Area of minimal submanifold in $S^3$

I am asked to prove that lower bound of area of compact minimal submanifold with no boundary in $S^3$ is $4\pi$. Only idea i have is Gauss equation using orthonormal frame $e_1,e_2$. We have $$1=K+|B_{12}|^2-\langle B_{11},B_{22}\rangle,$$ where $K$ is the Gauss curvature of $M$ and $B_{ij}=B(e_i,e_j)$ is the second fundamental form. Since $M$ is minimal, $B_{11}=-B_{22}$. We get: $1=K+||B||^2/2$. Thus by Gauss-Bonnet,

$$\mathrm{Vol}(M)=2\pi \chi(M) + \frac{1}{2}\int_{M}^{} \|B\|^2 \,d\mathrm{Vol}.$$ But I can not estimate the second fundamental form.

Solution 1:

First of all, the geodesic sphere in $\mathbb S^3$ has area $4\pi$. On the other hand, all others minimal surfaces has area bounded below by $4\pi$. The proof that I am aware of uses the Willmore energy. For any immersed surface $S$ of $\mathbb S^3$, the Willmore energy is given by $$W(S) = \int_S \left(1+\frac{|H|^2}{4}\right) d\mu.$$ (This is the area of $S$ when $S$ is minimal). It is known that the Willmore energy is invariant under conformal diffeomorphism.

The Theorem (by Li ad Yau in 1982 here) is:

Theorem Let $S$ be an immersed surface in $\mathbb S^3$ and $p\in S$. Then $W(S) \ge 4\pi n_p$, where $n_p$ is the multiplicity of $S$ at $p$.

The proof roughly goes like this: By a rotation if necessary, assume that the north pole of $\mathbb S^3$ is not in $S$. Then define for any $t>0$ the diffeomorphism: $\Phi : \mathbb S^3 \to \mathbb S^3$ by

• sending $\mathbb S^3\setminus \{N \} \to \mathbb R^3$ by the stereographic projection $\varphi$,
• pick any point $a \in \mathbb R^3$ and define $R_{t,a}(x) = t (x-a)$, and then
• goes back to $\mathbb S^3$.

(So the map is $ \Phi_{t,a} = \varphi^{-1} \circ R_{t,a} \circ \varphi$ (and $\Phi_{a,t} (N)=N$).

Which $\varphi, R_{t,a}$ are conformal, $\Phi$ is a conformal diffeomorphism. Now for each $p\in S$ and $t >0$, choose $\Phi_{t,a}$ with $a = \varphi (p)$. Then as $t\to +\infty$, $\Phi_{t,a} (S)$ converges to $n_p$ copies of geodesic spheres in $\mathbb S^3$ (away from any open neighborhood of $N$). Since $\Phi_{t,a}$ is conformal, and note that each geodesic sphere has Willmore energy $4\pi$, we conclude $W(S) \ge 4\pi n_p$.

Remark: I wonder if a proof not using Willmore energy is possible. But this is also the proof cited by Brendle here.