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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?

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    $\begingroup$ Doesn't your $ad$ maps need a time derivative? $\endgroup$ – Jason DeVito Mar 13 at 13:52
  • $\begingroup$ 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 $X\notin G$. $\endgroup$ – Berci Mar 13 at 15:53
  • $\begingroup$ @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? $\endgroup$ – Anju George Mar 14 at 12:44
  • $\begingroup$ Useful? $\endgroup$ – Cosmas Zachos Mar 19 at 14:46

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