If $X$ is a normal complex variety, and $D$ is an effective $\mathbb{Q}$-Cartier Weil divisor, then there is a natural map $\mathcal{O}(D) \otimes \mathcal{O}(-D) \rightarrow \mathcal{O}_X$. My question: under what conditions is this map surjective? Obviously this is true if $X$ is smooth or $D$ is Cartier; I am interested in the singular, non-Cartier case. I am happy to assume $X$ is $\mathbb{Q}$-factorial and log terminal, if that helps at all.

Here is what I know:

  1. It is not surjective in general. For instance, if $D$ is a ruling of the quadric cone, then this map is not surjective.

EDIT: Statement 2 is not correct:
I had convinced myself that the exact sequence below implies surjectivity, but this only holds if $\mathcal{O}_D(D)$ is isomorphic to $\mathcal{O}_D \otimes \mathcal{O}_X(D)$, which is not true in general, hence statement 2 is false!

  1. If $X$ is log terminal, and $D$ is Cartier in codimension 2 (meaning the non-Cartier locus of $D$ has codimension at least 3 in $X$), this map is surjective. This follows because the sequence $0 \rightarrow \mathcal{O}_X \to \mathcal{O}_X(D) \rightarrow \mathcal{O}_D(D) \to 0$ is exact in this case.

Are there any other conditions that guarantee surjectivity of this map? I am particularly interested in the case when $D$ happens to be the (reduced, but possibly not irreducible) exceptional locus of a birational morphism $f: X \rightarrow Y$.

EDIT: in the comments below, it appears that this only holds if $D$ is, in fact, a Cartier divisor.

  • $\begingroup$ 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? $\endgroup$ – Jason Starr Jun 5 '17 at 10:07
  • $\begingroup$ 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 . . . $\endgroup$ – Jason Starr Jun 5 '17 at 11:33
  • $\begingroup$ 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. $\endgroup$ – Jason Starr Jun 5 '17 at 11:42
  • 2
    $\begingroup$ 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. $\endgroup$ – Hacon Jun 5 '17 at 15:22
  • $\begingroup$ @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. $\endgroup$ – be928 Jun 5 '17 at 16:51

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