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Home , Burnside problem, Congruence subgroup

The Congruence problem

B.Sury Indian Statistical Institute Bangalore, India [email protected] 17th May 2011 St.Petersburg Steklov Institute Russia

B.Sury The Congruence subgroup problem If one views Z simply as a discrete subgroup of R, many of its number-theoretic properties lie hidden. For instance, to ‘see’ prime numbers, one needs to use some other topology on Z. This is the so-called profinite (or artihmetic) topology. Here, arithmetic progressions form a basis of open sets. The Chinese remainder theorem essentially tells us that the arithmetic topology is ‘built out of’ various p-adic topologies for all the primes p. The congruence subgroup problem is a question which asks whether this property generalizes to vastly more complicated groups. An affirmative answer would again amount to the ‘topology given by of finite index being built out of the p-adic topologies’.

B.Sury The Congruence subgroup problem If one considers a matrix group with integral entries, say, SLn(Z), for some n ≥ 2, the study of its of subgroups of finite index is a problem of on the face of it. But, one natural way of finding subgroups of finite index is number-theoretic; that is, to look at the natural ring homomorphism Z → Z/kZ for a natural number k and look at the of the corresponding SLn(Z) → SLn(Z/kZ). This is known as the principal congruence subgroup of level k.

Any subgroup of SLn(Z) which contains a principal congruence subgroup of some level is called a congruence subgroup.

B.Sury The Congruence subgroup problem Congruence subgroups show up in many situations in number theory. For instance, the main step in Wiles’s proof of Fermat’s last theorem is to obtain a non-constant meromorphic map from the quotient of the upper half-plane by a suitable congruence subgroup into an defined over Q. In fact, the classical theory of modular forms and Hecke operators which plays such a central role in number theory is a theory which ‘lives’ on congruence subgroups. Interestingly, the proof of this fact requires the solution of the congruence subgroup problem for the so-called Ihara SL2(Z[1/p]), as shown by Serre and Thompson.

B.Sury The Congruence subgroup problem CSP - Naive form

The first question one may ask is whether all subgroups of finite index in SLn(Z) are congruence subgroups. Note that the question is meaningful because of the existence of plenty of congruence subgroups in the sense that their intersection consists just of the identity element. Already in the late 19th century, Fricke and Klein showed that the answer to this question is negative if n = 2.

B.Sury The Congruence subgroup problem Indeed, since the of rank 2 is the principal congruence subgroup Γ (2) of level 2, any 2-generated finite group is a quotient of this group. It turns out that the finite, simple groups which can occur as quotients by congruence subgroups are the groups PSL2(Fp) for primes p.

But there are many finite, simple 2-generated groups (like An (n > 5), PSL3(Fq) for odd prime q) which are not isomorphic to PSL2(Fp) (the latter has abelian q-Sylow subgroups). So, the corresponding kernel cannot be a congruence subgroup of SL2(Z).

B.Sury The Congruence subgroup problem An explicit example of a noncongruence subgroup is the following. Let k be a positive which is not a power of 2. 1 2 1 0 In the (free) group generated by the matrices X = and Y = , 0 1 2 1 the group of all matrices which are words in X , Y such that the exponents of X , Y add up to multiples of k, is a subgroup of finite index in SL2(Z) but is not a congruence subgroup. In fact, it can be shown that there are many more noncongruence subgroups of finite index in SL2(Z) than there are congruence subgroups.

B.Sury The Congruence subgroup problem It was only in 1962 that Bass-Lazard-Serre - and, independently, Mennicke - showed that the answer to the question is affirmative when n ≥ 3. Later, in 1965, Bass-Milnor-Serre generalised this to the special linear and the symplectic groups over number fields.

Mennicke showed later that subgroups of finite index in SL2(Z[1/p]) are congruence subgroups and this is the principal reason behind the fact asserted above that the classical Hecke operators for SL(2, Z) live only on congruence subgroups.

B.Sury The Congruence subgroup problem Arithmetic, congruence groups in algebraic groups

How to pose the question for general (linear) algebraic groups over a number field k? The algebraic groups we consider are all linear (therefore, affine) algebraic groups; the formulation for abelian varieties is quite different.

B.Sury The Congruence subgroup problem Recall that an defined over k (considered as a subfield of C) is a subgroup G of GL(N, C) which is also the set of common zeroes of a finite −1 2 set of polynomial functions P(gij , det(g) ) in N + 1 variables with coefficients from k. The definition has to be slightly modified if k is a field of positive characteristic. The group G(k) = G ∩ GL(N, k) turns out to be defined independent of the choice of embedding G ⊂ GL(N, C) and, is called the group of k-points of G.

B.Sury The Congruence subgroup problem Here are some standard examples of algebraic groups defined over a subfield k of C. (i) G = GL(n, C). (ii) G = SL(n, C) = Ker(det : GL(n, C) → C∗). (iii) For any symmetric matrix M ∈ GL(n, k), the

O(M) = {g ∈ GL(n, C):t gMg = M}.

(iv) For any skewsymmetric matrix Ω ∈ GL(2n, k), the

Sp(Ω) = {g ∈ GL(2n, C):t gΩg = Ω}.

(v) Let D be a division algebra with center k (its dimension as a k-vector space 2 2 must be n for some n). Let vi ; 1 ≤ i ≤ n be a k-basis of D (then it is also a C-basis (as a vector space) of the algebra D ⊗k C).

The right multiplication by vi gives a linear transformation Rvi from D ⊗k C to 2 2 itself, and thus, one has elements Rvi ∈ GL(n , C) for i = 1, 2, ··· , n .

2 2 G = {g ∈ GL(n , C): gRvi = Rvi g ∀ i = 1, 2, ··· , n }. This last group has its G(k) = the nonzero elements of D.

B.Sury The Congruence subgroup problem Here, k will denote an algebraic number field.

Let G ⊂ SLn be a k-embedding of a . If Ok denotes the ring of algebraic in k, we denote by G(Ok ), the group G ∩ SLn(Ok ).

For any non-zero ideal I of Ok , one has the G(I ) = Ker(G(Ok ) → SLn(Ok /I )) of finite index in G(Ok ).

Note that it is of finite index because the ring Ok /I is finite.

B.Sury The Congruence subgroup problem Then, one could define a subgroup of G(Ok ) to be a congruence subgroup if it contains a subgroup of the form G(I ) for some non-zero ideal I .

The only problem is that, unlike G(k), the definitions of G(Ok ) and G(I ) etc. depend on the k-embedding we started with.

However, it turns out that for a new k-embedding, the ‘new’ G(Ok ) contains an ‘old’ G(I ) for some I 6= 0. Thus, the following definition is independent of the choice of the k-embedding : A subgroup Γ is a congruence subgroup if Γ ⊃ G(I ) for some I 6= 0.

B.Sury The Congruence subgroup problem The group G(Ok ) can be realized as a discrete subgroup of a product r1 r2 G(R) × G(C) of Lie groups, where k has r1 real completions and r2 non-conjugate complex completions. More generally, let S be any finite set of inequivalent valuations containing all the archimedean valuations.

The non-archimedean valuations in S correspond to nonzero prime ideals of Ok .

B.Sury The Congruence subgroup problem One has the bigger ring OS of S-units of k; these are the elements of k which admit denominators only from primes in S.

