I want to compute the Weingarten operator (shape) for the sphere $\{(x,y,z) \in \mathbb{R}^3 \ : \ x^2 + y^2 + z^2 = 1\}$. I am given the adapted frame: $$\left\{\begin{array}{l} E_1 = \cos \varphi \cos \theta \newcommand{\ddx}{\frac{\partial}{\partial x}} \ddx + \cos \varphi \sin \theta \newcommand{\ddy}{\frac{\partial}{\partial y}} \ddy + \sin \varphi \newcommand{\ddz}{\frac{\partial}{\partial z}} \ddz \\ E_2 = – \sin \theta \ddx + \cos \theta \ddy \\ E_3 = -\sin \varphi \cos \theta \ddx – \sin \varphi \sin \theta \ddy + \cos \varphi \ddz \end{array}\right.$$
Among other stuff, I have already computed the connection $1$-forms: $$\omega_{12} = \cos \varphi \newcommand{\d}{ \ \mathrm{d}} \d \theta, \qquad \omega_{13} = \d \varphi \quad \mbox{and}\quad \omega_{23} = \sin \varphi \d \theta$$
and the dual $1$-forms: $$\theta_1 = \d r, \quad \theta_2 = r \cos \varphi \d \theta \quad \mbox{and}\quad \theta_3 = r \d \varphi$$
I already know that $\newcommand{\s}{\mathcal{S}} \s(v) := – \nabla_v E_1$, since $E_1$ is normal to the sphere. I also know that considering the patch $\mathbf{x}(\theta,\varphi) = (\cos \varphi \cos \theta,\cos \varphi \sin \theta,\sin \varphi)$ and working with $\mathbf{x}_\theta, \mathbf{x}_\varphi$ and $\mathbf{x}_\theta \times \mathbf{x}_\varphi$, we get that $\s(v) = -\frac{1}{r}v$. But I want to understand what is going wrong in this other approach.
We have that $\s = – \nabla E_1 = -(\omega_{12}E_2 + \omega_{13}E_3) = \omega_{21}E_2 + \omega_{31}E_3$, fine. If I consider $$v = v^1 E_1 + v^2E_2 + v^3E_3$$
then we get: $$ \d \theta(v) = v^1 \d\theta(E_1) + v^2 \d \theta(E_2) + v^3 \d \theta(E_3) = \frac{1}{r \cos \varphi}v^2 \\ \mbox{and} \quad \d \varphi(v) = v^1 \d\varphi(E_1) + v^2 \d \varphi(E_2) + v^3 \d \varphi(E_3) = \frac{1}{r} v^3$$
That being said, we go for $\s(v)$:
$$\begin{align} \s(v) &= \omega_{21}(v)E_2 + \omega_{31}(v)E_3 \\ &= -\cos \varphi \d \theta(v) E_2 – \d \varphi(v) E_3 \\ &= -\frac{1}{r}v^2 E_2 – \frac{1}{r}v^3E_3 \\ &=-\frac{1}{r}\left(v^2E_2 + v^3E_3\right)\end{align}$$
I'm not exactly sure what is going on. As I am done typing this, it came to my mind: maybe this IS correct, since $\s$'s domain is a tangent space to the sphere, and so $v^1 = 0$? Maybe my problem is with the notation. If someone can explain me a little more about this, I'm very thankful.
Best Answer
Yes since {E2, E3} is an orthonormal basis of the tangent plane, $v^2 E_2 + v^3 E_3 = v$