A global Weinstein splitting theorem for holomorphic Poisson manifolds

This proof is a variant of the arguments in [5, Proposition 3.1] and [9, Theorem 5.6]. Suppose that $\dim\mathsf{L}=2k$ and let $\omega\in\mathsf{H}^{1}(\mathsf{X},\Omega^{1}_{\mathsf{X}})\cong\mathsf{H}^{1,1}(\mathsf{X})$ be any Kähler class. Note that since $\pi$ is holomorphic, the contraction operator $\iota_{\pi}$ on $\Omega^{\bullet}_{\mathsf{X}}$ descends to the Dolbeault cohomology $\mathsf{H}^{\bullet}(\mathsf{X},\Omega^{\bullet}_{\mathsf{X}})$. In particular, we have a well-defined class

We claim that $i^{*}\alpha\in\mathsf{H}^{2k}(\mathsf{L},\mathcal{O}_{\mathsf{L}})$ is nonzero. Indeed, since $\pi|_{\mathsf{L}}=\pi_{\mathsf{L}}$, we have the following commutative diagram:

The bottom arrow is an isomorphism since $\pi_{\mathsf{L}}^{k}$ is a trivialization of the anticanonical bundle of $\mathsf{L}$. Meanwhile $i^{*}\omega^{2k}\in\mathsf{H}^{2k}(\mathsf{L},\Omega^{2k}_{\mathsf{L}})$ is nonzero since $i^{*}\omega$ is a Kähler class on $\mathsf{L}$. It follows that $i^{*}\alpha=\iota_{\pi_{\mathsf{L}}}^{k}i^{*}\omega^{2k}\neq 0$ as claimed.

Using the Hodge symmetry $\overline{\mathsf{H}^{2k}(\mathsf{X},\mathcal{O}_{\mathsf{X}})}\cong\mathsf{H}^{0}(\mathsf{X},\Omega^{2k}_{\mathsf{X}})$, the complex conjugate of $\alpha$ gives a holomorphic $2k$-form

such that $i^{*}\mu\in\mathsf{H}^{0}(\mathsf{L},\Omega^{2k}_{\mathsf{L}})$ is nonzero. Since the canonical bundle of $\mathsf{L}$ is trivial, $i^{*}\mu$ must be a constant multiple of the holomorphic Liouville volume form associated with the holomorphic symplectic structure on $\mathsf{L}$. Hence by rescaling $\mu$, we may assume without loss of generality $i^{*}\mu=\frac{1}{k!}\eta^{k}$. Now consider the global holomorphic two-form

which is closed by Hodge theory, since $\mathsf{X}$ is compact Kähler. We claim that it restricts to the symplectic form on $\mathsf{L}$. Indeed,

as desired. ∎

Let $\sigma_{0}\in\mathsf{H}^{0}(\mathsf{X},\Omega^{2}_{\mathsf{X}})$ be any extension of the symplectic form on $\mathsf{L}$ to a global holomorphic two-form on $\mathsf{X}$, and let $\theta_{0}:=\pi^{\sharp}\sigma_{0}^{\flat}\in\mathcal{E}nd(\mathcal{T}_{\mathsf{X}})$. Since $\pi_{\mathsf{L}}=\pi|_{\mathsf{L}}$ is inverse to the symplectic form $\eta=i^{*}\sigma_{0}$ on $\mathsf{L}$, it follows easily that $\theta_{0}|_{\mathsf{L}}\in\mathcal{E}nd(i^{*}\mathcal{T}_{\mathsf{X}})$ is idempotent with image $\mathcal{T}_{\mathsf{L}}\subset i^{*}\mathcal{T}_{\mathsf{X}}$, giving a splitting

In particular, the characteristic polynomial of $\theta_{0}|_{\mathsf{L}}$ is given by

where $n=\dim\mathsf{X}$ and $k=\tfrac{1}{2}\dim\mathsf{L}$. But the coefficients of the characteristic polynomial of $\theta_{0}$ are holomorphic functions on $\mathsf{X}$, and since $h^{0}(\mathsf{X},\mathcal{O}_{\mathsf{X}})=1$, such functions are constant. We conclude that $P(t)$ is the characteristic polynomial of $\theta_{0}$ over all of $\mathsf{X}$.

