Timeline for Justification of the term "invertible sheaf"
Current License: CC BY-SA 2.5
9 events
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| Jul 27, 2010 at 15:12 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jul 27, 2010 at 14:47 | comment | added | Francesco Polizzi | Ad actually they were different... The behaviour of the sheaves $\mathcal{O}_X(n)$ in the weighted projective spaces is not the same as in the usual projective space, as you pointed out, so I had to be more accurate and take the double dual into account. I've edited the answer, anyway. Thank you for the remark. | |
| Jul 27, 2010 at 14:46 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jul 27, 2010 at 14:42 | comment | added | Donu Arapura | Right. I think that $ (\mathcal{O}_X(1) \otimes \mathcal{O}_X(1))^{**}\cong \mathcal{O}_X(2)$ should be fine, but it wouldn't be true without the double dual. | |
| Jul 27, 2010 at 14:36 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jul 27, 2010 at 14:02 | comment | added | Francesco Polizzi | I understand your doubt. Are you suggesting that $\mathcal{O}_X(1) \otimes \mathcal{O}_X(1)$ could be actually different from $\mathcal{O}_X(2)$, aren't you? Looking at the definitions, I know that these two sheaves agree in a dense open subset of $X$, but this is actually not enough to conclude. I will check it out... | |
| Jul 27, 2010 at 13:36 | comment | added | Donu Arapura | I have to apologize for the way this sounds, but this seems suspect. Are you sure there isn't a double dual in there somewhere? | |
| Jul 27, 2010 at 13:30 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Jul 27, 2010 at 13:23 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |