Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:47:02Z http://mathoverflow.net/feeds/question/96455 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96455/can-the-pushforward-of-a-very-ample-invertible-sheaf-under-a-birational-morphism Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive? Sue Sierra 2012-05-09T14:42:45Z 2012-05-10T02:07:12Z <p>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:</p> <ul> <li><p>$f^{-1}$ is defined away from a single point $x \in X$.</p></li> <li><p>$f_* \mathcal{O}_Y = \mathcal{O}_X$.</p></li> </ul> <p>Two questions:</p> <p>(1) If $V$ is a set of global sections of $\mathcal{M}$ that generate $\mathcal{M}$, consider the induced evaluation map <code>$V \otimes \mathcal{O}_X \to f_* \mathcal{M}$</code>. </p> <p>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.)</p> <p>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. </p> <p>(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?</p> <p>My chief interest in (2) is that a negative answer to (2) would be a cheap way to get a negative answer to (1). </p> http://mathoverflow.net/questions/96455/can-the-pushforward-of-a-very-ample-invertible-sheaf-under-a-birational-morphism/96496#96496 Answer by Karl Schwede for Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive? Karl Schwede 2012-05-09T19:21:52Z 2012-05-10T02:07:12Z <p>Hi Sue, I think it can be reflexive.</p> <p>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 $X$.</p> <p><strong>(Blow up a divisor:)</strong> 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.</p> <p>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 <strong><a href="http://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45354#45354" rel="nofollow">THIS ANSWER</a></strong> by Sándor Kovács).<br> 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.</p> <p>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).</p> <p><strong>(Define $\mathcal{M}$):</strong> It follows that $\mathcal{M} = O_Y(-F + n \pi^*A)$ is ample on $Y$ for $n \gg 0$. </p> <p>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.</p> <p>Ok, so this gives a negative answer for (2). </p> <p><strong>(Question 1.):</strong> 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.</p> <p>Thus the global section of $H^0(Y, \mathcal{M})$ already globally generate $\pi_* \mathcal{M}$, so I don't think (1) works either.</p> <p>Obviously if $X$ is Factorial then you won't run into this problem as you already pointed out.</p>