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.
 A: 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.
