4
$\begingroup$

Is there a manifold $M$ for which $\chi^{\infty} (M)$, the lie algebra of smooth vector fields on $M$ contains all finite dimensional Lie algebras(Up to isomorphism)?

A weaker question:

Is there a manifold $M$ such that for every $n>0$, $\chi^{\infty} (M)$ contains an n dimensional Lie subalgebra?

$\endgroup$
12
$\begingroup$

The answers to your questions are 'no' and 'yes'.

In the first place, there is no finite dimensional manifold whose Lie algebra of vector fields contains all of the Lie algebras ${\frak{sl}}(n,\mathbb{R})$ for all $n\ge 2$.

In fact, ${\frak{sl}}(n{+}2,\mathbb{R})$ cannot occur as a Lie algebra of vector fields on any $n$-manifold. The reason is that this Lie algebra does not have any nontrivial subalgebras of codimension less than $n{+}1$. If $\frak{g}$ were a subalgebra of the vector fields on $M^n$, then, for any $x\in M$, the subalgebra ${\frak{g}}_x$ consisting of the elements of $\frak{g}$ that vanish at $x$ would have codimension at most $n$ and hence would have to satisfy ${\frak{g}}_x={\frak{g}}$, i.e., all of the vector fields in ${\frak{g}}$ would have to vanish at all of the points of $M$, which is absurd.

In the second place, $M = \mathbb{R}^2$ has the property that there is a Lie subalgebra of dimension $n$ of the vector fields on $M$ for every $n\ge0$.

Technically, you can have Lie subalgebras of the vector fields on $\mathbb{R}^1$ of arbitrarily high dimension: Just let $X_k$ be any vector field on $\mathbb{R}^1$ with compact support contained in the interval $[k,k{+}1]$. Then the $X_k$ all Lie-commute and the algebra that they generate contains the $n$-dimensional (abelian) subalgebra spanned by $X_1,\ldots, X_n$.

However, this is kind of a cheating example since, at any one point, you can only 'see' a $1$-dimensional subalgebra. What you'd really like is an example in which the entire algebra injects into the vector fields on any open subset when you restrict to that open subset. With that extra assumption, you can't have finite dimensional Lie algebras of arbitrarily high dimension acting on $\mathbb{R}^1$; you have to go to dimension $2$ at least.

In dimension $2$, you can have arbitrarily high dimensional Lie subalgebras of, say the torus: Let $x$ and $y$ be $2\pi$-periodic coordinates on the torus, and consider the vector fields $$ X = \frac{\partial\ }{\partial x} \qquad\text{and}\qquad Y_k = \cos kx\ \frac{\partial\ }{\partial y} \qquad\text{and}\qquad Z_k = \sin kx\ \frac{\partial\ }{\partial y}. $$ Then, for any $n$, the vector fields $X,Y_0,\ldots,Y_n,Z_0,\ldots,Z_n$ form a (solvable) Lie algebra of vector fields on the torus. Moreover, they are the infinitesimal generators of a transitive Lie group action on the torus.

$\endgroup$
  • $\begingroup$ Prof. Bryant Thank you very much for your answer. May I ask you that you more explain?: Why it does not work for $M=\mathbb{R}$? as you started from $\mathbb{R}^{2}$. What is the reason for $\mathbb{R}^{2}$?Is this situation possible for a compact manifold for example spheres? $\endgroup$ – Ali Taghavi Mar 15 '14 at 19:54
  • $\begingroup$ I'll add the explanations in the answer above. $\endgroup$ – Robert Bryant Mar 15 '14 at 20:06
  • $\begingroup$ Prof. Bryant, My deep thanks and respect to you for your help and very interesting answer. $\endgroup$ – Ali Taghavi Mar 15 '14 at 22:10
  • $\begingroup$ Is there a nontrivial right action of $SL(n+2,\mathbb{R})$ on $\mathbb{R}^{n}$ or $S^{n}$? According to your answer, such actions can not be effective. So it would be interesting to find an element $g\neq e$ which fixss all elements. $\endgroup$ – Ali Taghavi May 11 '15 at 18:12
  • $\begingroup$ @AliTaghavi: No, there is not. By a theorem of Cartan, $\mathrm{SL}(n{+}2,\mathbb{R})$ cannot act nontrivially on any manifold of dimension less than $n{+}1$. $\endgroup$ – Robert Bryant May 11 '15 at 18:39

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.