# Which sheaves on a projective bundle are flat over the base scheme?

Assume $X$ is a noetherian scheme over $\mathbb{C}$ and $E$ a locally free sheaf of finite rank on $X$. Denote the the associated projective bundle by $f: \mathbb{P}(E)\rightarrow X$.

Are there any coherent sheaves on $\mathbb{P}(E)$ that are flat over $X$ except locally free ones?

I'm especially interested in such $G\in Coh(\mathbb{P}(E))$ with the property that the canonical morphism $f^{*}f_{*}G\rightarrow G$ is an isomorphism. For these $G$ we have $G_{|\mathbb{P}(E(x))}=\mathcal{O}_{\mathbb{P}(E(x))}^{r_x}$ for all closed points $x\in X$ and some $r_x\geq 1$. So we have $H^i(\mathbb{P}(E(x)),G_{|\mathbb{P}(E(x))})=\{0\}$ for all $i\geq 1$.

I'm trying to see that this implies $R^if_{*}G=0$ for all $i\geq 1$, for which flatness of $G$ over $X$ is needed. Maybe there is a version of the Cohomology and Base Change Theory that does not need a flatness assumption? Or is there another theorem which im implies the vanishing of the higher direct images in this case?

I asked this question in a similar way on Math.Stackexchange.

• There are many coherent sheaves on $\mathbb{P}(E)$ that are flat over $X$ yet not locally free. For instance, if $X$ is $\text{Spec}(\mathbb{C})$ itself, then every coherent sheaf on $\mathbb{P}(E)$ is flat over $X$. In another direction, for every section of $f$, $s:X\to \mathbb{P}(E)$, the coherent sheaf $s_*\mathcal{O}_X$ is flat over $X$. However, if a coherent sheaf $G$ is flat over $X$ and $f^*f_*G\to G$ is an isomorphism, then $G$ is of the form $f^*H$ for a locally free sheaf $H$ on $X$. Oct 5 '15 at 15:03
• Also, if you are just interested in sheaves such that $f^*f_*G\to G$ is an isomorphism, you can conclude that there are no higher direct images without needing flatness (in this situation). Namely, such sheaves are exactly sheaves of the form $f^*H$ for some $H$ (not necessarily locally free), and use the projection formula. Oct 5 '15 at 15:42
• Thank you both. The hint that such $G$ are of the form $f^{*}H$ was what I needed. This solves my problem completely. If someone will write a short answer, then I can accept it. Oct 5 '15 at 16:22