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As curvature can effect the topology and I learned from someone that the geometry of minimal surface is rigid, I want to study the geometry of minimal surface in a sphere. And here is my question:

$M \subseteq S^3$ is a minimal surface and with sectional curvature $K_M>0$. How to show $M$ is totally geodesic i.e the standard $S^2$ in $S^3$?

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Note that $M$ is orientable as a hypersurface. The conditions $K_M>0$ and Gauss Bonnet theorem imply that $M$ is diffeomorphic to a sphere. Then it is a general classical result of Almgrem that all minimal immersion $f : \mathbb S^2 \to \mathbb S^3$ must be totally geodesic. The proof can be found in here. The proof is roughly the following:

Let $n$ be a normal vector fields of the immersion and define

$$\Phi = f_{uu} \cdot n+ \sqrt{-1} f_{uv} \cdot n,$$

where $(u,v)$ is a local conformal coordinates of $\mathbb S^2$. Then it is shown that $\partial_{\bar z} \Phi = 0$ (where $z = u+ \sqrt{-1}v$) and $\Phi$ defines a quadratic differential. Since the space of holomorphic quadratic differential on $\mathbb S^2$ is trivial, we obtain $\Phi =0$.

On the other hand, we have

$$ |k_1 - k_2| = 2|f_u|^{-2} |\Phi|^2,$$

where $k_1, k_2$ are the principle curvatures. As $k_1 + k_2 =0$ (it's minimal), we have $k_1 = k_2 = 0$ and so the immersion is totally geodesic.

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