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? ∼ [email protected] 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.

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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). every finite-index subgroup of is a cocompact ∪ ≤ + If E and F are disjoint, and are either open or lattice in G. Γ closed, then vol(E F) vol(E) vol(F). 3 Show that the following are equivalent:Γ ∪ = + Furthermore, G is locally compact. This means that 1 G is compact. every closed ball Br (g) is compact. 2 is finite. 3 vol(G) is finite. Rem. Since G has a lattice, we also have vol(gE) vol(E) for = Γ all g G. However, for some Lie groups, it is only true that ∈ vol(E) vol(Eg) (or only that vol(E) vol(gE)), not both. = = Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 15 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 16 / 18

Deﬁnition Exercise An element u of SL(k, Q) is unipotent if the Prove ( ) of Mahler Compactness Criterion for k 2. ⇐ = following equivalent conditions are true: 1 1 is the only eigenvalue of u (in C). Hint: Let cn be a sequence of points in C. { } 2 (u I)k 0. Since C SL(2, R), there is a sequence γn in − = ⊆ 1 { } 3 SL(2, Z), such that cnγn 0 is bounded. (Why?) u is conjugate in SL(k, Q) to matrix that is upper-triangular with only 1s on the diagonal. By passing to a subsequence,( ) we may assume 1 2 c γ converges to some v1 R . n n 0 ∈ Exercise Note that v1 ≠0. (Why?) ( ) Assume G SO QS(x) , with QS deﬁned over Q. Now show there is a sequence in 1 , such = γn′ 01∗ Show that if has a unipotent element other 0 { } 2 GZ , - that cnγnγ′ converges to some v2 R . n 1 (∈ ) than I, then G/GZ is not compact. This implies c γ γ′ [v1 v2]. ( n) n n → k So cn SL(2, Z) has a convergent subsequence. Hint: Find u G and v1,v2 (Q )×, such that uv1 v1 { } ∈ ∈ T =T and uv2 v1 v2. Also note that (uv) S (uw) v Sw. = + = Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 17 / 18 Dave Witte Morris (U of Lethbridge) Lattices 1: Introduction CIRM, Jan 2019 18 / 18 Group theory = the study of symmetry

Example Lecture 2: Geometric aspects Symmetries of a tessellation (periodic tiling)

Dave Witte Morris University of Lethbridge, Alberta, Canada http://people.uleth.ca/ dave.morris [email protected]∼

Symmetric spaces symmetry group Z2 Z2 classical simple Lie groups = " Real rank and Q-rank Thm (Bieberbach,Γ 1910). tess of RΓ n, Zn. ∀ " Other spaces yield groups that are moreΓ interesting.

Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 1 / 12 Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 2 / 12

2 Other spaces yield groups that are more interesting. Eg. Tess’ns of hyperbolic plane h . (symmetric space) G SO(1, 2) SL(2, R). " " Rn is a symmetric space: y KaleidoTile homogeneous: x every pt looks like all other pts. x,y, isometry x ! y. 0 ∀ ∃ x0 reﬂection through a point y0 (x′ x) is an isometry. = −

Rem. Let G Isom(X). X homog G transitive c Wikipedia = ⇒ ⃝ X G/K with K Stab (p) cpct. = cocompact lattice SL(2, Z) SO(2, 1)Z ⇒ = = G = " Symmetry group of tessellation of X is lattice in G Γ Tessellations of otherΓ symmetric spaces if tiles are compact (or ﬁnite volume). correspond to lattices in other interesting groups.

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Other interesting groups G: simple Lie group. Proposition T SO(m, n) g SL(k, R) g Im,n g Im,n G simple Lie group SL(n, R), SO(m, n), etc ={ ∈ | = } = T = (lattice: GZ) such that G G (always true after conjugating) T T T = 1 SU(m, n): change R to C and g to g∗ g K k G k k− = max’l compact subgrp, = = { ∈ | = } (lattice: GZ Zi) ( SO SO SO , etc). + K (n), (m) (n) G/K is a" symmetric space. × Sp(m, n): change C to H ⇒ (lattice: GZ Zi Zj Zk GHZ ) + + + = Sketch of proof. SL(n, R), SL(n, C), SL(n, H) K cpct, so G-inv’t Riemannian metric on G/K. Sp(2n, R), Sp(2n, C), SO(n, H) ∃ T 1 Define φ(gK) (g )− K, so φ has order 2. = finitely many “exceptional grps” (E6,E7,E8,F4,G2) Average over 1, φ to make metric φ-invariant. { } So φ Isom(G/K). Theorem (Borel and Harish-Chandra) ∈ IK is an isolated fixed pt of φ DI φ(v) v Assume G simple Lie group, defined over Q. ⇒ K = − φ is the reflection through IK. Then GZ is a lattice in G. ⇒

Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 5 / 12 Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 6 / 12 2 SL(2, Z) acts on upper half-plane h by Defn. rankR G = max dim of R-split torus az b ab + . = max dim of ﬂat in G/K. cd z cz d ∗ = + " et 0# 2t So t i e i y-axis = geodesic. Eg. rankR SO(m, n) min(m, n) m if msectional curvature) connected subgroup of G that is conjugate " rankR G 1 via SL(k, R) to a group of diagonal matrices. = 20 " G SO(1, n), SU(1, n), Sp(1, n), “F4− ” Then p G/K, such that Ap is a ﬂat: " ∃ ∈ isometrically embedded copy of Rm in G/K. Always: G/K is CAT(0). (non-positive sectional curvature)

Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 7 / 12 Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 8 / 12

Defn. Let G SO Q(x) and GZ. Example = = n subspace V is totally isotropic if Q(V ) 0 . Let SO(1, n)Z. Then h not compact. & ' = \ Γ ={ } hn has ﬁnitely many cusps. Eg. For QI3,5 : (x, y, z, x,y,z,0, 0) . { } Γ \ Γ rankQ( ) = max dim tot isotrop Q-subspace. Γ Eg. rankQ( Γ) 0 Q(x) not isotropic over Q = ⇒ G/ compact. ⇒ Look at G/Γ from a large distance. Γ Γ Limit = “star" of ﬁnitely many rays. dimension of limit 1 rankQ( ). ∴ = = Limit is a point. Theorem (Hattori) dimension of limit 0 rankQ( ). Γ ∴ = = Asymptotic cone of G/ (or G/K) is a simplicial \ complex whose dimension is rank . This limit is the asymptotic coneΓof G/ . Q( ) Γ Γ Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 9 / 12 Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 10 / 12 Γ Γ

T T 1 Theorem (Hattori) Exer. Assume G G and K k G k k− . =T 1 ={ ∈ | = } Deﬁne φ(gK) (g )− K and let p I K G/K. Dimension of asymptotic cone of G/ is rankQ( ). = = ∈ Show:

1 For general G, rankQ GZ = max dimΓ of Q-split torusΓ : φ has order 2. connected subgroup of G that is conjugate 2 p is an isolated fixed point of φ: via SL(k, Q) to a group of diagonal matrices. If φ(g K) g K for all i, and g K p, i = i i → then g K for all large i. i ∈ Eg. rankQ SL(n, Z) n 1. 3 D φ(v) v for all v T (G/K). = − p = − ∈ p Assuming that φ is an isometry, this implies that φ SL(3, R)/ SL(3, Z): is the reflection through p (see below). Rem. Let γ be a geodesic through p. Since φ is an isometry, we know φ γ is the unique geodesic in the direction ◦ Dpφ γ′(0) γ′(0). Therefore φ γ(t) γ( t) for all t The defns of rankQ GZ agree when G SO Q(x) . = − = − = (and& all γ).' This means that φ is the& reflection' through p. & ' Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 11 / 12 Dave Witte Morris (U of Lethbridge) Lecture 2: Geometric aspects Marseille, Jan 2019 12 / 12 Lecture 3: Algebraic aspects We ignore finite groups. Dave Witte Morris University of Lethbridge, Alberta, Canada Assumps. http://people.uleth.ca/ dave.morris ∼ G = simple Lie group SL(n, R), SO(m, n), etc. [email protected] " = lattice in G. is infinite (i.e., G not compact). Γ Normal subgrps (center, commutator subgrp, and others) G is connected (will justify later). Γ Free subgroups and torsion-free subgroups Finitely presented and residually finite Borel Density Theorem

Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 1 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 2 / 13

Congruence Subgroups Proposition (Selberg’s Lemma) is virtually torsion-free : We prove two basic properties of . finite-index subgrp ′ < , ′ is torsion-free. ∃ Γ (no nontrivial elements of finite order) Proposition Γ Γ Γ Γ is residually finite : Proof.

