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 $k = 1$.

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ $\mathbb{RP}^2$ 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

2 Answers 2


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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

If $M$ is orientable and the codimension is $1$, the problem has been settled by Federer in

MR0348598 Reviewed Federer, Herbert Real flat chains, cochains and variational problems. Indiana Univ. Math. J. 24 (1974/75), 351–407. (Reviewer: F. J. Almgren Jr.) 49F22

The conclusion is that every codimension 1 homological area-minimizer on an orientable manifold is calibrated by a weakly closed measurable calibration form.

It is easy to construct smooth area-minimizing hypersurfaces so that any calibration form defined globally must be singular by the results of Yongsheng Zhang. Thus, do not expect a smooth calibration form.

For all codimensions larger than $1$, far more examples of non-calibrated area minimizers are available, even on manifolds with torsion-free homology in all dimensions. For instance, you can check out Almgren's example in Section 5.11 of the above paper. See also examples due to Lawson on a flat torus.

MR0460708 Reviewed Lawson, H. Blaine, Jr. The stable homology of a flat torus. Math. Scand. 36 (1975), 49–73. (Reviewer: Pierre Lelong) 32C30


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.