Let $(M^{n+k},g)$ be a Riemannian manifold. Call a surface $\Sigma^n \subset M$ calibrated if there is a closed $n$-form $\omega$ defined on a neighbourhood $U \subset M$ of $\Sigma$ so that $\omega \lvert \Sigma = \mathrm{vol}_\Sigma$ and for any $p \in U$ and $n$-tuples $(X_1,\dots,X_n) \in T_p M$ of orthonormal vectors $\omega(X_1,\dots,X_n) \leq 1$. (This is slightly different from the usual definition, where usually $\omega$ is defined on $M$.) A simple argument shows that a calibrated surface $\Sigma$ is area-minimising in the neighbourhood $U$, and a small perturbation of $\Sigma'$ of $\Sigma$ will have $\mathrm{Area}(\Sigma') \geq \mathrm{Area}(\Sigma)$. In particular a calibrated surface is minimal, that is stationary for the area functional, and has mean curvature $H_\Sigma = 0$.

There are many examples of calibrated area-minimising surfaces:

  1. linear subspaces of $\mathbf{R}^n$,
  2. minimal graphs of $u: \Omega \subset \mathbf{R}^n \to \mathbf{R}$, where $\Omega$ is an open domain in $\mathbf{R}^n$,
  3. special Lagrangian submanifolds $\Sigma \subset M$ in Calabi-Yau manifolds, that is Lagrangian submanifolds so that $\mathrm{Im} \, \Omega \lvert \Sigma = 0$ where $\Omega$ is the holomorphic volume form,
  4. holomorphic subvarieties of $\mathbf{C}^n$,
  5. area-minimising cones with an isolated singularity at the origin, for example the Simons cone $\mathbf{C}_S = \{ (X,Y) \in \mathbf{R}^n \times \mathbf{R}^n \mid \lvert X \rvert = \lvert Y \rvert \}$. (I believe these are calibrated because of the Hardt-Simon foliations.)

However I cannot think of any examples of area-minimising surfaces which are not calibrated.

Question: What are they? I am especially interested in the codimension one case, where $\Sigma^n \subset M^{n+1}$. In which settings, or under which hypotheses, is an area-minimising surface not be calibrated?

Remark: I can formulate a more technically precise question, at the price of using some terms from geometric measure theory. Let $B \subset \mathbf{R}^{n+k}$ be the unit ball, and $T \in \mathbf{I}_n(B)$ be an integral current with $\partial T = 0$ in $B$. Suppose that $T$ is area-minimising in the sense that for some $\epsilon > 0$ and all currents $S \in \mathbf{I}_{n+1}(B)$ with $\mathrm{spt} \, S \subset \subset B$ and $\mathrm{dist}(\mathrm{spt} \, S,\mathrm{spt} \, T) \leq \epsilon $, $\mathrm{Area} \, (T + \partial S) \geq \mathrm{Area} \, T$. Is there a neighbourhood of $T$ on which it admits a calibration? Here again I would be most interested in the case $k = 1$.

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    $\begingroup$ I think all minimal surfaces are locally calibrated, probably a result of Robert Bryant. $\endgroup$
    – Ben McKay
    Oct 9, 2020 at 15:28
  • $\begingroup$ @Ben I'm aware of this result, but I believe it's slightly different from what I am asking here. If I am not mistaken it says something like "if $\Sigma$ is a (regular) minimal surface, then for every point $p \in \Sigma$ there is $r > 0$ so that $\Sigma \cap B_r(p)$ is calibrated." However the local calibration forms cannot be patched to a global $n$-form. $\endgroup$
    – Leo Moos
    Oct 9, 2020 at 16:01
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    $\begingroup$ I might be mistaken but I think you should get some milage out of $RP^2\subset RP^3$. More precisely, I think it should be true that if it were calibrated, then so would the double cover $S^2 \to RP^3$. But this is not even stable. $\endgroup$ Oct 9, 2020 at 18:16
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    $\begingroup$ $\mathbb{RP}^2$ obviously can't be calibrated because it's not orientable, so maybe one should require orientability in the question. But it can still be homologically minimizing: Let $M^3$ be the quotient of the standard product $S^2\times \mathbb{R}$ by the involution $(u,t)\mapsto (-u,-t)$. The image of $S^2\times\{0\}$ in $M$ is an $\mathbb{RP}^2$ that is minimizing in its homology class: Anything it its homology class must meet all the images of the lines ${u_0}\times\mathbb{R}$, and the projection $M\to\mathbb{RP}^2$ given by $[u,t]\mapsto [u]$ is clearly area decreasing on any surface. $\endgroup$ Oct 9, 2020 at 21:18
  • $\begingroup$ By the way, because you have changed the definition of 'calibrated' (you now only require that it be calibrated in an open neighborhood $U$), it is no longer true that a calibrated submanifold need be area-minimizing in its homology class in $M$. Also, I think that, in your 'more precise' question, you meant to write '$\mathrm{Area}(T+\partial S) \ge \mathrm{Area}(T)$' instead of '$\mathrm{Area}(T+\partial S) \le \mathrm{Area}(T)$', didn't you? $\endgroup$ Oct 10, 2020 at 1:19

1 Answer 1


Actually, a better example along the lines Otis suggests would be the geodesic $\mathbb{RP}^1\subset\mathbb{RP}^2$. Of course, $\mathbb{RP}^1$ is orientable and it is homologically mass-minimizing, but it can't be calibrated on any open set $U\subset\mathbb{RP}^2$ containing $\mathbb{RP}^1$ because twice it is not even stable.

