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Say $X$ is a smooth algebraic variety, $U$ is a Zariski open set in $X$, $L$ is a local system on $U$, and $IC(L)$ the intersection cohomology sheaf on $X$ which restricts to $L$ on $U$. Then is:

$$\dim L_u \ge \sum_i \dim \mathrm{H}^i(\mathrm{IC}(L)_x) $$

for $u \in U$ and all $x \in X$?

If not, is it true after e.g. putting in some scalar depending only on the dimension of $X$?

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up vote 2 down vote accepted

The answer to the first question is certainly "no": you can easily find example of X whose IC complex (for the trivial local system of rank 1) has (total) stalk of dimension $\geq 1$ (for instance this is true for most nilpotent orbit closures in a simple Lie algebra).

Edit: Sorry I missed the assumption that X is smooth. But still I think the following is a counterexample. First, let $Y$ in ${\mathbb C}^n$ be a generic homogeneous hypersurface of degree d (it is smooth away from 0). Let $Z$be its projectivization. Then if I am not mistaken, the stalks of the IC sheaf of $Y$ at $0$ live in dimensions $-(n-1)$ and $-(n-2)$ and they are equal to $H^0(Z)$ and $H^1(Z)$ respectively (I am using perverse normalization). Now take $n=3$. Then $Z$ is a curve of degree $d$ in $\mathbb P^2$, so its genus is $g=\frac{(d-1)(d-2)}{2}$ and its $H^1$ has dimension $2g$. Now there is a finite map $\pi:Y\to \mathbb C^2$ of degree $d$ (take for example $Y$ to be given by the equation $x^d+y^d+z^d=0$ and consider the projection to $(x,y)$). Then $\pi_*$ of the constant sheaf is going to be equal to its Goresky-Macpherson extension from an open subset, where it will be equal to a local system $L$ of rank $d$. But the sum on the right hand side of your expression for $x=0$ is $1+(d-1)(d-2)$.

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All such X which I know, e.g. the nilpotent orbit closures you are talking about, are singular. If X is smooth the IC complex of the trivial local system on a Zariski open is certainly the constant sheaf on X! – Vivek Shende Feb 10 '12 at 22:20
Oh, sorry, I missed the assumption that $X$ is smooth – Alexander Braverman Feb 10 '12 at 22:25
I edited my answer: I think now I can construct and honest counterexample. – Alexander Braverman Feb 10 '12 at 22:59
Why should $\pi_*(IC(Y))$ be just the extension of the constant sheaf? BBD tells you that this extension is in there, but a priori couldn't there could be other summands supported inside $x^d + y^d = 0$? (I'll try actually computing the IC in this example though and see...) – Vivek Shende Feb 11 '12 at 0:57
by "extension of the constant sheaf" I meant "extension of the restriction to the open locus" which is of course not a constant sheaf. – Vivek Shende Feb 11 '12 at 0:58

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