Does push forward commute with taking dual?

Question

Let $f:X \to \mathbb{P}^1$ be a flat projective morphism between projective Noetherian schemes. Let $\mathcal{L}$ be line bundle on $X$ such that $f_*\mathcal{L}$ is locally free $\mathcal{O}_{\mathbb{P}^1}$-module. Then, is $f_*(\mathcal{L}^*) \cong (f_*\mathcal{L})^*$, where by $\mathcal{F}^*$ denotes the dual of a locally free sheaf $\mathcal{F}$?

Answer 1

Since this didn't already show up in the answers, we do have the following if we additionally assume that $R^i f_* L = 0$ for $i > 0$. Indeed, then we have

$$\mathcal{H}om_{\mathbb{P}^1}(f_* L, O_{\mathbb{P}^1}) \simeq R \mathcal{H}om_{\mathbb{P}^1}(R f_* L, O_{\mathbb{P}^1}) \simeq Rf_* R \mathcal{H}om_{X}(L, f^! O_{\mathbb{P}^1}).$$

The last isomorphism is Grothendieck duality.

Here $f^!O_{\mathbb{P}^1}$ can be viewed as simply $$(\omega_X \otimes f^* \omega_{\mathbb{P}^1}^*) [\dim X - \dim \mathbb{P}^1] \cong \omega_X \otimes (f^* O_{\mathbb{P}^1}(2))[\dim f]$$ where $[\dim f] = [\dim X - 1]$ is a shift (as a complex) by the dimension of the fibers of $f$. Since obviously $L$ is locally free, there are no higher Exts, and so here is the isomorphism we do obtain by taking the zeroth cohomology of $Rf_* R \mathcal{H}om_{X}(L, f^! O_{\mathbb{P}^1})$.

$$\begin{array}{rl} (f_* L)^* = & \mathcal{H}om_{\mathbb{P}^1}(f_* L, O_{\mathbb{P}^1}) \\ \cong & R^{\dim f} f_* \big(L^* \otimes \omega_X \otimes f^* O_{\mathbb{P}^1}(2)\big)\\ \cong & R^{\dim f} f_* \big(L^* \otimes \omega_X\big) \otimes O_{\mathbb{P}^1}(2). \end{array} $$

Let me know if this doesn't make sense.

Edit: Whoops, I just noticed that David Stapleton just posted the same thing in the comments.

Answer 2

No. Take $X= \mathbb{P}^1\times \mathbb{P}^1$ with $f$ the projection to the second factor, and $\mathcal{L}$ the locally free sheaf associated to the horizontal divisor $p\times\mathbb{P}^1$ for a point $p$. Then $f_*\mathcal{L}\cong \mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}$, but $f_*(\mathcal{L}^*)=0$.

Answer 3

This has little chance of being true, since we have $$\def\sh#1{\mathcal{#1}}\DeclareMathOperator{\Hom}{Hom}\def\P{\mathbb{P}} \sh{L}^* = \Hom(\sh{L}, \sh{O}_X) = \Hom(\sh{L}, f^* \sh{O}_{\P^1})$$ whereas the natural relationship is the sheafified adjunction $$f_*\Hom(f^* \sh{F}, \sh{G}) \cong \Hom(\sh{F}, f_* \sh{G}).$$ You would be asking for an isomorphism with the $f^*$ "in the wrong argument", so even constructing it would be highly serendipitous. It's easy to construct a counterexample (in fact, Ramsey's counterexample) from the principle that while a very ample line bundle will have many sections, its dual on $\P^1$ will have none at all.

So, say, with a point as base instead of $\P^1$ and $X = \P^1$, take $\sh{L} = \sh{O}_{\P^1}(-p)$ for any point $p$; then $f_* = \Gamma(\P^1, -)$ is the global sections functor, and $\Gamma(\P^1, \sh{L}) = 0$ while $\dim \Gamma(\P^1, \sh{L}^*) = 2$ (the vector space of homogeneous linear polynomials in two variables). Since the dual space of $0$ is still $0$, that contradicts your desired isomorphism.

To get Ramsey's example, just take the constant family over $\P^1$ with this example on each fiber.

Answer 4

Here is one more way to see that this can only happen accidentally:

Let $f:X\to Y$ be an arbitrary flat projective morphism and let $\mathscr L$ be a relatively ample line bundle. Then $\mathscr L^{\otimes n}$ has lots of sections for $n\gg 0$ while $R^if_*\mathscr L^{\otimes n}=0$ for $n\gg 0$. This means that $f_*\mathscr L^{\otimes n}$ is a locally free sheaf (of high rank) for $n\gg 0$. On the other hand, $\mathscr L^{\otimes (-n)}$ has no sections at all on open sets that are preimages of open sets in $Y$ for any $n>0$. In other words, $f_*\mathscr L^{\otimes (-n)}=0$, which is clearly not the dual of a high rank locally free sheaf.