Define G(OS ) = G ∩ SLn(OS ); then G(OS ) is a discrete subgroup of the Q product v∈S G(kv ) of real, complex and p-adic Lie groups, where kv denotes the completion of k with respect to the valuation v.

One calls a subgroup Γ ⊂ G(OS ) an S-congruence subgroup if and only if it contains G(I ) (the elements of G(OS ) which map to the identity element in G(OS /I )) for some nonzero ideal I of OS .

Of course, G(OS ) itself depends on the k-embedding of G chosen.

B.Sury The Congruence subgroup problem Hence, the most general definition is the following : A subgroup Γ ⊂ G(k) is an S-congruence subgroup if, for some (and, therefore, any) k-embedding of G, the group Γ contains G(I ) as a subgroup of finite index for some nonzero ideal I of OS . One also defines : A subgroup Γ ⊂ G(k) is an S-arithmetic subgroup if, for some (and, therefore, any) k-embedding, Γ and G(OS ) are commensurable (i.e., Γ ∩ G(OS ) has finite index in both groups).

B.Sury The Congruence subgroup problem The congruence kernel

Each of the two families (the S-arithmetic groups and the S-congruence groups) defines a topology on G(k) as follows. For any g ∈ G(k), one defines a fundamental system of neighbourhoods of g to be the cosets gG(I ) as I varies over the nonzero ideals of OS .

This topology Tc is called the S-congruence topology on G(k). If one takes a fundamental system of neighbourhoods of g to be the cosets gΓ as Γ varies over S-arithmetic subgroups of G(k), the corresponding topology Ta is called the S-arithmetic topology.

As S-congruence subgroups are S-arithmetic subgroups, Ta is finer. Note that Ta just gives the profinite topology on G(OS ).

B.Sury The Congruence subgroup problem If the resulting completions of (uniform structures on) G(k) are denoted by Gˆa and Gˆc , then there is a continuous surjective homomorphism π from Gˆa onto Gˆc .

The surjectivity of π is due to the fact that its contains an open compact subgroup (the closure of G(OS )) and a dense subgroup G(k).

Obviously, π is an isomorphism if all subgroups of finite index in G(OS ) are S-congruence subgroups. In general, the kernel C(S, G) := Ker(π), called the S-congruence kernel, measures the deviation.

B.Sury The Congruence subgroup problem A basic important property is that C(S, G) is a profinite group. ˆ ˆ Indeed, the closures Γca, Γcc of G(OS ) in Ga and Gc respectively, are profinite groups and C(S, G) ⊂ Γca. The congruence subgroup problem (CSP) is the problem of determining the group C(S, G) for any S, G.

B.Sury The Congruence subgroup problem Margulis-Platonov conjecture

Let us return to a general algebraic group now. There is a conjecture due to Serre which predicts precisely when the group C(S, G) is finite. The finiteness of C(S, G) has many interesting consequences as we shall see. Therefore, in the modern liberal sense one says that the congruence subgroup property holds for G with respect to S if C(S, G) is finite. For any S, the association G 7→ C(S, G) is a functor from the category of k-algebraic groups to the category of profinite groups.

B.Sury The Congruence subgroup problem Given G, a general philosophy (which can be made very precise) is that the larger S is, the ‘easier’ it is to compute C(S, G). In fact, when S consists of all the places v of k excepting the (finitely many possible) nonarchimedean places T for which G is kv -anisotropic (equivalently, G(kv ) is compact for the topology induced from kv ), the computation of C(S, G) amounts to a conjecture of Margulis & Platonov which has been proved in most cases. In fact, in most cases T is empty which means that the triviality of C(S, G) (for the S mentioned last) is equivalent to the simplicity of the abstract group G(k)/center.

B.Sury The Congruence subgroup problem In my thesis, I was able to prove the case when the complement of S is finite (and this implied the Margulis-Platonov conjecture) for many types - this has been given a completely uniform argument by Gopal Prasad and A.S.Rapinchuk. One important case where the Margulis-Platonov conjecture has still not been proved is that of the special of a division algebra with center k and with an involution of the second kind.

B.Sury The Congruence subgroup problem Necessary conditions for finiteness

First, we briefly indicate how the CSP for a general group reduces to CSP for some special types of groups. It reduces to connected algebraic groups G and, further, to the G/Ru(G) using nothing more than the Chinese remainder theorem.

Here, Ru(G) is the unipotent radical of G; it is the maximal, normal, connected, unipotent subgroup.

B.Sury The Congruence subgroup problem For tori T , the congruence kernel C(S, T ) is the trivial group - this is a nontrivial theorem due to Chevalley and is essentially a consequence of Chebotarev’s density theorem in global class field theory. Hence, the problem reduces to that for semisimple groups.

For such G, the group G(OS ) (for any embedding) can be identified with a Q lattice in the group GS := v∈S G(kv ) under the diagonal embedding of G(k) in GS .

B.Sury The Congruence subgroup problem Finally, it is necessary (as observed by Serre) for the finiteness of C(S, G) that G be simply-connected as an algebraic group - that is, there is no k-algebraic group G˜ admitting a surjective k-map π : G˜ → G with finite nontrivial kernel. This is actually equivalent to G(C) being simply-connected in the usual sense. Further, if one can compute C(S, G˜ ) for a simply-connected ‘cover’ G˜ of G, then one can compute C(S, G). Thus, the CSP reduces to the problem for semisimple, simply-connected groups.

B.Sury The Congruence subgroup problem Q If G is simply-connected and the group v∈S G(kv ) is noncompact (equivalently, G(OS ) is not finite), one has the strong approximation property.

This means that the closure Γcc of G(OS ) with respect to Tc can be identified Q with v6∈S G(Ov ).

Here Ov is the local in the p-adic field kv . Note that for the additive group, the strong approximation is a consequence of the Chinese remainder theorem.

Thus, if C(S, G) is trivial, the profinite completion of G(OS ) is also equal to Q v6∈S G(Ov ). Thus, roughly speaking, when the congruence kernel is trivial the topology given by subgroups of finite index is built out of the p-adic topologies as asserted in the introduction.

B.Sury The Congruence subgroup problem For G(k) itself, strong approximation means that Gc can be identified with the ‘S-adelic group’ G(AS ), the restricted direct product of all G(kv ), v 6∈ S with respect to the open compact subgroups G(Ov ). The reason that G must be simply-connected in that C(S, G) be finite, is as follows. Let there exist a k-map : G˜ → G with kernel µ and G˜ simply-connected.

Now, the S-congruence completion G˜c can be identified with the S-adelic group G˜ (AS ). If π is the homomorphism from the S-arithmetic completion G˜a to G˜c = G˜ (AS ), then it is easy to see that C(S, G) contains the infinite group −1 π (µ(AS ))/µ(k).

B.Sury The Congruence subgroup problem As mentioned at the beginning, subgroups of finite index in SL(n, Z) are congruence subgroups when n ≥ 3 while this is not so when n = 2. What distinguishes these groups qualitatively is their rank.

If G ⊂ SLn is a k-embedding, one calls a k-torus T in G to be k-split, if there −1 is some g ∈ G(k) such that gTg is a subgroup of the diagonals in SLn. The maximum of the dimensions of the various k-split tori (if there exist k-split tori) is called the k-rank of G.

So, k-rank of SLn is n − 1 and k-rank Sp2n = n. If F is a quadratic form over k, the group SO(F ) is a k-group whose k-rank is the Witt index of F .

