DISCRETE SUBGROUPS OF LIE GROUPS

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1 DISCRETE SUBGROUPS OF LIE GROUPS RONGGANG SHI Abstract. This is the course note of an introductory course on discrete subgroups of Lie groups. The contents contains basic facts about discrete subgroups, Borel density theorem, arithmeticity theorem of Margulis. We will review linear algebraic groups and give various examples. 1. Basics on lattice Locally compact topological spaces are assumed to be second countable and Hausdorff. Let X be a topological space, the Borel σ-algebra of X is the σ-algebra generated by open subsets of X. Let G be a locally compact topological group and λ G be a left Haar measure on G. Given a discrete subgroup Γ, a fundamental domain for Γ G is a measurable subset F of G such that for every element g G the set Γg F contains exactly one point. Now we show that a fundamental domain always exists. We cover G by F = {F 1, F 2, } where K i is open and any two points of K i represents different elements of Γ G. We can construct a fundamental domain for Γ G inductively by taking K 1 = F 1, K 2 = F 2 ΓF 1, K 3 = F 3 Γ(F 1 F 2 ),. Then it is easy to see that F = K i is a fundamental domain for Γ G. Remark: In some cases we allow the fundamental domain contain measure zero boundaries so that the shape of it behaves better geometrically. Definition 1.1. We call a discrete subgroup Γ of a locally compact group G lattice if for any fundamental domain F of Γ, the measure λ G (F ) is finite. Remark: if for some fundamental domain F one has λ G (F ) < then Γ is a lattice. To see this, take another fundamental domain F, then F = γ Γ (γf F ) which is a countable partition of F with F = γ Γ γ 1 (γf F ). Since λ G is left invariant, we have λ G (F ) = λ G (F ). Examples: 1. Z n is a lattice in R n. 2. Γ = SL(n, Z) is a lattice in G = SL(n, R). We will prove this and learn more about the quotient space later. 1

2 2 RONGGANG SHI Definition 1.2. The group G is said to be unimodular if the left Haar measure λ G is also right invariant. Proposition 1.3. If a locally compact group G has a lattice, then G is unimodular. Proof. Take g G. The measure gλ G defined by gλ G (A) = λ G (Ag) is also a left-haar measure. By the uniqueness of Haar measure, gλ G (A) = cλ G (A) for some c > 0. Let F be a fundamental domain for Γ, then F g is also a fundamental domain, therefore λ G (F ) = gλ G (F ) = cλ G (F ), hence c = 1. Since g is arbitrary, we see that G is unimodular. Proposition 1.4. Let H and L be closed subgroups of the Lie group G. Suppose G = HL and K = H L is compact, then under the appropriate normalization, for any f C c (G), we have f(g)dλ G (g) = f(hl)δ G (l) 1 dλ H (h) dρ L (l) G where Δ G (l) = detad(l). S L Sketch of Proof. Take h 0 H, l 0 L and let μ 1 = Δ G (l) 1 λ H ρ L. Then (1.1) dμ 1 (h 0 h, ll 1 0 ) = Δ(l 0)Δ G (l) 1 dλ H (h) ρ L (l) = Δ(l 0 )dμ 1 (h, l). By the uniqueness of left Haar measure for L ˆG where ˆG is the dual group of G, the measure (as μ 1 ) satisfying (1.1) is unique up to scalars. The group K acts on H L by sending (h, l) (hk, k 1 l). The quotient space H L/K of this action is diffeomorphic to G via the multiplication map. Therefore λ G induces a measure on H L/K and a measure μ 2 on H L by integration over K orbit. It is readily checked that μ 2 also satisfies (1.1). Therefore μ 1 and μ 2 are proportional and the pushforward of them are what we want. Example: Take the group {( ) } y 0 N = xy y 1 : x = 0. Then the λ G = ydxdy and the right Haar measure ρ G = dxdy. In fact solvable group is the only type of groups that is not unimodular. Now we describe the sequence of points in Γ G that tends to infinity. Here tends to infinity means leaves every compact subset when the index is sufficiently large. According to our assumption at the beginning, a locally compact space is metrizable by Urysohn s theorem. So a point tends to infinity if and only if it has no convergent subsequence. Theorem 1.5. Let Γ be a lattice in G and let π : G Γ G be the projection map. Let (g n ) be a sequence in G, then (π(g n )) has no convergent subsequence if and only if there exists a nonneutral sequence (γ n ) such that g 1 n γ n g n e where e is the neutral element of G.

