Snippets

# A quick and dirty definition of an arithmetic subgroup?

Or, at least, of an arithmetic subgroup of $G \leq \mathrm{SL}(n,\mathbb{R})$ a semisimple Lie group:

1. The “subgroup of integer points” $G_{\mathbb{Z}} = G \cap \mathrm{SL}(n,\mathbb{Z})$ is an arithmetic group …
2. … if embedded in a reasonable way which doesn’t distort the arithmetic structure—slightly more precisely, if it is [essentially] an algebraic group over the rationals.
3. Anything isomorphic to a finite extension is arithmetic too.

For a more precise definition—Witte Morris’ book is a good place to start; he fudges over a few things too, but actually has references to fill those in …

Standard
Overview / Outlines

# (Some) Lie theory for geometric topology / dynamics

(following an outline by Wouter van Limbeek; filling this outline is a work-in-progress. Sections 6 and 7 are particularly incomplete; sections 11 and 12—or 8 onwards—could [should?] be split to form its own post.)

Here we are working with connected Lie groups.

1) Classes of Lie groups: abelian, nilpotent, solvable, semisimple

Abelian Lie groups are completely classified: these all have the form $G \cong \mathbb{R}^k \times \mathbb{T}^e$ (where $\mathbb{T}^e$ denotes the e-dimensional torus.)

Nilpotent groups are those with a terminating lower central series $G_{(k+1)} = [G, G_{(k)}]$. Examples of nilpotent Lie groups: Heisenberg groups, unitriangular groups, the (3-by-3) Heisenberg group mod its center.

Solvable groups are those with a terminating upper central series $G^{(k+1)} = [G^{(k)}, G^{(k)}]$. There is an equivalent formulation in terms of composition series—in this sense solvable groups are built up from (simple) abelian groups. Examples of solvable Lie groups:

• $\mathrm{Sol} = \mathbb{R}^2 \rtimes \mathbb{R}$, where $t \in \mathbb{R}$ acts on $\mathbb{R}^2$ as the matrix $\left( \begin{array}{cc} e^t \\ & e^{-t} \end{array} \right)$.
• $\mathrm{Aff}(\mathbb{R})^2 \cong \mathbb{R} \rtimes \mathbb{R}_{> 0}$, where the action is by scalar multiplication.
• The group of all upper-triangular matrices.
• $\mathbb{R}^2 \rtimes \mathbb{R}$, lifted from $\mathbb{R}^2 \rtimes \mathrm{SO}(2)$, where the action is by rotation.

By definition, we have abelian $\subset$ nilpotent $\subset$ solvable

Simplicity and semisimplicity

A Lie algebra $\mathfrak{g}$ is simple if it has no proper ideals (i.e. proper subspaces closed under the Lie bracket.) $\mathfrak{g}$ is semisimple if it is a direct sum of simple Lie algebras.

A Lie group G is (semi)simple if its Lie algebra $\mathfrak{g}$ is (semi)simple.

Equivalently, G is simple if it has no nontrivial connected normal subgroups (what can go wrong if we don’t have that additional adjective?)

G is semisimple if its universal cover $\tilde{G}$ is a direct product of simple Lie groups. (Note that things can go wrong if we do not take the universal cover, e.g. the quotient of $\mathbb{SL}_2(\mathbb{R})$ by $\mathbb{Z}/2\mathbb{Z}$, where the action is the diagonalisation of $\pm 1 \curvearrowright \mathbb{SL}_2(\mathbb{R})$, is semisimple [in the sense that its Lie algebra is semisimple], but not a product.)

Relation between these types and the adjoint representation.

Recall the adjoint representation $\mathrm{Ad}: G \to \mathrm{GL}(\mathfrak{g})$ given by sending a group element g to the matrix representing $D_e c_g$, where $c_g$ denotes conjugation by g. Now

• G is abelian iff Ad(G) is the trivial representation
• G is nilpotent iff Ad(G) is unipotent
• G is solvable iff Ad(G) is (simultaneously) triangularizable (over an algebraically-closed field)
• G is semisimple iff Ad(G) is a semisimple representation (i.e. may be written as a direct sum of irreducible Ad-invariant representations.)
• G contains a lattice (see below) iff Ad(G) $\subset \mathrm{SL}(\mathfrak{g})$.

