Timeline for Defining Quotient Bundles
Current License: CC BY-SA 2.5
11 events
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| Jun 8, 2010 at 12:26 | comment | added | Andrew Stacey | (The style of the question is that of a beginner to DG so I'm being more pedantic than usual. I was sure you knew what was going on but couldn't be sure the questioner did.) You should be careful to distinguish between a (local) section of the vector bundle and a (local) section of the associated principal bundle. The latter is, of course, equivalent to a trivialisation of the vector bundle (by definition, if you set things up correctly) but the former is most definitely not. As originally phrased, your answer read as if you just needed a local section of the vector bundle. | |
| Jun 8, 2010 at 12:14 | comment | added | David Carchedi | @Andrew: I've changed the admitting a section bit to make it more clear, however, admitting a local section is the same as admitting a trivialization, no? Just think in terms of cocycle data for the associated principal bundle... | |
| Jun 8, 2010 at 12:12 | comment | added | David Carchedi | (Edited to reflect this) | |
| Jun 8, 2010 at 12:11 | history | edited | David Carchedi | CC BY-SA 2.5 |
changed "ker" to "coker"; added 7 characters in body
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| Jun 8, 2010 at 12:07 | comment | added | David Carchedi | Oh, I see what you mean about "coker", sorry. | |
| Jun 8, 2010 at 11:52 | comment | added | Andrew Stacey | "admit a section" -> "admit trivialisations", and I agree that "Ker" should be "coker" | |
| Jun 8, 2010 at 11:43 | comment | added | macbeth | Maybe this is just a misunderstanding of notation: as I've read your answer, $\sigma$ denotes the inclusion of fibres of $E'$ into fibres of $E$, and in particular has zero kernel. Sure, the ses $0\to E'_x\to E_x\to E_x/E_x'=coker(\sigma)\to 0$ splits, and I'm happy to believe that essentially our answers say the same thing. | |
| Jun 8, 2010 at 10:23 | comment | added | David Carchedi | No, no. Every short-exact sequence of vector spaces SPLITS, so I mean kernel, not cokernel. Now, to define the splitting in a natural way, you need to pick a basis this is true- but then this boils down to the same comment you made about the continuity of the determinant function. Everything is fine.. | |
| Jun 8, 2010 at 8:00 | comment | added | macbeth | (Reading Deane's comment, I notice my remarks might be ambiguous: by "canonical" I mean "uniquely specified from the data (including trivializations) so far given," not "independent of the choice of trivializations.") | |
| Jun 8, 2010 at 2:23 | comment | added | macbeth | Hi David, I'm a little worried about this. First, do you mean "coker" instead of "ker" throughout? Secondly and more importantly, I think as you've set this up the identification of $Im(\sigma_x)\oplus Coker(\sigma_x)$ with $\mathbb{R}^{n+k}$ isn't canonical. This is a problem because it then isn't clear that the identification on each fibre can be chosen so as to "vary continuously from fibre to fibre" (which is the key point for the local trivialization). | |
| Jun 7, 2010 at 23:34 | history | answered | David Carchedi | CC BY-SA 2.5 |