Computing the shape operator I am trying to compute the shape operator and Gaussian curvature for some smooth zero sets of polynomials $f$ in $\mathbb{R}^n$, oriented by $N = \nabla f / || \nabla f||$
The approach I thinking about is this (which I didn't learn from a text, so maybe there is a problem with it):


*

*Compute the normal at a point $p$, and pick some vectors $v, w \ldots $ that are a local frame for the tangent space. 

*Compute $\langle \nabla_{v} N, w \rangle = \frac{1}{||\nabla f||}\Sigma_{i,j=1}^{n+1} \frac{\partial^2 f}{\partial x_i \partial x_j} v_i w_j$. (The justification for this formula: $\nabla_v \frac{ \nabla f}{|| \nabla f ||} = (\nabla_v (\nabla f)) (1/ ||\nabla f||) + NormalComponent$)

*Deduce from this the matrix for $L_p(v) = - \nabla_v N$.


However, something seems to be wrong with this approach. For example, in my computation below for the sphere, I get a Gaussian curvature that is not constant.


*

*We pick some point where $y \not = 0$ and $x \not = 0$, then $((x,y,z),-y,x,0)$ and $((x,y,z),0,-z,y)$ describes a local frame. (The first triple is the point, the next three coordinates describe the vector in the tangent space of $\mathbb{R}^3$).

*Computing the matrix $L_p$ in this basis gives $1/r^2 \begin{pmatrix} x^2 + y^2 &  -xz \\ -xz & z^2 + y^2 \end{pmatrix}$, which has determinant $-y^2 / r^2$...


I am really confused. I would appreciate someone pointing out my mistake.

Okay - I figured out my mistake. Pretty silly - I was jumping from 2 to 3 as if my choice was of an orthonormal basis, but of course it was not.
