Why is Casimir operator to be an invariant of coresponding Lie Algebra?

[ndung200790]

Please teach me this:

Why is Casimir operator T^{a}T^{a} be an invariant of the coresponding Lie algebra? I know that Casimir operator commutes with all the group generators T^{a}.

Thank you very much for your kind helping.

 

[strangerep]

Why is Casimir operator T^{a}T^{a} be an invariant of the coresponding Lie algebra? I know that Casimir operator commutes with all the group generators T^{a}.

It's not clear what you're asking.

You said that you know a Casimir operator commutes with all the group generators T^{a}, so I guess you're asking why that means the Casimir is invariant under the group action?

If that's your question, the group action on an arbitrary quantity X is of the form
<br /> X ~\to~ e^{z_a T^a} X e^{-z_a T^a}<br />
where the z_a are ordinary (scalar) parameters, and I'm using implicit summation over the index a.

Expanding the exponentials as a Taylor series then shows that if X commutes with every T^a, then X also commutes with the exponentials. Thus you can pass X through one of the exponentials, and the exponentials cancel. Thus X is unchanged (i.e., invariant).

Or were you asking how to prove that T^a T^a commutes the other generators?
If that's your question, please give a specific example of the Lie algebra.

 

[samalkhaiat]

Please teach me this:

Why is Casimir operator T^{a}T^{a} be an invariant of the coresponding Lie algebra? I know that Casimir operator commutes with all the group generators T^{a}.

Thank you very much for your kind helping.

What exactly is your educational qualification? I ask this because you seem to ask question you know the answer to! If you KNOW that
[T^{a},T^{c}T^{c}] = 0,
then how can’t you know that
\delta^{a}\left(T^{c}T^{c}\right) = [T^{a}, T^{c}T^{c}] = 0
?

Sam