At this point on a Prof. Norman J Wildberger's presentation on Gauss's curvature and the the Theorema Egregium the curvature of a manifold $S$ at a point $p$ is written down as the determinant of the derivative map of Gauss-Rodrigues ($N$) from $S$ to the 2-sphere of radius $1,$ $S^2,$ at that point:
If $\vec r$ denotes the function from the parametric space to the surface $S,$ and $\mathrm dN$ the derivative of the Gauss map taking vectors from $T_pS$ to $T_{N(p)}S^2,$
the ratio of the determinant of the function composition $\mathrm dN\circ \vec r$ to the determinant of the function $\vec r$ is the Gauss curvature:
$$K(p) = \det \mathrm dN(p)=\frac{\mathrm dA(S^2)}{\mathrm dA(S)}=\frac{\det\left(\mathrm dN\circ \vec r\right)}{\det \vec r}=\frac{eg-f^2}{EG-F^2}$$
Or the Jacobian of the transformation between the two patches in the following diagram:
This is concordant with the first alternative formula:
$$K =\frac{\det \mathrm {II}}{\det \mathrm I}$$
of the determinant of the second fundamental form (i.e. the component along the normal vector of the second partial derivative of $\vec r$ with respect to the basis vectors in the tangent plane) to the first fundamental forms (i.e. the metric tensor).
Above the coefficients of the first fundamental form are $E=\langle r_u,r_u\rangle,$ $F=\langle r_u,r_v\rangle,$ $G=\langle r_v,r_v\rangle.$
The first fundamental form measures the distortion in lengths of the curves in the domain $(u,v),$ i.e. $L(\beta(t))=\int_0^t \vert \beta'(t)\vert dt=\int_0^t\left((u'(t)^2 + v'(t)^2\right)^{1/2}dt$ as compared to the charted curve on $S,$ i.e. $L(\alpha(t))=\int_0^t \vert\alpha'(t)\vert dt=\int_0^t \left(
u'(t)^2 E + 2 u'(t)v'(t) F + v'(t)^2 G \right)^{1/2}dt.$
Likewise it measures the distortion in the measurement of areas on $S$ through the determinant of the first fundamental form expressed as a matrix – $\det \begin{vmatrix}E&F\\F&G\end{vmatrix}=EG-F^2,$ i.e. the area of a $\vec r(V)$ patch on $S,$ corresponding to the image of a patch in $(u,v),$ is $A=\int_V \vert r_u \times r_v \vert du dv = \int_V \left(EG – F^2 \right)^{1/2}du dv.$
See here.
He writes $e,$ for example, as the image to a tangent vector $r_u$ on $S$ at point $p,$ i.e. $\mathrm dN_p(r_u)$ dotted with $r_u:$
$\bbox[5px,border:2px solid red]{ e =\mathrm dN_p(r_u)\cdot r_u = N \cdot r_{uu} }$
My question is why this last equality is true, including both the LHS and the RHS.
The other entries in the matrix being
$ f =\mathrm dN_p(r_u)\cdot r_v =\mathrm dN_p(r_v)\cdot r_u = N \cdot r_{uv}$
and
$g = \mathrm dN_p(r_v)\cdot r_u = N \cdot r_{vv} .$
I don't understand the application of the Gauss map to a tangent vector $r_u$ since the idea seems to be to apply it the normal vector to $S$ at the point $p.$ So applying it to a vector in the tangent space doesn't seem to make sense; yet it results in the dot product of the normal vector $N$ and the second derivative of $r$ with respect to $u.$
My bet is that $\mathrm dN$ may indicate the pushforward of the tangent space where $r_u$ is located. This wouldn't alter $r_u,$ given that the tangent space at $S^2$ is the same as at $S.$
Note to self: The first and second fundamental forms, as well as the differential of the Gauss map are all operators on the tangent space, and can be represented as matrices with respect to the basis of the tangent space $\{r_u,r_v\}.$
The matrix of the first fundamental form is:
$$M=\begin{bmatrix}\vert r_u \vert^2 & r_u\cdot r_v\\ r_v\cdot ru & \vert r_v\vert^2
\end{bmatrix}=\begin{bmatrix} E & F\\F &G
\end{bmatrix}$$
The matrix of the second fundamental form is composed of the following entries:
- Second fundamental form applied to $r_u,$ which is defined as:
$$e=\mathrm{II}_p(r_u,r_u)=- dN(r_u)\cdot r_u=-\langle dN(r_u),r_u\rangle=-\langle N_u,r_u\rangle$$
and since
$$\langle N,r_u \rangle=0$$
we can differentiate both sides with respect to $u:$
$$\begin{align}
\frac{d}{du}\langle N,r_u\rangle&=0\\
\langle N_u,r_u\rangle + \langle N,r_{uu}\rangle&=0
\end{align}$$
and hence,
$$e=\langle N,r_{uu}\rangle$$
Similarly,
$$f=\mathrm{II}_p(r_u,r_v)=-\langle dN(r_u),r_v\rangle=-\langle N_u,r_v\rangle=\langle N,r_{uv}\rangle$$
because $\langle N,r_v\rangle=0$ and
$$\begin{align}
\frac{d}{du}\langle N,r_v\rangle&=0\\
\langle N_u,r_v\rangle + \langle N,r_{uv}\rangle&=0
\end{align}$$
and $g=\mathrm{II}_p(r_v,r_v)=- dN(r_v)\cdot r_v=-\langle dN(r_v),r_v\rangle=-\langle N_v,r_v\rangle=\langle N,r_{vv}\rangle$
Therefore,
$$\Sigma=\begin{bmatrix}e&f\\f&g\end{bmatrix}=\begin{bmatrix}\langle N,r_{uu}\rangle&\langle N,r_{uv}\rangle\\\langle N,r_{uv}\rangle&\langle N,r_{vv}\rangle\end{bmatrix}$$
Best Answer
The main point here is that for a point $p\in S$, the tangent space $T_pS$ is the orthocomplement of the unit vector $N(p)$. But this coincides with the tangent space of the unit sphere in the point $N(p)$ as a subspace of $\mathbb R^3$. Viewing the Gauss map as a smooth map $S\to S^2$, you get a derivative $dN(p):T_pS\to T_{N(p)}S^2$, which you thus can also view as a linear map from $T_pS$ to itself. (Actually, this is already needed in order to have a determinant which is well-defined i.e. independent of the choice of basis.)
Edit (in view of your comment): The second fundamental form is then just given by $II(p)(v,w)=I(p)(dN(p)v,w)$ and if $r_u$ is the first vector in a basis for $T_pS$, you get $e=II(p)(r_u,r_u)=dN(p)(r_u)\cdot r_u$.
Second Edit (in view of your second comment): Sorry, I overlooked the last equation. You have to use the equation $N\cdot r_u=0$ here, which holds since $r_u$ is tangent to $S$. Taking the directional derivative of this in direction $r_u$, you get $0=dN(p)(r_u)\cdot r_u+N\cdot dr_u(r_u)$, so one actually should have $-N\cdot r_{uu}$ there. (I think that the sign is wrong in the talk, but this does not change the determinant.)