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.