B.Sury The Congruence subgroup problem Serre’s conjecture

Consider a general semisimple, simply-connected k-group G. Assume that G is absolutely almost simple (that is, G has no connected normal algebraic subgroups). For a finite S containing all the archimedean places, Serre formulated the characterization of the congruence subgroup property (that is, the finiteness of C(S, G)) conjecturally as follows. First, it is easy to see that a necessary condition for finiteness of C(S, G) is that for any nonarchimedean place v in S, the group G has kv -rank > 0 (equivalently, G(kv ) is not profinite by a theorem of Bruhat-Tits-Rousseau). Indeed, otherwise the whole of G(kv ) is a quotient of C(S, G).

B.Sury The Congruence subgroup problem Serre’s Conjecture: P C(S, G) is finite if, and only if, S-rank(G) := v∈S kv − rank(G) ≥ 2 and G(kv ) is noncompact for each nonarchimedean v ∈ S. When C(S, G) is finite, one says that the CSP is solved affirmatively or that the congruence subgroup property holds for the pair (G, S). As mentioned in the introduction, for k = Q, S = {∞, p} for some prime p and G = SL2, the CSP was solved affirmatively by Mennicke. The finitness of C(S, G) as against its being actually trivial, already has strong consequences.

B.Sury The Congruence subgroup problem One of them - as pointed out by Serre - is super-rigidity; this means:

If C(S, G) is finite, then any abstract homomorphism from G(OS ) to GLn(C) is essentially algebraic.

In other words, there is a k-algebraic group homomorphism from G to GLn which agrees with the homomorphism we started with, at least on a subgroup of finite index in G(OS ).

In particular, Γ/[Γ,Γ ] is finite for every subgroup of finite index in G(OS ). This last fact has already been used to prove in some cases that C(S, G) is NOT finite, by producing a Γ of finite index which surjects onto Z.

B.Sury The Congruence subgroup problem Relation with group cohomology

Consider the exact sequence defining C(S, G). Recall that the strong approximation theorem for G(k) means Gc can be identified with the ‘S-adelic group’ G(AS ), the restricted direct product of all G(kv ), v 6∈ S with respect to the open compact subgroups G(Ov ). Thus, we have the exact sequence

1 → C(S, G) → Gˆa → G(AS ) → 1

which defines C(S, G). By looking at continuous group cohomology Hi with the universal coefficients R/Z, one has the corresponding Hochschild-Serre spectral sequence which gives the exact sequence

1 1 1 G(AS ) 2 H (G(AS )) → H (Gˆa) → H (C(S, G)) → H (G(AS )).

B.Sury The Congruence subgroup problem Now, the congruence sequence above splits over G(k), the last map actually 2 2 goes into the kernel of the restriction map from H (G(AS )) to H (G(k)). So, if α denotes the first map, then we have an exact sequence

1 G(AS ) 2 2 1 → Cokerα → H (C(S, G)) → Ker(H (G(AS ) → H (G(k))).

Here G(k) is considered with the discrete topology. The last kernel is called the S-metaplectic kernel and it is a finite group. It has been computed in all cases by Gopal Prasad, Raghunathan and Rapinchuk. The cokernel of α is nothing but the S-arithmetic closure of [G(k), G(k)]; hence it is also a finite group since [G(k), G(k)] has finite index in G(k).

B.Sury The Congruence subgroup problem In fact, one expects the cokernel to be trivial, and this is known in most cases. Therefore, the middle term H1(C(S, G))G(AS ) is finite. But, this shows that the the quotient C(S, G)/[C(S, G), Gˆa] is finite. In fact, we have :

C(S, G) is finite if, and only if, it is contained in the center of Gˆa. The centrality of C(S, G) has been proved for all cases of S-rank at least 2 other than the important case of groups of type An which have k-rank 0.

B.Sury The Congruence subgroup problem Several people have proved vanishing results for cohomology of lattices in Lie groups and the techniques of the CSP can often be applied. For instance, the existence of cocompact arithmetic lattices ∆ in SO(n, 1) (n ≥ 5, n 6= 7) for which H1(∆, C) 6= 0 has been estabished in this manner. It is relevant to recall a famous old conjecture of Thurston to the effect that any compact (real) hyperbolic manifold admits a finite covering with non-vanishing first Betti number. The various results mentioned here prove it except for 3-manifolds.

B.Sury The Congruence subgroup problem The connection with CSP arises as follows. Embed the real hyperbolic manifold in a suitable complex hyperbolic manifold. That the latter admits a ‘congruence’ covering with non-vanishing first Betti number is a result of Kazhdan using his property T. More precisely: If H ⊂ G are Q-simple and H(R) = SO(n, 1) and G(R) = SU(n, 1), and Γ is a congruence subgroup of G(Q), then the restriction of H1(Γ, C) to the product of H1(gΓ g −1 ∩ H, C) over all g ∈ G(Q), is injective. The idea is to prove that the translates of C(S, H) under the group (AutG)(Q) topologically generate a subgroup of finite index in C(S, G).

B.Sury The Congruence subgroup problem There are other connections with cohomology in the form of Kazhdan’s property T, and Selberg’s property for minimal eigenvalue of Laplacian etc. For instance, Thurston’s famous conjecture on the first Betti number is solved for some cases using these techniques.

B.Sury The Congruence subgroup problem Profinite group theory

Since C(S, G) is a profinite group, it is conceivable that one can use general results on profinite groups profitably. This is indeed the case. In fact, one can characterize the centrality by means of purely profinite-group-theoretic conditions. Platonov & Rapinchuk proved that C(S, G) is finite if, and only if, the profinite group Γca has bounded generation.

In other words, there are elements x1, ··· , xr (not necessarily distinct) in Γ so that the set < x1 > ··· < xr > coincides with the whole group Γca.

B.Sury The Congruence subgroup problem The fore-runner of this result (which was used by them crucially) is the theorem of Lazard to the effect that among pro-p groups (for any prime p), the p-adic analytic groups are characterized as those admitting bounded generation as above. After the solution of the restricted Burnside problem, one also has other characterizations for analyticity of pro-p groups like the number of subgroups of any given index being a polynomial function of the index. These can also be adapted to characterize the finiteness of C(S, G) in terms of polynomial index growth for Γca.

B.Sury The Congruence subgroup problem Another characterization turns out to be true : a (topologically) finitely generated profinite group ∆ which can be continuously Q embedded in the S-adelic group p SL(n, Zp) for some n, has bounded generation. A special case of this was sufficient to prove Lubotzky’s conjecture that

C(S, G) is finite if, and only if, G\(OS ) can be continuously embedded in the Q S-adelic group p SL(n, Zp) for some n.

B.Sury The Congruence subgroup problem Aspects of subgroup growth have turned out to have deep relations with the CSP. For instance, apart from revealing finer structural facts like the number of noncongruence subgroups being a higher order function (in a more precisely known form) than the number of congruence subgroups (when CSP does not hold), the methods allow an analogue of the CSP to be asked for lattices (even nonarithmetic ones ! ) in semisimple Lie groups, viz., If Γ is a lattice in a semisimple over a local field (of characteristic 0), is the subgroup growth type of Γ strictly less than nlog n ?

