I'm trying to compute the principle curvatures of the saddle $M$ defined by $z= y^2 -x^2$ at the point $p = (0,0,0)$, but I know my computations are wrong. Maybe you can help to see where I went wrong.
I have computed the tangent plane as $z = 0$. So first take a vector $(0,1,0)= e_1$, corresponding to the $x=0$ part (so that $z = y^2$). Now, define a map $\gamma(s) = (0,s,s^2)$. Then, $\gamma : (-\epsilon,\epsilon) \rightarrow M$ and this satisfies $\gamma(0) = (0,0,0) = p$ and also $\gamma'(0) = (0,1,0) = e_1$.
Now, the first curvature should be the eigenvalue of the Weingarten map $L$. That is by definition $$L(e_1) = -\frac{d}{ds}n(\gamma(s))|_{s=0}$$ But $$n(\gamma(s)) = \frac{\gamma(s)}{|\gamma(s)|} = \frac{(0,s,s^2)}{\sqrt{s^2+s^4}}$$
When I take the derivative and evaluate at $s= 0$, I get $(0,0,1)$, which is not a multiple of $e_1$.
I know by a different method that I should get $L(e_1) = 2e_1$. Can you see where I went wrong? Thanks.
Consider the curve $\gamma(s) = (0,s,s^2)$. The tangent space $T_{\gamma(s)}M$ is spanned by the vectors $(1,0,0)$ and $\dot \gamma(s) = (0,1,2s)$. Hence a unitnormal field $n(s)$ along $\gamma$ is given by $$n(s) = (4s^2 + 1)^{-1/2}(0,2s,-1)$$ (it has length $1$ and is orthogonal to $(1,0,0)$ and $\dot \gamma(s)$). Thus for $e_1 = (0,1,0)$ (maybe a bad notation) the Weingarten map is $$L(e_1) = -d/ds_{s = 0} n(s) = (0,-2,0) = -2e_1.$$ The $-$ is a matter of choice of the sign of $n$.