I'm trying to understand the proof of

$Lie(Ker \pi) = Ker d \pi $


$\pi: G_1 \rightarrow G_2$

is a smooth homomorphism of two Lie groups $G_1, G_2$ and

$d\pi: g_1 \rightarrow g_2$

is it's derivative.

I understand that I can exchange $\pi$ and $d\pi$ in he following sense:

$\pi( exp(tv)) = exp(td\pi(v))$.

What I don't understand is how to prove the following step:

$exp(td\pi(v)) = e \ \forall t \in \mathbb{R} \Rightarrow d\pi(v) = 0$.

Thank you in advance!


1 Answer 1


You have the sequence of group morphisms $\ker\pi \hookrightarrow G_1 \rightarrow G_2$, where the first morphism is inclusion and the second one is $\pi$. Taking differentials, you obtain $Lie(\ker\pi) \hookrightarrow g_1 \rightarrow g_2$.

The differential of inclusion is injective, and its image is contained in $\ker d\pi$, obviously. Since they have the same dimension, they must be equal. Hence, the differential of inclusion is a natural isomorphism between $Lie(\ker\pi)$ and $\ker d\pi$.

  • $\begingroup$ Thank you for the response. I'm not sure I fully understand though. Can we assume the differential of the inclusion is injective since the inclusion is the identity into it's image? Also I don't see where we get that the image of $Lie(ker\pi)$ is in $kerd\pi$. $\endgroup$
    – diesmond
    Commented Aug 17, 2023 at 9:23
  • $\begingroup$ @diesmond About the first question: the differential of any injective smooth map is also injective. About the second one: you know that $\pi\circ i = 1$, the constant $1$ map. -The differential of a constant map is $0$, so $0 = d(\pi\circ i) = d\pi \circ di$, which tells us clearly that any vector in the image of $di$ is going to be in $\ker d\pi$ (the composition of both maps is 0) $\endgroup$
    – Compacto
    Commented Aug 17, 2023 at 9:52

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