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. 



Is there an explicit example of a coefficient sheaf for which hard Lefschetz fails?
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.




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**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.