γ ×, homo φ: F (finite), φ(γ) ≠ e. Define ρ3 : SL(n, Z3) and let ′ ker(ρ3). Γ ∀ ∈ ∃ → → = Proof. It suffices to show ′ is torsion-free. Γ Γ k Let γ ′, writeΓ γ I 3 T , T Γ 0 (mod 3). γ I ≠ 0, so (γ I) ≠ 0. Choose N " (γ I) . ∈ = + ̸≡ − ∃ − ij − ij γm (I 3kT)m Γ Ring homo Z Z yields SL(n, Z) SL(n, Z ). = + N N Γ k m 2k 2 → → I m(3 T) 2 3 T Let ρN : SL(n, ZN) be the restriction to . = + + +··· → I 3k mod" 3#k ℓ 1 if 3ℓ Since ZN finite, obvious that SL(n, ZN) is finite. mT ( + + ) m ≡ + k ℓ 1 ℓ 1 | By choice ofΓ , ≠ I , so ≠ΓI. I (mod 3 + + ) if 3 + " m. N ρN(γ)ij ρN( )ij ρN(γ) ̸≡

Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 3 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 4 / 13

Finitely generated Borel Density Theorem Proposition Obs. Classical simple Lie grp SL(n, R), SO(m, n), ... is finitely generated : S with S finite. is defined by polynomial equations: Zariski closed. = ⟨ ⟩

Proof for cocompact. Eg. SL(n, R) g Matn n(R) det(g) 1 Γ Γ = { ∈ × | = } G C with C open, and compact closure. and det g is a polynomial function of gi,j. = g1,1 g1,2 Let S γΓ Cγ C ≠ . ( discrete, so S finite!) det g1 1 g2 2 g1 2 g2 1. = { ∈ | ∩ ∅ } g2,1 g2,2 = , , − , , Γ $ % Let γ . Since G is connected,Γ chain quadratic ∈ Γ ∃ Eg. Entry of TI is sum of . C,Cγ1,Cγ2,...,Cγ Cγ, each intersects next. g m,ng gi,jgk,ℓ polynomial n = ± Γ 1 1 T So γ1 S, γk γk− 1 S, γγn− S. g Im,n g Im,n is system of quadratic equations. ∈ 1− ∈ 1 ∈ 1 = Hence γ (γγn− )(γnγn− 1) (γ2γ1− )γ1 S . = − ··· ∈⟨ ⟩ Theorem of real algebraic geometry In fact, is finitely presented: Set of real solns of any system of poly eqns (# vars < ) ∞ γ1, γ2,...,γ w1,w2,...,w . has only finitely many connected components. 0⟨ m | r ⟩ Dave Witte MorrisΓ (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 5 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 6 / 13 Γ By ignoring finite group, assume G is connected. Proposition (Borel Density Theorem) Homo ρ : G GL(n, C) Proposition (Borel Density Theorem) → 1 ρ( ) fixes vector v ρ(G) fixes v. is Zariski dense in G: ⇒ 2 ρ( ) fixes subspace W ρ(G)fixes W . homo ρ : G GL(n, C), ρ( ) fixes vector v Γ ⇒ → Γ ρ(G) fixes v. CorollaryΓ ⇒ Γ Idea of proof for cocompact. Z( ) is finite. G/ compact, ρ( )v v ρ(G)v bounded. Proof. Γ = ⇒ Γ Let A be any R-split torus of G. 1 Define ρ : G GL Matk k(R) by ρ(g)T gTg− . EigenvaluesΓ of elementsΓ of A are real. → × = If z Z( ), then ρ( )z z. So ρ(G)z z. Since Av bounded, conclude that A fixes v. ∈ & = ' = Therefore z Z(G). This is a finite group. ∈ Since ρ(G) is simple, it is generated by R-split tori. Γ Γ So all of ρ(G) fixes v. Cor. is not virtually abelian (or solvable).

Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 7 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 8 / 13 Γ

Normal subgroups Cor. If rankR G 2, then [ , ] has finite index in . ≥ is residually finite. Can replace rankR G 2 withΓ Γ weaker assump thatΓ ≥ So many normal subgroups of finite index. G SO(1, n), SU(1, n). Γ ∃ ̸≈ Theorem (Margulis Normal Subgroups Theorem) Conjecture. For G SO(1, n), [ ′, ′] has infinite If rankR G 2, and N is a normal subgroup of , = ≥ index, for some finite-index subgroup ′. then either is finite or is finite. /N N Γ Γ Γ Eg. Torsion-free lattice in SO 1 2 is Cor. If rankR G 2, then [ , ] has finite index in . ( , ) Γ ≥Γ I.e., the abelianization of is finite. free group or surface group. Γ Γ Γ Abelianization is infinite. Proof. Γ Suppose not. Then [ , ] is finite. So [ , ] Z( ). Theorem. If rankR G 1, then has many normal ⊆ = Therefore, after passing to a finite-index subgroup, subgroups of infinite index. we may assume [ , Γ] Γ I . So is abelian.Γ Γ Γ (True for every Gromov hyperbolicΓ group.) = { } →← Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 9 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 10 / 13 Γ Γ Γ

Free subgroups Borel Density Theorem

Cor. connected, proper subgroup H. ̸⊆ ProofΓ (requires Lie theory). Let h be the Lie algebra of . This is a subspace of g. Proposition H Since H, we know normalizes H. contains a nonabelian free subgroup. ⊆ So h is -invariant for the adjoint representation. Therefore,Γ h must be GΓ-invariant, so H ▹ G. ProofΓ is an application of the Ping-Pong Lemma. Since GΓ is simple, we conclude that H G. = Cor. is not contained in any Zariski-closed, proper subgroup H of G: is Zariski dense. Γ . Proof. H has < componentsΓ H◦ . ∞ ⇒ ⊃ Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 11 / 13 Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 12 / 13 Γ Exercise Let a be a diagonal matrix in SL(n, R), and v Rn. ∈ If akv k Z is bounded, show av v. { | ∈ } = Exercise Show that if N is a ﬁnite, normal subgroup of , then N Z(G). ⊆ Hint: Show that the centralizer of N is a ﬁnite-indexΓ subgrp of , and apply the Borel Density Theorem.

Dave Witte Morris (U of Lethbridge) Lecture 3: Algebraic aspects Marseille, Jan 2019 13 / 13 Lattices in Lie groups 4: Rigidity Observation Group homomorphism φ: Zk Rd → Dave Witte Morris φ extends to a continuous homo φˆ : Rk Rd. ⇒ → University of Lethbridge, Alberta, Canada http://people.uleth.ca/ dave.morris Proof. ∼ [email protected] k Standard basis e1,...,ek of R . ˆ { } Deﬁne φ(x1,...,x ) x1 φ(e1) x φ(e ). k = +··· + k k (“linear trans can do anything to a basis") Margulis Superrigidity Theorem. Linear transformation Mostow Rigidity Theorem. homomorphism of additive groups ⇒ Margulis Arithmeticity Theorem.

Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 1 / 10 Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 2 / 10

Obs. Group homomorphism φ: Zk Rd → φ extends to a continuous homo φˆ : Rk Rd. ˆ Want. φ: G′ φ “extends” to φ: G G′. ⇒ → → ⇒ → Want. Group homomorphism φ: G′ Not alwaysΓ true: " → ˆ φ “extends” to a continuous homo φ: G G′. ⇒ → lattice in SL(4, R), φˆ(γ) φ(γ) for γ ˙ (finite-indexΓ subgroup) = = ∈ Let’s say that is strictly* superrigid. ψ: SL(4, R) PSL(4, R) SL(4, R)/ I , Γ → = {± } Γ ψ( ) if I ∉ (torsion-free), + − Not always true:Γ φ: ψ( ) . Γ Γ+ Γ G SL(2, R), SL(2, Z), φ: Z G′. ˆ : PSL 4 R SL 4 R = & →→ ⊆ φ ( , ) ( , ) ker φˆ ker φ is infinite, G is simple Γ →Γ →← ⊃ ˆ ˆ φ is trivial Γ φ φ is trivial.Γ Sometimes need to replace with a finite cover. ⇒ ⇒ & | →← G Γ Thm. n 3 SL(n, Z) strictly* superrig. ≥ ⇒ Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 3 / 10 Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 4 / 10

