# Foliations by holomorphic curves on complex surfaces

On a complex surface, does there exist a non-singular foliation by holomorphic curves that is NOT a holomorphic foliation, i.e. a transversally holomorphic foliation?

The surface should be compact and can admit holomorphic foliations.

• Isn't the reference you mentioned available on rivista.math.unipr.it/vols/2010-1-2/indice.html EDIT (by Karl Schwede): In think the original poster is suggesting that you see the paper "On Kähler surfaces with semipositive Ricci curvature." by Marco Brunella. (I can't access it via the above link, even through my institutions library).
– home
Jul 30, 2013 at 22:28
• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. Jul 30, 2013 at 23:52
• Or at least gave a precise reference to the theorem in the paper and a statement of the title / authors of the paper (in case the link dies). Jul 31, 2013 at 0:40

If the curves (leaves of the foliation) are compact then the answer is no. If you have infinitely many pairwise disjoint compact curves then they are all contained in fibers of the same holomorphic fibration as I have proved in this paper. For another proof you can look at this paper by Winkelmann. There you will also find a non-holomorphic foliation by compact complex curves of a non-compact surface, see Section 3.

The same result holds true for codimension one foliations by compact leaaves, but it is not true for higher codimension foliations. The projective space admits a foliation by lines (twistor lines) which is not holomorphic. The leaf space is $S^4$.

If the general leaf is not compact then i do not know the answer.

• Shouldn't your theorem 2 say that $M$ is connected and that $\rho$ is not constant? Jul 26, 2013 at 9:56
• @BenMcKay: yes, probably this is fixed in the published version. Jul 26, 2013 at 10:57

Update: I have had a little time this weekend to think more deeply about the construction I outlined below and I have concluded that, unfortunately, the construction that I thought would work to produce a cocompact free action of a discrete subgroup of $G$ on the quotient $G/K$ does not seem to work. (At least, it would require a different method than what I proposed, starting with a cocompact discrete subgroup of $\mathrm{SL}(2,\mathbb{R})$.) Thus, constructing a compact example by this method will not work easily. I'm not removing the original answer entirely since some people have upvoted it (of course, they might want to reconsider their upvotes, given that this method didn't succeed).

Meanwhile, I have decided that there is a better method to try, but am still working on the details, which involve choosing a uniform (i.e., cocompact) discrete lattice in $\mathrm{SL}(3,\mathbb{R})$ with certain properties that I still hope will be possible.

The reason $\mathrm{SL}(3,\mathbb{R})$ enters the picture is this: If $X$ is a complex surface endowed with a foliation $\mathcal{F}$ by holomorphic curves, each point $x\in X$ has an open neighborhood $U$ on which there exists a nonvanishing $(1,0)$-form $\omega$ (unique up to a nonzero multiple) with the property that $\omega=0$ defines the complex tangent lines to the foliation $\mathcal{F}$ in $U$. The foliation is holomorphic in $U$ if and only if $\omega\wedge d\omega\equiv0$ (this is independent of the choice of $\omega$). Say that $\mathcal{F}$ is nowhere holomorphic if $\omega\wedge d\omega$ is nonvanishing for all such $x$ and $U$.

Now, by standard techniques, one can show that if $\mathcal{F}$ is nowhere holomorphic, then the sheaf of vector fields on $X$ whose flows preserve both the complex structure and $\mathcal{F}$ has finite dimensional stalks; in fact, the dimension can never be more than $8$, and it equals $8$ everywhere if and only if $(X,\mathcal{F})$ is locally equivalent to a standard model $(X_0,\mathcal{F}_0)$ defined as follows:

Let $\mathrm{SL}(3,\mathbb{R})\subset\mathrm{SL}(3,\mathbb{C})$ act in the usual way on $\mathbb{CP}^2$. Of course, this action preserves the real slice $\mathbb{RP}^2\subset\mathbb{CP}^2$ and is transitive on the (open) compliment of this real slice. In fact, this action commutes with the natural conjugation $[v]\mapsto[\bar v]$, and for any $[v]\in \mathbb{CP}^2\setminus\mathbb{RP}^2$, there is a unique complex line $[v\bar v]$ passing through both $[v]$ and $[\bar v]$, and $X_0=\mathbb{CP}^2\setminus\mathbb{RP}^2$ is foliated by such `real' lines (the leaves in $X_0$ are Poincaré disks). It is not difficult to see that this foliation $\mathcal{F_0}$ is nowhere holomorphic and that $\mathrm{SL}(3,\mathbb{R})$ acts transitively and holomorphically on $X_0$ preserving $\mathcal{F_0}$. Moreover, any local biholomorphism of a connected open subset $U\subset X_0$ into $X_0$ that preserves $\mathcal{F}_0$ is the restriction to $U$ of the action of an element of $\mathrm{SL}(3,\mathbb{R})$.

What seems possible (but, again, I need to check details and I'm not an expert on discrete subgroups of higher rank Lie groups of split type) is that there might exist a lattice $\Gamma\subset \mathrm{SL}(3,\mathbb{R})$ such that $\Gamma$ preserves a connected open subset $U\subset X_0$ on which it acts freely and properly discontinuously and such that the quotient $M=\Gamma\backslash U$ is compact. If so, that will provide an example of what we are looking for.