B.Sury The Congruence subgroup problem An unexpected connection of the CSP with cohomology of finite simple groups is revealed by some recent work.

dimH2(A,M) For a finite group A, denote by h(A), the supremum of the number dimM over all primes p and all simple Fp[G]-modules M. For any finite A, there is a bound h(A) = O(log|A|) and it was conjectured that there is a constant c > 0 such that h(A) ≤ c for all finite simple groups A. Partial results can be obtained from the methods of the CSP. For instance, if G is a fixed simple Chevalley scheme, then there is a constant c > 0 such that h(G(Fp)/center) ≤ c for all primes p.

The idea is to obtain the finite groups G(Fp) as quotients of the congruence completion of G(Z).

B.Sury The Congruence subgroup problem Status of finiteness of C(S, G)

Having talked about the general problem, we return back to the past briefly.

For the classical case of SLn(Z) with n ≥ 3, centrality can be proved with the help of elementary matrices and reduction to SLn−1. The key fact here is that any unimodular integral vector c which satisfies c ≡ en mod r must be in the orbit of En(r), the normal subgroup generated by those elementary matrices which are in the principal congruence subgroup Γ (r) of level r.

B.Sury The Congruence subgroup problem More precisely, look at G = SL3(Z) and the embedded SL2(Z) in its left hand top corner.

For any r, consider the normal subgroup E3(r) generated by U12(r) in G.

For any element g of Γ3(r) (for SL3), one may apply elementary row and column operations and get x, y in E3(r), with xgy in the principal congruence subgroup Γ2(r) of level r for the embedded SL2.

This implies easily that the action of G on Γ3(r)/E3(r) by conjugation is trivial.

Now, the first rows (a, b) of Γ2(r) for SL2(Z) give rise to corresponding a b 0 elements ∗ ∗ 0 of Γ3(r)/E3(r); these are the so-called Mennicke 0 0 1 symbols M(a, b) which will be shown to be trivial.

B.Sury The Congruence subgroup problem The Mennicke symbol has nice properties like: (i) M(a + tb, b) = M(a, b) for all t ∈ Z, (ii) M(a, rta + b) = M(a, b) for all t ∈ Z, (iii) M(a, b) is multiplicative in b, (iv) M(a, b) = 1 if b ≡ ±1 mod a. Using these, one can show that it is actually trivial.

Therefore, Γ3(r) = E3(r); so, a normal subgroup of finite index which contains E3(r) (as it must, for some r), also contains Γ3(r).

B.Sury The Congruence subgroup problem For general G whose k-rank is nonzero, there are unipotent elements in G(k) which play the role of the elementary matrices and which allow for similar (although much harder) proofs. Briefly, we recall the case of K-rank at least 2 which was dealt with by Raghunathan. For each parabolic k-subgroup Q with unipotent radical V , and each non-zero ideal a in OS , define V (a) = V ∩ G(a) and Q(a) = Q ∩ G(a).

B.Sury The Congruence subgroup problem Let EG(a) denote the subgroup of G(a) generated by V (a) for varying parabolic K-subgroups Q.

This is a generalization to arbitrary G of the elementary subgroups in SLN (a).

Obviously, EG(a) is normal in G(a) (in fact in the whole group G(OS )). Define similarly, the subgroups FG(a) as the ones generated by Q(a) as Q varies.

The groups FG(a) are also normal in G(OS ).

B.Sury The Congruence subgroup problem One innovation in Raghunathan’s proof is the introduction of these intermediate groups EG(a) and FG(a) for the proof of centrality. We have the inclusions EG(a) ⊂ FG(a) ⊂ G(a). Furthermore, given a and an element g ∈ G(K), there exists a non-zero ideal a’ such that gEG(a)g −1 ⊃ EG(a0). Similarly for FG(a).

We then obtain new completions GcF (resp. GcE ) by requiring FG(a) (resp. EG(a)) to be open in G(K).

We also get the kernels CE/F and CF of the corresponding extensions GcE of GcF and GcF of Gc (which are also split over G(K)).

B.Sury The Congruence subgroup problem To prove the centrality of CF , one proceeds as follows.

Fix the highest k- root β, and set Uβ to be the corresponding root subgroup in G.

Set ∆ = Uβ (OS ). One then makes the key observation that for any non-zero ideal a, the commutator subgroup [∆, G(a)] is contained in GF (a), as k-rank (G) ≥ 2. The reason for this is essentially the following. Since k-rank (G)≥ 2, for any two roots α and β with α 6= −β, the root groups Uα and Uβ lie in the unipotent radical of a parabolic subgroup of G defined over k.

Consequently, on the kernel CF , the action of ∆ is trivial, but the action extends to all of G(K); the simplicity of G(K) modulo its center now implies that the action of G(K) is also trivial.

B.Sury The Congruence subgroup problem Similarly, centrality of CE/F is proved using simplicity of M(K) for the groups M which are K-simple factors of the Levi part of Q.

One also uses the fact that there are only finitely many G(OS ) conjugacy classes of k-parabolic subgroups of G.

+ The final step that if CF and CE/F are centralized by G (K), then so also is the kernel C, is elementary but tedious.

B.Sury The Congruence subgroup problem For the groups of k-rank 1, the centrality was established by a rather different method which also works in several cases of k-rank 0. Note that for the groups of k-rank 0, there are no unipotents and one needs to work with concrete descriptions of them as unitary groups of hermitian forms etc. The idea of this method goes back to Raghunathan in 1986 and Rapinchuk which I was also able to carry out for a number of rank 0 groups. These proofs have been shortened and simplified by Gopal Prasad and Rapinchuk recently and we discuss them briefly now. What we discussed earlier for number fields is also valid for global fields of positive characteristic; these are finite extensions of Fq(t). Note that in the positive characteristic case, the valuations of k are non-archimedean. One must remember though that some of the proofs need to be more carefully done because lattices may not be finitely generated and the congruence subgroup property can fail for unipotent groups.

B.Sury The Congruence subgroup problem The centrality of the congruence kernel (under the hypotheses of Serre’s conjecture) was reduced to the following general statement. Let k be a global field and let G be an absolutely almost simple, simply conencted k-group satisfying the Margulis-Platonov conjecture. Let S be a finite set of places containing the archimedean ones, S is disjoint from T and S-rank(G) ≥ 2. Suppose, for each v 6∈ S, we have subgroups Gv in the S-arithmetic completion Gca, which pairwise commute and map onto G(kv ) under the map from Gca to Gcc = G(AS ). If the Gv (as v 6∈ S vary) generate a dense subgroup of Gca, then C(S, G) is central.

B.Sury The Congruence subgroup problem The proof of this reduction interestingly uses some abstract profinite group theory apart from some standard results on algebraic groups over local fields.

To use the above proposition, we need to locate such Gv ’s. We indicate in some detail how this is done when k-rank(G) ≥ 2. One crucial fact one needs to prove in this approach is that the S-arithmetic and S-congruence topologies on G(k) induce the same topology on U(k) where U is the k-subgroup generated by Uα as α varies over positive roots (for any ordering). This fact is trivial for number fields but not so in positive characteristic.

B.Sury The Congruence subgroup problem + Fix any ordering on the k-roots Φ (to be chosen later) and let U = UΦ+ and − U = UΦ− denote these unipotent k-subgroups normalized by the corresponding maximal k-split torus.

+ By the above remark, the map π : Gca → G(AS ) maps U\(k) isomorphically + + onto U (k) = U (AS ). Let θ denote its inverse map. For any root α, one may choose some ordering such that it is positive and, one can then define θα as the restriction of θ to Uα(AS ).

