Bounded cohomology is not a profinite invariant

Abstract We construct pairs of residually finite groups with isomorphic profinite completions such that one has non-vanishing and the other has vanishing real second bounded cohomology. The examples are lattices in different higher-rank simple Lie groups. Using Galois cohomology, we actually show that 
$\operatorname {SO}^0(n,2)$
 for 
$n \ge 6$
 and the exceptional groups 
$E_{6(-14)}$
 and 
$E_{7(-25)}$
 constitute the complete list of higher-rank Lie groups admitting such examples.


Introduction
A group invariant is called profinite if it agrees for any two finitely generated residually finite groups Γ, Λ with isomorphic profinite completions.A standard example is the abelianization H 1 (Γ), a more sophisticated example is largeness [20].It seems, however, that more often than not, group invariants fail to be profinite.Kazhdan's property (T ) [2], higher ℓ 2 -Betti numbers [15], Euler characteristic and ℓ 2 -torsion [14], amenability [17], finiteness properties [21], and most recently Serre's Property FA [6] are all known not to be profinite.This list is by no means exhaustive and bounded cohomology is another item: Lemma 1.Let Γ = Spin(7, 2)(Z) and Λ = Spin (3,6) For an ad hoc definition of the spinor group Spin(q)(O) of a quadratic form q over an integral domain O, we refer to [15,Section 3].The profinite completion Γ of a group Γ is the projective limit of the inverse system of finite quotient groups of Γ.The bounded cohomology H * b (Γ; R) with real coefficients of a discrete group Γ is the cohomology of the cochain complex ℓ ∞ (Γ * +1 , R) Γ , of bounded functions Γ * +1 → R which are constant on the orbits of the diagonal Γ-action on Γ * +1 , with the usual differential.The reader may consult [7] for further details.
Let us quickly prove the lemma.If the Witt index of an integral quadratic form q is at least two, then Spin(q)(Z) has the congruence subgroup property (CSP).This implies that Spin(q)(Z) ∼ = Spin(q)( Z) ∼ = Spin(q)( p Z p ) ∼ = p Spin(q)(Z p ).
The standard forms of signature (7,2) and (3,6) are isometric over Z p for all (finite) primes p because standard quadratic form theory shows Here we call two simple Lie groups isogenous if they have isomorphic Lie algebras.Note that a finite index subgroup of the group Γ from Lemma 1 is a lattice in SO 0 (7,2).The lemma actually provides the easiest possible example because the examples of lattices that we construct in SO 0 (6, 2) come from triality forms of type D 4 , as we will see.
Let us outline the proof of Theorem 2. The key result is the aforementioned theorem due to Monod-Shalom [24,Theorem 1.4] which extends a previous result of Burger-Monod [4,Corollary 1.6]: for a lattice Γ ≤ G in a higher rank Lie group, we have Here SO 0 (n, 2) is the identity component of the determinant one matrices that preserve the standard quadratic form of signature (n, 2).The group SO * (2n) is the quaternionic special orthogonal group as defined for instance in [25,Example A2.4.2,p. 430].Using the symbol "≈" for isogenous groups, we have the accidental isogenies SO 0 (3, 2) ≈ Sp(2; R), SO 0 (4, 2) ≈ SU(2, 2), and SO 0 (6, 2) ≈ SO * (8).Let G be one of the groups in the list and let Γ ≤ G be any lattice.By Margulis arithmeticity, we may assume Γ is an arithmetic subgroup of a simply-connected simple algebraic group G over some totally real number field k such that G ≈ G(k v ) for some real place v of k and such that G is anisotropic at all other infinite places of k.The congruence subgroup property translates the question whether there exists Λ ≤ H as in the theorem to whether there exists a simply-connected simple l-group H, an isomorphism l of topological rings between the finite adele rings of k and l, and a corresponding isomorphism ) such that H is anisotropic at all but one infinite place where it should be isogenous to a higher rank Lie group outside the list.The technical achievement of this paper, beside filling in the details of the arguments thus far, is to solve this problem by Galois cohomological methods.
