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# Isomorphic groups via hyperbolic geometry

$\mathrm{PSO}(1, 2)$ is the isometry group of hyperbolic 2-space under the hyperboloid model; $\mathrm{PSL}(2,\mathbb{R})$ is the isometry group of the upper half-plane; $\mathrm{PSU}(1,1)$ is the isometry group of the Poincaré unit disk. Since these are in fact models for the same space, all of these Lie groups are isomorphic.

Question: is there a proof of this isomorphism that doesn’t proceed through this identification as isometry groups? (Probably, via the Lie algebras for instance, but I’m a little too lazy to try and work it out at the moment.)

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# Mostow Rigidity: several proofs

Mostow rigidity (or, technically, the special case thereof, for $\mathrm{PO}(1,3)$) states that if M and N are two closed (i.e. compact and boundary-less) hyperbolic 3-manifolds, and $f: M \to N$ is a homotopy equivalence, then f is homotopic to an isometry $g: M \to N$.

This is a strong rigidity result. If M and N were hyperbolic surfaces, by contrast, there are plenty of homotopy equivalences between them which are not homotopic to isometries; indeed, this is the very subject of Teichmüller theory (technically, we’re looking at moduli space, not Teichmüller space, but eh.)

We can state it in algebraic terms, by identifying the hyperbolic 3-manifolds $M \cong \mathbb{H}^3 / \Gamma_M$ and $N \cong \mathbb{H}^3 / \Gamma_N$ with their respective fundamental groups $\Gamma_M = \pi_1(M)$ and $\Gamma_N = \pi_1(N)$, which are (uniform) lattices in $\mathrm{Isom}^+(\mathbb{H}^3) \cong \mathrm{PO}(1,3)$.

Mostow rigidity then states any isomorphism between the lattices is given by conjugation in $\mathrm{PO}(1,3)$. Again, contrast this with the case of lattices in $\mathrm{Isom}^+(\mathbb{H}^2) \cong \mathrm{PO}(1,2)$ (i.e. fundamental groups of hyperbolic surfaces), which have plenty of outer automorphisms (elements of the [extended] mapping class group.)

Here we describe, at a relatively high level, various proofs of this result.

### Step 1: lift and extend

Let $\tilde{h}: \mathbb{H}^3 \to \mathbb{H}^3$ be a lift of h to the universal cover(s). This map is a quasi-isometry, by a Milnor-Švarc argument:

• The Cayley graphs of $\Gamma_M$ and $\Gamma_N$ are both quasi-isometric to $\mathbb{H}^3$ by Milnor-Švarc, the quasi-isometry from the Cayley graphs into hyperbolic space being the orbit map.
• If we now construct quasi-inverses from hyperbolic space to the Cayley graphs, then $\tilde{h}$ is coarsely equivalent to a map between the Cayley graphs obtained by composing one orbit map with the quasi-inverse of the other. To wit: $\tilde{h}$ sends $\gamma x_0$ to $(h_* \gamma) x_0$, and the induced map $h_*$ is an isomorphism of the fundamental groups

The quasi-isometry $\tilde{h}$ from the hyperbolic space $\mathbb{H}^3$ to itself extends to a self-homeomorphism of the Gromov boundary $\partial \mathbb{H}^3$:

This is a standard construction which holds for any quasi-isometry between hyperbolic spaces in the sense of Gromov: the boundary extension is defined by $\partial\tilde{h}([\gamma]) = [\tilde{h} \circ \gamma]$, and Gromov hyperbolicity is used to show that this map is a well-defined homeomorphism; roughly speaking, the idea is that quasi-isometries are like “biLipschitz maps for far-sighted people”, and at the boundary at infinity, the uniformly bounded distortions become all but invisible.

### Step 2: boundary map is quasiconformal

Now we claim that $\partial\tilde{h}: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is quasiconformal.

This follows e.g. from the following geometric lemma: if L is a geodesic and P a totally-geodesic hyperbolic plane in $\mathbb{H}^3$ with $L \perp P$, then $\mathrm{diam}(\pi_{\hat(L)}(\tilde{h}(P)) \leq D$ for some constant D depending only on the quasi-isometry constants of $\tilde{h}$. (Here $\hat{L}$ is the unique geodesic uniformly close to the quasigeodesic $\tilde{h}(L)$.)

