Almgren's regularity Theorem ; a simple example?

Question

Let me remind Almgren's regularity Theorem: the singular set of area-minimizing surface has codimension at least $2$. I wish to share here a simple example in low dimension, although I don't know whether the manifold is really a minimizer (this is the question).

The dimension of the manifold is two. Its co-dimension is $2$ as well. The ambient space ${\mathbb R}^4$ is identified with the space $M_2({\mathbb R})$ of $2\times2$ matrices. The Euclidian norm is $$\|A\|=\left(\frac12{\rm Tr}A^TA\right)^{1/2}.$$ It is normalized so that $\|I_2\|=1$.

The orthogonal group $O_2({\mathbb R})$ is a curve contained in the unit sphere $S_3$. It is made of two unit circles $\cal R$ and $\cal S$, respectively that of rotations and that of symmetries (I like to recall that, within the $3$-dimensional sphere, $\cal R$ and $\cal S$ are linked).

We look for a minimizing surface $\Gamma$ with boundary $O_2({\mathbb R})$. If it was unique, then it would inherit the symmetry of $O_2({\mathbb R})$. In particular, it should satisfy $$O_2({\mathbb R})\cdot\Gamma=\Gamma=\Gamma\cdot O_2({\mathbb R}).$$ This means that $\Gamma$ would be a union of orbits under the action of $O_2({\mathbb R})\times O_2({\mathbb R})$. Equivalently, $\Gamma$ would be defined in terms of singular values only. However, a single orbit, when the singular values are given and distinct ($s_1>s_2>0$), is already a surface, a $2$-dimensional object. Since the set of pairs $(s_1,s_2)$ must vary continuously from $(1,1)$, it seems therefore impossible that $s_1\ne s_2$. In conclusion, the only symmetric candidate is the surface $\Gamma_0$ defined by $s_1=s_2$. This is the union of the unit disks whose boundaries are $\cal R$ and $\cal S$.

The surface $\Gamma_0$ is singular at the origin $0_2$. The singularity is of codimension $2$, according to Almgren's Theorem. Its area is $2\pi$. My question is

Does $\Gamma_0$ minimize the area, among all the surfaces whose boundary is $O_2({\mathbb R})$ ?

Remark that the distance of any two points taken in $\cal R$ and $\cal S$ equals $\sqrt2$. One may consider the not-that-much symmetric surface $\Gamma_1$, the runion of segments $[R_\theta,S_\theta]$ where $$R_\theta=\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix},\qquad S_\theta=\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}.$$ Notice that $\Gamma_1$ is not a critical point of the area functional. Actually it is contained in the unit sphere associated with the operator norm (definitely not a Euclidian one !) ; each point is therefore exposed. Its area, equal to $(\sqrt2+\log(1+\sqrt2)\pi$, exceeds that of $\Gamma_0$, because of $$\log(1+\sqrt2)=\int_1^{1+\sqrt2}\frac{ds}s>\frac{\sqrt2}{1+\sqrt2}=2-\sqrt2.$$

Answer 1

As I mentioned in my comment, a pair of orthogonal $2$-discs in $\mathbb{R}^4$ is known to be minimizing among all orientable $2$-currents with the same boundary by the technique of calibrations: One can assume that each $2$-disk lies in the $x_{2i-1}x_{2i}$-plane and satisfies $x_{2i-1}^2 + x_{2i}^2\le 1$ for $i = 1, 2$ and is oriented so that $\mathrm{d}x_{2i-1}\wedge\mathrm{d}x_{2i}$ is the positive area form on each disk. Then the closed $2$-form $\omega = \mathrm{d}x_1\wedge\mathrm{d}x_2 + \mathrm{d}x_3\wedge\mathrm{d}x_4$ pulls back to each disk to be its area form.

Because $|\omega(v_1,v_2)|\le |v_1\wedge v_2|$ for all vectors $v_1,v_2\in\mathbb{R}^4$, it is a calibration that calibrates the union of the two disks. The usual fundamental lemma of calibration theory then shows that the union of the two disks has minimal area among all all compact oriented surfaces with the same boundary.

