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Let $X$ be a projective variety of dimension $d$ over $k=\bar{k},$ with $L$ an ample line bundle on $X$ and $\eta=c_1(L).$ Hard Lefschetz gives an isomorphism (see BBD) $$ \eta^i:IH^{d-i}(X)\to IH^{d+i}(X) $$ with Tate twist ignored, which, together with the intersection pairing between $IH^{d-i}$ and $IH^{d+i},$ gives a non-degenerate bilinear form $$ IH^n(X)\times IH^n(X)\to(\mathbb Q,\mathbb Q_{\ell},\text{ or }\mathbb C...) $$ for each $n.$

Question: Is it $(-1)^n$-symmetric?

This is so when $X$ is non-singular (which follows from the general fact on "cup products"), or when $n=d.$ The question is related to this MO question Poincaré duality for intersection cohomology. I guess one can probably figured it out by doing some homological algebra on the level of complexes (i.e. before taking hypercohomology groups), and maybe it's written down somewhere.

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up vote 5 down vote accepted

You are right that this symmetry follows from a similar formula on the complex level. To begin with $\eta^i$ is induced from multiplication by $c_1(\mathcal L)^i$ in $H^\ast(X)$ and the $H^\ast(X)$-module structure on the intersection cohomology. Hence your result will follow from the fact that the Poincaré pairing is a module pairing ($\langle xy,z\rangle=\langle y,xz\rangle$) and the symmetry for the pairing itself $\langle y,z\rangle=\pm\langle z,y\rangle$. Now, the module pairing property is equivalent to $IH^\ast(X)\rightarrow IH^\ast(X)[-2n]^\vee$ being a module map. This in turn follows from the fact that the module structure is just induced from the action of $K$ (=$\mathbb Q$,...) on the complex $\mathcal{IH}_X$ and the fact that the duality map $\mathcal{IH}_X\rightarrow D(\mathcal{IH}_X)[-2n]$ is $K$-linear. Finally, the symmetry of the Poincaré pairing follows from the symmetry of the duality map $\mathcal{IH}_X\rightarrow D(\mathcal{IH}_X)[-2n]$. This latter fact is most easily seen by noting that any endomorphism $\mathcal{IH}_X\rightarrow\mathcal{IH}_X$ is determined by its restriction to the non-singular locus of $X$ and there it is, by the symmetry in the smooth case, equal to the identity map.

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Thanks, Torsten. Your answer seems to suggest that this commutativity also holds more generally, for symmetrically (or alternatively, if we change the sign appropriately) self-dual perverse sheaves for which hard Lefschetz applies (e.g. $\iota$-pure ones in char. $p,$ or those that underlie Hodge modules in char. 0). Am I right? – shenghao Mar 21 '11 at 2:09
Yes, most of the time. It depends on the self-duality in question. Note that I had to prove that the self-duality of $\mathcal{IH}_X$ was symmetric. Most of the time this will also be true for other self-dualities but it is not automtic. – Torsten Ekedahl Mar 21 '11 at 5:08
Just to take some silly example, if one modifies a symmetric self-duality by multipliying by a non-zero scalar one will in general not get a symmetric one. This example can be made worse by taking direct sums of examples using different scalars. – Torsten Ekedahl Mar 21 '11 at 11:41
Yes, but for an irreducible perverse sheaf, which is self-dual, isn't it either symmetrically or skew-symmetrically self-dual? I need to check the details, but it seems that one reduces to the local system, which is a self-dual irreducible representation of $\pi_1$ of some stratum, hence is either symmetric or skew-symmetric according to Schur's lemma, and so also is its $j_{!*}.$ – shenghao Mar 23 '11 at 13:47
You may be forced to extend scalars in order to be able to extract a square root but otherwise I agree. – Torsten Ekedahl Mar 23 '11 at 14:31

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