Are induced morphisms on cohomology strict with respect to the hodge filtration in the non Kähler case?

For a complex manifold $X$ there is the Hodge filtration on cohomology, induced by the filtration on the complex of holomorphic forms given by:

$$F^r\Omega_X^p:=\begin{cases}\{0\}\qquad\text{if }r>p\\\Omega^p_X\qquad \text{if } r\leq p. \end{cases}$$

If $X$ is compact Kähler, Hodge Theory tells us that this filtration induces a pure Hodge structure of weight $k$ on the $k$-th cohomology group. This implies, a posteriori, that morphisms respecting the hodge filtration, as given for example by the morphism $f^*$ induced on cohomology by a holomorphic map f, automatically respect it strictly, i.e. $F^r\cap im f^*=f^*F^r$.

This strictness is no longer given if we look at arbitrary maps between filtered vector spaces.

My question is if strictness still holds for morphisms induced by geometric maps if we drop the Kähler condition (and/or compactness). If not can you provide a counterexample? Is there a more general condition on the manifolds/maps that ensures strictness of induced maps?

• The pullback of a complex $k$ form of type $(p,q), p+q=k,$ on any complex manifold via any holomorphic map will still be of type $(p,q)$. – YangMills Sep 15 '16 at 13:41
• true, but I think this only implies that the filtration is respected, not that this is strictly so. – jorst Sep 15 '16 at 13:49
• yes, you are right, sorry. – YangMills Sep 15 '16 at 14:49

Let $X$ be a smooth projective variety and $D \subset X$ a very ample divisor. Since $U:= X- D$ is affine, the filtration $F^\bullet H^k(U,\mathbf{C})$ induced by the filtered complex in your question is trivial in the sense that $F^p H^k(U,\mathbf{C}) = H^k(U,\mathbf{C})$ for $p \le k$ and $F^p H^k(U,\mathbf{C}) = 0$ for $p > k$.
So if the pullback $i^* : H^k( X,\mathbf{C}) \to H^k(U,\mathbf{C})$ induced by $i : U \hookrightarrow X$ respect strictly the filtration $F^\bullet$, it would imply $\mathrm{Im}(i^*) = i^*(F^kH^k( X))$. The latter is false in general: take for example a variety $X$ such that $b_2(X) > 1$, $h^{2,0}(X) = 0$ and $D$ is irreducible. We have readily $i^*(F^2H^2( X)) = 0$, but since $\ker(i^*) \simeq H^0(D,\mathbf{C})$, for dimensional reason $H^2(X,\mathbf{C})\to H^2(U,\mathbf{C})$ is nonzero.
By the way, normally we don't call the filtration in your question the "Hodge filtration" for any complex manifold. For example, for an open variety $U$ which is the complement of the simple normal crossing divisor $D$ in a compact Kähler manifold $X$, what we call the Hodge filtration on $$H^k(U,\mathbf{C}) \simeq \mathbb{H}^k(X,\Omega^\bullet_X(\log D))$$ is the one induced by the naïve filtration on the logarithmic de Rham complex $\Omega^\bullet_X(\log D)$. Already when $U$ is affine, this filtration can be nontrivial, contrary to what happens to the filtration described in your question.
• Do you know an example where the naive filtration on the log $D$ de Rham complex on $X$ doesn't coincide with the naive filtration on the (usual) de Rham complex of $U$ (I think I can see in principle how it could happen)? I think taking the easiest examples $U = \mathbb{A}^1$ or $\mathbb{G}_m$ doesn't seem to work – Harrison Chen Apr 14 '17 at 4:50
• You can take $U = X-D$ where $X$ is smooth projective of dimension $n$ and $D$ is a very ample smooth divisor such that $j: H^{n-1}(D) \to H^{n+1}(X)$ is not injective. On the one hand since $U$ is affine, the filtration on $H^n(U)$ induced by the naive filtration on the classical de Rham complex is trivial. On the other hand, the naive filtration on the log-complex defines a mixed Hodge structure (MHS) on $H^n(U)$. Since the residue map $r: H^n(U) \to H^{n-1}(D)$ is a morphism of MHS's and since $Im(r) = \ker (j) \ne 0$, we see that the Hodge filtration on $H^n(U)$ is non-trivial. – HYL Apr 14 '17 at 6:13