Introduction to Arithmetic Groups 1 1. Rigidity of...

Introduction to arithmetic groups

Dave Witte Morris

University of Lethbridge, Alberta, Canadahttp://people.uleth.ca/!dave.morris

[email protected]

KAIST Geometric Topology Fair (January 11–13, 2010)

Abstract. Arithmetic groups are fundamental groups of locallysymmetric spaces. We will see how they are constructed, and discusssome of their important properties. For example, although the Q-rank ofan arithmetic group is usually defined in purely algebraic terms, we willsee that it provides important information about the geometry andtopology of the corresponding locally symmetric space. Algebraictechnicalities will be pushed to the background as much as possible.

Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 1 / 27

Introduction to Arithmetic Groups 1What is a superrigid subgroup?

1 rigidity of linkages2 group-theoretic superrigidity3 the analogy4 examples of superrigid subgroups5 why superrigidity implies arithmeticity6 some geometric consequences of superrigidity

Mostow Rigidity Theoremvanishing of the first Betti number

For further reading, see the references in [D. W. Morris, What is asuperrigid subgroup?, in Timothy Y. Chow and Daniel C. Isaksen, eds.:Communicating Mathematics. American Mathematical Society,Providence, R.I., 2009, pp. 189–206. http://arxiv.org/abs/0712.2299].


1. Rigidity of linkages

Example (two joined triangles)A

B

CD

B

CD

A

A B A CB C B D C D

This is not rigid.I.e., it can be deformed (a “hinge").


Example (Tetrahedron)A

B

C

D

A B A C A DB C B D C D

This is rigid (cannot be deformed).


Example (add a small tetrahedron)

A

E

DB

C

A

E

DB

C

A

BC

D

E

A

BC

D

E

A

BC

D

E

A B A C A DB C B D C DB E C E D E

This is rigid.

However, it is not superrigid:if it is taken apart, it can be reassembled incorrectly.


A

B

C

D

A B A C A DB C B D C D

A tetrahedron is superrigid: the combinatorialdescription determines the geometric structure.

Combinatorial superrigidityMake a copy of the object,

according to the combinatorial rules.The copy is the exact same shape as the original.

This talk: analogue in group theory


2. Group-theoretic superrigidity

Group homomorphism ! : Z" Rd

(i.e., !(m+n) = !(m)+!(n))# ! extends to a homomorphism !̂ : R" Rd.

Namely, define !̂(x) = x ·!(1).Check:

!̂(n) = !(n)!̂(x +y) = !̂(x)+ !̂(y)!̂ is continuous

(only allow continuous homomorphisms)


Group homomorphism ! : Zk " Rd

# ! extends to a homomorphism !̂ : Rk " Rd.

Proof.Use standard basis {e1, . . . , ek} of Rk.Define !̂(x1, . . . , xk) =

!xi!(ei).

(“linear trans can do anything to a basis")Linear transformation

# homomorphism of additive groups

Group Representation Theory:study homomorphisms into Matrix Groups.GL(d,C) = d$ d matrices over C

with nonzero determinant.This is a group under multiplication.


Group homomorphism ! : Z" GL(d,R)(i.e., !(m+n) = !(m) ·!(n))

%# extends to homomorphism !̂ : R" GL(d,R).(Only allow continuous homos.)

Proof by contradiction.

Spse & homo !̂ : R" GL(d,R)with !̂(n) = !(n) for all n ' Z.

!̂(0) = Id # det"!̂(0)

#= det(Id) = 1 > 0

R connected # !̂(R) connected# det

"!̂(R)

#connected in R$ = R( {0}

# det"!̂(R)

#> 0

# det"!(1)

#> 0

Maybe det"!(1)

#< 0. ")


Group homomorphism ! : Z" GL(d,R)%# extends to homomorphism !̂ : R" GL(d,R).Because: maybe det

"!(1)

#< 0.

det"!(2)

#= det

"!(1+ 1)

#=$det"!(1)

#%2> 0.

In fact, det"!(even)

#> 0.

May have to ignore odd numbers:restrict attention to even numbers.


May have to ignore odd numbers:restrict attention to even numbers.

