Timeline for surjectivity of double dual map for weil divisors on normal varieties
Current License: CC BY-SA 3.0
16 events
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| Jun 5, 2017 at 19:35 | history | edited | be928 | CC BY-SA 3.0 |
explained error in statement 2
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| Jun 5, 2017 at 19:29 | comment | added | be928 | Thank you! I had mistakenly convinced myself that the notation $\mathcal{O}_D(D)$ simply meant $\mathcal{O}_D \otimes \mathcal{O}_X(D)$, but this is not the case! | |
| Jun 5, 2017 at 18:00 | comment | added | Sándor Kovács | Here is why I think Christopher's argument works. Let $L$ be a reflexive sheaf of rank 1 and $\mathfrak m$ a maximal ideal. If $\dim L/\mathfrak m=1$, then $L$ is Cartier by Nakayama. If you have another $L'$, then $(L\otimes L')/\mathfrak m$ surjects onto $(L/\mathfrak m)\otimes (L'/\mathfrak m)$, so if $L\otimes L'$ is a line bundle, then so are $L$ and $L'$. | |
| Jun 5, 2017 at 17:57 | comment | added | Hacon | I agree that there is a short exact sequence $0\to \mathcal O _X\to \mathcal O_X(D)\to \mathcal O_D(D)\to 0$ and $0\to (\mathcal O _X(-D)\otimes \mathcal O_X(D))/torsion \to \mathcal O_X(D)\to \mathcal O_D\otimes \mathcal O_X(D)\to 0$. I don't yet see how this contradicts what I said above: why is $(\mathcal O _X(-D)\otimes \mathcal O_X(D))/torsion \cong \mathcal O _X$ or equivalently $\mathcal O_D\otimes \mathcal O_X(D)\cong \mathcal O_D(D)$? | |
| Jun 5, 2017 at 17:55 | comment | added | Sándor Kovács | @kdev: I don't see why $\mathscr O_D\otimes\mathscr O_X(D)$ would be $\mathscr O_D(D)$. Restriction of a reflexive sheaf is not necessarily reflexive, so very likely you'd have to take the reflexive hull which screws up your short exact sequence. | |
| Jun 5, 2017 at 17:32 | comment | added | be928 | @JasonStarr yes, thank you! Do you see something wrong with the reasoning in my previous comment? | |
| Jun 5, 2017 at 17:19 | comment | added | Jason Starr | @kdev. You do have a short exact sequence as you write by Proposition 5.26, p. 163, of Koll'ar-Mori, "Birational Geometry of Algebraic Varieties". | |
| Jun 5, 2017 at 17:02 | history | edited | be928 | CC BY-SA 3.0 |
added 4 characters in body
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| Jun 5, 2017 at 17:00 | comment | added | be928 | @Hacon -- thank you for your answer, although it seems to contradict what I was saying in (2)! Here was my reasoning for (2): because $0 \rightarrow \mathcal{O}(-D) \rightarrow \mathcal{O} \rightarrow \mathcal{O}_D \rightarrow 0$ is exact, after tensoring with $\mathcal{O}(D)$ we get $ \mathcal{O}(-D) \otimes \mathcal{O}(D) \rightarrow \mathcal{O}(D) \rightarrow \mathcal{O}_D(D) \rightarrow 0$, which should be exact on the left after modding out by torsion. Because the sequence in (2) is exact, this identifies $\mathcal{O}_X$ with that kernel. Am I missing something subtle? | |
| Jun 5, 2017 at 16:51 | comment | added | be928 | @JasonStarr I was using the definition that $\mathcal{O}_X(D)(U) = \{ f \in K(X) | (f)|_U + D|_U \ge 0 \}$ (coming from Karl Schwede's notes--definition 3.3 here: math.utah.edu/~smolkin/schwede_generalized_divisors.pdf ), but by reflexivity it should be the same as what you suggested. I need to think a little more about your example first. | |
| Jun 5, 2017 at 15:22 | comment | added | Hacon | My instinct says that unless $D$ is Cartier, this map is not surjective. Does the following argument work? Suppose $\mathcal O_X(D)\otimes \mathcal O_X(-D)\to \mathcal O_X$ is surjective then working locally, let $\frak m$ the maximal ideal. Abusing notation, we have $1=\sum f_i g_i\in k=\mathcal O_X/\frak m$ for $f_i\in\mathcal O_X(D)$ and $g_i\in \mathcal O_X(-D)$ so $1=fg$ for $f\in\mathcal O_X(D)$, $g\in \mathcal O_X(-D)$. But then $0=(f)+(g)$ and $(f)+D\geq 0$, $(g)-D\geq 0$ imply $(f)=-D$ and $(g)=D$ so that $D$ is Cartier. | |
| Jun 5, 2017 at 11:42 | comment | added | Jason Starr | Regarding 2, you do have such a short exact sequence. However, $\mathcal{O}_X(D)\otimes_{\mathcal{O}_X} \mathcal{O}_D$ is typically not an invertible sheaf on $D$. I think this is the source of the trouble. | |
| Jun 5, 2017 at 11:33 | comment | added | Jason Starr | Are you sure that 2 is correct? Let $M$ be a Fano manifold of dimension $\geq 2$ such that $\omega_M \cong \mathcal{O}(-r \underline{E})$ for some effective Cartier divisor $\underline{E}$ and for an integer $r>1$, e.g., this holds for $M\cong \mathbb{P}^1\times \mathbb{P}^1$. Let $X$ be Spec of the section ring of $M$ with respect to the invertible sheaf $\omega_M^\vee \cong \mathcal{O}(r\underline{E})$. It seems to me that $X$ is even canonical. Let $D$ be the divisor of $E$. I do not believe that the pairing is surjective in this case . . . | |
| Jun 5, 2017 at 10:07 | comment | added | Jason Starr | It would help if you clarify your definitions. I assume that you are defining $\mathcal{O}(-D)$ to be the pushforward from the regular locus of $X$ of the invertible ideal sheaf of $D$ (considered as a closed subscheme). Is $\mathcal{O}(D)$ defined to be the pushforward from the regular locus of the dual invertible sheaf? | |
| Jun 4, 2017 at 22:49 | review | First posts | |||
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| Jun 4, 2017 at 22:45 | history | asked | be928 | CC BY-SA 3.0 |