Timeline for When can you reverse the orientation of a complex manifold and still get a complex manifold?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2010 at 21:19 | vote | accept | solbap | ||
| Dec 1, 2010 at 16:41 | history | edited | solbap | CC BY-SA 2.5 |
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| Dec 1, 2010 at 16:32 | history | edited | solbap | CC BY-SA 2.5 |
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| Dec 1, 2010 at 15:33 | history | edited | solbap | CC BY-SA 2.5 |
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| Dec 1, 2010 at 0:21 | answer | added | Dmitri Panov | timeline score: 14 | |
| Nov 30, 2010 at 23:13 | comment | added | BCnrd | What is canonical is that even-dim'l C-manifolds have an intrinsic orientation determined by C-structure: an orientation of $\mathbf{C}$ endows all C-manifolds with functorial orientation, and changing initial choice affects the orientation on $n$-dimensional C-manifolds by $(-1)^n$. So for even $n$ the question is well-posed. This has nothing to do with changing $i$ and $-i$, and your impression in the affine case is wrong. In any dim., can "twist" structure sheaf by C-conj. to get a new C-manifold (modelled on $\overline{f}(\overline{z})$), but that's a different beast. | |
| Nov 30, 2010 at 22:48 | comment | added | BCnrd | Fix an alg. closure $\mathbf{C}$ of $\mathbf{R}$, equipped with unique abs. value extending the one on $\mathbf{R}$, complex analysis is developed without needing a preferred $\sqrt{-1}$. The complex structure has no reliance on any orientation. The so-called canonical orientation on complex manifolds is just the functorial one arising from a choice of $\sqrt{-1}$; can make either choice, complex structure can't tell! Likewise, the analytification functor on locally finite type $\mathbf{C}$-scheme has nothing to do with any such choice. Note $p$-adic analysis goes the same way. | |
| Nov 30, 2010 at 22:40 | comment | added | Tim Perutz | @J.C. Ottern: Any almost complex structure compatible with the orientation on a closed 4-manifold $X$ satisfies $c_1^2[X]=2\chi+3\sigma$ ($\chi$=Euler char, $\sigma$=signature). This is by Hirzebruch's signature theorem. | |
| Nov 30, 2010 at 22:31 | comment | added | J.C. Ottem | Is there a simple reason for why $\overline{\mathbb{CP}^2}$ is not a complex manifold? | |
| Nov 30, 2010 at 22:12 | answer | added | Spinorbundle | timeline score: 7 | |
| Nov 30, 2010 at 21:41 | history | asked | solbap | CC BY-SA 2.5 |