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](https://mathoverflow.net/questions/57701/poincare-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.



intersection pairing on intersection cohomology
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.