2) Levi decomposition of Lie groups

Note that subgroups of solvable groups are solvable, and extensions of solvable groups by solvable groups are solvable. Putting all of these together, we obtain that any Lie group G has a unique maximal closed connected normal solvable subgroup R (called the solvable radical of G.)

The Levi decomposition (also called, in some contexts, the Levi-Malcev decomposition) of G is G = RS, where R is the solvable radical, $S \subset G$ is semisimple, and $R \cap S$ is discrete.

If G is simply-connected, then $G = R \rtimes S$.

3) Classification of compact Lie groups

Any compact connected solvable Lie group G is a torus (Proof sketch: Induct on length of upper central series. For n=1, G is abelian and we are done. Otherwise, we have the short exact sequence $1 \to [G,G] \to G \to \frac{G}{[G,G]} \to 1$; the outer terms are tori $T_1$ and $T_2$, and we have a map $T_2 \to \mathrm{Out}(T_1) = \mathrm{GL}(\dim T_1, \mathbb{Z})$. Since this is a group homomorphism from a connected group to a discrete group, it musth ave trivial image; hence $T_1 \subset G$ is central, and we can induce on $\dim T_2$ to produce a section $\frac{G}{[G,G]} \to G$.

Compact connected semisimple Lie groups can also be classified, but this story involves much more algebra (representation theory, highest weights, etc.

Using the Levi decomposition, we have the following

Theorem. Any compact connected Lie group G is isomorphic to $(\mathbb{T}^k \times G_1 \times \dots \times G_n) / F$, where F is a finite group and $G_1, \dots, G)n$ are simple.

4) When is a Lie group an algebraic group? When is a Lie group linear?

An algebraic group is a group with a compatible algebraic variety structure (i.e. multiplication and inversion are regular maps.) A linear group is (isomorphic to) a subgroup of GL(n,k) for some field k. Note GL(n,k) is algebraic, and thus linear groups defined by polynomial equations (but not all linear groups!) are algebraic.

A Lie group is algebraic if it is isomorphic to a linear algebraic group (but this is not an iff [?])

Many of the classical Lie groups are (linear) algebraic groups, but note that not all Lie groups are algebraic: e.g. $\widetilde{\mathrm{SL}_2(\mathbb{R})}$ is not algebraic because its center is not algebraic (“is too large to be algebraic.”)

5) When is the exponential map a diffeomorphism?

We have an exponential map $\exp: \mathfrak{g} \to G$. Its derivative at the identity, $\exp_*: \mathfrak{g} \to \mathfrak{g}$, is the identity map; exp therefore restricts to a diffeomorphism from some neighborhood of 0 in $\mathfrak{g}$ to a neighborhood of 1 in G.

If G is connected, simply-connected, and nilpotent, the exponential map exp is a (global) diffeomorphism (in fact, an analytic isomorphism of analytic manifolds, if G is linear algebraic.)

6) Topology of Lie groups: fundamental group, homotopy type, cohomology …

Theorem. Any connected Lie group has abelian fundamental group.

Proof sketch: In fact this is true for any connected topological group, because the group structure forces things to be nice that way. To make this precise can be mildly annoying though.

Theorem (Weyl). The fundamental group of a compact semisimple Lie group is finite

Theorem. Any connected Lie group has trivial second homotopy and torsionfree third homotopy.

Proof sketch: Since any Lie group retracts onto its maximal compact subgroup, WLOG we work with only connected groups.

From the long exact sequence of homotopy groups obtained from the path-loop fibration, $\pi_k(G) \cong \pi_{k-1}(\Omega G)$. By Morse theory, $\pi_1(\Omega G) = 0$ and $\pi_2(\Omega_G) = \mathbb{Z}^t$.

