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As part of a course I'm taking this semester, I am studying surfaces from this book http://www.math.brown.edu/~res/Papers/surfacebook.pdf.

On page 142, the author presents a proof of the fact that every compact hyperbolic surface has a geodesic triangulation. The second sentence of the proof says that

"By compactness, there is some $d \in (0,1)$ such that every disk of radius $d$ on the surface is isometric to a disk of radius $d$ in $\mathbb{H}^2$."

I don't understand how compactness implies the existence of such a $d$. Could someone please explain how the author arrived at the conclusion.

Also, in the very next sentence, the author speaks about disks of radius $D/K$. I understand that $K$ is a constant which is specified later on in the proof, but what is $D$ here? I thought it may have been a typo (and in fact $d/K$ was intended) seeing as this pdf is only a draft copy of the book, but I went to Amazon and had a look at a preview of the book here http://www.amazon.com/Mostly-Surfaces-Student-Mathematical-Library/dp/0821853686/ref=sr_1_1?ie=UTF8&qid=1351257505&sr=8-1&keywords=mostly+surfaces, and there the $D$ is unchanged. So now I'm left confused.

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1 Answer 1

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Your first question is answered by a general fact from Riemannian geometry. If $M$ is a closed (complete by compactness) Riemannian manifold, then the injectivity radius is bounded away from zero.

Proof: If the injectivity radius weren't bounded away from zero, there would exist a sequence $x_k$ with injectivity radius at each point limiting to $0$ as $k\to\infty$. By compactness, pass to a convergent subequence. Examine $x=\lim x_k$. The exponential map at $x$ is a diffeomorphism of a small ball about zero in $T_xM$ onto its image in $M$. This image contains all but finitely many $x_k$. Connect each $x_k$ to $x$ by a geodesic; then extend the geodesic beyond $x$ by some fixed amount less than the radius at $x$. This geodesic is less than the injectivity radius at each $x_k$ and has strictly positive length so the injectivity radii of the $x_k$ could not have gone to $0$ after all.

So, on your surface $S$, there is a positive infimal geodesic length. That's $2d$. In this case, let's now move to the universal cover $\mathbb{H}^2$. Take any $p$ and identify it with some point in $\mathbb{H}^2$. Since geodesics on $S$ are exactly projections of geodesics in $\mathbb{H}^2$, we see that we can travel at least $d/2$ away from $p$ in $\mathbb{H}^2$ before the restriction of the projection becomes noninjective. Since the projection is a local isometry, we have the desired isometry between a disc of radius $d$ in $\mathbb{H}^2$ and a disc of radius $d$ around $p$ in $S$.

Here are two more perspectives.

  1. This is a more concrete way of seeing the previous argument. Lift $p$ to $\mathbb{H}^2$ and take a fundamental domain for the action of $\pi_1$ on $\mathbb{H}^2$ centered at $p$. This fundamental domain is a compact polygon. The previous discussion on injectivity radius shows that the boundary of the polygon is at least $d$ away from $p$, so the disc of radius $d$ about $p$ in $\mathbb{H}^2$ projects isometrically to the disc of radius $d$ about $p$ in $S$.
  2. On negatively curved closed manifolds, each free homotopy class of curves has a minimal geodesic representative. It's a fact that on closed hyperbolic surfaces, there is a shortest nontrivial geodesic; the length of this geodesic is twice the radius $d$.

The Amazon link doesn't let me look at page 142, but it seems to me that $D$ should in fact be $d$. Here's my informal interpretation, for what it's worth. The point of choosing $K$ is to make $X_p$ "dense enough" in $S$ so that every $N_p$ is contained in a disk of radius at most $d/2$. This gives that each $N_p\cup X_p$ is contained in a disk of radius $d$ that is isometric to a disk in $\mathbb{H}^2$, so that previous results about polygons in $\mathbb{H}^2$ apply.

But let's make sure that the argument works if we were to replace $D$ by $d$. Then if we were to we pick that $K$ is greater than, say, $6$, we'd be able to fit $3$ disjoint disks of radius $d/6$ in a circle of radius $d$ about $p\in X$. The triangle inequality would then force $X_p$, hence $N_p$, to be contained in $B_d(p)$, so unless I'm extra-flawed today, the argument would carry.

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Sorry but I don't see how this implies what I need about $d$. From the definition of $d$ you've given, why must now every disk of radius $d$ on the surface be isometric to some disk of radius $d$ in the hyperbolic plane? –  HJSprime Oct 26 '12 at 14:11
I'll rework the answer to better answer your question in a few hours. –  Neal Oct 26 '12 at 16:12
@HJSprime I've reworked the answer. I think it should do better than my first one :) –  Neal Oct 26 '12 at 19:35
Finally makes sense to me, thanks for the help. –  HJSprime Oct 27 '12 at 7:57

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