We conclude the introduction with some comments on related work and open questions.By definition, bounded cohomology comes with a comparison map H * b (Γ; R) −→ H * (Γ; R).The kernel of this homomorphism is denoted by EH * b (Γ; R) and is called exact bounded cohomology.In degree two, we have the well-known interpretation that EH 2 b (Γ; R) detects non-trivial quasimorphisms: maps f : Γ → R for which there exists D > 0 with for all g, h ∈ Γ such that f is not at bounded distance from an honest homomorphism [7, Section 2.3].For lattices in higher rank linear Lie groups it was verified by M. Burger and N. Monod [5,Theorem 21] that the comparison map in degree two is injective.So the question whether the existence of non-trivial quasimorphisms is a profinite property remains open for now.
However, for lattices in rank one groups, the situation is different.On the one hand, a result of Fujiwara shows that such lattices admit many quasimorphisms [8].On the other hand, Serre conjectured that these lattices should not have the congruence subgroup property, so an important ingredient to construct groups with isomorphic profinite completions would be missing.Yet, lattices in the rank one groups Sp(n, 1) and F 4(−20) share many properties with higher rank lattices which might suggest that they in fact do have CSP [22,Section 4].
If F 4(−20) has CSP, then having non-trivial exact second bounded cohomology, or equivalently having non-trivial quasimorphisms, is not a profinite property.Indeed, let F 4 be the unique simply-connected absolutely almost simple Q-split linear algebraic Q-group of type F 4 .Since the Dynkin diagram of type F 4 has no symmetries and the center of F 4 is trivial, we have F 4 ∼ = Ad F 4 ∼ = Aut F 4 .Therefore, the Hasse principle for simply-connected groups gives H 1 (Q, Aut F 4 ) ∼ = H 1 (R, Aut F 4 ), meaning every real form of type F 4 comes with a unique Q-structure.Moreover, any two Q-groups of type F 4 are Q p -split and hence Q pisomorphic for all finite primes p by Kneser's theorem [19].Therefore, if F 4(−20) has CSP, we can find arithmetic lattices Γ ≤ F 4(−20) and Another notion from this circle of ideas is Ulam stability.Here, instead of R, we consider unitary groups U(n) = {A ∈ GL n (C) : A * = A −1 } and define an ε-homomorphism as a map f : Γ → U(n) such that f (gh) − f (g)f (h) ≤ ε holds for all g, h ∈ Γ where • denotes the operator norm in M n (C).We say that Γ is uniformly U(n)-stable if there exists a function δ = δ(ε) with lim ε→0 δ(ε) = 0 such that for all εhomomorphisms f : Γ → U(n), there exists an honest homomorphism It is a straightforward consequence of the Burger-Monod theorem that lattices in higher rank linear Lie groups are uniformly U(1)-stable [10, Theorem 1.0.11].Most recently, L. Glebsky, A. Lubotzky, N. Monod, and B. Rangarajan proved the far reaching generalization that lattices in many higher rank semisimple Lie groups G are Ulam stable [10, Theorem 0.0.5].In fact, they are even uniformly stable with respect to more general metrics on U(n).Interestingly, the technical condition "property-G(Q 1 , Q 2 )" that G needs to satisfy to conclude this stability of lattices fails for SO 0 (n, 2), E 6(−14) and E 7(−25) by [10, Theorem 0.0.6] but it holds true for E 6(2) and E 7 (7) [10, Proof of Proposition 6.3.6].(It seems to be open whether it holds for groups of type SO 0 (p, q) with p, q ≥ 3.) We will see below that E 6(2) and E 7(7) contain lattices which are profinitely isomorphic to lattices in E 6(−14) and E 7(−25) , respectively.The authors actually entertain the idea that property-G(Q 1 , Q 2 ) might be necessary for Ulam stability [10, Section 7].If that was true, our theorem would thus have the corollary that Ulam stability is not a profinite property.
