Spin connection in higher dimension I have a problem regarding computation of spin connection in the case where One or more dimension is compactified. For example if we take a $D+1$ dimensional bosonic string action and write the $D+1$ dimensional metric in terms of $D$ dimensional fields,and we want to compute the spin connection then how to exactly do it.
I am mainly referring to calculation of 1.10 in the following paper about Kaluza-Klein Theory by C. Pope.
Edit-I actually used Cartan's first structure equation with zero torsion to get something useful but it did not work. Like for $$\hat{\omega}^{ab}$$,I used
$$d\hat{e^a} + \hat{\omega}^a_b \hat{e}^b  = 0$$ and then,
$$d\hat{e^a}  = d(e^{\alpha \phi} e^a)  = e^{\alpha \phi} d(e^a) + d(e^{\alpha \phi}) e^a \\ \ \ \ \ \ = e^{\alpha \phi} d(e^a) + \alpha e^{\alpha \phi} \partial_b \phi dx^b \wedge e^a \\ 
\ \ \ \ \ = - e^{\alpha \phi}\omega^a_b \wedge e^b + \alpha e^{\alpha \phi} \partial_b \phi dx^b \wedge e^a \\ 
\ \ \ \ \ = - \omega^a_b \wedge \hat{e}^b + \alpha  \partial_b \phi dx^b \wedge \hat{e}^a \\ 
\ \ \ \ \ = -\hat{\omega}^a_b \hat{e}^b$$
But this is no good, not even slight nearer, After all the formula written in the answer below can be obtained from Cartan's structure equation.I have no idea how to get  $$F^{ab}e^{\beta - 2 \alpha} \hat{e}^z$$ type term.I had actually calculated some spin connections before but there I always used there component form. 
 A: Firstly you need to express spin connection in terms of the Christoffel symbols using vanishing torsion condition, that gives
\begin{equation}
\omega_\mu{}^{a}{}_{b}=e^a_\nu\Gamma_{\mu\rho}{}^{\nu}e^\rho_b-e_b^\nu\partial_\mu e_\nu^a.
\end{equation}
Now since you know the ansatz for the metric, you know the decomposition of the Christoffel symbols and the vielbein. After some algebra you should get exactly what is written in (1.10).
UPDATE: At the very end of your derivation you have lost one term. Actually, this comes from the very beginning where you confuse the indices of hatted quantities with unhatted ones. 
Hence, if we let capital Latin indices to run through $\{a,z\}$, then we can write the total vielbein as $\hat{e}^A$. The Cartan equation then can be written as
\begin{equation}
d\hat{e}^A+\hat{\omega}^A{}_B\hat{e}^B=0.
\end{equation}
Given this notation the correct equation reads
\begin{equation}
\begin{aligned}
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_\mu \phi dx^\mu \wedge e^a \\ 
& = - e^{\alpha \phi}\omega^a_b \wedge e^b + \alpha e^{\alpha \phi} \partial_\mu \phi dx^\mu \wedge e^a \\ 
& = - \omega^a_b \wedge \hat{e}^b + \alpha  \partial_\mu \phi dx^\mu \wedge \hat{e}^a \\ 
& = -\hat{\omega}^a_B \hat{e}^B=-\hat{\omega}^a_b \hat{e}^b - \hat{\omega}^a_z \hat{e}^z
\end{aligned}
\end{equation}
The component $\hat{\omega}^a{}_z$ can be derived from the vanishing torsion equation $\nabla\hat{e}^z=0$. That will give the desired term $d\mathcal{A}$ in the connection.