The proof of the lemma uses three key ingredients from the geometry of hyperbolic space: thinness of ideal triangles, quasigeodesic stability—which says that the image of any geodesic segment by a quasi-isometry is uniformly (depending only on the quasi-isometry constants) close to the actual geodesic with the same endpoints, which we will call the “straightening” of the quasigeodesic—, and the (1-)Lipschitz property of projection to a geodesic.

Let $x \in L \cap P$ be the unique point of intersection. If $y \in P$, let z be the point in $\partial_\infty P$ corresponding to the ray from x through y. Now consider the ideal triangle with L as one of its sides and as the third vertex. h(x) is uniformly close to the images of the two sides ending at z , and also to straightenings $\bar{M_1},\bar{ M_2}$ thereof; since projection to a geodesic is Lipschitz, the projection of h(x) to the straightening $\hat{L}$ of the quasigeodesic h(L) is uniformly close to the projections of $\bar{M_1}, \bar{M_2}$.

But now the image of the geodesic ray [x, z] lies uniformly close to some geodesic ray, whose projection onto $\bar{L}$ lies between the projections of h(x) and h(z). Hence any point on this geodesic ray has image which projects between h(x) and h(z); again using the Lipschitz property of the projection, we obtain uniform bounds on the diameter we are trying to control, as desired.

Now note that any circle in $\partial\mathbb{H}^3 \cong \hat{\mathbb{C}}$ is the boundary of a totally geodesic hyperbolic plane in $\mathbb{H}^3$, and our lemma then tells us that the image of the circle under $\tilde{h}$ has its outradius/inradius ratio bounded above by $e^D$.

### Step 3: boundary map is conformal

And then we would be done, for this implies $\tilde{h}$ is (up to homotopy) a hyperbolic isometry, which descends to an isometry $h: M \to N$.

#### An argument along the lines of Mostow’s original

Suppose $\tilde{h}$ is not conformal. Consider the $\Gamma_M$-invariant measurable line field $\ell_h$ on $\hat{\mathbb{C}}$ obtained by taking the direction of maximal stretch at each point.

Rotating this line field by any fixed angle $\phi$ yields a new measurable line field $\ell_h^\phi$. But now $\bigcup_{0 \leq \phi \leq \pi/2} \ell_h^\phi$ yields a measurable $\Gamma_M$-invariant proper subset of $T^1M$, which contradicts the ergodicity of the geodesic flow on $T^1 M$.

An argument following Tukia, in the spirit of Sullivan

Suppose $\tilde{h} =: \phi$ is differentiable at z. Let $\gamma$ be a Möbius transformation fixing z and $\infty$, with z an attracting fixed point and with no rotational component. Now, by the north-south dynamics that hyperbolic Möbius transformations exhibit, $\gamma^n \phi \gamma^{-n} \to d\phi_z$ as $n \to\ infty$; on the other hand, $\gamma^n \phi \gamma^{-n}$ conjugates $\gamma^n\Gamma_1\gamma^{-n}$ to $\gamma^n\Gamma_2\gamma^{-n}$.

Again, using convergence dynamics, we have $\gamma^n\Gamma_i\gamma^{-n} \to \Gamma_i^\infty$ as $n \to \infty$, for i = 1, 2, where both $\Gamma_i^\infty$ are cocompact and of equal covolume.

Now $d\phi_z$ conjugates $\Gamma_1^\infty$ to $\Gamma_2^\infty$; thus $\phi$ is conformal a.e., and hence conformal.

### Alternative perspective I: the Gromov norm

Gromov gives a proof which follows Step 1 as above, but then proceeds differently: he uses the Gromov norm to show that $\tilde{h}$ is conformal, without (separately) showing that it is quasiconformal. We sketch an outline of this here; see Calegari’s blog and/or Lücker’s thesis for details.

The Gromov norm is a measure of complexity for elements of the (singular) homology—more precisely, it is the infimum, over all chains representing the homology element, of $L^1$ norm of the chain.

It is a theorem of Gromov that the Gromov norm of (the fundamental class of) an oriented hyperbolic n-manifold M is equivalent, up to a normalization dependent only on n, to the hyperbolic volume of M. The proof proceeds by

• observing that it suffices to take the infimum over rational chains of geodesic simplices; this suffices to give the lower bound, with a constant given in terms of volume of an ideal simplex;
• using an explicit construction involving almost-ideal simplices with almost-equidistributed vertices to obtain a upper bound.

