Timeline for Isomorphism classes of line bundles with connections
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 2, 2018 at 20:45 | comment | added | Ben McKay | @Marion: Arapura is correct of course. Do you want a smooth connection on a holomorphic vector bundle or a holomorphic connection on that bundle? The question is whether the connection has holomorphic coefficients $A^{\mu}_{\nu}$ in a holomorphic trivialization. A Hermitian connection which is not flat cannot be holomorphic, because a holomorphic connection has $(2,0)$ type curvature, while a Hermitian connection has $(1,1)$ type curvature. Look at Huybrechts, Complex Geometry. | |
| Mar 4, 2018 at 0:43 | comment | added | Yosemite Sam | @Marion By (O,d) I think abx meant the usual de Rham differential on functions. | |
| Mar 3, 2018 at 19:21 | comment | added | Marion | @abx I should put "a variety X". But still, you have not explained the notation $d$ (is it degree?). If so, then there exists a line bundle at least for each $d$. | |
| Mar 3, 2018 at 19:17 | comment | added | abx | @Marion: your question starts with a scheme $X$. This is why I assumed you were talking about a holomorphic connection. Anyway, I suggest that you revise this material in a good book. | |
| Mar 3, 2018 at 18:49 | comment | added | Marion | No, I think you are wrong. A Hermitian connection is a map $\bar{\delta}: \Omega^0(L) \to \Omega^{0,1}(L)$. It's curvature is (1,1) type. | |
| Mar 3, 2018 at 18:45 | comment | added | Donu Arapura | No. The connection form for holomorphic connection is holomorphic, whereas for a hermitian connection it isn't ( its $\partial \log ||f||$ where $f$ is a $C^\infty$ section.) Anyway, that's all I have to say about this. | |
| Mar 3, 2018 at 18:38 | comment | added | Marion | But from Kobayashi-Hitchin a $C^{\infty}$ connection corresponds to a hermitian connection. And a hermitian connection is defined on a hermitian line bundle which is holomorphic. | |
| Mar 3, 2018 at 18:36 | comment | added | Donu Arapura | You seem to be talking about $C^\infty$ connection on a holomorphic line bundle. And that's fine. But as I said, you have to clear from the beginning about what the category is that you're working in. | |
| Mar 3, 2018 at 18:24 | comment | added | Marion | Sorry, physicist here. I am not sure that I agree that $c_1$ is zero even in the case of holomorphic line bundles with holomorphic connections. Assume $X = \mathbb{CP}^2$ and $L \to E$ a hermitian line bundle over $X$ and equip it with a hermitian connection $A$ (local one-form presented here) s.t. $F = dA$ is the curvature. Then over a cycle $S$ we can have $\int_S F \backsim c_1(L)$ and this does not have to be zero. This is what physicists call magnetic flux. | |
| Mar 3, 2018 at 18:17 | history | answered | Donu Arapura | CC BY-SA 3.0 |