3 Discrete Subgroup of Lie Group 3 Remark: The idea behind this theorem is that injectivity radius of a point tends to zero when the point tends to infinity. Proof. Suppose π(g n ) tends to infinity. Take a sequence (U k ) that forms a topological basis at the identity e. Finite measure of Γ G implies that for any k, there exists I k such that for n I k, the map π restricts to g n U k is not injective. So there exists γ n,k Γ different from e such that gn 1 γ n,k g n Uk 2. Now we may assume I k is increasing and take γ n = γ n,k for appropriate k. Conversely, suppose gn 1 γ n g n e and π(g n ) has a convergent subsequence. Then by passing to a subsequence and replacing g n by some γ n g n we may assume g n converges. Then γ n e and by the discreteness of Γ, we must have γ n = e eventually. Theorem 1.6. Let G be a locally compact group and Γ be a lattice in G. Let H be a closed subgroup of G. For x Γ G = X, let H x is the stabilizer of x in H. If H x is a lattice in H, then xh is closed. Remark: If G is a real or p-aidc Lie group, then the Borel σ-algebra of H x H is isomorphic to that of xh via the natural map. So the assumption H x is a lattice is equivalent to the orbit xh supports a finite H invariant measure. Proof. Suppose x = Γg for some g G, then H x = g 1 Γg H. Suppose xh is not closed. Then there is a sequence (h n ) in H such that the sequence (Γgh n ) in X converges to some element in X xh. So the sequence (H x h n ) H x H tends to infinity. Hence by Theorem 1.5 there exists a sequence (γ n ) (Γ ghg 1 ) such that g 1 γ n g = e and h 1 n g 1 γ n gh n e. Therefore Theorem 1.5 again implies that the sequence Γgh n does not converge in X. This contradiction completes the proof. 2. The space X = SL(m, Z) SL(m, R) In this section we study the discrete subgroup Γ = SL(m, Z) of G = SL(m, R) in detail. Later after defining algebraic groups some results of this section will be generalized to the so called arithmetic subgroups and the proof will usually be omitted there Unimodular lattice of R m. Here we consider R m as row vectors with the usual Euclidean inner product and norm. Let e 1,..., e m be the standard basis where e i only has i-th column entry nonzero and equals one. The SL(m, R) acts naturally on it from right by matrix multiplications. Let λ be the Lebesgue measure of R m. We call a discrete subgroup of R m with covolume one unimodular lattice. The discrete subgroup Z m is a unimodular lattice. For each g SL(m, R), the module Z m g = {vg : v Z m } is also a unimodular lattice. Note any unimodular lattice of R n is a free Abelian subgroup. If we use the basis of it as row vectors, then the determinant is the covolume of the lattice, hence equals one. Therefore we have:

4 4 RONGGANG SHI Lemma 2.1. The group SL(m, R) acts transitively on the set of unimodular lattices of R m. The stabilizer of Λ = Z m is SL(m, Z), so the space X can be identified with the space of unimodular lattices. A nonzero vector v Z m is called primitive if v = cw for any integer c > 1 and w Z m. It can be proved that SL(m, Z) acts transitively on the set of primitive vectors Iwasawa Decomposition. Let N SL(m, R) be the group of lower triangular matrices, and let A = {diag(a 1,, a m ) SL(m, R) : a i > 0}. Lemma 2.2. The product map from N A SO(m, R) to SL(m, R) is a diffeomorphism. Proof. Let g SL(m, R) and let v 1,, v m be row vectors of g. Then the standard Gram-Schmidt orthogonal process for these vectors shows that there exists a lower triangular matrix (a ij ) with a ii > 0 such that (a ij )g = k SO(m, R). Take na = (a ij ) 1 we get the Iwasawa decomposition of g. It is easy to see that this process is smooth and it is the inverse of the multiplication map Siegel Set. Let n N and let n ij be the ij-th entry of n. For s > 0 we use N s to denote the subset of N with n ij s for all i > j. Let a A and let a i be i-th diagonal the entry of a. For t > 0 we take A t = {a A : a i /a i+1 t}. A siegel set for SL(m, R) is a subset of the form S(s, t) = N s A t SO(m, R). Lemma 2.3. The measure of S(s, t) is finite with respect to the left Haar measure on SL(m, R). Proof. It is not hard to see that N, A, SO(m, R) are unimodular and NA is a closed subgroup. Let dn, da, dk be some fixed Haar measures on the three groups. Proposition 1.4 implies that where Δ(a) = det(ad NA (a)). suffices to show that dλ G (nak) = Δ(a) 1 dndadk A t Δ(a) 1 da <. We leave the concrete calculation to the reader. Since N s and SO(m, R) are compact, it Theorem 2.4. For s 1/2 and t 2/ 3, the Siegel set S(s, t) contains a fundamental domain for SL(m, Z) Therefore the group SL(m, Z) is lattice in SL(m, R). We need a couple of lemmas to deal with N and A part respectively. Lemma 2.5. The group N = (N Γ)N 1/2. One can see this by considering elements of N with essential entries parallel to the diagonal.

5 Discrete Subgroup of Lie Group 5 Lemma 2.6. Let g G and let nak be the Iwasawa decomposition of g. If e 1 g is the shortest nonzero vector in Z m g, then a 1 /a 2 2/ 3. Proof. According to Lemma 2.5, possibly by replacing g by γg with γ (N Γ) we can assume that n ij 1/2. We know that e 1 g 2 = a 2 1 e 2 g 2 = a 2 e 2 + n 21 a 1 e 1 a a2 1. The result follows from this inequality. Proof of Theorem 2.4. We prove it by induction on the number m 2. Let g G and v 0 Z m with gv 0 = min{ gv : v Z m 0}. The vector v 0 must be primitive, so there exists γ 1 Γ such that e 1 γ = v 0. Let g = γg, then the e 1 g is a shortest nonzero vector for Z m g. Now Lemma 2.5 and 2.6 proves the Theorem for m = 2 In general let g = nak be the Iwasawa decomposition. Then an induction process for the submatrix of na with order n 1 and Lemma 2.6 complete the proof Mahler s Compactness Criterion. Theorem 2.7. A subset M of X is relatively compact iff there exists ε > 0 such that the length of any nonzero vector of any lattice Λ M is bigger than ε. Proof. Let φ : X R be the map that sends each lattice to the length of length of the shortest nonzero vector. Since φ is continuous, if M is relatively compact, so is φ(m) considered as a subset of R >0, hence we can find the ε. Conversely, we show that every element x of M can be represented by an element of a bounded subset of S(1/2, 2/ 3). It follows from the proof of Theorem 2.4 that we can find a representative of x in S(1/2, 2/ 3) such that a 1 is the absolute value of the shortest nonzero vector of the lattice Z n g. So a 1 ε and hence M is relatively compact. 3. Borel density theorem Let G be a connected semisimple Lie group, then G = G 1 G r where G i is a closed connected simple Lie group. This decomposition is unique up to permutations. We say G has no non-compact factors if each G i is non-compact. The proof of the following Theorem is taken from [3]. Theorem 3.1. Let G be a connected semisimple Lie group without compact factors and let Γ be a lattice in G. Suppose that V is a finite dimensional real vector space and T be a right action of G on V. Then any T (Γ)-invariant linear subspace W of V is also G-invariant.

6 6 RONGGANG SHI Proof. Step one:some observations. Without loss of generality we assume: 1. W is one dimensional. This is by considering exterior products of order dim(w ) for the space V. 2. The linear span of W T (G) is V. Suppose the W is not G-invariant, or equivalently (since G is connected and semisimple) the action of G is not trivial. Since G has no noncompact factors, T (G) GL(V ) must be unbounded. Let T be the induced action of G on P(V ). We show that Step two: There is a sequence (g n ) G and a map π : P(V ) P(V ) such that T (g n ) π pointwise and the image of π is a finite union of proper linear subvarieties. Let s fix a basis for V and consider GL(V ) as matrices. We take a sequence g n G such that the sequence T (g n ) is unbounded. There is a sequence of positive real numbers c n such that maximum value of the absolute value of the entries of c n T (g n ) is bounded above and below. Therefore there is a convergence subsequence and without loss of generality, we may assume c n T (g n ) converges to L End(V ) which is nonzero. Therefore for v ker L, vt (g n ) converges to L(v). Since T (g n ) is unbounded, we have c n 0. The determinant is continuous and det T (g n ) = 1 since we assume that G is semisimple, so L is degenerate. Therefore the image of L is a proper linear subvariety. Now we consider the sequence of linear maps T (g n ) ker L. No matter whether it is bounded, a similar process as above and the fact the dimension of the corresponding kernel is decreasing allows us to find the required transformation π and the subsequence. Step three: dynamics of the action G on P(V ). We take w to be the image of W in P(V ). The map G P(V ) which sends g wt (g) factors to a map Γ G P(V ). Let μ be the pushforward of the probability Haar measure on Γ G by this map. Let a sequence (g n ) and π be as in step two. We claim that the support of μ is contained in the image Y of π. Let us fix a metric on P(V ) and denote by D(x) the distance from x P(V ) to Y. Note that μ is G invariant, so we have P(V ) D(x) dμ(x) = P(V ) D(xT (g n )) dμ(x) 0. This proves the claim. Note that for every g G, the point wt (g) is in the support of μ. So wt (G) Y and G must permutes the irreducible linear subvarieties of Y. Since G does not have finite index subgroups, Y has only one connected component and hence Y is a proper linear subvariety. This contradicts assumption 2 in Step One and completes the proof.

