Given $f(x,y) = x^2 – y^2$ and the corresponding surface
\begin{align}
\Sigma = \left\{ (x,y,z) \in \mathbb{R}^{3}: x \in \mathbb{R}, y \in \mathbb{R}, z = x^2 – y^2\right\}
\end{align}
prove that the vector $u = (1,0,0)$ belongs to the vector space $T_O \Sigma$. Then compute the Weingarten operator $L_O: T_O \Sigma \to T_0 \Sigma$ along with $L_O (u)$.
Attempted solution
First off, we can see that given the fact $z=x^2 -y^2$ we can write the original surface in terms of $u,r$ as
\begin{align}
\mathcal{X}(u,r) = \left( u,r,u^2-r^2\right).
\end{align}
It's easy to compute that
\begin{align}
\mathcal{X}_u &= (1,0,2u)\\
\mathcal{X}_r &= (0,1,-2r)\\
\mathcal{X}_u \times \mathcal{X}_{r} &= (-2u, 2r,1).
\end{align}
From that we have
\begin{align}
N(u,r) = \dfrac{\mathcal{X}_u \times \mathcal{X}_{r}}{||\mathcal{X}_u \times \mathcal{X}_{r}||} = \dfrac{(-2u, 2r,1)}{\sqrt{4u^2 + 4r^2 +1}}.
\end{align}
Differentiating $N$ gives (skipping painful calculations)
\begin{align}
&N_u = (-2,0,0) \cdot (4u^2 + 4r^2 + 1)^{-1/2} – (-2u, 2r, 1) \cdot 4u \cdot (4u^2 + 4r^2 +1)^{-3/2}\\
&N_{r} = (0,2,0) \cdot (4u^2 + 4r^2 + + 1)^{-1/2} – (-2u,2r,1) \cdot 4r \cdot (4u^2 + 4r^2 + 1)^{-3/2}.
\end{align}
Now to prove what is required, I noticed that $(1,0,0)$ is equivalent with $(1,0)$ using our projection above. So
\begin{align}
T_p \Sigma &= \mathrm{span} \left\{ \mathcal{X}_{u}(1,0), \mathcal{X}_{r}(1,0)\right\}\\
&= \mathrm{span} \left\{ (1,0,2), (0,1,0)\right\}
\end{align}
which means (???) that $(0,0,1) \in T_O \Sigma$.
As for the Weingarten operator, we know that
\begin{align}
L_p (w_1) &= -N_u (1,0)\\
&=\cdots\\
&=(0,0,1/2)
\end{align}
and we do the same for $L_p (w_2) = -N_r(1,0)$.
Is my method correct? Is there anything that I missed or forgot entirely? Thank you.
Best Answer
The question is asking to find $T_O\Sigma$ where $O=(0,0,0)=(u,r,u^2-r^2)$. Thus you should plug in $u=r=0$ in your expression for $T_p\Sigma$ (and using $O$ instead of $p$ to indicate the point) i.e. $$T_O \Sigma = \mathrm{span} \left\{ \mathcal{X}_{u}(0,0), \mathcal{X}_{r}(0,0)\right\}=\mathrm{span} \left\{ (1,0,0), (0,1,0)\right\}$$ Then you see that the vector $u=(1,0,0)$ is in $T_O\Sigma$. (Another notation issue, $u$ is already used denote the vector $(1,0,0)$ in the question, you shouldn't use it to denote the coordinates of $x,y$. I think using $x,y$ instead of $u,r$ is better.)
For the Weingarten operator, by definition and the fact that $u=r=0$, we have $$L_O(\mathcal X_u(0,0))=L_O((1,0,0))=-N_u(0,0)=(2,0,0)$$ and $$L_O(\mathcal X_r(0,0))=L_O((0,1,0))=-N_r(0,0)=(0,-2,0).$$ Therefore, $L_O=\left(\begin{matrix}2&0\\0&-2\end{matrix}\right)$. Finally, $L_O(u)=2u=(2,0,0)$.