Timeline for Poincare pairing and polarization of Hodge structure. Kuga-Satake construction.
Current License: CC BY-SA 3.0
11 events
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| Mar 13, 2012 at 5:06 | comment | added | naf | 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.) | |
| Mar 12, 2012 at 18:18 | comment | added | Rogelio Yoyontzin | 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. | |
| Mar 12, 2012 at 5:49 | comment | added | naf | 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}).) | |
| Mar 11, 2012 at 21:49 | vote | accept | Rogelio Yoyontzin | ||
| Mar 11, 2012 at 21:44 | vote | accept | Rogelio Yoyontzin | ||
| Mar 11, 2012 at 21:45 | |||||
| Mar 11, 2012 at 21:43 | comment | added | Rogelio Yoyontzin | 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. | |
| Mar 10, 2012 at 8:11 | comment | added | naf | 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). | |
| Mar 10, 2012 at 4:40 | history | edited | Rogelio Yoyontzin | CC BY-SA 3.0 |
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| Mar 9, 2012 at 19:20 | vote | accept | Rogelio Yoyontzin | ||
| Mar 9, 2012 at 19:20 | |||||
| Mar 9, 2012 at 12:35 | answer | added | Donu Arapura | timeline score: 8 | |
| Mar 9, 2012 at 6:35 | history | asked | Rogelio Yoyontzin | CC BY-SA 3.0 |