Analogously, may need to restrict to multiples of 3(or 4 or 5 or . . . )

Restrict attention to multiples of N.{multiples of N} is a subgroup of Z

“Restrict attention to a finite-index subgroup"

PropositionGroup homomorphism ! : Zk " GL(d,R)# ! “almost" extends to homo !̂ : Rk " GL(d,R)such that !̂(Rk) * !(Zk). (“Zariski closure")

This means Zk is superrigid in Rk.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 11 / 27

Lagrange interpolation& polynomial curvey = anxn + an(1xn(1 + · · · + a0

through any n+ 1 points.

Idea: Zariski closure is like convex hull.


Proposition (“Zk is superrigid in Rk”)Group homomorphism ! : Zk " GL(d,R)# ! “almost" extends to homo !̂ : Rk " GL(d,R)such that !̂(Rk) * !(Zk). (“Zariski closure")

!̂(Rk) * !(Zk): image of ! controls image of !̂.Good properties of !(Z) carry over to !̂(R).

Example: If all matrices in !(Z) commute,then all matrices in !̂(R) commute.

Example: If all matrices in !(Z) fix a vector v ,then all matrices in !̂(R) fix v .

Generalize to nonabelian groups.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 13 / 27

3. The analogyCombinatorial superrigidityMake copy of object, obeying combinatorial rules.The copy is the exact same shape as the original.

Maybe not exactly the same object:may be rotated from the original position;may be translated from original position.

These are trivial modifications:rotations and translations are symmetries of thewhole universe (Euclidean space R3).

Same result can be obtained with the original objectby moving the whole universe to a new position.


Combinatorial superrigidity“If the object can be moved somewhere,then the whole universe can be moved there."

& is a superrigid subgroup of the group G means:homomorphism ! : & " GL(d,R) extendsto homomorphism !̂ : G " GL(d,R)

Group-theoretic superrigidityMake a copy of & as a group of matrices.The same copy of & can be obtained by moving allof G into a group of matrices.


4. Superrigid subgroups

Example. Zk is superrigid in Rk.Generalize to nonabelian groups.

Zk is a (cocompact) lattice in Rk. I.e.,Rk is a (simply) connected group (“Lie group")Zk is a discrete subgroupall of Rk is within a bounded distance of Zk

&C, +x ' Rk, &m ' Zk, d(x,m) < C .

If can replace Zk with & and Rk with G,then & is a (cocompact) lattice in G.


Lie groups are of three types:solvable (many normal subgrps, e.g., abelian)simple (“no" normal subgroups, e.g., SL(k,R))combination (e.g., G = Rk $ SL(k,R))

More or less: & = Zk $ SL(k,Z)(& has a solvable part and a simple part)

Today: we consider solvable groups.


DefinitionA connected subgroup G of GL(d,C) is solvable if itis upper triangular

G *

'()C$ C C0 C$ C0 0 C$

*+,

(or is after a change of basis).

ExampleAll abelian groups are solvable.

Proof.Every matrix can be triangularized over C.Pairwise commuting matrices can besimultaneously triangularized.


Examples of lattices

G =

'((()

1 R R R0 1 R R0 0 1 R0 0 0 1

*+++, & =

'((()

1 Z Z Z0 1 Z Z0 0 1 Z0 0 0 1

*+++,

& = G superrigid

G =

'()R+ 0 00 R+ 00 0 R+

*+, & =

'()

2Z 0 00 2Z 00 0 2Z

*+,

& = G superrigid


G =

'()

1 R C0 1 00 0 1

*+, & =

'()

1 Z Z+ Zi0 1 00 0 1

*+,

G = G & = G superrigid

G, =

'()

1 t C0 1 00 0 e2"it

*+, G, =

'()

1 R C0 1 00 0 T

*+,

& , =

'()

1 Z Z+ Zi0 1 00 0 1

*+, = & . & is a lattice in

both G and G,.

& = G %= G, so & %= G,. & is not superrigid in G,.E.g., id map ! : & " & does not extend to !̂ : G, " & .

(& = G is abelian but G, is not abelian.)Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 20 / 27

Proposition& superrigid in G # & = G (mod Z(G)).

Proof.The inclusion & ! GL(d,R))must extend to G ! GL(d,R) with G * & .

Theorem (converse)Lattice & in solv grp G is superrig i! & = G (mod Z(G)).

& %= G,: some of the rotations associated to G, donot come from rotations associated to & .

rot

-# -0 $

.=- #|#| 00 $

|$|

.


CorollaryA lattice & in a Lie group G is “superrigid” i!

& = G (mod Z(G) · (cpct ss normal subgrp))and simple part of & is “superrigid."

