# Closed orientable surfaces have even Euler characteristic

It is of course completely standard that closed orientable surfaces have even Euler characteristic. What is the most elementary proof of this?

More specifically, suppose I have a finite simplicial complex $K$ with vertices $V$, edges $E$ and faces $F$. I suppose that each edge is contained in precisely two faces, and that the link of each vertex is a cycle of size at least three. This implies that $|K|$ is a closed surface, and it is not hard to see that $3|F|=2|E|$, so $|F|$ is even. We want to show that $|V|-|E|+|F|$ is even, or equivalently that $|V|=|E|\pmod{2}$. This is not true in general if $K$ is not orientable, so we need to say something about orientations.

Let $D$ be the set of directed edges, and let $\chi\colon D\to D$ reverse direction. Let $S$ be the set of permutations $\sigma\colon D\to D$ such that

• For any $(u,v)\in D$ we have $\sigma(u,v)=(v,w)$ and $\sigma(v,w)=(w,u)$ and $\sigma(w,u)=(u,v)$ for some $w$ such that $\{u,v,w\}\in F$
• If $u,v,w$ are as above, and $x$ is the other vertex such that $\{u,v,x\}\in F$, then $\sigma(v,u)=(u,x)$ and $\sigma(u,x)=(x,v)$ and $\sigma(x,v)=(v,u)$.

We find that $S$ bijects with the set of orientations of $|K|$. Also, we have $\sigma^3=1$ and we can describe the composite $\rho=\sigma\chi$ as follows: the edge $\rho(u,v)$ starts at $u$, and is the next edge round $u$ after $(u,v)$ in clockwise order.

Given $K$ and $\sigma$ as above, it seems that there should be some very direct combinatorial argument to show that $|V|=|E|\pmod{2}$, but I am not seeing one.

• You want a direct proof avoiding the use of the classification, right? – Francesco Polizzi Sep 11 '18 at 13:06
• @FrancescoPolizzi Yes. – Neil Strickland Sep 11 '18 at 13:08
• I do not think this is so easy. You must use orientability in some essential way, and at the and of the story this could be more or less equivalent to the classification. A possible strategy of proof is noticing that (1) since $S$ is orientable than $H^2(S, \, \mathbb{R})$ is one-dimensional, generated by the orientation class and (2) $H^1(S, \, \mathbb{R})$ carries a symplectic form given by the intersection of $1$-cycles, hence it is even-dimensional. – Francesco Polizzi Sep 11 '18 at 13:32
• Is $(2)$ independent on the classification (I mean, the part about the non-degenericity of the intersection form)? – Francesco Polizzi Sep 11 '18 at 13:33
• Here is the opposite of what you asked for, the least elementary proofs I know: qchu.wordpress.com/2014/10/14/… – Qiaochu Yuan Sep 11 '18 at 20:58

The set of vertices can be split as the union $V=V_{\text{odd}}\cup V_{\text{even}}$ of vertices with odd and even degrees, respectively. Since the sum of degrees over all vertices is the same as twice the number of edges, we know that $|V_{\text{odd}}|= 0\pmod 2$. Therefore we want to establish that $|V_{\text{even}}|=|E|\pmod 2$.
The permutation $\sigma$ is the product of $|F|$ cycles of length 3, so $\operatorname{sign}(\sigma)=1$. The permutation $\chi$ is the product of $|E|$ cycles of length 2 so $\operatorname{sign}(\chi)=(-1)^{|E|}$. And finally the permutation $\rho$ is the product of $|V_{\text{odd}}|$ cycles of odd length and $|V_{\text{even}}|$ cycles of even length, therefore $\operatorname{sign}(\rho)=(-1)^{|V_{even}|}$. Since $\rho=\sigma \chi$ we must have $$\operatorname{sign}(\rho)=\operatorname{sign}(\sigma)\operatorname{sign}(\chi)\implies |E|=|V_{\text{even}}|\pmod 2$$
• @FrancescoPolizzi Orientability is equivalent to the existence of $\sigma$ as is explained in the question. – Gjergji Zaimi Sep 12 '18 at 6:47
• Here's another way to say it: as $|D|$ is even, cycle decomposition shows that the parity of any permutation $\alpha$ is $|D/\alpha|\pmod{2}$. Thus $|D/\alpha\beta|=|D/\alpha|+|D/\beta|\pmod{2}$. Now take $\alpha=\sigma$ and $\beta=\chi$ so $\alpha\beta=\rho$, and note that $D/\rho\simeq V$ and $D/\sigma\simeq F$ and $D/\chi\simeq E$. This gives $|V|+|E|+|F|=0\pmod{2}$, as required. – Neil Strickland Sep 12 '18 at 23:10