# intersection pairing on intersection cohomology

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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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.
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
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