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Let $X$ be a projective complex algebraic variety of dimension $d.$ Does it make sense to ask if properties like $$ (x,y)=(-1)^i(y,x) $$ holds, for $x\in IH^i(X,\mathbb Q)$ and $y\in IH^{2d-i}(X)?$ And if it makes sense, is it true?

Edit: For the first question, my concern is that, if the self-duality $D_XIC_X\cong IC_X$ of the intersection complex only ensures that $IH^i(X)$ and $IH^{2d-i}(X)$ are dual, without specifying a particular duality, then it makes no sense to ask if the "product" is graded-commutative. Of course, if it does give a particular isomorphism, then it makes sense.

Edit: As I learned from Gabber, the answer is yes (of course I'd take the responsibility for misunderstanding his comment if it turns out to be ...), and it follows from the symmetric pairing $$ IC_X[-d]\otimes IC_X[-d]\to\mathbb Q_{\ell} $$ normalized so that on the smooth locus, it is the natural identification. I'll be appreciated if anyone can give a reference on this.

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I'm not sure it does make sense unless $i=d$ (even for ordinary cohomology). In this middle case, I would imagine it is true, although I don't have a reference on hand. –  Donu Arapura Mar 7 '11 at 22:52
Dear Donu Arapura: Thank you for the comment. Unless I miss something, isn't the cup product for singular cohomology always graded-commutative? Especially in this situation where we have de Rham isom: exterior product of diff forms is graded-comm. I'll be greatly appreciated if you could let me know when you have a reference or a proof on this. Thanks! –  shenghao Mar 8 '11 at 12:27
Dear Shenghao, you're right, not sure what I was thinking.... –  Donu Arapura Mar 8 '11 at 13:06
It seems to me that you are almost answering your own question: On the smooth locus the pairing coincides with the ordinary Poincare duality isomorphism: If $U$ denotes the smooth locus of $X$, then $IC_{U}=\mathbb{Q}_{U}[d]$ and so the duality isomorphism from $IC_X$ to $IC_X^{\wedge}$ is just the unique morphism in $Hom(IC_Y, IC_Y^{\wedge})=\mathbb{Q}$ extending the duality isomorphism for $\mathbb{Q}_{U}[d]$. You may want to see the and the references in there. –  J.C. Ottem Mar 10 '11 at 15:48
Thanks, Ottem, for giving me this paper, which is very helpful. –  shenghao Mar 11 '11 at 14:38

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