1

I encountered a theorem about Lie algebra of Lie subgroups. It reads (for $G$ a Lie group and $H$ a Lie subgroup) :

${\rm Lie}(H)=\{X \in {\rm Lie}(G):\exp(tX) \in H\}$.

First, I don't really understand this intuitively. Second, when I think about it, the inclusion "$\subset$" I think I understand : $X\in {\rm Lie}(H) \iff tX \in {\rm Lie}(H)$ since ${\rm Lie}(H)$ is a $\mathbb{R}$-vector space so by definition of the exponential mapping, we must have $\exp(tX) \in H$. But if $\exp(tX)\in H$ why must we have $X\in {\rm Lie}(H)$?

cqfd
  • 12,219
  • 6
  • 21
  • 51
roi_saumon
  • 4,196
  • 1
    The condition should be "there exists $\varepsilon >0$ such that $\mathrm{exp}(tX) \in H$ for all $-\varepsilon < t < \varepsilon$". Intuitively, $\mathrm{Lie}(G)$ is the tangent space of $G$ at the identity element, and the $\mathrm{log}, \mathrm{exp}$-correspondence is an actual isomorphism near $0$ of the tangent space and near identity on $G$. Given a Lie subgroup, in particular closed submanifold, $H$, its tangent space $\mathrm{Lie}(H)$ gets identified with a vector subspace of $\mathrm{Lie}(G)$. To find which one, one uses the fact that the $\mathrm{log}, \mathrm{exp}$-correspondence... – Pavel Čoupek Jun 19 '20 at 04:00
  • 1
    .. one uses the fact that the $\mathrm{log}, \mathrm{exp}$-correspondence should be an isomorphism near $0$ for $H$ also, so the "directions" of the tangent subspace can be recognized as those vectors that are mapped to $H$ when scaled to all sufficiently small magnitudes. – Pavel Čoupek Jun 19 '20 at 04:03

1 Answers1

1

We can view $\operatorname{Lie}(H)=\mathfrak{h}$ as $T_eH\subseteq T_eG=\mathfrak{g}$. That is, we can view the Lie algebra as the tangent space at the identity. The tangent space consists of the trajectories of curves through $e$. That is, in general we can write for $p\in M$ a manifold $$ T_pM=\{\gamma'(0)\:|\:\gamma:(-\varepsilon,\varepsilon)\to M\:\text{is smooth}, \gamma(0)=p\}.$$ So, in this case we can see \begin{equation} T_eH=\{\gamma'(0)\:|\:\gamma:(-\varepsilon,\varepsilon)\to H\:\text{is smooth}, \gamma(0)=e\}. \end{equation} Now, choose $X\in \mathfrak{h}$. $L(X)=\{tX:t\in\Bbb{R}\}\subseteq \mathfrak{h}$ is a one-dimensional linear subspace. Define $\gamma_X: L(X)\to H$ by sending $tX\mapsto \exp(tX)$. By the properties of the exponential, $$ \gamma_X'(0)=\frac{d}{dt}\bigg|_{t=0}\exp(tX)=X$$ and it follows that $X\in \frak{h}$. So, every element of $\mathfrak{h}$ is the exponential of an element of $\mathfrak{g}$ (in fact of an element of $\mathfrak{h}$). Conversely, we claim that given an element of $Y\in\mathfrak{g}$ so that $\gamma(t)=\exp(tY)\in H$ for some $t\in (-\varepsilon,\varepsilon)$ and $\gamma(0)=e$, we have by the characterization above that $Y=\gamma'(0)\in \mathfrak{h}$, so the converse follows.

Intuitively, this is saying that at $e\in G,$ the tangent directions corresponding to $T_eH\subseteq T_eG$ are exactly those directions along which you can walk for a short period of time without leaving $H\subseteq G$. As a simple example, you can think of trying this with $S^1\subseteq T^2$, the torus $S^1\times S^1$.

Alekos Robotis
  • 26,184
  • Tank you. I don't really the forward direction. Can't we just say : "if $X \in \mathfrak{h}$ then $exp(tX)\in H$"? – roi_saumon Jun 19 '20 at 10:38
  • Also I am reading from introduction to smooth manifolds from Lee and there is not this $(-\epsilon,\epsilon)$ thing.-------------------------------------------------------------------------- Proposition 20-9. Lert $G$ be a Lie group, and let $H \subset G$ be a Lie subgroup. With Lie(H) considered as a subalgebra of $Lie(G)$ in the usual way, the exponential map of $H$ is the restriction to $Lie(H)$ of the exponential map of $G$, and $Lie(H)={X \in Lie(G) : exp(tX)}\in H \text{for all } t \in \mathbb{R}}$ – roi_saumon Jun 19 '20 at 13:42
  • Why in this case we have $\forall t\in \mathbb{R}$? – roi_saumon Jun 26 '20 at 12:31