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

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    $\begingroup$ Explain the notation please. Does it mean the smooth vector fields? $\endgroup$ – Ben McKay Jul 23 '16 at 20:27
  • $\begingroup$ @BenMcKay yes I mean the space of smooth vector fields.I revise it. Thanks for your comment. $\endgroup$ – Ali Taghavi Jul 23 '16 at 20:31

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 $\dim M \leq \dim N$.

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.

  • $\begingroup$ Thank you very much for your elegant strategy. May be the extra assumption "L is an ideal " makes some facilities? $\endgroup$ – Ali Taghavi Aug 6 '16 at 18:43
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    $\begingroup$ The second step seems fishy, even in finite dimensions; if we integrate an irrational line in $\Bbb R^2$ to a subgroup in the torus, it needn't be closed, so we don't get a manifold structure on the quotient. $\endgroup$ – Mike Miller Aug 7 '16 at 12:16
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    $\begingroup$ @MikeMiller Agreed. I was actually thinking of coming back today to remark on this. There are a few ways we could try to get around this: (1) restrict to cases where $\pi_1(\mathrm{Diff}(M))$ is trivial (2) replace $\exp(L)$ by its closure (but then we have to argue that its closure isn't the whole group) (3) See if Hurtado's proof adapts to the universal cover of $\mathrm{Diff}(M)$. $\endgroup$ – DES-SupportsMonicaAndTransfolk Aug 7 '16 at 14:14
  • $\begingroup$ @DavidSpeyer What about the following alternative opposit approach for compact symplectic manifold $(M, \omega)$ with volum form $\Omega=\omega^{n}$: $\endgroup$ – Ali Taghavi Aug 7 '16 at 14:56
  • $\begingroup$ 1)Is the space $S$ of all $f$ with $\int_{M} fd\Omega =0$ a Lie algebra (with poisson structure on $C^{\infty}(M)$? \new line 2) Is there a non zero Lie algebra morphism from $\chi^{\infty}(M) $ to $C^{\infty}(M)$? Is there a Lie algebra Morphism whose image is not contained in the above $S$. If the answers to the above questions would be positive then we have a codimension one Lie subalgebra of $\chi^{\infty}(M)$. $\endgroup$ – Ali Taghavi Aug 7 '16 at 14:57

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 $\theta_0$.

  • $\begingroup$ Prof. Michor thank you for your answer. I think for every n dim manifold there is a codimension n subliealgebra consist all vector fields vanishing at a given point p.But in my question I did not assume any restriction on M. So I rwvise the question. $\endgroup$ – Ali Taghavi Jul 24 '16 at 14:32

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