Then, since $\tilde{h}$ is a homotopy equivalence (and by looking at this last construction carefully), it—or, rather, its extension to $\overline{\mathbb{H}^3}$—sends regular simplices close to regular ideal simplices. Simplices in the chains from the upper bound construction above are (almost) equidistributed, and the error can be made arbitrarily small as we use the construction to get arbitrarily close to the bound.

Translating the vertices of our ideal simplices, we get a dense set of equilateral triangles that are sent by $\partial\tilde{h}$ to equilateral triangles on the boundary. This implies $\partial\tilde{h}$ is conformal a.e., and hence conformal.

(In short: Gromov’s theorem and its proof allows us to control volume in terms of explicitly-constructed singular chains and to control what happens to simplices in those chains under our homotopy equivalence.)

### Alternative perspective II: dynamically characterizing locally-symmetric Riemannian metrics

Besson, Courtois and Gallot have an entirely different proof, which uses a characterization of locally-symmetric Riemannian metrics in terms of entropy to conclude that, since f is a degree-1 map between locally-symmetric spaces, M and N are homothetic. It then follows from e.g. an equal-volume assumption that M and N would be isometric.

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# Two (co)compactness criteria for lattices

Here and H are Lie groups, $\Gamma$ and $\Lambda$ are lattices in one of those Lie groups, and C is a subset of the Lie group.

### Detecting (non)compactness via accumulation at the identity

Conjugates by compact sets do not accumulate: The image of $C \subset H$ in $H / \Lambda$ is precompact iff the identity element $\mathrm{id}_H$ is not an accumulation point of the conjugated lattice ${}^C\Lambda$.

[[ some pictures to be inserted here ]]

Proof: ${}^C\Lambda \cap U^{-1}U = \{e\}$ iff we cannot find distinct $u_1, u_2 \in U$ s.t. $u_1c \Gamma = u_2c \Gamma$ for all $c \in C$, iff the translated orbit maps (covering maps!) $H \to H/\Lambda$ given by $h \mapsto hc\Lambda$ are injective on U for every $c \in C$.

Now if our image is precompact, then we may cover it by finitely many pancakes (i.e. neighborhoods in $H / \Lambda$ diffeomorphic to each connected component of their preimage in the cover H); by this finiteness, and H-equivariance, we can find some small open neighborhood U so that the desired injectivity condition on our translated orbit maps holds.

If our image is not precompact, then given any small open neighborhood U we can find a sequence of elements in C escaping any compact translate $U^{-1}K$. By an inductive / diagonal argument (involving larger and larger compact sets, each of which contains a previous $Uc_n\Lambda$) we have a sequence $(c_n) \subset C$ with $Uc_n\Lambda$ pairwise disjoint. Since $\Lambda$ has finite covolume, these cannot all have the same (nonzero) volume, and so (by fundamental domain considerations) not all of the translated orbit maps are injective on U.

Mahler’s compactness criterion is the special case where $H = \mathrm{SL}(\ell,\mathbb{R})$ and $\Lambda = \mathrm{SL}(\ell,\mathbb{Z})$.

This can be proven without using that $\Lambda$ is a lattice, by considering $\mathrm{SL}(\ell,\mathbb{R}) / \mathrm{SL}(\ell,\mathbb{Z})$ as the moduli space of unit-covolume lattices in $\mathbb{R}^\ell$ (or: “wow, the lattices are back at a whole new level!”): then the result says that a set of unit-covolume lattices is compact iff it does not contain arbitrarily short vectors.

The forward implication is fairly immediate after a moment’s thought; the reverse implication follows by considering the shortest vectors which appear in the lattices of any given sequence from our set, observing that those vectors form bounded orbits, taking (sub)sequential limits in those orbits using compactness in $\mathbb{R}^\ell$, and finally passing back to (sub)sequential limits in the sequence of lattices.

### Detecting (non)compactness with unipotents

Cocompact lattices have no unipotents: The homogeneous space $G / \Gamma$ is compact only if $\Gamma$ has no nontrivial unipotents.

One proof [of the contrapositive] follows from (the general version of) Mahler’s criterion above, via the Jacobson-Morosov Lemma which gives a map $\phi: \mathrm{SL}(2,\mathbb{R}) \to G$ sending our favorite unipotent $\left( \begin{array}{cc} 1 & 1 \\ & 1 \end{array} \right)$ to a nontrivial unipotent $u \in \Gamma$. Call a the image under $phi$ of our favorite hyperbolic element $\left( \begin{array}{cc} 2 \\ & 1/2 \end{array}\right)$. Then $a^nua^{-n}$ accumulates to the identity, and so $G / \Gamma$ is not precompact by the first criterion above.