Theorem (Margulis Superrigidity Theorem)All lattices in SL(n,R) are “superrigid” if n . 3.Similar for other simple Lie groups, R-rank . 2.

Corollary (Margulis Arithmeticity Theorem)Every lattice in SL(n,R) is “arithmetic" if n . 3.

(like SL(n,Z))Only way to make a lattice: take integer points

(and minor modifications)

Similar for other simple groups with R-rank . 2.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 22 / 27

5. Why superrigidity implies arithmeticityLet & be a superrigid lattice in SL(n,R).We wish to show & * SL(n,Z),

i.e., want every matrix entry to be an integer.

First, let us show they are algebraic numbers.Suppose some %i,j is transcendental.Then & field auto ! of C with !(%i,j) = ???.

Define /!-a bc d

.=-!(a) !(b)!(c) !(d)

..

So /! : & " GL(n,C) is a group homo.Superrigidity: /! extends to !̂ : SL(n,R)" GL(n,C).There are uncountably many di!erent !’s,but SL(n,R) has only finitely many n-dim’l rep’ns.


& is a superrigid lattice in SL(n,R)and every matrix entry is an algebraic number.

Second, show matrix entries are rational.Fact. & is generated by finitely many matrices.Entries of these matrices generate a field extensionof Q of finite degree. “algebraic number field"So & * SL(n, F). For simplicity, assume & * SL(n,Q).

Third, show matrix entries have no denominators.Actually, show denominators are bounded.(Then finite-index subgrp has no denoms.)

Since & is generated by finitely many matrices,only finitely many primes appear in denoms.

So su"ces to show each prime occurs to bdd power.


& is a superrigid lattice in SL(n,R)and every matrix entry is a rational number.Show each prime occurs to bdd power in denoms.

This is the conclusion of p-adic superrigidity:

Theorem (Margulis)If & is a lattice in SL(n,R), with n . 3,and ! : & " SL(k,Qp) is a group homomrphism,then !(&) has compact closure.

I.e., &k, no matrix in !(&) has pk in denom.

Summary of proof:1 R-superrigidity # matrix entries “rational”2 Qp-superrigidity # matrix entries ' Z


Some consequences of superrigidityLet & be the fundamental group of a locallysymmetric space M that has finite volume.

We always assume 0M has no compact or flat factors, and is complete.Also assume M is irreducible: M " M1 $M2.

Assume & is superrigid.

Topology of M determines its geometry:

Corollary (Mostow Rigidity Theorem)If & / fund group & , of finite-vol loc symm space M,,then M is isometric to M, (up to normalizing constant).

More generally, if & ! & ,, then M ! M, as a totallygeodesic subspace (up to finite covers).


& = superrig fund grp of fin-vol loc symm space M .

CorollaryThe first Betti number of M vanishes: H1(M ;R) = 0.

RemarkIt is conjectured [Thurston] that if M is afinite-volume hyperbolic manifold, then

H1(1M ;R) " 0, for some finite cover 1M .

So it is believed that the fundamental group of ahyperbolic manifold is never superrigid(although most hyperbolic mflds are Mostow rigid).


Introduction to Arithmetic Groups 2examplesrelation to locally symmetric spaces M = X/!compactness criterion (two versions)basic group-theoretic properties

congruence subgroupsresidually finitevirtually torsion free (Selberg’s Lemma)

finitely presented

For further reading, see D. W. Morris, Introduction to Arithmetic Groups.http://people.uleth.ca/!dave.morris/books/IntroArithGroups.html

More advanced:M. S. Raghunathan, Discrete Subgroups of Lie Groups, Springer, 1972.V. Platonov and A. Rapinchuk, Algebraic Groups and Number Theory,

Academic Press, 1993.


Examples of arithmetic groupsLet ! be an arithmetic group :

! = SL(3,Z) = {3" 3 integer matrices of det 1 }(or subgroup of finite index)

or ! # SO(1,3;Z) = SO(1,3)$ SL(4,Z)= {g % SL(4,Z) | g I1,3 gT = I1,3 } I1,3 =

"#

1 &1&1&1

$%

or . . .

! = GZ := G $ SL(n,Z) for suitable G ' SL(n,R).

Theorem (“Reduction Theory”)For suitable G ' SL(n,R), ! is a lattice in G:

1 ! is discrete, and2 !\G has finite volume (maybe compact).


