Group of Hodge classes

Question

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. :-)

Answer 1

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.