It can be checked that θα is independent of the ordering for which α > 0 because it is also the inverse of the isomorphism from U\α(k) to Uα(AS ).

As Uα(kv ) is naturally contained in Uα(AS ), the images θα(Uα(kv )) generate (as α varies over roots) a subgroup Gv of Gca.

B.Sury The Congruence subgroup problem For k-rank(G) ≥ 2, the Kneser-Tits conjecture is valid for G over k as well as over all kv (note T is empty).

This means that G(k) is generated by Uα(k) and, for each v, G(kv ) is generated by Uα(kv ) as α varies over all roots.

This immediately implies that Gv maps onto G(kv ) for each v 6∈ S.

For each α, by strong approximation, Uα(kv )’s generate a dense subgroup of Uα(AS ) as v 6∈ varies.

Thus, the closure in Gca of the subgroups generated by Gv ’s (as v varies) contains each Uα(k) and thus, the closure is the whole of Gca.

B.Sury The Congruence subgroup problem The most interesting and difficult statement to prove is the one which is left-over; viz., to show Gv and Gw commute in Gca for v 6= w.

For this, it suffices to prove that Uα,v := θα(Uα(kv )) and Uβ,w := θβ (Uβ (kw )) commute element-wise for any roots α, β and for v 6= w with v, w 6∈ S. If these two roots are independent or if one of them is a positive, integral multiple of the other, there is an ordering of Φ such that both are positive. In + this case, since Uα,v , Uβ,w are contained in the images under θ of U (kv ) and + U (kw ) respectively, we will be through. Finally, we have the trickiest case when one of the roots is a negative integral multiple of the other. In this case, using the fact that for any root γ, the unipotent root subgroup U2γ ⊂ Uγ , the assertion reduces to the case when α is simple and β = −α.

B.Sury The Congruence subgroup problem For this, we proceed as follows. Fix an ordering so that α is simple and let β = −α. As k-rank(G) ≥ 2, there exists a simple root γ 6= α.

Look at the torus Tγ which is the connected component at the identity of the intersection of the kernels of all simple roots other than γ. Let P+, P− be the parabolic k-subgroups generated by the reductive group + − M := ZG (Tγ ), the unipotent group U and, respectively, by ZG (Tγ ), U . + − Note that M normalizes both Ru(P ) and Ru(P ), and contains the subgroup Gα generated by Uα, U−α.

B.Sury The Congruence subgroup problem By a remark before beginning the proof, both the S-congruence and + S-arithmetic topologies on G(k) induce the same topology on Ru(P )(k) and − similarly on Ru(P )(k).

+ − Thus, the restriction of π to Ru(\P )(k) as well as to Ru(\P )(k) are injective. + − Since M(k) normalizes Ru(P )(k) and Ru(P )(k), we have

+ + [M\(k) ∩ C(S, G), Ru(\P )(k)] ≤ C(S, G) ∩ Ru(\P )(k) = {1}

− − [M\(k) ∩ C(S, G), Ru(\P )(k)] ≤ C(S, G) ∩ Ru(\P )(k) = {1} + − As Ru(P )(k) and Ru(P )(k) generate the whole of G(k), we have that M\(k) ∩ C(S, G) commutes with G(k) and is, hence, contained in the center of Gca.

In particular, the restriction of π to G\α(k) gives a central extension of Gα(AS ) which implies that the inverse images of Gα(kv ) and Gα(kw ) commute in Gca for v 6= w and v, w 6∈ S.

Thus, further particularly, the inverse images of Uα(kv ) and of U−α(kw ) commute in Gca for v 6= w.

B.Sury The Congruence subgroup problem Let us indicate how the centrality of congruence kernel is proved for SL2(Z[1/p]).

That is, we are considering the case k = Q, S = {∞, p} and G = SL2.

As before, the map π : Gca → G(AS ) gives an isomorphism

+ + + U\(Q) → U (Q) = U (AS ) and − − − U\(Q) → U (Q) = U (AS ) where U+, U− are the upper and lower triangular unipotent subgroups.

As Q is dense in AS by strong approximation, the maps 1 a u+ : Q → U+(Q); a 7→ 0 1 1 0 u− : Q → U−(Q); b 7→ b 1

± ± give rise to continuous homomorphisms uc : AS → U\(Q) by composing the above inverses.

B.Sury The Congruence subgroup problem ± For all q 6= p, Gq is defined to be the subgroup of Gca generated by uc(Qq).

As before, the only non-obvious part is to check that Gq’s commute element-wise. The basic relation we need is:

2 ! !     a b  1  1 0 1 a 1 0 1 1−ab 0 [ ] = ab−1 2 0 1 b 1 ab 0 1 0 1 − ab 1−ab 1

B.Sury The Congruence subgroup problem That is, writing h(t) = diag(t, t−1), we have

+ − + a2b 1 − ab2 [u (a), u (b)] = u ( ab−1 )h( 1−ab )u ( 1−ab ).

+ − We need this while checking that [uc(Qq1 ), uc(Qq2 )] = 1 in Gca for any primes q1, q2 6= p.

B.Sury The Congruence subgroup problem Let us enumerate all the primes other than p as q1, q2, ··· ∗ ∗ We first observe that it suffices to get hold of one a0 ∈ Qq1 and one b0 ∈ Qq2 + − such that [uc(a0), uc(b0)] = 1. To see this, note that for any t ∈ Q∗, the inner conjugation on G(Q) by diag(t, 1) also induces an automorphism σt of Gca.

+ − + − −1 Now, 1 = σt ([uc(a0), uc(b0)]) = [uc(ta0), uc(t b0)].

∗ ∗ ∗ + − As Q is dense in Qq1 × Qq2 , it follows that [uc(Qq1 ), uc(Qq2 )] = 1.

B.Sury The Congruence subgroup problem To get hold of a0, b0 we proceed as follows. For any m ≥ 2, there exists n(m) ≡ 0 mod m! so that pn(m) ≡ 1 mod 2m (q1q2 ··· qm) . n(m) Therefore, we can get 1 − p = ambm with (am, q1) = 1 = (bm, q2) and m am ≡ 0 mod (q2q3 ··· qm) m bm ≡ 0 mod (q1q3 ··· qm) .

There exists a subsequence {mk } such that ∗ amk → a0 ∈ Qq1 and ∗ bmk → b0 ∈ Qq2 .

B.Sury The Congruence subgroup problem + − + 2 −n(m) −n(m) − 2 −n(m) Now [u (am), u (bm)] = u (−ambmp )h(p )u (ambmp ). 2 2 Since Qq’s commute in AS , we have that ambm, ambm → 0 in AS .

+ 2 −n(m) − 2 −n(m) So, u (−ambmp ), u (ambmp ) → 1 in Gca.

n(m) Also, h(p ) → 1 in Gca since m! divides n(m).

+ − Thus, we have proved that [uc(a0), uc(b0)] = 1 and we are through.

B.Sury The Congruence subgroup problem This idea has also been exploited by Andrei Rapinchuk & Igor Rapinchuk to investigate the centrality of the congruence kernel for the elementary subgroups over any commutative Noetherian ring. They prove: Let G be a universal Chevalley-Demazure group scheme corresponding to a reduced, irreducible root system of rank ≥ 2. Let R be a commutative Noetherian ring where 2 is assumed to be a unit if the root system is of type Cn(n ≥ 2) or G2. Consider the corresponding elementary subgroup E(R). Then, the congruence kernel of E(R) is central.

