^r(G(\mathbb {F}_p),\mathbb {Q}/\mathbb {Z})=0\) for \(r=1,2\). We recall from [22] that this condition is satisfied for all but finitely many of the groups \(G(\mathbb {F}_p)\).
5 A lemma on the cohomology of finite Lie groups
In this section we shall prove Theorem 10, which completes the proof of Theorem 4.
Before stating the theorem, we note that if G is an algebraic group over \(\mathbb {Q}\), then we may write G in the form \(\mathscr {G}\times _\mathbb {Z}\mathbb {Q}\), for some group scheme \(\mathscr {G}\) over \(\mathbb {Z}\). The group \(\mathscr {G}(\mathbb {F}_p)\) depends on the choice of \(\mathscr {G}\), not just on G. Nevertheless if we alter the group scheme \(\mathscr {G}\) then only finitely many of the groups \(\mathscr {G}(\mathbb {F}_p)\) will change. Because of this, the following statement makes sense, where we are writing \(G(\mathbb {F}_p)\) in place of \(\mathscr {G}(\mathbb {F}_p)\) for some fixed choice of \(\mathscr {G}\).
Theorem 10
Let \(G/\mathbb {Q}\) be a simple, simply connected algebraic group. For every positive integer n there are infinitely many prime numbers p, such that \(H^3(G(\mathbb {F}_p),\mathbb {Z}/n)\) contains an element of order n.
I assume this sort of result is known to experts, and many special cases are consequences of results in algebraic K-theory (for example the results of [19] imply the case \(\mathrm {SL}_r\)).
Proposition 5
As a corollary to this, we note the following.
Corollary 4
Let T be a subgroup of a finite group G. Let d be a positive integer and l a prime number, such that \(\big |[G:T]\big |_l=\big |d\big |_l\). If \(H^r(T,\mathbb {Z})\) contains an invariant class of order \(dl^t\) then \(H^r(G,\mathbb {Z})\) contains an element of order \(l^t\).
Proof
Let \(\tau = \mathrm {CoRest}^T_G(\sigma )\), where \(\sigma \) is the invariant class on T of order \(dl^t\). By Proposition 5, the restriction of \(\tau \) to T has order \(\frac{dl^t}{\gcd ( dl^t, [G:T])}\). The condition on d implies that the order of \(\mathrm {Rest}^G_T(\tau )\) is a multiple of \(l^t\). Hence the order of \(\tau \) is a multiple of \(l^t\), so some multiple of \(\tau \) has order \(l^t\). \(\square \)
In order to apply the corollary, it will be useful to note the following.
Lemma 3
Proof
Proof of Theorem 10
By the Chinese remainder theorem, it is sufficient to prove the theorem in the case \(n=l^t\), where l is a prime number.
Let e be the largest positive integer, such that Q is a multiple of e in the lattice \(\mathrm {Sym}^2 (P)\). Also let \(d_1,\ldots , d_r\) be the degrees of the basic polynomial invariants of the Weyl group of G / k (where r is the rank of G / k). The smallest of these degrees is \(d_1=2\), and the others depend on the root system (see [22]). By extending the number field k if necessary, we may assume that k contains a primitive root of unity of order \(d_1\cdots d_r\cdot e\cdot n\). By the Chebotarev density theorem, there are infinitely many prime numbers which split in k; we’ll show that each of these prime numbers has the desired property.
From now on we fix a prime number p which splits in k, and we are attempting to show that \(H^3(\mathscr {G}~(\mathbb {F}_p),\mathbb {Z}/n)\) contains an element of order n. By abusing notation slightly we shall write \(G(\mathbb {F}_p)\) for the group \(\mathscr {G}~(\mathbb {F}_p)\). We may identify \(G(\mathbb {F}_p)\) with \(\mathscr {G}~(\mathscr {O}_k/\mathfrak {p})\) for some prime ideal \(\mathfrak {p}\) above p). We shall also write \(T(\mathbb {F}_p)\) for the subgroup \(\mathscr {T}~(\mathscr {O}_k/\mathfrak {p})\).
