Timeline for Hopf fibration extended to bundle over $\mathbb{C}^2$
Current License: CC BY-SA 4.0
11 events
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| May 10, 2020 at 3:02 | comment | added | Guest123412341234 | I think I may have (somewhat) sorted this out. Although hard to find in the paper I believe the curvature form I mentioned above is scaled by a function depending on $r$ (radial coordinate). The bundle I described above ends up coming through in the limit $r\to\infty$, where $\mathbb{C}^2$ approaches its asymptotic boundary $\mathbb{S}^2$. Although I am still not sure explicitly what the bundle on the bulk $\mathbb{C}^2$ is, I think it is well defined and a proposed "extension" in the way I was describing about is a bit naive. Thanks for t comments, I welcome any more thoughts on the question. | |
| May 10, 2020 at 2:34 | comment | added | David Roberts♦ | Are you sure that form extends continuously to the origin? I'm not. | |
| May 10, 2020 at 1:33 | comment | added | Guest123412341234 | @DeaneYang, I may be wrong about the extension. What I am certain of is that he considers the curvature form $F=d\omega$ which should be over $\mathbb{S}^2$, however, he treats it as if it were over $\mathbb{C}^2$ (or $\mathbb{C}P^2\cong\mathbb{C}^2\cup\mathbb{S}^2$). The context is that he is considering a twisted Dirac operator, where the twisting is coming from a scaled version of the connection one form $\omega$. | |
| May 9, 2020 at 17:29 | comment | added | Deane Yang | If this is in fact what the author does, then it is indeed incorrect. Unless the author is using some kind of generalized covariant derivative, extending it implies extending the bundle, which is not possible here. As you say, it is indeed possible to extend it to the punctured plane but no further. | |
| May 9, 2020 at 13:18 | comment | added | Guest123412341234 | @DavidRoberts , this makes sense, thanks you. In essence when I say `extend' what I want is the following; I have a connection from $\mathbb{S}^2$ to some some line bundle, this connection has a curvature form $F$ associated to it, I want to somehow get a bundle over $\mathbb{C}^2$ with the same curvature form (which does seem like an inclusion actually). | |
| May 9, 2020 at 10:48 | comment | added | David Roberts♦ | If I understand correctly what you are asking, one thing to point out is that the bundle over $S^2$ cannot extend to a bundle over $\mathbb{C}^2$, since then the Hopf bundle would be trivial—if by 'extend' you mean along some inclusion $S^2 \hookrightarrow \mathbb{C}^2$. | |
| May 9, 2020 at 10:23 | history | edited | Guest123412341234 | CC BY-SA 4.0 |
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| May 9, 2020 at 9:38 | comment | added | Liviu Nicolaescu | The bundle in question is the tautological line bundle. Its total space is $\mathbb{C}^2\setminus 0$ and the projection is the natural projection $\mathbb{C^2}\setminus \{0\}\to\mathbb{CP}^1$ that associates to a nonzero vector the line it determines | |
| May 9, 2020 at 4:09 | history | edited | Guest123412341234 | CC BY-SA 4.0 |
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| May 9, 2020 at 3:55 | history | edited | Guest123412341234 | CC BY-SA 4.0 |
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| May 9, 2020 at 3:49 | history | asked | Guest123412341234 | CC BY-SA 4.0 |