Where have I gone wrong in the following computation of the shape operator of surface?
Suppose we have a surface $M = \{(x,y,f(x,y)) \: | \: (x,y) \in \mathbb{R}^2 \}$ for some nice $f:\mathbb{R}^2 \to \mathbb{R}$. Let $\phi: \mathbb{R}^2 \to M$ be the obvious parametrisation. The tangent space at a point are given by the span of $\phi_x = (1,0,f_x)$ and $\phi_y = (0,1, f_y)$ (evaluated at that point). The unit normal vector is $N = \frac{\phi_x \times \phi_y}{\|\phi_x \times \phi_y \|} = \gamma (-f_x, -f_y,1)$, where $\gamma = (1+f_x^2+f_y^2)^{-1/2}$.
The coefficients of the first fundamental form are
$E = \phi_x \cdot \phi_x = 1 + f_x^2$, $F = \phi_x \cdot \phi_y = f_x f_y$, and $G = \phi_y \cdot \phi_y = 1 + f_y^2$. Note that $EG – F^2 = 1 + f_x^2 + f_y^2 = \gamma^{-2}$. The coefficients of the second fundamental form are
$e = N \cdot \phi_{xx} = \gamma f_{xx}$, $f = N \cdot \phi_{xy} = \gamma f_{xy}$, and $g = N \cdot \phi_{xx} = \gamma f_{yy}$.
By the Weingarten equations, we have the shape operator given as
$\frac{1}{EG -F^2} \begin{bmatrix} e & f \\ f & g\end{bmatrix}
\begin{bmatrix} G & -F \\ -F & E\end{bmatrix} =
\gamma^3 \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix}
\begin{bmatrix} 1 + f_y^2 & -f_x f_y \\ -f_x f_y & 1 + f_x^2\end{bmatrix}
$
This matrix is supposed to be symmetric ([sic] SEE BELOW). However, even choosing a simple $f$, like $f = x^2 – y^2$ yields something not symmetric:
$\gamma^3 \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}
\begin{bmatrix} 1 + 4y^2 & -4xy \\ -4xy & 1 + 4x^2\end{bmatrix}
= 2\gamma^3 \begin{bmatrix} 1 + 4y^2 & -4xy \\ 4xy & -(1 + 4x^2)\end{bmatrix}
$
Please set me straight.
ADDENDUM:
I have conflated a few things and tripped over some basic linear algebra. The matrix above need not by symmetric in general. Rather, it is self-adjoint, and the corresponding bilinear form is what is symmetric.
Best Answer
The shape operator at a point $p$ is a linear map $S_p : T_p M \to T_p M$, so a priori it is meaningless to ask whether it is symmetric. The correct thing to say (as discussed in comments) is that it is self-adjoint with respect to the metric/first fundamental form $g$; that is, $$g(S_p(X),Y) = g(X,S_p(Y)).$$
This is equivalent to the bilinear form associated to $S$ by the metric (the second fundamental form) $$A(X,Y) = g(S_p(X), Y)$$ being symmetric, which you have established is true.
Once you have fixed a basis, the map $S_p$ can be considered as a matrix, at which point it makes sense to ask whether or not it is symmetric. If the basis is orthonormal then the answer will be yes, but it general it will not. Thus the source of your confusion is the fact that the first fundamental form is not (a multiple of) the identity matrix in your chosen coordinates.