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Thanks to Choi-Schoen theorem, we know that the space of embedded minimal surfaces into $S^3$ of fixed genus is compact. My question are simples:

Can we remove the embeddness assumption?

Can we replace $S^3$ by $S^n$ for $n\geq3$, still for surfaces (not hypersurfaces)?

Else what are the most reasonable results of "weak" compactness we can expect. For instance, is that reasonable to expect that if we have $\displaystyle \Sigma_\infty =\lim_{n\rightarrow +\infty} \Sigma_n$ in a reasonable sense, do we have $\displaystyle Area(\Sigma_\infty)= \lim_{n\rightarrow +\infty} Area(\Sigma_n)$?

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Without embeddedness, the Choi--Schoen theorem is false.

For example, there is a huge family of rotationally symmetric immersed tori in $\mathbb{S}^3$ (the only embedded one is the Clifford torus, by Brendle's solution to the Lawson conjecture). See Brendle's survey, Theorem 1.4, where he shows that there is an infinite family of immersed minimal tori of the form $$ \left(\sqrt{1-r(t)^2} \cos s, \sqrt{1-r(t)^2} \sin s, r(t) \cos t, r(t) \sin t\right) \in \mathbb{S}^3\subset\mathbb{R}^4 $$ for $r(t)\in (0,1)$ a smooth function. I'll sketch the construction and argue that compactness fails.

Thanks to the symmetry, the minimal surface equation for $r(t)$ can be integrated once, to become $$ \frac{r'(t)^2}{r(t)^4(1-r(t)^2)^2}+\frac{1}{r(t)^2(1-r(t)^2)}=\frac{4}{c^2} $$ for some constant $c\in (0,1]$. The solution $r(t) = \frac{1}{\sqrt{2}}$ (i.e. $c=1$) is the embedded Clifford torus.

You can argue that the maximum and minimum values of $r(t)$ (depending on $c$) are given by $$ \overline{x}(c)=\sqrt{\frac{1-\sqrt{1-c^2}}{2}}, \underline{x}(c)=\sqrt{\frac{1+\sqrt{1-c^2}}{2}}. $$ Furthermore, the period of the solution is given by $$ T(c) = \int_{\underline{x}(c)}^{\overline{x}(c)}\frac{c}{x\sqrt{1-x^2}\sqrt{4x^2(1-x^2)-c^2}}dx. $$

For this to be an immersed torus, one needs that $\frac{2\pi}{T(c)}$ is rational, so that it closes up. You can check that $T(c)\to\sqrt{2}\pi$ as $c\nearrow 1$ and $T(c)\to\infty$ as $c\searrow 0$.

Now, suppose that $c$ is chosen so that $\frac{2\pi}{T(c)}$ is rational. Choose $t$ so that $r'(t) =0$ and $r(t) = \underline{x}(c)$. Then, as computed in Brendle's survey, we have that the second fundamental form at this point satisfies $$ h(\partial_s,\partial_s) = r(t)\sqrt{1-r(t)^2} $$ On the other hand, the metric satisfies $$ g(\partial_s,\partial_s) = 1-r(t)^2, $$ so the norm of the second fundamental form at this point is at least $$ \frac{r(t)}{\sqrt{1-r(t)^2}}=\sqrt{\frac{1+\sqrt{1-c^2}}{1-\sqrt{1-c^2}}}\to\infty $$ as $c\searrow 0$.


As for minimal surfaces in $\mathbb{S}^n$ for $n>3$, I think that the space of even minimal spheres is quite big. See, e.g. Calabi's paper or e.g. this paper. I'm not completely sure about embeddedness of these examples, or whether or not their second fundamental forms blow up (certainly their areas do).


If you have that $\Sigma_n$ converges to $\Sigma_\infty$ as varifolds (up to a subsequence, this always holds assuming a priori area estimates on $\Sigma_n$), then $area(\Sigma_n) \to area(\Sigma_\infty)$.

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  • $\begingroup$ @ Otis, Thank you very much for the counter example. For your last remark, I image that the limit object can be singular?I mean, a priori you can converge to a union of smooth and singular mesure, no? To be more precise, is that know that there is a no neck energy property in the bubbling process? $\endgroup$
    – Paul
    Jun 7, 2016 at 11:11
  • $\begingroup$ Hi Paul, the "varifold statement" is essentially a beefed up version of the fact that Radon measures (think of the Hausdorff measure restricted to $\Sigma_j$ with uniform mass bounds admit a convergent subsequence to some measure (the measure has no structure, a priori). No area can be bubbled off (essentially, because area is a subcritical quantity). Another way to see this is via the monotonicity formula: if a minimal surface concentrates a lot of area in a small region, then its overall area must be really huge. $\endgroup$ Jun 7, 2016 at 11:26
  • $\begingroup$ @Paul If you treat the minimal surfaces as conformal harmonic mappings, that there is no neck and no energy loss follows from the fact that the mapping are conformal. $\endgroup$ Jun 7, 2016 at 21:16

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