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.

 

[blue_raver22]

Yes, it suffices to show this to first order. This is because the matrix exponentials A and B are defined as the solutions to the differential equations dA/dt = XA and dB/dt = YB, respectively. Therefore, to first order, we can approximate A and B as A = I + tX and B = I + tY, where I is the identity matrix. Substituting this into the equation B^{-1}MB = AM, we get:

(I - tY)M(I + tY) = (I + tX)M

Expanding this out and keeping only terms up to first order, we get:

M - tYM + tMY = M + tXM

Simplifying and canceling out the M terms, we get:

-YM + MY = XM

which is the equation we wanted to show. Therefore, we can conclude that B^{-1}MB = AM holds to first order and, by extension, for all values of t. This is because the Lie group representations A and B are defined as the solutions to the same differential equation, and thus their first-order approximations will always be equal.