Timeline for Vector bundles on affine scheme
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2011 at 15:58 | comment | added | Andrew Parker | Keeping in mind, of course, that you cannot twist beyond the top rank = dimension of your variety, since any vector bundle whose rank exceeds the dimension necessarily splits a trivial line bundle. | |
| Apr 21, 2011 at 8:11 | comment | added | Francesco Polizzi | Of course you can. The point of $(2)$ is that in dimension $3$ you find indecomposable vector bundles of any rank, in contrast with the case of affine ruled surfaces, where every vector bundle splits as a sum of line bundles. | |
| Apr 20, 2011 at 21:47 | comment | added | Sasha | If you already have infinite number of line bundles you can twist any given bundle of higher rank to produce infinite number of vector bundles of that rank, can't you? | |
| Apr 20, 2011 at 19:12 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| Apr 20, 2011 at 19:05 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |