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 $\mathcal{F}$?
[Math] Does push forward commute with taking dual
ag.algebraic-geometry
Best Answer
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.