# $Ker(exp_G )$ is discrete

$$G ⊂ GL_n (\Bbb R)$$ is an abelian connected Lie group, $$\mathfrak{g}$$ its Lie algebra and $$exp_G : \mathfrak{g} → G$$ the exponential map.

Prove that $$Ker(exp_G )$$ is discrete.

My attempt:

$$Lie(Ker(exp_G))=Ker(d_0\ exp_G)$$ since $$exp_G$$ is a morphism of topological groups $$(\mathfrak{g},+) → (G,.)$$, where $$d_0$$ is the differential at $$0 \in \mathfrak{g}$$.

Since $$d_0\ exp_G$$ is the map $$X\to X$$, its kernel is $$\{0\}$$, which gives $$Lie(Ker(exp_G))=\{0\}$$, therefore $$Ker(exp_G)$$ is discrete as $$exp_G$$ is a local homeomorphism at $$0\in \mathfrak{g}$$

Is this a sound proof?

Thank you for your help.

## 1 Answer

As the prove above isn't satisfactory, I am suggesting a new one:

$$Ker\ exp_G = \{γ ∈ g : exp_G γ = e_G \}$$

$$exp_G$$ being a local injection (by restriction of the exponential from $$M_n(\Bbb R)$$), we see that $$0_{\mathfrak{g}}$$ is isolated.

Translation being a homeomorphism of $$Ker\ exp_G$$, every element is isolated and $$Ker\ exp_G$$ is discrete.