Let $G$ be a Lie group and $g$ its Lie algebra. We denote by $\exp{X}$ the exponential map for an element $X \in g$ and by $\mbox{Ad}$ the adjoint representation of $G$ on $g$ and $\mbox{ad}$. Let $m \subset g$ be a subspace that is invariant under all $\mbox{ad}(X)$ for $X \in g$. Can you give me a proof that it is invariant $\mbox{Ad}(\mbox{exp}(tX)$ for $X \in g$ and $t \in \mathbb{R}$.
[Math] Adjoint representation of exponential on Lie group
lie-groups
Best Answer
I believe the proof just comes down to the formula $${\rm Exp}\,({\rm ad\,}X) = {\rm Ad}(\exp X),$$ where ${\rm Exp}$ denotes the exponential map from ${\rm End}(g)\to {\rm GL}(g)$. This is proven in most texts on Lie groups. Fix some $X\in g$ and $t\in\mathbb{R}$. Then if $Y\in m$, $$ {\rm Ad}(\exp tX)Y = {\rm Exp}\,({\rm ad\,}tX)Y = \sum_{k=0}^\infty \frac{({\rm ad}\,tX)^k}{k!}Y.$$ It should be straightforward to check $({\rm ad}\,tX)^kY/k!\in m$ for each $k\geq 0$, so the partial sums are in $m$ and hence the entire series is in $m$ by closedness. Thus ${\rm Ad}(\exp tX)m\subset m$, as desired.
It might be worth noting that I have implicitly assumed the Lie algebra is finite dimensional here.