Variation on Gauss-Bonnet Theorem (disjoint discs) In lecture the other day we were talking about the (local) Gauss-Bonnet Theorem: Which states: $$\int_{U} K \cdot dA + \int_{0}^{L} \kappa_{g} \cdot dt = 2\pi$$ where K is the Gauss Curvature of the manifold and $\kappa_g$ is the geodesic curvature. This is for a manifold, U, which is diffeomorphic to a closed disc with a smooth boundary.

The variation is as follows:
Now imagine $V$ is diffeomorphic to a closed disc. Let $U = V - \cup_{i=1}^{k} D_{i}$ where $D_{i} \subset V$ are disjoint open discs. What would the Guass Bonnet formula for U be?

So what's being added here is "boundaries" on the inside of the manifold, k of them to be specific, which don't overlap (as they are disjoint) tus we may assume each one traces out an entire disc unimpeded. Could we then say each disc contributes 2$\pi$ and thus we would modify our GB equation by changing the RHS from $2\pi$ to $2k\pi$? That seems really simplistic though.. I saw something about a global version with the Euler Characteristic $\chi$ = $2 - 2g$ where $g$ is the "genus" or "number of handles" could this be the solution?
 A: Yes. The genus is only defined for surfaces without boundary, but the Euler characteristic does exactly what you think it does (except it's half what you've described). You can define a triangulation for your surface (some triangles will have curved sides), then count the number $V$ of vertices, plus the number $F$ of faces, minus the number $E$ of edges. The difference $\chi:=V+F-E$ is the Euler characteristic. This is a topological invariant (the same for any two homeomorphic regions), and it's equal to 1 for a simply connected subset of $\mathbb{R}^2$. Now, imagine you have some such triangulation, and you remove $k$ non-adjacent, interior triangles. The Euler characteristic decreases by $k$, since $F$ decreases by $k$ and $V$ and $E$ are unchanged. Thus, a region with $k$ holes has Euler characteristic $\chi=1-k$. Your disc with discs removed is homeomorphic to this region, so it also has Euler characteristic $\chi=1-k$. Then the Gauss-Bonnet theorem (see e.g. Chavel's Riemannian Geometry: A Modern Introduction, Theorem V.2.7) states
$$\int_{\partial M} \kappa\, ds + \iint_M \mathcal{K} \,dA = 2\pi \chi.$$
