Let $\varphi:G\times M\rightarrow M$ be a Lie group $G$ (effective) action on a smooth manifold $M$.

Fix $g\in G$ and let $\varphi_g(x):=\varphi(g,x)$. Then $\varphi $ induces a smooth homomorphism $\lambda: G\rightarrow Aut(M);g\mapsto \varphi_g$.

Now, if $\frak g$ is the Lie algebra of $G$ and $\Gamma(M,TM)$ the algebra of smooth vector-fields on $M$. Then, is it true that $\lambda $ induces a homomorphism $\alpha:\mathfrak g \rightarrow \Gamma(M,TM)$? How $\alpha $ is defined? Is $\alpha$ injective? Can we say that the Lie algebra of $G$ is embedded in the Lie algebra of smooth vector-fields of $M$?


You get a linear map $\alpha\colon \mathfrak{ g} \to \Gamma(M,TM)$ as follows: For each $x\in M$ and $X\in \mathfrak g$, $$ \alpha(X)_x = \left.\frac{d}{dt}\right|_{t=0} (\exp tX)\cdot x. $$ However, in the case of a left action, this is a Lie algebra anti-homomorphism, not a homomorphism. In the case of a right action, it's a homomorphism.

This is proved in my Introduction to Smooth Manifolds, Theorems 20.15 and 20.18.

  • 1
    $\begingroup$ @Ronald: You're welcome, and thank you!! $\endgroup$ – Jack Lee Oct 8 '17 at 22:37

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