I Derivative of the Ad map on a Lie group

[eipiplusone]

Hi,

let $G$ be a Lie group, $\varrho$ its Lie algebra, and consider the adjoint operatores, $Ad : G \times \varrho \to \varrho$, $ad: \varrho \times \varrho \to \varrho$.

In a paper (https://aip.scitation.org/doi/full/10.1063/1.4893357) the following formula is used. Let $g(t)$ be a smooth curve on $G$, with $\frac{d}{dt}|_{t=0} g(t) = v$. And let $u$ be some arbitrary element in $\varrho$. Then,

$$\frac{d}{dt}|_{t=0} Ad_{g(t)} u = ad_{\frac{d}{dt}|_{t=0} g(t)} u = ad_{v} u .$$

I know that this identity holds for Matrix groups, but the present setup is a general Lie group.

Furthermore, in the book "Dynamical systems and geometric mechanics", the above property is actually used as a definition of the $ad$-map, for any $v \in \varrho$ and any curve $g$ with $g'(0) = v$.

Any hints as to why the formula is true would be greatly appreciated.

 

[fresh_42]

Apart from the fact that your formulas doesn't display, you might want to read
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/where it is described without any general linear group, just manifolds and vector fields.
Beside that, there is also Ado's theorem.

 

[eipiplusone]

(hopefully, the equations display now).

I don't see anything about the Ad-map in Pantheon-link - am I missing something?

Regarding Ado's theorem. As I understand it, it can be used to show that any finite dimensional Lie group is locally isomorphic to a Linear group (a matrix group). I guess that might allow us to "transfer" the formula from the Linear group (on which it holds) to the abitrary Lie group. I will think about that. If you have any hints as to how the argument would go, I would love to hear them.

 

[fresh_42]

I don't see anything about the Ad-map in Pantheon-link - am I missing something?

$\mathcal{L}_X(Y) = \operatorname{ad}X (Y) = [X,Y]$.

The most important formula is $\operatorname{Ad} exp (\rho) = exp (\mathfrak{ad}\rho)$.

$\operatorname{Ad}$ is induced by the conjugation in the group, leading to a conjugation with group elements on its tangent space, which if differentiated results in the left multiplication $\mathfrak{ad}$ in the Lie algebra.

You don't need Ado, I just mentioned it to emphasize that "not a general linear group" is relative. Of course there are local Lie groups as well, which do not naturally allow a matrix representation. Here's a nice example:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/which can also serve as an easy to calculate example of the curves $g(t)$ you mentioned.

If you really want to dive in the subject, then I recommend Varadarajan's book on Lie groups. But for a quick look, the explanation of Lie derivatives and the examples there will do.