Taking generalized eigenspaces of $\theta_{0}$, we obtain a decomposition

where $i^{*}\mathcal{F}=\mathcal{T}_{\mathsf{L}}$. Projecting the two-form $\sigma_{0}$ to $\wedge^{2}\mathcal{F}^{\vee}$ we therefore obtain a new holomorphic two-form $\sigma_{1}$ such that $i^{*}\sigma_{1}=\eta$, which has the additional property that $\mathcal{G}\subset\ker\sigma_{1}$. Then $\theta_{1}=\pi^{\sharp}\sigma_{1}^{\flat}$ also has characteristic polynomial $P(t)$, giving a splitting $\mathcal{T}_{\mathsf{X}}=\mathcal{F}^{\prime}\oplus\mathcal{G}$ with respect to which $\theta_{1}$ takes on the Jordan–Chevalley block form

where $\phi\in\mathsf{End}(\mathcal{F}^{\prime})$ is nilpotent. Using the identity $\theta_{1}\pi^{\sharp}=\pi^{\sharp}\theta_{1}^{\vee}$ as maps $\Omega^{1}_{\mathsf{X}}\to\mathcal{T}_{\mathsf{X}}$, one calculates that $\pi$ must be block diagonal, i.e. equal to the sum of its projections to $\wedge^{2}\mathcal{G}$ and $\wedge^{2}\mathcal{F}^{\prime}$. The latter projection, say $\pi^{\prime}\in\wedge^{2}\mathcal{F}^{\prime}$, is then nondegenerate because $\theta_{1}=\pi^{\sharp}\sigma_{1}^{\flat}$ is invertible on $\mathcal{F}^{\prime}$. We may therefore define a two-form $\sigma:=(\pi^{\prime})^{-1}\in\wedge^{2}\mathcal{F}^{\prime}\subset\Omega^{2}_{\mathsf{X}}$. By construction, the endomorphism $\theta:=\pi^{\sharp}\sigma^{\flat}$ preserves the decomposition $\mathcal{T}_{\mathsf{X}}=\mathcal{F}^{\prime}\oplus\mathcal{G}$, acts as the identity on $\mathcal{F}^{\prime}$, and has $\mathcal{G}$ as its kernel. In particular, $\theta$ is idempotent and $\sigma$ restricts to the symplectic form on $\mathsf{L}$. Moreover, $\sigma$ is closed by hypothesis. Thus $\sigma$ is a subcalibration of $\pi$ compatible with $\mathsf{L}$, as desired. ∎

The proof of Frejlich–Mărcu\cbt makes elegant use of the notion of a Dirac structure. For completeness, we present here an essentially equivalent argument based on the related notion of a gauge transformation of Poisson structures.

We recall from [13, Section 4] that if $B$ is a closed holomorphic two-form such that the operator $1+B^{\flat}\pi^{\sharp}\in\mathcal{E}nd(\Omega^{1}_{\mathsf{X}})$ is invertible, then the gauge transformation of $\pi$ by $B$ is a new Poisson structure $B\star\pi\in\mathsf{H}^{0}(\mathsf{X},\wedge^{2}\mathcal{T}_{\mathsf{X}})$ whose underlying foliation is the same as that of $\pi$, but with the symplectic form on each leaf modified by adding the pullback of $B$. Equivalently, $B\star\pi$ is determined by its anchor map, which is given by the formula $(B\star\pi)^{\sharp}=\pi^{\sharp}(1+B^{\flat}\pi^{\sharp})^{-1}:\Omega^{1}_{\mathsf{X}}\to\mathcal{T}_{\mathsf{X}}$. The skew symmetry of $B\star\pi$ follows from the skew symmetry of $\pi$ and $B$, while the identity $[B\star\pi,B\star\pi]=0$ is deduced using the closedness of $B$ and the equation $[\pi,\pi]=0$.

We apply this construction to the family of two-forms $B(t):=t\sigma$. Note that since $\theta:=\pi^{\sharp}\sigma^{\flat}$ is idempotent, the operator $1+B^{\flat}(t)\pi^{\sharp}=1+t\theta^{\vee}\in\mathcal{E}nd(\Omega^{1}_{\mathsf{X}})$ is invertible for all $t\neq-1$. We therefore obtain a family of holomorphic Poisson bivectors

with anchor map

Since $\theta$ is idempotent, we have $(1+t\theta^{\vee})^{-1}=1-\tfrac{t}{1+t}\theta^{\vee}$, which implies that

for all $t\neq-1$. Since $[\pi(t),\pi(t)]=0$ for all $t\neq-1$, the bilinearity of the Schouten bracket therefore implies the desired identities (1). This implies immediately that $\mathcal{F}=\operatorname{img}\pi_{\mathcal{F}}$ is the tangent sheaf of the symplectic foliation of the Poisson bivector $\pi_{\mathcal{F}}$; in particular, it is involutive.