Relation to locally symmetric spacesRecallX = Rn is a symmetric space.

1 X is homogeneous : Isom(X) is transitive on X2 ( ! % Isom(X), ! has an isolated fixed point:

!(x) = &x fixes only 0.

ExampleX = Hn is also a symmetric space.

(Isom(Hn) ) SO(1, n))

ExerciseX = symmetric space (connected), G = Isom(X)*

!+ X = G/K, K compact.


symmetric space X = G/K, K compact

DefinitionM is a locally symmetric space (complete):

universal cover of M is a symmetric space.I.e., M = !\X, ! ' Isom(M) discrete (& torsion-free).

ExampleG = Isom(Rn)* = Rn " SO(n), K = SO(n)

!+ X = Rn.Let ! = Zn ' G, so M = Zn\Rn = Tn.

ExampleG = SO(1, n)*, K = SO(n) !+ X = Hn.Let ! = SO(1, n;Z), so M = !\Hn = hyperbolic mfld.


locally symmetric space M = !\X, X = G/K

Best mflds are compact. Next best: finite volume.

Recall! is a lattice in G: ! is discrete, and !\G has finite volume.

Proposition! is a torsion-free lattice in G, X = G/K!+ M = !\X is locally symmetric of finite volume.

So lattices are the fundamental groupsof finite-volume locally symmetric spaces.

And arithmetic groups are the latticesthat are easy to construct.


M = !\X, X = G/K, ! a lattice in G

Henceforth, assume X has:no flat factors: X " X1 "Rnno compact factors: X " X1 " compact

Then G is semisimple with no compact factors and trivial center.

Fundamental grp provides topological info about any space.For locally symmetric spaces, it does much more:usually completely determines all of the topology and geometry.

Mostow Rigidity Theorem


Theorem (Mostow Rigidity Theorem)Suppose M1 and M2 are finite-volume locally symm.Assume

dimM1 > 2, andM1 is irreducible: M1 ,) M- "M-- (up to finite covers).

If !1 . !2, then M1 . M2 (modulo a normalizing constant).

Every aspect of the geometric structure of M isreflected as an algebraic property of thefundamental group ! .The geometric category of irreducible locallysymmetric spaces with dim > 2 is equivalent tothe algebraic category of “irreducible” latticesin appropriate semisimple Lie groups.


Compactness criterionObservationFor M = !\X with X = G/K:

M is compact i! !\G is compact.(Because M = !\G/K and K is compact.)

Say ! is a cocompact lattice in G.

Proposition!\G is not compact i!

(g1, g2, g3, . . . % G and(#1,#2,#3, . . . % !" = ! & {e}, such that

g&1i #i gi / e.


!\G not compact i! (gi,#i, g&1i #i gi / e

Proof (0).Suppose !\G is compact.For a,b % G, let ab = a&1ba (“conjugation of b by a”).

Since !\G is compact, ( cpct C with !C = G.

C is compact and !" is discrete, hence closed,so (!")C is closed.

Therefore

e % g&1i #i gi ' (!")G = (!")!C= ((!")!)C = (!")C = (!")C ,1 e. /2


!\G not compact i! !gi,!i, g"1i !i gi # e

CorollarySL(2,Z)\ SL(2,R) is not compact.

Proof.

Let gi ="i

1/i

#and !i =

"1 1

1

#. Then

g"1i !i gi =

"1/i

i

#"1 1

1

#"i

1/i

#

="

1 1/i21

##"

1 01

#= e.

Similarly, SL(n,Z)\ SL(n,R) is not compact (if n $ 2).


!\G not compact i! !gi,!i, g"1i !i gi # e

Corollary (Godement Criterion)!\G is compact i! ! has no unipotent elements.

Definitionu % SL(n,R) is unipotent: (u" Id)n = 0.

Equivalently, 1 is the only eigenvalue of u, so

u is conjugate to

$%&

1 &. . .0 1

'()


Corollary (Godement Criterion)!\G not cpct i! ! has nontrivial unipotent elements.

Proof (').! gi % G and !i % !( with g"1

i !i gi # Id.1 Char poly of Id is det("" Id) = ("" 1)n.2 !i % SL(n,Z) !' char poly of !i has Z coe!s.3 Similar matrices have same characteristic poly,

so the char poly of g"1i !i gi also has Z coe!s.

