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The motivation for considering the Hodge filtration as opposed to just the grading is supposed to be that the filtration varies holomorphically in a family, whereas the grading does not always.

Can someone provide an example of a smooth morphism $f \colon X \to S$ such that the fibers are smooth projective complex manifolds where the grading on $ R^k f_* \mathbb{Z}$ does not vary holomorphically over $S$? Ideally, the example would be a family of abelian varieties.

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up vote 5 down vote accepted

Almost any interesting example that you can write down will satisfy this property. E.g. consider a non-trivial family of elliptic curves over some base $S$ (the family $y^2 = x(x-1)(x-\lambda)$ over $\mathbb P^1\setminus \{0,1,\infty\}$ will do).

Then you get a two-dimensional variation of Hodge structure $\mathcal V$ which is a rank two holomorphic bundle over $S$. The first step in the Hodge filtration gives a rank one subbundle $\mathcal F \subset \mathcal V$. This interpolates the $(1,0)$ part of the grading. It is spanned by the holomorphic differential $\omega := dx/2y$.

If we restrict to a small open subset $U$ of $S$, over which we can trivialize the family of elliptic curves in the smooth category, then we can choose a constant basis for the homology of the family, and integrate $dx/2y$ over it, to get a pair of functions $\omega_1$ and $\omega_2$, which are the periods of the elliptic curves in the family; they vary holomorphically, which reflects the fact that $\mathcal F$ is a holomorphic bundle. But if we integrate a basis for the $(0,1)$ part of the grading over this homology basis, we get the pair of anti-holomorphic functions $\overline{\omega}_1, \overline{\omega}_2$. If the family of elliptic curves is non-trivial, then these will really form non-constant holomorphic functions on $U$, implying that the $(0,1)$ part of the grading cannot be interpolated holomorphically.

Summary: The basic point is that the $(q,p)$ part of the grading is obtained from the $(p,q)$ part by applying complex conjugation, and this is not a process that will preserve the holomorphicity of a bundle in general. I.e. the Hodge filtration varies holomorphically but the conjugate filtration $\overline{F}$ does not, and hence neither does the intersection $F^p\cap \overline{F}^q$, in general.

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