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). 



Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive?
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](https://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45354#45354)** 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.


