2
$\begingroup$

Let $G$ be a closed connected subgroup of $\operatorname{GL}(n, \mathbb{R})$. Denote $\mathfrak{g} \subseteq \operatorname{M}(n, \mathbb{R})$ its Lie algebra.

It is known that $X \in \mathfrak{g}$ if and only if $\exp(tX) \in G$ for all $t \in \mathbb{R}$. I'm wondering if it is true that $X \in \mathfrak{g}$ if and only if $\exp(X) \in G$.

$\endgroup$

1 Answer 1

4
$\begingroup$

If $X\in\mathfrak g$ then $\exp(X)\in G$. The converse is not true. For instance take $$ X=\begin{pmatrix}0 & 2\pi \\ -2\pi & 0 \end{pmatrix}. $$ Then $\exp(X)$ is the identity matrix, so it belongs to any closed connected subgroup $G$. However the Lie algebra $\mathfrak g$ does not necessarily contain $X$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .