Euler characteristic of sphere with a hole

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 $$\mathrm{d}s^2=\mathrm{d}\theta^2 + \sin^2{\theta} \mathrm{d}\phi^2$$ 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.

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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. – Aaron Mazel-Gee Oct 26 '11 at 11:10
Your formula for $\chi$ in the first paragraph is only true when the boundary curves are geodesic, no? – Mariano Suárez-Alvarez Oct 26 '11 at 11:18
@Aaron: That comment looks like an answer to me. – joriki Oct 26 '11 at 16:21
Alright, I will promote it. I was worried that I had been too lazy to look up the precise general analytic formula, but I guess that's not really the point here. – Aaron Mazel-Gee Oct 27 '11 at 15:58

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.
Unlike the analytic formula, formula $2-2g-b$ for Euler char works for surfaces without boundary nicely, missing condition here is compactness. – Grigory M Nov 26 '11 at 17:42