4 char poly of g"1i !i gi # char poly of Id

and both have Z coe"cients.So the two char polys are equal (if i is large).

5 Therefore, the char poly of !i is ("" 1)n.So the only eigenvalue of !i is 1.


Corollary (Godement Criterion)!\G not cpct i! ! has nontrivial unipotent elements.

The proof of the other direction ()) depends ona fundamental fact from Lie theory:

Theorem (Jacobson-Morosov Lemma)If u is any unipotent element of G

(connected semisimple Lie group in SL(n,R))then there is a continuous homomorphism

# : SL(2,R)# G with #*"

1 11

#+= u.

Proof ofGodement ()) Let !i = u % ! and gi = #

*"i

1/i

#+.


Congruence Subgroups

We will use a construction known as“congruence subgroups”

to prove two basic properties of arithmetic groups.

1 ! is residually finite2 ! is virtually torsion-free


Proposition! is residually finite :

*! % !(, ! finite-index subgrp H < ! , ! " H.

Proof.! " Id # 0, so !(! " Id)ij # 0.

Choose N $ (! " Id)ij = !ij " Idij.Ring homo Z# ZN yields SL(n,Z)# SL(n,ZN).

Let #N : ! # SL(n,ZN) be the restriction to ! .Since ZN finite, obvious that SL(n,ZN) is finite.

Let H = ker#N . Then !/H + img(#N) is finite.So H is a finite-index subgroup.

By choice of N, #N(!)ij # #N(Id)ij, so #N(!) # Id.So ! " ker(#N) = H.

Terminology: ker(#N) is a (principal) congruence subgroup of ! .Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 15 / 18

Proposition (Selberg’s Lemma)! is virtually torsion-free :

! finite-index subgrp H < ! , H is torsion-free.(no nontrivial elements of finite order)

Proof.Define #3 : ! # SL(n,Z3) and let H = ker(#3).

It su"ces to show H is torsion-free.Let h % H, write h = Id+ 3kT , T ,- 0 (mod 3).

hm = (Id+ 3kT)m

= Id+m(3kT)+,m2

-32kT 2 + · · ·

- Id+ 3kmT (mod 3k+$+1) if 3$ |m,- Id (mod 3k+$+1) if 3$+1 $m.


Finite presentationProposition! is finitely presented :

! = .!1,!2, . . . ,!m | w1,w2, . . . ,wr /.

Proof of finite generation.Fund grp of any compact mfld is finitely generated:

% : .M # M , ! cpct C 0 .M , %(C) = M .Let S = {! % ! | !C 1 C #2 }. (! prop disc, so S finite!)Then ! = .S/:Given ! % ! . Since M is connected, ! chainC,!1C,!2C, . . . ,!nC = !C , with !kC 1 !k+1C #2.So !1 % S, !"1

1 !2 % S, . . . , !"1n ! % S.

Therefore ! = !1(!"11 !2) · · · (!"1

n !) % .S/.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 17 / 18

! is finitely generated (if M is compact)because S = {! % ! | !C 1 C #2 } is finite.

For noncompact case, can construct a nicefundamental domain for ! in X:

!C = X, {! % ! | !C 1 C #2 } is finite.Furthermore, C is open.

Finite generation follows from above argument.Finite presentation follows froma more sophisticated argument

(since X is connected, locally connected,and simply connected)

[Platonov-Rapinchuk, Thm. 4.2, p. 195]


Introduction to Arithmetic Groups 3

noncompactness via isotropic vectorsQ-rank and the asymptotic conecocompact arithmetic subgroups of SO(1, n)

(restriction of scalars)

For further reading, see D. W. Morris, Introduction to Arithmetic Groups.http://people.uleth.ca/!dave.morris/books/IntroArithGroups.html

More advanced:V. Platonov and A. Rapinchuk, Algebraic Groups and Number Theory,

Academic Press, 1993.C. Maclachlan and A. Reid, The Arithmetic of Hyperbolic 3-Manifolds,

Springer, 2002.


Noncompactness via isotropic vectorsExample! = SO(1,3;Z) is an arithmetic subgroup of SO(1,3).

(provides hyperbolic 3-manifold M = !\H3)

Is M compact?

PropositionSpse Q( !"x) is a (nondegenerate) quadratic form over Z

(e.g., Q(x1, x2, x3) = x21 # 3x2

2 # 7x23)

Let G = SO(Q) and ! = SO(Q;Z).Then !\G is not compact $%

& isotropic Q-vectors: Q(v) = 0 with v " 0.