It remains to show that $\mathcal{G}$ is involutive if and only if $\sigma_{\mathcal{F}}$ is closed. To this end, observe that if $\sigma_{\mathcal{F}}$ is closed, then its kernel $\mathcal{G}=\ker\sigma_{\mathcal{F}}$ is involutive by elementary Cartan calculus. Conversely, if $\mathcal{G}$ is involutive, then both $\mathcal{F}^{\vee}$ and $\mathcal{G}^{\vee}$ generate differential ideals in $\Omega^{\bullet}_{\mathsf{X}}$. This implies that the exterior derivatives of $\sigma_{\mathcal{F}}\in\wedge^{2}\mathcal{F}^{\vee}$ and $\sigma_{\mathcal{G}}\in\wedge^{2}\mathcal{G}^{\vee}$ lie in complementary subbundles of $\Omega^{2}_{\mathsf{X}}$, namely

Since $\mathrm{d}\sigma_{\mathcal{F}}+\mathrm{d}\sigma_{\mathcal{G}}=\mathrm{d}\sigma=0$, we conclude that $\mathrm{d}\sigma_{\mathcal{F}}=0$, as desired. ∎

For the first statement, note that $\operatorname{rank}\mathcal{G}=\dim\mathsf{X}-\operatorname{rank}\mathcal{F}$ and the fibres of $t$ have dimension equal to $\operatorname{rank}\mathcal{F}$. Therefore $t^{-1}\mathcal{G}$ has rank equal to $\dim\mathsf{X}$. Note that $\mathcal{G}$ is identified with the normal bundle of the foliation $\mathcal{F}$ and therefore $t^{-1}\mathcal{G}$ surjects by $s$ onto the normal bundle of every leaf. Meanwhile every $t$-fibre surjects by $s$ onto the corresponding leaf. Considering the ranks, it follows that $t^{-1}(\mathcal{G})$ is complementary to the fibres of $s$, defining a flat Ehresmann connection. Since $s$ is proper, this connection is complete, as desired.

For the second statement, note that the horizontal lifts of a leaf $\mathsf{L}\subset\mathsf{X}$ of $\mathcal{F}$ are, by definition, given by intersecting the preimage $s^{-1}(\mathsf{L})$ with the leaves of $t^{-1}\mathcal{G}$. But $t$ projects $s^{-1}(\mathsf{L})$ onto $\mathsf{L}$, which is complementary to $\mathcal{G}$. It follows that the intersection of the leaves of $t^{-1}\mathcal{G}$ with $s^{-1}(\mathsf{L})$ are the fibres $t^{-1}(y)$ for $y\in\mathsf{L}$, as claimed. The third statement then follows immediately from the description of the holonomy representation (2) above. ∎

Suppose that $\mathsf{X}$ is a compact Kähler manifold and $\mathcal{F}$ is a regular holomorphic foliation having a compact leaf with finite holonomy group. Then by the global Reeb stability theorem for compact Kähler manifolds [12, Theorem 1], every leaf of $\mathcal{F}$ is compact with finite holonomy group. As observed in [11, Example 5.28(2)], this implies that the submersions $s,t:\mathsf{Hol}(\mathcal{F})\to\mathsf{X}$ are proper maps.

If $\mathcal{G}$ is any foliation complementary to $\mathcal{F}$, we obtain from Section 3.2 a flat Ehresmann connection on the fibration $s:\mathsf{Hol}(\mathcal{F})\to\mathsf{X}$ whose monodromy representation induces the holonomy representation of every leaf. Let us choose a base point $x\in\mathsf{X}$, and consider the monodromy representation

Note that by Section 3.2, $\mathsf{Hol}(\mathcal{F})$ is a Kähler manifold. If we choose a Kähler class on $\mathsf{Hol}(\mathcal{F})$, its restriction to $s^{-1}(x)$ gives a Kähler class $\omega\in\mathsf{H}^{1,1}(s^{-1}(x))$ that is invariant under the monodromy action. Hence $\rho$ factors through the subgroup $\mathsf{Aut}_{\omega}(s^{-1}(x))$ of biholomorphisms of $s^{-1}(x)$ that fix the class $\omega$.

