I am given the following metric

$$ds^2 = \frac{dr^2}{1-2m/r} + r^2dS,$$

where $dS$ is the standard metric on the unit sphere $S^2$. I am told that this is isometric to $\mathbb{R}^3$ or (taking its exterior region) to one-half of the spatial Schwarzschild manifold.

The spatial manifold is the $\{t=0\}$- slice of the Schwarzschild manifold so I get the absence of the $dt^2$ term, but I don't understand the "one-half"?

This comes from The Inverse Mean Curvature Flow and the Riemannian Penrose Inequality by Huisken and Ilmanen.


1 Answer 1


I don't have the book, but I would guess that it's referring to the maximally extended Schwarzschild metric. If you write the metric in Kruskal-Szekeres coordinates you find it in effect consists of two copies of the Schwarzschild coordinate. This is because the KS $u$ coordinate (in the exterior region) is defined by:

$$ u = \left(\frac{r}{2M}-1\right)^{1/2} \,\, e^{r/4GM} \cosh\left(\frac{t}{4GM}\right) $$

So for every value of $r$ there are two values of $u$ corresponding to the two square roots.


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