By definition, the action of a Lie algebra $\mathfrak g$ on a manifold $M$ is a Lie algebra homomorphism, $\mathcal A: \mathfrak g\rightarrow\mathfrak X(M), \xi\mapsto\xi_M$ such that the action map $\mathfrak g\times M\rightarrow TM, (\xi,m)\mapsto\xi_M(m)$ is smooth.

Now, if we know the action of a Lie group $G$ on the manifold $M$, $G\times M\rightarrow M, (g,m)\mapsto gm$, then an action of the corresponding Lie algebra $\mathfrak g$ is thus


First, I would like to show that this action is linear in $\xi$. Pick $a\in\mathbb R$, one obtains $\frac{d}{dt}|_{t=0}\exp(-ta\xi)m=a\frac{d}{dt'}|_{t'=ta=0}\exp(-t'\xi)m=a\xi_M$. Then, if you pick a second vector $\eta\in\mathfrak g$, form a 3rd one $\xi+\eta=\zeta$ and consider the induced vector on $M$,


I want to prove $\zeta_M=\xi_M+\eta_M$, but got stuck. How can I proceed from here? Thank you very much!


If $\rho: G\times M\to M$ denotes the action map, the chain rule gives you

$$\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\text{exp}_{\mathfrak g}(-t\xi).m = \left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\rho(\text{exp}_{\mathfrak g}(-t\xi),m)\\ = \text{d}\rho_{(e,m)}\left[\left(\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\text{exp}_{\mathfrak g}(-t\xi),0_{\text{T}_mM}\right)\right]=\text{d}\rho_{(e,m)}(-\xi,0_{\text{T}_mM}).$$

since $\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\text{exp}_{\mathfrak g}(-t\xi)=-\xi$ by definition of the exponential function. From this description the linearity in $\xi$ follows from the linearity of $\text{d}\rho_{(e,m)}$.

For the smoothness of this action, you can you use the functoriality of the tangent bundle construction: If $f: M\to N$ is a smooth map, then $\text{d}f: \text{T}M\to\text{T}N$ is smooth as well. Apply this this to $\rho: G\times M\to M$ and use that $\text{T}G\cong \text{T}_eG\times G = {\mathfrak g}\times G$ through translation to construct the desired map ${\mathfrak g}\times M\to M$.


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