I Derivative of the Ad map on a Lie group

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

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Apart from the fact that your formulas doesn't display, you might want to read
where it is described without any general linear group, just manifolds and vector fields.
Beside that, there is also Ado's theorem.
 
(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

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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:
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.
 
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