7 Discrete Subgroup of Lie Group 7 4. Linear Algebraic groups In this section we give a non-intrinsic definition of algebraic groups defined over fields of characteristic zero. Here non-intrinsic refers to that we always embed it to GL(n, C) which has a structure of an affine variety defined over Q. A field k in this note should be between the bottom field Q and the top field C Some concepts. Let D is the determinant polynomial in n n matrix and let A = k[x 11,, X nn, D 1 ] which is the ring of k-regular functions on GL(n, C). We call a subgroup G of GL(n, C) given by the common loci of a finite subset of the ring A linear algebraic k-group or linear algebraic group defined over k. So a linear algebraic group G is an abstract group with a variety structure which is encoded in the ring of k-regular functions on G denoted by k[g]. In fact k[g] = A/a where a is the ideal of functions in A vanishing on the algebraic set G. Given two linear algebraic k-groups G and H, a k-rational morphism is a group homomorphism φ : G H such that for any f k[h], the pullback function f φ k[g]. For a field extension k /k, the k points of G is G(k ) = G GL(n, k ). The Lie algebra g of G is a C vector subspace of M(n, C) with a k-structure defined by g = {X M(n, C) : exp tx G for t R}. Here the k-structure of the vector space means a k vector space g(k) = g M(n, k) with the property that g(k) k C = g. It can be proved that that g(k) {X M(n, k) : df e (X) = 0 for f F} where e denotes the identity. Most of the time, as in the examples below we get an equality. The algebraic group G is called semisimple if g is a semisimple Lie algebra over C. This is equivalent to g(k) is a semisimple Lie algebra over k. If k R and we give M(n, R) the usual Euclidean topology, then G(R) is a real Lie group and the algebraic Lie algebra defined above is the same as the one for Lie groups. Examples: 1. SL(n) = {g GL(n) : det(g) = 1}. The Lie algebra is sl(n) = {X M(n) ( : tr(x) = ) 0}; 0 In 2. Let J n = be the standard symplectic matrix. The I n 0 symplectic group is the isotropy group of the corresponding symplectic form. We denote it by Sp(2n) = {g GL(2n) : gj n g t = J n }. It has Lie algebra sp(n) = {X M(2n) : XJ n + J n X t = 0}. 3. SO(m, n) = {g SL(m + n) : gi m,n g t = I m,n } where m n 0 and I m,n = diag(i m, I n ). It has Lie algebra so(m, n) = {X M(m + n) : XI m,n + I m,n X t = 0}. 4. We may generalize example 3 as follows. Let φ be a nondegenerate quadratic form in n variables with coefficients in the field k. Then the linear

8 8 RONGGANG SHI algebraic group SO(φ) = {g SL(n) : φ g = φ} is defined over k. Let B φ M(n, k) be the symmetric matrix for φ. The Lie algebra of SO(φ) is so(φ) = {X M(n) : XB + BX t = 0} Representation. A k-rational representation of a linear algebraic kgroup G is a k-rational morphism of algebraic groups from G to some GL(V ) where V is a finite dimensional vector space with a k-structure. Equivalently, Under a basis of V (k), the entries of the representation (in GL(n)) are regular functions of G defined over k. We need two kinds of representations. The first is the right regular representation ρ of G on k[g]. For each f k[g] and g, h G, let (ρ h f)(g) = f(gh). It is not hard to see that have f(, t X it Y tj, ) = s b s c s k[g] k[g] where b s, c s k[g] and the expression is minimal. Therefore ρ h f = c s (h)b s. Hence the C-linear span of ρ(g)f is finite dimensional vector space with a k-structure. It is not hard to see that any finite elements of C[G] is contained in some finite dimensional k-rational right regular representation of G. Another representation is the adjoint representation. This time the space is g. For each g G and X g, we have Ad g X = gxg 1. It is easy to see that Ad is a k-rational representation of G Restriction of scalars. Let K/k be a finite field extension of degree d. Fix a k-basis of K, the multiplication map embeds K into M(d, k) as algebraic ring defined over k. In this embedding the image is a d-dimensional k-subspace so as a variety it is given by d 2 d linear forms. If we view the coefficients of a polynomial f K[, X ij, ] as an element of M(d, k), we get d 2 polynomials in k[gl(nd)]. If G is defined by a finite set F K[, X ij, ], then a finite set of induced polynomials from elements of F and polynomials restricts each entry X ij to be in K M(n, d) defines an algebraic k-group which is denoted by Res K/k G. It is not hard to see from this construction that the natural map R : G(K) Res K/k G(k) is an isomorphism and Res K/k is functorial from the category of linear K-groups to the category of linear k-groups. Let σ : K C be a k-embedding. For f F, take f σ to be the polynomial with every coefficient c changed to σ(c). Let G σ be the group defined by F σ = {f σ : f F}. Then K[G σ ] = K[G] K σ(k) and G σ is a K-group with K-structure comes from the isomorphism σ : K σ(k). Let Σ(K/k) be the set of k embedding of K into C. Recall that after restriction of scalars K is an affine k-variety. Its K-points are K k K = σ Σ(K/k) σ(k) as K-algebraic rings via the morphism φ which sends a b to σ(ab). Therefore by comparing the equations one can see that the group Res K/k G = σ Σ(K/k) Gσ. It is an isomorphism of algebraic groups over K if K/k is Galois. Let R : Res K/k G G be the natural map defined

