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Let $f:X \to Y$ be a proper morphism between smooth algebraic varieties (say over $\mathbb{C}$), let me write $A_X$ for the constant sheaf on $X$ with coefficients of the appropriate type. Then the decomposition theorem of [Beilinson-Bernstein-Deligne] tells me that $f_* A_X$ splits as a direct sum of shifted semisimple perverse sheaves.

So suppose I have collected some of the summands, say into $\mathcal{F}$, and want to know if I have found everything. Invoking proper base change, it is enough to ask, for every point $y \in Y$, whether

$\dim H^*(y,\mathcal{F}_y) = \dim H^*(X_y)$,

since any missing summands would have to contribute something somewhere. But computing the RHS requires actually thinking about the topology of $X_y$. Something which is easier to compute are the virtual Betti numbers, defined for an arbitrary variety in terms of the weight filtration as

$b^j_{vir}(Z) = \sum_i (-1)^{i+j} \dim Gr^j_W H^i_c(Z)$

At first glance these may not look easier, but the point is that they are motivic, i.e., additive under cutting into constructible pieces (and multiplicative under Zariski locally trivial fibrations); evidently they agree with the ordinary Betti numbers for smooth proper varieties. Thus for instance since a rational curve with a single node can be reassembled into $\mathbb{A}^1$, it has a single nonvanishing virtual Betti number, $b^2_{vir} = 1$.

To determine whether $f_* A_X = \mathcal{F}$, the latter known to be a summand of the former, is it enough to compare virtual Betti numbers? To be precise, does it suffice to know, for every $y \in Y$ and all $j$, that $b^j_{vir}(X_y) = \sum_i (-1)^{i+j} \dim Gr^j_W H^i(y,\mathcal{F}_y)$?

I sketch an argument in the affirmative which unfortunately due to my lack of knowledge about what weights are, I cannot seem to make precise. It is enough to show that for any nonzero summand $G$ of $f_* A_X$, there is some $y \in Y$ where not all the virtual Betti numbers of $G_y$ vanish. Let me take $y$ which has a neighborhood where $G$ is a sum of shifted local systems supported on some smooth subvariety $Y'$, and here it seems to me that restricting to $y$ must act as some global shift on the weights. Since $G$ was pure to begin with, being a summand of $f_* A_X$, so is $G_y$ (up to a shift) and hence its virtual Betti numbers are its actual Betti numbers, which see $G$.

I have put the "reference request" tag in the hopes that if the answer to the question is yes, it is some well known thing, maybe somewhere in [BBD], to which someone can point me.

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Maybe I'm missing something here, but haven't you reduced to the case where $G$ is a sum of (pure) variations of Hodge structure (shifted into appropriate degree) on some neighborhood of $y$ in $Y^\prime$. Taking the stalk of such a VHS at $y$ is a pure Hodge structure. I don't see where a change in weights comes in. –  Sam Gunningham Jul 9 '11 at 3:01
    
Ah, I see now you wanted to work in the BBD setting... –  Sam Gunningham Jul 9 '11 at 3:02
    
I'm happy to work in any setting where the answer to the question is yes. If it's true in the MHM setting that's certainly good enough for me... I just have even less idea how those work than the BBD formalism. –  Vivek Shende Jul 9 '11 at 4:15
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1 Answer

(I'll just note beforehand that is this is a "I don't know the answer to your actual question, perhaps you'd rather know the answer to a slightly different one" answer). The answer is yes if you replace virtual Betti number with Frobenius trace; these are also motivic, if admittedly slightly harder to calculate, as they involve counting points. This is covered, for example, in Section 3.4 of my second paper with Geordie Williamson.

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Thanks Ben, that's indeed something I wanted to know as well. –  Vivek Shende Jul 9 '11 at 4:25
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