Many more things can be said: see e.g. this survey.

Relation with the Lie algebra

7) Geometry of Lie groups: relation between geometry of an invariant metric and algebraic structure.

Note that a Lie group structure yields a natural left-invariant (or right-invariant) metric, given by propagating the inner product at the identity by group multiplication. With additional hypotheses, we can make this bi-invariant:

Theorem. Every compact Lie group G admits a bi-invariant metric.

Idea of proof Use the natural Haar measure on G to average the left-invariant metric.

We can play this invariant metric and the group structure off against each other: e.g. the geodesics of G coincide with the left-translates of 1-parameter subgroups of (at the identity); as a corollary of this, we obtain that any Lie group G is (geodesically) complete.

8) Selberg’s lemma

Lemma (Selberg 1960). A finitely generated linear group over a field of zero characteristic is virtually torsion-free.
(cf. Theorem (Malcev 1940). A finitely generated linear group is residually finite.)

Proof: Using number fields (local fields, see Cassels or Ratcliffe), or Platonov’s theorem, which seems to be a bunch of commutative algebra (see Bogdan Nica’s paper.)

9) Finite generation/presentation of lattices: Milnor-Svarc and Borel-Serre.

A lattice is a discrete group $\Lambda \leq G$ with $G / \Lambda$ of finite volume (as measured by the natural Haar measure on G.) The prototypical example is $\mathrm{SL}(2,\mathbb{Z}) \subset \mathrm{SL}(2, \mathbb{R})$, which of course is the isometry group of the torus.

Facts

1. Any lattice in a solvable group is co-compact (also called uniform.)
2. (Borel, Harish-Chandra) Any noncompact semisimple group contains a lattice that is not co-compact (also called non-uniform.)

Theorem (Milnor-Svarc). If $\Gamma$ acts on a proper geodesic metric space X properly discontinuously and cocompactly, then $\Gamma$ is finitely-generated.

Proof: Since $\Gamma \curvearrowright X$ cocompactly, the action has a compact fundamental domain $Z = X / \Gamma$. By the proper discontinuity of the action, the vertices of Z cannot accumulate. But compactness then implies Z has finitely many vertices, and hence finitely many sides. Since each algebraically independent generator would add a side to Z, this implies that $\Gamma$ is finitely generated.

In fact, many non-uniform lattices are also finitely-generated.

The $\mathbb{R}$-rank of a Lie group G is the dimension of the largest abelian subgroup of Z(a) simultaneously diagonalizable over $\mathbb{R}$, as a varies over all semisimple elements of G ($a \in G$ is semisimple if Ad(a) is diagonalizable over $\mathbb{R}$. Geometrically, we may interpret the $\mathbb{R}$-rank as the dimension of the largest flat in G.

A locally-compact group G is said to have property (T) if for every continuous unitary representation $\rho: G \to \mathcal{U}(\mathcal{H})$ into some Hilbert space there exists $\epsilon > 0$ and compact $L \subset G$ s.t. if $\exists v \in \mathcal{H}$ with $\|v\| = 1$ s.t. $\|\rho(l) v - v\| < \epsilon$ for all $l \in L$, then $\exists v' \in \mathcal{H}$ with $\|v'\| = 1$ s.t. $\rho(G)$ fixes v‘. (i.e., the existence of an almost-invariant vector implies the existence of a fixed point.)

Theorem (Kazhdan, 1968): If G is a simple Lie group of $\mathbb{R}$-rank $\geq 2$, and $\Gamma \subset G$ is a lattice, then $\Gamma$ has property (T), and hence is finitely-generated.

Alternative proof: every such lattice is arithmetic, then use a fundamental domain and argue [as before]

10) Levi-Malcev decomposition of lattices

The Levi-Malcev decomposition as applied to a lattice $\Lambda \subset G = RS$ tells us that $\Lambda = \Psi \Sigma$ where $\Psi = \Lambda \cap R$ is a lattice in a solvable group and $\Sigma = \Lambda \cap S$ is a lattice in a semisimple group: note that each of these intersections remains discrete, and has finite co-volume for if not $\Lambda$ would not have finite co-volume.

