I have two basic questions regarding hyperkaehler manifolds.

  1. A holomorphic symplectic manifold is a complex manifold $X$ endowed with a $(2,0)$-form $\omega$. I know that a Hyperkaehler manifold can be seen as a holomorphic symplectic manifold. But is the converse also true? Is every holomorphic symplectic manifold always Hyperkaehler? If not, can you give a counterexample?

  2. Does a Hyperkaehler manifold possess a natural polarization? What is it?

Thanks in advance for your answers.


1) The converse is not true. There has to be some compatibility between the complex structure and the symplectic structure. There are manifolds that admit both complex structures and symplectic structures, but without any such pair of structures being compatible. An example is the Kodaira-Thurston manifold. Let $K=T^3 \times \mathbb{R}/\mathbb{Z}$ where $j \in \mathbb{Z}$ acts by

$$\psi: (x,y,z,t) \mapsto (x,y + jx, z, t+j).$$

The symplectic form $dx \wedge dy + dz \wedge dt$ is invariant under $\psi$. The projection $\pi : K \to T^2$ given by $(x,y,z,t) \mapsto (z,t)$ has fibers that are symplectic $T^2$ with symplectic form the restriction of $dx \wedge dy$. The universal cover is $\mathbb{R}^4$ with deck group

$$G = \{ (j_1,j_2,k_1,k_2) | \mbox{ in } \mathbb{Z}^4 , (\mathbb{j}',\mathbb{k}') * (\mathbb{j},\mathbb{k}) = (\mathbb{j}+\mathbb{j}',A_{\mathbb{j}'} \mathbb{k} + \mathbb{k}')\}$$

where $\mathbb{j}=(j_1,j_2)$ and $A_{\mathbb{j}}$ is the matrix

$$A_{\mathbb{j}} = \begin{pmatrix} 1 & j_2 \\ 0 & 1 \end{pmatrix}$$

So, $\pi_1(K)=G$. Note that the subgroup $H=\langle (j_1,0,0,0) \rangle \cong \mathbb{Z} \leq G$ is a normal subgroup, and is the smallest normal subgroup such that the quotient $G/H$ is abelian, thence $H_1(K)$, which is the abelianization of $\pi_1(K)$ , is equal to $\mathbb{Z}^3$. Hence, the first Betti number is 3. But the first Betti number of a compact Kaehler manifold must be even, hence the manifold $K$ is not Kaehler and, in particular, not hyper-Kaehler.

2) Any Kaehler manifold $M$ (and therefore, any hyper-Kaehler manifold) has a polarization given by: $\mathcal{P}:=T_{(1,0)}(M)=\{ (x,v) \in T_x(M)^{\mathbb{C}} | J_x(v)=iv \}$, where $J : TM^{\mathbb{C}} \to TM^{\mathbb{C}}$ is the integrable almost complex structure induced by the complex structure on $M$.

  • $\begingroup$ Example 1) is very nice, indeed. Thanks a lot. As for comment 2) I have objections. A Hyperkaehler manifold is a complex manifold which means that its tangent bundle is holomorphic. You cannot define a polarisation for the holomorphic tangent bundle the way you do it in your comment. Do you know some example of a holomorphic symplectic manifold with a natural Lagrangian foliation (polarisation) other than the cotangent bundle? $\endgroup$ – Flavius Aetius Apr 30 '18 at 16:07
  • $\begingroup$ One example is the complex torus $T^{2n}$ with canonical coordinates $(q_i,p_i)$, and symplectic form $\sum_i dq_i \wedge dp_i$, where you have Lagrangian foliations given by equations $q_i - \alpha_i p_i=\mbox{const}$ for $i=1,\cdots,n$ and $\alpha_i \in \mathbb{R}$. If all $\alpha_i$ are irrational then the leaves are dense, otherwise are compact. $\endgroup$ – DMG May 1 '18 at 17:39
  • $\begingroup$ The problem of finding examples of Lagrangian foliations on Kaehler manifolds other than the complex torus seems a difficult problem. Under the (strong) hypothesis that the foliation is parallel to the Levi-Civita connection, I. Vaisman proves that the manifold has to be the complex torus: jstor.org/stable/… $\endgroup$ – DMG May 1 '18 at 17:39
  • $\begingroup$ The complex torus is not Hyperkaehler, right? In other words it is not a holomorphically symplectic symplectic manifold, that is the symplectic form is a real $(1,1)$-form, not a $(2,0)$-form. I do not understand the meaning of $q_i-a_ip_i=const$. Do these equations define submanifolds in the torus. What is the meaning of the parameter $a_i$. Are these parameters free? Do I choose them according to some rule? $\endgroup$ – Flavius Aetius May 1 '18 at 20:27
  • $\begingroup$ According to Wikipedia, 1. is indeed true. I guess your symplectic form is not holomorphic? $\endgroup$ – red_trumpet Oct 1 '20 at 9:35

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