Note that in particular, the theorem applies to SLn(Z[t1, t2, ··· , tr ]) for n ≥ 3.

B.Sury The Congruence subgroup problem We indicate the main steps in the proof of centrality. Note that unlike the classical situation where G(k) has an action on the congruence kernel, there is no such k here.

As usual, let E(R) denote the subgroup of G(R) generated by Uα(R) := eα(R) as α varies in Φ. One defines the arithmetic and congruence topologies on any subgroup of G(R) - we are interested in Γ := E(R). We have the corresponding exact sequence

π 1 → C(Γ ) → Γb → Γ → 1 Note that the congruence topology on E(R) is induced by that on G(R) but the arithmetic one may not be for all we know.

B.Sury The Congruence subgroup problem Step I. The profinite completion Rb of the ring R can be naturally identified Q topologically with m Rm where m runs through maximal ideals of finite index and Rm is the m-adic completion of R. Proof uses only Chinese remainder theorem. Step II. For some N > 0, every element of G(Rb) is a product of at the most N elements of the set Sb = {eα(t): t ∈ Rb, α ∈ Φ}. The proof of this reduces to local rings by step I. For such rings, one reduces to the big cell and to G over the residue field.

B.Sury The Congruence subgroup problem Step III. The congruence completions G(R) and E(R) are both equal to G(Rb). Further, there is N > 0 so that each element of this congruence closure is a product of at the most N elements of the set S which is the closure in congruence topology of {eα(t): α ∈ Φ, t ∈ R}. This just follows from step II.

B.Sury The Congruence subgroup problem Step IV. The profinite and congruence topologies on E(R) induce the same topology on each Uα(R). This means that for each normal subgroup N of finite index in E(R), there is an ideal of finite index in R such that eα(I ) ⊂ N ∩ Uα(R) for each root α. The proof involves explicit computations with commutation relations and is, therefore, case by case.

B.Sury The Congruence subgroup problem The proof of the theorem proceeds similarly to the number theoretical case above.

Since π gives isomorphisms from U\α(R) to Uα(R), and since we have a continuous homomorphism eα : Rb → G(Rb) = G(R), one has isomorphisms ecα : Rb → U\α(R) for each α of topological groups.

Define, for each maximal ideal m of finite index in R, the subgroup Γbm of Γb which is generated by the image of Rm under ecα for varying α. Then, Q i j (i) [ecα(s), ebβ (t)] = e\iα+jβ (Nαβij s t ) for α 6= −β and s, t ∈ R;b 0 (ii) Γbm and Γbm commute element-wise in Γb.

B.Sury The Congruence subgroup problem Combining this with an old result of M.Stein asserting that the kernel of π restricted to Γbm is in its center for any maximal ideal m of finite index, the centrality of Ker(π) in Γb can be deduced using some observations about topological groups. We remark that in the number-theoretic situation there is an action of G(k) on C which makes it easier; here, we needed Stein’s result as well as the bounded generation property of E(R) mentioned in step III.

B.Sury The Congruence subgroup problem One may pose an analogue of the congruence subgroup problem for the groups Aut(Fn), where Fn is the free group of rank n. Note that this group does not have faithful finite-dimensional representations if n ≥ 3 but does possess such representations when n = 2. M.Asada, K-U.Bux, M.Ershov and A.Rapinchuk have proved also an analogue of the congruence subgroup property for the group Aut(F2). The problem is still open for n ≥ 3.

B.Sury The Congruence subgroup problem For any finitely generated, residually finite group G, one may pose the problem for Aut(G) as follows. Note that Aut(G) ia also finitely generated and residually finite. Note also that G injects into its profinite completion Gb and Aut\(G) is finitely generated as a . For any normal subgroup N of finite index in G, consider the subgroup

ΓN := {σ ∈ Aut(G): σ(N) = N, σ|G/N = I } of finite index in Aut(G). Note that if N is taken to be characteristic, then this is a normal subgroup.

B.Sury The Congruence subgroup problem One may ask if every subgroup of finite index in Aut(G) contains Γ (N) for some N as above.

As before, one may define the topologies Ta, Tc and get a continuous homomorphism π from the completion Aut\(G) to Aut(G). We note that Aut(G) = Aut(Gb) where the latter is the group of continuous automorphisms. Thus the problem is : Is π : Aut\(G) → Aut(Gb) is injective?

B.Sury The Congruence subgroup problem Look at G = Fn for any n ≥ 2. It is known that Aut(Fbn) is not finitely generated as a topological group; hence π cannot be surjective unlike the classical situation!

M.Asada had shown that any normal subgroup of finite index in Aut(F2) which contains Int(F2), contains some ‘congruence’ subgroup Γ (N). Later, the other 3 authors re-interpreted and simplified Asada’s proof and showed that the injectivity of π holds for Aut(F2) without any assumption.

B.Sury The Congruence subgroup problem The proof involves explicit calculations in the free groups. We merely mention the fact that the proof of the congruence subgroup property for Aut(F2) uses strongly the fact that Out(F2) = GL2(Z) is virtually free and, very interestingly, this fact is what prevents the latter group from having the congruence subgroup property!

B.Sury The Congruence subgroup problem Coming to the classical cases of Dedekind rings, the centrality/finiteness of C(S, G) (for the cases where it is conjectured to be finite) is still open for two very important cases where new ideas may be required but these cases have a lot of relevance to the theory of automorphic forms. These two cases are: (i) G = SL(1, D) for a division algebra with center k, and (ii) G = SU(1, D) where D is a division algebra with a center K which is a quadratic extension of k and D has a K/k-involution.

B.Sury The Congruence subgroup problem Structure of infinite C(S, G)

Let us turn to the case when Serre’s conjecture has a negative answer; that is, when C(S, G) is infinite. One theme which has been exploited in proving finiteness of C(S, G) is : If C(S, G) is not finite, it has to be really huge. This can be used to show that if C(S, G) is infinite, it is infinitely generated as a profinite group. Melnikov used results on profinite groups to prove that, for SL2(Z), one has C({∞}, SL2) =∼ Fˆω, the free profinite group of countably infinite rank. This proof uses the existence of a central element of order 2 in SL2(Z).

B.Sury The Congruence subgroup problem In joint work with Alec Mason, Alexander Premet and Pavel Zalesskii (appeared in Crelle’s Journal), we completely determine the structure of C(S, G) when G(OS ) can be identified with a lattice in G(kv ) for some nonarchimedean completion kv . We use group actions on trees and profinite analogues as well as a detailed analysis of unipotent radicals in every rank 1 group over a local field of positive characteristic, to deduce the following structure theorem :

B.Sury The Congruence subgroup problem Theorem

Let G, k, S be as above, with S-rank (G) = 1. Then,

(i) if G(OS ) is cocompact in G(kv )(in particular, if char. k = 0), then C(S, G) =∼ Fˆω, and

(ii) if G(OS ) is nonuniform (therefore, necessarily char. k = p > 0), then C(S, G) =∼ Fˆω t T , a free profinite product of a free profinite group of countably infinite rank and of the factor T which is a free profinite product of groups, each of which is isomorphic to the direct product of 2ℵ0 copies of Z/pZ.

