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Given a parameterization of hyperbolic paraboloid, $X(u,v)=(u,v,v^2-u^2)$ $$X_u = (1,0,-2u) \space \space X_v=(0,1,2v)$$ $$N(u,v)=\frac{(2u,-2v,1)}{\sqrt{4u^2+4v^2+1}}$$ is the Gauss map (unit normal).

The differential of Gauss map at point $p$ ($dN_p$, also known as the negative Weingarten map) satisfy the following $$dN_p(X_u)=N_u=\frac{(8v^2+2, 8uv,-4u)}{(4u^2+4v^2+1)^{3/2}}$$

I am able to get the $N_u$ but I do not know what is the explicit expression for $dN_p$ ? and how it is calculated from $dN_p ( (1,0,-2u) )$ to get $N_u$ ?

I learned that the Weingarten map ($-dN_p$ in matrix form) can be calculated using first fundamental and second fundamental form : $$-dN_p = \left(\begin{matrix} E & F \\ F & G \end{matrix} \right)^{-1} \left(\begin{matrix} L & M \\ M & N \end{matrix} \right) $$ where $N$ is coefficient of second fundamental form (not to confuse with the Gauss map above). I obtained for Weingarten map as: $$-dN_p=\frac{1}{(4u^2+4v^2+1)^{3/2}} \left( \begin{matrix} -2(1+4v^2) & 8uv \\ -8uv & 2(1+4u^2) \end{matrix} \right)$$ This is in matrix form which i cannot see how it acts on $X_u=(1,0,-2u)$ to give $-dN_p(X_u)=N_u$ since it is a $2 \times 2$ matrix, while $X_u$ is $1 \times 3$ matrix.

So how to get the explicit local expression of $-dN_p$ NOT in matrix form ? and the way how it acts on $X_u$ ?

Is it differential of Gauss map $dN_p = \frac{\partial N}{\partial u} du + \frac{\partial N}{\partial v}dv$ ??

Please help me. Thank you so much.

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The matrix form tells you that $dN_p(X_u) = aX_u + bX_v$, where $a$ and $b$ are the components of the first column vector. You can certainly differentiate $N$ with respect to $u$ and get this same vector in $\Bbb R^3$, but for purposes of differential geometry we want to understand $dN_p$ as a linear map from the tangent plane $T_pM$ to itself.

P.S. You might find my (free) differential geometry text, linked in my profile, helpful.

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  • $\begingroup$ thank you so much for your explanation and advices. Sorry, I still didn't get it. I know $dN_p(X_u) = aX_u + bX_v$ is the expansion based on the $\{X_u, X_v\}$ as the basis of the tangent plane . How is the actual expression for $dN_p$ alone without acting on the $X_u$ ? and how it acts explicitly on $X_u$ taking the above case as an example e.g. $X_u = (1,0,-2u) ^{T} $? Thank you in advance. $\endgroup$
    – JR makoto
    Commented Jan 5, 2023 at 5:40
  • $\begingroup$ I’m not sure why you want this. But if you view $N$ as a function from $\Bbb R^2$ ($uv$-space) to $\Bbb R^3$, its derivative at $(u,v)$ is a $3\times 2$ matrix whose columns are $\partial N/\partial u$ and $\partial N/\partial v$. You get the value on $X_u$ by multiplying by $(1,0)^\top$. $\endgroup$
    – Ted Shifrin
    Commented Jan 5, 2023 at 5:47
  • $\begingroup$ thank you so much for comments. I get the $dN$ which is the Jacobian of $N$, and multiplying $(1,0)^T$ to get $N_u$. But where does the third $-2u$ go, as $X_u = (1,0,-2u)^T$ ? Can I understand it this way: $$dN= \frac{\partial N}{\partial u } du + \frac{\partial N}{\partial v} dv$$ $$X_u = (1,0,-2u)= 1 \frac{\partial}{\partial u} + 0 \frac{\partial}{\partial v}$$ $$dN(X_u) = \left(\frac{\partial N}{\partial u } du + \frac{\partial N}{\partial v} dv \right) \left( 1 \frac{\partial}{\partial u} + 0 \frac{\partial}{\partial v} \right) =\frac{\partial N}{\partial u} = N_u $$ ? $\endgroup$
    – JR makoto
    Commented Jan 5, 2023 at 8:30
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Ted Shifrin
    Commented Jan 5, 2023 at 17:59
  • $\begingroup$ Once you are satisfied with the answer, you should accept it so that the question does not stay on the unanswered list. $\endgroup$
    – Ted Shifrin
    Commented Jan 10, 2023 at 6:04

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