# Help With Kaluza–Klein reduction using Cartan

1. Apr 19, 2014

### smiller

Hey everyone. I found a pretty good paper on Kaluza–Klein reduction here. However, I am having a hard time deriving eq. (1.10) which is sort of fundamental. (1.10) are the spin connections via the Cartan structure equations.

I was wondering if anyone could be kind enough to fill in the steps to arrive at (1.10).

What I have so far, just to show where I am stuck, is this. Let's try and get $$\hat{\omega}^{ab}$$

So I start with the basic Cartan structure equation for no torsion: $$d\hat{e^a} + \hat{\omega}^a_b \hat{e}^b = 0$$. Now I try and calculate the first term of the left hand side:

$$d\hat{e^a} = d(e^{\alpha \phi} e^a) = e^{\alpha \phi} d(e^a) + d(e^{\alpha \phi}) e^a \\ \ \ \ \ \ = e^{\alpha \phi} d(e^a) + \alpha e^{\alpha \phi} \partial_b \phi dx^b \wedge e^a \\ \ \ \ \ \ = - e^{\alpha \phi}\omega^a_b \wedge e^b + \alpha e^{\alpha \phi} \partial_b \phi dx^b \wedge e^a \\ \ \ \ \ \ = - \omega^a_b \wedge \hat{e}^b + \alpha \partial_b \phi dx^b \wedge \hat{e}^a \\ \ \ \ \ \ = -\hat{\omega}^a_b \hat{e}^b$$

So here I am stuck. First, how do I get $$dx^b \rightarrow \hat{e}^b$$ so I can combine everything to get the spin connection? And from here, where do I get the term:

$$1/2 F^{ab}e^{\beta - 2 \alpha} \hat{e}^z$$? I have tried many guesses for how to get these to match (1.10) and nothing works out quite right.

Could someone be kind and show me the additional steps needed to get (1.10)? I know this is sort of tedious to write out but I am stuck and so would really appreciate it. Thank you.

Last edited: Apr 19, 2014