# Cohomology of a hypersurface in a projective bundle

Let $Y$ be a smooth hypersurface in the total space of a projective bundle $\pi:\mathbb{P}(\mathscr{E})\to B$ (here $\mathbb{P}(\mathscr{E})$ denotes the projective bundle of lines in the vector bundle $\mathscr{E}\to B$ and we are working over $\mathbb{C}$). Then $[Y]=aH+D\in A_*\mathbb{P}(\mathscr{E})$, where $H=c_1(\mathscr{O}_{\mathbb{P}(\mathscr{E})}(1))$, $a\in \mathbb{N}$ and $D=\pi^*D'$ for some divisor $D'$ in $B$. I would like to relate the cohomology of $Y$ to the cohomology of $\mathbb{P}(\mathscr{E})$ via the Lefschetz hyperplane theorem but the problem is that $Y$ is not an ample divisor as is required to apply Lefschetz. I have looked at cases where the cohomology of Y is known and in these cases only the middle cohomology differs from that of $\mathbb{P}(\mathscr{E})$ just as if $Y$ were in fact an ample divisor. And after going through the proof of the hyperplane theorem, if $V$ is a hypersurface in a complex manifold $M$ what's really needed is not necessarily the ampleness of the line bundle corresponding to $V$, but we need

$$\tag{\dagger} H^q(M,\Omega_{M}^p(-V))=H^q(V,\Omega_{V}^{p-1}(-V))=0$$ for $p+q<\text{dim}(M)$ (which is implied by ampleness via Kodaira's vanishing theorem). So what I'm thinking is that if $Y$ is a hypersurface in a projective bundle then perhaps condition ($\dagger$) is still satisfied even though $Y$ is not an ample divisor, and this could explain why Lefschetz holds in the known cases I am familiar with. As such, I would like to know if anyone knows whether or not a divisor corresponding to a hypersurface in a projective bundle satisfies condition ($\dagger$), and more generally if there is any sort of criterion for determining if a divisor satisfies condition ($\dagger$). Thank you in advance.

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Does the Leray spectral sequence + Lefschetz hyperplane theorem on every fiber help? –  Piotr Achinger Jun 1 '12 at 19:11
There could be singular fibers so perhaps not. –  James Fullwood Jun 4 '12 at 16:20