Timeline for Linearly trivial bundles on hypersufaces in $\mathbb CP^n$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2013 at 23:47 | vote | accept | aglearner | ||
| Feb 3, 2013 at 1:45 | answer | added | Mohan | timeline score: 2 | |
| Feb 2, 2013 at 20:49 | answer | added | Mohan | timeline score: 1 | |
| Feb 1, 2013 at 14:56 | history | edited | aglearner | CC BY-SA 3.0 |
Replaced n! by 4n and made the idea of proof more affirmative
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| Feb 1, 2013 at 1:07 | comment | added | aglearner | Serge, in fact this space will be connected. I just realised that every two points on $V_n$ can be joined by a chain of just two lines (and not $n!$ :) as I wrote previously, I changed now $n!$ for two since this works). Indeed, the dimension of the space of lines through a generic point on $V_n$ is $n!-n-1$. So the space of all paths between $x$ and $y$ can be identified with the intersection of two subvarieties of $\mathbb CP^{n!}$ of dimensions $n!-n$. All irreducible components of this intersection have dim at least $n!-2n$ hence it is connected. | |
| Feb 1, 2013 at 0:54 | history | edited | aglearner | CC BY-SA 3.0 |
added 122 characters in body
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| Feb 1, 2013 at 0:17 | history | edited | aglearner | CC BY-SA 3.0 |
added 52 characters in body
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| Jan 31, 2013 at 23:48 | answer | added | Sándor Kovács | timeline score: 1 | |
| Jan 31, 2013 at 22:38 | comment | added | aglearner | Mahdi, my question is about holomorphic bundles. Linearly trivial bundle is the bundle that restricts to each $\mathbb CP^1$ in $\mathbb CP^n$ as $O+O+...+O$ ($k$ copies of the structure sheaf of $\mathbb CP^1$). So for example according to my definition $O(1)$ is not linearly trivial ($T\mathbb CP^n$ neither). Serge, I agree, probably connectedness is is the most complicate thing to understand | |
| Jan 31, 2013 at 21:59 | history | edited | Mahdi Majidi-Zolbanin |
edited tags
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| Jan 31, 2013 at 21:52 | comment | added | Mahdi Majidi-Zolbanin | Do you mean projective plane instead of projective line? Because all bundles on $\mathbf{CP}^n$ are linearly trivial, so then you are saying all bundles on $\mathbf{CP}^n$ split? Or did I misunderstand? | |
| Jan 31, 2013 at 21:11 | history | edited | aglearner | CC BY-SA 3.0 |
added 6 characters in body
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| Jan 31, 2013 at 18:51 | comment | added | Serge Lvovski | I wonder how you are going to prove connectedness... | |
| Jan 31, 2013 at 10:42 | history | asked | aglearner | CC BY-SA 3.0 |