Morphism of Lie algebras giving an action of Lie group on manifolds 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 $M$?

 A: 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. 
