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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?

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  • $\begingroup$ For any smooth projective variety $X$, the cup product on cohomology is a morphism of Hodge structures $H^*(X) \otimes H^*(X) \to H^*(X)$ and the same holds in the $p$-adic setting. This should imply the orthogonality you want (but I am not sure exactly what you mean by that). $\endgroup$
    – naf
    Commented Mar 10, 2012 at 8:11
  • $\begingroup$ what do you mean by "the same hodls in the p-adic setting"? we do not have a Hodge strcuture anymore on the p-adic case since we do not have an action of $'mathfrac C^*$. I mean that since the cup product on $H_{et}(X_{\bar K,\mathbb Q_p))$ is bilinear pairing its extension to $H_{et}(X_{\bar K,\mathbb Q_p)\otimes \mathb C_p$ is also biiinear and we have the notion of orthogonal vector spaces over $\mathbb C_p$. I wonder if the decomposition given by Faltings is orthogonal. $\endgroup$
    – Rogelio Yoyontzin
    Commented Mar 11, 2012 at 21:43
  • $\begingroup$ Well, by the "same" in the p-adic setting I mean that the cup product is a map of Galois modules. Knowing this and the Galois module structure of $H^2_{et}(X_{\bar{K}},Q_p)$ gives restrictions on the Poincare duality pairing with respect to the Hodge-Tate decomposition of Faltings. (If you mean by "orthogonality" that the cup product of elements in distinct summands is zero, then this is not true (but neither does this hold for the Hodge decomposition over $\mathbb{C}).) $\endgroup$
    – naf
    Commented Mar 12, 2012 at 5:49
  • $\begingroup$ Good, thank you. However if I have a polarization on a Hodge structure, then the Hodge decomposition is orthogonal. I understand the poincare pairing dose not give a polarization on all $H^2(X,/mathbb Q)$ and this is why you said it is not true that we hava an ortogonal decomposition on $H^2(X,/mathbb \mathbb C)= H^{2,0}/oplus H^{1,1}\oplus H^{0,2}$ right? what is a good reference for this on the p-adic settings for etale cohomology? Thanks once again. $\endgroup$
    – Rogelio Yoyontzin
    Commented Mar 12, 2012 at 18:18
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    $\begingroup$ No, even on the primitive cohomology the pairing is not orthogonal in the naive sense. What one gets is that the pairing of $H^{2,0}$ with itself is trivial, similarly for $H^{0,2}$, the pairing on $H^{2,0} \otimes H^{0,2}$ is non-degenerate as is the pairing on $H^{1,1}$ with itself and $H^{1,1}$ is orthogonal to the other summands. This is also what happens in the $p$-adic setting. (Over $\mathbb{C}$ you can change the pairing to a Hermitian one using complex conjugation and this will be orthogonal in the usual sense, but I don't think there is any $p$-adic analog of this.) $\endgroup$
    – naf
    Commented Mar 13, 2012 at 5:06

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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.

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  • $\begingroup$ I wonder if the same happend for etale cohomology... $\endgroup$
    – Rogelio Yoyontzin
    Commented Mar 11, 2012 at 21:47

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