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?

  • $\begingroup$ Well, there's the obvious answer $[\frac{d}{dt} e^X] Y e^{-X} + e^X Y [\frac{d}{dt} e^{-X}] $ (and you know how to write down those two derivatives) ... is that not an acceptable form? $\endgroup$ Oct 6, 2021 at 4:26
  • $\begingroup$ At least somethings based on $d/dt X(t)$ but not $d/dt \exp X(t)$. $d/dt \exp X(t)$ is not easy/obvious to compute/approximate numerically, for example. $\endgroup$
    – Po C.
    Oct 6, 2021 at 5:18
  • $\begingroup$ Sure, that's what I mean when I say that "you know how to write down those two derivatives": Just insert the standard formula that you quote after your sentence "I start with the original formula for the derivative of the exponential map" $\endgroup$ Oct 6, 2021 at 5:33
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    $\begingroup$ @LSpice - yes, that would be nice in general; for present purposes, since the OP states that "actually I just need finite-dimensional matrices", something more pedestrian might suffice ... $\endgroup$ Oct 7, 2021 at 2:12
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    $\begingroup$ @LSpice I am actually doing numerical optimization that's why I need the derivative formula very badly. That simple formula has a great advantage that you can iteratively go higher order derivative. My codes is in finite dimensional of course, but indeed I have sparse linear operator of arbitrary dimension (maximum frequency for my choice of precision) in the first place. $\endgroup$
    – Po C.
    Oct 11, 2021 at 1:51

1 Answer 1


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.


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