Equality involving matrix exponentials / Lie group representations

[MisterX]

We have that A and B belong to different representations of the same Lie Group. The representations have the same dimension. X and Y are elements of the respective Lie algebra representations.
A = e^{tX}
B = e^{tY}

We want to show, for a specific matrix M

B^{-1} M B = AM

Does it suffice to show this to first order?

\left(1 -tY + \dots \right)M \left(1 + tY + \dots \right) = \left(1 + tX + \dots \right)M

In other words is

-YM + MY = XM

sufficient to show

B^{-1} M B = AM

for all t?

I have seen this used in physics derivations, but it's not clear to me if and why this is sufficient.

 

[MisterX]

Never mind, this came as a result of misreading something.

 

[jostpuur]

I've been wondering the same thing, so I wouldn't mind following some discussion.

According to my experience with theoretical physics, usually in particular situations it is quite easy to prove

B^{-1} M B = AM

directly, and therefore the "Lie algebra way" isn't critical. Of course it would still be interesting to know some theory about the topic.

 

[MisterX]

actually the expression I am trying to prove is

B^{-1} M^{(\mu)} B = \sum_{\nu} A^\mu{}_\nu M^{(\nu)}
That is, A and B are from four dimensional representations, and there are four different 4x4 matrices M^\mu, which form a space closed under conjugation by B. The matrix A is the corresponding element of a different representation, but marvelously A gives the components of B^{-1} M^{(\mu)} B within that subspace spanned by the M^\mu. I think this is not a general problem, we have to get into the specifics. I may follow up in the quantum mechanics forum, in case you are interested.