Hence the general scheme for understanding lattices in Lie groups: first understand lattices in solvable groups and in semisimple groups, then piece them together using the Levi-Malcev decomposition …

11) (Non-)arithmetic lattices. When is a lattice arithmetic?

Most generally, an arithmetic subgroup of a linear algebraic group G defined over a number field K is a subgroup Γ of G(K) that is commensurable with $G(\mathcal{O})$, where $\mathcal{O}$ is the ring of integers of K.

Hence we may define more abstractly a lattice $\Gamma$ in a connected (solvable) Lie group G is said to be arithmetic if there exists a cocompact faithful representation i of G into $G^*_{\mathbb{R}}$ (where $G^* \subset \mathrm{GL}(n, \mathbb{C})$ is an algebraic subgroup) with closed image and s.t. $i(\Gamma) \cap G^*_{\mathbb{Z}}$ has finite index in both $\Gamma$ and $G^*_{\mathbb{Z}}$.

We have that

Theorem (Borel, Harish-Chandra). Arithmetic subgroups are lattices.

Conversely,

Theorem (Mostow). (4.34 in Raghunathan.) Let G be a simply-connected solvable Lie group and $\Gamma \subset G$ a lattice. Then G admits a faithful representation into $\mathrm{GL}(n, \mathbb{R})$ which sends $\Gamma$ into $\mathrm{GL}(n, \mathbb{Z})$.

Note, however, the counterexample on pp. 76-77 of Raghunathan.

Much stronger results are true in semisimple Lie groups:

Theorem (Margulis arithmeticity). Any irreducible lattice in a semisimple Lie group with no rank one factors is arithmetic.

(More precisely, see statement of Selberg’s conjectures in Section 7 of this article.)

Theorem (Margulis’ commensurator criterion.) Let $\Gamma < G$ be an irreducible lattice (where G may have rank 1). Then $\Gamma$ is arithmetic iff the commensurator of $\Gamma$ is dense in G.

Recall the commensurator of $\Gamma < G$ is the subset of G consisting of elements g s.t. $\Gamma$ and $g \Gamma g^{-1}$ are commensurable

12) Rigidity of lattices

(i.e. when can you deform a lattice in the ambient Lie group?)

Theorem (Weil local rigidity). Let $\Gamma$ be a finitely generated group, G a Lie group and $\pi: \Gamma \to G$ be a homomorphism. Then $\Gamma$ is locally rigid if $H^1(\Gamma, \mathfrak{g}) = 0$. Here $\mathfrak{g}$ is the Lie algebra of G and $\Gamma$ acts on $\mathfrak{g}$ by $\mathrm{Ad}_G \circ \pi$.

Theorem (Mostow rigidity). Let Γ and Δ be discrete subgroups of the isometry group of $\mathbb{H}^n$ (with n > 2) whose quotients $\mathbb{H}/\Gamma$ and $\mathbb{H}/\Delta$ have finite volume. If Γ and Δ are isomorphic as discrete groups, then they are conjugate.

i.e. lattices in hyperbolic space are pretty darn rigid. But then there’s even more. A lattice is caled irreducible if no finite index subgroup is a product;

Theorem (Margulis superrigidity). Let Γ be an irreducible lattice in a connected semisimple Lie group G of rank at least 2, with trivial center, and without compact factors. Suppose k is a local field.

Then any homomorphism π of Γ into a noncompact k-simple group over k with Zariski dense image either has precompact image, or extends to a homomorphism of the ambient groups.

Remark. Margulis superrigidity implies Margulis arithmeticity (how?)

Margulis’ normal subgroup theorem. Let G be a connected semisimple Lie group of rank > 1 with finite center, and let $\Gamma < G$ be an irreducible lattice.

If $N \leq \Gamma$ is a normal subgroup of $\Gamma$, then either N lies in the center of G (and hence $\Gamma$ is finite), or the quotient $\Gamma / N$ is finite.