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?
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
$$\int_{\partial M} \kappa\, ds + \iint_M \mathcal{K} \,dA = 2\pi \chi.$$