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 usual 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 $$\zeta(X)_x = T_e \bar{\lambda}(-, x).X$$ 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, but it seems that there should be some simple relationship between these two mappings.

(X-post to mathoverflow)


1 Answer 1


So I have discovered that indeed that indeed $\zeta = T\lambda: \frak{g} \to \frak{X}$$(M)$.

There is a simple reason why this mapping is an antihomomorphism, even though the Lie functor should take Lie group homomorphism to Lie algebra homomorphisms.

The Lie bracket of most Lie algebras is defined via left-invariant vector fields on the Lie group, but the bracket on $\frak{X}$$(M)$ is usually defined in such a way that it corresponds to right-invariant vector fields on $\text{Diff}(M)$. Thus, if we chose to define Lie brackets for all of our Lie algebras consistently using the right-invariant convention (as Olver does, for example), then the fundamental vector field mapping will be a homomorphism.


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