$G ⊂ GL_n (\Bbb R)$ 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.
Best Answer
As the prove above isn't satisfactory, I am suggesting a new one:
$\ker\ exp_G = \{γ ∈ \mathfrak{g} : exp_G γ = e_G \}$
$exp_G$ 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.