Gaussian curvature as a smooth function

Question

Why is it that for a regular surface S in $\bf{R}^3$ the Gaussian curvature is a smooth function over S? Does this also hold true for $\bf{R}^n$?

Intuitively it makes sense, but I am having trouble formally proving it.

Answer 1

Here's an answer for a regular surface:

Say that we have a patch $\textbf{x}$ with coordinates $u$ and $v$. I'm going to give a formula for the Gaussian curvature $K$, then we'll see why regularity is important. First, we obtain the components of the first fundamental form as \begin{align*}E&=\textbf{x}_u\cdot\textbf{x}_u\\F&=\textbf{x}_u\cdot \textbf{x}_v\\ G&=\textbf{x}_v\cdot \textbf{x}_v.\end{align*} The unit normal is given by $$U=\frac{\textbf{x}_u\times \textbf{x}_v}{|\textbf{x}_u\times \textbf{x}_v|}=\frac{\textbf{x}_u\times \textbf{x}_v}{\sqrt{EG-F^2}}.$$ Since our surface is regular, we know that $\sqrt{EG-F^2}=|\textbf{x}_u\times \textbf{x}_v|>0$ which makes $U$ smooth. Next, we define the quantities \begin{align*} L&=\textbf{x}_{uu}\cdot U\\ M&=\textbf{x}_{uv}\cdot U\\ N&=\textbf{x}_{vv}\cdot U. \end{align*} Finally, we can just use the formula $$K=\frac{LN-M^2}{EG-F^2}.$$ Once again, this is a smooth function since each term in the expression is smooth and the denominator is never zero.

You can also see it directly from $K$ being the determinant of the shape operator (which is where these formulas really come from).

Reference for formulas: Barrett O'Neil, Elementary Differential Geometry