most recent 30 from http://mathoverflow.net 2013-05-24T13:38:07Z http://mathoverflow.net/feeds/question/90664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90664/poincare-pairing-and-polarization-of-hodge-structure-kuga-satake-construction Yoyontzin 2012-03-09T06:35:34Z 2012-03-10T04:40:24Z

<p>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?</p> <p>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 $K$:</p> <p>$H_{et}^2(X_{\bar K},Q_p)\otimes \mathbb C_p = H^2(X,\mathcal{O}_X)(0) \oplus H^1 (X, \Omega^1)(-1) \oplus H^0(X,\cal O_X)(-2)$ Is this decomposition orthogonal respect to the Poincare pairing on etale cohomology? </p>

http://mathoverflow.net/questions/90664/poincare-pairing-and-polarization-of-hodge-structure-kuga-satake-construction/90690#90690 Donu Arapura 2012-03-09T12:35:27Z 2012-03-09T12:35:27Z

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