This passage is part of the solution of an exercise in Differential Geometry. Let $S$ be a surface, and let $X: U \longrightarrow S$ be an orthogonal parametrization. If $N^X = \frac{X_u \wedge X_v}{|X_u \wedge X_v|}$, then
$$
\langle X_{uu}, N^X \rangle \langle X_{vv}, N^X \rangle = \langle X_{uu}, X_{vv} \rangle – \langle X_{uu}^T, X_{vv}^T\rangle, \qquad \qquad \qquad (*)
$$
where $T$ denotes the "part tangent to the surface", according to the book.
The exercise asks the reader to compute the Gaussian curvature for a surface parametrized by an orthogonal parametrization.
My questions are the following:
What does "part tangent to the surface" mean? What is the image of the second partial derivative, say $X_{uu}$? How to prove $(*)$?
Thanks in advance and kind regards
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Best Answer
I got this with a hint by Rodrigo Dias.
$X_{uu}$ lives in $\Bbb{R}^3$, so we can write it as $$ X_{uu} = \langle X_{uu}, X_u \rangle X_u + \langle X_{uu}, X_v \rangle X_v + \langle X_{uu}, N^X \rangle N^X, $$ and analogously for $X_{vv}$: $$ X_{vv} = \langle X_{vv}, X_u \rangle X_u + \langle X_{vv}, X_v \rangle X_v + \langle X_{vv}, N^X \rangle N^X. $$ Recall that $\Bbb{R}^3 = T_XS \oplus \langle N^X \rangle$.
From these expressions, $(*)$ easily follows, sufficing to write the inner product $\langle X_{uu}, X_{vv} \rangle$.