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Let $T$ be an $n$-dimensional area-minimising hypersurface in $\mathbf{R}^{n+1}$. If $T$ has bounded area growth in the sense that there is a constant $C > 0$ so that $\mathcal{H}^n(T \cap B_R) \leq C R^n$ for all $R > 0$, then there are rigidity theorems for $T$. For example, when $n \leq 6$ then the work of Simons [1] implies that $T$ must be an $n$-dimensional plane. (In larger dimensions there are singular area-minimising hypercones.)

Question. Is there an example of an area-minimising hypersurface with unbounded growth? Could such an example exist in low dimensions, when $n \leq 6$? What about $n = 2$?

[1] James Simons. Minimal varieties in Riemannian manifolds. Annals of Mathematics, Second Series, Vol. 88, No. 1 (1968), pp. 62-105.

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    $\begingroup$ What definition of "area minimizing" are you using? $\endgroup$
    – Ryan Budney
    Commented Jun 23, 2021 at 18:42
  • $\begingroup$ @RyanBudney Any surface that coincides with $T$ outside a compact set, and that has no boundary, has larger area. $\endgroup$
    – Leo Moos
    Commented Jun 23, 2021 at 18:50
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    $\begingroup$ Isn't this a straightforward comparison argument? I.E. for generic $R$, $\partial B_R \cap \Sigma=\sigma$ is smooth and just pick the appropriate subset (which exists by Alexander duality), $\Omega$, of $\partial B_R \setminus \sigma$ so $\partial \Omega=\sigma$. Then $\Sigma'=(\Sigma\setminus B_R)\cup \Omega$ has more area, but the area in the closed ball is Euclidean. Am I missing something? $\endgroup$
    – RBega2
    Commented Jun 23, 2021 at 20:22
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    $\begingroup$ Just a small comment that if the minimizing condition included "oriented" then this result fails: take the planes $\mathbb{R}^2 \times \mathbb{Z} \subset \mathbb{R}^3$ with the $e_3$ orientation. It's a nice exercise to check that this is area-minimizing as a oriented surface (i.e., as a current with $\mathbb{Z}$ coefficients) but of course the area growth is like $R^3$ instead of $R^2$. It's interesting to spot the point where RBega2's construction fails in this setting. $\endgroup$
    – Otis Chodosh
    Commented Jun 24, 2021 at 1:16
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    $\begingroup$ @LeoMoos It is a bit hard to see and some illustrations omit the second half of it, but away from the axis, the helicoid consists of two spirals alternating in orientation, so you can decrease the area using a wide neck while still preserving the orientation. $\endgroup$
    – mlk
    Commented Jun 24, 2021 at 11:09

2 Answers 2

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This is a straightforward comparison argument. Let $\omega_n$ be the volume of $\partial B_1\subset \mathbb{R}^{n+1}$.

For generic $R$, one has $\partial B_R \cap T=\tau$ a smooth submanifold. By Alexander duality, there is a subset, $\Omega$, of $ \partial B_R\setminus \tau$ so $\partial \Omega=\tau$. Clearly, $\mathcal{H}^n(\Omega)\leq \omega_n R^n$. In fact, up to replacing $\Omega$ by it's complement one has $\mathcal{H}^n(\Omega)\leq \frac{1}{2}\omega_n R^n$.

By the area minimization property, $$\mathcal{H}^n(\Sigma\cap B_R)\leq \mathcal{H}^n(\Omega)\leq \omega_n R^n.$$

The monotonicity formula ensures the bound holds for all $R$.

In fact, this argument should work (using slicing) for an area minimizing integral $\mathbb{Z}_2$ current.

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  • $\begingroup$ I'm just curious: why would slicing be required when working with currents? The singular set has codimension seven, so $\partial B_R \cap T$ (or $\partial B_R \cap \mathrm{spt} \lVert T \rVert$ to be precise) would still be regular for generic $R$, no? Mind you I've never seen slicing 'in the wild', so I'm not positive how it's usually applied. $\endgroup$
    – Leo Moos
    Commented Jun 23, 2021 at 23:33
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    $\begingroup$ You're probably right, though I guess one would need to be a little worried about the argument being circular (i.e., the area bound is almost certainly used implicitly in the regularity theory). Slicing (of currents) is useful as then you would obtain a slice that is a closed current in the boundary of the ball, so the region $\Omega$ could also be a current whose boundary was a slice $\endgroup$
    – RBega2
    Commented Jun 24, 2021 at 0:20
  • $\begingroup$ I think you can disregard my previous comment, at least when the dimension is large enough, say $n \geq 8$ that the singular set would be larger than a set of isolated points. I appreciate your explanation regarding the slicing argument. $\endgroup$
    – Leo Moos
    Commented Jun 24, 2021 at 9:45
  • $\begingroup$ Personally in the case of currents I think one would simply prefer slicing because it is the more natural operation. In particular you'd get some continuity properties for free. That being said though, for currents and in fact for any orientable surfaces, you'd need some more work, because while by standard arguments there is always a minimal current on the sphere with the prescribed boundary, it may have multiplicity larger than 1 in some places (e.g. if $\tau$ consists of several identically oriented circles on the same hemisphere). $\endgroup$
    – mlk
    Commented Jun 24, 2021 at 9:48
  • $\begingroup$ @mlk Doesn't Otis's comment above point to the fact that the comparison argument is inapplicable for currents? As far as I understand when $T$ is the union of countably many parallel planes, then the comparison gives a bound like $R \mathcal{H}^n(\partial B_R)$ because there are roughly $R$ planes intersecting $B_R$. $\endgroup$
    – Leo Moos
    Commented Jun 24, 2021 at 10:56
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When $n = 2$ the sort of examples you require does not exist. This is due to Fischer-Colbrie and Schoen, "The structure of complete stable minimal surfaces in 3-manifolds of non-negative scalar curvature".

https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160330206

Essentially: requiring just that the minimal surface is area minimizing for compact perturbations up to second order (which is weaker than the strict area-minimizing condition you asked for), they prove (among other things) that the only "stable" minimal surface in $\mathbb{R}^3$ is the plane.

The reason is basically that the second variation of the area gives a Laplace equation with a potential, which in $n = 2$ can be related to the scalar curvature of the minimal surface. And non-trivial minimal surfaces in $\mathbb{R}^3$ all have negative scalar curvature.

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  • $\begingroup$ Is the helicoid area-minimising? I'm thinking one could 'plug in' a catenoidal neck into the planar ends to decrease area, somewhere far from the axis. $\endgroup$
    – Leo Moos
    Commented Jun 23, 2021 at 18:47
  • $\begingroup$ Ah, you actually want area minimizing. Okay. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2021 at 19:45
  • $\begingroup$ In that case, you have no examples in n = 2. not sure about 3--6. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2021 at 19:54
  • $\begingroup$ Fair enough, I'd forgotten that Fischer-Colbrie--Schoen don't need any hypothesis regarding the area growth. By the way, when you write 'second variation of the mean curvature' do you mean 'second variation of the area' - it seems like you're talking about the Jacobi operator? $\endgroup$
    – Leo Moos
    Commented Jun 23, 2021 at 20:05
  • $\begingroup$ Oops.. stupid typo. Fixed. $\endgroup$
    – Willie Wong
    Commented Jun 23, 2021 at 20:06

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