## Simultaneous Uniformization TheoremOctober 23, 2010

Posted by Sarah in Uncategorized.
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The other day, in the graduate student talks, Subhojoy was talking about the Simultaneous Uniformization Theorem. It was a nice treat because I used to be really into geometric topology (or at least as much as an undergrad with way too little background could.)

The big reveal is
$QF(S) \simeq T(S) \times T(S)$
but most of the talk, naturally, goes into defining what those letters mean.

The Riemann Mapping Theorem says that any simply connected set $U \subset \mathbb{C}$ is conformally equivalent to the disc $\mathbb{D}$. Conformal maps are angle-preserving; in practice, they’re holomorphic functions with derivative everywhere nonzero. Conformal maps take round circles to round circles if the circles are small enough.

A Riemann Surface is a topological surface with conformal structure: a collection of charts to the complex plane such that the transition maps are conformal.

The first version of the Uniformization Theorem says that any simply-connected Riemann surface is conformally equivalent to the Riemann sphere (or complex plane or unit disc; these are all equivalent.)

The second, more general version of the Uniformization Theorem says that any Riemann surface of genus $\ge 2$ is conformally equivalent to $\mathbb{H}^2/\Gamma$ where $\mathbb{H}^2$ is the hyperbolic plane and $\Gamma$ is a discrete subgroup of $PSL_2 \mathbb{R}$.

To understand this better, we should observe more about the universal cover of a Riemann surface. This is, of course, simply connected. Its deck transformations are conformal automorphisms of the disc. But it can be proven that conformal automorphisms of the disc are precisely the Mobius transformations, or functions of the form

$\frac{az + b}{cz+d}$

This implies that the automorphism group of $\mathbb{D}$ is $PSL_2\mathbb{R}$.

Now observe that there’s a model of the hyperbolic plane on the disc, by assigning the metric

$\frac{4(dr^2 + r^2 d\theta^2)}{(1-r^2)^2}.$

And, if you were to check, it would turn out that conformal transformations on the disc preserve this metric.

So it begins to make sense; Riemann surfaces are conformally equivalent to their universal covering space, modulo some group $\Gamma$ of relations, a subgroup of the group of deck transformations of the universal cover.

$\Gamma$ are called Fuchsian groups — these define which Riemann surface we’re talking about, up to a conformal transformation.

Now we can define Fuchsian space as

$F(S) = \{ \rho: \pi_1(S) \to PSL_2\mathbb{R} | \rho \mathrm{discrete, injective} \}$

It’s the set of maps from the fundamental group of a surface to $PSL_2\mathbb{R}$.

And we can define Teichmuller space as the space of marked conformal structures on the surface S.
This is less enormously huge than you might think, because we consider these up to an equivalence relation. If $f: S \to X$ and $g: S \to Y$ are conformal structures, and there exists a conformal map $h: X \to Y$ such that $h \circ f = g$ then we consider $(f, X) \sim (g, Y)$ equivalent structures.

In fact, Teichmuller space is not that enormously huge: $T(S) \simeq \mathbb{R}^{6g-6}$. It turns out that Teichmuller space is completely determined by what happens to the boundary circles in a pair of pants decomposition of the surface.

Here’s a picture of a pair of pants (aka a three-punctured sphere):

Here’s a picture of a decomposition of a Riemann surface into pairs of pants:

(Here’s a nice article demonstrating the fact. It’s actually not as hard as it looks.)

Now we generalize to Quasi-Fuchsian spaces. For this, we’ll be working with hyperbolic 3-space instead of 2-space. The isometries of hyperbolic 3-space happen to be $PSL_2 \mathbb{C}$.
Instead of a Poincare Disc Model, we have a ball model; again, $PSL_2 \mathbb{C}$ acts by Mobius transformations, functions of the form $\frac{az+b}{cz+d}$.

A quasiconformal function takes, infinitesimally, circles to ellipses. It’s like a conformal map, but with a certain amount of distortion. The Beltrami coefficient definds how much distortion:

$\mu = \frac{f_{\bar{z}}}{f_z}$

Quasi-Fuchsian space, QF(S), is the set of all quasiconformal deformations of a Fuchsian representation. In other words, this is the space of all representations to $PSL_2 \mathbb{C}$ preserving topological circles on the boundary.

Now, the Simultaneous Uniformization Theorem can be stated:
the Quasi-Fuchsian space of a surface is isomorphic to the product of two Teichmuller spaces of the surface.

One application of this theorem is to hyperbolic 3-manifolds.
If $M = \mathbb{H}^3/\Gamma$ is a hyperbolic 3-manifold, and if $\Gamma \simeq S$ then $M \simeq S \times \mathbb{R}$.

In other words, we can think of a hyperbolic three-manifold as the real line, with a Riemann surface at each point — you can only sort of visualize this, as it’s not embeddable in 3-space.

