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Assume $(M,g)$ is Schwarzschild manifold, $A$ is the minimum area in the homology class of the boundary. How to prove that $A=$ the area of the horizon.

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David, the underlying manifold $M$ of the Schwarzschild spacetime is $\mathbb{R}^2\times S^2$. It's an open manifold with no boundary. So what do you mean by that? Also, with no motivation, this comes off a bit like homework. – Igor Khavkine Jul 3 '13 at 14:25
@Igor, I think the question is appropriate if it is edited to be the question in my answer (which I am pretty sure is the only reasonable question one could ask). I doubt it is homework (I might hazard a guess that it comes from reading Bray's thesis, where he asserts it is true, but leaves it to the reader, because it is not essential to his main goal). – Otis Chodosh Jul 3 '13 at 15:07

I'm interpreting your question in the only reasonable manner that I can think of:

$M= \mathbb{R}^3\backslash \{0\}$ and $g_m = \left( 1 + \frac{m}{2r}\right)\delta$ is the "doubled" Schwarzschild. Of course it is easy to prove that $S_{m/2}(0)$ is a minimal surface. Is it true that $S_{m/2}(0)$ is actually area minimizing in its homology class?

This clearly is a stronger than the same question in "exterior" Schwarzschild, i.e. $M\backslash B_{m/2}(0)$ with the same metric there.

The answer is yes. I'll give a sketch because I'm short on time:

Let $w=1+\frac{m}{2r}$. $\Delta_\delta w = 0$ by computation (it has to hold if you look up the formula for conformal change of scalar curvature and know that Schwarzschild is scalar flat).

Then, if $\Sigma$ is a "competitor" which is homologous to $S_{m/2}$, an appropriate Green's theorem shows that $$ \int_{S_{m/2}} D_\nu w = \int_\Sigma D_\nu w $$ These integrals are all with respect to the $\delta$-quantities. Here is where I will skip the details. Multiply by some constant $C$ (which you should explicitly determine) so that the LHS becomes $area(S_{m/2},g_m)$. That is $$ area(S_{m/2}(0),g_m) = C\int_{S_{m/2}} D_\nu w = C\int_\Sigma D_\nu w $$ Now, I'll leave it up to you to check that $CD_\nu w$ is bounded from above by $w^4$ (you just need to compute the various quantities). Thus, we've shown that $$ area(S_{m/2}(0),g_m) \leq \int_\Sigma w^4 = area(\Sigma,g_m). $$ You may further check that equality implies that $\Sigma = S_{m/2}(0)$.

This is a classic example of a "calibration" argument.

I'll remark that Hugh Bray, in his thesis gave a beautiful argument to show that the other centered spheres are isoperimetric. These ideas have had a good deal of influence in related works, e.g. Eichmair--Metzger proved a quantitative isoperimetric inequality in Schwarzschild and used this to study large isoperimetric surfaces in initial data which is $C^0$-asymptotic to Schwarzschild. Also, Brendle--Eichmair used this, along with other results to study isoperimetric surfaces in the "doubled" Schwarzschild, as I defined above.

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Thanks for your detailed answer and remark~ – david Jul 4 '13 at 5:36

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