If $X$ is a K3 surface over the complex (algebraic) then I wonder if the poincare pairing induces a polarization on the Hodge structure $H^2(X,\mathbb Z)$? The point is that I see that when constructing the Kuga-Satake abelian variety, people take the primitive cohomology in order to have a polarization on the induced Hodge structure, but I wonder if the reason they do not take the hole $H^2$ is because we do not have a polarization induced by the Poincare paring... I know that if this induces a polarization of the Hodge structure, then the Hodge decomposition is orthogonal but if this is not a polarization, is the Hodge decomposition stil orthogonal?


On the other hand, on etale cohomology, Faltings theorem give an analogous to the Hodge decomposition for (say) a $K3$-surface over a $p$-adic field $K$:


$H\_{et}^2(X\_{\bar K},Q\_p)\otimes \mathbb C\_p = H^2(X,\mathcal{O}\_X)(0) \oplus H^1 (X, \Omega^1)(-1) \oplus H^0(X,\cal O\_X)(-2)$ Is this decomposition orthogonal respect to the Poincare pairing on etale cohomology? 



Poincare pairing and polarization of Hodge structure.  Kuga-Satake construction.
The point is this. For a polarization on a weight $n$ Hodge structure $H$,
you need a bilinear form $\langle,\rangle$ so that $i^n\langle x, Cy\rangle$ is positive
definitie, where the Weil operator $C$ is multiplication by $i^{p-q}$ on $H^{pq}$.
If you take $H$ to be the primitive second cohomology $PH^2(X)$ of a K3 or any surface, then the standard intersection pairing will work thanks to the Hodge index theorem. 
However, on the full
cohomology $H^2(X)= PH^2(X)\oplus \mathbb{Q}(-1)$, the resulting pairing won't be positive
definite. If you want to use $H^2$ you certainly can, provided you change the sign of
the pairing on the second factor.


I'm afraid I don't know enough about the $p$-adic Hodge story, but I imagine the issues are similar.


