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How can we tell if a given tensor is a curvature tensor?

  1. May 15, 2008 #1
    Under what circumstances do we know whether a given tensor of 4th rank could be the curvature tensor of a manifold? For instance, if I specify some arbitrary functions R_{ijkl} (with the necessary symmetries under interchange i<->j, k<->l, and ij<->kl), is it necessarily the case that there is a manifold for which R_{ijkl} is the curvature? I guess that satisfying the Bianchi identities is a necessary precondition.

    And if so, is there any way to construct the metric explicitly?


  2. jcsd
  3. May 15, 2008 #2
    I'm in a hurry so I'll give you a very dirty answer to decode.
    The Riemann tensor is a gl(N,R) Lie-algebra valued 2-form, just like the field strength in Yang Mills theories (there, the algebra is su(N)). In the latter case, one may impose the recover the gauge connection from the field strength in a particular gauge, namely the Fock-Schwinger gauge (the calculation is done here: http://www.physics.thetangentbundle.net/wiki/Quantum_field_theory/gauge_theory/gauge_fixing [Broken])

    Now, if you have some experience dealing with gravity as a gauge theory (see e.g. Nash & Sen, or Nakahara), then you have a connection from which you have enough data to reconstruct almost anything else.
    Last edited by a moderator: May 3, 2017
  4. May 15, 2008 #3
    Hi lbrits,

    Thanks for the pointer to differential forms, it certainly seems to offer a new angle to the question. However, it seems to me that the link you provided about Fock-Schwinger gauge assumes the gauge group is Abelian, since it constructs a potential A which satisfies F=dA. For Yang-Mills or the gauge formulation of gravity, the field strength/curvature is related to the connection by R = dA + A^A.

    In the non-abelian case, the problem is how to solve the equation R=dA + A^A for A, which I don't know how to do. I'm not even sure what condition to put on R to guarantee that a solution exists. (dR=0 is not generally true. Contrary to my earlier post, the Bianchi identities can't even be applied since we need A to do this.)


  5. May 15, 2008 #4
    The trick is that [tex]x^\mu A_\mu = 0[/tex], which is the actual gauge condition. Thus the equation presented is both Abelian and non-Abelian.

    You had me worried there for a second =)
    Last edited: May 15, 2008
  6. May 16, 2008 #5
    Brilliant! Thanks! Following your lead, I've also located a few papers which have set me moving again. For anyone following the thread, [1] appears to be the paper which first demonstrated this solution, and it's very understand-able. [2] is where I found [1], although I didn't understand much of it.


    [1] Cronstrom, Phys Lett B90 (1980) 267
    [2] Kummer & Weiser, Z. Phys C31 (1986) 105
  7. May 16, 2008 #6
    Strange, the Fock-Schwinger gauge is ancient... it is, after all, named after Fock and Schwinger =)
  8. May 17, 2008 #7
    Perhaps you're right there, I'm not sure. Well, it's a reference, if not the original reference.

    It occurred to me that I'm not even sure if the set of admissible curvatures forms a vector space. By this I mean that if R and R' are curvature tensors corresponding to metrics g and g' respectively, it's not clear to me that R+R' is also a curvature tensor, and if so, how its metric relates to g and g'. Perhaps its possible to prove or disprove this using the Fock-Schwinger gauge - I'll see whether I can get anywhere with it.

    This is quite a contrast to electromagnetism, because in that context the sum of two EM fields also satisfies Maxwell's equations, and the set of EM field configurations is a vector space (a Hilbert space). I was interested to see whether a similar analysis can be done for gravitation; but this one may not go through quite as cleanly as I hoped!


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