1. Nov 26, 2011

### unchained1978

In working with these vierbein fields I've come across these terms such as $e^{aμ}$$e^{μ}_{a}$ where the e's are vierbein fields. The thing is I have no idea what this represents because of the repeated μ's.You can rewrite this with the local lorentz metric to raise and lower the a's and b's but you're still left with identical greek indices. Any help would be greatly appreciated.

2. Nov 26, 2011

### George Jones

Staff Emeritus
Could you give a reference to where you see this?

3. Nov 26, 2011

### Matterwave

I think Wald will use, for example, the notation $(e_\mu)^a(e_\mu)_a$ I wasn't able to figure that out too well either.

I think the latin index should just tell you that it's a vector or a one form, and the Greek index should tell you which basis vector he's talking about.

4. Nov 26, 2011

### unchained1978

I was decomposing the curvature tensor in terms of vierbeins, and when I contracted it with the vierbein I ended up with a term like the one mentioned. I can't get rid of it.

5. Nov 26, 2011

### Bill_K

If it arose legitimately, the value is eμa eμa = δμμ = 4.

6. Nov 26, 2011

### dextercioby

It must have been a typo or something, both Lorentz and world indices must be used repeatedly, only if summed over and summation should be <covariant vs contravariant>, so that

$$e^{\mu}_{~a} e^{\mu a}$$

is wrong, while

$$e^{\mu}_{~a} e_{\mu}^{~a}$$

is correct.

7. Nov 28, 2011

### unchained1978

Not to disagree with you, but I've come at this problem a few different ways now and I always get stuck on the same term. It's not a typo, it somehow represents the inner product of a vierbein field with itself.

8. Nov 29, 2011

### dextercioby

The 'inner product' is still expressible in terms of the metric, therefore covariant, so that my statement from point 6 applies.