Timeline for Definition of a complex structure on a vector bundle
Current License: CC BY-SA 3.0
15 events
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| S Mar 29, 2014 at 9:14 | history | suggested | Riccardo | CC BY-SA 3.0 |
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| Mar 29, 2014 at 8:59 | review | Suggested edits | |||
| S Mar 29, 2014 at 9:14 | |||||
| Oct 27, 2010 at 19:58 | vote | accept | TonyS | ||
| Mar 17, 2010 at 16:37 | history | edited | Andrea Ferretti | CC BY-SA 2.5 |
Closed LaTeX environment
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| Mar 17, 2010 at 15:10 | answer | added | Sebastian | timeline score: 4 | |
| Mar 10, 2010 at 10:33 | comment | added | Joel Fine | @Andrew Stacey, I agree, if $E$ is a holomorphic bundle and $f$ anti-holomorphic then $f^*\bar E$ is again a holomorphic bundle (e.g., look at what happens to the transition functions in a holomorphic local trivialisation of $E$). | |
| Mar 10, 2010 at 10:11 | history | edited | TonyS | CC BY-SA 2.5 |
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| Mar 10, 2010 at 10:09 | comment | added | TonyS | Thanks for your comments. I edited the question. Your answers show me, that this subject is more delicate than I thought it is. | |
| Mar 10, 2010 at 10:04 | history | edited | TonyS | CC BY-SA 2.5 |
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| Mar 10, 2010 at 8:14 | comment | added | Andrew Stacey | To return to the actual question, you certainly get a complex structure on $f^*E$ - simply pull-back the complex structure on $E$! Then the morphism $f^*E \to E$ is a morphism of complex bundles (note that this covers $f : S \to S$ rather than the identity). But this identifies $(f^*E)_p$ with $E_{f(p)}$ as complex vector spaces so to get the identity that TonyS wants, one should take the pull-back of $\overline{E}$. I wouldn't be surprised to learn that $f^*\overline{E}$ was then a holomorphic vector bundle over $S$: the two antiholomorphic factors should combine to a holomorphic one. | |
| Mar 10, 2010 at 8:10 | comment | added | Andrew Stacey | I don't know much about the subtleties of complex geometry so this might be obvious, but I don't see why $E$ and $f^*E$ are isomorphic as smooth bundles. Is complex conjugation on $S$ always homotopic to the identity? Thinking about the real situation, it's not true that if I have a diffeomorphism $\alpha : M \to M$ and a vector bundle $E \to M$ that $\alpha^*E$ and $E$ are isomorphic: take $M$ to be a coproduct of two identical manifolds and $E$ trivial over one factor, non-trivial over the other, and $\alpha$ the swap. What am I missing? | |
| Mar 9, 2010 at 22:37 | comment | added | Joel Fine | @Andrew Stacey, ah that would make more sense since, of course, $\bar E \cong E^*$ as smooth bundles. What confused me is that $f^*E$ is already naturally a complex vector bundle. Reading the last sentence of the question again, it seems that what you suggest is what TonyS actually means. | |
| Mar 9, 2010 at 22:25 | comment | added | Andrew Stacey | Given the third paragraph, I'd guess the questioner means "smooth real bundles" but it would be useful to have this confirmed. | |
| Mar 9, 2010 at 21:06 | comment | added | Joel Fine | I'm a little confused by the statement that $E$ and $f^*E$ are necessarily isomorphic as smooth bundles. An anti-holomorphic $f$ involution of a Riemann surface reverses its orientation. It follows that given a line bundle $L$ of degree $d$, the bundle $f^*L$ has degree $-d$ and so is rather isomorphic to $L^*$. | |
| Mar 9, 2010 at 17:39 | history | asked | TonyS | CC BY-SA 2.5 |