# $\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$

Question statement is as in the title:

$\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$

Since $\varphi$ is a Lie group homomorphism, in particular it is continuous. As $H$ must be Hausdorff, every point set is closed, and hence $\{e\}$ is closed in $H$, which implies that $\ker\varphi = \varphi^{-1}(e)$ is closed as well.

So $N=\ker\varphi$ is an embedded Lie subgroup of $G$. Therefore, the inclusion $j:N\to G$ is a Lie Group homomorphism, and induces a Lie algebra homomorphism $D_e j :T_e N \to T_e G$. So I only need to show that $D_e j$ is in fact an isomorphism between $T_e N$ and $\ker D_e \varphi$.

But I'm stuck at this point. I know that for $X\in T_e N$, we have, for $f\in C^{\infty}(H)$,

$$D_e\varphi (D_e j(X))(f)=D_e j (X)(f\circ \varphi)=X(f\circ \varphi\circ j)=0$$

So $D_ej(T_e N) \subset \ker D_e \varphi$.

At this point, I can either show inverse inclusion: that is, for every $X\in \ker D_e \varphi$, we can find $Y\in T_e N$ such that $D_e j (Y)=X$, or for all $f\in C^{\infty}(G)$, $Y(f\circ j)=X(f)$

Or I can somehow show that dimensions of the two spaces are the same.

But I can't proceed from here.

It'd be great if someone could help me

## 1 Answer

Hint: Use the characterization of the Lie algebra of a closed subgroup i.e. that $X$ lies in the Lie algebra of $N$ if and only if $exp(tX)\in N$ for all $t\in\mathbb R$. Then use how $\varphi(exp(tX))$ can be expressed using $D_e\varphi$.

• Got it, Thanks! – user160738 Feb 20 '17 at 9:48