In Section 2, we prove the "if part" of Theorem 2 and in Section 3, we prove the "only if" part.The authors acknowledge financial support from the German Research Foundation via the Research Training Group "Algebro-Geometric Methods in Algebra, Arithmetic, and Topology", DFG 284078965.The second author is additionally grateful for financial support within the Priority Program "Geometry at Infinity", DFG 441848266.We are indebted to Francesco Fournier-Facio, Ryan Spitler, and the referee for helpful discussions and suggestions.
Fix a prime number p 0 .Let E 7 be the unique simply-connected absolutely almost simple Q-split linear algebraic Q-group of type E 7 .The center Z(E 7 ) ∼ = µ 2 is isomorphic to the algebraic group of "second roots of unity" [27, Table on p. 332], so that µ 2 (K) = {±1} for any field extension K/Q.By [34, Section I.5.7], the corresponding equivariant short exact sequence of Gal(Q)-groups and functoriality yield a commuting diagram of Galois cohomology sets which is exact at the middle term of each row.The direct sums in the upper row denote the subsets of the cartesian products consisting of elements with all but finitely many coordinates equal to the unit class.We agree that the infinite prime ∞ with Q ∞ = R is included.Let us collect some information on this diagram Proposition 3. The map f is surjective and ∆ p 0 has trivial kernel.
Proof.The surjectivity of f is implicit in the work of Borel-Harder [3] but can be cited explicitly from Prasad-Rapinchuk [29, Proposition 1].The map ∆ p 0 has trivial kernel by exactness of the lower sequence and because H 1 (Q p 0 , E 7 ) = 0 by a result of M. Kneser [19].
) consists of two elements, corresponding to the real forms E 7( 7) and E 7(−25) .
Proof.This was already observed as part of the investigation in [16] but let us give a direct argument for the convenience of the reader.By [34, Section I.5.7], the map π R sits in the exact sequence By [ .Of course, the unit class 1 ∈ H 1 (R, Ad E 7 ) corresponds to the split form E 7 (7) .
The surjectivity of f according to Proposition 3 lets us find classes α, β ∈ H 1 (Q, Ad E 7 ) such that both α and β split at all finite primes p except possibly at p 0 and such that α corresponds to the real form E 7(−25) at ∞, whereas β corresponds to the real form E 7(7) at ∞.As the Dynkin diagram of type E 7 comes with no symmetries, the set H 1 (Q, Ad E 7 ) classifies all Q-forms of type E 7 .The upshot of Proposition 5 is that the simply-connected Q-forms G 1 and G 2 defined by α and β, respectively, are split and thus isomorphic over Q p for all finite primes p.It then follows from [13, Lemma 2.5 and Lemma 2.6] that we also have an isomorphism G 1 ∼ = G 2 of group schemes over the finite adele ring A f Q .Moreover, neither G 1 nor G 2 is topologically simply-connected at the infinite place.Indeed, we have [26,Table 10,p.321], the simply-connected real Lie group of type E 7 (7) has cyclic center of order four while the simply-connected real Lie group of type E 7(−25) has infinite cyclic center.(The latter is actually a consequence of E 7(−25) giving rise to a hermitian symmetric space.)By [28,Main Theorem], this implies that the metaplectic kernels of G 1 and G 2 are trivial.The surjectivity result for the map f comes with an additional statement on the existence of isotropic preimages [29, Theorem 1 (iii)] which allows us to assume that rank Q G 1 = 3 and rank Q G 2 = 7.Therefore, the centrality of the congruence kernels of G 1 and G 2 follows from [30].Together with [27, Theorem 9.1 and Theorem 9.15], we conclude that the congruence kernels of G 1 and G 2 are in fact trivial.