Identifying \(H^3(G(\mathbb {F}_p),\mathbb {Z}/n)\) with the n-torsion in \(H^4(G(\mathbb {F}_p),\mathbb {Z})\), we see that it’s sufficient to prove there is an element of order n in \(H^4(G(\mathbb {F}_p),\mathbb {Z})\).
It remains to determine the order of q in \(H^4(T(\mathbb {F}_p),\mathbb {Z})\otimes (\mathbb {F}_p^\times )^{\otimes 2}\), or equivalently the order of q in the subgroup \(\mathrm {Sym}^2(P)/(p-1)\). By definition, q is the the reduction modulo \(p-1\) of the element \(Q\in \mathrm {Sym}^2(P)\). We defined e to be the largest integer such that Q is a multiple of e. Since we are assuming that \(p\equiv 1 \bmod e\), the order of q in \(\mathrm {Sym}^2(P)/(p-1)\) is precisely \(\frac{p-1}{e}\).
To summarize, we have shown that \(H^4(T(\mathbb {F}_p),\mathbb {Z})\) has an invariant element of order \(\frac{p-1}{e}\). Therefore \(H^4(G(\mathbb {F}_p),\mathbb {Z})\) has an element of order \(\frac{p-1}{d_1 \cdots d_r \cdot e}\). Since \(p \equiv 1 \bmod (d_1 \cdots d_r \cdot e \cdot n)\) it follows that \(H^4(G(\mathbb {F}_p),\mathbb {Z})\) has an element of order n. \(\square \)
6 Proof of Theorem 6
6.1 The groups \({\bar{H}}^2({\mathbb {Z}_l})\)
Proposition 6
- 1.
The groups \({\bar{H}}^\bullet ({\mathbb {Z}_l})\) do not depend on the open subgroup L in their definition (Equation 3).
- 2.
\({\bar{H}}^0({\mathbb {Z}_l})={\mathbb {Z}_l}\),
- 3.
\({\bar{H}}^1({\mathbb {Z}_l})=0\),
- 4.
\( {\bar{H}}^2({\mathbb {Z}_l}) = \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} {\bar{H}}^2(\mathbb {Z}/l^t). \)
- 5.
The group \({\bar{H}}^2({\mathbb {Z}_l})\) is torsion-free and contains no non-zero divisible elements.
- 6.
For any open subgroup L of \({\widehat{G(\mathbb {Q})}}\) we have \( \bar{H}^2({\mathbb {Z}_l})^L = \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} \left( {\bar{H}}^2(\mathbb {Z}/l^t)^L \right) \).
Proof
- 1.
Suppose M is an open subgroup of L. Then we have an isomorphism of \({\varGamma (L)}\)-modules \(\mathcal {C}(L,{\mathbb {Z}_l})= \mathrm {ind}_M^L \mathcal {C}(M,{\mathbb {Z}_l})\). The result follows from this by Shapiro’s Lemma (Theorem 7).
- 2.
Since \({\varGamma (L)}\) is dense in L, it follows that the \({\varGamma (L)}\)-invariant continuous functions on L are constant. This shows that \({\bar{H}}^0({\mathbb {Z}_l})={\mathbb {Z}_l}\).