Defn. is strictly superrigid: φ: G′ Margulis Superrigidity Theorem ˆ ˆ → φ: G G′, φ(γ˙) φ(γ˙) for γ˙ ˙. is superrigid if G ≠ SO(1, n), SU(1, n): ⇒ ∃ → = ∈ ˆ ˆ Γ Γ φ: G′ φ: G G′, φ(γ˙) φ(γ˙) C. (C cpct) Theorem (Margulis" Superrigidity Theorem) Γ Γ → ⇒ ∃ → ∈ is strictly superrigid if G ≠ SO(1, n), SU(1, n) MostowΓ Rigidity Theorem" (weak form) and not cocompact. The lattice determines the group: Γ 1 2 G1 G2 if Z(Gi) I . CounterexamplesΓ for cocompact lattices: + ⇒ + = { } S diag(1, 1, α, α, α), α √2, Margulis Arithmeticity Theorem = − − − = Γ Γ G SO(S) SO(2, 3), Every lattice in G can be obtained by taking Z-points = + GZ is a cocompact lattice in G, unless G SO(1, n) or SU(1, n). = [α] & σ (a bα) : a bα, Warning: Some cocpct latts come from cpct . Γ + = − (G )Z σ : SO(Sσ ) SO(5) compact. × → + σˆ : G SO(5) # homo G compact group. Cor. List of all latts in G (except SO(1, n), SU(1, n)). → Γ →← → Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 5 / 10 Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 6 / 10 " " Superrigidity implies arithmeticity is a superrigid lattice in SL(n, R) and every matrix entry is an algebraic number. Let be a superrigid lattice in SL R . (n, ) Γ We wish to show SL(n, Z), ⊂ Second, show matrix entries are rational. Γi.e., want every matrix entry to be an integer. Recall. is generated by ﬁnitely many matrices. Γ First, let us show they are algebraic numbers. Entries of these matrices generate a ﬁeld extension

Suppose some γi,j is transcendental. of Q ofΓ finite degree. “algebraic number field" Then field auto φ of C with φ(γ ) ???. So SL(n, F). For simplicity, assume SL(n, Q) ∃ i,j = ⊂ ⊂ ab φ(a) φ(b) Define φ . Third, show matrix entries have no denominators. #cd$ = #φ(c) φ(d)$ Γ Γ Actually, show denominators are bounded. So φ: " GL(n, C) is a group homo. (Then finite-index subgrp has no denoms.) → Superrigidity: φ extends to φˆ : SL(n, R) GL(n, C). Since is generated by finitely many matrices, " → There are uncountablyΓ many different φ’s, only finitely many primes appear in denoms. but SL(n, R) has" only finitely many n-dim’l rep’ns. So suffiΓ ces to show each prime occurs to bdd power.

Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 7 / 10 Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 8 / 10

is a superrigid lattice in SL(n, R) and every matrix entry is a rational number. ShowΓ each prime occurs to bdd power in denoms. Exercise Show that if is strictly superrigid in G, then the This is the conclusion of -adic superrigidity: p abelianization of is ﬁnite: /[ , ] is ﬁnite. Theorem (Margulis) Γ Exercise Γ Γ Γ Γ φ: SL(k, Q ) φ( ) has compact closure → p ⇒ Prove the Mostow Rigidity Theorem under the (unless G SO(1, n), SU(1, n)) = additional assumption that is strictly superrigid. I.e.,Γ ℓ, no matrix in φ( Γ) has pℓ in denom. 1 ∃ Hint: Use a corollary of the Borel Density Theorem ˆ Γ Summary of proof: Γ to prove that φ is onto. 1 R-superrigidity matrix entries “rational” ⇒ . 2 Q -superrigidity matrix entries Z p ⇒ ∈

Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 9 / 10 Dave Witte Morris (U of Lethbridge) Lattices 4: Rigidity U of Lethbridge, Oct 2018 10 / 10