The minimum codimension of Lie subalgebra of $\chi^{\infty}(M)$

Question

Assume that $M$ is an arbitrary manifold.

Is there a Lie subalgebra of $\chi^{\infty}(M)$, the space of smooth vector fields on $M$, whose codimension is equal to one?

If not, what is a counter example?

In particular what is the answer to this question for $M=\mathbb{R}^{2}$ or $M=S^{2}$?

The question is a particular case of the following general question:

Question: For a $n$ dimensional manifold $M$, is it true to say that the minimal codimension of Lie sub algebras of $\chi^{\infty}(M)$ is equal to $n$?

The above question serachs for obstruction for finite codimensionality of Lie subalgebras (codimension equal to one) of $\chi^{\infty}(M)$ while the following post concerns diversity and variation of finite dimensional Lie subalgebras.

A Manifold for which $\chi^{\infty}(M)$ is rich

Answer 1

Let $L$ be a sub-algebra of $\mathrm{Vect}(M)$. I think one might be able to prove $\mathrm{codim}\ L \geq \dim M$ by using a recent result of Hurtado. Here is a sketch of the proposed proof, every step of which is difficult:

(1) Let $G \subset \mathrm{Diff}(M)$ be the group generated by $\exp(L)$. Problem: The exponential need not converge.

(2) Let $N = \mathrm{Diff}(M)/G$. One hopes that $N$ is a manifold of dimension $\mathrm{codim}(L)$. Problem It is not clear how to put a manifold structure on $N$.

We then get a homomorphism $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ by the action of $\mathrm{Diff}(M)$ on the quotient $N$. Hurtado shows that such maps only exist if $\dim M \leq \dim N$.

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We can, at least momentarily, dodge discussions of quotients and exponentials of infinite dimensional Lie groups by defining $N$ to be be the set of subalgebras of $\mathrm{Vect}(M)$ conjugate to $L$ by an element of $\mathrm{Diff}(M)$ (or perhaps $\mathrm{Diff}_0(M)$.)

Hopefully, the set of all codimension $n$ subspaces of $\mathrm{Vect}(M)$ is some sort of manifold by a Grassmannian like construction, and then $N$ can be a closed submanifold of this. If we are lucky, then the tangent space to $N$ at $[L]$ is $\mathrm{Vect}(M)/L$.

We clearly have a map $\mathrm{Diff}(M) \to \mathrm{Bijections}(N)$ and, presumably, once we have built the smooth structure on $N$ this will be a map $\mathrm{Diff}(M) \to \mathrm{Diff}(N)$ and Hurtado's result will tell us that $\dim M \leq \dim N = \dim(\mathrm{Vect}(M)/L)$.

All of this needs to come with the disclaimer that I am a simple minded algebraic geometer, who doesn't like these infinite dimensional objects.

Answer 2

Let $M=S^1=\mathbb R/2\pi \mathbb Z$. Then $\mathfrak X^\infty(M) = \{ f\partial_\theta: f\in C^\infty(M)\}$. Fix a point $\theta_0\in S^1$. The subalgebra $\{f\partial_\theta: f\in C^\infty(S^1): f(\theta_0)=0\}$ has codimension 1. The corresponding Lie subgroup is the group of all diffeomorphisms which fix $\theta_0$.