Quick tetrad/vierbein question

[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.

 

[George Jones]

Could you give a reference to where you see this?

 

[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.

 

[unchained1978]

Could you give a reference to where you see this?

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.

 

[Bill_K]

If it arose legitimately, the value is e^μ_a e_μ^a = δ^μ_μ = 4.

 

[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.

 

[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.

 

[dextercioby]

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