# 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.

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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.
Thanks for the answer, Donu. A somehow related question: what about varieties over $\overline{\mathbb F}_p?$ Namely $X$ is projective (smooth or not) and $L$ is semisimple local system (or perverse sheaf). Without being of geometric origin, they won't have a model over a finite subfield, so that the usual proof breaks. – shenghao Aug 5 '12 at 15:57