Is there an explicit example of a coefficient sheaf for which hard Lefschetz fails? Recall that if $X$ is a projective algebraic complex manifold, $L$ is a semisimple $\mathbb C$-local system on $X$ of geometric origin (roughly speaking, this means that $L$ is a cohomology sheaf $R^if_*\mathbb C$ for some algebraic morphism $f:Y\to X;$ see BBD for the precise definition), and $\eta\in H^2(X,\mathbb C)$ is an ample class, then 
$$
\eta^i\cup-:H^{\dim X-i}(X,L)\to H^{\dim X+i}(X,L)
$$
is an isomorphism. This also holds when the projective variety $X$ is allowed to have singularities and $L$ is a perverse sheaf (again semisimple of geometric origin), appropriately shifted.
I'd like to know an example for which this fails; of course, $L$ is no longer of geometric origin. Over finite field, as long as $L$ is assumed semisimple, hard Lefschetz always holds, and conjecturally $L$ is of geometric origin. 
 A: The good news (or is it bad news?) is that there is no counterexample in the semisimple case. Simpson proved that hard Lefschetz holds for any semisimple local system on a smooth complex projective variety. To be more precise, apply theorem 1 and lemma 2.6 of his paper on Higgs bundles and local systems. More recently, Sabbah and Mochizuki have extended this to some (or perhaps all, I can't quite remember) semisimple perverse sheaves in accordance with a conjecture of Kashiwara.

Addendum  Here is  a counterexample for nonsemisimple local systems (which was something that I had wondered about  myself). Let $X$ be a smooth projective curve with genus $g>1$. Let $\pi=\pi_1(X)$.
Let $\mathbb{Q}_\rho$ denote a nontrivial rank one $\pi$-module with character $\rho$,
and $\mathbb{Q}$ the trivial module. We can see, using Euler characteristics, that
$$Ext^1_\pi(\mathbb{Q},\mathbb{Q}_\rho)\cong H^1(X,\mathbb{Q}_\rho)\not=0$$
Thus we can form a nonsplit extension
$$0\to \mathbb{Q}_\rho \to L\to \mathbb{Q}\to 0$$
We necessarily have $H^0(X,L) = H^0(\pi, L)=0$. On the other hand, by Poincaré duality
$$H^2(X,L) = H^0(X,L^*)^*= H^0(\pi,L^*)^*\not=0$$
So $H^0(X,L)\not= H^2(X,L)$, i.e. hard Lefschetz fails.
