Volume-minimizing submanifold implies calibrated?

Question

Let $X$ be a smooth manifold of dimension $d$ and $M$ an oriented submanifold of dimension $p < d$ so that the multiples k⋅M are absolutely minimizing $p$-volume in their integral homology classes for all k∈Z . Is $M$ calibrated by some $p$-form w? (Thanks to Robert Bryant for correcting my initial question.)

Definitions: A $p$-form $w$ is called a calibration if it is closed and its evaluation on every geometric $p$-vector $v$ of norm 1 is at most 1 in norm, $|w(v)| \le 1$. One says a $p$-dimensional submanifold $M$ is calibrated (by a calibration $w$) if $w$ evaluates to 1 on each unit tangent $p$-vector to $M$. The word "geometric" above is used to distinguish primitive (or geometric) $p$-vectors from linear combinations of these; "geometric" means "rank one" in tensor language.

Note: It is elementary that if $M$ is calibrated by any $w$ then $M$ minimizes $p$-volume in its homology class, so I'm asking for a kind of converse. In the case $p+1=d$ the converse amounts to a continuum version of Max Flow/Min Cut which is discussed in John Sullivan's 1990 Princeton Ph.D. thesis. I tried asking a form of this question last year, but I did not use the term "calibration". I'm hoping with the correct language an expert will notice the question and be able to answer.

The question below is a good one. I not even know if one can "locally calibrate" some neighborhood of stable oriented minimal submanifold with oriented normal bundle. The local question may contain the important difficulties. A simple closed stable geodesics on a surface can run through areas of positive curvature, so even in that case, the local calibration is not found simply by pulling the length form of the geodesic back along normal coordinates.

Do you know if you can construct the calibration locally -- i.e. in a regular tubular neighborhood of the submanifold? — Rbega, Jun 24 '15 at 19:25
@Rbega: As far as I know, this is an open question even when one assumes that $M$ is compact and the Jacobi operator (which controls the second variation) has strictly positive eigenvalues. — Robert Bryant, Jun 25 '15 at 15:14
Michael Freedman answered his own question in a 1604.1604.00354 where it looks like the claim is that this holds for $p = d-1$. — zzz, Apr 05 '16 at 01:55

score 12 · Answer 1 · answered Jun 24 '15 at 13:46

I think you need to change your hypotheses somewhat if you want a 'yes' answer. Consider the case of $X=\mathbb{RP}^2$ with its standard Riemannian metric of constant curvature, and let $M = \mathbb{RP}^1$ be the (totally) geodesic generator of $H_1(X,\mathbb{Z})=\mathbb{Z}_2$. Then $M$ is diffeomorphic to $S^1$ and so it orientable, and it is easy to show that $M$ has least length in its integral homology class, but it can't be calibrated: If it could, its double cover (i.e., $2M$) would be calibrated, even though the homology class of its double cover is trivial. If you don't like $\mathbb{RP}^2$ because it's not orientable, replace it with $S^3/\mathbb{Z}_2 = \mathrm{SO}(3)$.

By taking $X = \mathbb{RP}^2 \times S^1$ and $M$ to be a closed geodesic whose homology class is the sum of the homology class of the $S^1$ factor and the generator of $H_1(\mathbb{RP}^2,\mathbb{Z})$, you can construct an example of this kind whose homology class is not torsion.

Maybe what you want to assume is that $k\cdot M$ is absolutely minimizing in its homology class for all $k\in\mathbb{Z}$. After all, this would be true if $M$ were calibrated by some calibration.

score 4 · Answer 2 · answered May 16 '23 at 14:38

4

For the corrected version of the question imposing area-minimization of all multiples, the answer is yes. This is a combination of 4.12 and 5.8 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

Unfortunately, the calibration form produced is a priori only measurable. This cannot be improved all the way to smoothness. Explicit constructions can show that even when one has a smooth minimizer, it can happen that all calibration forms have non-continuous points.

answered May 16 '23 at 14:38

Zhenhua Liu

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Could you give an example (or reference?) of a smooth minimizer which does not admit a continuous calibration? – Aidan Backus May 18 '23 at 23:06
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To Aidan, sure, I'll take some time to write it up and put it somewhere. I'll inform you when I finish it. – Zhenhua Liu Jun 05 '23 at 16:22
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To Aidan, I've written it up in https://arxiv.org/abs/2310.19860 – Zhenhua Liu Nov 01 '23 at 10:26
Zhenhua, thanks! At some point down the road I may email you with some comments. – Aidan Backus Nov 01 '23 at 15:41

score 3 · Answer 3 · answered Jun 29 '15 at 13:39

Just a related comment:

Not long ago Dima Burago and Sergei Ivanov showed that if one considers $\mathbb{R}^n$ with a translation-invariant $k$-area integrand (a $k$-area density), then the condition that all flat $k$-discs AND their multiples are area-minimizing is equivalent to the ellipticity (convexity) of the integrand (= all $k$-discs are calibrated by constant-coefficient forms).

http://link.springer.com/article/10.1007%2Fs00039-004-0465-8?LI=true#page-1

Volume-minimizing submanifold implies calibrated?

3 Answers3