Derivative map for connected Lie groups is injective

Question

Suppose that $G$ and $H$ are connected Lie groups. I want to show that the map $$\text{Hom}_{\text{ LieGp }}(G,H) \rightarrow \text{Hom}_{ \mathbb{R} }(T_1G, T_1H): \phi \mapsto D\phi $$ is injective. I really have no idea how to approach this one, have been trying for a bit but I'm stuck.

Thanks!

Answer 1

The key here is to use the Lie group exponential map $\exp:\operatorname{Lie}(G) \to G$, which takes 1-dimensional subspaces of the Lie algebra (tangent space at the identity) to 1-parameter subgroups of $G$. A few facts are needed:

1. The exponential map is natural, which means $\phi \circ \exp = \exp \circ d\phi$.
2. The exponential map is a diffeomorphism onto its image when restricted to some open neighborhood of $0$ in $\operatorname{Lie}(G)$.
3. A neighborhood of the identity in $G$ generates $G$ (by connectedness).

What this all tells you is that if you know what $d\phi$ does on the tangent space $\operatorname{Lie}(G)$, then you can use the exponential map to discover what $\phi$ does on a neighborhood of the identity in $G$. But such a neighborhood generates $G$, so you have recovered $\phi$.

Remark: You only need $G$ to be connected here. The image of $\phi$ will have to land in the identity component of $H$.