Non-residually finite extensions of arithmetic groups

The aim of the article is to show that there are many finite extensions of arithmetic groups which are not residually finite. Suppose $G$ is a simple algebraic group over the rational numbers satisfying both strong approximation, and the congruence subgroup problem. We show that every arithmetic subgroup of $G$ has finite extensions which are not residually finite. More precisely, we investigate the group \[ \bar H^2(\mathbb{Z}/n) = direct limit ( H^2(\Gamma,\mathbb{Z}/n) ), \] where $\Gamma$ runs through the arithmetic subgroups of $G$. Elements of $\bar H^2(\mathbb{Z}/n)$ correspond to (equivalence classes of) central extensions of arithmetic groups by $\mathbb{Z}/n$; non-zero elements correspond to extensions which are not residually finite. We prove that $\bar H^2(\mathbb{Z}/n)$ contains infinitely many elements of order $n$, some of which are invariant for the action of the arithmetic completion $\widehat{G(\mathbb{Q})}$ of $G(\mathbb{Q})$. We also investigate which of these (equivalence classes of) extensions lift to characteristic zero, by determining the invariant elements in the group \[ \bar H^2(\mathbb{Z}_l) = projective limit \bar H^2(\mathbb{Z}/l^t). \] We show that $\bar H^2(\mathbb{Z}_l)^{\widehat{G(\mathbb{Q})}}$ is isomorphic to $\mathbb{Z}_l^c$ for some positive integer $c$. When $G(\mathbb{R})$ has no simple components of complex type, we prove that $c=b+m$, where $b$ is the number of simple components of $G(\mathbb{R})$ and $m$ is the dimension of the centre of a maximal compact subgroup of $G(\mathbb{R})$. In all other cases, we prove upper and lower bounds on $c$; our lower bound (which we believe is the correct number) is $b+m$.


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.
Arithmetic groups are residually finite. Indeed, if Γ is an arithmetic group and 1 = γ ∈ Γ, then there is even a congruence subgroup which does not contain γ. On the other hand, Deligne wrote down a central extensionΓ of Sp 2n (Z) (n ≥ 2) by Z, such thatΓ is not residually finite. More precisely, the groupΓ fits into an exact sequence: and any subgroup of finite index inΓ contains 2Z.
In this note, we show that a weaker version of Deligne's result holds for a large class of arithmetic groups.
1.1. We briefly recall Deligne's construction. The Lie group Sp 2n (R), is not simply connected. In fact, its fundamental group is isomorphic to Z. We shall writeSp 2n (R) for the universal cover of Sp 2n (R), so we have an exact sequence: 1 → Z →Sp 2n (R) → Sp 2n (R) → 1.
One definesΓ to be the preimage of Sp 2n (Z) inSp 2n (R). Note thatSp 2n (R) is a Lie group, but is not the group of real points of an algebraic group; in fact Sp 2n is simply connected as an algebraic group. ThusΓ is not an arithmetic group.
1.2. There are some cases for which Deligne's argument generalizes easily. Suppose G is an algebraic group over Q, which is simple and simply connected. As we have seen above, the group G(R) may fail to be simply connected with the archimedean topology; this happens whenever a maximal compact subgroup of G(R) has infinite centre. We shall assume that fundamental group π 1 (G(R)) has more than 2 elements. We can define just as before an extension 1 → π 1 (G(R)) →Γ → Γ → 1, where Γ is an arithmetic subgroup of G(Q). There is also a canonical double coverG(R) met of G(R), called the metaplectic cover: To show that this generalization is not vacuous, we remark that π 1 (G(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. Let G be a simple algebraic group over Q, which is algebraically simply connected, and has positive real rank. Assume also that G and has finite congruence kernel. Let Γ be an arithmetic subgroup of G(Q). Then there is a finite abelian group A and an extension of groups 1 → A →Γ → Γ → 1, such thatΓ is not residually finite. Gromov-hyperbolic group is residually finite (see for example [10], [1], [11], [26], [15]). This question turns out to be related to the following conjecture of Serre [21]. Conjecture 1. Let G/Q be a simple, simply connected algebraic group of real rank 1. Then the congruence kernel of G is infinite.

It is an important open question in geometric group theory whether every
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 Γ be an arithmetic subgroup of a Lie group with real rank 1. It is known that Γ is Gromov-hyperbolic (see chapter 7 of [9]). Since hyperbolicity is invariant under quasi-isometry, every finite extension of Γ is also hyperbolic, and hence by assumption residually finite. If the congruence kernel were finite, then the groupsΓ 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/Q is a simple, simply connected group of real rank 1 then for every positive integer n there is a surjective homomorphism from the congruence kernel of G to Z/n. Acknowledgement. I'd like to thank Lars Louder for many useful discussions.