Let Γ 0 and Λ 0 be arithmetic subgroups of G 1 and G 2 , respectively, which we may assume intersect the center trivially.Since the congruence kernel of G 1 is trivial, the profinite completion Γ 0 agrees with the congruence completion Γ 0 .The latter is an open subgroup of G 1 (A f Q ) by strong approximation.Similarly, under the isomorphism Γ is a lattice in the Lie group G 1 (R) while Λ is a lattice in G 2 (R).Since Γ does not meet the center of G 1 , Γ is also a lattice in any central quotient of G 1 (R).As we explained in the introduction, the Burger-Monod-Shalom theorem gives H 2 b (Γ; R) ∼ = R and H 2 b (Λ; R) ∼ = 0 because G 1 (R) has type E 7(−25) whereas G 2 (R) has type E 7 (7) .This completes the proof that Lie groups of type E 7(−25) exhibit non-profinite second bounded cohomology.
Remark 7. We are grateful to the anonymous referee who suggested to us the following alternative way to construct a Q-form G 1 of type E 7 which has type E 7(−25) at the real place and splits at all finite places.Consider the standard quadratic form q = x 2 1 + • • • + x 2 8 of rank eight.It corresponds to an element α q ∈ H 1 (Q, SO(4, 4)) because the discriminant of q and the standard form of signature (4, 4) are both trivial.The boundary map is given by the Hasse-Witt invariant [34, III.3.2.b), p. 141] so that δ 2 (α q ) = w 2 (q) = 1.Thus α q has a preimage β q ∈ H 1 (Q, Spin(4, 4)).Via the obvious Dynkin diagram inclusion D 4 ⊂ E 7 , the class β q maps to a class in H 1 (Q, E 7 ).The image in H 1 (Q, Ad E 7 ) thus defines a Q-form G 1 of type E 7 which splits at every finite place and has Satake-Tits index at the real place by construction.This shows that G 1 (R) is the Lie group E 7(−25) .
We now turn our attention to the group E 6(−14) .Again, we fix a rational prime p 0 .Let 2 E 6 be any simply-connected absolutely almost simple quasisplit Q-group which neither splits at p 0 nor at ∞.Such a form exists according to [3,Proposition p. 58].The Hasse principle for adjoint groups [27,Theorem 6.22,p. 336] shows that the diagonal map is injective.But f is also surjective by [29, Proposition 1] because at p 0 , the group 2 E 6 has no inner twist, meaning the set The set H 1 (R, Ad 2 E 6 ) has three elements corresponding to the quasisplit form E 6(2) , the hermitian form E 6(−14) , and the compact form E 6(−78) as we infer one more time from [1, Table 3, p. 1094 and remark in 10.3].Let α, β ∈ H 1 (Q, Ad 2 E 6 ) be the unique classes corresponding to E 6(−14) and E 6(2) at the infinite place, respectively, and which are trivial at all finite places.We denote the simply-connected Q-forms of 2 E 6 corresponding to α and β by G 1 and G 2 .By construction, G 1 and G 2 are isomorphic over Q p for all finite primes p. From the tables [27, p. 332] and [26,Table 10,p.321], we infer that the centers of G 1 (R) and G 2 (R) have order three whereas the centers of the simply-connected real Lie groups of type E 6(−14) and E 6(2) are infinite cyclic and order six, respectively.So again, the metaplectic kernels of G 1 and G 2 are trivial.It follows anew from [29,Theorem 1 (iii)] that rank Q G 1 = 2 and rank Q G 2 = 4 so that both congruence kernels are central by [30].Finally the case of type 2 E 6 , which was still excluded in [27, Theorem 9.1, p. 512], was meanwhile settled by P. Gille [9].So [27,Theorem 9.15] shows that both G 1 and G 2 have trivial congruence kernel.With these remarks, the rest of the argument goes through as before and we conclude that E 6(−14) exhibits non-profinite second bounded cohomology.
