As I recall (from Guillemin and Pollack "Differential Topology") the Euler characteristic of a (for my purposes, compact and oriented) smooth manifold X is defined as $\chi(X)=I(\Delta,\Delta)$, where $I(\Delta, \Delta)$ is the intersection number of the diagonal in $X\times X$ with itself. I have recently become interested in characteristic classes, and in the first page of the notes I'm reading, the author states this definition, and says $\chi(X)$ may equivalently be defined to be $I(X_0,X_0)$, the intersection of the zero-section in the tangent bundle of $X$ with itself. Can anyone explain how these definitions are equivalent?
[Math] Euler characteristic is equal to self-intersection number of zero-section
characteristic-classesdifferential-geometrydifferential-topologymanifolds
Best Answer
Let's look at first definition. To compute the self-intersection number one needs to deform one of two copies of $\Delta$ slightly to make intersection transversal. It can be done in a small neighborhood of the diagonal — i.e. in the normal bundle of the diagonal in $X\times X$. But this normal bundle is isomorphic to the tangent bundle of $X$.