Statement of Results
Throughout this section, we fix a simple algebraic group G/Q, such that (1) G is (algebraically) simply connected; (2) G has positive real rank (i.e. G(R) is not compact, and arithmetic subgroups of G are infinite); (3) The congruence kernel of G/Q is finite (and hence conjecturally the real rank of G is at least 2). We do not assume that G is absolutely simple.
We'll show that Theorem 1 is a consequence of the following result.
Theorem 2. Let G/Q be as described above and let Γ be an arithmetic subgroup of G(Q). For every positive integer n there is a subgroup Υ of finite index in Γ and a central extension such thatΥ is not residually finite. More precisely, every subgroup of finite index inΥ contains the subgroup Z/n.
Proof of Theorem 1. We'll now show that Theorem 1 is a consequence of Theorem 2. Let Γ be an arithmetic group with finite congruence kernel. By Theorem 2, there is a subgroup Υ of finite index in Γ and a central extensionΥ of Γ by Z/n, such that every subgroup of finite index inΥ contains Z/n. Let σ ∈ H 2 (Υ, Z/n) be the cohomology class corresponding to this extension. By Shapiro's lemma, there is an isomorphism We'll write Σ for the image of σ in H 2 (Γ, A). Corresponding to the cohomology class Σ, there is a (non-central) extensionΓ of Γ by A. These two group extensions are related by the following commutative diagram: Suppose for the sake of argument thatΓ is residually finite. Hence the subgroup pr −1 Υ is residually finite. There is therefore a subgroup Φ ⊂Υ of finite index, such that Φ ∩ A is trivial. The image of Φ inΥ is then a subgroup of finite index inΥ, whose intersection with Z/n is trivial. This is a contradiction.
2.1. Some refinements of Theorem 2. Let G/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 Γ of G.
Suppose that we have a central extension of Γ by Z/n as follows: We shall write σ ∈ H 2 (Γ, Z/n) for the cohomology class of this extension. Suppose for a moment thatΓ is residually finite. We can then find a subgroup Υ ⊂Γ of finite index, such that the intersection of Υ with Z/n is trivial. Hence Υ projects bijectively onto a subgroup of Γ, which we shall also call Υ. The preimage of Υ inΓ is the direct sum Z/n ⊕ Υ. As a result of this, we know that the restriction of σ to Υ is trivial. This means that in order to construct a non-residually finite extension of Γ, we need a non-zero element of the direct limit where Υ runs over subgroups of finite index in Γ. The argument above shows that Theorem 2 is implied by the following result.
Theorem 3. For every positive integer n, there are infinitely many elements of order n inH 2 (Z/n).
We shall actually prove a stronger result, which needs a little more notation to state. We shall write G(Q) for the arithmetic completion of the group G(Q), i.e.
where Υ runs through the subgroups of finite index in Γ. There is a natural projection G(Q) → G(A f ), and the congruence kernel is, by definition, the kernel of this map. This means that we have an extension of topological groups The group G(Q) acts smoothly onH 2 (Z/n). Let S be a finite set of prime numbers. By an S-arithmetic level, we shall mean an open subgroup L of G(Q) of the form Theorem 4 will be proved in section 4. The proof requires a technical result on the cohomology of finite groups of Lie type, which is proved in section 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 σ of order n inH 2 (Z/n) with the following property. There is a prime number p depending on σ, such that for all primes q = p the element σ is fixed by pr −1 (G(Q q )).

2.2.
Virtual lifting to characteristic zero. Let l be a prime number. Any central extension of Γ by Z/l t+1 gives rise to a central extension by Z/l t . We'll say that the extension of Γ by Z/l r virtually lifts to characteristic zero if for every t > r there is a arithmetic subgroup Υ t of Γ and a central extension of Υ t by Z/l t , such that the extensions fit into a commutative diagram of the following form.
Here the map Z/l t → Z/l r is the usual reduction map, and the map Υ t → Γ is the inclusion.
Equivalently, an element ofH 2 (Z/l r ) virtually lifts to characteristic zero if it is in the image of the following group.H There is a continuous action of G(Q) on the cohomology groupH 2 (Z l ). Our next result will show that there are indeed families of non-residually finite central extensions, which virtually lift to characteristic zero. Before stating the result we'll need a little notation. The group G × R is semi-simple over R, and decomposes as a product of finitely many simple groups G i /R. We'll say that a simple group G i /R is of complex type if G i is the restriction of scalars of a group defined over C, or equivalently if G i × C is a product of two simple groups; otherwise we say that G i is of real type. We'll write b R for the number of simple factors if G × R of real type and b C for the number of simple factors if G × R or complex type. We'll also write m for the dimension of the centre of a maximal compact subgroup K ∞ ⊂ G(R).
