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Let $f: Y \to X$ be a birational morphism of projective varieties. Let $\mathcal{M}$ be a very ample invertible sheaf on $Y$. Suppose also that:

  • $f^{-1}$ is defined away from a single point $x \in X$.

  • $f_* \mathcal{O}_Y = \mathcal{O}_X$.

Two questions:

(1) If $V$ is a set of global sections of $\mathcal{M}$ that generate $\mathcal{M}$, consider the induced evaluation map $ V \otimes \mathcal{O}_X \to f_* \mathcal{M}$.

Let $\mathcal{N}$ be the image of this map. Is it possible for $\mathcal{N}$ to be reflexive? (Full disclosure: I would like the answer to be no.)

If we know that $X$ is smooth at $x$, or more generally that $(f_* \mathcal{M})^{\vee\vee}$ is invertible, then the answer is no. Let $E$ be the exceptional locus of $f$. Then $E$ is positive-dimensional, so any hyperplane section meets $E$ nontrivially, and thus any section of $f_* \mathcal{M}$ must vanish at $x$. But if $(f_* \mathcal{M})^{\vee\vee}$ is only reflexive, then I don't see how to generalise this argument.

(2) Is it possible for $f_* \mathcal{M}$ to be reflexive? If so, what is the weakest possible condition that will guarantee that $f_* \mathcal{M}$ is not reflexive?

My chief interest in (2) is that a negative answer to (2) would be a cheap way to get a negative answer to (1).

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A stupid comment, you need $X$ to be normal to guarantee that $E$ is positive dimensional. –  Karl Schwede May 9 '12 at 17:01
    
If $E$ is 0-dimensional then we can work locally and suppose that $f$ is finite; but then $f_* \mathcal{O}_Y = \mathcal{O}_X$ implies that $f$ is an isomorphism. So I guess I should have said that $f$ is not an isomorphism. –  Sue Sierra May 10 '12 at 10:00

1 Answer 1

up vote 3 down vote accepted

Hi Sue, I think it can be reflexive.

Let's consider $X = \text{Proj} k[x,y,u,v,t]/\langle xy - uv \rangle$. This has only an isolated singularity at $x=y=u=v=0$ (it's the simplest non-Q-factorial singularity I know of). Fix $U$ to be the regular locus of $X$.

(Blow up a divisor:) Set $\pi: Y \to X$ to be the blowup of the prime divisor $D = V(x,u)$. This is a small resolution of $X$ (and also contains $U$ as an open set). Lets define $F$ on $Y$ to be such that $O_Y(-F) = O_X(-D) \cdot O_Y$. In other words, $F$ is the inverse image of $D$. Notice that $\pi_* O_Y(-F) = O_X(-D)$ since $\pi$ is a small resolution. Let me explain this point.

Choose $V \subseteq X$ an open set. Then $$ \Gamma(V, \pi_* O_Y(-F)) = \Gamma(\pi^{-1} V, O_Y(-F)) = \Gamma((\pi^{-1} V) \cap U, O_Y(-F)) $$ since $\pi^{-1}V \setminus U$ is codimension 2 (see for example THIS ANSWER by Sándor Kovács).
But $\Gamma((\pi^{-1} V) \cap U, O_Y(-F)) = \Gamma(V \cap U, \pi_* O_Y(-F))$. It follows that $\pi_* O_Y(-F)$ is determined away from the singularity, and is therefore S2 / reflexive.

Since $Y$ is smooth, $F$ is a Cartier divisor and we know that $-F$ is $\pi$-ample by construction. Set $A = V(t)$ to be an ample Cartier divisor on $X$ (the choice of $A$ doesn't matter so much here).

(Define $\mathcal{M}$): It follows that $\mathcal{M} = O_Y(-F + n \pi^*A)$ is ample on $Y$ for $n \gg 0$.

Then we know $$ \pi_* O_Y(-F + n \pi^* A) = (\pi_* O_Y(-F) ) \otimes O_X(nA) = O_X(-D + nA). $$ which is reflexive. Note that you can also consider $-mF + mn \pi^* A$ to make $\mathcal{M}^m$ as ample as you'd like. The same sort of computation still holds.

Ok, so this gives a negative answer for (2).

(Question 1.): Let's now consider (1). Note $\pi_* \mathcal{M}$ is globally generated (since $n \gg 0$). But on the other hand, $$H^0(X, \pi_* \mathcal{M}) = H^0(U, \mathcal{M}) = H^0(Y, \mathcal{M})$$ again since $X \setminus U$ and $Y \setminus U$ are codimension 2.

Thus the global section of $H^0(Y, \mathcal{M})$ already globally generate $\pi_* \mathcal{M}$, so I don't think (1) works either.

Obviously if $X$ is Factorial then you won't run into this problem as you already pointed out.

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Thanks, Karl. This is the kind of thing I was worried about, but hadn't written an example down so was hoping it didn't exist. –  Sue Sierra May 10 '12 at 10:15

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