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Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra.

Let $ M$ be a manifold and $\mathfrak{X}(M)$ be its Lie algebra of vector fields on $M$.

Let $G\times M\rightarrow M$ be an action of the Lie group $G$ on the manifold $M$. Let $A\in \mathfrak{g}$. Then we have a vector field $A^*:M\rightarrow TM$ defined as $m\mapsto (\delta_m)_{*, e}(A)$ where $\delta_m:G\rightarrow M$ is given by $g\mapsto gm$. This gives a map $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ defined as $A\mapsto A^*$. It can be checked that this is a morphism of Lie algebras. So, given an action $G\times M\rightarrow M$, one can associate a morphism of Lie algebras $\mathfrak{g}\rightarrow \mathfrak{X}(M)$.

I was thinking of the converse : When can we trace back from a morphism of Lie algebras $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ to get an action $G\times M\rightarrow M$ of the Lie group $G$ on the manifold $M$. More precisely, the question is as follows :

When does a morphism, $\mathfrak{g}\rightarrow \mathfrak{X}(M)$, of Lie algebras gives an action $G\times M\rightarrow M$ of the Lie group $G$ on the manifold $M$?

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  • $\begingroup$ If some one ask me, I would say this should always be the case.. any morphism $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ should give an action of $G$ on $M$ whose associated map of Lie algebras is precisely the one I have started with... But, I do not have any explanation supporting my claim.. $\endgroup$
    – Praphulla Koushik
    Commented Sep 15, 2019 at 17:53
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    $\begingroup$ At best you expect to recover an action of the universal cover of the connected component of $G$. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 15, 2019 at 18:59
  • $\begingroup$ You can think of this as follows: you have a map $G \to Diff(M)$, and differentiating you get a map $\mathfrak{g} \to \mathfrak{X}(M)$. So it's just the usual question of, given a map between Lie algebras, when does it lift to the Lie groups (except here the second one is infinite-dimensional). $\endgroup$
    – Kevin Casto
    Commented Sep 15, 2019 at 19:45
  • $\begingroup$ @QiaochuYuan you can make any assumption on $G$ and/or on $M$ that does not make the question super trivial... how does one recover an action when $G$ is connected? $\endgroup$
    – Praphulla Koushik
    Commented Sep 16, 2019 at 2:52
  • $\begingroup$ @KevinCasto I am not very comfortable with infinite dimensional manifolds/Lie groups, forget about their corresponding Lie algebras :D... Is this comment just to get some intuition or this is actually the case. Can you give some reference that says $\mathfrak{X}(M)$ is precisely the Lie algebra (in the usual sense ?) of the infinite dimensional Lie group $Diff(M)$.. $\endgroup$
    – Praphulla Koushik
    Commented Sep 16, 2019 at 2:56

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Here is a partial answer. Observe that whenever you have a (right) action $\theta$ of $G$ on $M$, your Lie algebra morphism $\hat{\theta} \colon \mathfrak{g} \to \mathfrak{X}(M)$ (it is usually called the infinitesimal generator of $\mathbf{\theta}$) sends every $X$ to a complete vector field (you may check that the flow of $\hat{\theta}(X)$ is given by: $(t, p) \mapsto p \cdot exp(tX)$; this flow is definitely global, which means that the vector field is complete).

Necessary condition: A Lie algebra morphism $\mathfrak{g} \to \mathfrak{X}(M)$ must send every element of $\mathfrak{g}$ to a complete vector field in order to be an infinitesimal generator of some (right) action of $G$ on $M$.

This condition is also sufficient if $G$ is simply-connected:

Let $G$ be a simply-connected Lie group with a Lie algebra $\mathfrak{g}$, and let $\varphi \colon \mathfrak{g} \to \mathfrak{X}(M)$ be a Lie algebra morphism sending every element of $\mathfrak{g}$ to a complete vector field. The there is a unique smooth (right) action of $G$ on $M$ whose infinitesimal generator is $\varphi$.

See Theorem 20.16 in Introduction to Smooth Manifolds by John Lee.

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    $\begingroup$ This is definitely a good observation. Thank you. I do not recall seeing this result when I read Lee... Now I see... Thanks again.. $\endgroup$
    – Praphulla Koushik
    Commented Sep 16, 2019 at 11:48

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