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Lattices in Lie Groups Lattices in Lie groups Dave Witte Morris Lattices in Lie groups 1: Introduction University of Lethbridge, Alberta, Canada http://people.uleth.ca/ dave.morris What is a lattice subgroup? ∼ Dave.Morris@uleth.ca Arithmetic construction of lattices. Abstract Compactness criteria. During this mini-course, the students will learn the Classical simple Lie groups. basic theory of lattices in semisimple Lie groups. Examples will be provided by simple arithmetic constructions. Aspects of the geometric and algebraic structure of lattices will be discussed, and the Superrigidity and Arithmeticity Theorems of Margulis will be described. Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 1 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 2 / 18 A simple example. Z2 is a cocompact lattice in R2. 2 2 R is a connected Lie group Replace R with interesting group G. group: (x1,x2) (y1,y2) (x1 y1,x2 y2) + = + + Riemannian manifold (metric space, connected) is a cocompact lattice in G: (discrete subgroup) distance is right-invariant: d(x a, y a) d(x, y) + + = Every pt in G is within bounded distance of . group ops (x y and x) continuous (differentiable) Γ + − g cγ where c is bounded — in cpct set 2 = Z is a discrete subgroup (no accumulation points) compact C G, G C . Γ Every point in R2 is within ∃ ⊆ = G/ is cpct. (gn g ! γn , gnγn g) 2 bdd distance C √2 of Z → Γ ∃{ } → = 2 Z is a cocompact lattice is a latticeΓ in G: Γ Γ ⇒ 2 in R . only require C (or G/ ) to have finite volume. Γ Replace R2 with interesting group G. Usual way to make a lattice:Γ let Z-points GZ. = { } = Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 3 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction Γ CIRM, Jan 2019 4 / 18 2 2 T 2 2 2 Z is a cocompact lattice in R . x I1 2 x x x x , = 1 − 2 − 3 Replace R2 with interesting group G. For k k matrix S (invertible, symmetric) and x Rk: × T ∈ Eg. G Isom( hyperbolic n-space ) SO(1, n) QS(x) : x S x (quadratic form) n= n 1 2 2 2 ( 2 = 2 2 2 2 h x R + x1 x2 x3 xn 1 + QI(x) x1 x2 xk x = { ∈ | − − − ··· − = } = +2 +···2 =2 ∥ ∥ 2 QIm,n (x) x1 xm xm 1 xm n = +··· − + −···− + More general: G SO(m, n) SO(I ) = = m,n (special orthogonal group) SL(k, R) k k, det 1 where I diag 1 1 1 1 1 1 . ={ × = } m,n ( , ,..., , , ,..., ) SO(S) : g SL(k, R) gT Sg S = − − − = { ∈ | = } g SL(k, R) Q (gx) Q (x) Example = { ∈ | S = S } SO QS(x) . 1 x1 x1 = T T Q is defined, over- Q if coefficients are in Q. x I1,2 x x ⎡ 1 ⎤ ⎡x2⎤ x1 x2 x3 ⎡ x2⎤ S = − = − k 1 x3 x3 QS is isotropic over Q if v (Q )×, Q(v) 0. ⎢ − ⎥ ⎢ ⎥ ( ) ⎢ − ⎥ ∃ ∈ = 2 2 2 2 2 2 x x⎣ x ⎦ ⎣x ⎦ x x . ⎣ ⎦ T 1 = 1 − 2 − 3 = 1 − 2 − 3 Exer. S′ cT ST SO(S′) T − SO(S)T SO(S) * + = ⇒ = - Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 5 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 6 / 18 Z2 is a cocompact lattice in R2. More general Replace R2 with interesting group . Assume G SO Q(x) with Q(x) defined over Q. G = Usual way to make a lattice: let GZ. G/GZ is cpct, ! Q(x)- is not isotropic over Q. = Proof ( ) for SO(1, 2). Eg. G SO(m, n) SO (Im,n) G/GZΓ is not cpct ⇒ $ = = ⇒ compact Z3 closed →0 ∉ Z3 . unless m 0 or n 0. (G cpct GZ finite boring) G/GZ G( )× G( )× ⇒ ⇒ ⇒ ⇒ = = 2 SO(1, 2) Q SO Q(x) with Q(x) x1x2 x3. More general - T = + 1/21/20, 1-/21/20 01/20 1/2 1/20 I 1/2 1/20 1/200 Assume G SO Q(x) with Q(x) defined over Q. ⎡ − ⎤ 1,2 ⎡ − ⎤ ⎡ ⎤ = 001 001= 001 is cpct ! is not isotropic over Q. ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ G/GZ , Q(x)- a⎣ diag(1/t,⎦ t,1⎣) Q(a x)⎦ Q(x)⎣ a ⎦ G. • t = ⇒ t = ⇒ $ t ∈ v : (1, 0, 0) Z3 & a v (1/t, 0, 0) →0 . Eg. Here is a cocompact lattice in SO(1, 2): • = ∈ t = → 2 2 2 Let G SO (7x1 x2 x3) SO(1, 2). Fact. G/GZ always has finite vol: GZ is a lattice in G. = − − - Then GZ is a cocompact lattice in G. (Since 7 ≠# #.) + (if Q(x) is defined over Q) Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 7 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 8 / 18 More general More general Assume G SO Q(x) with Q(x) defined over Q. = Assume G SO Q(x) with Q(x) defined over Q. G/GZ is cpct, ! Q(x)- is not isotropic over Q. = G/GZ is cpct, ! Q(x)- is not isotropic over Q. Part 1: G/GZ is closed in SL(k, R)/ SL(k, Z). Proof ( ) ⇐ Proof. G/GZ & SL(k, R)/ SL(k, Z): gGZ ' g SL(k, R) Suppose gnγn g with gn G and γn SL(k, Z). n → ∈ ∈ Proof has 2 parts: For x Z : Q(γnx) Q(gnγnx) Q(gx), ∈ =n → 1 but Q(γ x) Q(Z ) Z. G/GZ is closed in SL(k, R)/ SL(k, Z). n ∈ ⊂ So Q(γ x) Q(gx) is eventually constant: (because Q(x) is defined over Q) n = Q(γnx) Q(γ x). 2 G/GZ is bounded in SL(k, R)/ SL(k, Z). = ∞ Then Q(gx) Q(γnx) Q(γ x). (because Q(x) is not isotropic over Q) = 1 = ∞1 So Q(g γ− x) Q(γ γ− x) Q(x). · ∞1 = ∞ · ∞ = Therefore gγ− SO Q(x) G. ∞ ∈ = Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 9 / 18 Dave Witte Morris (U of Lethbridge) Lattices, 1: Introduction- CIRM, Jan 2019 10 / 18 Part 1: G/GZ is closed in SL(k, R)/ SL(k, Z). More general Part 2: G/GZ is bounded in SL(k, R)/ SL(k, Z). Assume G SO Q(x) with Q(x) defined over Q. = Lemma (Mahler Compactness Criterion) G/GZ is cpct, ! Q(x)- is not isotropic over Q. Let C SL(k, R). ⊂ The image of C in SL(k, R)/ SL(k, Z) is bounded Eg. 7x2 x2 x2 is not isotropic over Q. $ 1 − 2 − 3 →0 is not an accumulation point of C Zn. ⇐⇒ Proof ( ). Theorem of Number Theory ⇒ $ n Q(x) isotropic over R with at least 5 variables Spse cnzn →0 and cnγn h. Since hZ is discrete, 1 → $→ $ Q(x) is isotropic over Q. and h(γ− z ) c z →0 , we have z →0 . n n ≈ n n ≈ n = ⇒ Converse is an exercise (for k 2). Generalization of fact that every positive integer is a sum of 4 squares. = Part 2 of the proof. n n So More general never provides a cocompact lattice gnzn G(Z )× Q(gnzn) Q(zn) Q (Z )× Z× ∈ ⇒ = ∈ ⊆ in SO(m, n) with m n 5 (unless m 0 or n 0). Q(gnzn) 0 gnzn 0. + ≥ = = ⇒ ̸→ ⇒ ̸→ , - Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 11 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 12 / 18 Cocompact lattices in SO(m, n) Replace R2 with interesting G: simple Lie group. Q(x) x2 x2 αx2 αx2 αx2, α √2, SO(m, n) g SL(k, R) gT I g I = 1 + 2 − 3 − 4 − 5 = ={ ∈ | m,n = m,n } G SO Q(x) SO(2, 3), (lattice: GZ) = - T T GZ , (Z- Zα)-points. SU(m, n): change R to C and g to g∗ g = [α] = + = (lattice: GZ Zi) Then is a cocompact lattice in G. + Γ Sp(m, n): change C to H Idea ofΓ proof. σ (a bα) : a bα. (Galois aut) (lattice: ) + = − GZ Zi Zj Zk GHZ Gσ : SO(Qσ) SO(x2 x2 αx2 αx2 αx2) SO(5). + + + = = = 1+ 2+ 3+ 4+ 5 - SL(n, R), SL(n, C), SL(n, H) Map ω ' (ω, ωσ ) embeds Z[α] & R R. ⊕ Sp(2n, R), Sp(2n, C), SO(n, H) (1, 1σ ), (α, ασ ) are lin indep so image discrete. finitely many “exceptional grps” ( ) So image of in G Gσ is discrete. Cocpct lattice! E6, E7, E8, F4, G2 × Q(x) not isotropic over Q[α]: Theorem (Borel and Harish-Chandra) Q(v) 0Γ Qσ (vσ ) 0 vσ 0 v 0. Assume G simple Lie group, defined over Q. = ⇒ = ⇒σ = ⇒ = Can mod out compact group G . Then GZ is a lattice in G. Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 13 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 14 / 18 Hint: You may assume the following basic facts Exercise about volume (for subsets of G): Assume is a lattice in G. vol(Eg) vol(E) for all g G. = ∈ 1 Show that every finite-index subgroup of is a If E is compact, then vol(E) < . ∞ latticeΓ in G. If U is open and nonempty, then vol(U) > 0. 2 Γ If E F, then vol(E) vol(F). Show that if is a cocompact lattice in G, then ⊆ ≤ vol(E F) vol(E) vol(F).
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