More generally, there is the Angle Criterion for any pair of oriented $p$-planes in $\mathbb{R}^{2p}$ that says when the union of the two unit balls in the two planes has least area among all oriented $(p{+}1)$-currents with the same boundary: First one shows that, given such a pair $E_1$ and $E_2$ of oriented $p$-planes, then there is an oriented orthonormal basis $e_1,\ldots e_{2p}$ in $\mathbb{R}^{2p}$ and angles $\theta_,\ldots,\theta_p$ with $$ 0\le \theta_1 \le \theta_2 \le \theta_p\le \pi-\theta_{p-1} $$ such that $e_1,\ldots,e_p$ is an oriented basis of $E_1$ while $e_1^*,\ldots,e_p^*$ is an oriented basis of $E_2$, where $e_i^* = \cos\theta_i\,e_i + \sin\theta_i\,e_{i+p}$.

Frank Morgan conjectured that the union of the two oriented unit disks in these planes is $p$-area minimizing among all compact oriented $(p{+}1)$-manifolds with the same boundary) if and only if $\theta_p\le \theta_1+\cdots+\theta_{p-1}$.

Then Dana Mackenzie showed that the union of $E_1$ and $E_2$ can be calibrated whenever $\theta_p\le \theta_1+\cdots+\theta_{p-1}$, verifying one direction of Morgan's Angle Conjecture.

Finally, when $\theta_p > \theta_1+\cdots+\theta_{p-1}$, Gary Lawlor explicitly constructed a compact oriented $(p{+}1)$-manifold whose boundary is the same as the boundary of the two disks and that has less area than that of the two disks, thus verifying the other direction of Morgan's Conjecture.

See G. Lawlor, The Angle Criterion, Invent. Math. 95 (1989), 437–446 for details.

Answer 2

The following answer about the unorientable case is to complement Robert Bryant's detailed explanation of the orientable case.

If you allow competitors for the area-minimization to be unorientable, the formulation is mod 2 flat chains/currents. The regularity problem is much easier and the singular set is again of codimension at least two. The main reference is

MR0260981 - The singular sets of area minimizing rectifiable currents with codimension one and of area minimizing flat chains modulo two with arbitrary codimension Federer, Herbert Bull. Amer. Math. Soc. 76 (1970), 767–771.

For the fact that orthogonal planes are area-minimizing mod $2$ the relevant references are

MR0675210 - On the singular structure of two-dimensional area minimizing surfaces in Rn Morgan, Frank Math. Ann. 261 (1982), no. 1, 101–110.

MR0735418 - Examples of unoriented area-minimizing surfaces Morgan, Frank Trans. Amer. Math. Soc. 283 (1984), no. 1, 225–237.

MR0879855 - Calibrations modulo ν Morgan, Frank Adv. in Math. 64 (1987), no. 1, 32–50.

The 2-d case is handled in the Corollary 7 of the first reference. (Note that the original statement of Corollary 7 itself is fine but the restatement in the introduction has a typo of using oriented instead of unoriented.) Then in the second paper, the fact is proved for all dimensions. The final one gives a vast generalization.

The rough idea is to use constancy theorem and orthogonal projections, to reduce the unorientable case to the orientable case. This is one of the common techniques to prove mod $2$ minimizing. The great part is that it can handle some complex analytic varieties like hyperbolics and (the degenerate case) two or more orthogonal planes.

Another powerful approach is by Gary Lawlor, which handles area-minimizing cones better. On the other hand, it cannot handle the complex curve and hypersurface case as Morgan's work above. Thus, both are complementary to each other.

MR1073951 - A sufficient criterion for a cone to be area-minimizing Lawlor, Gary R. Mem. Amer. Math. Soc. 91 (1991), no. 446, vi+111 pp.

Most of the mod $2$ area-minimizers known to date are proven using these two methods. There is some further work by Morgan and Lawlor, like slicing, etc., that can also handle the mod $2$ case.

Remark: the easiest examples to test Almgren's theorem is definitely complex analytic curves. As already pointed out by Robert Bryant, the calibration form is rescaled powers of Kahler forms. Thus any complex analytic subvarieties on Kahler manifolds are area-minimizing among orientable competitors. If you want really exotic examples, e.g., fractal singular sets, non-smoothable singular sets, etc., I will shamelessly refer you to my recent works: https://arxiv.org/abs/2110.13137 https://arxiv.org/abs/2206.08315 and https://arxiv.org/abs/2306.07271