- (3,4)For each \(r>0\) we have by Theorem 9 a short exact sequenceIn the notation of the previous section, we have$$\begin{aligned}&0 \rightarrow \left( \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} \right) ^1 H^{r-1}(\varGamma (L),\mathcal {C}(L,\mathbb {Z}/l^t)) \rightarrow {\bar{H}}^{r}({\mathbb {Z}_l})\\&\qquad \quad \quad \rightarrow \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} H^{r}(\varGamma (L),\mathcal {C}(L,\mathbb {Z}/l^t)) \rightarrow 0. \end{aligned}$$By Lemma 2, \({\bar{H}}^0(\mathbb {Z}/l^t)=\mathbb {Z}/l^t\) and \(\bar{H}^1(\mathbb {Z}/l^t)=0\). Both of these projective systems consist of finite groups, so satisfy the Mittag–Leffler condition. Therefore \(\left( \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} \right) ^1\) vanishes on both of them. As a result of this we have for \(r=1,2\):$$\begin{aligned} 0 \rightarrow \left( \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} \right) ^1 {\bar{H}}^{r-1}(\mathbb {Z}/l^t) \rightarrow {\bar{H}}^{r}({\mathbb {Z}_l}) \rightarrow \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} {\bar{H}}^{r}(\mathbb {Z}/l^t) \rightarrow 0. \end{aligned}$$In particular \({\bar{H}}^1({\mathbb {Z}_l})=0\).$$\begin{aligned} {\bar{H}}^{r}({\mathbb {Z}_l}) = \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} {\bar{H}}^{r}(\mathbb {Z}/l^t). \end{aligned}$$
- (5)Consider the short exact sequence of modules:This gives the exact sequence in cohomology$$\begin{aligned} 0 \rightarrow \mathcal {C}(L,{\mathbb {Z}_l}){\mathop {\rightarrow }\limits ^{\times l^t}} \mathcal {C}(L,{\mathbb {Z}_l})\rightarrow \mathcal {C}(L,\mathbb {Z}/l^t)\rightarrow 0. \end{aligned}$$We already saw in Lemma 2 that \({\bar{H}}^1(\mathbb {Z}/l^t)=0\). This shows that \({\bar{H}}^2({\mathbb {Z}_l})\) is torsion-free. Suppose \(\sigma \) is a divisible element in \({\bar{H}}^2({\mathbb {Z}_l})\). Then the image of \(\sigma \) in \({\bar{H}}^2(\mathbb {Z}/l^t)\) is a divisible element for each t. Since \(\bar{H}^2(\mathbb {Z}/l^t)\) is a \(\mathbb {Z}/l^t\)-module, the image of \(\sigma \) in \(\bar{H}^2(\mathbb {Z}/l^t)\) must be zero. By (4) it follows that \(\sigma =0\).$$\begin{aligned} \cdots \rightarrow {\bar{H}}^1(\mathbb {Z}/l^t) \rightarrow {\bar{H}}^2({\mathbb {Z}_l}) {\mathop {\rightarrow }\limits ^{\times l^t}} {\bar{H}}^2({\mathbb {Z}_l}) \rightarrow \cdots . \end{aligned}$$
- (6)
This follows because the functor \( \begin{array}{c} \lim \\ {\mathop {\scriptstyle t}\limits ^{\textstyle \leftarrow }} \end{array} \) commutes with the functor \(-^L\) of L-invariant elements. \(\square \)
Proposition 7
Remark 3
It is tempting to suggest that the exact sequence of the proposition follows from a spectral sequence of the form \(H^r_\mathrm {cts}(L,\bar{H}^s({\mathbb {Z}_l})) \implies H^{r+s}({\varGamma (L)},{\mathbb {Z}_l})\), which would be proved in the same way as in the finite coefficient case (Proposition 4). Unfortunately this is not quite so simple. The problem is that the groups \({\bar{H}}^r({\mathbb {Z}_l})\) will probably not be Hausdorff for \(r \ge 3\), and so there is no off-the-shelf spectral sequence for us to use. Admittedly we could truncate at \({\bar{H}}^2({\mathbb {Z}_l})\) to obtain a spectral sequence with three rows, or we could try to work with the more general spectral sequence constructed in [8]. Instead we’ve gone for a more elementary approach, and we prove the exact sequence of the proposition by taking the projective limit of such exact sequences in the finite coefficient cases.
Proof
6.2 The groups \(H^\bullet _\mathrm {cts}(L,{\mathbb {Z}_l})\)
We shall next concentrate on the continuous cohomology groups in the exact sequence of Proposition 7.
- 1.
\(p \not \in S\).
- 2.
\(p \ne l\).
- 3.
G is unramified over \(\mathbb {Q}_p\).
- 4.
