Is there an example of a real analytic (compact) manifold $M$ such that the following two lie algebras are isomorphic Lie algebras:

  1. $\chi^{\infty}(M)$, the Lie algebra of all smooth vector fields on $M$.

  2. $\chi^{\omega}(M)$, the Lie algebra of all analytic vector fields on $M$.

(Before any lie algebraic obstruction, Is there a trivial obstruction as cardinality,etc?)

Motivations: Lets reduce (and change) the question to a "ring" setting." Are the ring of smooth and analytic real functions on $M$ isomorphic?" This is a trivial question because the analytic ring is an integral domain but the smooth one is not. So in this trivial case the "zero divisor" is a ring theoretical obstruction for such ismorphism. Now it would be interesting to search for some Lie algebraic obstructions for isomorphicity of $\chi^{\infty}(M)$ and $\chi^{\omega}(M)$. This "ring" situation which I mentioned, shows that (perhaps) no $\mathbb{S}^{n}$ can be an example for my question. Because two isomorphic Lie algebras have isomorphic complexification. on the other hand the complexification of $T\mathbb{S}^{n}$ is the trivial bundle. Moreover sections of the trivial bundles corresponds to "functions".

As my final question:

What are the structures of the enveloping algebras of two Lie algebras under my questions? How can we compare them?


1 Answer 1


The Lie algebra of analytic vector fields on a compact real analytic manifold is simple —this was proved in [Grabowski, J. Isomorphisms and ideals of the Lie algebras of vector fields. Invent. Math. 50 (1978/79), no. 1, 13--33. MR0516602 (80g:57036)]—while the algebra of smooth vector fields has lots and lots of ideals. Indeed, in this last case the algebra has a maximal ideal per point of the manifold —see [Shanks, M. E.; Pursell, Lyle E. The Lie algebra of a smooth manifold. Proc. Amer. Math. Soc. 5, (1954). 468--472. MR0064764 (16,331a)]

  • $\begingroup$ Thank you very much for your answer and very interesting information. $\endgroup$ Mar 22, 2014 at 20:20

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.