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

1. Jan 4, 2012

### 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.

2. Jan 4, 2012

### strangerep

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
$$X ~\to~ e^{z_a T^a} X e^{-z_a T^a}$$
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.

3. Jan 4, 2012

### samalkhaiat

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