More is true if we have more structure on our lattices:

Godement: Suppose G is defined over a real algebraic number field F (e.g. $\mathbb{Q}$), and let $\mathcal{O}$ be the ring of integers in F. The homogeneous space $G / G_{\mathcal{O}}$ (in the case of $F = \mathbb{Q}$, $G / G_{\mathbb{Z}}$) is compact iff $G_{\mathbb{Z}}$ has no nontrivial unipotents.

[[ some more pictures here ]]

Proof of the new (reverse) direction uses the arithmetic structure on $G_{\mathbb{Z}}$. Again we prove the contrapositive. Suppose $G / G_{\mathbb{Z}}$ is not compact. By (Mumford’s generalisation of) the Mahler criterion above, this implies ${}^gG_{\mathbb{Z}}$ accumulates at the identity for some $g \in G$. By continuity, the characteristic polynomials of the elements in an accumulating sequence $g\gamma_n g^{-1}$ converge (in the sense of coefficients; this makes sense if we are working within some fixed ambient $\mathrm{SL}(\ell, \mathbb{R})$, for example) to $(x-1)^\ell$.

But this implies, by matrix similarity, that the characteristic polynomials of the $\gamma_n$ converge to the same; and now we use the fact that $\gamma$ is an integer matrix to say that this implies the $\gamma_n$, for n sufficiently large, has characteristic polynomial $(x-1)^\ell$, and hence is unipotent.

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# The many faces of the hyperbolic plane

$\mathbb{H}^2$ is the unique (up to isometry) complete simply-connected 2-dimensional Riemann manifold of constant sectional curvature -1.

1. It is diffeomorphic to $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ as a topological space (or, indeed, isometric as a Riemannian manifold, if we give $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ a left-invariant metric): to show this, we note that $\mathbb{H}^2$ has isometry group $\mathrm{SL}(2,\mathbb{R}) / \pm\mathrm{id}$, and the subgroup of isometries which stabilize any given $p \in \mathbb{H}^2$ is isomorphic to $\mathrm{SO}(2)$.
2. Since any positive-definite binary quadratic form is given by a symmetric 2-by-2 matrix with positive eigenvalues, and since the group of linear transformations on $\mathbb{R}^2$ preserving any such given form is isomorphic to $\mathrm{O}(2)$, $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the space of positive-definite binary quadratic forms of determinant 1, via the map from $\mathrm{SL}(2,\mathbb{R})$ to the symmetric positive-definite 2-by-2 matrices given by $g \mapsto g^T g$.
3. $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2) \cong \mathbb{H}^2$ is also the moduli space of marked Riemann surfaces of genus 1, i.e. the Teichmüller space Teich(S) of the torus S. One way to prove this is to note that any such marked surface is the quotient of $\mathbb{R}^2$ by a $\mathbb{Z}^2$ action; after a suitable conformal transformation, we may assume that the generators of this $\mathbb{Z}^2$ act as $z \mapsto z + 1$ and $z \mapsto z + \tau$ for some $z \in \mathbb{H}^2$ (in the upper half-plane model.) But now $\tau$ is the unique invariant specifying this point in our moduli space.
4. Since any marked Riemann surface of genus 1 has a unique flat metric (inherited as a quotient manifold of the Euclidean plane), $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the moduli space of marked flat 2-tori of unit area.
5. Since there is a unique unit-covolume marked lattice associated to each marked complex torus in the above, $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the space of marked lattices in $\mathbb{R}^2$ with unit covolume.
6. Note we may go directly between marked Riemann surfaces and quadratic forms by considering intersection forms.
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# Tits Alternative for abstract Lie groups …

… with finitely many connected components follows from the Tits alternative for linear groups. The reduction appears to be standard (and is used e.g. in the proof of Gromov’s polynomial growth theorem) but not written down anywhere; I am recording it here for future reference.

Let be a Lie group with finitely many connected components. Ad(G) is a subgroup of a linear group; by the Tits Alternative for linear groups, it is either virtually solvable or contains a free group on two generators. In the former case, G, being (a finite extension of) an extension of Ad(G) by Z(G), is virtually solvable as well. In the latter case, the two generators of the free group have pre-images which are generators of a free group in G.

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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.