Today I Learned: the Infinite Nielsen Extension

I just saw something so cool that I had to write it down.

Grab a bordered Riemann surface \(R_0\).

Done it yet?

Goodo.

Now uniformize it with the unique (compatible) hyperbolic metric with geodesic boundaries; the universal cover of \(R_0\) will sit as some (strict) subset of the hyperbolic plane \(\mathbb{H}\). The fundamental group of \(R_0\) acts discontinuously on \(\mathbb{H}\), and the the quotient space under this action is a hyperbolic surface \(R_1\supset R_0\). Now forget about the hyperbolic metric on \(R_1\) and just remember its Riemann surface structure. And...

...uniformize it with a hyperbolic metric that has geodesic boundaries; ...etc.
(Just repeat what we did in the previous paragraph to get a \(R_2\) and so forth.)

By the end of this process, we wind up with whole bunch of Riemann surfaces\begin{align}R_0\subset R_1\subset R_2\subset R_3\subset\ldots \end{align}so that when each one is uniformized to have a hyperbolic metric with geodesic boundaries, the lengths of these boundaries are strictly decreasing! In fact, take the Riemann surface obtained as the "union" of all of these and you'll have something that uniformizes to a hyperbolic surface with cusps.

And that's the infinite Nielsen extension.

It's...it's actually kinda obvious when you think about it, but I still reckon it's pretty sweet. =P

In summary:

Maan...pictures are too efficient. Gotta nerf them.

Today I Learned: the Poincare Inequality

It's been a while since I've last written on this blog - and I've been learning some pretty cool stuff. Yes, I've had to concede that analysis is cool. Well, at least it's functional analysis.

Here's an awesome result that I saw yesterday:

Theorem of win (*): let \(\overline{M}\) be a connected compact Riemann surface with non-empty boundary, and \(K\) a smooth nonpositive function on \(\overline{M}\), then there exists a \(\mathbb{C}\)ompatible Riemannian metric \(g\) on \(\overline{M}\) which has \(K\) as its Gaussian curvature! Moreover, this metric \(g\) becomes unique (and still exists) if you designate a (smooth) metric on the boundary.

It's paraphrased slightly because I wanted to avoid asserting that every smooth Riemannian metric in a conformal class of metrics on a manifold can be written in the form: \[g=e^{2\varphi}g_0\] for some base metric \(g_0\) and any smooth function \(\varphi: \overline{M}\rightarrow\mathbb{R}\). In retrospect, maybe I should have just gone with that. Oh well, it's too late Liz Lemon, it's too late (Blast! Can't find a referencing Youtube clip for this 30 Rock quote).

On the plus side, I managed to introduce \(\mathbb{C}\)ompatible instead of saying compatible with its Riemann surface structure. Pretty Stokes'd with that...

Basically, this theorem tells us that as long as you don't deal with positive curvature, you can impose any kind of Gaussian curvature on your surface. This result isn't (to my understanding) all that hard to prove, although I still find some of the language a bit intimidating. But the fact that it's solving for a Dirichlet problem (i.e.: you specify the metric on the boundary) as opposed to something that let's you specify the geodesic curvature on the boundary, is unfortunate. Well, at least for me. Because I really want to prove stuff like, you can uniformize a bordered Riemann surface uniquely so that it's hyperbolic and has constant boundary geodesic curvature blah.

So I tried to prove this instead...and I think that I've done it. Almost all of the arguments are based on what Osgood, Phillips and Sarnak used in this little paper, all I did was that I:

1. Came up with a family of functionals that link together the two that they study for surfaces with negative Euler characteristic, and found appropriate constraints on the domain of these functionals.
2. Worked out a tweaked Poincare inequality that makes proves the boundedness of a sequence of functions tending to the infimum of one of our given functionals (step 3 of this).
3. Emailed Ben Andrews at the ANU to see if elliptic regularity would hold for one of the steps that I needed. He was really quick about responding to my emails and politely nudged me towards a pretty sweet reference when I asked for one (+we both know that I know waaaay too little analysis, and need to do a bit more ground-work before I can make use of his suggestions/hints/glaringly obvious solution to anyone who's moderately okay with analysis).

So let's try to prove this tweaked Poincare inequality, like, now.

Claim: given a Riemannian surface \((R,g)\), let \(W^{1,p}(R)\) denote its \((1,p)\)-th  Sobolev space. Then for a constant \(C\) dependent only upon \(R\) and \(p\), \begin{align}\|\varphi\|_p=\left(\int_R|\varphi|^pdA\right)^{\frac{1}{p}}\leq C\left(\int_R|\nabla\varphi|^pdA\right)^{\frac{1}{p}}=\|\nabla\varphi\|_p\end{align} is true for functions \(\varphi\) in any subspace \(W^{1,p}(R)^{\tau}\subset W^{1,p}(R)\) constrained by: \begin{align}(1-\tau)\int_R\varphi dA+\tau\int_{\partial R}\varphi|_{\partial R} ds=0,\,0\leq\tau\leq1,\end{align}where \(\varphi|_{\partial R}\) denotes the trace of \(\varphi\) - the notion agrees with what you think it means when \(\varphi\) is continuously defined over all of \(R\).

