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. 
 A: 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 $G\stackrel{\text{Ad}}{\longrightarrow}\text{GL}({\mathfrak g})$. 
A: First suppose $G$ is a matrix lie group. $\mathrm{ad}_{g^{-1}}X = g^{-1}Xg\,,$ so $e^{t(\mathrm{ad}_{g^{-1}}X)} = e^{tg^{-1}Xg}\,.$ 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 $\frac{d}{dt}e^{t\mathrm{ad}_{g^{-1}}X}\vert_{t=0} = \mathrm{ad}_{g^{-1}}X\,.$
Now from the definition of $\mathrm{ad}_{g^{-1}},$ we have that $\mathrm{ad}_{g^{-1}}X = \frac{d}{dt}g^{-1}e^{tX}g\vert_{t=0}\,.$ Hence they have the same derivative at $t = 0\,,$ so it follows immediately that $g^{-1}e^{tX}g = e^{t\mathrm{ad}_{g^{-1}}X}\,.$
