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There is only one differentiable structure permitted in R^2, meaning, I think, that all atlases in R^2 are diffeomorphic to the Cartesian atlas. But, doesn't the polar coordinate system represent an atlas that is not truly diffeomorphic to the Cartesian atlas, due to the coordinate singularity it has at its origin?

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    $\begingroup$ Polar coordinates give a single chart (not an entire atlas) for exactly the reason you mention: the coordinates don't cover the whole plane. $\endgroup$ Jul 27, 2010 at 20:30
  • $\begingroup$ Polar coordinates aren't differentiable across the origin, but don't they include it? To put it another way, why can't the polar coordinate chart be considered a C^0 atlas on R^2 (even though it's not C^1)? $\endgroup$
    – Outis
    Jul 27, 2010 at 20:54
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    $\begingroup$ Outis, can you give even a $C^0$ map based on polar coordinate that is a homeomorphism to a neighbourhood of the origin in $\mathbb{R}^2$? $\endgroup$ Jul 27, 2010 at 21:24
  • $\begingroup$ Robin, I presume I cannot give such a map. Can it be shown that the conventional map of (r,$\theta$) to (x,y) is discontinuous across the origin? $\endgroup$
    – Outis
    Jul 28, 2010 at 15:00

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I think Matt Noonan's comment technically answered my question. The polar coordinate chart is not a valid atlas on $\mathbb{R}^2$ because the angular coordinate is discontinuous across the non-negative x-axis.

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  • $\begingroup$ You don't seem to distinguish clearly between an atlas (which is a collection of co-ordinate charts) and a co-ordinate chart. Polar co-ordinates can be used to define co-ordinate charts that cover $R^2\backslash\{0\}$, but none of them will cover the origin, because the domain of the co-ordinate chart is always assumed to be open. So at best polar co-ordinates can be used to define an atlas of $R^2\backslash\{0\}$. $\endgroup$
    – Deane Yang
    Jul 29, 2010 at 2:34
  • $\begingroup$ Point taken: Multiple polar coordinate charts can eliminate the discontinuity along the positive x-axis. I understand that a single chart may or may not constitute an atlas. Sometimes it does, as in the Cartesian case for $\mathbb{R}^2$. Sometimes multiple charts are required, as in the polar case for $\mathbb{R}^2\backslash {0}$. My original question asked, among other things, whether a single polar coordinate chart constitutes an atlas for $\mathbb{R}^2$. The answer is negative. $\endgroup$
    – Outis
    Jul 29, 2010 at 13:52

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