# Exponential map to simply connected abelian Lie group is an isomorphism.

In my notes I have the following claim.

Let $$G$$ be a simply connected abelian Lie group. Since $$G$$ is abelian, we have that $$exp(A+B) = exp(A) exp(B)$$ for all $$A,B∈Lie(G)$$. Therefore, the exponential map is an isomorphism between the additive group $$Lie(G)$$ and $$G$$.

I think I understand surjectivity because we know $$exp : Lie(G) \to G$$ restricts to a diffeomorphism on a neighbourhood $$U\ni 0$$. And since $$G$$ is connected, we have that $$\langle U \rangle=G$$ so any element $$g\in G$$ can be written $$g=exp(X_1)^{\pm 1}\ldots exp(X_n)^{\pm 1}=exp(\pm X_1 \pm \ldots \pm X_n)$$ (because $$G$$ is $$G$$ is abelian so $$Lie(G)$$ is abelian as well). But why is it injective and what does it has to do with simple connectivity?

• I'd try to argue that $\exp: \mathfrak{g}\to G$ is the universal cover of $G$ (for example, see this). But $G$ is assumed simply connected. So... Commented Jul 17, 2020 at 23:29

The lie algebra $$\mathfrak{g}$$ is an abelian Lie group, and as you pointed out, $$\exp:\mathfrak{g}\to G$$ is a homomorphism. Since there is a neighbourhood of $$0$$ on which $$\exp$$ is a diffeomorphism, the kernel $$\ker (\exp)$$ of this homomorphism is discrete. Therefore $$\exp$$ is a covering map. Since $$G$$ is simply-connected, it must be a homeomorphism, and in particular bijective. So it is an isomorphism.
Let $$U\ni 0$$ be a neighbourhood of zero such that $$\exp:U\to \exp(U)$$ is a diffeomorphism. In particular, for $$v\in U\setminus\{0\}%$$, we have $$\exp(v)\neq\exp(0)=1$$, since $$\exp$$ is injective on $$U$$. So $$K\cap U = \{0\}$$. But $$K\cap U$$ is open in $$K$$ in the subspace topology by definition, so $$\{0\}$$ is open in $$K$$, which means $$K$$ is discrete.
• Thank you. Why the fact that there is a neighbourhood of $0$ on which $\operatorname{exp}$ is a diffeomorphism implies that the kernel is discrete? Commented Jul 18, 2020 at 16:18
• @roi_saumon Let $U\ni 0$ be a neighbourhood of zero such that $\exp:U\to \exp(U)$ is a diffeomorphism. In particular, for $v\in U\setminus\{0\}%$, we have $\exp(v)\neq\exp(0)=1$, since $\exp$ is injective on $U$. So $K\cap U = \{0\}$. But $K\cap U$ is open in $K$ in the subspace topology by definition, so $\{0\}$ is open in $K$, which means $K$ is discrete. Commented Jul 18, 2020 at 16:53
• Oh, nice! I didn't knew that you just need to show that ${0}$ is open because topological group are homogeneous spaces. Commented Jul 18, 2020 at 17:05