Exercise
I'm attempting to find the Gaussian curvature of the catenoid $M$ parametrized by $$ f(u,v)=(a\cosh v\cos u,a\cosh v\sin u,a v). $$ I've run through the typical computations of $E,F,G,e,f,g$ to get $$ K(u,v)=\frac{eg-f^2}{EG-F^2}=-a^{-2}\text{sech}^4(v), $$ which I believe to be right.
Problem
My problem is that before going through the computations described above, I attempted to find $K(u,v)$ by computing the actual differential of the Gauss map (so I could then take its determinant) as follows: I computed that the Gauss map $N$ is $$ N(u,v)=\frac{\big(\cos(u),\sin(u),-\sinh(v)\big)}{\cosh(v)}. $$ This is a map from a subset $U\subset\mathbb{R}^2$ into the two-sphere $S^2$, so if $v\neq 0$, then $\sinh(v)\neq 0$ and the projection given by $\pi\,:\,(x,y,z)\mapsto (x,y)$ is a local coordinate system for $S^2$. Then $$ d\big((\pi\circ N)(u,v)\big)=\begin{pmatrix}-\tfrac{\sin u}{\cosh v} & -\frac{\cos u\sinh v}{\cosh^2 v}\\ \frac{\cos u}{\cosh v} & -\frac{\sin u\sinh v}{\cosh^2 v}\end{pmatrix} $$ and hence $$ \det\Big(d\big((\pi\circ N)(u,v)\big)\Big)=\frac{\sin^2 u\sinh v}{\cosh^3 v}+\frac{\cos u ^2\sinh v}{\cosh^3 v}=\frac{\sinh v}{\cosh^3v} $$ should be the Gaussian curvature of the catenoid (since this should work for $v=0$ as well via continuity), but evidently it isn't (it doesn't even involve $a$).
Questions
Apparently I have a misunderstanding of how differentials work, which I would like cleared up if possible. So
- Why doesn't this method work?
- Is there any way to make this method work?
- What is it exactly that I am misunderstanding in terms of differentials here? I thought that the differential of a map $F:M\to N$ between manifolds could be computed via taking the total derivative of the coordinate expression of $F$. Is this not the case?
Any help would be greatly appreciated. Hopefully a question of this sort hasn't been asked before, but I'm not sure how I'd even check, since its more of a "proof verification" type problem. Thanks in advance.