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, 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 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 in which an answer 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) $$\mathrm{Ad}_{e^{X}} = e^{\mathrm{ad}_{X}},$$ 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?
