Proof of the formula $\chi=2-2g$ for closed orientable surfaces

Question

Is there any simple proof for the formula $\chi=2-2g$, where $\chi$ is the Euler characteristic and $g$ the genus? This is likely to be answered before, but I cannot find it. I use the definitions $\chi = \sum (-1)^k \text{rk} (H_k(X;\mathbb Z))$ and $g$ the maximum number of cuttings along non-intersecting closed simple curves without disconnecting the space $X$.

At the moment, the only method I can think of is the following steps:

1. Prove the classification theorem that every orientable closed surface is $S^2$ or a sum of tori.
2. Compute the homology groups $H_n(X;\mathbb Z)$ using 1.
3. Plug in the definition $\chi = \sum (-1)^k \text{rk} (H_k(X;\mathbb Z))$ to calculate the Euler characteristic.

This seems quite long and could take 20 pages.

Question: Is there a more direct way to prove $\chi=2-2g$ without going through this long process?

Answer 1

The hard part seems to be proving that $H_2(\Sigma;\Bbb Z) = \Bbb Z$ for any closed connected oriented surface $\Sigma$. This requires to my eye either Poincare duality or the following triangulation proof.

0) $\Sigma$ is triangulable. Consider the graph whose vertices are the 2-simplices of $\Sigma$ and whose edges are between adjacent 2-simplices. One may pick a maximal sub-tree of this graph; it will contain every vertex, and some subset of the edges. One may use this maximal tree of 2-simplices to build a triangulation of the disc and a quotient map $D^2 \to \Sigma$ which is injective on the interior; use the 0-simplices and leftover 1-simplices to build a regular CW structure on $\Sigma$ with exactly one 2-cell. Now use cellular homology; the top boundary operator is zero because the boundary of this disc traverses each remaining 1-cell exactly twice, once in each direction; because these have opposite orientation, they cancel out. Thus the chain complex looks like $0 \to \Bbb Z\xrightarrow{0} C_1 \xrightarrow{\partial_1}\to C_0\to 0$, and the second cellular homology is thus $\Bbb Z$, as desired.

Here is a proof that $H_1(\Sigma;\Bbb Z) = \Bbb Z^{2g}$ by induction.

1) A surface of genus 0 is one such that every embedded curve separates the manifold, and hence is null-homologous. Every curve can be represented (after a homotopy) by one with a finite set of transverse self-intersections. Now trace out the curve; when it self-intersects for the first time, that portion may be thought of as an embedded curve; because that portion is null-homologous, our homology class can be represented by a curve with fewer self-intersections. Inducting downwards on the number of self-intersections, we see that $H_1(\Sigma;\Bbb Z) = 0$.

2) A surface of genus $g$ has a collection of $g$ curves you may cut along without disconnecting it. Cut along one of those to get a surface of genus $g-1$ with two boundary components; one may pick a path between the boundary components which does not intersect the given $(g-1)$ curves. Cut along this path to obtain a surface of genus $g-1$ with one boundary component. Reversing this procedure, one finds that we just showed that $\Sigma \cong \Sigma' \# T^2$ where $g(\Sigma) = g$ and $g(\Sigma') = g-1$. Now Mayer-Vietoris gives (using orientability!) that $H_1(\Sigma;\Bbb Z) = H_1(\Sigma';\Bbb Z) \oplus \Bbb Z^2$.

Inductively the calculation follows.

Secretly I just proved the classification theorem --- you only need the small additional info that $H_1(\Sigma;\Bbb Z) = 0 \implies \Sigma \cong S^2$, which follows similar lines as the triangulation argument in (0).

Answer 2

I know this is a very old question so you may have found your answer already but will share this anyway since I think its quite a nice idea.

This does still require you to first prove that any surface is the sum of tori but does not require any homology at all.

I think you can see that its true using the poincare-hopf theorem though this is my own idea so it may have errors. Also this is very visual and non rigorous so apologies of its not well explained.

Take your surface (make it a torus for now) then take a plane with a vortex vector field on it spinning around the origin. Place the torus in $\mathbb{R}^3$ with the hole pointing along the z axis. Then sweep the plane perpendicular to the x axis with the origin on the x axis. As it sweeps over the torus we look at the intersection of the plane with the torus and imagine leaving the vector field behind then projecting each vector onto the tangent plane at its base thus defining a vector field on the torus.

Now we look at where the zeros of the vector field are. Obviously there are 2 index 1 zeros, one as the plane first hits the torus and one as the plane leaves. However while the plane is passing along the hole the intersection will split into 2 disjoint circles, and at the exact point of splitting i.e. when the intersection looks like a figure 8 we will get another 0. If you draw it, it should be clear that this is zero has index -1.

Clearly we get another similar zero when the intersection merges back into 1 circle on the other side.

So in total you have $\sum(index)= 1-1-1+1=0$ this is expected since $\chi(T^2)=0$. But try this now with an n hole torus S it should be clear that for every hole you get 2 zeros of index -1 and then 2 of index 1 (when the plane enters and exits the surface). Therefore since g is the number of holes $\chi(S)=\sum(index)=2-2g$.