I am wondering about the following situation. Suppose $X$ is a compact Kahler manifold and $F \subset T_X$ is a holomorphic foliation. Suppose that the ``normal bundle'' $T_X / F$ of the foliation is trivial. Suppose that I have a compact leaf $L \subset X$. Is it true that the holonomy of $L$ is trivial?

My guess is that the holomorphicity of $F$ and $X$ being Kahler is irrelevant but this is the situation I am considering.

Since $L$ has a trivial normal bundle, I want to say that there is a tubular neighborhood of $L$ such that $F$ is taken to the trivial horizontal foliation in $N_L = L \times \mathrm{C}^k$. I do not see how to do this directly.

Choose a basis $\omega_1, \dots, \omega_k$ giving a frame of $(T_X / F)^*$ and choose a ``constant metric'' on $T_X/F$ meaning,

$$ g(X,Y) = \sum_{i = 1}^k \omega_i(X) \omega_i(Y) $$

I claim that this is a bundle-like metric for $F$. Indeed,

$$ (\mathcal{L}_Z g)(X,Y) = Z(g(X,Y)) + g([Z,X], Y) + g(X, [Z,Y]) $$

but the forms $\omega_i$ are harmonic so closed meaning,

$$ Z(\omega_i(X)) - X(\omega_i(Z)) = \omega_i([Z,X]) $$

and if $Z$ is a section of $F$ then $X(\omega_i(Z)) = 0$ so we see that $\mathcal{L}_Z g = 0$ along $F$.

Therefore, since the foliation is Riemannian, every compact leaf should have finite holonomy. Then by Reeb stability, I get a tubular neighborhood $U$ of $L$ which is diffeomorphic to $\tilde{L} \times_H T$ where $\tilde{L} \to L$ is the holonomy cover, $H$ is the holonomy groupoid and $F$ is taken to the standard foliation arising from the fiber bundle structure over $L$. Can I argue that this fiber bundle is trivial if I know $T_X / F$ is trivial?

  • 1
    $\begingroup$ Do you consider the normal bundle with the flat connection induced from the foliation? Then the answer is trivially "yes". Otherwise, without assuming that the foliation is holomorphic, there are counterexamples. The easiest is the Reeb foliation on $T^2$ which has a compact leaf $S^1$, and the holonomy is multiplication with some real number $\lambda\in(0,1)$. I don't know about the holomorphic situation. $\endgroup$ Commented Feb 13, 2023 at 14:52
  • $\begingroup$ @SebastianGoette I am just asking that the normal bundle is trivial as a holomorphic vector bundle. Yes, I think you are right that there are many counterexamples if the bundle is just required to be topologically trivial. $\endgroup$
    – Ben C
    Commented Feb 15, 2023 at 9:13
  • $\begingroup$ Sorry for the wrong counterexample. Do I understand correctly that you prove that the foliation is Riemannian? If so, then indeed the holonomy of L should be trivial. Indeed, if it is finite and non-trivial, consider its lift to a finite cover of L where the monodromy is trivial. Then on this finite cover you get two different holomorhic trivializations of your holomorphically trivial bundle (one coming from the fact that normal bundle is trivial, other from the connection given by the foliation). This is impossible since a holomorphically trivial bundle has unique holomorphic trivialisation. $\endgroup$ Commented Apr 12, 2023 at 19:53

1 Answer 1


I think the answer is yes and the Kähler hypothesis (or some variation) is necessary.

Proof in the Kähler case

Say $F$ is a foliation of codimension $k$ on a closed Kähler manifold $X$ of dimension $n$. If the normal bundle to $F$ is trivial, then there should be holomorphic linearly independent $1$-forms $\omega_1, \ldots, \omega_k$ vanishing along $F$. (I'm not sure what "trivial normal bundle" means if the foliation is singular, but I would guess that it is the same, allowing possibly the forms to be only generically independent).

Now, since $X$ is Kähler, these forms are closed. Integrating these forms locally along a holomorphic transversal (a k-dimensional complex manifold transverse to the foliation), one gets a "transverse translation structure", i.e. local biholomorphisms from transversals to $\mathbb C^k$ defined up to translation and invariant under holonomy (because the forms vanish along the foliation). In particular, the holonomy along a closed leaf is a translation fixing $0$, hence it is trivial.

More concretely, if $c(t)$ is a closed loop tangent to the foliation and $c'(t)$ a path from a transversal to itself following $c(t)$ in a nearby leaf, then $c'$ has to come back to the same point because the integral of the forms $\omega_i$ between $c(t)$ and $c'(t)$ are independent of $t$.

A non-kähler counterexample

Let $\Gamma$ be a uniform lattice in the complex semisimple Lie group $G = \mathrm{SL}(2,\mathbb C)$ and $X = G/\Gamma$. Note that $TX$ has a trivialization given by (quotients of) right invariant vector fields. Let $u$ be the vector $$\left( \begin{matrix}1&0\\0&-1 \end{matrix}\right) \in \mathrm{Lie}(G)$$ and let $V$ be the corresponding right-invariant vector field. The flow of $V$ defines a smooth holomorphic foliation of dimension $1$ whose normal bundle is again trivialized by right-invariant vector fields. (It is isomorphic to $X\times \mathrm{Lie}(G)/\mathbb Cu$).

The flow generated by $V$ is the flow by left multiplication by $\exp(tu)$. Up to conjugating $\Gamma$, we can assume that $\exp(\lambda u)\in \Gamma$ for some $\lambda \notin 2i\pi \mathbb Z$. Then the leaf trough identity is closed (and isomorphic to the elliptic curve $\mathbb C/ \lambda\mathbb Z \oplus 2\pi i \mathbb Z$). One can verify that the holonomy of the foliation along the closed loop $\exp(t\lambda u), 0\leq t \leq 1$ is given by the adjoint action of $\exp(\lambda u)$ on $\mathrm{Lie}(G)/\mathbb C u$, which has infinite order.


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