# 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.