Thanks to [Choi-Schoen theorem](http://link.springer.com/article/10.1007%2FBF01388577), 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 $\displaystyle Area(\Sigma\_\infty)= \lim\_{n\rightarrow +\infty} Area(\Sigma\_n)$?



minimal surfaces in $S^n$
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](http://download.springer.com/static/pdf/626/art%253A10.1007%252Fs11511-013-0101-2.pdf?originUrl=http%3A%2F%2Flink.springer.com%2Farticle%2F10.1007%2Fs11511-013-0101-2&token2=exp=1465224146~acl=%2Fstatic%2Fpdf%2F626%2Fart%25253A10.1007%25252Fs11511-013-0101-2.pdf%3ForiginUrl%3Dhttp%253A%252F%252Flink.springer.com%252Farticle%252F10.1007%252Fs11511-013-0101-2*~hmac=7a33bef3f2dbcaa719fa276e8768ecdf78c2a1850d24e77a1036570e06d6b4a6) to the Lawson conjecture). See Brendle's [survey](http://download.springer.com/static/pdf/667/art%253A10.1007%252Fs13373-013-0034-2.pdf?originUrl=http%3A%2F%2Flink.springer.com%2Farticle%2F10.1007%2Fs13373-013-0034-2&token2=exp=1465224022~acl=%2Fstatic%2Fpdf%2F667%2Fart%25253A10.1007%25252Fs13373-013-0034-2.pdf%3ForiginUrl%3Dhttp%253A%252F%252Flink.springer.com%252Farticle%252F10.1007%252Fs13373-013-0034-2*~hmac=02236f5e5a1d6d0a6e9649f0291ecdf62bc9aa8e19fd38bc1038fe735cf52ea3), 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 $c\searrow 0$.




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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](http://www.ams.org/mathscinet-getitem?mr=0233294) or e.g. [this](http://www.ams.org/mathscinet-getitem?mr=2912703) 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).




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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 $area(\Sigma\_n) \to area(\Sigma\_\infty)$. 


