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Let $(M^{n+1}, \langle \cdot, \cdot \rangle)$ be a parallelizable Riemannian manifold with a vector bundle isomorphism

$$\varphi : TM \to M \times \mathbb{R}^{n+1}.$$

For $x \in M$, denote by $\varphi_x : T_x M \to \mathbb{R}^{n+1}$ the restriction of $\varphi$ to the tangent space $T_x M$; it is a linear isomorphism.

Given a vector $v \in \mathbb{R}^{n+1}$, define a vector field $V = V_v$ on $M$ by

$$V(x) = \varphi_x^{-1}(v), \quad x \in M.$$

Question: when is the normal distribution associated to $V$ integrable? That is, if $\mathcal{D} = (\mathcal{D}_x)_{x \in M}$ is defined by $\mathcal{D}_x = \{ V(x) \}^{\perp}$, when is $\mathcal{D}$ integrable? What conditions can we impose on $\varphi$ for this to be true?

Eduardo Longa
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1 Answers1

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If you look on foliations from a differential form perspective then parallelizability helps you quite a lot in finding nowhere vanishing one forms $\alpha \in \Omega^1(M)$ on $M$. This allows you easily to find regular hyperplane distributions, i.e. a smooth distribution of constant rank $n$, namely the kernel of your nowhere vanishing one form, $\mathcal{D}_x=\ker \alpha_x $. This distribution $\mathcal{D}$ defines a foliation iff your one form $\alpha $ is closed. So it is a question on what the exterior differential does to the trivializing sections of the cotangent bundle.

In terms of the vector bundle isomorphism you are describing, finding a regular foliation translates into understanding the induced Lie algebra structure on the sections of this trivial bundle, from the Lie bracket on the tangent bundle on $M$. This is due to the Frobenius theorem. For example if you would ask for $\phi$ to be a Lie algebra isomorphism on the sections of the corresponding bundles, where on the trivial bundle the constant sections commute, then any such $v$ would do the job. Though in general that would be a very (probably way too) restrictive condition to ask for.

Florian
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