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Please, how does one show that up to diffeomorphisms there are exactly three finite dimensional Lie algebras of vector fields on the real line $\mathbf R$, namely $\{\partial_x\},\{\partial_x, x\partial_x\},\{\partial_x, x\partial_x, x^2 \partial_x\}$.

Peter
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Nate
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  • Does anyone have a reference at least? – Peter Oct 27 '16 at 14:02
  • Actually, I'm no longer sure I understand the question. For any function $f(x)$, ${f \partial_x}$ is a $1$ dim Lie algebra. Of course, it's isomorphic (as Lie algebra) to ${\partial_x}$. So is the claim that any finite dimensional subalgebra of the vector fields on $\mathbb{R}$ is isomorphic to one of the three you listed? – Jason DeVito - on hiatus Oct 27 '16 at 14:39
  • Yes, that is the claim. So any finite dimensional Lie subalgebra of vect(R) is is diffeomorphic to one of the three. – Peter Oct 27 '16 at 15:40

1 Answers1

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The result is not true. In fact, for any natural number $N$, there is an abelian subalgebra of dimension $N$.

First, every smooth vector field is of the form $f(x)\partial_x$ for some smooh function $f:\mathbb{R}\rightarrow\mathbb{R}$. We compute $[f\partial_x, g \partial_x] = (fg' - f'g)\partial_x$. Also, I will stop writing $\partial_x$ everywhere, simply because it is redundant.

Now, suppose $f_i$ for $i=1$ to $i = N$ is a collection of smooth bump functions where $f_i$ is supported on $[i,i + 1/2]$. This also implies that $f_i'$ is supported in $[i, i+1/2]$.

Now $[f_i, f_j] = f_i f_j' - f_i ' f_j$. Of course, if $i = j$, then this vanishes automatically. On the other hand, if $i\neq j$, then both $f_i f_j'$ and $f_i' f_j$ vanish identically. This follows because because $[i,i+1/2]\cap [j,j+1/2] = \emptyset$ for $i\neq j$.

Thus, $[f_i,f_j] =0$ for any $i$ and $j$, so the subalgebra is abelian.

Jason DeVito - on hiatus
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  • What if you replace smooth by analytic? – Peter Oct 27 '16 at 19:54
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    @Peter: I am not sure what happens in that case, except that the every abelian subalgebra of analytic vector fields is at most 1 dimensional. If you replace "analytic" by "polynomial", then I can prove the result is true: every finite dimensional Lie subalgebra of vector fields on $\mathbb{R}$ with polynomial coefficients is one of the three listed. – Jason DeVito - on hiatus Oct 27 '16 at 19:57
  • Yes, I can also prove it for polynomials. Does it extend to analytic functions as well? – Peter Oct 27 '16 at 21:59