7
$\begingroup$

Let $X \stackrel{\pi}{\to} \mathbb{D}$ be a proper holomorphic family with fibres $X_t = \pi^{-1}(t)$. Siu proved, when the $X_t$'s are projective, that the plurigenera $h^0(X_t, mK_{X_t})$ are constant. It is conjectured that the same will hold true when the fibres $X_t$ are Kähler.

Are there examples known where this fails when the fibres are not Kähler?

My usual example of a compact non-Kähler manifold is the Hopf surface, but if the fibres of such holomorphic family are all Hopf surfaces, then the plurigenera are all zero (in particular, they are constant).

$\endgroup$

2 Answers 2

6
$\begingroup$

The first example of this phenomenon was discovered by Iku Nakamura in his 1975 paper Complex parallelisable manifolds and their small deformations, J. Differential Geom. 10 (1975), 85-112.

The manifold $X$ is a $3$-dimensional solvmanifold, and he writes down its small deformations explicitly, see his Theorem 2 and the discussion on pages 96-99.

$\endgroup$
1
$\begingroup$

Edit. Something is wrong with the following examples (the Hopf surface has no nontrivial pluricanonical forms, so also "twists" have no nontrivial pluricanonical forms). I will try to fix it soon (maybe using twists of the Iwasawa manifold).

Here is the simplest example that I can think of, but there may be simpler examples. These examples are "bundles" of Hopf surfaces, and they are "universal counterexamples" in non-Kähler geometry. In particular, note that any counterexample must have that the canonical bundle is not positive (or the manifolds would be projective, where Siu's theorem applies). In the following examples, the canonical bundles are all homologically equivalent to the structure sheaf / the trivial divisor class.

Let $(E,0)$ be an elliptic curve over $\mathbb{C}$. On the surface $E\times E$, form the Cartier divisor $$D = \underline{\Delta} - \underline{\{0\}\times E} - \underline{E\times\{0\} },$$ where $\Delta$ is the diagonal. Denote by $\mathcal{L}$ the associated invertible sheaf, $\mathcal{O}_{E\times E}(D)$. Denote by $$\text{pr}_1:E\times E\to E,$$ the projection onto the first factor. For $p\in E(\mathbb{C})$, the restriction of $\mathcal{L}^{\otimes 2}$ to the fiber $E$ of $\text{pr}_1$ over $p$ is either isomorphic to $\mathcal{O}_E$ if $p$ is a $2$-torsion point, or it is an invertible sheaf of degree $0$ on $E$ that has only the zero global section if $p$ is not a $2$-torsion point.

Form the geometric vector bundle $\pi:V\to E\times E$ such that there is a universal $\mathcal{O}_V$-module homomorphism, $$s:\pi^*(\mathcal{L}\oplus \mathcal{L})\to \mathcal{O}_V,$$ i.e., $V$ is $\text{Spec}_{E\times E}\text{Sym}^\bullet(\mathcal{L}\oplus \mathcal{L})$. The dualizing sheaf of $V$ is isomorphic to $\pi^*( \mathcal{L}^{\otimes 2} )$. Denote by $U\subset V$ the open complement of the zero section of $\pi$. There is a scaling action of $\mathbb{G}_m$ on $V$, compatible with the projection to $\pi$: just scale $s$ by the universal invertible global section $t$ of $\mathcal{O}_{\mathbb{G}_m}$. Let $q\in \mathbb{C}^*$ be an element of complex modulus different from $1$. Then the scaling of $U(\mathbb{C})$ by $q\in \mathbb{G}_m(\mathbb{C})$ is a free, proper, discontinuous action with respect to the Euclidean / analytic topology on $U(\mathbb{C})$. Thus, there is a well-defined quotient in the category of complex analytic spaces, and this quotient is even a compact complex manifold with a projection to the underlying analytic space of $E\times E$, $$\rho:M\to (E\times E)^{\text{an}}.$$ In particular, there is a projection of $(E\times E)^{\text{an}}$ to $E^{\text{an}}$ by projection on the first factor, $$M \xrightarrow{\rho} (E\times E)^{\text{an}} \xrightarrow {\text{pr}_1} E^{\text{an}}.$$ The relative dualizing sheaf of $\text{pr}_1\circ \rho$ is still $\rho^*(\mathcal{L}^{\otimes 2})$. Of course the restriction of $\mathcal{L}^{\otimes 2}$ to the fiber over $0\in E^\text{an}$ is the structure sheaf, which has nonzero global sections. But for $p\in E^{\text{an}}$ not a $2$-torsion point, the restriction of $\mathcal{L}^{\otimes 2}$ to the fiber over $p$ has only the zero section.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .