Gauge Curvature by example of the curvature of the graphs of $y=f(x), z=F(x,y)$ and $w=H(x,y,z)$.

Question

I am trying to understand Guage Curvature - or perhaps, more precisely, the Gaussian curvature of a Riemannian manifold in gauge coordinates.

The best way I can think is by example: finding the curvatures of a graph of a curve on a 2D graph: $y=f(x)$; the graph of a surface on a 3D graph: $z=F(x,y)$; and the graph of a volume on a 4D graph (or the analytical version because obviously we cannot graph it): $w=H(x,y,z)$. I tried following the examples on Wikipedia in the following, but the explanations are sloppy and the examples hard to follow (or without insulting the contributors to the page, it is written in a way that only people who already know the topic could infer an answer from what is written):

https://en.wikipedia.org/wiki/Curvature

https://en.wikipedia.org/wiki/Gaussian_curvature

To get a clear answer, I would like to:

1. Understand (or confirm) a good general definition of curvature in gauge coordinates. So far I understand it as follows: $$ \kappa = \frac{\det(\textbf{II})}{\det(\textbf{I})} $$ where $\textbf{I}$ and $\textbf{II}$ are the first and second fundamental forms.
   I understand the first fundamental form as: $\textbf{I}=(\nabla L)^T\cdot\nabla L$; where $L$ is the explicitly defined manifold - e.g. $L(u)=[f(u),u]^T$ for the graph of $y=f(x)$. However, I'm still struggling with the second fundamental form.

2. Apply the correct definition to the examples above - i.e., how do I end up with: $$ \kappa_{1} = \frac{f_{xx}}{{(1+{f_x}^2)}^{3/2}} \\ \kappa_{2} = \frac{F_{xx}F_{yy}-{F_{xy}}^2}{{(1+{F_x}^2+{F_y}^2)}^2} $$ where $\kappa_{1}$ and $\kappa_{2}$ are the curvatures for the respective curve and surface. And with this understanding, I would want to then be able to figure out $\kappa_{3}$ for the volume.

Thanks in advance!

Answer 1

Thanks to Modern Geometry by Dubrovin et el I managed to figure this out.

The second fundamental form would be given by $\textbf{II}=(\nabla\otimes\nabla L)^T\cdot(\nabla L^\perp/ ||{\nabla L^\perp}||)$.

What this all means computationally for the explicitly defined manifold $L(x_1...x_n)=[f(x_1...x_n),x_1...x_n]^T$, or the n+1 dimensional graph, is as follows: $$ L=\begin{pmatrix}f(x_1...x_n) \\ x_1 \\ \vdots \\ x_n \end{pmatrix} \\ \nabla L = \begin{pmatrix}f_{x_1} & \cdots & f_{x_n} \\ 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{pmatrix} \\ \textbf{I}=\begin{pmatrix}1+{f_{x_1}}^2 & \cdots & f_{x_1}f_{x_n} \\ \vdots & \ddots & \vdots \\ f_{x_n}f_{x_1} & \cdots & 1+{f_{x_n}}^2 \end{pmatrix} \\ \det\textbf{I}=1+\sum_{i=1}^{n}{f_{x_i}}^2 $$ Then the second fundamental: $$ \nabla\otimes\nabla L := \frac{\partial^2L_i}{\partial x_j\partial x_k} = \left\{ \begin{array}{lcr} \textbf{Hessian}(f) & \mbox{if $i = 1$};\\ \textbf{0} & \mbox{if $i > 1$}.\end{array} \right. \\ \nabla L^\perp = \begin{pmatrix}1 \\ -f_{x_1} \\ \vdots \\ -f_{x_n} \end{pmatrix} \textrm{ for right-handed coordinate}\\ \textbf{II}=\frac{\textbf{Hessian}(f)}{\sqrt{1+\sum_{i=1}^{n}{f_{x_i}}^2}} \\ \det\textbf{II} = \frac{\det\textbf{Hessian}(f)}{\left( 1+\sum_{i=1}^{n}{f_{x_i}}^2 \right)^{\frac{n}{2}}} $$ And finally, the curvature: $$ \kappa = \frac{\det\textbf{II}}{\det\textbf{I}} = \frac{\det\textbf{Hessian}(f)}{\left( 1+\sum_{i=1}^{n}{f_{x_i}}^2 \right)^{\frac{n+2}{2}}} $$