2
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.