Fundamental forms of a sphere are proportional

Question

I'm trying to prove that the first and second fundamental forms of the sphere are proportional to each other, regardless of the parametrization. I was trying to exploit the fact that the normal to a sphere has the same direction as the radius-vector. If the radius of the sphere is $a>0$ and it is parametrized by some function $\mathbf{r}(u,v) = (x(u,v),y(u,v),z(u,v))$, the normal vector field becomes $\mathbf{n}(u,v) = \frac{\mathbf{r}(u,v)}{a}$. Then the forms become $$\mathfrak{G} = \begin{bmatrix} \langle\mathbf{r}_u,\mathbf{r}_u\rangle & \langle\mathbf{r}_v,\mathbf{r}_u\rangle \\ \langle\mathbf{r}_u,\mathbf{r}_v\rangle & \langle\mathbf{r}_v,\mathbf{r}_v\rangle \end{bmatrix}, \quad Q=\frac{1}{a}\begin{bmatrix} \langle\mathbf{r}_{uu},\mathbf{r}\rangle & \langle\mathbf{r}_{vu},\mathbf{r}\rangle \\ \langle\mathbf{r}_{uv},\mathbf{r}\rangle & \langle\mathbf{r}_{vv},\mathbf{r}\rangle \end{bmatrix}$$ I can't find a way to show the final step that $\langle\mathbf{r}_u,\mathbf{r}_u\rangle \propto \langle\mathbf{r}_{uu},\mathbf{r}\rangle$ (and equivalently for the other entries) however. I might be missing something obvious, but any help is appreciated!

Answer 1

All you need is that $|\mathbf{r}|^2 =a$, so

\begin{align*} 0&= \partial _u |\mathbf{r}|^2 =2 \langle \mathbf{r}, \mathbf{r}_u\rangle \\ 0&= \partial _v |\mathbf{r}|^2 =2 \langle \mathbf{r}, \mathbf{r}_v\rangle \end{align*}and thus

\begin{align} \langle \mathbb r_{uu} , \mathbb r\rangle &= \langle \mathbb r_u , \mathbb r\rangle _u - \langle \mathbb r_u, \mathbb r_u\rangle = - \langle \mathbb r_u, \mathbb r_u\rangle, \\ \langle \mathbb r_{uv} , \mathbb r\rangle &= \langle \mathbb r_u , \mathbb r\rangle _v - \langle \mathbb r_u, \mathbb r_v\rangle = - \langle \mathbb r_u, \mathbb r_v\rangle,\\ \langle \mathbb r_{vv} , \mathbb r\rangle &= \langle \mathbb r_v , \mathbb r\rangle _v- \langle \mathbb r_v, \mathbb r_v\rangle = - \langle \mathbb r_v, \mathbb r_v\rangle. \end{align}

Answer 2

You can also prove it without any computation.

The metric on the sphere is homogeneous (under the action of $SO(3)$ acting on $\bf R^3$). So are the first and second fundamental forms. Note that the isotropy group of a point $s$ is $O(2)$ acting on the tangent space $T_xS$.

Now, up to a constant there exists a unique quadratic form on this plane invariant by $O(2)$. So at this point $II_x= \lambda _x I_x$, where $I$ and $II$ are the first and second fundamental forms. By homogeneity, $\lambda$ is constant, and the result follows.