Use the Jacobi identity to show Lie algebra structure constant id.

[pdxautodidact]

Homework Statement

Use the Jacobi identity in the form
$$ \left[e_i, \left[e_j,e_k\right]\right] + \left[e_j, \left[e_k,e_i\right]\right] + \left[e_k, \left[e_i,e_j\right]\right] $$

and $\left[e_i,e_j\right] = c^k_{ij}e_k$ to show that the structure constants $c^k_{ij}$ satisfy the identity
$$ c^h_{im}c^m_{jk} + c^h_{km}c^m_{ij} + c^h_{jm}c^m_{ki} = 0 $$

Homework Equations

The Attempt at a Solution

Not sure where to start with this one. Using the definition of the structure constant I can show the Jacobi identity equals zero, but does this imply the structure constant identity is equal to zero? I don't see it, if so. Anyway, still not homework, I'm doing this stuff by myself. Also, Einstein convention is enforced (as always).


Any advice for this would be better than the solution, I just started working on it today and moved on so I could sleep on it.

cheers.

 

[pdxautodidact]

Solved!

I see now, $c^h_{im}c^m_{jk}$ is really equivalent to $\left [e_i,c^m_{jk}e_m\right]$. i suspect the following two follow suit. Still could use help on that commutation bit! Or any advice on how to write this one out without using an entire page of my legal pad.

 

[fzero]

I see now, $c^h_{im}c^m_{jk}$ is really equivalent to $\left [e_i,c^m_{jk}e_m\right]$. i suspect the following two follow suit. Still could use help on that commutation bit! Or any advice on how to write this one out without using an entire page of my legal pad.

You are on the right track. Just note that $\left [e_i,c^m_{jk}e_m\right]= c^m_{jk} \left [e_i,c^m_{jk}e_m\right]$ and apply the expression for the commutator once again. If you do this for the other terms, you should find that the Jacobi identity is equivalent to the expression with the structure constants times a basis element $e_h$. If you choose dummy indices consistently, it should just take a couple of lines to establish the result.

 

[pdxautodidact]

You are on the right track. Just note that $\left [e_i,c^m_{jk}e_m\right]= c^m_{jk} \left [e_i,c^m_{jk}e_m\right]$ and apply the expression for the commutator once again. If you do this for the other terms, you should find that the Jacobi identity is equivalent to the expression with the structure constants times a basis element $e_h$. If you choose dummy indices consistently, it should just take a couple of lines to establish the result.

Did you mean that $c^m_{jk}\left[e_i,e_m\right] = [e_i,c^m_{jk}e_m]$?
Because of bilinearity? If so, thanks it's really easy, if not could you please explain?

cheers

 

[fzero]

Did you mean that $c^m_{jk}\left[e_i,e_m\right] = [e_i,c^m_{jk}e_m]$?
Because of bilinearity? If so, thanks it's really easy, if not could you please explain?

cheers

Yes, I forgot to cut after I pasted. It is indeed because of the bilinearity of the Lie bracket.