I'm trying to solve the following problem of the book "Grupos de Lie - Luiz A. B. San Martin":

Question: Let $G$ be a connected Lie Group with Lie algebra $\mathfrak{g}$. Suppose that $X,Y \in \mathfrak{g}$ generate $\mathfrak{g}$ (i.e. $X$ and $Y$ are not contained in any proper subalgebra of $\mathfrak{g}$, or in a equivalent way, the successive brackets of $X$ and $Y$ generates the Lie Algebra $\mathfrak{g}$). Show that the $1$-parameter groups $\text{exp}(tX)$ and $\text{exp}(tY)$ generate $G$.

Consider $ B = \langle \exp (t X), \exp (s Y)\rangle$ as the smallest group $\subset$ $G$, such that $\exp (tX)$ and $\exp (sY)$ belongs to $B$ , $\forall\ t,s \in \mathbb{R}$.

Some ideas.

Once $G$ is connected the problem is equivalent to showing that in a neighborhood $V$ of $1$ in $G$, the result holds. So I'm trying to find a smart way to generate the brackets of $X$ and $Y$, by derivating curves contained in $\langle \exp (tX), \exp (s Y)\rangle $ in $t=0$. For example, something like the curve $\exp(X) \exp(tY) \exp(-X)$, in this specific case

\begin{align*}\left.\frac{d}{dt}\exp(X) \exp(tY) \exp(-X) \right|_{t=0} &=\left.\exp(te^{ad(X)}Y) \right|_{t=0}\\ &= e^{ad(X)}Y\\ &= Y + [X,Y] + \frac{1}{2}[X,[X,Y]] + ..., \end{align*}

If I would be able to construct curves such that $\alpha_i: \mathbb{R} \to G $, such that $\text{Im}(\alpha_i)\subset \langle \exp (tX), \exp (sY)\rangle, $ $\alpha_i(0) = 1$ and $$\left\{\left.\frac{d}{dt} \alpha_1 (t) \right|_{t=0},..., \left.\frac{d}{dt} \alpha_n (t)\right|_{t=0} \right\},$$

is a basis of $\mathfrak{g}$, then the local diffeomorphism at $0$, \begin{align*} \varphi: \mathbb{R}^n &\to G\\ (t_1,...,t_n)&\mapsto \alpha_1(t_1)\cdot ...\cdot \alpha_n(t_n), \end{align*}

would solve the problem, however, this idea has not been fruitful.

Can anyone help me?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.