Timeline for Some questions on the intersection theory on a Hilbert scheme of points of a surface.
Current License: CC BY-SA 2.5
16 events
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| Feb 18, 2010 at 15:09 | history | edited | James O | CC BY-SA 2.5 |
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| Feb 17, 2010 at 0:01 | comment | added | James O | But the function $$\alpha \rightarrow exp\left( \sum \frac{z_i P_\alpha[-i]}{(-1)^{i-1}i} \right) \cdot 1$$ is well defined, no? This should be the case since $P_\alpha[-i]$ is defined for any class $\alpha$. Thank you again! | |
| Feb 16, 2010 at 2:10 | answer | added | Dmitri Panov | timeline score: 2 | |
| Feb 16, 2010 at 2:01 | comment | added | Hiraku Nakajima | I mean 'defined' instead of 'well-defined' in the above comment. | |
| Feb 16, 2010 at 0:34 | comment | added | Hiraku Nakajima | The left hand side is not well-defined for $\alpha$. For the right hand side, you should remind yourself of the definition of $P_\Sigma[i]$ in Chapter 8 of the book. | |
| Feb 16, 2010 at 0:15 | history | edited | James O | CC BY-SA 2.5 |
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| Feb 16, 2010 at 0:09 | comment | added | James O | Dear Professor Nakajima, would you agree that the formula on page 99 of your book $$\sum z^n [S^n \Sigma] = exp\left( \sum \frac{z_i P_\sigma[i]}{(-1)^{i-1}i} \right) \cdot 1 $$ is well defined if we replace $\Sigma$ by any 2-homology class $\alpha$? If so, I presume this satisfy the posed question. | |
| Feb 15, 2010 at 22:18 | history | edited | James O | CC BY-SA 2.5 |
Maths don't display in comment section when comments get hidden, i copied them here.
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| Feb 14, 2010 at 22:47 | comment | added | Hiraku Nakajima | You should not delete the comments. Just add further comments. I have used your definition given in the deleted comment to conclude $\Phi_n(2H)$ is $2(line)^{[n]}$, and get a confusion. I do not understand why you choose $(quadric)^{[n]}$ yet. How about $\Phi_n(-2H)$ ? Is it $-2(line)^{[n]}$ or $-(quadric)^{[n]}$ ? Not only giving the answer, please explain the reason why you choose. | |
| Feb 14, 2010 at 19:31 | comment | added | James O | Yes, $2H$ should be sent to the symmetric product of the quadric. Thank you Dmitri. I guess my previouss comments were not well written, so I deleated them. What I was trying to say is that I completely agree with that $\Phi_n$ could not be a homomorphism and that I am just looking for a good extension of the function $[\Sigma] \mapsto [\Sigma^{[n]}]$ to any 2-homology class. Thanks a lot Professor Nakajima and Dmitri. | |
| Feb 14, 2010 at 14:06 | comment | added | Dmitri Panov | Dear Professor Nakajima, for James's definiton, $\Phi_n$ for $2H$ is the symmetric product of the quadric. | |
| Feb 14, 2010 at 4:45 | comment | added | Hiraku Nakajima | Let $X$ be $P^2$ and $H$ be the hyperplane class. What is your definition of $\Phi_n$ for $2H$ ? Is it the symmetric product of a quadric or twice of the symmetric product of a line ? | |
| Feb 13, 2010 at 21:53 | history | edited | James O | CC BY-SA 2.5 |
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| Feb 13, 2010 at 5:15 | comment | added | Hiraku Nakajima | What do you mean by `a homomorphism $\Phi_n$' ? The map $\Sigma\to \Sigma^{[n]}$ is not linear in $\Sigma$. | |
| Feb 12, 2010 at 15:59 | history | edited | James O | CC BY-SA 2.5 |
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| Feb 12, 2010 at 15:27 | history | asked | James O | CC BY-SA 2.5 |