G = SO(Q) and ! = SO(Q;Z).!\G is not compact $% &v ' (Qn)(, Q(v) = 0.

Proof (%).Godement Criterion: & unipotent u ' ! .Jacobson-Morosov (over Q):

& ! : SL(2,Q)" GQ with !"#

1 11

$%= u.

Let a = !"#

21/2

$%.

Algebraic Group Thry: a is diagonalizable over Q,so a has eigenvector v ' Qn with eigenval " " ±1.

Then Q(v) = Q(akv) = Q("kv) = "2k Q(v)" 0.So Q(v) = 0.



Example! = SO(1,3;Z) #% !\H3 is not compact.

Proof: Q(x0, x1, x2, x3) = x20 # x2

1 # x22 # x2

3,so Q(1,1,0,0) = 11 # 12 # 02 # 02 = 0.

ExampleLet Q( !"x) = 7x2

0 ##x21x2

2 # x23 and ! = SO(Q;Z).

Then !\H3 is compact.

Proof: 7 is not a sum of 3 squares (in Q),so Q has no isotropic vectors.


ExampleLet Q( !"x) = 7x2

0 # x21 # x2

2 # x23 and ! = SO(Q;Z).

Then !\H3 is compact. (bcs 7 is not a sum of 3 squares)

Recall: Every positive integer is a sum of 4 squares.So this method will not constructcompact hyperbolic n-manifolds for n > 3.

Fact from Number TheoryIf Q(x1, . . . , xn) is a quadratic form over Z,

with n ) 5, and Q is isotropic over R,then Q is isotropic.

Need to do something more sophisticated.(Return to this later: “restriction of scalars”)


Q-rank and the asymptotic cone


DefinitionLet ! = SO(Q;Z).

V * Qn is totally isotropic if Q(V) = 0.Q-rank(!) = max dim tot isotrop Q-subspace.

ExampleSuppose Q-rank(!) = 0. Then $ isotropic Q-vector.

so M = !\X is compact.


ExampleSuppose Q-rank(!) = 0. Then $ isotropic Q-vector.

so M = !\X is compact.Look at !\X from a large distance.

!\X compact % limit is a point.+ dimension of limit = 0 = Q-rank(!).

Definitionasymptotic cone of metric space (M,d)

= limt",

&(M, 1

t d),m0'.


ExampleSpse ! * SO(1, n), and !\Hn not compact.

!\Hn has finitely many cusps.

Limit = “star" of finitely many rays.+ dimension of limit = 1 = Q-rank(!).

Theorem (Hattori)Asymptotic cone of !\X is a simplicial complexwhose dimension is Q-rank(!).


Theorem (Hattori)Asymptotic cone of !\X is a simplicial complexwhose dimension is Q-rank(!).

More precisely, the asymptotic cone of !\X is equal to thecone on the “Tits building” of parabolic Q-subgroups of G.

Remark!\X is quasi-isometric to its asymptotic cone.

RemarkAnother important application of Q-rank:cohomological dimension of ! = dimX #Q-rank(!)

(if ! is torsion-free)


Cocompact lattices in SO(1, n)ExampleLet

! =!

2,Q( !"x) = x2

0 #!x21 #!x2

2 # · · ·#!x2n,

G = SO(Q) $ SO(1, n),! = GZ[!] = G % SL

"n+ 1,Z[!]

#.

Then ! is a cocompact arithmetic subgroup of G.

Later: why ! is an arithmetic subgroup.

Key observation for compactnessQ has no isotropic Q(!)-vectors.


! =!

2, Q( !"x) = x20 #!x2

1 #!x22 # · · ·#!x2

n

Key observation for compactnessQ has no isotropic Q(!)-vectors.

Proof.Suppose Q(v) = 0.Galois auto of Q(!): (a+ b!)" = a# b!.

Q"( !"x) = x20 +!x2

1 +!x22 + · · · +!x2

n.0 = Q(v)"= (v"0 )2 +!(v"1 )2 + · · · +!(v"n )2= Q"(v").

Since all coe!cients of Q" are positive,must have v" = 0.

So v = 0.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 11 / 19

More general!0,!1, . . . ,!n algebraic integers, s.t.

!0 > 0 and !1,!2, . . . ,!n < 0,& Galois aut " of Q(!0, . . . ,!n),

!"0 , . . . ,!"n all have the same sign(all positive or all negative).

ThenG = SO(!0x2

0 + · · · +!nx2n;R) $ SO(1, n),

! = GZ[!0,...,!n] is a cocpct arith subgrp of G.

If n is even, this construction provides all of thecocompact arithmetic subgroups of SO(1, n).

For n odd, also need SO(Q; quaternion algebra)(And n = 7 has additional “triality” subgroups.)


Definition of arithmeticityRecallArithmetic subgroup :! = GZ := G % SL(n,Z) for suitable G ' SL(n,R).

We always assume G is semisimple, connected.

TheoremG is (almost) Zariski closed

(defined by polynomial functions on Matn(n(R))

ExampleSL(2,R) = {A ) Mat2(2(R)

$$ detA = 1 }and detA = a1,1a2,2 # a1,2a2,1 is a polynomial.


DefinitionG is defined over Q:G is defined by polynomial funcs with coe"s in Q.

ExampleSL(n,R) is defined over Q.SO(1, n) is defined over Q.Let ! =

!2 and Q( !"x) = x2

0 #!x21 #!x2

2Then G = SO(Q) is not defined over Q.

(It is defined over Q[!].)

Definition (starting point)GZ is an arithmetic subgroup of G

if G is defined over Q.


Definition (starting point)GZ is an arithmetic subgroup of G

if G is defined over Q.

Complete definition is more general; it ignores:compact factors of G, anddi"erences by only a finite group.

DefinitionSpse G (K " SL(n,R), defined over Q, with K cpct.

Let ! * = image of (G (K)Z in G.Any subgroup ! of G that is commensurable with ! *

(! % ! * has finite index in both ! and ! *)is an arithmetic subgroup of G.


Restriction of scalars

Recall! =

!2, Q( !"x) = x2

0 #!x21 #!x2

2 # · · ·#!x2n,

G = SO(Q;R), ! = SO"Q;Z[!]

#.

Want to show ! is an arithmetic subgroup of G.

As an warm-up, let us show SL"2,Z[!]

#is $

an arithmetic subgroup of SL(2,R)( SL(2,R).


Example! = SL

"2,Z[!]

#, with ! =

!2,

G = SL(2,R)( SL(2,R)," = Galois automorphism of Q[!],% : ! " G : # #

"#,#"

#.

Then !% is an arithmetic subgroup of G.

Outline of proof.Since {1%,!%} =

&(1,1), (!,#!)

'is linearly indep,

+ T ) GL(2,R) with T"Z[!]%

#= Z2.

(T = T , T ) GL(4,R) has (T"Z[!]%

#2 = (Z2)2 = Z4.So ( (T !% (T#1)(Z4) = Z4, so (T !% (T#1 ' SL(4,Z).

In fact, (T !% (T#1 = SL(4,Z)% (T G (T#1, and (T G (T#1 isdefined over Q

So !% is an arithmetic subgroup of G.Dave Witte Morris (Univ. of Lethbridge) Introduction to arithmetic groups KAIST Geom Topology Fair 17 / 19

! =!

2, Q( !"x) = x20 #!x2

1 #!x22 # · · ·#!x2

n,G = SO(Q;R), ! = SO

"Q;Z[!]

#.

Want to show ! is an arithmetic subgroup of G.

Idea of proof.Galois aut of Q(!): (a+ b!)" = a# b!.

G" = SO(x21 + x2

2 +!x23 +!x2

4 +!x25) $ SO(5).

Map % : z # (z, z") gives !% ' G (G" .After change of basis (mapping

"Z[!]%)n to Z2n),

we have !% = SL(n,Z)% (G (G")so !% is an arithmetic subgroup of G (G" .

Can mod out the compact group G" ,so ! is arithmetic subgroup of G.


Restriction of scalarsSuppose

G is defined over Q(!) (algebraic number field),"1, . . . ,"n : Q(!)! C are the noncon-

jugate embeddings.

ThenG" = G"1 #G"2 # · · ·#G"n is defined over Q,andGZ[!] is isomorphic to (G")Z

via the map ! : # ! (#"1,#"2, . . . ,#"n).We assume here that Z[!] is the entire ring of integers of Q(!).

Example: If Q( "!x) has coe!cients in Q(!),then SO

"Q;Z[!]

#is an arithmetic subgroup

of a product of orthogonal groups.


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