Timeline for Two questions on canonical line bundle over $\mathbb{C}P^{n}$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2014 at 16:57 | comment | added | abx | This is not an elementary result, you need some basic facts on cohomology, Serre duality, etc. These notes which I found through Google seems as accessible as possible. | |
| Sep 30, 2014 at 16:26 | comment | added | Ali Taghavi | @abx could you please give me an elementary reference for classification of algebraic vector bundle and the uniquness of decomposition which you mentioned?Thanks | |
| Sep 21, 2014 at 17:48 | comment | added | Ali Taghavi | @abx thank you very much for your comments in particular your comment on algebra geometric part of my question. | |
| Sep 21, 2014 at 17:46 | comment | added | Ali Taghavi | @Sasha Thank you very much for your comments. I was incorrect in computation of Chern character. What about the "difference element" as an element of relative k theory? | |
| Sep 21, 2014 at 17:40 | comment | added | Sasha | $(e^x)^2+1 = e^{2x}+1 = 2 + 2x + 2x^2 + \frac43 x^3 + \dots$ while $2e^x = 2 + 2x + x^2 + \frac13 x^3 + \dots$ | |
| Sep 21, 2014 at 17:23 | comment | added | abx | As algebraic vector bundles, $(\ell_{1}\otimes \ell_{1})\oplus1$ and $\ell_{1}\oplus\ell_{1}$ -- or, in algebraic geometry language, $\mathcal{O}_{\mathbb{P}^1}(2)\oplus \mathcal{O}_{\mathbb{P}^1}$ and $\mathcal{O}_{\mathbb{P}^1}(1)^2$ -- are not isomorphic. Any algebraic vector bundle on $\mathbb{P}^1$ can be written $\mathcal{O}_{\mathbb{P}^1}(d_1)\oplus \ldots \oplus \mathcal{O}_{\mathbb{P}^1}(d_r)$ with $d_1\leq \ldots \leq d_r$, and this decomposition is unique. | |
| Sep 21, 2014 at 17:16 | comment | added | abx | No, this is false -- compare the coefficients of $x^2$. | |
| Sep 21, 2014 at 16:12 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
added 361 characters in body
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| Sep 21, 2014 at 15:47 | comment | added | Ali Taghavi | @Sasha could you please more explain: in $\mathbb{Z} [x]/x^{n+1}$ , we have ${(e^{x})}^{2}+1=2e^{x}$ so what is the contradiction? | |
| Sep 21, 2014 at 14:50 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
edited title
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| Sep 21, 2014 at 10:46 | comment | added | Sasha | Compare Chern characters. | |
| Sep 21, 2014 at 10:24 | history | asked | Ali Taghavi | CC BY-SA 3.0 |