Sigh...such a mouthful. Here comes the proof (**):

Let's prove this by contradiction. Assume that there is sequence of functions \(\{\varphi_k\}\) in \(W^{1,p}(R)^{\tau}\) so that \begin{align}\int_R|\varphi_k|^pdA>k\int_R|\nabla\varphi_k|^pdA.\end{align} This means that \(\varphi_k\) can't be \(0\), hence the norms \(\|\varphi_k\|_p\) are nonzero and we may assume without loss of generality that the \(\varphi\) were normalized to begin with. Note that the normalization of a function still lies in \(W^{1,p}(R)^\tau\) because it's a vector subspace of the Sobolev space \(W^{1,p}(R)\).

The normalization means that \(\|\varphi_k\|_p=1\) and \(\|\nabla\varphi_k\|_p<\frac{1}{k}\). The former fact, combined with the Rellich-Kondrachov theorem tells us that a subsequence of the \(\{\varphi_k\}\) must converge strongly (and hence a subsequence of that subsequence converges pointwise) to some \(\varphi\in L^p(R)\). Relabel \(\{\varphi_k\}\) as this subsequence, coz we can.

Okay, I lied slightly. It doesn't follow from the most broad versions of Rellich-Kondrachov theorem straight off. The idea is that for bounded domains and \(p\geq n\), we know that \(W^{1,p}(\Omega)\subset W^{1,n-\epsilon}(\Omega)\) and so if you bring \(\epsilon\) close enough to \(0\), you can get that \(W^{1,n-\epsilon}(\Omega)\) is compactly embedded in \(L^p(\Omega)\). Hence, \(W^{1,p}(\Omega)\) is compactly embedded in \(L^p(\Omega)\) for all \(p\in[1,\infty]\). But to be honest, I'm a bit worried about this argument, because it's based on my understanding of the \(n=p\) case argument presented in a remark at the end of section 5.7 in Evans (**). And maybe I've misunderstood him, because he invokes Morrey's inequality and Arzela-Ascoli to handle when \(p>n\). Worrisome.

Anyhoo, back to the proof.

Let's show that the \(\varphi\) we found actually lies in \(W^{1,p}(R)^{\tau}\). We're gonna do this by constructing a weak derivative in \(L^p(R)\). Take an arbitrary compactly supported smooth function \(\phi\in C^{\infty}_c(R)\) (i.e.: a test function), then we have that: \begin{align}\int_R \varphi\,\partial_{x_i}\phi dA=\lim_{k\to\infty}\int_R\varphi_k\,\partial_{x_i}\phi dA=-\lim_{k\to\infty}\int_R\partial_{x_i}\varphi_k\,\phi=0,\end{align}
where the last limit is because \(\|\partial_{x_i} \varphi_k\|_p\leq\|\nabla \varphi_k\|_p<\frac{1}{k}\). Therefore, we may take the weak derivative \(\partial_{x_i}\varphi\) to be the zero function, and so we see that any first order weak derivatives of \(\varphi\) are zero, which means that \(\varphi\in W^{1,p}(R)\). Now invoking either problem 11 of Evans (**) or corollary 2.1.9 of (***), we know that \(\varphi\) has to be constant. At the same time, we know that \(\varphi\) has to also satisfy the constraint that we initially placed upon \(W^{1,p}(R)\). Therefore, \(\varphi=0\), which contradicts the fact that \(\|\varphi\|_p=1\).

Hooray! 

Or rather, maybe...hooray? I guess in the course of writing this up, I've realised that there are a few details here and there that I'm a tad worried about. For example, do I know that the constraint here in this last step extends to \(\varphi\)? I don't see why not, and it certainly does on the area integral, but I'm not so comfortable with what's happening on the boundary with the trace operator and jazz. (Update: yea, it works out because the trace operator is continuous.) Plus there's that whole business with showing that \(\varphi\) is in \(W^{1,p}(R)\). I'm not really sure why we had to play with those integrals. Is it because we cannot conclude from \(\|\nabla\varphi_k\|_p\) tending to \(0\) that \(\|\nabla\varphi\|_p\) is zero and we just had to go through that rigmarole? (Update: I now think that it's to show that zero is a valid candidate for the weak derivative.) I'm not comfortable enough with analysis to say for sure yet. But whatever the case, I've learned stuff man. I've learned stuff.

