How to calculate the mean curvature of clifford torus immersed in the standard sphere $\mathbb{S}^3$?

Question

We know that clifford torus is a minimal surface immersed in $\mathbb{S}^3$, but how to express this immersion in parametrization and how to calculate the mean curvature of this immersion? If anyone can help me, thanks very much!

Answer 1

There is absolutely no need to parametrize the Clifford torus to compute its mean curvature. One only needs to find a normal vector field in $S^3$ and an orthonormal frame of its tangent bundle. The Clifford torus $T$ is $T = \left\{(x_1,y_2,x_2,y_2)\in \Bbb R^4 \mid {x_1}^2+ {y_1}^2 = {x_2}^2 + {y_2}^2 = \frac{1}{2} \right\}$. Consider the two vector fields $E_1$ and $E_2$ defined by $$ E_1(x_1,y_1,x_2,y_2) = \sqrt{2}(-y_1,x_1,0,0) \quad \text{and} \quad E_2(x_1,y_1,x_2,y_2) = \sqrt{2}(0,0,-y_2,x_2). $$ One readily checks that $\{E_1,E_2\}$ induces an orthonormal frame on $T$. The vector fields $$ N(x_1,y_1,x_2,y_2)=(x_1,y_1,-x_2,-y_2) \quad \text{and} \quad \bar{N}(x_1,y_1,x_2,y_2) = (x_1,y_1,x_2,y_2) $$ are such that $\{E_1,E_2,N,\bar{N}\}$ is an orthonormal frame of $\iota^*(T\Bbb R^4)$ where $\iota\colon T\to \Bbb R^4$ is the inclusion. Moreover, $\bar{N}$ is the normal vector field to $S^3$ in $\Bbb R^4$, and $N$ is the normal vector field to $T$ in $S^3$.

Let $W$ be the Weigarten map of $T\subset S^3$, defined by $W(X) = \nabla^{S^3}_X N$, where $\nabla^{S^3}$ is the Levi-Civita connection of $S^3$. By definition, the mean curvature $H$ of $T\subset S^3$ is the trace of $W$, that is $$ H = \langle W(E_1),E_1\rangle + \langle W(E_2),E_2\rangle. $$ Recall that $\nabla^{S^3}$ is the orthogonal projection of the ambient directional derivative onto $TS^3 = \bar{N}^{\perp}$, so that for $i\in\{1,2\}$, one has $W(E_i)=\nabla^{S^3}_{E_i}N = D_{E_i}N - \langle D_{E_i}N,\bar{N}\rangle \bar{N}$. From the expression of $N$, one has $D_{E_1}N = E_1$ and $D_{E_2}N = -E_2$. (Coincidentally, these values have no component in the direction of $\bar{N}$: this is not a general fact.) It follows that $W(E_1) = E_1$ and $W(E_2)=-E_2$. Finally, $$ H = \langle W(E_1),E_1\rangle + \langle W(E_2),E_2\rangle = \|E_1\|^2 - \|E_2\|^2 = 1-1 = 0. $$ Note that we have shown more than the fact that $T$ is minimal in $S^3$: we have shown that its principal curvatures are $1$ and $-1$.

If one really wants to use a parametrization, then consider the one given by $f(s,t) = \frac{1}{\sqrt{2}}(\cos s, \sin s, \cos t, \sin t)$. One has $E_1 = \sqrt{2}\partial_s f$, $E_2 = \sqrt{2}\partial_t f$, etc.