# Do minimal submanifolds minimize area locally?

A few days ago I asked this question on math.stackexchange:

Consider $(\tilde{M},g)$ a riemannian manifold and $M \subset \tilde{M}$ riemannian submanifold.

Is it true that if $M$ is a minimal submanifold of $\tilde{M}$ then for every $p \in M$ there exists a neighborhood $W$ of $p$ in $\tilde{M}$ such that $V=W\cap M$ has least area among every $\Omega \subset W$ with $\partial \Omega = \partial V$?

I've been thinking about it, I think it is true but I don't know how to prove.

If it's true, how should I go about proving it?

As asked there, we say a submanifold is minimal if the mean curvature vanishes identically, or equivalently, it's a critical point of the area functional.

Thanks in advance!

## 1 Answer

Yes, this is true, but finding an explicit general proof in the literature seems to be a challenge.

I think that many people just believe it without having seen an actual proof. One reason is that it's very easy to prove that small pieces of minimal submanifolds are strongly stable, i.e., small nontrivial perturbations of a minimal submanifold supported in sufficiently small domains must strictly increase the area. While this is pretty convincing, it doesn't actually prove local minimizing, though, because you don't know that everything with the same boundary is a perturbation of the thing you start with.

For a proof that works when the ambient manifold $\tilde M$ is Euclidean space, see, for example, Theorem 2.1 in Curvy slicing proves that triple junctions locally minimize area, by Gary Lawlor and Frank Morgan, J. Diff. Geom 44 (1996), 514–528. I think that standard techniques can be used to extend their proof to cover general ambient manifolds, since all you want is the local result.

If you are willing to assume that the competitor $\Omega$ is an oriented submanifold, there is an easier proof by the method of calibrations. The strategy is this: With $p\in M^n$ given, one constructs a closed $n$-form $\phi$ on $\tilde M$ with the property that $|\phi(e_1,\ldots,e_n)|\le 1$ for every orthonormal $n$-tuple $e_1,\ldots,e_n\in T_x\tilde M$ for every $x\in M$ and such that $\phi$ pulls back to equal the area form on $M\cap V$ for some open convex $p$-neighborhood $V\subset\tilde M$. [This construction of the calibration $\phi$ is a little tricky, since the first thing you would think of is to try to do this by pulling back the volume form on $M$ to a tubular neighborhood of $M$ by the 'closest point on $M$' projection, which works when $M$ has dimension $1$ or codimension $1$ but not otherwise, even when the ambient space is flat. Of course, you will need to use the fact that $M$ is a minimal submanifold (i.e., its mean curvature vanishes) in order to construct $\phi$.] Anyway, once $\phi$ is constructed, the rest is easy: If $\Omega\subset \tilde M$ is $n$-dimensional and oriented and satisfies $\partial\Omega = \partial(M\cap V)$ (in the oriented sense), then because $\Omega\subset V$ and $V$ is convex (which implies that $\Omega \cup (-(M\cap V)$ is homologous to $0$), we have $$\mathrm{vol}(\Omega) \ge \int_\Omega \phi = \int_{M\cap V} \phi = \mathrm{vol}(M\cap V).$$ (The first inequality follows because, by construction, $\phi$ pulls back to $\Omega$ to be no more than the volume form on $\Omega$. The second equality follows by Stokes' Theorem, since $\mathrm{d}\phi=0$ and $\Omega \cup (-(M\cap V))$ has no boundary and is null-homologous. The third equality follows because $\phi$ pulls back to $M\cap V$ to be its volume form.)