Original Attempt (modified): A locally homogeneous (noncompact) example can be constructed as follows: First, let $G\subset\mathrm{SL}(3,\mathbb{C})$ be the $5$-dimensional, connected Lie subgroup whose Lie algebra is $${\frak{g}} = \left\{\ \begin{pmatrix}ir & z & w\\ \bar z & -ir & \bar{w}\\0&0&0\end{pmatrix}\ \ : \ r\in\mathbb{R},\ z,w\in\mathbb{C}\ \right\},$$ let $H\subset G$ be the connected subgroup whose Lie algebra is as above with $w=0$, while $K\subset H$ is the connected Lie subgroup whose Lie algebra is as above with $z=w=0$. Then $H\simeq\mathrm{SL}(2,\mathbb{R})$ while $K\simeq S^1 = \mathrm{SO}(2)$.

Now consider the (non compact) $G$-homogeneous space $M = G/K$. I claim that there is a $G$-invariant complex structure on $M$ and a $G$-invariant nonholomorphic foliation of $M$ by complex curves. To see this, consider the canonical left invariant $1$-form on $G$, written in the form $$\gamma = g^{-1}\ dg = \begin{pmatrix}i\rho & \zeta & \omega\\ \bar\zeta & -i\rho & \bar{\omega}\\0&0&0\end{pmatrix},$$ where $\rho$ is $\mathbb{R}$-valued and $\zeta$ and $\omega$ are $\mathbb{C}$-valued. The structure equation $d\gamma = -\gamma\wedge\gamma$ gives $$d\begin{pmatrix}\zeta \\\omega\end{pmatrix} = -\begin{pmatrix}2i\rho&0 \\-\bar\omega& i\rho\end{pmatrix}\wedge \begin{pmatrix}\zeta \\\omega\end{pmatrix}$$ as well as $d\rho = i\ \zeta\wedge\bar\zeta$.

From these equations it follows immediately that there is unique complex structure $J$ on $M=G/K$ for which the $(1,0)$-forms on $M$ pullback to $G$ to be linear combinations of $\zeta$ and $\omega$, that this complex structure is $G$-invariant, and that there is a unique foliation $\mathcal{F}$ of $M$ by $J$-holomorphic curves such that preimages in $G$ of the leaves of $\mathcal{F}$ are the leaves of the codimension $2$ foliation defined by $\omega=\bar\omega=0$. Moreover, because $\omega\wedge d\omega\not=0$, the foliation $\mathcal{F}$ is not holomorphic.

Now, as I wrote, $M$ is not compact, so it's not the example that you are looking for. However, if we now use the fact that $H\simeq\mathrm{SL}(2,\mathbb{R})$, we can find a discrete subgroup $\Gamma\subset H=\mathrm{SL}(2,\mathbb{R})$ such that $\Gamma$ acts properly and discontinuously on $H/K = \Delta$ (i.e., the Poincaré disk) and such that the quotient $C = \Gamma\backslash (H/K)$ is a compact Riemann surface.

[Here is the step that does not work to produce a desired $\hat\Gamma$:] Choosing $\Gamma$ to be an appropriate congruence subgroup, for example, we can even assume that $\Gamma$ preserves a lattice in $\mathbb{R}^2\simeq\mathbb{C}$. In other words, there exists a lattice $\Lambda\subset\mathbb{C}$ such that the abelian discrete subgroup of $G$ defined as $$\hat\Lambda = \left\{\ \begin{pmatrix}1 & 0 & w\\ 0 & 1 & \bar{w}\\0&0&1\end{pmatrix}\ \ : w\in\Lambda\ \right\},$$ is stable under conjugation by $\Gamma\subset H$.

Finally, let $\hat\Gamma\subset G$ be the subgroup generated by $\Gamma\subset H\subset G$ and $\hat\Lambda$. Then I believe (but haven't had time to check the details) that $\hat\Gamma$ acts properly and discontinuously on $M = G/K$ with compact quotient. Since $\hat\Gamma$, being a subgroup of $G$, preserves the complex structure on $M$ and the foliation $\mathcal{F}$, it follows that these pass to the quotient $X = \hat\Gamma\backslash(G/K)$ and give an example of the desired kind.

This is a question which is related to Lagrangian foliations on hyperkaehler manifolds, but much of the results are conjectures, or unpublished. Let $M$ be a K3, and $\omega$ a cohomology class on a boundary of a Kaehler cone, satisfying $\omega^2=0$ (such clases are called nef). There are quite a few of such classes on K3; for a general non-algebraic K3, for all classes with $\omega^2=0$, either $\omega$ or $-\omega$ is nef.

It follows from the general compactness arguments that $\omega$ is represented by a closed, positive (1,1)-current. If this current is continuous, it has exactly one positive and one vanishing eigenvalue, hence its kernels give a foliation by holomorphic curves.

The tangent bundle of a general non-algebraic K3 has no rank 1 coherent subsheaves, hence this K3 has no holomorphic foliations. However, in Serge Cantat's thesis (unpublished) he showed an existence of a Holder continuous positive closed (1,1)-form representing a class
with $\omega^2=0$, using McMullen's exotic holomorphic automorphisms on non-algebraic K3. This should give such a non-holomorphic foliation by holomorphic curves.

I think that such foliations (without relation to K3) were also discussed in Marco Brunella's paper Brunella, Marco On Kähler surfaces with semipositive Ricci curvature. Riv. Math. Univ. Parma (N.S.) 1 (2010), no. 2, 441–450. but it is not available anywhere, so I cannot check.