Coordinate charts and open sets

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Ibix
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I have a (probably trivial) question about coordinate charts. I've been studying Sean Carroll's lecture notes on General Relativity. I'm on my second re-read and I'm trying to make sure I understand the basics properly. I hope the terminology is correct - this is my first use.

Carroll cites the circle as an example of multiple charts being necessary. The obvious coordinate system using the angular co-ordinate has an image in R1 bounded by 0 and [itex]2\pi[/itex], and you need to use one of those to describe the complete circle. The image is therefore not open, so you need multiple charts. Fair enough.

He then mentions that the cylinder can be represented by a single co-ordinate chart. I've seen answers to this on this on this very forum, such as mapping cylindrical polars [itex](z,\theta)[/itex] on to an annulus in the Euclidean plane, using the same angular coordinate and something like [itex]z=\tan(r)[/itex], with the boundary being the circles defined by [itex]r=\pi/2, 3\pi/2[/itex]. Again, fair enough.

But this annular section of the plane has a boundary, and r ranges up to it. I eventually concluded that this is fine because you can only reach the boundary as [itex]z\rightarrow\pm\infty[/itex], so any finite z maps to a point that is within the annulus not on its boundary.

Is that last paragraph correct? As long as the image of the map has no boundary (like the trivial case of Cartesian coordinates on a Euclidean plane), or the boundary maps back to infinity in the original set, it's a valid chart? (Subject, of course, to the other conditions on the map that I haven't mentioned.)
 

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pasmith
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He then mentions that the cylinder can be represented by a single co-ordinate chart. I've seen answers to this on this on this very forum, such as mapping cylindrical polars [itex](z,\theta)[/itex] on to an annulus in the Euclidean plane, using the same angular coordinate and something like [itex]z=\tan(r)[/itex], with the boundary being the circles defined by [itex]r=\pi/2, 3\pi/2[/itex]. Again, fair enough.

But this annular section of the plane has a boundary, and r ranges up to it. I eventually concluded that this is fine because you can only reach the boundary as [itex]z\rightarrow\pm\infty[/itex], so any finite z maps to a point that is within the annulus not on its boundary.

Is that last paragraph correct?
Yes. The image of the cylinder is [itex]\{x \in \mathbb{R}^2 : \frac\pi2 < \|x\| < \frac{3\pi}2 \}[/itex], which is open as the inequalities are strict.
 
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Ibix
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Thank you! More dumb questions to follow, I'm sure.
 

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