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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?)

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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. –  ulrich Jul 5 '11 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? –  Vivek Shende Jul 5 '11 at 10:48
    
Sorry, I misread the "at least" as "equal". You are right (though I think the twist should perhaps be $k$?). –  ulrich Jul 5 '11 at 11:07
    
oh yes, I'll fix the twist –  Vivek Shende Jul 5 '11 at 11:35

2 Answers 2

up vote 6 down vote accepted

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.

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Ah, thanks for the counterexample. –  Vivek Shende Jul 5 '11 at 14:01

I'm not sure that I would call it a bundle if the fibre dimensions vary, but that's a minor quibble.

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.

Postscript: In view of ulrich's answer, the above decomposition seems to be the best one can hope for in general.

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