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Let $M^{n-1}$ be a smooth closed manifold, embedded into the round sphere $\mathbf{S}^n$ via a regular map $\Phi$. Using tools from geometric measure theory, one can prove the existence of a $n$-dimensional current $T$ in the ball $B \subset \mathbf{R}^{n+1}$ with $\partial T = \Phi(M)$, which has the least area among all such surfaces.

Question. What is known regarding the generic uniqueness of $T$? How does this relate to the regularity of $\Phi$: for example if it is in $C^{k,\alpha}$, $C^\infty$ or $C^\omega$?

  • I vaguely remember seeing some results asserting the generic uniqueness in the Baire sense, but I can't find the paper now. The literature on the Plateau problem is honestly too vast...
  • I'd be most interested in results that establish uniqueness in a stronger sense, for example unless $\Phi$ belongs to some space of embeddings $M \to \mathbf{S}^n$ with quantifiable codimension.
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  • $\begingroup$ By quantifiable codimension, do you mean finite co-dimension? Because I would assume that as long as there aren't two solutions with the same tangent-space at almost all boundary points (which I guess is even more rare than non-uniqueness), I can always slightly wiggle around one of these boundary points to favor one or the other solution and thus gain uniqueness in an infinite dimensional number of ways. $\endgroup$
    – mlk
    Commented Sep 3, 2021 at 13:55
  • $\begingroup$ @mlk Loosely speaking I meant that $T$ be unique unless $\Phi$ belongs to some 'bad' set of maps $M \to \mathbf{S}^n$ which is 'small in some strong sense'. It would have been more precise to write finite codimension; I was trying to express some flexibility. As to the rest of your comment, I'm not sure I completely follow the point you're making. Could you reformulate it or elaborate on it? $\endgroup$
    – Leo Moos
    Commented Sep 3, 2021 at 14:20
  • $\begingroup$ I think this might be the paper you are remembering: mathscinet.ams.org/mathscinet-getitem?mr=653546 . Im not sure if the paper specifically answers your question but I think it will be relevant. $\endgroup$
    – Otis Chodosh
    Commented Sep 3, 2021 at 17:13
  • $\begingroup$ @LeoMoos Wouldn't the exceptional set then be of finite dimension and not co-dimension? In any case, I think that is also not true, even if you discount reparametrization: Take a non-unique situation (such as the two circles for which the corresponding catenoid and two-disc solutions have the same area) and then add a small closed curve somewhere far away. Now the minimal surface will always split into one of the old solutions and a small surface for the curve. Thus you have non-uniqueness for any such curve, of which there are infinite-dimensionally many. $\endgroup$
    – mlk
    Commented Sep 3, 2021 at 17:32
  • $\begingroup$ @OtisChodosh Thanks Otis, I'll take a look. $\endgroup$
    – Leo Moos
    Commented Sep 3, 2021 at 18:39

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