# How to find the differential of the Gauss map explicitly?

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 $$dN_p = \frac{\partial N}{\partial u} du + \frac{\partial N}{\partial v}dv$$ ??

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.
• 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. Commented Jan 5, 2023 at 5:40
• 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 $(1,0)^\top$. Commented Jan 5, 2023 at 5:47
• 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: $$dN= \frac{\partial N}{\partial u } du + \frac{\partial N}{\partial v} dv$$ $$X_u = (1,0,-2u)= 1 \frac{\partial}{\partial u} + 0 \frac{\partial}{\partial v}$$ $$dN(X_u) = \left(\frac{\partial N}{\partial u } du + \frac{\partial N}{\partial v} dv \right) \left( 1 \frac{\partial}{\partial u} + 0 \frac{\partial}{\partial v} \right) =\frac{\partial N}{\partial u} = N_u$$ ? Commented Jan 5, 2023 at 8:30