B.Sury The Congruence subgroup problem A surprising consequence of the theorem is that C(S, G) depends only on the characteristic of k when G(OS ) is a lattice in a rank 1 group over a nonarchimedean local field. As a by-product of the above result, we also obtain the following surprising result which is of independent interest :

B.Sury The Congruence subgroup problem Theorem

Let U be the unipotent radical of a minimal kv -parabolic subgroup of G where G is as in (ii) of the above theorem. Then either U is abelian or is automatically defined over k.

B.Sury The Congruence subgroup problem We note that although the profinite completion of a free group of finite rank r is the free profinite group of rank r, the profinite completion of a free group of countably infinite rank is not the group Fˆω appearing here.

The profinite completion is much larger and, the group Fˆω can be constructed from a countably infinite set X by looking at the free group F (X ) and at only those normal subgroups of finite index in F (X ) which contain all but finitely many elements of X . One further distinction between the case of free profinite groups and those of (discrete) free groups is that closed subgroups of free profinite groups may not be free profinite.

For instance, Zp is a closed subgroup of Zˆ.

B.Sury The Congruence subgroup problem We recall the two open problems which may be termed outstanding here. The first is the finiteness of the C(S, G) in Serre’s conjecture is still very much open for the groups of type An with k-rank 0. Solving them would lead to applications to the theory of automorphic forms. These G, S are described as follows :

B.Sury The Congruence subgroup problem (i) Let D be a division algebra with center k and the k-group G is SL(1, D). For S with S-rank(G) ≥ 2, is C(S, G) finite? (ii) Let D be a division algebra with center K which is a quadratic extension of k and suppose D admits an involution of the 2nd kind whose fixed field is k. Then the k-group G is SU(1, D). For S so that S-rank(G) ≥ 2, is C(S, G) finite?

B.Sury The Congruence subgroup problem The second question is when k is a number field, S consists only of the archimedean places, and G is a k-group such that S-rank of G is 1, then what is the structure of C(S, G) ? Recently, it has been proved by Lubotzky that

C(S, G) is finitely generated as a normal subgroup of G\(OS ). However, the structure is unknown as yet. This is unknown excepting the case of SL(2, Z); our theorem above deals with all the cases where S contains a nonarchimedean place.

B.Sury The Congruence subgroup problem Clopen subgroups

I don’t give any more details of the proof of the above results. However, I discuss a recent result which leads to a simplification in the proof of one of the principal results there. Specifically Zel’manov’s solution of the restricted Burnside problem for topological groups is no longer required there. The result we will prove is already stated as a lemma by Raghunathan in 1986 but he merely gave a very brief sketch of the proof and it was not completely clear that it works. We work out a detailed proof especially for the benefit of group-theorists who are not experts in algebraic groups. By applying the lemma to S-arithmetic lattices in G of K-rank one, where char(K) 6= 0 and |S| = 1, we provide a lower estimate for the number of subgroups of a given index in such a lattice which are not S-congruence.

B.Sury The Congruence subgroup problem Raghunathan’s Lemma: Let K be a global field and let S be a finite nonempty set of places of K containing all those of archimedean type. Suppose that G is connected, simply-connected and K-simple with strictly positive S-rank. Let Γ be an S-arithmetic subgroup of GK , the K-rational points of G. If N is any noncentral subgroup of GK which is normalized by Γ then the closure of N in the S-congruence topology is also open.

B.Sury The Congruence subgroup problem Applying this to the classical case of an S- which occurs as a lattice Λ in GKv , where char(K) 6= 0, G has K-rank one and S = {v}, we prove a result on the ubiquity of finite index subgroups of Λ which are not S-congruence.

To describe it, recall that associated with GKv is its Bruhat-Tits tree T . Bass-Serre theory shows how a presentation for Λ can be inferred from its action on T , via the structure of the quotient graph Λ\T .

B.Sury The Congruence subgroup problem It is known that the first Betti number of Λ\T , b1(Λ\T ), is finite. We have:

Theorem. Let Fr be the free group on r generators, where r = b1(Λ\T ) and let f (r, n) denote the number of index n subgroups of Fr . Let nc(Λ, n) be the number of S-noncongruence subgroups of index n in Λ. Then there exists a constant n0 = n0(Λ) such that, if n > n0, then

nc(Λ, n) ≥ f (r, n).

Moreover, if r ≥ 1, then for all n > n0, there exists at least one normal, S-noncongruence subgroup of index n in Λ.

B.Sury The Congruence subgroup problem In some important special cases, it is possible prove an explicit version of Raghunathan’s lemma in an elementary way. Here K has any characteristic and S is arbitrary.

Let H be a subgroup of SL2(O(S)). Recall that the order of H, o(H), is the O(S)-ideal generated by all h12, h21, h11 − h22, where (hij ) ∈ H. For each O(S)-ideal q, define

 12q , char(K) = 0 ψ(q) = q4 , char(K) 6= 0

Lemma. Let N be a noncentral normal subgroup of SL2(O(S)), with o(N) = n(6= {0}). Let q be any nonzero O(S)-ideal. If n0 = n + q, then

0 SL2(ψ(n )) ≤ N.SL2(q).

B.Sury The Congruence subgroup problem Theorem.

Let N be a noncentral normal subgroup of SL2(O(S)). Then \ N¯ = N.SL2(q) = N.SL2(ψ(n)), q6={0}

where n = o(N). Under further restrictions, this can be improved. For example, from the results of Mason it follows that, if o(N) is prime to 6, then \ N¯ = N.SL2(q) = N.SL2(n). q6={0}

In particular, if o(N) = O(S), then N¯ = SL2(O(S)).

B.Sury The Congruence subgroup problem We indicate now some pieces of the proofs of our results on the structure of the congruence kernel. We recall what a free profinite product is and also some important facts about profinite groups which we need.

The free profinite product G1 t · · · t Gn of profinite groups G1, ··· , Gn can be defined by an obvious universal property but it can also be constructed as the completion of the free product G1 ∗ · · · ∗ Gn with respect to the topology given by the collection of all normal subgroups N of finite index in the free product such that N ∩ Gi is open in Gi for each i.

B.Sury The Congruence subgroup problem One has the profinite counterpart of Schreier’s formula :

Let H be an open subnormal subgroup of a free profinite group Fˆr of rank r. Then, H is a free profinite group of rank 1 + (r − 1)[Fˆr : H]. In general, one says that a profinite group ∆ satisfies Schreier’s formula if the following holds. Let d(∆) denote the smallest cardinality of a set of generators of ∆ converging to 1. Here, a subset X of a profinite group ∆ is said to be a set of generators converging to 1 if every open subgroup of ∆ contains all but finitely many elements in X and < X > is dense in ∆. Then, ∆ is said to satisfy Schreier’s formula if, for every open normal subgroup H, one has

d(H) = 1 + (d(∆) − 1)[∆ : H].

Free profinite groups satisfy Schreier’s formula.

B.Sury The Congruence subgroup problem One also uses the following results : Let ∆ be a finitely generated pro-p group, for some prime p. Then, ∆ is free pro-p if, and only if, it satisfies Schreier’s formula. On the other hand, a free pronilpotent group ∆ with d(∆) ≥ 2 never satisfies Schreier’s formula. Note that this is not true if d(∆) = 1. In fact, any direct product ∆ = ∆1 × ∆2 with d(∆) ≥ 2 does not satisfy Schreier’s formula. This is one of the key (although simple to prove) facts we use in our proof.

