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Timeline for Group of Hodge classes

Current License: CC BY-SA 3.0

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Nov 26, 2013 at 15:11 vote accept Bryan261
Nov 26, 2013 at 15:05 answer added Dan Petersen timeline score: 3
Nov 26, 2013 at 13:16 review Close votes
Nov 26, 2013 at 14:35
Nov 26, 2013 at 13:06 comment added Bryan261 I'm from a foreign country, i don't speak well engish. sorry .. $ X $ and $ Y $ are subvarieties of a smooth projective variety $ M $ such that $ M = X \bigcup Y $. I would like to know if we can construct a short exact sequence $ 0 \to \mathrm{Hdg} ( X \bigcup Y ) \to \mathrm{Hdg} ( X ) \oplus \mathrm{Hdg} ( Y ) \to \mathrm{Hdg} ( X \bigcap Y )$. Thanks a lot.
Nov 26, 2013 at 12:59 comment added Dan Petersen I voted to close as "unclear" - I don't think the question makes any sense as written. Are $X$ and $Y$ subvarieties of some other variety? Open or closed? Constructible?
Nov 26, 2013 at 12:55 comment added jmc Actually, it is not so clear to me what you want with this question. Could you sketch more context? For example, do you consider my previous comment trivial, or is that the sort of thing you are looking for. If you think it is trivial, then maybe you can include it as a sort of context/introduction in your question.
Nov 26, 2013 at 12:54 comment added jmc If $f \colon X \to Y$ is a morphism of smooth projective varieties, then the induced map $f^{*} \colon H(Y, \mathbb{Q}) \to H(X, \mathbb{Q})$ is a morphism of Hodge structures. So $H^{2k}(Y, \mathbb{Q})$ mapst to $H^{2k}(X, \mathbb{Q})$, and $H^{k,k}(Y)$ maps to $H^{k,k}(X)$. Therefore, there is an induced map $f^{*} \colon \textrm{Hdg}_{k}(Y) \to \textrm{Hdg}_{k}(X)$.
Nov 26, 2013 at 12:46 review First posts
Nov 26, 2013 at 12:47
Nov 26, 2013 at 12:27 history asked Bryan261 CC BY-SA 3.0