Theorem 6. The groupH 2 (Z l ) G(Q) is isomorphic to Z l c for some positive integer c. More precisely, c is in the range where b R , b C and m as the integers defined above. In particularH 2 (Z l ) is non-zero.
For comparison, we note that the construction of Deligne implies the bound c ≥ m; this is because π 1 (G(R)) has a finite index subgroup isomorphic to Z m .
As an easy consequence, of the theorem, we obtain the following: Let G/Q be simple and simply connected with finite congruence kernel.
There is a subgroup ofH 2 (Z/l t ) G(Q) isomorphic to (Z/l t ) c , all of whose elements virtually lift to characteristic zero, where c is the positive integer in Theorem 6. Table 1.
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. We stress that Theorem 6 impliesH 2 (Z l ) G(Q) is non-zero even in cases where H 2 (Γ, C) = 0 for all arithmetic subgroups Γ of G(Q). This happens when G has large real rank and the symmetric space associated to G has no complex structure, for example when G = SL 5 /Q. The extensions constructed by the method of Deligne exist only in the case m > 0; our result shows thatH 2 (Z l ) G(Q) is non-zero even in cases where m = 0.
Remark. The author believes that rank Z l H 2 (Z l ) G(Q) = b R + b C + m. Proving this would amount to showing that the restriction map H 3 cts (G(Q l ), Q l ) → H 3 (G(Q), Q l ) is surjective. As long as G(R) has no simple factors of complex type, Theorem 6 tells us precisely the rank ofH 2 (Z l ) G(Q) . Some examples are given in Table 1. In this table, Spin(r, s) denotes the Spin group of an arbitrary quadratic forms over Q of signature (r, s). The congruence subgroup property for such groups was established by Kneser [12].
The case SL 2 /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 havē H 2 (Z/n) = 0 andH 2 (Z l ) = 0.

Background material
3.1. Continuous cohomology. We shall make use of the continuous cohomology groups H • 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 × A → A.
In all cases under consideration here, the group G will be metrizable, locally compact, totally disconnected, separable and σ-compact. The coefficient group A will always be Polonais. Under these restriction, the continuous cohomology groups defined in [7] (based on continuous cocycles) are the same as those defined in [16], [17], [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 ind G H (A) for the induced module, consisting of all continuous functions f : G → A satisfying f (hg) = h · f (g) for all g ∈ G and h ∈ H. This agrees with the notation of [7] but not [16], [17], [18]. The following version of Shapiro's lemma holds for these induced representations.
Theorem 7 (Shapiro's Lemma). Let H be a closed subgroup of G, where G satisfies the conditions above. For any continuous H-module A, there is a canonical isomorphism of topological groups: Proof. This follows Propositions 3 and 4 of [7] in view of the remark following Proposition 4.
We shall also make frequent use of the following. 3.2. The derived functor of projective limit. By a projective system, we shall mean a sequence of abelian groups A t , indexed by t ∈ N, and connected by group homomorphisms as follows: We shall write lim ← t A t for the projective limit of the system. The functor lim ← t is leftexact from the category of projective systems of abelian groups to the category of abelian groups. As such, it has derived functors lim ← t • A t . It is known (Corollary 3.5.4 of [24]) that the higher derived functors lim ← t n A t for n ≥ 2 are all zero.
The projective system (A t ) is said to satisfy the Mittag-Leffler property if for every t ∈ N, there is a j ∈ 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 Theorem 9 (Theorem 3.5.8 of [24]). Let · · · → C r 2 → C r 1 be a projective system of cochain complexes of abelian groups, each indexed by r ≥ 0. Assume that this projective system has the Mittag-Leffler property, and let C r = lim ← t C r t be the projective limit of the As a simple example, we show how to express the cohomology of G(Q) in terms of the cohomology of its S-arithmetic subgroups. As before, we let G/Q be a simple, simply connected algebraic group, and K f = p K p a compact open subgroup of G(A f ). We shall write Γ for the arithmetic group G(Q) ∩ K f . More generally, if S is a finite set of prime numbers, then we use the notation Γ S for the corresponding S-arithmetic group, i.e.
where H • (g, k, C) are the relative Lie algebra cohomology groups studied in [5].
