Let $M$ be a smooth manifold, and $G$ a Lie group with Lie algebra $\mathfrak{g}$. The Lie algebra of the diffeomorphism group of $M$ is the Lie algebra of vector fields on $M$; that is $\text{Lie}(\text{Diff}(M)) = \Gamma(TM)$. 


Given a smooth left-action of $G$ on $M$ expressed as a Lie group homomorphism $\lambda : G \to \text{Diff}(M)$, we obtain a Lie algebra homomorphism $\text{Lie}(\lambda): \mathfrak{g} \to \Gamma(TM)$. 


What is the relation between this Lie algebra homomorphism and the fundamental vector field mapping $\zeta: \mathfrak{g} \to \Gamma(TM)$, which is an anti-homomorphism of Lie algebras? Recall that the fundamental vector field mapping is given by
$$\zeta(X)\_x = T\_e \lambda(-, x).X$$
for any $x \in M$ and $X \in \mathfrak{g}$. Here I have written the group action as a smooth map $\bar{\lambda}: G \times M \to M$.


I don't know much about infinite-dimensional Lie groups and manifolds, personally, but it seems that there should be some simple relationship between these two mappings.


([X-post to stackexchange](https://math.stackexchange.com/questions/1513576/relation-between-the-lie-functor-applied-to-a-lie-group-action-and-the-fundamen))



Relationship between the Lie functor applied to a Lie group action, and the fundamental vector field mapping?
There are some sign issues for Lie algebra actions on manifolds. I will follow the conventions of [[BGV, section 1.1]](http://www.springer.com/us/book/9783540200628).
There is a natural left action on functions by
$$(g.f)(x)=f(\lambda\_{g^-1}x)\;.$$
If you differentiate it, you get
$$(X.f)(x)=\frac d{dt}\Bigr|\_{t=0}f(\lambda\_{e^{-tX}}x)=-df\_x(\zeta(X))$$
for $X\in\mathfrak g$. The right hand side is the usual derivative of $f$ by a vector field. This explains why $\zeta$ is an anti-homomorphism.