By Section 3.3 below applied to the monodromy action of $\Pi:=\pi_{1}(\mathsf{X},x)$ on $\mathsf{Z}:=s^{-1}(x)$, there exists a finite-index subgroup $\Gamma<\pi_{1}(\mathsf{X},x)$ whose image $\rho(\Gamma)<\mathsf{Aut}_{\omega}(s^{-1}(x))$ contains no finite subgroups that act freely on $s^{-1}(x)$. Let $\phi:\widetilde{\mathsf{X}}\to\mathsf{X}$ be the covering space corresponding to $\Gamma$, with base point $\widetilde{x}\in\widetilde{\mathsf{X}}$ chosen so that $\phi_{*}\pi_{1}(\widetilde{\mathsf{X}},\widetilde{x})=\Gamma$. Then $\phi$ is a finite étale cover since $\Gamma$ has finite index.

We claim that the pair $(\phi^{-1}\mathcal{F},\phi^{-1}\mathcal{G})$ is a suspension. Note that to prove this, it suffices to show that the holonomy groups of the leaves of $\phi^{-1}\mathcal{F}$ are trivial, for in this case the holonomy groupoid of $\phi^{*}\mathcal{F}$ embeds as the graph of an equivalence relation in $\widetilde{\mathsf{X}}\times\widetilde{\mathsf{X}}$, which in turn implies that $\phi^{-1}\mathcal{F}$ is a fibration whose quotient map is proper. Then the complementary foliation $\phi^{-1}\mathcal{G}$ is a flat Ehresmann connection, as desired, and this establishes part 1 of the theorem.

To see that the holonomy groups of the leaves of $\phi^{-1}\mathcal{F}$ are indeed trivial, suppose that $\widetilde{\mathsf{L}}\subset\widetilde{\mathsf{X}}$ is a leaf of $\phi^{-1}\mathcal{F}$. Then $\phi$ restricts to an étale cover $\widetilde{\mathsf{L}}\to\mathsf{L}$ where $\mathsf{L}$ is a leaf of $\mathcal{F}$. If we choose a base point $\widetilde{y}\in\widetilde{\mathsf{L}}$ with image $y=\phi(\widetilde{y})\in\mathsf{L}$, then the germ of the leaf space of $\phi^{-1}\mathcal{F}$ at $\widetilde{y}$ is identified with the germ of the leaf space of $\mathcal{F}$ at $y$, so that the holonomy group of $\widetilde{\mathsf{L}}$ at $\widetilde{y}$ is canonically identified with the image of the composition

Choose a homotopy class of a path $\widetilde{\gamma}$ from $\widetilde{x}$ to $\widetilde{y}$, let $\gamma$ be its projection to a homotopy class from $x$ to $y$, and let $\operatorname{hol}_{\gamma}:s^{-1}(y)\to s^{-1}(x)$ be the isomorphism given by parallel transport of the Ehresmann connection on $\mathsf{Hol}(\mathcal{F})\to\mathsf{X}$. The adjoint actions of $\widetilde{\gamma},\gamma$ and $\operatorname{hol}_{\gamma}$ then fit in a commutative diagram

Note that since $\mathsf{Hol}(\mathcal{F})_{y}$ acts freely on $s^{-1}(y)$, and the holonomy $\operatorname{hol}_{\gamma}$ is an isomorphism between the fibres, the conjugate of $\mathsf{Hol}(\mathcal{F})_{y}$ by $\operatorname{hol}_{\gamma}$ acts freely on $s^{-1}(x)$. We conclude that the dashed arrow embeds the holonomy group of $\widetilde{\mathsf{L}}$ as a finite subgroup in $\mathsf{Aut}_{\omega}(s^{-1}(x))$ that acts freely on $s^{-1}(x)$. Since the diagram is commutative, this subgroup lies in the image of $\Gamma$ and hence by our choice of $\Gamma$ it must be trivial, as claimed. ∎

By part 1 of the theorem, we may assume without loss of generality that $(\mathcal{F},\mathcal{G})$ is a suspension with base $\mathsf{Y}$ and typical fibre $\mathsf{L}^{\prime}$, where $\mathsf{L}^{\prime}$ is a covering space of $\mathsf{L}$. Moreover the monodromy representation of this suspension takes values in the group $\mathsf{Aut}_{\omega^{\prime}}(\mathsf{L}^{\prime})$ of biholomorphism of $\mathsf{L}^{\prime}$ that fix some Kähler class $\omega^{\prime}\in\mathsf{H}^{1,1}(\mathsf{L}^{\prime})$.

