# Projection formula for flat morphisms

Let $$f\colon X\to Y$$ be a flat morphism between two smooth projective varieties. Let $$L$$ be a locally free sheaf on $$X$$ and $$\mathcal{F}$$ a coherent sheaf on $$Y$$. How to prove $$f_*(L\otimes f^*\mathcal{F})\cong f_*L\otimes\mathcal{F}$$? I think it is a well-known result but I couldn't find a reference. You can assume $$f$$ is smooth and surjective if you want. Thank you.

• Tag 0B54 gets pretty close, but not quite all the way there. Because $f$ is flat, you may write $f^*$ for $Lf^*$; because $Y$ is smooth quasi-projective, $\mathscr F[0]$ is perfect (you can find a finite locally free resolution). Because $L$ is locally free, you can write $\otimes$ on the RHS instead of $\otimes^{\mathbf L}$. This only leaves the derived tensor product on the LHS, but I don't see a reason why that should be usual tensor product in this situation. (This could be a way to look for a potential counterexample.) – R. van Dobben de Bruyn May 9 '20 at 1:19
• By the way, this is true with no assumption on $f$ and $L$ (in fact, for any morphism of ringed spaces) if $\mathcal{F}$ is locally free. – abx May 9 '20 at 4:46

Based on my comment, I constructed the following counterexample (which I believe is standard):

Example. Let $$(E,O)$$ be an elliptic curve, let $$Y = E$$ and $$X = E \times E$$, with $$f \colon X \to Y$$ the first coordinate projection. Let $$\mathscr L = \mathcal O_{E \times E}(\Delta - E \times O)$$, and let $$\mathscr F = \mathcal O_O$$.

Then $$f_* \mathscr L = 0$$, since $$H^0(U \times E, \mathcal O_{U \times E}(\Delta|_U - U \times O)) = 0$$ for every open $$U \subseteq E$$ as $$\mathscr L|_U$$ is a (fibrewise) degree $$0$$ line bundle that is not trivial.

On the other hand, $$\mathscr L \otimes f^* \mathscr F = \mathcal O_{O \times E}$$ since $$\Delta|_{O \times E} = (O,O) = (E \times O)|_{O \times E}$$. The short exact sequence $$0 \to \mathcal O_X(-O \times E) \to \mathcal O_X \to \mathcal O_{O \times E} \to 0$$ gives a long exact sequence $$0 \to \mathcal O_E(-O) \to \mathcal O_E \to f_*\mathcal O_{O \times E} \to \mathcal O_E(-O) \to \mathcal O_E \to R^1f_* \mathcal O_{O \times E} \to 0$$ since $$R^if_* \mathcal O_X(-O \times E) = \mathcal O_E(-O)$$ for $$i \in \{0,1\}$$ by the usual (derived) projection formula. Thus, \begin{align*} & & & & f_*\big(\mathscr L \otimes f^*\mathscr F\big) = \mathcal O_O \neq 0 = f_*\mathscr L \otimes \mathscr F. & & & & \square \end{align*} Remark. What's going on is that $$Rf_* \mathscr L = \mathcal O_O[-1]$$ is not flat over $$Y$$ even though $$\mathscr L$$ is. There is a $$\mathscr Tor_1$$ term interfering in the derived projection formula of [Tag 0B54]. To see the above formula for $$Rf_* \mathscr L$$, use the short exact sequence $$0 \to \mathcal O_X(-E \times O) \to \mathscr L \to \mathscr L|_{\Delta} \to 0.\tag{1}\label{1}$$ Since $$\mathcal O_X(\Delta)|_\Delta = T_E = \mathcal O_E$$, we get $$\mathscr L|_\Delta = \mathcal O_E(-O)$$. Note that $$f$$ induces an isomorphism $$\Delta \to E$$, so the long exact sequence of \eqref{1} reads $$0 \to f_*\mathscr L \to \mathcal O_E(-O) \to \mathcal O_E \to R^1f_* \mathscr L \to 0.$$ Above we computed $$f_* \mathscr L = 0$$, so the map $$\mathcal O_E(-O) \to \mathcal O_E$$ is the natural inclusion, so $$R^1f_* \mathscr L = \mathcal O_O$$. $$\square$$

On the other hand, $$Rf_* \mathcal O_X = \mathcal O_E \oplus \mathcal O_E[-1]$$ is a complex of free modules, so the derived tensor product of the LHS of [Tag 0B54] is just a usual tensor product, as we saw implicitly in the computation of $$Rf_* \mathcal O_{O \times E} = Rf_* f^* \mathcal O_O$$ above.

• Dear Remy, I think your counterexample is right and thank you. However, I am actually in a special situation that I demand the $L$ to be a torsion line bundle on an abelian variety. The $L$ in your example is not a torsion line bundle. In this case if $f_*L\neq0$, then the projection formula is true by base change theorem. If $f_*L=0$, then i don't know how to do it. I wish to prove that $h^0$ of $L$ restricted to each fibre would be 0 but I couldn't. – Rick Sanchez May 9 '20 at 3:04
• Even if $\mathscr L = \mathcal O_X$ I am not sure what to expect. For a smooth morphism in characteristic $0$ it is a highly nontrivial theorem that $R^if_* \mathcal O_X$ is locally free for all $i$ (uses Hodge theory), but in positive characteristic this is unknown as far as I'm aware. (Warning: in the case above the derived pushforward $Rf_* \mathcal O_X$ splits because this is true for $R\Gamma(E,\mathcal O_E)$, but in general it will be a little more complicated and you need to think about a spectral sequence for Tor as you filter $Rf_* \mathscr L$ by cohomological degree.) – R. van Dobben de Bruyn May 9 '20 at 3:13
• That said, if your interest is specifically in characteristic $0$, then maybe the Hodge theory literature might be of help ― sometimes results for $\Omega^i_{X/Y}$ can be extended to torsion line bundles. I'm not super familiar with this, so I don't know any good references off hand. – R. van Dobben de Bruyn May 9 '20 at 3:19
• Let's work with a fibration betweem abelian varieties over $\C$. It needs not to be complicated. When $L=\mathcal{O}_X$, then $h^0$ is always 1. And you will have that the projection formula is true for any quasi-coherent $\mathcal{F}$ by base change. See Qing Liu's book algebraic geometry and arithmetic curves Remark 5.3.21(c) for example. When $L$ is only torsion and $f_*L\neq0$, it's similar since in this case $L$ restricted onto fibre is trivial. – Rick Sanchez May 9 '20 at 3:23
• I think a more precise version well deserves a new question, so that other people can share their thoughts! – R. van Dobben de Bruyn May 9 '20 at 3:29