Euler characteristic for topological surfaces and triangulations

algebraic-topologygeneral-topology

I've been trying to understand the Euler characteristic of surfaces. Let's define the Euler characteristic of a (regular, closed) surface $S$ as $\chi(S)=V-E+F$, where $V$, $E$ and $F$ are, respectively, the number of vertices, edges and faces of a given triangulation of $S$. Then we should prove this does not depend on the given triangulation. This is kind of straightforward, using induction if you will, if we can take common refinements of two triangulations. Of course we need to consider finite triangulations in order to $\chi(S)$ to be computable.

I got to this question:

Prove Euler characteristic is a homotopy invariant without using homology theory

where a comment links to the Wikipedia article Hauptvermutung. This article's name is the conjecture that any two triangulations of a surface have a common refinement, and it states that "The manifold version is true in dimensions $\displaystyle m\leq 3$."

So this is the problem: consider any geodesic triangle $T$ on a sphere $S$. Then $T$ determines a triangulation of $S$. Consider a new triangle $T'$, obtained from $T$ by keeping two of its edges the same, but changing the other edge by a ''wave of increasing period'', i.e., something that resembles the graph of $\sin^2(1/x)e^{-1/x^2}$ on the interval $[0,1/2\pi]$ (which is even smooth).

Then the finite triangulations determined by $T$ and $T'$ do not admit a common finite refinement, which seems to contradict the Wikipedia's article.

Of course, there are other ways of proving invariance of $\chi$ by the chosen triangulation in different contexts, such as using Gauss-Bonnet in the case of reguar surfaces, or homology theory for CW complexes.

But how would one go to prove that the Euler characteristic of a compact topological surface does not depend on the given triangulation (which seems to always exist by Theorem 6.2.8 of this book? Or how could one define the Euler characteristic of arbitrary topological manifolds in better ways?

Best Answer

As often happens, that Wikipedia article took shortcuts that confused the issue, and in fact the statement of the Hauptvermutung that was in that article was flat out wrong. I've rewritten the Wikipedia page to correct this error.

In this case, the correct statement of the Hauptvermutung is a bit more complicated, as you can see if you check an actual mathematical reference such as The Hauptvermutung Book. Here's the statement. Let $X$ be a topological space which possesses a triangulation. We say that $X$ satisfies the Hauptvermutung if for any two triangulations $T,T'$ of $X$ there exist refinements $S,S'$ of $T,T'$ such that $S,S'$ are combinatorially equivalent; this means that there exists a homeomorphism $h : X \to X$ such that $h$ is a simplicial isomorphism from the triangulation $S$ to the triangulation $S'$.

The Wikipedia article (before I edited it) gave the impression that the homeomorphism $h$ must always be the identity, and as your example shows, this impression is not correct.

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