9 Discrete Subgroup of Lie Group 9 over K corresponding to the projection to the identity σ. Then R maps Res K/k G(k) isomorphically to G(K) and the inverse is given by R above. From this we can see that the pair then (Res K/k G, R) has the following universal property: for any linear k-group H and K-morphism φh G, there is a unique k-morphism φ : H R K/k G such that R φ = φ. So Res K/k is right adjoint to the functor of extension of scalars from k to K between the category of corresponding linear algebraic groups. Example: 3. Let m n 0 with m 1. We can view GL(m + n, C) as ( Im 0 real points of the Q-group R Q[i]/Q GL(m + n). Take I m,n = 0 I n ). Then the isotropy group of the Hermitian form defines a Q-group SU(m, n) whose real points are given by SU(m, n, R) = {g SL(m + n, C) : gi m,n g t = I m,n } Chevalley s Theorem. Theorem 4.1. Let G be a linear algebraic group defined over k and let H be a closed subgroup defined over k. Then there is a k-embedding φ : G GL(V ) and a line L V defined over k such that H = {g G : φ(g)l = L}. Proof. Let F = {f 1,, f m } k[g] which general the ideal of functions in k[g] vanishing on H. We take W to be a finite dimensional subspace of C[G] defined over k and invariant under the right regular representation of G. Let L F be the minimal H invariant k-rational subspace of V. The following should be clear: H = {g G : ρ(g)w = W }. Now we only need to modify the above representation to get a rational line and injectivity. The former is by considering exterior products of order dim(l). The latter is by adding an injective direct factor, e.g. the identity map would work k-rank. The k-rank of a semisimple (or reductive) algebraic group defined over k is the common dimension of the maximal k-split torus. The k-rank of above examples can be explicitly calculated. 1. For any field k, the k ranks of SL(m) and Sp(2m) are m 1 and m respectively. 2. Let φ be a nondegenerate quadratic form in m variables defined over k. The Witt index of φ is the dimension of a maximal k-isotropic subspace of k m. This number is equal to the k rank of SO(φ). In particular the k rank of SO(m, n) where m n 0 is n. 3. The Q rank of SU(m, n) where m n 0 is equal to its R rank and it is equal to n.

10 10 RONGGANG SHI 5. Arithmetic Group 5.1. Remarks concerning topology. Let G be a linear algebraic group defined over k and K is a local field containing k. The group G(K) with the smallest topology such that K[G] are continuous with respect to the locally compact topology of K is a locally compact topological group. This is our default topology for the group G(K). When we think of G as an algebraic variety we will emphasize Zariski. For example we say a set M G(k) is Zariski dense if the closure of M under the Zariski topology (denoted by M Zar ) is G. There is a coarser Zariski topology on G coming from the k-structure and we call it Zariski k-topology. It is known that M Zar is kclosed (hence defined over k since chark = 0). Hence the Zariski closure of M under the Zariski k-topology is the same as that in the geometric Zariski topology defined here A theorem of Borel and Harish-Chandra. We have seen that SL(n, Z) is a lattice in SL(n, R). In general we take GL(N, Z) = {g GL(N, Q) : g and g 1 have entries in Z}. If G GL(N) is an algebraic group defined over Q, then we take G(Z) = G(Q) GL(N, Z). The proof of the following Theorem and its generalizations can be found in [2] or [6] 4.6. Theorem 5.1. Let G be a semisimple algebraic group defined over Q, then G(Z) is a lattice in G(R). Moreover, G(Z) is cocompact if and only if the Q-rank of G is zero. We call a non-discrete absolute value of Q a place of Q. Let R be the set of inequivalent places of Q. It is well-known that R = {, prime numbers}. Let S be a finite set of places of Q containing. The S-adic integer is defined by Z S = {x Q : x Z p for p S}. Corollary 5.2. Let G be a semisimple algebraic group defined over Q, then G(Z S ) is a lattice in v S G(Q v). Moreover, G(Z S ) is cocompact if and only if the Q-rank of G is zero. We can extend the above results to number fields. Theorem 5.3. Let G be a semisimple algebraic group defined over a number field k. Let S be a finite set of places of k containing the Archimedean ones. Then G(o k,s ) is a lattice in v S G(k v). Moreover, G(o k,s ) is cocompact iff k-rank of G is zero. Remark: It is not hard to see that if we get rid of some compact factors from v S G(k v), then we still get a lattice. We call such kind of lattices arithmetic. It can be proved that Proposition 5.4. An arithmetic lattice is finitely presented.

