# Equality involving matrix exponentials / Lie group representations

1. Apr 22, 2014

### 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.

Last edited: Apr 22, 2014
2. Apr 23, 2014

### MisterX

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

3. Apr 23, 2014

### 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

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

4. Apr 23, 2014

### 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.