B.Sury The Congruence subgroup problem Let Fˆr be the free profinite group of finite rank r ≥ 2. Let H be a closed normal subgroup of Fˆr of infinite index. If Fˆr /H does not satisfy Schreier’s formula, then H is a free profinite group of countably infinite rank. The analogue of Kurosh subgroup theorem holds for open subgroups of free profinite products but not for closed subgroups. For closed normal subgroups of free profinite products, one has a similar statement which involves projective profinite groups in place of free profinite groups.

B.Sury The Congruence subgroup problem Projective profinite groups are closed subgroups of free profinite groups. We use the following result : Let F be a free profinite group of infinite rank and let P be a projective profinite group with d(P) ≤ d(F ). Then the free profinite product F t P =∼ F . Note that the above result also shows that free factors of a free profinite group need not be free profinite groups. Another fact we use is : Let F be a free profinite group of rank r (any cardinal number) ≥ 2. Then, a closed normal subgroup H of infinite index with d(F /H) < r must be a free profinite group and it has rank max(r, N0). Some of these facts that we use are proved using the notion of a profinite graph of profinite groups and a notion of the of such a profinite graph of groups.

B.Sury The Congruence subgroup problem Crucial for our purposes is the following surprising result. It ensures that the structure of any C(U) can deduced from a detailed description of one particular U. (This result in fact holds for any global field.) If U is nonabelian, then U is defined over k.

B.Sury The Congruence subgroup problem Proof.

Let P = P(K) and U = U(K). Obviously, P is a parabolic subgroup of G and U = Ru(P), the unipotent radical of P. Choose a maximal torus T of G contained in P and let Φ denote the root system of G relative to T . Let ∆ be a basis of simple roots in Φ.

Adopt Bourbaki’s numbering of simple roots and denote by αe the highest root of Φ with respect to ∆. Let α∨ denote the coroot corresponding to α ∈ Φ, an element in X∗(T ) ⊂ X∗(G). ∨ × Recall α (K ) is a 1-dimensional torus in T . As usual, we let Uα = {xα(t) | t ∈ K} denote the root subgroup of G corresponding to α. Given x ∈ G we denote by ZG (x) the centraliser of x in G.

B.Sury The Congruence subgroup problem We discuss here one main case; viz., we suppose that G is not of type C3; then G is of type A2, A3, D4 or D5. A quick look at the Tits indices reveals that P is G-conjugate to the normaliser in G of the 1-parameter unipotent subgroup U . αe From this it follows that in our present case the derived subgroup of U has dimension 1 as an algebraic group and coincides with the centre Z of U. Moreover, Z is G-conjugate to U . αe

B.Sury The Congruence subgroup problem By our assumption, the derived subgroup [U, U] contains an element u 6= 1. Then u ∈ [U, U] ⊂ [U(kv ), U(kv )] ⊂ [U, U] = Z. Since the subgroup Z is T -invariant, the preceding remark implies that there is a long root β ∈ Φ such that U = Uβ .

Then u = xβ (a) for some nonzero a ∈ K.

We claim that the centraliser ZG (u) is defined over k. To prove this claim it suffices to verify that the orbit morphism g 7−→ gug −1 of G is separable.

The latter amounts to showing that the of ZG (u) coincides with gu := {X ∈ g | (Ad u)(X ) = X }.

B.Sury The Congruence subgroup problem After adjusting ∆, possibly, we can assume that β = αe.

For each α ∈ Φ we choose a nonzero vector Xα in gα = Lie Uα and let t denote the Lie algebra of T .

0 Denote by g the K-span of all Xγ with γ 6∈ {±αe} and set g(α) := g ⊕ t ⊕ g . e −αe αe 0 Clearly, g = g ⊕ g(αe).

B.Sury The Congruence subgroup problem 0 It is easy to observe that both g and g(αe) are (Ad u)-stable and one can choose X such that αe (Ad u)(X ) = X + a[X , X ](∀ X ∈ g0). γ γ αe γ γ

Since αe is long, standard properties of root systems and Chevalley bases imply that if γ ∈ Φ is such that γ 6= −α and γ + α ∈ Φ, then [X , X ] = λ X for e e αe γ γ αe+γ some nonzero λγ ∈ K. u 0 From this it follows that g ∩ g coincides with the K-span of all Xγ such that γ 6∈ {±αe} and αe + γ 6∈ Φ.

On the other hand, the commutator relations imply that each such Xγ belongs u 0 to Lie ZG (u). Therefore, g ∩ g ⊂ Lie ZG (u).

B.Sury The Congruence subgroup problem The differential dαe is a linear function on t. Since G is simply connected, we have that dα 6= 0 and [X , X ] 6= 0 (it is e αe −αe well-known that dαe = 0 if and only if p = 2 and G is of type A1 or Cn). As (Ad u)(h) = h − a(dα)(h)X (∀ h ∈ t), e αe this implies that gu ∩ g(α) = g ⊕ ker dα. e αe e u u But then g ∩ g(αe) ⊂ Lie ZG (u), forcing g ⊆ Lie ZG (u). x Since Lie ZG (x) ⊆ g for all x ∈ G, we now conclude that the group ZG (u) is defined over k.

◦ Hence the connected component ZG (u) is defined over k, too.

B.Sury The Congruence subgroup problem Let C denote the connected component of the centraliser Z (U ). G αe

The argument above shows that Lie C = Lie ZG (u). ◦ ◦ Since C ⊆ ZG (u) , we must have the equality ZG (u) = C. Then C is a k-group, hence contains a maximal torus defined over k, say T 0.

0 As ker αe ⊂ ZG (u), the torus T has dimension l − 1, where l = rk G. Let H denote the centraliser of T 0 in G. By construction, H is a connected reductive k-group of semisimple rank 1 containing U . αe

B.Sury The Congruence subgroup problem Since G is simply connected, so is the derived subgroup of H. As U is unipotent, it lies in [H, H]. αe Since 1 6= u ∈ G(k) ∩ [H, H], the group [H, H] is k-isotropic.

The classification of simply connected k-groups of type A1 now shows that ∼ [H, H] = SL2(K) as algebraic k-groups. As a consequence, u belongs to a k-defined Borel subgroup of [H, H]; call it B. Since u commutes with U , it must be that U = R (B). αe αe u

B.Sury The Congruence subgroup problem Let S be a k-defined maximal torus of B.

Since [H, H] is k-isomorphic to SL2(K), there exists a k-defined cocharacter × µ: K → [H, H] such that S = µ( ×), µ(t)x (t0) µ(t)−1 = x (t2t0)(∀ t, t0 ∈ ). K αe αe K

Then S is k-split in G, and hence it is a maximal kv -split torus of G (recall that G has kv -rank 1).

B.Sury The Congruence subgroup problem Since S normalises U , it lies in P. αe × As P is defined over kv , there exists a kv -defined cocharacter ν : K → P such that P = P(ν).

× × Since µ(K ) and ν(K ) are maximal kv -split tori in P, they are conjugate by an element of U. In conjunction with the earlier remarks this yields that rν = Int x ◦ sµ for some x ∈ U and some positive integers r and s. But then P(µ) = P(sµ) = P(Int x ◦ sµ) = P(rν) = P. Since µ is defined over k, so are P and U.

B.Sury The Congruence subgroup problem