Of course, this also works for any $\mathbb{RP}^{2n-1}\subset\mathbb{RP}^{2n}$, and there are higher codimension examples of closed geodesics in (orientable) lens spaces that are homologically mass-minimizing but that cannot be calibrated on any open neighborhood of the geodesic. One can even foliate $\mathbb{RP}^3$ by homologically mass-minimizing geodesics that cannot be calibrated on any open neighborhood.

What one probably needs to assume, at least, is that every multiple of $\Sigma$ is homologically area-minimizing in some neighborhood before one could hope to construct a 'neighborhood' calibration.

Remark (10/12/20): I just recalled one example of possible interest for this question, since the OP is interested in what can happen in Euclidean space. A student of mine, Timothy Murdoch, in his PhD thesis "Twisted calibrations and the cone on the Veronese surface" (Rice University, 1988), showed that the $3$-dimensional cone in $\mathbb{R}^5$ on the Veronese surface in $S^4$ is area-minimizing, but, of course, it's not orientable. However, its 'double cover' is a cone on the $2$-sphere and so is orientable. I don't know whether this double cover is area-minimizing in $\mathbb{R}^5$ or not. It obviously cannot be calibrated, even if it is area-minimizing.

Explicitly, here is the example: Think of $\mathbb{R}^5$ as $S^2_0(\mathbb{R}^3)$, the traceless $3$-by-$3$ matrices with real entries endowed with the quadratic form $\langle a,b\rangle = \mathrm{tr}(ab)$, which is invariant under $\mathrm{SO}(3)$ with the irreducible action $A\cdot a = AaA^{-1}$ for $A\in\mathrm{SO}(3)$ and $a\in S^2_0(\mathbb{R}^3)$. Then the Veronese cone $C\subset S^2_0(\mathbb{R}^3)$ is the set of matrices $a$ with eigenvalues $t^2,t^2, -2t^2$ for some $t\ge0$. It is a cone on an $\mathrm{SO}(3)$-homogeneous minimal surface $\mathbb{RP}^2\subset S^4$ known as the Veronese surface. (Note that $C$ and $-C$ intersect only at the origin.) $C$ is smooth except at the origin, and, if you define the 'double cover' by counting each smooth point as two points with different orientations, then the double cover is homeomorphic to $\mathbb{R}^3$, parametrized by the quadratic map $s:\mathbb{R}^3\to S^2_0(\mathbb{R}^3)$ defined by $$ s(x) = |x|^2\, I_3 - 3\,x\,x^T\quad\text{for}\ x\in\mathbb{R}^3. $$ Tim showed that, if you take the (literal) Riemannian double cover of $S^2_0(\mathbb{R}^3)\setminus (-C)$, then the double cover of $C\setminus\{0\}\simeq \mathbb{R}^3\setminus\{0\}$ can be calibrated in the ambient double cover as a Riemannian manifold.

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    $\begingroup$ This is very informative. I am especially surprised by your remark on foliations of $\mathbf{R} P^3$ by mass-minimising geodesics, as it almost seemed to contradict a theorem of Rummler-Sullivan cited in Haefliger's Some remarks on foliations with minimal leaves. I remembered this as saying that the leaves of a minimal foliation (of some $X$) are necessarily calibrated, but checking back it only gives a form $\omega$ on $X$ which is relatively closed. A difference I admit had gone over my head... $\endgroup$
    – Leo Moos
    Oct 11, 2020 at 17:49
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    $\begingroup$ I'll add one small question. Do you expect this phenomenon, that an area-minimising surface has a non-minimising multiple to possibly also arise when working in Euclidean space? $\endgroup$
    – Leo Moos
    Oct 11, 2020 at 17:53
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    $\begingroup$ Yes, this is much better than what I had suggested, thanks Robert! Is there any evidence that the "all multiples are minimizing" condition does imply the existence of a calibration (even, say for geodesics on a surface)? $\endgroup$ Oct 11, 2020 at 18:25
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    $\begingroup$ @LeoMoos: Response to first comment: The fibers of the Hopf map $\pi:S^3\to S^2$ are geodesics, so this is a foliation with minimal leaves that cannot be calibrated. It can't even be calibrated locally, since the only possible candidate 1-form (up to sign) isn't closed. But this is a problem only for curves. If the leaves have dimension $k>1$ and are minimal, then every point has an open neighborhood $U$ on which there is a calibration that callibrates the leaves within $U$. (However, the 'obvious' candidate $k$-form is only closed when the orthogonal plane field is integrable.) $\endgroup$ Oct 11, 2020 at 18:56
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    $\begingroup$ @LeoMoos: Response to second comment: I don' expect it in Euclidean space, but I don't immediately see how to rule it out. $\endgroup$ Oct 11, 2020 at 19:42

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