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Let $f :X \to Y$ be a submersion between smooth projective varieties over $\mathbb{C}$ and let $\alpha \in Z^k(X)$ be an algebraic cycle of $X$. Is is true that for all odd numbers $p$ and $q$ such that $p+q = 2k$, $\alpha$ lives in the kernel of $H^{2k}(X,\mathbb{Q}) \to H^p(Y,R^qf_*\mathbb{Q})$?

The reason why I ask this question is the following:

Since $f$ is a submersion, the class $[\alpha]$ in $H^p(Y,R^qf_*\mathbb{Q})$ can be represented by an element in

$$f^*\Omega^p_{Y,\mathbb{R}} \otimes \Omega^{q-p}_{X/Y,\mathbb{R}}.$$

However, it seems to me that any class which can be represented by one of above elements cannot be of type $(k,k)$. Can someone explain what happens here? Thank you.

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up vote 4 down vote accepted

No, $\alpha$ doesn't have to lie in that kernel. To construct an explicit counterexample, take a surface $X$ mapping to a curve $Y$, $k=1$, and let $\alpha$ be nontrivial divisor on $X$ not supported on the fibres such that $\alpha\cdot (fibre)=0$. Then $[\alpha]$ will be supported in $H^1(Y, R^1f_\ast\mathbb{Q})$. To make this more explicit, you can even take $X$ to be a product of $Y$ with another curve.

To resolve the seeming paradox, note that $R^qf_\ast\mathbb{Q}$ carries a variation of Hodge structure, and its cohomology $H^p(Y, R^qf_\ast\mathbb{Q})$ carries a a Hodge structure of weight $p+q$ compatible with the one on $H^*(X)$. The Hodge filtration is not completely obvious, and in particular, if $p,q$ are odd and $k=(p+q)/2$, then the $(k,k)$-part need not vanish. The details of the construction can be found in the beginning sections of Zucker, Hodge theory with degenerating coefficients..., Annals (1979).

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