Let $X(t)$ be a $C^1$ (continuously differentiable) path in the Lie algebra (actually I just need finite-dimensional matrices). It is well-known (from [Wikipedia page of Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map), also in many Lie algebras/groups textbooks) that
$$\mathrm{Ad}\_{e^{X}} = e^{\mathrm{ad}\_{X}}$$
and that
$$
\frac{d}{dt}e^{X(t)} = e^{X(t)}\frac{1 - e^{-\mathrm{ad}\_{X}}}{\mathrm{ad}\_{X}}\frac{dX(t)}{dt}.
$$


I am wondering, **is there a formula of the adjoint action on exponential map**
$$
\frac{d}{dt} \mathrm{Ad}\_{e^{X(t)}} Y= {?}
$$
where $Y$ is in Lie algebra (or just a matrix).


Please refer to [Wikipedia page of Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map) for the notations for exponential map $e^X$ and adjoint action $\mathrm{Ad}\_{e^X}$:


* $e^X = \sum\_{k=0}^\infty \frac{1}{k!} X^k$
* $\mathrm{Ad}\_{e^X} Y= e^X Y e^{-X}$
* $\mathrm{ad}\_{X} Y= X Y - Y X$.




---


I found in a previous question [On the derivative of the exponential of adjoint action on a Lie algebra](https://mathoverflow.net/questions/270460/on-the-derivative-of-the-exponential-of-adjoint-action-on-a-lie-algebra) in which an [answer](https://mathoverflow.net/a/298882) stated without derivation that (rephrased in notations):
$$
\frac{d}{dt} e^{\mathrm{ad}\_{X(t)}}Y = e^{\mathrm{ad}\_{X(t)}} \left( \mathrm{ad}\_{\frac{d}{dt}X(t)} Y \right)
$$
If such formula is correct, then by the equation (proved as a Lemma in [Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map))
$$\mathrm{Ad}\_{e^{X}} = e^{\mathrm{ad}\_{X}},$$
the answer to my question would simply be:
$$
\frac{d}{dt} \mathrm{Ad}\_{e^{X(t)}} Y
=
\mathrm{Ad}\_{e^{X(t)}} \left( \mathrm{ad}\_{\frac{d}{dt}X(t)} Y \right).
$$


However, I am wondering, is such simple formula too good to be true? **Is there any reference asserts this formula?**




---


I am trying to derive this formula, since this formula was stated without derivation. I start with the original formula for the derivative of the exponential map:
$$
\frac{d}{dt}e^{X(t)} = e^{X(t)}\frac{1 - e^{-\mathrm{ad}\_{X}}}{\mathrm{ad}\_{X}}\frac{dX(t)}{dt}
$$
Let $\tilde{X}(t) = \mathrm{ad}\_{X(t)}$ which is a linear operatior on Lie algebra. Then, with direct substitution:  

$$
\begin{aligned}
\frac{d}{dt}e^{\tilde{X}(t)} 
&= 
e^{\tilde{X}(t)}\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}\frac{d\tilde{X}(t)}{dt}
\\
&=
e^{\mathrm{ad}\_{X(t)}}\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}} \mathrm{ad}\_{\frac{d}{dt}X(t)}
\end{aligned}
$$
The middle term is explicitly
$$
\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}
\mathrm{ad}\_{\frac{d}{dt}X(t)}
=
\sum\_{k = 0}^\infty \frac{(-1)^k}{(k + 1)!}(\mathrm{ad}\_\tilde{X})^k
\mathrm{ad}\_{\frac{d}{dt}X(t)}
$$
If this middle term is indeed identity, we would have the previous simple formula $\frac{d}{dt} e^{\mathrm{ad}\_{X(t)}} = e^{\mathrm{ad}\_{X(t)}} \mathrm{ad}\_{\frac{d}{dt}X(t)}$. In other word, the composition $\mathrm{ad}\_\tilde{X} (\mathrm{ad}\_{\frac{d}{dt}X(t)})$ is zero. To see when it is zero, I expand this composition:
$$
\begin{aligned}
\left(\mathrm{ad}\_\tilde{X} (\mathrm{ad}\_{\frac{d}{dt}X(t)})
\right)Y
&=
\left(
\mathrm{ad}\_{X(t)} \circ \mathrm{ad}\_{\frac{d}{dt}X(t)} - \mathrm{ad}\_{\frac{d}{dt}X(t)} \circ \mathrm{ad}\_{X(t)}
\right)Y
\\
&=
[X,[\frac{d}{dt}X, Y]]
-
[\frac{d}{dt}X,[X, Y]]
\\
&=
[X,[\frac{d}{dt}X, Y]]
+
[\frac{d}{dt}X,[Y, X]]
\\
&=
-
[Y,[X, \frac{d}{dt}X]] \text{ by Jacobi identity}.
\end{aligned}
$$
which requires $[Y,[X, \frac{d}{dt}X]]$ is zero. I guess it is generally not true, unless, for example, $X(t) = t X$, or we can restrict the $X$ and $Y$ satisfy this equation.


At this point, I know that if $[Y,[X, \frac{d}{dt}X]] = 0$, then we have that simple formula, otherwise, I am not sure $\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}\frac{d\tilde{X}(t)}{dt}$ could be simplified. Did I go into a bad direction in deriving the formula?



Derivative of adjoint action of exponential map
Long story short, here is the answer:
$$
\begin{aligned}
\frac{d}{dt}\exp{X(t)}
&=
\exp{X}\frac{1 - \exp(-\mathrm{ad}(X))}{\mathrm{ad}(X)}\frac{dX}{dt}
\\
\frac{d}{dt} \mathrm{Ad}(\exp{X(t)})
&=
\mathrm{Ad}(\exp{X}) \mathrm{ad} \left(\frac{1 - \exp(-\mathrm{ad}(X))}{\mathrm{ad}(X)}\frac{dX}{dt}\right)
\end{aligned}
$$


It is indeed a beautiful formula. I will come back to add more details on the derivation.


