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First let me state a result of Kazdan and Warner

Let $M$ be a compact orientable two dimensional manifold. Let $f:M \rightarrow \mathbb{R}$ be a function that has the same sign as the Euler characteristic of $M$ at some point $p$. Then $M$ admits a Riemannian metric such that the Gaussian Curvature is equal to $f$.

I want to know if there is a combinatorial analogue of this statement. Namely, is the following statement true:

Let $M$ be a compact orientable two dimensional manifold. Let $f:M \rightarrow \mathbb{R}$ be a function that is zero except at a finite set of points $p_1 \ldots p_n$. At one of those points, $f$ has the same sign as the Euler characteristic of $M$. Then does $M$ admit a triangulation with $p_1 \ldots p_n$ as vertices such that the ``Curvature'' at each of these points is same as $f$?

Here by curvature we mean the angle deficit from $2 \pi$. And the curvature of $M$ at any point that is not a vertex is defined to be zero.


I understand from the counter example Liviu gave that I have to relax my definition of triangulation. Let me explain the motivation for my question. What I was wondering is that can one hope to prove the Kazdan Warner theorem this way.... .....the basic idea is this.......given a smooth function $f$ that satisfies the Gauss Bonnet sign condition, construct a sequence of discrete functions $f_n$ (which are zero at all but finitely many points) that converge in some appropriate sense to $f$ (pointwise?) Then argue that the singular metrics $g_n$ converge in some sense to the desired solution. Is there any chance such an approach can be made precise? To start with one can just ask for solutions in a ``weak'' sense.

Another question I have is......are there any questions in geometric analysis...particularly questions of this nature ``find a metric satisfying some condition'' where such an approach actually works.......ie first solve a
discretized version of the problem and then show that the desired solution is in some sense an appropriate limit?

A particular example I had in mind while asking this question is the following : A proof that the De Rham Cohomolgy is isomorphic to the singular cohomology using combinatorial hodge theory. The idea of the proof roughly is that one defines a discrete analogue of the $d$ operator and discretizes the De Rham Cohomology and shows that in an appropriate sense this converges to the singular cohomology (and hence giving the isomorphism).

Of course this particular example has nothing to do with finding metrics on manifolds, but I stated that example to illustrate that general idea......you prove some statement by solving a discrete version and then try to show the desired solution is some appropriate limit. Are there other instances where such an idea has been carried out (particularly for the type of questions I am asking .......finding metrics satisfying some condition ....or may be even isometric embedding problems)?

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2 Answers

First, as Liviu noted, you need to relax the concept of triangulation, you need to allow pseudo-triangulations (take a simplicial complex and identify some faces via homeomorphisms). The question was analyzed in detail by Marc Troyanov, starting in

M. Troyanov, "Les surfaces Euclidiennes a singularites coniques", Ens. Math. 32 (1986) 79–94.

You definitely need more conditions than you stated in the question (Gauss-Bonnet formula). Here is the precise statement.

Let $S$ be a compact Riemann surface with marked points $p_i$ and cone angles $\theta_i$. For each $i$ define $\beta_i= (2π)^{-1}\theta_i -1$. Clearly, $-1<\beta_i<\infty$. Then:

There exists a flat metric on $S$ with conical singularities at $p_i$ and cone angles $\theta_i$ (in the given conformal class) provided that the (necessary) Gauss-Bonnet condition holds: $$ \chi(S) +\sum_i \beta_i = 0.$$

Note that Troyanov did not discuss promoting this metric to a (pseudo) triangulation, but that is easy once you have a flat metric with singularities.

A similar result holds if you use hyperbolic metrics on each simplex. It is an outstanding open problem to do the same when triangles are equipped with spherical metrics (and some cone angles are greater than $2\pi$). The best result in this direction, so far, is due to Alex Eremenko, see here, who gave a solution in the case of metrics on $S^2$ with three singular points.

I am unaware of anybody using this theorem to recover Kazdan-Warner, this does not seem easy and I am not sure it would be very useful. I do know that people are interested in giving a combinatorial description of Pontryagin classes and (I think, it was first done by Jeff Cheeger) it could be done by using piecewise-flat metrics.

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I have expanded on my question and explained the motivation for it. –  Ritwik Jan 9 '13 at 17:49
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I think that you need to state very clearly what you mean by triangulation of the manifold, and what do you mean by "angles". (On a manifold that would require a choice of metric.) If by triangulation you mean the affine realizeation ofba finite simplicial complex inside some Euclidean space, and you measure the angles using the induced metric then the answer is no.

Consider the case $n=4$. The affine simplicial complex with $4$ vertices and homeomorphic to a manifold is given by the boundary of a $3$-dimensional simplex. In this case the curvature can only be positive, and clearly one can produce metrics on the $2$-sphere that are negative somewhere.

On the other hand, if you allow the faces of this tetrahedron to be curved then the answer could be positive. Here is one plausible solution.

On $S^2$ fix four points $p_1,\dotsc, p_4$ and a metric whose curvature at $p_i$ is $f(p_i)$. Such a metric exists by Kazhdan-Warner. Next connect the points by geodesic arcs, and assume you can do this so that these arcs do not intersect in the interior. You now have a curved triangulation with the properties you desire.

I think that the question needs to be formulated more carefully.

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