The Simultaneous Uniformization Theorem implies that there is a hyperbolic metric on this 3-manifold for any choice of conformal structure at infinity.

This contrasts with the Mostow Rigidity Theorem, which states that a closed 3-manifold has at most one hyperbolic structure.

Together, these statements imply that any hyperbolic metric on $S \times \mathbb{R}$ is determined uniquely by the choice of conformal structures at infinity.

## Bulldog! Bulldog! Bow wow wow!August 29, 2010

Posted by Sarah in Uncategorized.
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Just a few notes as the week goes by:

1. I’ve got classes picked out now: modern algebra, measure theory and integration, algebraic topology, and harmonic analysis. And the new requirement, “ethical conduct of research” (which I assume means “don’t plagiarize and don’t torture monkeys.”)

2. I love Ikea. I know everybody loves Ikea, but there’s no reason to depart from convention here. Everything is beautiful and cheap and quite a lot of things come in orange.

3. The international students, apparently, were taken aside to learn Yale football chants and associated paraphernalia (including the “Bulldog! Bulldog! Bow wow wow!” cheer.) We Americans were never told anything of the kind. I guess this goes along with the tradition of immigrants knowing the Constitution better than we do.

4. Thanks to a fellow grad student, I’m remembering how much I like geometry and topology. Here’s something I learned (at a pub night, no less!) A hyperbolic manifold is the hyperbolic plane modulo a discrete group of isometries. That means, if you look at the Poincare disc model, some of the points on the circular boundary are limit points of the orbit of some point in the interior under isometries in the group. Consider the set of such points. Apparently: if it’s not the whole circle, it is a Cantor set, and its Hausdorff dimension equals the first eigenvalue of the Laplacian on the manifold. WHOA.

## Convergence of the Discrete Laplace-Beltrami OperatorJune 15, 2010

Posted by Sarah in Uncategorized.
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Via the Geomblog, here’s a paper about the convergence of discrete approximations to the Laplace-Beltrami operator.

What is the Laplace-Beltrami operator? It’s a generalization of the Laplacian for use on surfaces. Like the Laplacian, it’s defined as the divergence of the gradient of a function.
But what does this really mean on a surface?
Given a vector field X, the divergence is defined as
$div X = \frac{1}{\sqrt{|g|}} \partial_i (\sqrt{|g|}X^i)$
in Einstein notation, where g is the metric tensor associated with the surface. (Compare to the ordinary divergence of a vector field, which is the sum of its partial derivatives.)
$(grad (f))^i = g^{ij} \frac{\partial f}{\partial x^j}$
Combining the definitions,
$\Delta f = \frac{1}{\sqrt{|g|}} \partial_i(\sqrt{|g|} g^{ij} \partial_j f)$
Euclidean space is a special case where the metric tensor is just the Kronecker delta.

Why do we care about this operator? Well, you can analyze surfaces with it; on the applied end this means it’s important for signal and image processing. (The Laplace-Beltrami operator is essential to diffusion maps, for instance.) The thing is, in computational applications we generally want a discrete version of the operator to converge to the real McCoy. Hence the importance of convergence algorithms.

Taubin’s discrete version of $\Delta f$ is defined as
$f(v) = \sum_i \omega_i(f(v_i - f(v))$
an averaging operator over the neighbors. This is a weighted graph Laplacian. But this cannot be an approximation to the Laplace-Beltrami operator because it goes to 0 as the mesh becomes finer.

The authors use a different discretization. We assume a triangular mesh on the surface. The local tangiential polygon at a point v is defined to be the polygon formed by the images of all the neighbors upon projection onto the tangent space at v.

A function can be lifted locally to the local tangiential polygon, by defining it at the points on the surface, and making it piecewise linear in the natural way.

After a few lines of manipulation of the Taylor series for a $C^2$ function on the plane, it follows that
$\Delta f(0, 0) = f_{xx}(0,0) + f_{yy}(0, 0)$
$= \frac{2 \sum_{i = 1}^n \alpha_i (f(x_i, y_i) - f(0, 0))}{\sum_{i = 1}^n \alpha_i x_i^2} + O(r)$

Now you can use this as a discrete Laplacian applied to the local lifting function. The authors show this converges to the Laplace-Beltrami operator by using the exponential map from the tangent space into the surface. The Laplace-Beltrami operator can be computed from the second derivatives of any two perpendicular geodesics; let the geodesics be the images under the exponential map of the basis elements of the tangent space.

We use the fact that the normal to a sufficiently fine triangular mesh approximates the normal to the surface according to
$N_\Sigma(v) = N_A v + O(r^2)$
where r is the size of the mesh.
This shows that the basis for the approximating tangent plane is close to the basis for the true tangent plane with error only O(r^2).
Calculating the approximate Laplacian gives us then that the error is only O(r).
There are also some numerical simulations in the paper giving evidence that this approximation is faster and more accurate than a competing approximation.