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.

  • $\begingroup$ What exactly the problem is? If you have a metric, you have to make some choice for the vielbein that reproduces such a matric. Then, there is an explicit formula for spin connection in terms of first derivatives of the vielbein. There is no conceptual problem in any dimension since the solution is explicit. $\endgroup$
    – John
    Commented Apr 20, 2014 at 20:57
  • $\begingroup$ The problem is relating the spin connection of D+1 dimension to D dimension, see the first term on right hand side.And how to get that last term. $\endgroup$
    – user44895
    Commented Apr 20, 2014 at 21:17
  • $\begingroup$ Comment to the question (v3): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. $\endgroup$
    – Qmechanic
    Commented Apr 21, 2014 at 6:21
  • $\begingroup$ @user44895 As it said in the paper, the computations are mechanical but tedious. One just takes the vielbein from the paper, the formula for spin-connection and expands index-contractions. Do you want somebody to write down the intermediate steps here? $\endgroup$
    – John
    Commented Apr 21, 2014 at 7:03

1 Answer 1


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.

  • $\begingroup$ How did you get $$e^b \wedge \hat{e}^a$$ term in second last line above, it seems wrong.How will that equate to the spin connection mentioned in the paper. $\endgroup$
    – user44895
    Commented Apr 21, 2014 at 14:11
  • 1
    $\begingroup$ By the way, if anyone knows a good reference regarding some brute calculation like this one about spin connection and so then I will really appreciate it. Because there seems to be a big gap in between (I have seen Nakahara level book). $\endgroup$
    – user44895
    Commented Apr 21, 2014 at 17:13
  • $\begingroup$ This term again comes from your bad notations. You should distinguish between flat and curved indices. I changed the expression according to that. $\endgroup$
    – Edvard
    Commented Apr 22, 2014 at 8:07
  • $\begingroup$ Yes, I see that. Do you know a reference regarding some good calculation stuff like this one, because I don't feel very confident with these type of calculations.And One more thing.. Thanks. $\endgroup$
    – user44895
    Commented Apr 22, 2014 at 8:59
  • $\begingroup$ I do not know a reference, where you can find some explicit calculations. I would recommend to search for some lectures on spin-geometry with problems inside. Actually, you already know everything that is needed for such kind of calculations, just take care about indices. $\endgroup$
    – Edvard
    Commented Apr 22, 2014 at 14:45

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