Shape operator and orthogonality of eigenvectors When studying differential geometry (at a hobby level) I always run into problems when it comes to the varying notations and statements about the shape operator $S_p(\mathbf{x})=I^{-1}_pII_p(\mathbf{x})$. 
More specifically, my problems have to do with the orthogonality of the eigenvectors of $S_p$. 
The cause of some problems might relate to uncertainty "where" (in which coordinate space) stuff is happening, so 


*

*first of all, considering a surface $X:\Omega\rightarrow \mathbb{R}^3$, am I right in thinking that the parameters $\mathbf{x}$ in $S_p(\mathbf{x})$ are "from" the parameter domain, that is $\mathbf{x}\in\Omega$?


Now, let $\mathbf{u}_1$ and $\mathbf{u}_2$ be the two eigenvectors with $S_p\mathbf{u}_i=k_i\mathbf{u}_i$.
Many authors state: 
"The eigenvectors of $S_p$ are called principal directions. ...  Recall that it also follows from the Spectral Theorem that the principal directions are orthogonal..."


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*How can this be? As noted above, in my opinion $S_p$ is "evaluated" on $\Omega$. The $S_p$-Matrix is not symmetric. And hence (according to maple) $<\mathbf{u}_1,\mathbf{u}_2>\neq0$. 


However, it seems that when speaking about orthogonality of the eigenvectors it is always implied that they are first transferred via the Jacobian $J_X$ to the tangent plane: $<J_X\mathbf{u}_1,J_X\mathbf{u}_2>=<\mathbf{u}_1,\mathbf{u}_2>_I=0$.
So, my intuition seems to be wrong, can someone point me in the right direction?
Finally, I believe prove orthogonality $<\mathbf{u}_1,\mathbf{u}_2>_I=0$ should be easy using simple substitutions, but I seem to be missing some linear algebraic argument... any idea?
\begin{equation}
<\mathbf{u}_1,\mathbf{u}_2>_I = \mathbf{u}_1^T I_p \mathbf{u}_2 = \\
\textit{using } S_p\mathbf{u}_2=I^{-1}_pII_p\mathbf{u}_2=k_2\mathbf{u}_2 \textit{ we get }\\
\frac{1}{k_2}\mathbf{u}_1^T I_p I^{-1}_pII_p\mathbf{u}_2=\frac{1}{k_2}\mathbf{u}_1^T II_p\mathbf{u}_2\\
\text{ but I cannot see how this is zero }
\end{equation}
Any pointers?
 A: You make an excellent point. The matrix representation of the shape operator $S_p$ with respect to the coordinate basis may very well not be a symmetric matrix. [See, for example, the remark on p. 47 and Example 6 on p. 49 of my text.]
In your question, the vectors $\mathbf u_i$ are in the tangent plane of the surface, not back in $\Omega$. To use the first fundamental form matrix $I_p$, you would need their coordinate representations in terms of the parametrization.
Nevertheless, the shape operator, as a linear map from the tangent space of the surface to itself, is symmetric or self-adjoint. Recall that $S_p(\mathbf u) = -D_{\mathbf u}\mathbf n(p)$, where $\mathbf n$ is the Gauss mapping (i.e., unit normal of the surface) and $D_{\mathbf u}$ denotes the directional derivative (or, more fancily, the covariant derivative) in the direction of $\mathbf u$. If the coordinates in $\Omega$ are given by $u_1,u_2$ and we denote $X_i = \partial X/\partial u_i$, etc., then $X_1$ and $X_2$ give a basis for the tangent plane, and so we only need to check that $S_p(X_i)\cdot X_j = X_i\cdot S_p(X_j)$ for all $i,j=1,2$. The only nontrivial check is this:
$$S_p(X_1)\cdot X_2 = -D_{X_1}\mathbf n \cdot X_2 = -\mathbf n_1\cdot X_2 = \mathbf n\cdot X_{21} = \mathbf n\cdot X_{12} = -\mathbf n_2\cdot X_1 = S_p(X_2)\cdot X_1.$$
The two crucial ingredients here are the symmetry of second-order mixed partial derivatives and the observation that $\mathbf n\cdot X_i = 0$ implies $(\mathbf n \cdot X_i)_j = 0$.
