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.

  • 3
    $\begingroup$ 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)... $\endgroup$ – YCor Jul 31 '16 at 18:56
  • $\begingroup$ @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. $\endgroup$ – Ali Taghavi Jul 31 '16 at 20:06
  • 2
    $\begingroup$ 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? $\endgroup$ – Paul Reynolds Aug 3 '16 at 22:24
  • $\begingroup$ @PaulReynolds very interesting comment.Some thing to the maximal Ideal spaces of $C(X)$ which corresponds to $X$. $\endgroup$ – Ali Taghavi Aug 4 '16 at 15:04


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.

Old, wrong answer:

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.

| cite | improve this answer | |
  • $\begingroup$ 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? $\endgroup$ – Ali Taghavi Aug 3 '16 at 14:29
  • 1
    $\begingroup$ 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."? $\endgroup$ – David E Speyer Aug 6 '16 at 15:24
  • 1
    $\begingroup$ 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? $\endgroup$ – David E Speyer Aug 6 '16 at 15:30
  • 1
    $\begingroup$ 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. $\endgroup$ – David E Speyer Aug 6 '16 at 16:03
  • 1
    $\begingroup$ 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. $\endgroup$ – David E Speyer Aug 6 '16 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.