11 Discrete Subgroup of Lie Group Examples of lattice. 1. The first group are noncompact lattices: The group has the property that the Q rank is equal to R rank and bigger than zero. SL(m, Z) in SL(m, R) where m 2; Sp(2m, Z) in Sp(2m, R) where m 2; SO(m, n, Z) in SO(m, n, R) where m n 1; SU(m, n, Z) in SU(m, n, R) where m n Let the quadratic form φ = x x2 n 1 + 2x2 n where n 4 be defined over k = Q( 2). Then φ is anisotropic over k and hence the k rank of SO(φ) is positive. So SO(φ, Z[ 2]) is a non cocompact lattice in SO(φ, C). 3. Let φ = x x2 n 1 2x 2 n where n 3 be a quadratic form over k = Q( 2). Then SO(Z[ 2]) is a cocompact lattice in SO(n 1, 1, R). 4. The group SL(2, Z[ 2]) is a non cocompact lattice in SL(2, R) SL(2, R). 5. Let p be a prime number. The group SL(2, Z[1/p]) is a non cocompact lattice in SL(2, R) SL(2, Q p ) Density of lattice. Let G be an algebraic group defined over k R and let K be a local field containing K. Theorem 5.5. Let G be a Zariski connect semisimple algebraic group defined over R. Suppose that G(R) has no compact factors and Γ is a lattice in G(R). Then Γ is Zariski dense. Proof. Let H be the Zariski closure of Γ, then H is an algebraic subgroup of G defined over R. It follows from Chevalley s theorem that there is a R-rational representation T : G GL(V ) and a one-dimensional R-rational line L V such that H = {g G : T (g)l = L}. Since Γ G(R) is a lattice, the Borel density theorem 3.1 implies that G(R) H. By assumption that G is connected so G = H Superrigidity. Margulis superrigidity theorem says that under some conditions, a group homomorphism from a lattice in a semisimple Lie group extends to a continuous homomorphism of the whole group. We will state a simple version that will be used in the next section. Theorem 5.6 (Margulis). Let G be an almost simple algebraic R-group with R-rank greater than or equal to 2 and let Γ be lattice in G(R). Let H be a k-simple adjoint algebraic group where k is either R or Q p. Suppose there is a homomorphism T : Γ H(R) such that T (Γ) is Zariski dense in H. Then (1) if k = R and H(R) is not compact, there is a unique algebraic group homomorphism φ : G H defined over R such that φ Γ = T. (2) if k = Q p, the group T (Γ) is relatively compact in H(Q p ). 6. Arithmeticity of lattice in simple Lie groups of higher rank Let G be a semisimple algebraic R-group and let Γ be lattice in G(R). We can ask whether Γ comes from a Q-structure of G.

12 12 RONGGANG SHI Definition 6.1. Let G be a Zariski connected semisimple algebraic group defined over R and let Γ be a lattice in G(R). We call the lattice Γ arithmetic if there is a semisimple Q-group H and an R-epimorphism τ : H G such that ker τ(r) is compact and τ(h(z)) is commensurable with Γ. Now the question is whether every lattice in a semisimple algebraic group arithmetic? We only discuss the case for absolutely almost simple R-groups. Theorem 6.2 (Margulis). Let G is an absolutely simple algebraic group defined over R. If the real rank of G is greater than or equal to 2, then any lattice Γ in G(R) is arithmetic. Here real rank at least two is essential since in this case we have super rigidity theorem of Margulis Theorem 5.6. In real rank one case, some groups such as SO(n, 1) and SU(m, 1) where n 2 and m = 2, 3 there are nonarithmetic lattices both cocompact or non cocompact. It is still unknown whether there are non-arithmetic lattices in SU(m, 1) for m > 3. In other real rank absolutely simple algebraic groups all the lattices are arithmetic. Before going into the proof, let us see some examples of G(R): SL(m, R) for m 3; Sp(2n, R) for n 2; SO(m, n, R) for m n 2; SU(m, n, R) for m n 2. The remaining parts of this section are devoted to the proof of Theorem 6.2. We make some reductions and assumptions first. We will assume that G is adjoint or equivalently that G has trivial center. The general case actually follows from this case by considering algebraic simply connected covering group of H. We take k to be the field generated by T r(adγ) for γ Γ. Lemma 6.3. There is a faithful m-dimensional representation φ : G GL(V ) such that under some basis of V we have φ(γ) GL(m, k) for every γ Γ. Proof. Let T C[G] be the function of trace of the adjoint representation. Let ρ denote the right regular representation of G. It follows from 4.2, that V = Span C ρ(g)t is finite dimensional. Since Γ is a lattice in G(R), Borel density theorem implies that there is a finite set {γ 1,, γ m } such that {ρ(γ i )T : 1 i m} is a basis for V. Now it suffices to prove two facts: (1) Under this basis ρ(γ) has entries in k; (2) ρ is injective. (1): The Borel density theorem and a induction process shows that there exists α 1,, α m Γ such that the vectors ( ρ(γi )T (α 1 ),, ρ(γ i )(α(m) ) t = ( T (α1 γ i ),, T (α m γ i ) ) t : 1 i m in k m are linear independent. Since we have ρ(γ) GL(m, k). ρ(γ) ( T (α j γ i ) ) = ( T (α j γ i γ) ),

