# Lie subalgebras of $\chi^{\infty}(M)$ of codimension $n = \dim M$

For a connected $n$-manifold $M$, the Lie algebra of all smooth vector fields is denoted by $\chi^{\infty}(M)$. For a point $p\in M$ we define $L_{p}=\{X\in \chi^{\infty}(M)\mid X(p)=0\}$. Of course $L_{p}$ is a Lie subalgebra of $\chi^{\infty}(M)$ whose codimension is equal to $n$.

Is it true that every codimension-$n$ Lie subalgebra of $\chi^{\infty}(M)$ is necessarily in the form of (or isomorphic to) $L_{p}$ for some $p\in M$?

Here is a related post.

• The question whether every such subalgebra is equal to some $L_p$ is very natural. The question whether every such subalgebra is isomorphic to some $L_p$ sounds more artificial (at least thus isolated)... – YCor Jul 31 '16 at 18:56
• @YCor I agree that the "Isomorphic" part of my question is not so natural.But on ther extrem if one can find an example of a Lie subalgebra of codimension n but its structure is different from $L_{p}$, then such Lie subalgebra would be very strange. – Ali Taghavi Jul 31 '16 at 20:06
• This makes me want to know what the space of all codimension $n$ Lie subalgebras of $\mathfrak{X}(M)$ looks like, e.g. is it much bigger than $M$ or just a bit bigger? – Paul Reynolds Aug 3 '16 at 22:24
• @PaulReynolds very interesting comment.Some thing to the maximal Ideal spaces of $C(X)$ which corresponds to $X$. – Ali Taghavi Aug 4 '16 at 15:04

# Correction:

The argument I gave initially is wrong. I treated $\mathfrak X(M)'$ like the space of differential forms. Only operations on $\mathfrak X(M)$ go over to the dual as (negative) adjoint operations, so $\mathcal L_X$ makes sense but $i_X$ and $d$ do not. Since it created some interest I leave the old answer.

# Corrected argument:

Let $\alpha$ be a 1-form on $M$ which is non-zero and closed near a point $p$ in $M$. Then consider the current $\alpha. \delta_p \in \mathfrak X(M)'$ where $\delta_p$ is the Dirac delta. Consider the the space $$L^{\alpha}_p = \{X\in \mathfrak X(M): \mathcal L_X(\alpha. \delta_p)=0\}$$ Let us compute this space. The question is local, so we assume that we are in $\mathbb R^n$ and $p=0$. Since $\mathcal L_X$ acts as the negative adjoint, for an arbitrary field $Y$ we have $$0=\langle Y, \mathcal L_X(du^1.\delta_p)\rangle = -\langle \mathcal L_XY, du^1.\delta_p\rangle = -[X,Y]^1(0) = \big(-(\partial_iY^1(0))X^i(0) + (\partial_iX^1(0))Y^i(0)\big)\partial_1$$ Running $Y$ through a basis of $T_0M$ with $\partial_i Y(0)=0$ for all $i$ implies that $\partial_i X^1(0)=0$ $\forall i$. Choosing $Y$ with $Y(0)=0$ in such a way that $dY^1$ runs through a basis of $T_0^*M$, implies $X^i(p)=0$. The converse is also true, thus $$L^{\alpha}_0 = \{X\in \mathfrak X(\mathbb R^n): dX^1(0)=0, X(0)=0 \}$$ which has codimension $2n$, SIGH.

# Remark, and proof of the statement in the question:

In a related question it was hinted that the determination of all the maximal ideals of $\mathfrak X(M)$ would be of interest. These are all of infinite codimension and are of the form: For a point $p$ in $M$ consider all vector fields $X$ which vanish at $p$ of infinite order. Here $M$ should be compact or $\mathfrak X(M)$ should be replaced by the space of vector fields with compact support. This is proved by Purcell and Shanks: For a related result and references see

• MR0516602 Grabowski, J. Isomorphisms and ideals of the Lie algebras of vector fields. Invent. Math. 50 (1978/79), no. 1, 13–33.

In fact, this paper contains a proof of your question: Let $M$ be compact or replace $\mathfrak X(M)$ by the Lie algebra $\mathfrak X_c(M)$ of vector fields with compact support. So let $A=C^\infty_c(M)$ and let $\mathcal L = \mathfrak X_c(M)$. By Proposition 3.6, they satisfy the assumtions of the following theorem.

Theorem 5.1. Let $A$ be an $I$-algebra and let $\mathcal L$ be an admissible $A$-Lie module. Then for each maximal-prime finite-codimensional ideal $J$ of $A$ the Lie subalgebra $\mathcal L_J$ of $\mathcal L$ is maximal finite-codimensional and the mapping $\mathfrak M_A \ni J \mapsto \mathcal L_J\in \mathfrak M_{\mathcal L}$ is a bijection.

