I'm trying to find the Weingarten map of the surface parametrized by $\sigma(u, v) = (u, v, u^2 + v^2)$. Now, one can easily find that the Gauss map (which is just the surface normal $\bf N$) is given by
$$
{\bf N}(\sigma(u, v)) = -\frac{(u,v,1/2)}{\sqrt{u^2+v^2+1/4}}
$$
The Weingarten map is the negative derivative of this, $\mathcal{W} = -D_{\vec p}{\bf N}$. To calculate the derivative of a smooth function $f$ from a surface $S$ parametrized by $\sigma$, onto a surface $\tilde{S}$ parametrized by $\tilde{\sigma}$, one first needs to find functions $\alpha$ and $\beta$ such that
$$
f(\sigma(u, v)) = \tilde{\sigma}(\alpha(u, v), \beta(u, v))
$$
so that, for a curve on $S$, $\gamma(t) = \sigma(u(t), v(t))$, the tangent vector $(f\circ\gamma)'$ is given by
$$
\tilde{\sigma}_\alpha(\alpha_uu'+\alpha_vv') + \tilde{\sigma}_\beta(\beta_uu'+\beta_vv')
$$
In this case, since the Gauss map maps from the tangent space to the surface, $T_{\vec p}S$, onto itself, we need to find $\alpha$ and $\beta$ such that
$$
{\bf N}(\sigma(u,v)) = \sigma(\alpha(u,v),\beta(u,v))
$$
but
$$
\sigma(\alpha(u,v),\beta(u,v)) = (\alpha(u,v),\beta(u,v),\alpha^2(u,v)+\beta^2(u,v))
$$
and so clearly, we must have
$$
\alpha(u,v) = -\frac{u}{\sqrt{u^2+v^2+1/4}} \\
\beta(u,v) = -\frac{v}{\sqrt{u^2+v^2+1/4}}
$$
but then $\alpha^2(u,v) + \beta^2(u,v)$ doesn't equal the last component, so there aren't functions $\alpha$ and $\beta$ that satifsy the above equality.
What is happening? I can't figure this out. From this, it seems like the Weingarten map shouldn't exist, but clearly it does. Any help here would be greatly appreciated!
Best Answer
To know $-dN_{\sigma(u,v)}$, it suffices to know its values in a basis for $T_{\sigma(u,v)}\sigma(\Bbb R^2)$. Compute $$-dN_{\sigma(u,v)}\left(\frac {\partial \sigma}{\partial u}(u,v)\right) = -\frac{\partial (N\circ \sigma)}{\partial u}(u,v) $$using the expression for $(N\circ \sigma)(u,v)=N(\sigma(u,v))$ you have, and similarly for the other derivative.