Let $$G$$ be a Lie group, and let $$g,h\in G$$. Suppose we have the map $$\Lambda_g:G\to G$$ such that $$h\to ghg^{-1}$$ This induces a map $$\mathfrak{ad}_g$$ on the tangent spaces such that $$\mathfrak{dg}_g: X\to \frac{\partial}{\partial t}ge^{tX}g^{-1}$$ Why do we consider $$e^{tX}$$ instead of $$X$$ directly?
• Doesn't your $ad$ maps need a time derivative? – Jason DeVito Mar 13 at 13:52
• If it's a matrix group, then the $1$-parameter subgroup generated by $X$ is $\gamma:t\mapsto e^{tX}$ (check that $\gamma'(0)=X$), and in particular, we have $e^{tX} \in G$, though it might be the case that $X\notin G$. – Berci Mar 13 at 15:53
• @Berci- The fact that $e^{tX}\in G$ is surprising to me. How do we prove that? Is this just the exponential map that maps tangent vectors to curves in the Lie group? – Anju George Mar 14 at 12:44