December 14, 2017 / / In news

### A fun fact about the Euler characteristic

There are many equivalent definitions of the Euler characteristic $\chi(Y)$ of a space $Y$. I recently learned another of these, valid in the case that $Y$ is a compact, orientable, smooth manifold.

Suppose $Y$ is a compact and orientable manifold.

First, orient $Y \times Y$ with the product orientation. Define the diagonal $\Delta := {(y,y) \in Y \times Y }$; orient $\Delta$ via the identification with $Y$ given by $y \mapsto (y,y)$.

Next, perturb $\Delta$ to a submanifold $\Delta’$ via a generic and sufficently $C^1$-small perturbation. If this perturbation is sufficiently small, $\Delta’$ and $\Delta$ are isotopic, which induces an orientation on $\Delta’$. Since the perturbation is generic and $Y$ is compact, $\Delta$ and $\Delta’$ intersect transversely in a finite number of points.

Define the “intersection number” to be $I(\Delta, \Delta’):= \Sigma_{p \in \Delta \cap \Delta’} \sigma_p$, where $\sigma_p := +1$ if the orientation of $T_p (Y\times Y)$ matches the direct sum orientation of $T_p \Delta \oplus T_p \Delta’$, and $\sigma_p := -1$ otherwise.

Finally, the Euler characteristic is actually equal to $\chi(Y) = I(\Delta, \Delta’)$!

The idea of the proof is as follows. The map $F:TY \to N\Delta$, $F:(x,v)\mapsto (x,x,v,-v)$ is a vector bundle isomorphism between $TY$ and $N\Delta$, the normal bundle of $\Delta$ in $Y\times Y$. Using the tubular neighborhood theorem, this yields a diffeomorphism between neighborhoods of the zero section of $TY$ and $\Delta$. Pulling back $\Delta$ and $\Delta’$ via this diffeomorphism, keeping track of orientations, and applying the Poincar'{e}-Hopf Theorem yields the desired result.