# Triangulation of Surfaces without Jordan-Schoenflies

Does anyone know of a proof of the fact that any 2-manifold can be triangulated that does not use the Jordan-Curve Theorem or the Jordan-Schoenflies Theorem? Thanks for your help

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Are your surfaces topological or smooth ? –  Mohan Ramachandran Feb 29 '12 at 14:02

Take the proof that any compact smooth manifold admits triangulations, and set the dimension to two.

The idea goes like this:

• Embed your surface (or $n$-manifold) in $\mathbb R^5$ ($\mathbb R^{2n+1}$ in general).

• Triangulate $\mathbb R^5$, and make the surface transverse to the triangulation. If the surface does not intersect each simplex in a locally linear manner, subdivide the triangulation and repeat this step until it does.

• The pull-back of the triangulation to the surface is a decomposition into convex polyhedra. A subdivision turns this into a triangulation.

Paraphrasing Allen Hatcher:

If you're interested in topological surfaces, the paper

A.J.S. Hamilton, The triangulation of 3-manifolds, Oxford Quart. J. Math. 27 (1976), 63-70

takes the Kirby-Siebenmann machinery and scales it down to 3 dimensions where it becomes somewhat simpler, so one can prove existence and uniqueness of triangulations of 3-manifolds using only standard PL techniques, such as results of Waldhausen. Presumably the same approach would work for surfaces. Since the method works in 3 dimensions it can't be using the topological Shoenflies theorem since this fails in 3 dimensions. On the other hand, it would use some PL (or smooth) surface theory so it wouldn't be entirely "from scratch".

edit: Allen wrote this argument up in a recent paper. See this thread for details http://mathoverflow.net/a/151760/353

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Moise:Geometric Topology in dimensions 2 and 3 has a proof using PL versions of JCT and Schoenflies theorem .Also see Siebenmann Russian Math Surveys pages 645-672 2005 titled Osgood-Schoenflies revisited where he gives a proof of triangulation of surfaces in the spirit of his work with Kirby . –  Mohan Ramachandran Feb 29 '12 at 16:10
Hi Ryan. Would you be able to explain why the Kirby-Siebenmann technique doesn’t work for dimensions $\geq 4$? What prevents the technique from being scaled down to these higher dimensions? Thanks! –  Leonard Jul 14 at 18:59