Now note that if the fundamental group of $\mathsf{L}$ is finite and the universal cover of $\mathsf{L}$ has no nonvanishing vector fields, then $\mathsf{H}^{0}(\mathsf{L}^{\prime},\mathcal{T}_{\mathsf{L}^{\prime}})=0$ as well. Therefore $\mathsf{Aut}_{\omega^{\prime}}(\mathsf{L}^{\prime})$ is finite by [8, Proposition 2.2]. The kernel of the monodromy representation therefore gives a finite index subgroup of $\pi_{1}(\mathsf{Y})$, and passing to the corresponding finite étale cover of $\mathsf{Y}$, we trivialize the suspension, giving an étale map $\mathsf{L}^{\prime}\times\mathsf{Y}\to\mathsf{X}$ that implements the desired splitting of the tangent bundle. Then passing to the universal cover of $\mathsf{L}^{\prime}$, we obtain the desired statement. ∎

By taking preimages under $\rho$, we reduce the problem to the case where $\rho$ is injective, so we may assume without loss of generality that $\Pi<\mathsf{Aut}_{\omega}(\mathsf{Z})$ is a subgroup. Moreover, by [8, Proposition 2.2], the neutral component $\mathsf{Aut}\left(\mathsf{Z}\right)_{0}$ of the group of all biholomorphisms of $\mathsf{Z}$ is a finite index subgroup in $\mathsf{Aut}_{\omega}(\mathsf{Z})$. In particular, $\Pi\cap\mathsf{Aut}\left(\mathsf{Z}\right)_{0}$ has finite index in $\Pi$, and is therefore finitely generated by Schreier’s lemma. Hence we may assume without loss of generality that $\Pi<\mathsf{Aut}\left(\mathsf{Z}\right)_{0}$.

By [8, Theorems 3.3, 3.12 and 3.14], there is an exact sequence

where $\mathsf{N}$ is the closed subgroup exponentiating the Lie algebra of holomorphic vector fields with nonempty zero locus, and $\mathsf{T}$ is a compact complex torus (a finite connected étale cover of the Albanese torus of $\mathsf{Z}$).

Since $\Pi$ is finitely generated, its image in $\mathsf{T}$ is a finitely generated abelian group. Therefore, by the classification of finitely generated abelian groups, there is a finite-index subgroup $\Gamma<\Pi$ whose image in $\mathsf{T}$ is torsion-free. Suppose that $\mathsf{G}<\Gamma$ is a finite subgroup that acts freely on $\mathsf{Z}$. We claim that $\mathsf{G}$ is trivial. Indeed, by construction, the image of $\mathsf{G}$ in $\mathsf{T}$ is a torsion-free finite group, hence trivial. We must therefore have that $\mathsf{G}<\mathsf{N}$. But by Section 3.3 below, every element of $\mathsf{N}$ has a fixed point, and therefore the only subgroup of $\mathsf{N}$ that acts freely is the trivial subgroup. ∎

Let $\mathsf{U}\subset\mathsf{N}$ be the set of elements with at least one fixed point. Note that $\mathsf{U}=p(a^{-1}(\Delta))$, where $\Delta\subset\mathsf{X}\times\mathsf{X}$ is the diagonal and $p:\mathsf{N}\times\mathsf{X}\to\mathsf{N}$ is the projection. Since $a$ is holomorphic, $a^{-1}(\Delta)\subset\mathsf{N}\times\mathsf{X}$ is a closed analytic subspace. Since $p$ is proper, the Grauert direct image theorem implies that the image $p(a^{-1}(\Delta))=\mathsf{U}\subset\mathsf{N}$ is a closed analytic subvariety. But we assume that every generating vector field for the action has at least one zero; therefore $\mathsf{U}$ contains the image of the exponential map of $\mathsf{N}$, and in particular it contains an open neighbourhood of the identity. It follows that $\dim\mathsf{U}=\dim\mathsf{N}$, and therefore $\mathsf{U}=\mathsf{N}$ since $\mathsf{N}$ is connected. ∎

By Theorem 3.1, part 2, it suffices to show that $h^{0}(\widetilde{\mathsf{L}},\mathcal{T}_{\widetilde{\mathsf{L}}})=0$ where $\widetilde{\mathsf{L}}$ is the universal cover of $\mathsf{L}$, but this vanishing is well known. Indeed, in this case, the canonical bundle of $\widetilde{\mathsf{L}}$ is trivial, so that $\mathcal{T}_{\widetilde{\mathsf{L}}}\cong\Omega^{n-1}_{\widetilde{\mathsf{L}}}$ where $n=\dim\mathsf{L}$. We then have