Assume that $M$ is an arbitrary manifold.



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> Is there a Lie subalgebra of $\chi^{\infty}(M)$, the space of smooth vector fields on $M$, whose codimension is equal to one?
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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](https://mathoverflow.net/questions/160449/a-manifold-for-which-chi-inftym-is-rich?rq=1)



The minimum codimension of Lie subalgebra of $\chi^{\infty}(M)$
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](http://arxiv.org/abs/1307.4447). 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.


