# Derivative map for connected Lie groups is injective [closed]

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!

• Do you mean to ask if any linear map $T_1G\to T_1H$ can be realized as the differential of a group homomorphism $G\to H$? If so, the answer is no. Commented Sep 4, 2020 at 20:31
• I think that would mean that the map is surjective, wouldn't it? What I'm asking is if the derivative of two group homomorphisms can have the same derivative, given that G and H are connected. Commented Sep 4, 2020 at 20:41

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 $$H$$.