# Finding the Lie group when the corresponding Lie algebra is known

Let $M$ be a smooth manifold. Then there is a Lie algebra structure on $C^{\infty}(M)$ (considered as a vector space).

I want to find the corresponding Lie group, $G$. I'm interested in how this works in general, so I will not specify a specific Lie bracket.

I know that there is a correspondence between Lie groups and Lie algebras given by the exponential map.

$$\exp: C^{\infty}(M)\to G$$ For each $f\in C^{\infty}(M)$, $\exp(f):=\gamma(1)$ where $\gamma:\mathbb{R}\to G$ is the unique $1$-PSG of $G$ (that is, continuous homomorphism of groups) such that $\dot{\gamma}(0)=f$.

How can I use this to find $G$?

Obviously, $G$ will depend on the Lie bracket, but I don't see how to use this. Do I need to find a basis for $C^{\infty}(M)$ and try to work out how $\exp$ acts on this basis? I don't really see how this could be done, since nothing is known about $G$.

• $C^{\infty}(M)$ is an algebra (of smooth functions), which is infinite-dimensional. – Dietrich Burde Dec 6 '17 at 20:44
• Start by defining what you mean by a manifold and a Lie algebra. – Moishe Kohan Dec 7 '17 at 3:36
• What Lie algebra structure do you think there is on $C^\infty(M)$? – Mariano Suárez-Álvarez Dec 9 '17 at 17:02
• In any case, the corresponence between Lie groups and Lie algebras breaks down for infinite dimensional Lie algebras and infinite dimensional Lie groups, so it is very hard to make sense of your question. – Mariano Suárez-Álvarez Dec 9 '17 at 17:03
• @MarianoSuárez-Álvarez Any Poisson structure is a Lie algebra structure on $C^{\infty}(M)$. But I didn't know the correspondence breaks down in infinite dimensions. Thanks for your help. – A. Goodier Dec 9 '17 at 17:08