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the conformal structure does not see the conical singularities of a polyhedral surface.

This is a quote from the Preface of Quantum Triangulations (eds.: Carfora, Marzuoli). The sentiment is attributed to Marc Troyanov, in a 1991 Trans. AMS paper, "Prescribing Curvature on Compact Surfaces with Conical Singularities," that I cannot access easily.

Can anyone provide a brief explication or unpacking of the meaning of this intriguing statement? I'd appreciate it—thanks!

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2 Answers 2

up vote 7 down vote accepted

Consider the following pair of surfaces:

  1. $P\;$ is the plane with the origin removed.

  2. $C$ is the cone $z = \sqrt{x^2+y^2}$ in $\mathbb{R}^3$, with the origin removed.

There are several possible structures we could put on these surfaces. For example, both of the surfaces support a differentiable structure. As differentiable manifolds, $P\;$ and $C$ are diffeomorphic. Thus the differentiable structure cannot "see" the conical singularity at the tip of the cone.

Each of these spaces also has a Riemannian metric. Both Riemannian metrics are flat (i.e. Euclidean), but the two surfaces are not isometric. In particular, a loop on $C$ that surrounds the origin will have a holonomy of $(2-\sqrt{2})\pi$ radians, while a loop on $P\;$ that surrounds the puncture has zero holonomy. Thus the metric can "see" the conical singularity.

You do not need the full metric structure to detect the conical singularity. For example, if we give each surface a Euclidean similarity structure, then it is still possible to define the holonomy of a closed curve, and we can detect the difference between $P\;$ and $C$.

However, it turns out that $P\;$ and $C$ are equivalent as conformal surfaces. For example, the map $P\to C$ that maps the point in the plane with polar coordinates $(r,\theta)$ to the point on the cone with cylindrical coordinates $\bigl(r^{1/\sqrt{2}},\theta,r^{1/\sqrt{2}}\bigr)$ is a conformal equivalence. Thus the conformal structures cannot "see" the conical singularity.

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Thank you, Jim, for the lucid explanation! –  Joseph O'Rourke Feb 25 '12 at 12:51
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One way to construct a cone is to take a sector in $\mathbb C$ with angle $2\pi\alpha$ and identify its rays. The function $z^{\frac{1}{\alpha}}$ is then well defined on the cone, holomorphic outside of its origin and continuous in the origin of the cone. Moreover this function defines a bijection between the cone and $\mathbb C$. So we can postulate that $z^{\frac{1}{\alpha}}$ is a complex coordinate on the cone. And so from this point of view the cone is just $\mathbb C$. The origin is not different from other points anymore :)

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