Finding Gauss curvature of surface

Question

Consider the surface $S=F(\mathbb{R}^2)$ where $F:\mathbb{R}^2 \to \mathbb{R}^3$ is defined by $$(r, \varphi) \mapsto ( r \cos \varphi, r \sin \varphi, \varphi).$$

I would like to find the Gauss curvature of $S$.

We defined the Gauss curvature of $S$ at the point $p$ as $\det(W_p)$ where $W_p: T_p S \to T_p S$ is the Weingarten-map defined by $W_p: X\mapsto - d_p N(X)$. Here $N: S \to \mathbb{R}^3$ assigns to each $p\in S$ a vector $N(p)$ that is orthogonal on the tangential space $T_p S$.

Working with this definition seems terribly complicated. How do I find the map $N$ ?

Is there a nicer way of doing this?

Answer 1

Work locally. Call ${\bf x}(r, \varphi) = (r \cos \varphi, r \sin \varphi, \varphi)$, and write ${\bf p} = {\bf x}(r,\varphi)$. One way is computing: $${\bf N}(r, \varphi) = \frac{\frac{\partial {\bf x}}{\partial r}\times \frac{\partial {\bf x}}{\partial \varphi}(r,\varphi)}{\left\|\frac{\partial {\bf x}}{\partial r}\times \frac{\partial {\bf x}}{\partial \varphi}(r,\varphi)\right\|}.$$

From now on I'll drop the pair $(r,\varphi)$ from things and I'll abreviate partial derivatives. Let: $$E = \langle {\bf x}_{r},{\bf x}_r\rangle,\quad F = \langle {\bf x}_r,{\bf x}_\varphi\rangle, \quad G = \langle {\bf x}_\varphi,{\bf x}_{\varphi}\rangle.$$

Also write: $$e = \langle {\bf N},{\bf x}_{rr}\rangle, \quad f = \langle {\bf N},{\bf x}_{r\varphi}\rangle, \quad g = \langle {\bf N},{\bf x}_{\varphi\varphi}\rangle.$$

We have that the following holds: $$e = \langle -{\rm d}_{\bf p}{\bf N}({\bf x}_r),{\bf x}_r\rangle, \quad f = \langle -{\rm d}_{\bf p}{\bf N}({\bf x}_r),{\bf x}_\varphi\rangle = \langle -{\rm d}_{\bf p}{\bf N}({\bf x}_\varphi),{\bf x}_r\rangle,\quad g = \langle -{\rm d}_{\bf p}{\bf N}({\bf x}_\varphi),{\bf x}_\varphi\rangle.$$

Write the matrix of $-{\rm d}_{\bf p}{\bf N}$ in the basis $\{{\bf x}_r,{\bf x}_\varphi \}$: $$\begin{cases} -{\rm d}_{\bf p}{\bf N}({\bf x}_r) &= a_{11}{\bf x}_r + a_{12}{\bf x}_\varphi \\ -{\rm d}_{\bf p}{\bf N}({\bf x}_\varphi) &= a_{21}{\bf x}_r+a_{22}{\bf x}_\varphi\end{cases} \implies [-{\rm d}_{\bf p}{\bf N}] = (a_{ij})_{1\leq i,j\leq 2}$$

Applying $\langle \cdot, {\bf x}_r\rangle$ and $\langle \cdot, {\bf x}_\varphi \rangle$ to the above, we get: $$\begin{cases} e &= Ea_{11}+Fa_{12} \\ f &= Ea_{21}+Fa_{22}\end{cases}\quad \text{and}\quad \begin{cases} f &= Fa_{11}+Ga_{12} \\ g &= Fa_{12}+Ga_{22}\end{cases},$$ which we can read as: $$\begin{pmatrix} e & f \\ f & g\end{pmatrix} = \begin{pmatrix} E & F \\ F & G\end{pmatrix}\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}.$$ Hence: $$\begin{align}\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix} &= \begin{pmatrix} E & F \\ F & G\end{pmatrix}^{-1}\begin{pmatrix} e & f \\ f & g\end{pmatrix}\\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix} &= \frac{1}{EG-F^2} \begin{pmatrix}G & -F \\ -F & E\end{pmatrix}\begin{pmatrix} e & f \\ f & g\end{pmatrix}\end{align}.$$

Since $K = \det(a_{ij})$, we obtain: $$K = \frac{eg-f^2}{EG-F^2}.$$ Bonus: since you've probably defined $H = \frac{1}{2}{\rm trace}(-{\rm d}_{\bf p}{\bf N})$, we get: $$H = \frac{1}{2}\frac{eG+Eg-2fF}{EG-F^2}.$$

Now you just have to apply the formulas. If you use the first expressions for $e,f,g$ that I told you, you can avoid computing $-{\rm d}_{\bf p}{\bf N}$ directly. I took the time to do all these computations because I didn't want just to throw them at your face, but that's pretty much it: once you know this, it becomes mechanical. You can see this in Do Carmo's book, too.

Answer 2

This well known theorem of Gauss is well explained in text books of differential geometry, I shall not repeat it here. Starting as you did, follow how subsequently $K$ is computed in terms of coefficients of second ( numerator L ,N and M ) and first (denominator E,F and G) fundamental forms of surface theory. Please note that $L N -M^2$ depends on first fundamental form only.

$$K = \frac{LN -M^2}{EG-F^2}.$$

You can apply this relation to find $K$ of the given helicoid.

Answer 3

$\frac{\partial}{\partial r}=(\cos\phi,\sin\phi,0)$, $\frac{\partial}{\partial \phi}=(-r\sin\phi,r\cos\phi,1)$, so $E=1, F=0, G=r^2+1$, plug into $$K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right).$$ where $u=r, v=\phi$ we got $$K = -\frac{1}{2\sqrt{1+r^2}}\left(\frac{d}{d r}\frac{2r}{\sqrt{1+r^2}}\right)=-\frac{1}{(1+r^2)^2}.$$