Proof. For each r ≥ 0 we shall write C r (Γ S , F) for the usual (inhomogeneous) cochain complex, consisting of all functions f : Γ S r → F. Since G(Q) is the union of the groups Γ S , it follows that The maps in this projective system are restrictions of functions, and they are obviously surjective. Therefore the projective system satisfies the Mittag-Leffler condition. As a consequence, we have short exact sequences By the theory of the Borel-Serre compactification (see [4]), the cohomology groups H r (Γ S , F) are finite dimensional vector spaces. Therefore the projective system H r (Γ S , F) S satisfies the Mittag-Leffler condition, so we have lim ← In the case F = C, the theorem of [2] implies that H r (Γ S , C) = H r (g, k, C) whenever S contains more than r primes. Hence the projective limit (over S) is in this case H r (g, k, C).
3.3. The congruence kernel. Let G/Q be a simple, simply connected group with real rank at least 1. By Kneser's strong approximation theorem (see [13] , where A f is the ring of finite adèles of Q. It follows that there is an isomorphism of topological groups: where Γ runs over the congruence subgroups of G(Q). Recall that an arithmetic subgroup of G is any subgroup of G(Q), which is commensurable with a congruence subgroup. The arithmetic completion G(Q) is defined to be the completion of G(Q) with respect to the arithmetic subgroups of G, i.e.
There is a canonical surjective homomorphism G(Q) → G(A f ). The congruence kernel Cong(G) is defined to be the kernel of this map, so we have a short exact sequence: The congruence kernel is trivial if and only if every arithmetic subgroup of G is a congruence subgroup. If G(R) is simply connected as an analytic group, then the congruence kernel is never trivial, but may still be finite. It has been conjectured by Serre [21], that the congruence kernel is finite if and only if each simple factor of G over Q has real rank at least 2. In the case that Cong(G) is finite, it is known that Cong(G) is contained in the centre of G(Q), and is a cyclic group.

Proof of Theorem 4
In this section, we assume that the group G/Q is a simple, simply connected algebraic group with positive real rank. We shall also assume that the congruence kernel Cong(G) is finite. Hence, conjecturally that the real rank of G is at least 2.
If L is compact and open then Γ(L) is an arithmetic group and L is its profinite completion. If L is an S-arithmetic level, then Γ(L) is an S-arithmetic group.
We shall write C(L, Z/n) for the group of continuous functions f : L → Z/n. We regard C(L, Z/n) as a Γ(L) × L-module, in which (for the sake of argument) Γ(L) acts by left-translation and L acts by right-translation. We regard Γ(L) as a discrete topological group, and L as a topological group with the subspace topology from G(Q). We do not assume that elements of C(L, Z/n) are uniformly continuous, and so the action of L is not smooth unless L is compact. The action is continuous, where C(L, Z/n) is equipped with the compact-open topology.
We shall also use the following notation, which was introduced earlier: where Υ ranges of the arithmetic subgroups. It's therefore sufficient to consider the case that the level L is compact and open. Under this assumption, we have (as Γ(L)-modules): where Υ ranges over the arithmetic subgroups of Γ(L). Since direct limits commute with cohomology, this implies If we choose L to be compact, then C(L, Z/n) is discrete, and therefore H • (Γ(L), C(L, Z/n)) is discrete.

4.2.
Low degree terms. We shall now describe some of the low degree terms of the spectral sequence of Proposition 4. Lemma 2. With the notation introduced above, Proof. ForH 0 , note that for any arithmetic group Υ, H 0 (Υ, Z/p r ) = Z/n. Furthermore the restriction maps from one of these groups to another, are all the identity map. ForH 1 , we must show that for every element σ ∈ H 1 (Υ, Z/n), there is an arithmetic subgroup Υ ′ ⊂ Υ, such that the restriction of σ to Υ ′ is zero. Any such σ is a homomorphism Υ → Z/n, so we may simply set Υ ′ = ker σ.
By Lemma 2, we know that E r,0 2 = H r cts (L, Z/n) and E r,1 2 = 0. Therefore the bottom left corner of the E 2 sheet of the spectral sequence looks like this:  Theorem. Let S be a finite set of prime numbers and let L be an S-arithmetic level. Then the groupH 2 (Z/n) L contains infinitely many elements of order n.
Proof. Let L be an S-arithmetic level. In this case the group Γ(L) is an S-arithmetic group. By the theory of the Borel-Serre compactification, there is a resolution of Z as a Γ(L)-module consisting of finitely generated Z[Γ(L)]-modules. This implies that the cohomology groups H r (Γ(L), Z/n) are all finite. In view of this, the sequence in Equation 2 has the form finite → H 2 (Z/n) L → H 3 cts (L, Z/n) → finite. To prove the theorem, it is therefore sufficient to show that H 3 cts (L, Z/n) contains infinitely many elements of order n.
