3
$\begingroup$

Suppose we have the metric $ ds^2 = dr^2 + \alpha^2 d\phi^2$, where $\alpha$ is a constant, $0 \leq r \leq \infty$, $ 0 \leq \phi \leq 2 \pi$ and we identify points $\phi = 0$ with points $\phi = 2\pi$.

Now since we have a constant metric the Christoffel symbols are zero and thus so to is the curvature tensor. So this should be flat space. When I try to find the length along radial paths from $0$ to $R$ I find the length to be $R$, but when I find the length of fixed radius paths from $0$ to $2\pi$ I find them to be of length $2\pi \alpha$. So regardless of the position in the $r$ direction we have fixed circumference, thus it seems like this space can be embedded in 3D cylindrical coordinates as a cylinder.

Question 1: Is the above correct? If it is, why are cylinders flat but not spheres or saddle points? Intuitively is it because flattening a cylinder only introduces folds where as a sphere would have complex deformations?

Question 2: If you draw a sphere in spherical coordinates where $r$, $\theta$ and $\phi$ are all draw at right angles a sphere would look like a rectangle with relevant sides of $\theta$ and $\phi$ identified. Since a rectangle looks flat to me, how can you be sure that you are embedding in such a way as to visualise the curvature? Is it that any coordinates where the sides are identified will reveal the curvature?

$\endgroup$
  • 1
    $\begingroup$ This is simple differential geometry: a cylinder has only one principal curvature non-zero, a sphere has both of them and so do saddle points, therefore Gauss curvature, which is the product of principal curvatures, must be zero only for the cylinder. $\endgroup$ – auxsvr Jun 15 '14 at 7:46
  • $\begingroup$ Partially regarding question two: There are two notions of curvature, extrinsic and intrinsic. The latter is described by the Riemann curvature, and attempting to visualize why a sphere is curved in that respect, and a cylinder is not is somewhat difficult, as our intuitive notion of curvature we were taught was extrinsic. $\endgroup$ – JamalS Jun 15 '14 at 8:05
2
$\begingroup$

I answered an imported version of this question on physics overflow. I decided to post that answer here as well. Due to formatting differences between the sites it will be a inconvenient to maintain up to date versions of both answers, therefore I will only update the original posting of the answer unless a major modification is necessary.


To answer the first question:

This is the line element for a cylinder though the variable names might make more sense when the line element is written as, $ds^2 = dz^2 + r^2 d\phi^2$, where $r$ is a constant. In this context $z$ is the elevation on the cylinder, $r$ is the radius of the cylinder, $\phi$ is the angle which wraps around the cylinder.

Cylinders are flat because they can be cut and unfolded, making them into a rectangular sheet, without stretching the material of the cylinder. More formally there exists an isometric transformation from the cylinder to the plane. The identification of $\phi=0$ with $\phi=2\pi$ is a topological property of the cylinder which doesn't necessarily say anything about its curvature.

To be clear when I say that a transformation is isometric I meant that it preserves the relative distances of points as measured locally within the manifold. Distances measured through the embedding space are irrelevant.

Now suppose we wanted to map the plane to the sphere. We might start by turning the plane into a cylinder since we know we can do isometrically. To turn the cylinder into a sphere we would have to identify all the points on the rim at the top of the cylinder. The problem with this last identification is that it sends points that were formerly a finite distance apart to the same location meaning that the metric cannot have been preserved.

The above example isn't really a proof that there is not isometric mapping from a plane to a sphere, but I think it at least gives some insight into what is going on.

To answer the second question:

It looks like you are thinking of Identifying $\theta=0$ with $\theta=\pi$ and $\phi=0$ with $\phi=2\pi$ which would not form a sphere. Instead this would form a torus.

To form a sphere from $0 \leq \theta \leq \pi$ and $0\leq \phi \leq 2\pi$:

  • For $ 0 < \theta < 2\pi $ identify $\phi=0$ with $\phi = 2\pi$.
  • For $\theta = 0 $ identify all values of $\phi$ with a single point.
  • For $\theta = \pi $ identify all values of $\phi$ with a single point.

An important thing to realize at this point is that we still don't have a proper sphere because we haven't defined a metric. As it stands our manifold could just be some deflated beach ball.

One line element which would be consistent with the boundary conditions is,

$$ds^2 = d\theta^2 + \theta(\pi-\theta) d\phi^2, $$

which is not the metric of a $2$-Sphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.