# Exact sequence of relative differential forms

Let $$f : Y \to X$$ be a morphism of projective manifolds, with $$(X,D)$$ log smooth. Consider the exact sequence which defines the relative (logarithm) cotangent bundle (where $$\Delta := f^{\ast} D$$):

$$0 \to f^{\ast}\Omega_X^1(\log D) \to \Omega_Y^1(\log \Delta) \to \Omega_{Y/X}^1(\log \Delta) \to 0.$$

Let $$\mathscr{M}$$ be a line bundle on $$Y$$ with $$H^0(Y, \mathscr{M}^m) \neq 0$$, i.e., some multiple of $$\mathscr{M}$$ admits a section. Let $$\psi : Z \to Y$$ be a desingularization of the finite cyclic covering associated to taking roots of a section $$\sigma \in H^0(Y, \mathscr{M}^m) \backslash \{ 0 \}$$; so $$H^0(Z, \psi^{\ast} \mathscr{M}) \neq 0$$ also.

Set $$h : = f \circ \psi$$ and consider the pullback of the above exact sequence. I want to show that this induces a (decreasing) filtration $$\mathscr{F}_k^{\bullet}$$ ($$k \geq 0$$) of the bundle $$\psi^{\ast} \Omega_Y^{\bullet}(\log \Delta))$$ with

$$\mathscr{F}_k^{\bullet}/\mathscr{F}_{k+1}^{\bullet} \simeq \psi^{\ast} \left( \Omega_{Y/X}^{\bullet -k}(\log \Delta) \right) \otimes h^{\ast} \Omega_X^k(\log D).$$

I don't understand how such a filtration is constructed, is there some general result that I am missing? I apologise in advance if this result is well-known.

This problem arises from the study of hyperbolicity, specifically from the works of Eckhart Viehweg and Kang Zuo (2002, 2003). A more detailed treatment was given in section 2.2 of this article - https://arxiv.org/pdf/1809.05616.pdf

• Welcome to Mathoverflow! Is the dot in the exponent just a variable, i.e. you don't mean to work with all the wedge powers at once? Jan 18, 2021 at 2:43
• @WillSawin I assumed that it meant work with all the exterior powers at once. Jan 18, 2021 at 20:52

For $$0 \to A \to B \to C \to 0$$ an exact sequence of modules, we obtain a filtration on $$\wedge^d B$$ where $$F^k / F^{k+1}$$ is $$\wedge^k A \otimes \wedge^{d-k} C$$.
To see this, recall that $$\wedge^d B$$ is defined as the quotient of $$B\otimes \dots \otimes B$$ by relations of the form $$x \otimes y = - y \otimes x$$. We define $$F^k$$ to be the image in $$\wedge^d A$$ of the submodule generated by tensors $$x_1 \otimes \dots \otimes x_d$$ where at least $$k$$ of the factors $$x_1,\dots, x_d$$, come from $$A$$.
Using the relations, we can ensure that the first $$k$$ factors come from $$A$$, which gives a map to $$\wedge^k A \otimes \wedge^{d-k} C$$ by projecting the last $$d-k$$ factors from $$C$$ to $$B$$, and one can check this is an isomorphism.
Glancing at the paper, I think that they mean for the exact sequence to exist one $$\cdot$$ at a time, so this is a complete explanation, plugging in the terms of your exact sequence for $$A,B, C$$, but I didn't check 100%.