Riemann tensor in 3d Cartesian coordinates

[Orodruin]

With one chart that is an open set, you could cover everything on the cylinder except for one line.

As already stated several times in this thread, you can cover the entire cylinder with a single chart. Nothing requires the open set to be simply connected.

 

[pervect]

As already stated several times in this thread, you can cover the entire cylinder with a single chart. Nothing requires the open set to be simply connected.

I went back and reviewed my physics textbook (Wald) on that at point, and couldn't find anything definitive.

I suppose I'd need a textbook reference and/or (probably and) a lot more thinking to consider the case about a single chart overlapping itself, having never considered anything so strange before.

If I'm understanding you correctly, you're saying that we can apply the compatibility conditions that are required if any two charts O_a and O_b overlaping even if a=b, so that a chart is allowed to overlap itself as log as it satisfies the same compatibility conditions that we'd have for two different charts? At the moment, the idea is making my brain melt. I will say I am a lot more comfortable with having two different charts overlap than having a chart overlap itself.

 

[Ibix]

I went back and reviewed my physics textbook (Wald) on that at point, and couldn't find anything definitive.

Carroll's lecture notes imply it's possible to cover a cylinder with a single chart (last paragraph on p39, which is the 9th page of chapter 2: https://preposterousuniverse.com/wp-content/uploads/grnotes-two.pdf). He doesn't offer a solution, though (I think Orodruin posted the solution I'd seen before higher up this thread).

 

[PeterDonis]

(I think Orodruin posted the solution I'd seen before higher up this thread).

stevendaryl gave it in post #69.

 

[PeterDonis]

a chart is allowed to overlap itself

I don't think that's what Orodruin is saying. He's just saying that the open subset of $\mathbb{R}^N$ that is used in a coordinate chart does not need to be simply connected. For example, in the single-chart solution for the cylinder that stevendaryl posted in post #69, the open subset of $\mathbb{R}^2$ used in the chart is the plane minus a single point (i.e., all 2-tuples $(x, y)$ except $(0, 0)$), which is an open subset but is not simply connected. But the chart is still one-to-one; it doesn't overlap itself anywhere. (Note that you have to use the Cartesian coordinates $(x, y)$ on $\mathbb{R}^2$ for this to be true; it won't work if you use polar coordinates on $\mathbb{R}^2$, as Orodruin noted some posts ago.)

 

[Orodruin]

I would say something but Peter covered most of it in his post so let me just offer an explicit embedding of the cylinder in $\mathbb R^3$ using coordinates $(s,t)\neq (0,0)$:
$$
x = \frac{s}{r}, \quad y = \frac tr, \quad z = \ln(r)
$$
with $r = \sqrt{s^2+t^2}$. I think we can all agree that this is a continuous map from an open subset of $\mathbb R^2$ to the submanifold of $\mathbb R^3$ that we would typically refer to as a cylinder of radius one.

Just to add: The point is that the cylinder is homeomorphic to an open set in the plane while the circle is not homeomorphic to an open set in one dimension.

 

[stevendaryl]

I suppose I'd need a textbook reference and/or (probably and) a lot more thinking to consider the case about a single chart overlapping itself, having never considered anything so strange before.

If you look back over the posts, you will see that the one-chart coverage of the cylinder is just the Euclidean plane minus one point. There is no weirdness of a chart overlapping itself.

You map the point on the cylinder $z, \theta$ to the point $x = e^{z} cos(\theta), y = e^{z} sin(\theta)$.