# Cohomology of projective space bundles

Suppose $Y$ is an algebraic variety and $\mathcal{E}$ a coherent sheaf on $Y$. Suppose $f:X=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E})) \to Y$ is a morphism of algebraic varieties with all fibres scheme theoretically projective spaces.

If the fibres all had the same dimension, I would have $\mathrm{R}f_* \mathbb{C}_X = \mathrm{H}^*(\mathbb{P}^n) \otimes \mathbb{C}_Y$.

In the case that the fibre dimension varies, let $Y_k$ be the locus where the fibre dimension is at least $k$. Then is it true that $\mathrm{R}f_* \mathbb{C}_X = \bigoplus \mathbb{C} _{Y_k}$-2k$(k)$?

(and if not in general, are there any reasonable assumptions which make it true?)

• There seems to be a missing term in your formula since it does not specialize to the one in the case where the fibres are equidimensional.
– naf
Jul 5, 2011 at 9:44
• Really? If the fibres are of dimension $n$, then $Y_0=Y_1=...=Y_n = Y$ and (hopefully) the shifts and twists are put in the right places so that the $\mathbb{C}_{Y_k}$ give the cohomology of projective space? Jul 5, 2011 at 10:48
• Sorry, I misread the "at least" as "equal". You are right (though I think the twist should perhaps be $k$?).
– naf
Jul 5, 2011 at 11:07
• oh yes, I'll fix the twist Jul 5, 2011 at 11:35

I think what you want is true if $X$ and all the $Y_k$ are smooth (or have some very mild singularities e.g. quotient singularities) but I don't know many such examples. In general it appears to be false as shown by the following example:
Let $Y$ be the quadric cone given by $x_1x_2 - x_3x_4 = 0$ in $\mathbb{A}^4$. If we blow up the vertex the exceptional divisor is isomorphic to the quadric in $\mathbb{P}^3$ given by the same equation. $Y$ has a small resolution $f:X \to Y$ which is given (in the fibre over the vertex) by projecting the quadric onto one of its factors. The fibre of $f$ over the vertex is $\mathbb{P}^1$ and $f$ is a morphism of the type you want.
Since $f$ is birational and $X$ is smooth, it follows from Verdier duality that $Rf_*({\mathbb{C}}_X)$ is self dual (with a shift, depending on your conventions). However, one can see that the object in the derived category given by your formula is not self dual.
My guess is that if you apply the decomposition theorem of BBD (or the version for Hodge modules due to Saito), and play around a bit, then you would get something like $$\mathbb{R}f_*\mathbb{Q}=\bigoplus IC(\mathbb{Q}_{Y_k})(k)[-2k]$$ where $IC$ is the intersection cohomology complex normalized suitably. If the singularities of the $Y_k$ aren't too bad then it would reduce to what you are claiming.