An Update On Kaluza-Klein Theory

jonah
Kaluza Klein Image from Wikipedia
Kaluza-Klein Theory unifies four-dimensional electromagnetism and general relativity by studying five-dimensional gravity. Image from wikipedia.

I recently posted an article on Kaluza-Klein theory. This was partly because I was working a paper on it as a final project in my second semester of general relativity. The paper is finished, and I thought I’d upload it for the more mathematically inclined of my readers. If you’re interested, you can find it here.

8 thoughts on “An Update On Kaluza-Klein Theory”

  1. Thanks for publishing this, it is very useful now that I need to learn about Kaluza-Klein reduction for my thesis. I’m stuck defining the curvature form, isn’t it d\omega + \omega \wedge \omega with everything hatted? In particular, I don’t understand why the third term \omega\wedge e, is of this form. Could the fact that the manifold comprises two disjoint ones, M^4\times M^1 be related? If so, how is the spin connection decomposed in such a manifold?

  2. Hey Peter, I’m glad I could help! As you guessed, I’m taking advantage of the fact that the manifold can be decomposed. I think you actually caught a typo in that line. The index c should run only from 0 to 3.

    RIght, in five dimensions, it’s
    \hat{R} = d\hat{\omega} + \hat{\omega} \wedge\hat{omega}

    So what I’ve done is decomposed my last \hat{\omega}, \hat{\omega}^c_{\ b} into the four dimensional one plus an additional term:
    \hat{R} = d\hat{\omega} + \hat{\omega}^A_c \wedge\hat{\omega}^c_B + \hat{omega}^A_z \wedge\hat{\omega}^z_B

    Then I’m just using how the spin connections are defined. Recall that
    \omega^a_b = \omega_\mu^a_b dx^\mu
    if you wedge the five dimensional one with the four dimensional one in the z direction, e^z pops out.

    Hope this helps!

  3. The definition of the spin connection I know is \hat{\omega}_{M AB} = \hat{e}_A{}^N \nabla_M \hat{e}_{BN}, is this what you mean?

    I assume that \eta_{ab} = diag(-1,1,1,1), \hat{\eta}_{AB} = diag(-1,1,1,1,1). Then, \hat{\omega}^z{}_a = -\hat{\omega}^z{}_a only if a\neq 0, otherwise it equals \hat{\omega}^z{}_a, right? But if this is so, shouldn’t \hat{\omega}^{az} depend on whether a=0 or not?

    Thanks for your help, it is much appreciated. I’ve been able to do most of the calculations, but I’m still stuck with these problems.

      1. Oh sorry, yes. I had a brain fart. All I meant was that if you wedge a unit one form e_A with itself in the “tetrad” basis, it goes to zero. So if you wedge a five-dimensional spin connection with a four-dimensional one, the four e_a one-forms cancel with themselves, leaving just e_z and the component of the spin connection multiplying it.

        Also sorry for taking so long to get back to you. I’m at a summer school at the moment, and as you probably know, those are pretty intense.

        (I also apologize in advance if I say something very wrong. It’s been about a year since I’ve thought about Kaluza-Klein stuff!)

  4. Hi Jonha, great effort!

    I am also following Pope in his example, however as he has left out the details from spin-connection to the curvature I was pleased to see that someone like yourself has taken the time to go through it in a little more detail.

    However, firstly I cannot make sense of your decomposition in the Cartan structure equations, and secondly, the one calculation that you have done for R_{zz} seems rather spooky. What do I mean, well, take a look at your dilaton term (\sim \partial\phi\, \partial\phi) in (23) clearly this vanishes since you are taking the wedge product between \hat e^z and itself, no?

    Also, I fail to see how \hat R^{zz}{}_{mu\nu} is what determines \hat R_{zz}, obviously you need more terms?

        \[\hat R_{zz} = \hat R^M{}_{z M z}\]

    So you also need \hat R^{az}.

      1. Hi Isak,

        I’m sorry for not responding earlier… somehow I missed your comment! I haven’t looked at this paper in a long time. Are you still stuck? If so, I’ll try and go back through my derivation.

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