Timeline for Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections?
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2011 at 14:48 | vote | accept | Qiaochu Yuan | ||
| Nov 30, 2009 at 13:25 | comment | added | Andrew Stacey | I have a suspicion that editing an answer doesn't provoke an automatic notification to the original questioner that the answer's been edited. So I'm leaving this comment in case Qiaochu hasn't noticed the revision. | |
| Nov 22, 2009 at 21:14 | comment | added | Andrew Stacey | @Gian: the point is that f is fixed and you have complete freedom in choosing h. So you can choose it to go as fast as you need. | |
| Nov 21, 2009 at 17:14 | comment | added | Agustí Roig | @Yemon. Ow! I must have been blind: I didn't see the "X" in "g,h: X ...". Thanks. | |
| Nov 21, 2009 at 1:21 | answer | added | Jose Capco | timeline score: 0 | |
| Nov 20, 2009 at 15:25 | history | edited | Qiaochu Yuan | CC BY-SA 2.5 |
Fixed typo.
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| Nov 20, 2009 at 12:01 | answer | added | Andrew Stacey | timeline score: 6 | |
| Nov 20, 2009 at 11:31 | comment | added | Gian Maria Dall'Ara | Even for manifolds I don't see if one can find, for example, a non zero function h on a disk, wich goes so fast to zero at the boundary, such that fh goes to zero at the boundary too. | |
| Nov 20, 2009 at 11:27 | comment | added | Gian Maria Dall'Ara | For locally compact hausdorff spaces this is equivalent to finding h continuous on X such that fh extends to a continuous function up to the boundary, since than you can extend the product to a g continuous on the full space. | |
| Nov 20, 2009 at 11:10 | comment | added | Yemon Choi | @Agusti: the point is that f might not extend continuously to all of X. For instance, take X=[0,1], U to be (0,1], and $f(x)= 1/x$. | |
| Nov 20, 2009 at 8:21 | comment | added | Yemon Choi | @Theo: while it's always nice to see love for Gelfand-Naimark, I don't quite understand its relevance to Qiaochu's question... (Also, C*-algebras are very rigid and non-sheafy objects) | |
| Nov 20, 2009 at 8:15 | comment | added | Theo Johnson-Freyd | @Yemon: I thought about that after I made my post, when I reread the original question. But it still doesn't make a difference: g/h = gh*/|h|^2, where h* is the complex conjugate of h. So complex division is really just real division. | |
| Nov 20, 2009 at 8:14 | comment | added | Theo Johnson-Freyd | Incidentally, there is another interesting way to define topological spaces in terms of their functions. Recall that on a locally compact Hausdorff space the continuous (complex) functions that vanish at infinity form a c-star algebra, and all commutative c-star algebras arise this way. So sometimes you should only consider functions that vanish at infinity. | |
| Nov 20, 2009 at 8:07 | comment | added | Yemon Choi | @Theo: we are of course using the multiplicative structure on K, at least in the question. So while there might turn out to be no difference I'm not convinced it's for the reason you give. | |
| Nov 20, 2009 at 8:03 | comment | added | Theo Johnson-Freyd | As topological spaces $\mathbb C = \mathbb R \times \mathbb R$, and so for the purposes of detecting spaces there is no difference. | |
| Nov 20, 2009 at 6:40 | comment | added | Yemon Choi | Don't you mean that $h$ is nowhere vanishing on U? | |
| Nov 20, 2009 at 6:24 | history | asked | Qiaochu Yuan | CC BY-SA 2.5 |