For regular surface $S\subset \mathbb{R}^3$,we have differential of normal vector $dN_p:T_p(S) \to T_p(S)$.
What's the size of linear map $dN_p$?I was a bit confuse here.
Since we know dimension of $T_p(S)$ is 2,so $dN_p$ should be a $2\times 2$ matrix,but in another view,each point in $T_p(S)$ is tangent of curve on $S$,so it's $v\in \mathbb{R}^3$ so $dN_p$ is a $3\times 3 $matrix.
Can someone explain a bit more clear?
Best Answer
$T_pS$ is a $2$-dimensional vector space, so any matrix representation of $dN_p$ will be $2\times 2$. In order to obtain one, you need a basis for the tangent plane. Typically, you have a parametrization of $S$ by $\mathbf x(u,v)$, and a basis for the tangent plane will be $\left\{\dfrac{\partial\mathbf x}{\partial u}(u_0,v_0), \dfrac{\partial\mathbf x}{\partial v}(u_0,v_0)\right\}$ when $\mathbf x(u_0,v_0)=p$.