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I would like to know if there exists a relation between $ \mathrm{Hdg}_k(X \bigcup Y ) $, $ \mathrm{Hdg}_k(X) $, $ \mathrm{Hdg}_k( Y) $ and $ \mathrm{Hdg}_k(X \bigcap Y )$ ( short exact sequence, or direct sum or something like that ), such that $ \mathrm{Hdg}_k ( X ) = H^{k,k} ( X ) \bigcap H^{2k} ( X , \mathbb{Q} ) $ is the group of Hodge classes. Can you tell me if you know some references about this subject ? Thanks a lot. :-)

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  • $\begingroup$ 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 $f^{*} \colon \textrm{Hdg}_{k}(Y) \to \textrm{Hdg}_{k}(X)$. $\endgroup$
    – jmc
    Nov 26, 2013 at 12:54
  • $\begingroup$ 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. $\endgroup$
    – jmc
    Nov 26, 2013 at 12:55
  • $\begingroup$ 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? $\endgroup$ Nov 26, 2013 at 12:59
  • $\begingroup$ 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. $\endgroup$
    – Bryan261
    Nov 26, 2013 at 13:06

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I'll assume that $X$ and $Y$ are open subvarieties of $M$, so that this looks like a Mayer-Vietoris sequence; I think this is the only sensible interpretation of the question. Since $X$ and $Y$ are not going to be projective I also interepret $\newcommand{\Hdg}{\operatorname{Hdg}}\Hdg(-)$ in the sense of mixed Hodge theory, that is, $\Hdg^k(X)$ denotes the Hodge classes in $\mathrm{gr}^W_{2k} H^{2k}(X)$. Once we're working in mixed Hodge theory there's no longer any reason to assume $M$ smooth and projective either, so let's just consider an arbitrary algebraic variety covered by two Zariski opens.

Then there is the usual Mayer-Vietoris sequence $$ \ldots \to H^n(M) \to H^n(X) \oplus H^n(Y) \to H^n(X \cap Y) \to H^{n+1}(M) \to \ldots$$ where the morphisms are maps of mixed Hodge structures and in particular restrict to $$ \Hdg^k(M) \to \Hdg^k(X) \oplus \Hdg^k(Y) \to \Hdg^k(X \cap Y). $$ However contrary to your guess it will usually not be exact on the left. For instance, take $M = \mathbf P^1$, and $X, Y$ the standard covering by two copies of $\mathbf A^1$, and $k=1$. On the other hand, if $M$ is smooth then it is exact on the right (i.e. $\Hdg^k(X) \oplus \Hdg^k(Y) \to \Hdg^k(X \cap Y)$ is onto) simply because $H^{2k+1}(M)$ has weights $\geq 2k+1$. More generally, if $U \subset V$ is an open immersion of smooth varieties then $W_nH^n(V) \to W_nH^n(U)$ is onto.

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  • $\begingroup$ I'm sorry, can I formulate my question differently ?! sorry, because i need to have $ \mathrm{Hdg}_k ( X ) \oplus \mathrm{Hdg}_k ( Y ) \to \mathrm{Hdg}_k ( X \bigcup Y ) $ surjective. In other words : do we have $ \mathrm{Hdg}_k ( X \bigcap Y ) \to \mathrm{Hdg}_k ( X ) \oplus \mathrm{Hdg}_k ( Y ) \to \mathrm{Hdg}_k ( X \bigcup Y ) \to 0 $ a exact sequence ? Thanks a lot. $\endgroup$
    – Bryan261
    Nov 26, 2013 at 15:31
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    $\begingroup$ If you think about the example I already gave you with $\mathbf P^1$ it will be clear that no exact sequence of the kind you want can exist. $\endgroup$ Nov 26, 2013 at 18:12
  • $\begingroup$ Thank you very much. :-) $ \\ $ Can you tell me, please, in which books can i find somethings about this subject ? Thank you. $\endgroup$
    – Bryan261
    Nov 26, 2013 at 18:29

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