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Let $X$ be a proper singular variety over $k=\overline{\mathbb F}_p,$ irreducible of dimension $d.$ Let $H^*(X)$ and $IH^*(X)$ be the $l$-adic cohomology groups and $l$-adic intersection cohomology groups of $X,$ resp. Then, is the natural map $H^*(X)\to IH^*(X)$ compatible with the cup-product on $H^*(X)$ and the intersection product on $IH^*(X)?$

Background and Motivation: Given $X_0/\mathbb F_q$ an $\mathbb F_q$-structure of $X,$ one deduces a Galois action on $H^*(X)$ and $IH^*(X),$ with respect to which they are "mixed" (the 2nd one being pure), and the weight filtrations $W$ on both of them are independent of the choice of $X_0.$ One has a natural morphism $$ H^n(X) \to IH^n(X) $$ which factors as $$ Gr^W_n H^n(X) \hookrightarrow IH^n(X), $$ and this turns out to be injective.

As $X$ is singular, Poincaré duality might fail, i.e. the cup-product $$ H^i(X)\otimes H^{2d-i}(X) \to H^{2d}(X), $$ which is Galois equivariant, may be degenerate. This is the case when $H^i(X)$ (resp. $H^{2d-i}(X)$) is not pure of weight $i$ (resp. $2d-i$) for the reason of Galois, and $W_{i-1}H^i(X)$ (resp. $W_{2d-i-1}H^{2d-i}(X)$) is contained in the left kernel (resp. right kernel) of the cup-product pairing. I would like to know if this is the only obstruction for Poincaré duality to hold, namely they are exactly the left/right kernel.

Since the intersection pairing $$ IH^i(X)\otimes IH^{2d-i}(X)\to\mathbb Q_l(-d) $$
is perfect (I don't know if the pairing has a geometric definition in char. $p$ --- $D_XIC_X\simeq IC_X$ is the only reason I know), if my question in the beginning has a positive answer, then the cup-product pairing on $Gr^W_*H^*(X)$ will be perfect.

Correction: The argument above for non-degeneracy on $Gr^W_*H^*(X)$ is wrong. Here's a counter-example. Let $X$ be the projective cone of a projective smooth curve of genus $g,$ either over char. 0 or $p.$ Then $H^1(X)=0$ but $H^3(X)$ is of dimension $2g$ and pure of weight 3.

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up vote 3 down vote accepted

I think the answer to your question is yes, at least if $X$ is irreducible.

One may think of cup product in cohomology arising from the obvious pairing in the derived category $\mathbb{Q}_{l,X} \otimes^L \mathbb{Q}_{l,X} \to \mathbb{Q}_{l,X}$. The pairing on intersection cohomology comes from a pairing $IC_X \otimes^L IC_X \to \omega_X$, where $\omega_X$ is the dualising complex on $X$. There are natural maps $\mathbb{Q}_{l,X}[d] \to IC_X$ and $\mathbb{Q}_{l,X}[2d] \to \omega_X$; the first comes from the construction of $IC_X$ and the second, by adjunction, from the natural surjective map $H^{2d}(X) \to \mathbb{Q}_l(-d)$. This gives rise to a commutative diagram involving the two pairings which gives the commutativity at the level of cohomology that you wanted. (Note that one can check the commutativity of the diagram after restricting to the smooth locus of $X$.)

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Thanks ulrich. So the claim holds on the level of complexes, as in my previous question… – shenghao Jun 17 '11 at 11:15

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