I was reading differential geometry form Andrew Pressley In that I had following doubt
I understand everything except one thing why Weingarten map $\omega (\gamma^.)=(\gamma^.)$
Form definition Weingarten map is just a negative of differential of Gauss map, where Gauss map gives normal .
Please Help me to understand above proposition
Any Help will be appreciated
Pressley has some garbage here. What you didn't tell us (perhaps because you don't remember it) is that Pressley must have defined $\langle u,v\rangle$ to be $\text{II}(u,v) = \mathcal W(u)\cdot v$. (This is confirmed by the second part of the proposition.) I don't own the book, and so I cannot check. However, there's a mistake since the equation should be $$-\dot{\mathbf N}\cdot\dot\gamma = \mathcal W(\dot\gamma)\cdot\dot\gamma = \langle \dot\gamma,\dot\gamma \rangle,$$ having the usual dot product in the first spot.
By the way, $\mathcal W(\dot\gamma) = \lambda\dot\gamma$ if and only if $\gamma$ is a line of curvature (i.e., $\dot\gamma$ is an eigenvector of $\mathcal W$).