\(K_p\) is a maximal hyperspecial compact open subgroup of \(G(\mathbb {Q}_p)\). This implies that if we let \(K_p^0\) be the maximal pro-p normal subgroup of \(K_p\), then the group \(G(\mathbb {F}_p)=K_p/K_p^0\) is a product of some of the simply connected finite groups of Lie type described in detail in [22].
- 5.
\(H^r(G(\mathbb {F}_p),\mathbb {Q}/\mathbb {Z})=0\) for \(r=1,2\). We recall from [22] that this condition is satisfied for all but finitely many of the groups \(G(\mathbb {F}_p)\).
Lemma 4
With the notation introduced above, \(H^r_\mathrm {cts}(K_\mathrm {tame}, {\mathbb {Z}_l})=0\) for \(r=1,2,3\).
Proof
(It might be tempting to imagine that the result above can be extended further in a simple way. However, we note that the projective system in Eq. 7 does not satisfy the Mittag–Leffler condition, so we do not expect \(H^4_\mathrm {cts}(K_\mathrm {tame},{\mathbb {Z}_l})\) to be finitely generated as a \({\mathbb {Z}_l}\)-module).
Lemma 5
Proof
Lemma 6
We have \(H^0(\mathfrak {g}\otimes \mathbb {Q}_l,\mathbb {Q}_l)=\mathbb {Q}_l\), \(H^1(\mathfrak {g}\otimes \mathbb {Q}_l,\mathbb {Q}_l)=0\), \(H^2(\mathfrak {g}\otimes \mathbb {Q}_l,\mathbb {Q}_l)=0\) and \(H^3(\mathfrak {g}\otimes \mathbb {Q}_l,\mathbb {Q}_l)=\mathbb {Q}_l^b\), where b is the number of simple components of \(G \times _\mathbb {Q}\mathbb {C}\). (Note that in the notation of the introduction we have \(b=b_\mathbb {R}+2b_\mathbb {C}\)).
Proof
6.3 The end of the proof
The dimensions of the groups \(H^r(G(\mathbb {Q}),\mathbb {Q}_l)\) are the same as those of \(H^r(G(\mathbb {Q}),\mathbb {C})\), and by Proposition 2 these are the same as the the relative Lie algebra cohomology groups \(H^r(\mathfrak {g},\mathfrak {k},\mathbb {C})\). Here \(\mathfrak {k}\) is the Lie algebra of a maximal compact subgroup \(K_\infty \) of \(G(\mathbb {R})\).
Since \({\bar{H}}^2({\mathbb {Z}_l})\) is torsion-free, it follows that \(\bar{H}^2({\mathbb {Z}_l})^{\widehat{G(\mathbb {Q})}}\) is a torsion-free \({\mathbb {Z}_l}\)-module, which spans \(\bar{H}^2(\mathbb {Q}_l)^{\widehat{G(\mathbb {Q})}}\). On the other hand, \({\bar{H}}^2({\mathbb {Z}_l})\) has no non-zero divisible elements. This implies that \({\bar{H}}^2({\mathbb {Z}_l})^{\widehat{G(\mathbb {Q})}}\cong {\mathbb {Z}_l}^{c}\), where \(c= \dim {\bar{H}}^2(\mathbb {Q}_l)^{\widehat{G(\mathbb {Q})}}\). This finishes the proof of the Theorem 6.
Corollary
There is a subgroup of \({\bar{H}}^2(\mathbb {Z}/l^t)^{\widehat{G(\mathbb {Q})}}\) isomorphic to \((\mathbb {Z}/l^t)^{c}\), all of whose elements virtually lift to characteristic zero, where \(c=\mathrm {rank}_{{\mathbb {Z}_l}} \left( {\bar{H}}^2({\mathbb {Z}_l})^{\widehat{G(\mathbb {Q})}}\right) \).
Proof
Acknowlegements
I’d like to thank Lars Louder for many useful discussions. I’d also like to thank the anonymous referee for suggesting several improvements.
Declarations
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Authors’ Affiliations
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