Timeline for Differentiable structures on R^3
Current License: CC BY-SA 3.0
14 events
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S Dec 2, 2015 at 11:37 | history | suggested | hrkrshnn | CC BY-SA 3.0 |
Formatting.
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Dec 2, 2015 at 11:25 | review | Suggested edits | |||
S Dec 2, 2015 at 11:37 | |||||
May 16, 2010 at 21:42 | history | edited | Henri | CC BY-SA 2.5 |
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May 16, 2010 at 21:42 | comment | added | Qfwfq | But the fact that two atlases $\mathfrak{A}$ and $\mathfrak{B}$ are not compatible just means that $id:(\mathbb{R},[\mathfrak{A}])\rightarrow (\mathbb{R},[\mathfrak{B}])$ is not a diffeomerphism, but it does'n exclude that there are other diffeomerphisms between the differentiable structures: in fact, up to diffeomorphism, there's only one smooth structure on $\mathbb{R}$. | |
May 16, 2010 at 21:42 | comment | added | Qfwfq | What Henri means, I think, is that on $\mathbb{R}$ there are infinitely many non-compatible (maximal) atlases each giving the topological space $\mathbb{R}$ the structure of a differentiable manifold. | |
May 16, 2010 at 21:28 | comment | added | Peter Luthy | The inverse of $x^3$ is not smooth, but the term diffeomorphism as it pertains to differentiable structures does not require this: a bijective map $f:\mathbb{R}\rightarrow\mathbb{R}$ is a diffeomorphism of $(\mathbb{R},A)$ and $(\mathbbf{R},B)$ if $x\in A$ iff $x(f)\in B$. Here A and B are maximal atlases. | |
May 16, 2010 at 21:19 | comment | added | Akela | I have also specified that the structure is $C^\infty$, in response to your "even better:----" addition. Re: your EDIT -- The point is that there are uncountably many structures on $\mathbb R^4$. | |
May 16, 2010 at 21:16 | comment | added | Henri | Ok, I didn't understand what you meant! I edited my post though. | |
May 16, 2010 at 21:14 | history | edited | Henri | CC BY-SA 2.5 |
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May 16, 2010 at 21:13 | comment | added | Akela | $x \mapsto x^3$ is a diffeomorphism from $\mathbb R$ with the usual differentiable structure to $\mathbb R$ with the differentiable structure corresponding to the atlas you gave above. | |
May 16, 2010 at 21:09 | comment | added | Henri | Yes, but $x\mapsto x^3$ is no diffeomorphism! You surely mean "homeomorphism". | |
May 16, 2010 at 21:08 | history | edited | Henri | CC BY-SA 2.5 |
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May 16, 2010 at 21:06 | comment | added | Akela | Oops! Now I have added "upto diffeomorphism" in the criterion. Sorry for the lapse! | |
May 16, 2010 at 21:02 | history | answered | Henri | CC BY-SA 2.5 |