Just trying to understand the exponential map in Lie groups. How do I answer this question:

Show that the Lie algebra of a Lie subgroup $H \subset G$ can be identified with $$\mathfrak{h}=\{X\in \mathfrak{g}:\exp(tX)\in H\text{ for all sufficiently small }t\in \mathbb{R}\}$$


I understand that there is a correspondence between Lie subalgebras $\mathfrak{h}\subset \mathfrak{g}$ and connected Lie subgroups $H \in G$, but what's the relationship with $\exp$ here?


1 Answer 1


Let $X\in \mathfrak{g}$, $X$ defines on $G$ a vector field invariant by left multiplication $\hat X(g)=dL_g.X$ where $L_g(u)=gu$. $exp(X)$ is the value of the flow $\phi_X$ of $\hat X$ at $1$. $dexp_0(\hat X)=X$, thus if $exp(tX)\in H$, ${d\over{dt}}exp(tX)=X\in T_IH=\mathfrak{h}$.

  • $\begingroup$ I still don't quite fully understand. If $X \in \mathfrak{g}$, then isn't $X$ itself a left invariant vector field? Why do we need $\hat{X}$? Also why is $L_g(u)=gu.exp(X)$ instead of just $gu$? Sorry if this is trivial, my understanding of this topic is pretty poor at the moment. $\endgroup$
    – elDin0
    May 11, 2018 at 19:02
  • 1
    $\begingroup$ For me, $X$ is an element of the tangent space at the identity $T_IG$ and $\hat X$, the left-invariant vector associated to $X$: $\hat X(I)=X$. $\endgroup$ May 11, 2018 at 20:14

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