It will be useful to have the following notation. A prime number p will be called a tame prime if it satisfies all of the following conditions: (1) p is not in the finite set S; (2) p is not a factor of |Cong(G)|; (3) p is not a factor of n; (4) G is unramified over Q p .
(5) The group K p is a maximal hyperspecial compact open subgroup of G(Q p ) (see [23]). This implies that if we let K 0 p be the maximal pro-p normal subgroup of K p , then the quotient G(F p ) = K p /K 0 p is a product of some of the simply connected finite Lie groups described in [22]. (6) H r (G(F p ), Q/Z) = 0 for r = 1, 2. We recall from [22] that this condition is satisfied for all but finitely many of the groups G(F p ). We note that all but finitely many primes are tame. For each tame prime p, we shall write K * p for a lift of K p to G(Q); note that such a lift exists and is unique by conditions (2) and (6). The group L contains the following subgroup Evidently, pr(K tame ) is a direct summand of pr(L); since K tame ∩ Cong(G) is trivial, it follows that K tame is a direct summand of L. Hence by the Künneth formula, H 3 cts (K tame , Z/n) is a direct summand of H 3 cts (L, Z/n). It is therefore sufficient to prove that H 3 cts (K tame , Z/n) contains infinitely many elements of order n.
Since the coefficient ring Z/n is finite, we have (by Proposition 8, section 2.

of [20]) a decomposition
By the Künneth formula, the group on the right contains a subgroup of the form p tame H 3 cts (K p , Z/n). (3) and (5) for tame primes p, we may identify H • cts (K p , Z/n) with H • (G(F p ), Z/n). To prove the theorem, it is therefore sufficient so show that there are infinitely many tame primes p, such that H 3 (G(F p ), Z/n) contains an element of order n. This follows from Theorem 10, which will be proved in the next section.

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 Q, then we may write G in the form G × Z Q, for some group scheme G over Z. The groups G(F p ) depend on the G, not just on G. Nevertheless if we alter the group scheme G then only finitely many of the groups G(F p ) will change. Because of this, the following statement makes sense, where we are writing G(F p ) in place of G(F p ) for some fixed choice of G.
Theorem 10. Let G/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(F p ), 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 SL r ).
In the proof we shall use the Cartan-Eilenberg theory of invariant cohomology classes, which we recall now. Let T be a subgroup of a finite group G, and let A be a G-module. We shall write Rest G T and CoRest T G for the restriction and corestriction maps between H • (G, A) and H • (T, A). A cohomology class σ ∈ H r (T, A) is called invariant if for every g ∈ G, Rest T T ∩T g (σ) = Rest T g T ∩T g (σ g ). We'll use the following result.  In order to apply the corollary, it will be useful to note the following. Lemma 3. Let l be a prime number, and let x be an integer such that x ≡ 1 mod 2l. Then for every integer d we have Proof. We recall that the l-adic logarithm function log l converges on the multiplicative group 1 + 2lZ l . If log l converges at an element x, then we have | log(x)| l = |x − 1| l . Our congruence condition implies that log l (x) and log l (x d ) both converge, so we have 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. We shall introduce some notation. We fix a semi-simple model G of G over Z, and let k be a number field such that G splits over O k . Let T be a maximal torus in G, defined and split over O k . Let P be the lattice of algebraic characters T → GL 1 /O k . The roots of G with respect to T are elements of the lattice P . Consider the element where Φ is the set of roots. If we identify elements of P with a group of characters of the Lie algebra t of T , then we may similarly identify elements of Sym 2 (P ) with quadratic forms on t. The element Q corresponds to the restriction of the Killing form to t. Therefore Q is non-zero.
Let e be the largest positive integer, such that Q is a multiple of e in the lattice Sym 2 (P ). Also let d 1 , · · · , 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 · · · d r · e · 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 (G(F p ), Z/n) contains an element of order n. By abusing notation slightly we shall write G(F p ) for the group G(F p ). We may identify G(F p ) with G(O k /p) for some prime ideal p above p). We shall also write T (F p ) for the subgroup T (O k /p).
Identifying H 3 (G(F p ), Z/n) with the n-torsion in H 4 (G(F p ), Z), we see that it's sufficient to prove there is an element of order n in H 4 (G(F p ), Z).
