# Is every conformal manifold equivalent to a flat one with cone singularities?

Consider a polytope with a $$2$$-dimensional surface and the corresponding metric on this surface (coming from the embedding in $$3$$-dimensional Euclidean space). Intrinsically the metric is flat everywhere apart from the vertices of the polytope, where one has cone-like singularities if the angle sum does not equal $$2\pi$$.

Is every conformal manifold equivalent to such a flat cone-manifold? More precisely, is there a sequence of equivalent conformal manifolds that approximates such a singular manifold?

• Are you asking about manifolds of dimension 2 or higher ? – Alexandre Eremenko Nov 17 '19 at 14:19
• Oh, yes, dimension 2 only. Then the conformal moduli space is finite-dimensional, and so is the space of different cone-manifolds (parametrized by the distances and angles of the cone singularities), so the question seems to make sense. – Andi Bauer Nov 17 '19 at 15:27

The answer is yes, and there are several ways to prove it. The result can be restated as "on every Riemann surface there exists a flat conformal metric with conic singularities". In fact the singularities can be prescribed, the only condition is that Gauss Bonnet holds. References: For compact surfaces:

MR1005085 Troyanov, Marc, Prescribing curvature on compact surfaces with conical singularities. Trans. Amer. Math. Soc. 324 (1991), no. 2, 793–821.

For open surfaces:

MR1166122 Hulin, D., Troyanov, M. Prescribing curvature on open surfaces. Math. Ann. 293 (1992), no. 2, 277–315.

At page 10 of

http://page.math.tu-berlin.de/~bobenko/Lehre/Skripte/RS.pdf

it is claimed that the answer is yes.