The topologically invariant Euler characteristic of a 2-surface is given by $\chi=\frac{1}{4\pi}\int\sqrt{g}\mathcal{R}$ (where $\mathcal{R}$ is the scalar curvature) and is equivalent to $\chi=2-2g – b$ for a surface with $g$ handles and $b$ boundaries.
For a sphere this is simple, $\sqrt{g}=R^2$ and $\mathcal{R}=\frac{2}{R^2}$, where $R$ is the radius. Therefore, $\chi=2$, as suggested by the second formula above. If we wanted to add a boundary, we could take the usual metric
\begin{equation}
\mathrm{d}s^2=\mathrm{d}\theta^2 + \sin^2{\theta} \mathrm{d}\phi^2
\end{equation}
and limit $\theta$ to $0\leq\theta\leq\theta_0$, with $\theta_0<\pi$. We have added a hole in the sphere, thus adding a boundary, so the Euler charateristic should be 1. By the first formula, however, we get $\chi=(1-\cos^{-1}{\theta_0})$.
Moreover, this should be topologically equivalent to a flat disc, on which we may take a flat metric, so $\mathcal{R}=0$ and $\chi=0$.
Where have I gone wrong?
Many thanks in advance.
Best Answer
Your formula $\chi=2−2g−b$ only works for surfaces with boundary. As a manifold, a punctured sphere has no boundary (locally, the boundary of a surface looks like a chunk of the closed upper half-plane which contains a piece of the x-axis, and the punctured sphere has no such points). Rather, you should punch out an open disk to obtain a manifold with boundary. The general analytic formula for Euler characteristic includes terms involving integrals along the boundary, and this should exactly counteract the fact that you're integrating scalar curvature over less of the whole sphere.