# Lie Groups/Exponential map identity

I have come across this identity a few times and I have absolutely no idea why it holds.

$g^{-1}\exp(tX)g=\exp(t(\text{ad}_{g^{-1}}X))$

Would any one be able to explain exactly why this holds or point me to a resource that would explain.?

Thanks.

That's a special case of the naturality of the exponential map: If $\varphi: G\to H$ is a morphism of Lie groups with derivative $\text{d}\varphi:{\mathfrak g}\to{\mathfrak h}$, then the following diagram commutes: $$\begin{array}{ccc} G & \stackrel{\varphi}{\longrightarrow} & H \\ \uparrow &&\uparrow\\ {\mathfrak g} & \stackrel{\text{d}\varphi}{\longrightarrow} &{\mathfrak h}\end{array};$$ here, the left and right vertical maps are the exponential maps of $G$ and $H$, respectively.
Applying this to the conjugation automorphism $G\xrightarrow{\text{c}_{g^{-1}}: h\mapsto g^{-1}hg} G$, you have $\text{d}(\text{c}_{g^{-1}})\stackrel{\text{def}}{\equiv}\text{Ad}_{g^{-1}}: {\mathfrak g}\to{\mathfrak g}$, so $$\begin{array}{ccc} G & \stackrel{\text{c}_{g^{-1}}}{\longrightarrow} & G \\ \uparrow &&\uparrow\\ {\mathfrak g} & \stackrel{\text{Ad}_{g^{-1}}}{\longrightarrow} &{\mathfrak g}\end{array}$$ commutes, which is what you were asking for.
Note that $\text{d}\varphi$ is always a linear map, so you can move the scaling by $t$ outside in your equation. Also, one should write $\text{Ad}$ uppercase here, as $\text{ad}$ is usually reserved for the adjoint representation ${\mathfrak g}\to{\mathfrak g}{\mathfrak l}({\mathfrak g})$, which is itself the derivative of $G\stackrel{\text{Ad}}{\longrightarrow}\text{GL}({\mathfrak g})$.
First suppose $G$ is a matrix lie group. ad$_{g^{-1}}X=g^{-1}Xg\,,$ so $e^{t(ad_{g^{-1}}X)}=e^{tg^{-1}Xg}\,.$ So $\displaystyle e^{tg^{-1}Xg}=\sum_n \frac{(tg^{-1}Xg)^n}{n!}=\sum_n g^{-1}\frac{(tX)^n}{n!}g=g^{-1}e^{tX}g\,.$
For a general Lie group, $e^{tY}=\gamma(t)$ where is $\gamma(t)$ is the unique one parameter subgroup with $\gamma(0)=e$ and where $\gamma'(0)=Y\,.$ So it follows from the definition that $\frac{d}{dt}e^{tad_{g^{-1}}X}\vert_{t=0}=ad_{g^{-1}}X\,.$ Now from the definition of $ad_{g^-1},$ we have that $ad_{g^{-1}}X=\frac{d}{dt}g^{-1}e^{tX}g\vert_{t=0}\,.$ Hence they have the same derivative at $t=0\,,$ so it follows immediately that $g^{-1}e^{tX}g=e^{tad_{g^{-1}}X}\,.$