13 Discrete Subgroup of Lie Group 13 (2) Take g H such that ρ(g)(t ) = T. That is T (hg) = T (h) for all h G. Let the ring R be the C linear span of Ad(h) End(g) for h H. Then g is a semisimple R module, that is g = L n i i where various L i are pair vise non isomorphic simple R modules. Now Corollary A.4 implies that B = i End C (L i ). Since all the r R satisfies T r(rad(g)) = T r(r), we have Ad(g) is the identity transformation. In view of Lemma 6.3, we assume from now on in addition that that G is defined over k and Γ G(k). Lemma 6.4. Let σ : k R be an inclusion of field. Then either σ(γ) is relatively compact, or there is an algebraic group isomorphism φ : G G σ defined over R such that γ Γ is sent to γ σ. Proof. Since G is connected and absolutely simple, so is G σ. Let σ : Γ G σ be the map induced by σ. If σ(γ) is not relatively compact, case (1) of superrigidity theorem implies that σ extends to an R morphism of algebraic groups which has to be an isomorphism since the groups involved are absolutely simple. Lemma 6.5. The lattice Γ is a finitely generated group. The proof of this lemma uses representation theory. Roughly speaking the G(R) has Kazhdan s property T since G is R simple and the real rank of G is bigger than or equal to 2. As a lattice in G(R), the group Γ inherits the property T. Now is a standard fact that a discrete group with property T is finitely generated. Lemma 6.6. The the field k generated by all T r(ad(γ)) for γ Γ is a number field. Proof. According to Lemma 6.5, we can find a finite generating set of Γ. Then the field k is the same as the field generated by the entries of this generating set. Therefore the field k is a finitely generated field. So it suffices to prove that for each γ Γ, the trace of Ad(γ) is an algebraic number. Let σ : k R be an embedding of field. If σ(γ) is relatively compact, then T r(ad(σ(γ))) is bounded by the dimension of of the group G. Otherwise Lemma 6.4 shows that it is equal to T r(ad(γ)). Anyway it does not depend on the choice of the embedding σ. Of course this is impossible if T r(ad(γ)) is transcendental over Q. Proof of Theorem 6.2. Recall that we assume that G has trivial center and there is number field k such that G is an algebraic k-group in GL(n) and Γ GL(n, k). We can get a Q-group Res k/q G by restriction of scalars and let R : Res k/q G G be the canonical surjective morphism defined over k. We know that Res k/q G(Q) is mapped isomorphically to G(K) by R and we take R to be the inverse of this isomorphism. Let H 1 be the Zariski closure of R (Γ) in Res k/q G, then H 1 is defined over Q. By possibly passing to a finite index subgroup of Γ, we can assume that H 1 is connected. By dividing