We shall use the following formula for the order of the the group G(F p ) (see Theorem 25, in Chapter 9 of [22]) In this formula, N is the number of positive roots; r is the rank and d 1 , . . . , d r are the degrees of the fundamental invariants on the Weyl group. Note also that since T is a split torus of rank r, we have |T (F p )| = (p − 1) r .
Since p splits in k and k contains an primitive 2l-th root of unity (because d 1 = 2), we have p ≡ 1 mod 2l. Hence by Lemma 3, We therefore have [G(F p ) : T (F p )] l = |d 1 · · · d r | l .
By Corollary 4, it is sufficient to show that H 4 (T (F p ), Z) has an invariant element of order d 1 · · · d r · n. It will actually be more convenient to find an invariant element of H 4 (T (F p ), Z) ⊗(F × p ) ⊗2 ; this is because such an element is canonical, whereas the invariant element of H 4 (T (F p ), Z) would depend on a choice of primitive root modulo p.
We shall construct our invariant element of H 4 from elements of H 2 . Since T (F p ) is a finite group, we have canonical isomorphisms: where as before, P is the lattice of algebraic characters of T .
Recall that the cohomology ring of the cyclic group F × p is the symmetric algebra on H 2 (F × p , Z). The group T (F p ) is a product of copies of F × p , and so by the Künneth formula, H • (T (F p ), Z) contains as a subring the algebra H • (F × p , Z) ⊗r , which is isomorphic to the symmetric algebra on H 2 (F × p , Z) r . More canonically, this subring is the symmetric algebra on H 2 (T (F p ), Z). In particular, H 4 (T (F p ), Z) contains Sym 2 (H 2 (T (F p ), Z)) as a subgroup; this is the subgroup generated by cup products of elements of H 2 (T (F p ), Z) 1 . From this, we see that H 4 (T (F p ), Z) ⊗ (F × p ) ⊗2 contains as a subgroup the group We claim that the following element of H 4 (T (F p ), Z) is an invariant cohomology class: where Φ is the root system of G with respect to T . The element q is evidently in the subgroup Sym 2 (P )/(p − 1). Equivalently, we can regard q as the quadratic function q : Here we are writing the group (F × p ) ⊗2 additively. Suppose g is an element of G(F p ) and suppose that both t and g −1 tg are in T (F p ). To show that q is an invariant class, we must show that q(t) = q(g −1 tg). Evidently we have The numbers α(t) are the non-zero eigenvalues in the action of t on the Lie algebra g ⊗ F p . These eigenvalues are the same as those of g −1 tg, and so the numbers α(t) are the same (possibly in a different order) as the numbers α(g −1 tg). From this it follows that q(t g ) = q(t), so q is an invariant class in H 4 (T (F p ), Z) ⊗ (F × p ) ⊗2 . It remains to determine the order of q in H 4 (T (F p ), Z) ⊗ (F × p ) ⊗2 , or equivalently the order of q in the subgroup Sym 2 (P )/(p − 1). By definition, q is the the reduction modulo p − 1 of the element Q ∈ 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 ≡ 1 mod e, the order of q in Sym 2 (P )/(p − 1) is precisely p−1 e . To summarize, we have shown that H 4 (T (F p ), Z) has an invariant element of order p−1 e . Therefore H 4 (G(F p ), Z) has an element of order p−1 d 1 ···dr·e . Since p ≡ 1 mod (d 1 · · · d r · e · n) it follows that H 4 (G(F p ), Z) has an element of order n.
6. Proof of Theorem 6 6.1. The groupsH 2 (Z l ). As before, we let L be an open subgroup of the arithmetic completion G(Q), and we shall now fix a prime number l. We introduce a new module Again, we regard the group C(L, Z l ) as a Γ(L) × L-module. We have We define, analogously to the notationH • (Z/l t ), The main focus of this section is to determine the groupH 2 (Z l ) G(Q) . We begin by establishing some easy properties of the modulesH • (Z l ). (5) The groupH 2 (Z l ) is torsion-free and contains no non-zero divisible elements. In the notation of the previous section, we have By Lemma 2,H 0 (Z/l t ) = Z/l t andH 1 (Z/l t ) = 0. Both of these projective systems consist of finite groups, so satisfy the Mittag-Leffler condition. Therefore lim ← t 1 vanishes on both of them. As a result of this we have for r = 1, 2: In particularH 1 (Z l ) = 0. (5) Consider the short exact sequence of modules: This gives the exact sequence in cohomology We already saw in Lemma 2 thatH 1 (Z/l t ) = 0. This shows thatH 2 (Z l ) is torsionfree. Suppose σ is a divisible element inH 2 (Z l ). Then the image of σ inH 2 (Z/l t ) is a divisible element for each t. SinceH 2 (Z/l t ) is a Z/l t -module, the image of σ inH 2 (Z/l t ) must be zero. By (4) it follows that σ = 0. Proposition 7. For any S-arithmetic level L ⊂ G(Q), there is an exact sequence as follows: It is tempting to suggest that the exact sequence of the proposition follows from a spectral sequence of the form H r cts (L,H s (Z l )) =⇒ H r+s (Γ(L), 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 groupsH r (Z l ) will probably not be Hausdorff for r ≥ 3, and so there is no off-the-shelf spectral sequence for us to use. Admittedly we could truncate atH 2 (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. For any t ≥ 0 the spectral sequence in Equation 1 gives rise to an exact sequence: . We shall write A t for the image of the map H 2 (Γ(L), Z/l t ) →H 2 (Z/l t ) L and B t for the image the mapH 2 (Z/l t ) L → H 3 cts (L, Z/l t ). We therefore have three exact sequences: (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 Equation 7 does not satisfy the Mittag-Leffler condition, so we do not expect H 4 cts (K tame , Z l ) to be finitely generated as a Z l -module).
Lemma 5. Let L be an S-arithmetic level in G(Q). For r = 0, 1, 2, 3 we have Here g is the Lie algebra of G over Q.
Proof. Recall that we have a decomposition of the group L/Cong(G) in the form L S × K W × K tame . This gives rise to the following spectral sequence H r cts (L S × K T , H s (K tame , Z l )) =⇒ H r+s cts (L/Cong(G), Z l ). From Lemma 4, we see that the bottom left corner of the E 2 -sheet is as follows: Therefore the spectral sequence H r cts (L/Cong(G), H s (Cong(G), Q l )) collapses to the bottom row, so we have H r cts (L, Q l ) = H r cts (L/Cong(G), Q l ). In particular for r ≤ 3 we have We'll calculate these cohomology groups above using the Künneth formula. Suppose first that p is a prime in W , which is not equal to l. The group K p contains a normal pro-p subgroup of finite index. From this it follows that Next, suppose that p is a prime in S which is not equal to l. We recall from [7] that there is a spectral sequence which calculates the cohomology of G(Q p ) in terms of the cohomology of its compact open subgroups. Let K 0 p be a maximal pro-p subgroup of G(Q p ). The subgroup K 0 p is compact and open in G(Q p ). There are finitely many maximal compact subgroups of G(Q p ), which contain K 0 p ; we call these subgroups K 1 , . . . , K n . In the spectral sequence, the E 1 -sheet is given by Here we are using the notation The map E r−1,s is an alternating sum of restriction maps; in other words, its (i 0 , . . . , i r )-component is equal to As we are assuming here that p = l, the cohomology groups H s cts (K i 0 ,...,is , Q l ) are zero for s > 0. Therefore the spectral sequence consists of a single row in E 1 ; this row is the simplicial cochain complex of a simplex with n vertices. As this simplex is contractable, we have From the Künneth formula we have for r ≤ 3: H r cts (L, Q l ) = H r cts (L l , Q l ), where L l is either K l or G(Q l ), depending on whether the prime l is in S or not. In either case we have H • cts (L l , Q l ) = H • (g ⊗ Q l , Q l ); this is proved in [14] for K l and in [7] for G(Q l ). As a result of this we have H r cts (L, Q l ) = H r (g ⊗ Q l , Q l ) for r ≤ 3. Lemma 6. We have H 0 (g ⊗ Q l , Q l ) = Q l , H 1 (g ⊗ Q l , Q l ) = 0, H 2 (g ⊗ Q l , Q l ) = 0 and H 3 (g ⊗ Q l , Q l ) = Q b l , where b is the number of simple components of G × Q C. (Note that in the notation of the introduction we have b = b R + 2b C ).
Proof. The dimension of the Lie algebra cohomology does not depend on the base field, so we may instead calculate the cohomology of g ⊗ C. There is a decomposition: where each g i is a complex simple Lie algebra. By Whitehead's first and second lemmas we have H r (g i , C) = 0 for r = 1, 2, and H 0 (g i , C) = C. It is well known (see for example section 1.6 of [5]) that H • (g i , C) is isomorphic to the singular cohomology of a compact connected Lie group with Lie algebra g i . Hence by Proposition 8 each group H 3 (g i , C) is 1-dimensional. The lemma follows from the Künneth formula.
Corollary. There is a subgroup ofH 2 (Z/l t ) G(Q) isomorphic to (Z/l t ) c , all of whose elements virtually lift to characteristic zero, where c = rank Z l H 2 (Z l ) G(Q) .
Proof. We have a short exact sequence This gives a long exact sequence containing the following terms We've shown thatH 1 (Z/l t ) = 0, so we have We'll write A for the subgroup of elements inH 2 (Z/l t ) which virtually lift to characteristic zero. By definition, A is the image ofH 2 (Z l ) in H 2 (Z/l t ). We therefore have a short exact sequence Taking G(Q)-invariants, we have an exact sequence The result follows becauseH 2 (Z l ) G(Q) is isomorphic to Z l c .

Appendix: A result on compact symmetric spaces
In this appendix, we shall calculate the low dimensional cohomology of the compact simple symmetric spaces. Such spaces have the form G/K, where G is a compact, connected, simple Lie group and K is a closed, connected subgroup. In the following result, we shall use the shorthand H • (X) for the singular cohomology on the topological space X with coefficients in R.
Proposition 8. Let G be a compact, connected simple Lie group and K a closed, connected subgroup. There are isomorphisms: Here z(k) * denotes the dual space of the centre of the Lie algebra k of K.
Remark. One might expect to be able to look this result up in tables; however I didn't manage to find such tables, so I am including a proof. The result is a straightforward consequence of results in Borel's thesis [3].
Proof. We shall first review some results from [3] on the cohomology of compact Lie groups and their classifying spaces. Suppose that G is compact connected Lie group. We shall write BG for the classifying space of G. This is a space with a fibre bundle such that the cohomology of EG is the cohomology of a point. The real singular cohomology of G is (as a ring) an exterior algebra whose generators are cohomology classes x 1 , . . . , x n in odd dimensions s 1 , . . . , s n . For each generator x i ∈ H s i (G), there is an element y i ∈ H s i+1 (BG) called the transgression of x i . Furthermore the cohomology ring H • (BG) is equal to the polynomial ring R[y 1 , . . . , y n ]. In particular BG only has non-zero real cohomology in even dimensions.
As an example of this, let T be an n-dimensional compact torus, and let t be its Lie algebra. Recall that the cohomology of T is exactly the exterior algebra of H 1 (T ). Furthermore, we may identify H 1 (T ) with the dual space of t. As a result of this, we know that H • (BT ) is the algebra R[t] of polynomial functions on t. Furthermore, H 2n (BT ) is the space of homogeneous polynomials on t of degree n.
Suppose now that G is a compact, connected Lie group and T is a maximal torus in G. We shall write W for the Weyl group of G with respect to T . We have a fibre bundle G/T → BT ↓ BG (BT = EG/T ).
Corresponding to this there is a spectral sequence H r (BG, H s (G/T )) ⇒ H r+s (BT ).
As G is connected, it follows easily that BG is simply connected. This implies that H s (G/T ) is a trivial bundle on BG, and so the spectral sequence takes the form H r (BG) ⊗ H s (G/T ) ⇒ H r+s (BT ).
In particular we have an edge map H • (BG) → H • (BT ). We'll use the following result, which describes this edge map. As a result of this proposition, we know that for semi-simple G we have H 2 (BG) = 0. This is because there are no W -invariant linear forms on t. For simple G we have H 4 (BG) = R.
Recall that H 4 (BG) is the space of W -invariant quadratic forms on t. The restriction of the Killing form is one such form, and any other is a constant multiple of this. As a consequence, we see that H 1 (G) = H 2 (G) = 0 and H 3 (G) = R. This proves Proposition 8 in the case that K is trivial.
Assume now that K is a non-trivial closed, connected subgroup of G. We shall use the spectral sequence of the following fibration: That is: (9) H r (BG) ⊗ H s (G/K) ⇒ H r+s (BK).
Let S be a maximal torus in K and T ⊃ S be a maximal torus in G, and let W G and W K be the corresponding Weyl groups. From the spectral sequence in Equation 9 we have an edge map H • (BG) → H • (BK). By Proposition 9, we may interpret this as a map where s is the Lie algebra of S. This map has been determined by Borel: Furthermore there is an exact sequence: We've seen that H 3 (BK) = 0 (because 3 is odd) and the edge map H 4 (BG) → H 4 (BK) is injective, so it follows that H 3 (G/K) = 0.
We finally translate the result above into a more usable form. In the following corollary, G is a semi-simple, simply connected algebraic group over R and K ∞ is a maximal compact subgroup of G(R). We shall write b R and b C for the number of simple components of G of real and of complex type. We write g and k for the Lie algebras of G(R) and K ∞ respectively, and we write z(k) for the centre of k.