For the group SO 0 (6, 2), we can actually argue similarly.We let 3 D 4 be a simply-connected absolutely almost simple quasisplit Q-group that localizes to the quasisplit triality form of type 3 D 4 at a fixed prime p 0 and that splits at ∞.Note that the Q-group 3 D 4 can either have outer type 3 D 4 or 6 D 4 .Since again H 1 (Q p 0 , Ad 3 D 4 ) is trivial according to [27,Proposition 6.15.(1), p. 334], we have a pointed bijection Hence there exist Q-forms G 1 and G 2 of 3 D 4 that localize to the inner forms SO 0 (6, 2) and SO 0 (4, 4) at the real place, respectively, and to the trivial inner twist of 3 D 4 at all finite places so that they are isomorphic over Q p for all p.We have rank Q G 1 = 1 and rank Q G 2 = 2, so both groups have central congruence kernel, this time by [31].The tables [27, p. 332] and [26,Table 10,p. 320] show that G 1 (R) and G 2 (R) have trivial center whereas the corresponding topological universal coverings have center isomorphic to Z⊕Z/2 and (Z/2) 3 , respectively.This implies that the metaplectic kernels of G 1 and G 2 are trivial and so are the congruence kernels.Hence we can once more construct profinitely isomorphic lattices in G 1 (R) and G 2 (R) as we did in type E 7 and we conclude that SO 0 (6, 2) ≈ SO * (8) exhibits non-profinite second bounded cohomology.
Finally, we observe that Lemma 1 generalizes effortlessly to the groups Γ = Spin(n, 2)(Z) and Λ = Spin(n − 4, 6)(Z) for n ≥ 7. Let Γ 0 ≤ Γ be a finite index subgroup which intersects the center of The group Γ 0 is a lattice in every quotient group of Spin(n, 2)(R) by a central subgroup so that all Lie groups isogenous to SO 0 (n, 2) exhibit non-profinite second bounded cohomology for n ≥ 7.

Proof of Theorem 2-"only if part"
In this section, we show that the remaining higher rank Lie groups defining hermitian symmetric spaces do not exhibit non-profinite second bounded cohomology.Recall that the group SO 0 (3, 2) is isogenous to Sp(2, R) and the group SO 0 (4, 2) is isogenous to SU(2, 2).
As preparation, let k and l be totally real number fields and let G and H be simply-connected absolutely almost simple groups defined over k and l, respectively.Assume that G is anisotropic at all infinite places of k except one which we call v and that H is anisotropic at all infinite places of l except one which we call w.Suppose moreover that rank kv G ≥ 2 and rank lw H ≥ 2 and that there exist arithmetic subgroups Γ ≤ G(k) and Λ ≤ H(l) such that Γ ∼ = Λ.By adelic superrigidity [13,Theorem 3.4], we have an isomorphism j : A f l → A f k of topological rings and a group scheme isomorphism η : By the proof of [18, Proposition 2.5 (a), p. 238], the isomorphism j induces a bijection u → u ′ of the finite places of k and l and isomorphisms k u ∼ = l u ′ .Correspondingly, the isomorphism η over j splits into a family of isomorphisms G(k u ) ∼ = H(l u ′ ).In particular, G and H have the same Cartan Killing type.
Proposition 8.The R-groups G v and H w are inner twists of one another.
