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Let $G$ a compact connected Lie group. Then, the exponential map $\exp: LG \rightarrow G$ is surjective. (where $LG$ is the Lie Algebra of $G$).

$\textbf{Proof:}$ For any torus $T' \subset G$ we have that $\exp: LT' \rightarrow T'$ is surjective. So we have that \begin{align} T = \text{im}(\exp: LT \rightarrow T) \subset \text{im}(\exp: LG \rightarrow G) \end{align} Let $g \in G$. Then it exists a (maximal) torus $T \subset G$ such that $g \in T$, so we have that $g \in \text{im}(\exp)$ and hence surjectivity. $ \, \, \, _\blacksquare$

I have some problem to understand how can we say directly that for any torus the exponential map is surjective.

Any suggestions? Thanks in advance!

userr777
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1 Answers1

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An $n$-dimensional torus is a commutative Lie group whose Lie algebra is $\mathbb{R}^n$. Consider $e_1,...,e_n$ a basis of $\mathbb{R}^n$, and $\Gamma$ the commutative group generated by $e_1,...,e_n$, $T$ is the quotient $\mathbb{R}^n/\Gamma$. The neutral of $T$ is $p(0)$, and for every $u\in\mathbb{R}^n$, $exp_0(tu)=p(tu)$ where $p$ is the quotient map $\mathbb{R}^n\rightarrow T$. Thus the exponential is just the quotient map $p$.

Tsemo Aristide
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