I am new to differential geometry and Riemannian geometry.
I have on two separate occasions (separated by 6 months) encountered exercises where I feel like I am not giving a complete answer.
Problem 1:
Show that the Gaussian curvature of the surface of a cylinder is zero.
Problem 2:
Use Cartesian coordinates to write out and solve the geodesic equations for a two-dimensional flat plane and show the solutions are straight lines.
In both cases, my argument went something like
- Define the metric.
- Show the Christoffel symbols are zero.
- Use this to show there is no curvature.
But I feel like this is just me avoiding the real stuff. In other words, I resort to calculus because I don't actually know what I'm doing. So I make a roundabout argument rather than diving right into the mathematics itself.
Is my argument sensible? Is there a more formal way to approach this type of problem? If you could do an example that would be fantastic.
Best Answer
A few words.. (too long to be a comment, I think)
Suppose we have a parametrization ${\bf x}(u,v) = (x^1(u,v),x^2(u,v),x^3(u,v))$ and a unit normal vector $\bf N$ all the way.
For problem $1$, if you call the coefficients of the first fundamental form $$E = \langle {\bf x}_u, {\bf x}_u \rangle, \quad F = \langle {\bf x}_u,{\bf x}_v\rangle, \quad G = \langle{\bf x}_v,{\bf x}_v\rangle$$ and the coefficients of the second fundamental form $$e = \langle{\bf N},{\bf x}_{uu}\rangle, \quad f = \langle {\bf N},{\bf x}_{uv}\rangle, \quad g = \langle {\bf N},{\bf x}_{vv}\rangle,$$ then we can prove that $$H = \frac{1}{2}\frac{Eg+eG-2Ff}{EG-F^2}, \qquad K = \frac{eg-f^2}{EG-F^2}.$$ You can see Do Carmo's book for proofs of this, for instance. If we have a cylinder parametrized by ${\bf x}(u,v) = (r\cos u, r\sin u, v)$, then a unit normal vector is ${\bf N}(u,v) = (\cos u, \sin u, 0)$. We compute straightly $$\begin{align*}{\bf x}_{uu}(u,v) &= (-r\cos u, -r\sin u, 0) \\ {\bf x}_{uv}(u,v) &= {\bf 0} \\ {\bf x}_{vv}(u,v) &= {\bf 0}\end{align*},$$ so that $f = g = 0$, and follows that $K = 0$.
If you look, for example, in O'Neill's Elementary Differential Geometry, he goes on and on using differential forms, but really, in the end you will be doing the exact same calculations. My point is, when I'm working in $\Bbb R^3$, I just suck it up and go for the calculations. Once you understand what really is going on behind it, you wont't be so bothered by doing this.
Sure, another way is noticing that one of the principal curvatures is zero (along the $z-$axis), and hence $K = 0$, but then you would have to prove indeed that this last thing I said is true.
For problem $2$, you can avoid covariant derivatives and Christoffel symbols, if you're lucky and know your way around ODE's. Just remember that a curve $\alpha$ in a surface $M$ is given by $\alpha(t) = {\bf x}(u(t),v(t))$, and $\alpha$ is a geodesic if its acceleration is normal to $M$, that is, if exists a function $\lambda$ such that $$\alpha''(t) = \lambda(t){\bf N}(u(t),v(t)).$$
You can take a look at my answer here, about finding the geodesics of the... cylinder.
Another way, more specific, is using Clairaut parametrizations. In details, ${\bf x}$ is Clairaut in $u$ if $E$ and $G$ don't depend on $v$ and $F = 0$. This way the (pre)-geodesics are given by $\alpha(u) = {\bf x}(u,v_0)$, for fixed $v_0$, and all the other ones are given by $\alpha(u) = {\bf x}(u,v(u))$, where $v(u)$ satisfies $$\frac{{\rm d}v}{{\rm d}u} = \frac{\pm c\sqrt{E}}{\sqrt{G}\sqrt{G-c^2}},$$and I again point you to O'Neill for this part. I don't know if you can always find such a parametrization. Maybe locally.
Certainly there are better explanations and hints out here, but I hope this does helps you.