# Failure of Borel-Schmid quasi-unipotency theorem in non-algebraic case

Let $$\mathcal{X} \to \Delta^{\times}$$ be a smooth family of compact Kähler manifolds over a punctured disc. When this family is algebraic, the celebrated quasi-unipotency theorem (I think, due to Borel) states that the monodromy action of a generator of $$\pi_1(\Delta^{\times}, o) \simeq \mathbb{Z}$$ on the Betti cohomology of a fibre is quasi-unipotent (recall that an operator $$T$$ is called quasi-unipotent if $$T^N$$ is unipotent for some $$N$$).

Both of the proofs which I know sufficiently use the algebraicity of the family.

The first one is due to Grothendieck and is based on the action of an absolute Galois group in étale cohomology. The first step is to choose a number field $$K$$, on which our family is defined.

The second one is due to Schmid and works for arbitrary polarised variations of Hodge structures. It intensively uses the geometry of polarised period domains and I am not sure if non-polarised period domains share the same properties.

Thus, my questions are:

1)Are there known examples of non-polarisable variations of pure Hodge structures over a punctured disc with non-quasi-unipotent monodromy?

2)If yes, are there any of those of geometric origin? More precisely, is there a known example of family of non-projective Kähler manifolds over a punctured disc, for which the quasi-unipotency theorem fails?

One universal cover of the punctured disk $$\Delta^*$$ is the upper half plane $$\mathcal{H}$$, $$\pi:\mathcal{H}\to \Delta^*, \ \ z=\pi(w) = e^{2\pi iw}.$$ There is a natural translation action of the integers $$\mathbb{Z}$$ on $$\mathcal{H}$$, and $$\pi$$ is a quotient of this free action.
Let $$(E,0)$$ be a general elliptic curve with its usual (additive) group structure. Consider the product complex manifold with its projection to $$\mathcal{H}$$, $$X:=\mathcal{H} \times E \times E, \ \ \text{pr}_1:X\to \mathcal{H}.$$ Consider the following lift of the $$\mathbb{Z}$$-action to $$X$$ lifting "translation by $$1$$" to the following biholomorphism, $$\tau:\mathcal{H}\times E\times E \to \mathcal{H}\times E \times E, \ \ \tau(w,x,y) = (w+1,2x+3y,x+2y).$$ The projection $$\text{pr}_1$$ is equivariant for this action.
Define $$\mathcal{X}$$ to be the quotient of the free action of $$\mathbb{Z}$$ on $$X$$, $$\rho:X\to \mathcal{X}.$$ The composition $$\pi\circ \text{pr}_1$$ is $$\mathbb{Z}$$-invariant. Thus, there is a unique induced holomorphic submersion, $$p:\mathcal{X}\to \Delta^*, \ \ p\circ \rho = \pi \circ \text{pr}_1.$$ Every fiber of $$p$$ is biholomorphic to the Kähler manifold $$E\times E$$.
With respect to a standard symplectic basis $$(\alpha,\beta)$$ for $$H_1(E;\mathbb{Z}) \cong \mathbb{Z}^2$$, a corresponding basis for $$H_1(E\times E;\mathbb{Z})$$ equals $$(\alpha\otimes 1,1\otimes \alpha,\beta\otimes 1,1\otimes \beta)$$. With respect to this basis, the monodromy operator action on $$H_1(E\times E;\mathbb{Z})$$ is given by a $$4\times 4$$ matrix that is in block form $$(2|2)$$ where each of the $$2\times 2$$ blocks is the following integer entry matrix, $$\left[ \begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array} \right].$$ In particular, this matrix is not quasi-unipotent: the eigenvalues are $$2\pm \sqrt{-3}.$$
Note, there is no positive $$(1,1)$$ class on $$E\times E$$ that is preserved by this monodromy action. Thus, although $$p$$ is a proper holomorphic submersion whose fibers are Kähler manifolds, it is not a proper holomorphic submersion between (noncompact) Kähler manifolds.
• Thank you! That is a nice example, however I was more interested in the case when the total space of $\mathcal{X}$ is Kähler, however one cannot find a monodromy invariant rational Kähler class; it is especially interesting if there are examples when (generic) doesn't possess a rational Kähler class Jan 29 '19 at 15:43