Proof.The case k = l = Q was handled in [14, Proposition 2.7] (and it is apparent that the same arguments give the case of k = l if v = w).For the general case, we argue as follows.Let G 0 be the up to Q-isomorphism unique Q-split simply-connected Q-group of the same Cartan Killing type as G and H. Let T ⊂ G 0 be a maximal Q-split torus, pick a set ∆ of simple roots of G 0 with respect to T, and let Sym ∆ be the subgroup of the permutation group of ∆ given by Dynkin diagram symmetries.Then we have a split short exact sequence (9) Let us now first suppose that G and hence G 0 does not have type D 4 .We may then assume that Sym ∆ ∼ = Z/2Z because the proposition is trivial if Sym ∆ is.Note that the first Galois cohomology with coefficients in Z/2Z classifies quadratic field extensions.So we conclude that π * α and π * β correspond to quadratic extensions K/k and L/l, respectively, such that [18,Theorem 2.3,p. 237]).This shows in particular that K and L have the same number of real embeddings [18, Theorem 1.4 (h), p. 79].It follows that the number of real places in k extending to complex places in K equals the number of real places in l extending to complex places in L. Translating back from field extensions to Galois cohomology classes, this shows that α and β localize to outer forms at the same number of infinite places.Since α and β localize to the compact real form (which may be inner or outer depending on the Cartan Killing type) at all other infinite places, α v and β w must be either both outer or both inner forms.In any case, they are inner twists of each other.Now if G and hence G 0 does have type D 4 , then Sym ∆ ∼ = S 3 ∼ = Z/3 ⋊ Z/2.Note that subgroups of the same order are conjugate in this group.Therefore, the first Galois cohomology with coefficients in the trivial Galois module S 3 classifies Galois extensions with Galois groups either Z/2, or Z/3, or S 3 (or trivial).So π * α and π * β correspond to Galois extensions K/k and L/l of one and the same of these types, again such that L .So once more, K and L have the same number of real embeddings.As K and L are Galois over k and l, respectively, real places extend either only to real places or only to complex places in these extensions.Correspondingly, α and β localize to outer forms again at the same number of infinite places of k and l.As all nontrivial homomorphisms Gal(R) → Sym ∆ are conjugate, any two real outer forms are inner twists of each other even in type D 4 .It follows again that α v and β w must be inner twists of one another.Now let G be one of the groups listed at the beginning of this section and let Γ ≤ G be a lattice.Let Λ ≤ H be a lattice in another higher rank Lie group and assume that Γ ∼ = Λ.We have to show that H also defines a hermitian symmetric space.By Margulis arithmeticity [23, Chapter IX, Theorem 1.11, p. 298, ( * * ) and Remark 1.6.(i),pp.293-294], there exists a k-group G and an l-group H with k, l, G, and H as above such that G v is isogenous to G, such that H w is isogenous to H, and such that G(O k ) is commensurable with Γ while H(O l ) is commensurable with Λ.Since Γ and Λ are profinitely isomorphic, so are suitable finite index subgroups of G(O k ) and H(O l ).Therefore, we have the conclusion from above that there exists an isomorphism j : A f l → A f k and an isomorphism η : G × k A f k −→ H × l A f l over j and Proposition 8 holds true for G and H. Proposition 10.The groups G ≈ SU(n, m) with n, m ≥ 2 do not exhibit non-profinite second bounded cohomology.
Proof.By Proposition 8, the group H w is an inner twist of G v , hence H is isogenous to SU(n ′ , m ′ ) for some n ′ , m ′ ≥ 2 because the generalized special unitary groups are the only outer real forms of type A n in the classification of real simple Lie groups.So H defines a hermitian symmetric space, too.Proposition 11.The groups G ≈ Sp(n, R) with n ≥ 2 do not exhibit non-profinite second bounded cohomology.
Proof.Let G 0 = Sp n be the unique Q-split simply-connected Q-group of type C n .Then Sym ∆ = 1, so Aut G 0 = Ad G 0 and we have Z(G 0 ) = µ 2 .From the short exact sequence of Gal(Q)-groups we obtain a boundary map δ K : H 1 (K, Ad G 0 ) → H 2 (K, µ 2 ) for any field extension K/Q.As we saw in [16,Section 5], the kernel of δ R consists of the class corresponding to the split form Sp(n, R) only, while the other real forms of type C n are the groups Sp(p, q) with p + q = n and they form precisely the fiber under δ R of the nontrivial element in SO 0 (4, 3).However, this possibility can be excluded because the symmetric space defined by G is 10-dimensional whereas the symmetric space defined by H is 12-dimensional and the dimension mod four is a profinite invariant by [14,Theorem 2.1].