14 14 RONGGANG SHI H 1 by the radical and then the center (both of them are defined over Q), we get a Q-group H. Since G is absolutely simple, the map R H1 factors to a surjective map φ : H G. Let Γ be the image of Γ 1 in H, then Γ is mapped isomorphically under φ to Γ. We claim that it suffices to prove the following two facts using the superrigidity theorem of Margulis. (1) ker(φ) is compact; (2) There is a finite index subgroup Γ Γ such that Γ H(Z). Assume we (1) and (2) holds for the moment. The surjectivity and (1) implies that Γ is lattice. According to Theorem 5.1, we know that H(Z) is a lattice in H(R). So the group Γ in (2) must have finite index in H(Z). In view of Definition 6.1 and (1), we are done. (1): Since H has trivial center, as an algebraic group defined over R it is a direct product of R simple groups and one factor of them is isomorphic to G. Therefore H = G H i where H i is R simple algebraic group. It suffices to prove that each H i (R) is a compact group. Let π i : H H i be the projection map, then π i R is Zariski dense in H i. Suppose that H i (R) is noncompact, then by case (1) of super rigidity Theorem 5.6, the map π i R extends to a algebraic group morphism f i : G H i. Take i = 1 for simplicity, then the Zariski closure of Γ is contained in the image of the map G H which sends g G to (g, f 1 (g), ). This contradicts the Zariski density of the Γ in H. (2) Since Γ is finitely generated and H is semisimple, the primes in the denominator of entries of γ Γ are finite. So it suffices to prove that for each prime p there is a finite index subgroup Γ p of Γ such that Γ p H(Z p ). We consider the natural map ΓtoΓ. Note that H is adjoint and by possibly factorizing H into Q p simple factors, we can use (2) of superrigidity theorem to get that Γ is relatively compact in the open compact subgroup H(Z p ). So the desired result follows. Appendix A. Noncommutative algebra Let R be a ring and let M be an R module. We say M is a simple module if the only R submodules of M are 0 and M. If M is a direct sum of simple submodules then M is said to be semisimple. The set S = End R (M) has a natural ring structure where the multiplication is the composition of maps. The R module M is also an S module. Shur s lemma tells us that S is a division ring if M is a simple R module. If we consider r R as maps of M, then we get a natural map from R to End S (M). This map is called the structure map of M. Lemma A.1. Let M be a semisimple R module, then any submodule M of M is a direct summand of M. We use Zorn s lemma to choose a maximal submodule of M of M such that M M = 0 and show that M M = M.

15 Discrete Subgroup of Lie Group 15 Proposition A.2 (Density theorem for semisimple modules). Let M a semisimple R module and let S = End R (M). Then for any φ End S (M) and any finite set x 1,..., x m M there is r R such that φ(x i ) = rx i for any i. Proof. Suppose m = 1. By Lemma A.1 there exists a R-submodule M such that M = Rx 1 M. Then the projection map π : M Rx 1 belongs to S. Since φ is an S-module homomorphism, it commutes with π. Therefore φ maps x 1 to Rx 1 and we are done. In general case we take V = M m and x = (x 1,, x m ) V. We claim that φ m End EndR (V )(V ). The existence of the required r follows from the m = 1 case for the R module V. Let k be a field. A k algebra is a ring R with a fixed subring k contained in the center of R. In this case an R module is a k vector space V with a k-algebra map R End k (V ). The following is a direct consequence of the above proposition. Corollary A.3 (Burnside). Let R be a k algebra and M is a simple R module. Suppose dim k (M) < and S = End R (M) = k, then the structure map R End k (M) is surjective. Remark: if k is algebraically closed, then End R (M) is a division ring, hence equals k. Corollary A.4. Let R be a C algebra. Suppose V = m i=1 V n i i where each V i is a simple R module. Suppose dim k V i < for each i and V i is not isomorphic to V j if i = j. Then the image of R in End C V is B = m i=1 End C(V i ) where each End C (V i ) acts on V n i i as scalars. Proof. It is clear that the image of R in End C (V ) is contained in B since V is semisimple. So we only need to prove the surjectivity. By the remark following Proposition A.3 we know that End R (V ) = m i=1 GL(n i, C). Now every element of R acts on M as linear transformations which is determined by a finite basis set. The density of the ring in Proposition A.2 gives what we need. References [1] A. Borel Linear algebraic groups, Second edition. Graduate Texts in Mathematics, 126. Springer-Verlag, New York, [2] A. Borel and Harish-Chandra, Arithmetic subgroups of algebraic groups. Ann. of Math. (2) C535. [3] H. Furstenberg, A note on Borel s density theorem, Proc. Amer. Math. Soc. 55 (1976), no. 1, 209C212. [4] B. Farb and K. Dennis, Noncommutative algebra, Graduate Texts in Mathematics, 144. Springer-Verlag, New York, [5] G. A. Margulis, Discrete subgroups of Semisimple Lie groups. Ergebnisse der Mathematik und ihrer Grenzgebiete (3) [Results in Mathematics and Related Areas (3)], vol. 17, Springer-Verlag, Berlin, 1991.

16 16 RONGGANG SHI [6] V. Platonov and A. Rapinchuk, Algebraic Groups and Number Theory, Pure and Apllied Mathematics vol 139. [7] M. S. Raghunathan, Discrete subgroups of Lie groups, Springer, New York, [8] R. Zimmer,Ergodic theory and semisimple groups, Monographs in Mathematics, 81. Birkhauser Verlag, Basel, School of Mathematical Sciences, Xiamen University, Xiamen , PR China address: