Metric of Manifold with Curled up Dimensions

1. Aug 29, 2012

Markus Hanke

Would someone here be able to write down for me an example of a metric on a manifold with both macroscopic dimensions, and microscopic "curled up" dimensions with some radius R ? Number of dimensions and coordinates used don't matter.

Not going anywhere with this, I am just curious as to how such a metric could look like, mathematically.

2. Aug 29, 2012

Tinyboss

A metric is a locally-defined structure--that is, when defining one, we may consider regions as small as we like. So if one dimension is curled up into a circle of radius R, we could define our metric on balls of radius R/100000 for instance. At that scale, we can't even detect the curling.

The "curling-up" of dimensions is a global property of the space, and so (at least in the category of Riemannian manifolds) it is not reflected by the metric, which is local. There are geometric structures you can put on a manifold that, I think, do force a connection between the global and local structures. But I'm only starting to learn about that, so I'll let someone else jump in and say whether that's the right way to think about it.

3. Aug 29, 2012

Markus Hanke

I am not an expert in this ( which is why I asked the question in the first place ), but somehow that statement does not appear to make sense. My thinking is that the extra dimension(s) would be a direct product between some "normal" manifold and the extra one, in which case they must appear in the metric tensor in some form.

Last edited: Aug 29, 2012
4. Aug 29, 2012

HallsofIvy

The simplest example would be an infinitely long cylinder with a very small radius.
Take the axis of the cylinder to be the x-axis, the radius R, and $\theta$ the angle a line, pependicular to the axis, from the axis to a point on the cylinder makes with the z-axis. Then we can write $(x, y, z)= (x, R cos(\theta), R sin(\theta))$.

Then $dy= -Rsin(\theta)d\theta$ and $dz= Rcos(\theta)d\theta$ and then $ds^2= dx^2+ dy^2+ dz^2= dx^2+ R^2cos^2(\theta)d\theta+ R^2sin^2(\theta)d\theta= dx^2+ R^2d\theta^2$.
The metric tensor is
$$\begin{bmatrix}1 & 0 \\ 0 & R^2\end{bmatrix}$$
and the "curling up" will be for R very small.

More generally, given any metric tensor, you can "add" curled up dimensions by adding rows and columns with all 0s except that the values on the diagonal are very small numbers- almost 0.

Last edited by a moderator: Aug 29, 2012