Do minimal submanifolds minimize area locally?

Question

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?

Link: Do minimal submanifolds minimize area locally?

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!

Answer 1

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.)

Answer 2

For orientable competitors (integral currents), Federer proved a much more general result in

MR0388226 Reviewed Federer, Herbert A minimizing property of extremal submanifolds. Arch. Rational Mech. Anal. 59 (1975), no. 3, 207–217. (Reviewer: William K. Allard) 49F20

His result concerns more general elliptic integrands and proves the minimizing property of any region $M$ of a small enough area.

For the area-integrand, a sketch of proof for orientable and non-orientable competitors is as follows. The proof is essentially due to Robert Bryant and Gary Lawlor.

Observation 1: for a map $F:W\to V,$ with $W$ a region in $\tilde{M}$ and $V$ a region of $M,$ $F$ is area-non-increasing if and only if $F^\ast dvol_M$ is a calibration. Here $dvol_M$ is any choice of volume form locally if $M$ is non-orientable.

Proof: straightforward calculation.

Observation 2: any multiplicity $1$ integral or mod $p$ ($p\ge 2$ integers) current $T$ that admits an area-non-increasing map $F$ onto it is area-minimizing. Here we require $F$ restricted to the support of $T$ to be identity.

Proof: straightforward calculation.

Remark: mod $2$ currents include unorientable submanifolds and mod $p$ currents for $p\ge 3$ can admit more exotic singularities like triple-junctions, etc.

Thus, to prove the conclusion you want, it suffices to find a suitable map $F$ onto a neighborhood of $p$ in $M,$ so that $F^\ast dvol_M$ is a calibration.

Unfortunately, as pointed out by Robert Bryant, the usual normal bundle projection $\pi$ does not work. Straightforward calculations in Fermi coordinates show that $\pi^\ast dvol_M$ has comass $1+O(y^2),$ with $y$ the distance to $M.$

It is the idea of Robert Bryant and Gary Lawlor to overcome this by tweaking the projection to dominate the $y^2$ term.

One route is as follows. Set up a normal coordinate $(x_1,\cdots, x_m)$ centered at $p$ on $M.$ Take orthonormal frames $\nu_1,\cdots,\nu_m$ in the normal bundle of $M$ near $p,$ so that they are parallel with respect to the intrinsic geodesics on $M$ starting at $p$. Set up a Fermi coordinate using this frame, with coordinate label $(x_1,\cdots,x_m,y_1,\cdots,y_n).$ Now consider a coordinate transformation $x_j=(1+C\sum_iy_i^2)u_j,v_j=y_j$, with $C>0$ a constant. It is straightforward to verify that $(u_j,v_j)$ serves as a new coordinate system.

Define $F(u_1,\cdots,u_m,v_1,\cdots,v_n)=(u_1,\cdots,u_m,0,\cdots,0)$. Straightforward calculations back in $(x,y)$ coordinate show that $F^\ast dvol_M$ is a calibration that calibrates $M$ near $p$ for $C$ large enough.

Remark 1: At every step, we need to shrink our coordinate chart a bit. The transformation introduces a $(1+Cy^2)^{-m}$ factor and some other $1+O(y^2)$ factors in the comass, while not introducing any new $1+cy$ factor. A tedious calculation can verify that the $(1+Cy^2)^{-m}$ eventually dominates. See Tubes by Alfred Gray on how to do the calculations.

Remark 2: One might be curious about where is the minimality used. It is hidden in the expansion of comass $\pi^\ast\omega=1+O(y^2).$ If $M$ is not minimal, then the comass will be $1+O(y).$ The above construction breaks down in this case.

Answer 3

Just a comment to supplement both nice answers. Robert Bryant alludes to the commonly held idea that since minimal surfaces are strictly stable on small scales (this is easy to prove) then the result follows. Like Robert said, this does not directly imply the result since not all competitors are graphical! One can make this idea into a proper argument but it needs some GMT/regularity theory (these are deep results, much more difficult than the techniques used in the other two answers, I think, but assuming these things hold, the proof is easy).

I will prove:

If $M\subset (\tilde M,\tilde g)$ is a smooth minimal surface then $M\cap B_\delta(p)$ is uniquely minimizing for $\delta>0$ sufficiently small.

I'll allow $\delta$ to depend on $p$, but a simple modification would show that $\delta$ can be chosen independent of $p$ if $M$ is compact (or more generally, $M$ and $\tilde g$ have appropriate uniform estimates).

Assume this fails. Then, there are currents $\Sigma_j$ minimizing area among $\Sigma'$ with $\partial\Sigma' = M\cap \partial B_{1/j}(p)$. Rescale around $p$ by $j$, so $M_j : = j(M\cap \partial B_{1/j}(p)-p)$ converges to a flat $k$-disk $M_\infty$. Let $\hat \Sigma_j$ denote the corresponding rescaled surface.

Using the projection onto the $k$-plane containing $M_\infty$ we see that $M_\infty$ is uniquely minimizing. Thus, passing to a subsequence so that $\hat\Sigma_j$ converges, it must converge (in the flat topology) to $M_\infty$.

Regularity theory (interior and boundary) applies to show that for $j$ large, there's a section $s_j$ of the normal bundle $\hat M_j$ so that $\Sigma_j = M_j + s_j$ (really I mean use the exponential map of the rescaled metric here). PDE estimates give $s_j\to 0$ in $C^\infty$.

Thus we can write $$ |\hat \Sigma_j| = |M_j| + Q_1(s_j) + Q_2(s_j,s_j) + Q_E(s_j) $$ where $Q_1$ is the first variation of area, $Q_2$ is the second and $Q_E$ is the error term in the Taylor expansion. $Q_1(s_j) = 0$ since $M_j$ is minimal. Moreover, $Q_2(s_j,s_j)\geq c\Vert s_j\Vert^2_{W^{1,2}}$ for $c>0$. This is because $M_\infty$ is strictly stable (the Jacobi operator is just the $\mathbb{R}^{n-k}$-valued Laplacian) and thus the $M_j$ are uniformly strictly stable as well for $j$ large. The Taylor series remainder is $O(\Vert s_j\Vert_{W^{1,2}}^3)$. Putting this together, we find that $$ |\hat \Sigma_j| \geq |M_j| + (c/2) \Vert s_j\Vert_{W^{1,2}}^2. $$ for $j$ large. Since $|\hat\Sigma_j| \leq |M_j|$ by assumption, we see that $\hat\Sigma_j = M_j$.

[This proof is an example of a general philosophy in this sort of area that the "worst case" competitor will solve a PDE and thus will be forced to be graphical. (See, for example, this paper of Brian White).]