Since $\mathfrak M_A = \{A_p: p\in M\}$ and $\mathcal L_J =\{X\in \mathcal L: X(A)\subset J\}$, the result follows. For notation see Grabowski's paper.

Here is a counterexample: Let $\alpha$ be a 1-form on $M$ which is non-zero and closed near a point $p$ in $M$. Then consider the current $\alpha\otimes \delta_p \in \mathfrak X(M)'$ where $\delta_p$ is the Dirac delta. Consider the space of all $X\in\mathfrak X(M)$ with $i_X(\alpha\otimes\delta_p) = 0$ and $\mathcal L_X(\alpha\otimes\delta_p) = 0$. Since $i_{[X,Y]} = [i_X,\mathcal L_Y]$ and $\mathcal L_{[X,Y]} =[\mathcal L_X,\mathcal L_Y]$, this space is a Lie algebra. Its codimension is 2.

More detail: Choose a Riemannian metric $g$ on $M$ and and consider a chart $(U,u)$ centered at $p$ such that $\alpha|_U = du^1$. Then $0 = i_X(du^1\otimes\delta_p) = du^1(X)(p)= X^1(p)$ and $0=\langle Y, \mathcal L_X(du^1\otimes\delta_p)\rangle = \langle Y, (i_Xd + di_X) (du^1\otimes\delta_p)\rangle = -\langle Y, i_X(du^1\wedge d\delta_p)\rangle = -X^1(p) \text{div}(Y)(p) + Y^1(p) \text{div}(X)(p).$

Since $X^1(p)=0$ and $Y$ is arbitrary, we see that the Lie subalgebra is given by $\{X: X^1(p)=0, \text{div}(X)(p)=0\}$ and thus has thus has codimension $2$. The divergence is with respect to the density of $g$.

One can also use $\text{div}(f.X) = f\text{div}(X) + g(\text{grad}^g(f),X)$ to make a more local computation.

• Prof. Michor Thank you very much for your answer. May be I am missing some thing to understand your answer: In the usual metric of $\mathbb{R}^{2}$ is it obvious that the space of all vector field $X$ with $\{X: X^1(p)=0, \text{div}(X)(p)=0\}$ is a Lie algebra? – Ali Taghavi Aug 3 '16 at 14:29
• While I agree with the verification, I am confused by the result. I would have thought that a codimension $k$ subalgebra of $\mathrm{Vect}(M)$ would correspond to a codimension $k$ subgroup of $\mathrm{Diff}(M)$, and thus to a $k$ dimensional space on which $\mathrm{Diff}(M)$ acts. But, if we allow $\mathrm{Diff}(M)$ to act on pairs $(p, a)$ where $a$ is in $T^{\ast}_p M$, we expect an orbit of dimension $2 \dim M$, not $2$. Is there some way to correct my intuition, beyond just saying "infinite dimensional Lie groups are hard."? – DES-SupportsMonicaAndTransfolk Aug 6 '16 at 15:24
• Here is a more concrete question, although it may simply be an ignorance of how currents work: Choose $p$ and $\alpha$ as you say, and let $H$ be the hypersurface $u^1=0$ near $p$. If $X$ is a vector field tangent to $H$, then you show that $i_X(\alpha \otimes \delta_p)=0$. But I think it makes sense to flow $\alpha \otimes \delta_p$ along $X$ for a positive amount of time and, if I do, I think I get a different current, supported at some $p' \in H$ other than $p$. How can the current be killed by the Lie algebra action, but moved by its exponential? – DES-SupportsMonicaAndTransfolk Aug 6 '16 at 15:30
• Actually, I think the computation must be wrong somehow. In $\mathbb{R}^2$, let $X = \partial/\partial x$ and $Y = x \partial/\partial y$. Then $X$ and $Y$ both have coefficient of $\partial/\partial y$ equal to $0$ at the origin, and both have divergence $0$, but $[X,Y] = \partial/\partial y$, for which the coefficient of $\partial/\partial y$ at the origin is nonzero. – DES-SupportsMonicaAndTransfolk Aug 6 '16 at 16:03
• I can't think of any finite codimension subalgebra $L$ which is not contained in some $L_p$. One wants to prove this by reducing to the analogous result for $C^{\infty}(M)$. A subgoal I've been thinking about is to take $M = \mathbb{R}^n$, in which case $\mathrm{Vect}(M) \cong C^{\infty}(M)^{\oplus n}$, and try to show that finite codimension sub-Lie-algebras of $\mathrm{Vect}(M)$ must be $C^{\infty}(M)$ submodules. But no success so far, and I wouldn't be amazed if the reason is that there really are some very unusual sub-Lie-algebras which I haven't thought of. – DES-SupportsMonicaAndTransfolk Aug 6 '16 at 17:25