Oh and as per usual, I lament the fact that I still haven't finished writing up about Lie's line-sphere correspondence, and resolve to do it soon.

*: Prop 1.7 in Ch 14 of Taylor's "Partial Differential Equations Vol. 3". It's kind of a long chapter. In fact, chapter 14 is kind of the whole book minus some backgroundy stuff, I think?
**: the proof is adapted from Theorem 1 in section 5.7 of Evans's "Partial Differential Equations". It's a pretty thick book, but the writing is beautiful.
***: Ziemer's "Weakly Differentiable Functors".

Today I learned: A Pain in the Anal-ysis

I'm currently watching the Aussie Open men's final between Djokovic and Murray, and it's shaping out to be a reasonably long match (I may have written this when they were still at 6-7,7-6). Speaking of long matches, I hear that the longest women's match is six and a half hours long and featured a half-hour 643 point rally! And they only played two sets.

So let's take our minds off such painful, laborious sports and do something relaxing, like analysis. Yes. Analysis.

I feel that most pure mathematicians that I know have very little tolerance or appreciation for analysis, and I am certainly no hipster-analysis-lover. But I've been looking at Osgood, Phillips and Sarnak's paper on finding the (unique) nicest metric compatible with the complex structure of a Riemann surface \(R\). To be bluntly honest, I haven't been exposed to very much functional analysis, and the idea here is sorta cool. Let's try to really crudely summarise the schematic of what I suspect is a generic strategy in this area:

1. Take a big vector space \(W\) of functions, like a Sobolev space, and define a functional \(F:W\rightarrow\mathbb{R}\) on it. Often, the functional is measuring some sort of energy these functions defined on a Riemannian manifold.
2. Show that \(F:W\rightarrow\mathbb{R}\) is strictly convex (concave), this means that the global minimum (maximum) of \(F\) is unique.
3. Plonk \(W\) in a bigger space \(H\) - often a \(L^p\) space, something like the Rellich-Kondrachov theorem is then used to show that a sequence of functions in \(W\) whose energies tend to the infimum of \(F\) has a convergent subsequence tending to some \(\varphi\in H\).
4. Use stuff like the elliptic regularity theorem to show that \(\varphi\) is a lot smoother than your average function in \(H\) and is in fact in \(W\) and hence unique by point 2.

Well, that's sorta my take on what appears to be important. I'll probably realise at some point that I'm pretty wrong.

During the process of going through and verifying each step of this proof there were two things that I couldn't get my head around. The first I'm still paranoid about, and I'll probably ask about it on MathOverflow in a day or two. It has to do with whether elliptic regularity holds for sufficiently nice nonlinear elliptic operators. The second had to do with why a strongly convergent sequence of measurable functions in \(L^p(\Omega\subset(\mathbb{R}^n,\mathrm{d}V))\) always has a subsequence that converges pointwise amost everywhere (a.e.) to a measurable function.

Let's prove this jazz.

So we've got a sequence of measurable real functions \[f_n:\Omega\rightarrow\mathbb{R}\] in \(L^p(\Omega)\) and we want to show that there's a subsequence that converges pointwise a.e.. Let's begin by figuring out what pointwise convergence actually means.

In order for the sequence \(\{f_n\}\) to converge to some function \(f\), it needs to get arbitrarily close to \(f\) as \(n\) blows up. That is: for any \(x\in\Omega\) and any \(\epsilon>0\), \begin{align}\limsup_{n\rightarrow\infty}|f_n(x)-f(x)|\ngtr\epsilon. \end{align} And since we're dealing with a.e. pointwise convergence, let's just turn this into a measure theoretic statement: for any \(\epsilon>0\), \begin{align}\mathrm{Vol}(\{x\in\Omega:\,\limsup_{n\rightarrow\infty}|f_n(x)-f(x)|>\epsilon\})=0.\end{align}
Okay, good, we've heuristically justified why this condition is equivalent to pointwise a.e. convergence.

Now, just by interpreting what \(\limsup\) is actually doing, we see that \begin{align}\mathrm{Vol}(\{x\in\Omega:\,\limsup_{n\rightarrow\infty}|f_n(x)-f(x)|>\epsilon\})\\=\lim_{n\rightarrow\infty}\mathrm{Vol}\left(\bigcup_{k=n}^{\infty}\{x:\,|f_k(x)-f(x)|>\epsilon\}\right). \end{align} So if we can show that \begin{align}\mathrm{Vol}\left(\bigcup_{k=n}^{\infty}\{x:\,|f_k(x)-f(x)|>\epsilon\}\right)\leq\sum_{k=n}^{\infty}\mathrm{Vol}(\{x:\,|f_k(x)-f(x)|>\epsilon\})\end{align} tends to \(0\) then we're done. But of course, one way of showing that the tail of a series converges to \(0\) is to show that the series converges.

