G Lie connected group, show that exp: Lie(G) $\rightarrow$ G is an Homomorphism of group iff G is abelian.

I think I've made the $\Leftarrow$:

If G is abelian $\Rightarrow$ Lie(G) is abelian $\Rightarrow$ $[X,Y]=0$ $\Rightarrow$ thanks to the Backer-Hausdorf-Campbell formula $exp(X+Y)=exp(X)exp(Y)$ $\Rightarrow$ exp is an Homomorphism (since G is connected and therefore exp is surjective).

How can I see the $\Rightarrow$ implication?


Assume $\exp : \text{Lie}(G) \to G$ is a morphism. Then it's clear that elements in the image of $\exp$ commute, since $\exp(X) \exp(Y) = \exp(X + Y) = \exp(Y + X) = \exp(Y) \exp(X)$. Since $\exp$ is a local diffeo at $0$, it follows the identity $e \in G$ has an abelian neighborhood $U$ (i.e. one in which every two elements commute). It's well-known that, since $G$ is connected, any element of $G$ is a product of elements in $U$, and from this it follows that $G$ is abelian.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.