Now, if we knew that \(\mathrm{Vol}(\{x:\,|f_k(x)-f(x)|>\epsilon\})\) tended to \(0\) we could easily take a subsequence \(\{f_{n_m}\}\) so that \begin{align}\mathrm{Vol}(\{x:\,|f_{n_m}(x)-f(x)|>\epsilon\})<2^{-m}\end{align} and this subsequence would have to converge pointwise a.e. to \(f\) - which is precisely the type of conclusion that we'd like. So everything boils down to using strong convergence to get that \begin{align}\mathrm{Vol}(\{x:\,|f_k(x)-f(x)|>\epsilon\})\rightarrow0\text{ as }n\rightarrow\infty.\end{align} This, however, can be shown as follows:\begin{align}&\mathrm{Vol}(\{x:\,|f_k(x)-f(x)|>\epsilon\})=\mathrm{Vol}(\{x:\,\epsilon^{-1}|f_k(x)-f(x)|>1\})\\=&\int_{\{x:\,\epsilon^{-1}|f_k(x)-f(x)|>1\}}\mathrm{d}V\\ <&\int_{\{x:\,\epsilon^{-1}|f_k(x)-f(x)|>1\}}\epsilon^{-p}|f_k(x)-f(x)|^p\mathrm{d}V\\\leq&\int_{\Omega}\epsilon^{-p}|f_k(x)-f(x)|^p\mathrm{d}V\\=&\epsilon^{-p}\|f_n-f\|_p^p.\end{align}Since strong convergence is defined to mean that \(\|f_n-f\|_p\) tends to \(0\) and \(\epsilon\) is just some fixed constant, this proves our claim. Oh, and I should probably point out that \(\|\cdot\|_p\) is notation for the norm on \(L^p(\Omega)\).

I really ought to have finished writing this post before the match ended. Oh well. =)
 

A First Post and a Problematic Problem

I made my first MathOverflow post today.

I really should be posting MathOverflow a lot more. I mean, I check it out every so often, usually when I'm Googling for a solution to a problem that I can't be bothered solving. In fact, that's pretty much how I was motivated into responding to a two year old unanswered question about solving Beltrami equations.

The problem is: given a complex-valued function \(\mu:U\rightarrow\mathbb{C}\), can you find a solution \(f:U\rightarrow\mathbb{C}\) to the Beltrami equation:\begin{align*}\frac{\partial f}{\partial\bar{z}}=\mu\frac{\partial f}{\partial z}?\end{align*}
 According to Wikipedia, Gauss has a way of doing it for a nice neighbourhood around \(z=0\) if \(\mu\) can be written as a power series in \(z\) and \(\bar{z}\), although for the life of me I can't get their algorithm to work - in fact, I'm fairly convinced that the current Wikipedia entry is wrong. So, I sat down, tried to bash things out by writing down power series and stuff and saw that the following construction should work:

Step 1: Let's write \(\mu=\mu(z,\bar{z})\) and mean that it's a power series in \(z\) and \(\bar{z}\). Then try to solve for the following differential equation (which isn't too bad, or rather, you kinda know that a solution exists):\[\frac{\partial F}{\partial z}(z,w)=-\mu(w, F(z,w)),\;\text{ where } F(0,w)=w.\]

Step 2: Find \(G(z,w)\) such that \(G(z,F(z,w))=w\). That is, the functions\[(z,w)\mapsto (z,F(z,w))\text{ and }(z,w)\mapsto (z,G(z,w))\]are inverse to each other.

Step 3: Take \(f(z)=G(\bar{z},z),\) and it should work.

Note that the construction of the above \(F(z,w)\) and \(G(z,w)\) is strongly motivated by the Wikipedia entry on how Gauss solved this problem.

The annoying thing is that you get a (local) solution to the problem using this construction, but in the game of solving Beltrami equations, we're really looking for quasiconformal maps - that is: homeomorphisms of \(U\) which are quasi-angle-preserving (in the sense that the angles are off, but the ratio of the "off-ness" is bounded).

So sad. =(

On a slightly lighter note, check out this quotation taken from the shortest published paper (which was later retracted for reasons that will be obvious to any sane person) I've ever seen. Go read it, it's only four paragraphs.

Seriously, go read it.

Orrite, if you really can't be bothered reading the whole paper, at least read the following line taken from the conclusions section:

"In brief an impossible proposition was proved as possible. This is a problematic problem."

Someone needs to make a T-shirt based on this, stat.

On a slightly darker (?) note, I really ought to write up a bit more of that Klein-Lie stuff.