What symmetries are in the following action:

  1. [tex] S=\int d^4x\frac{m}{12}A_μ ε^{μ \nu ρσ} H_{\nu ρσ} + \frac{1}{8} m^2A^μA_μ [/tex]
    [tex] H_{\nu ρσ} = \partial_\nu B_{ρσ} + \partial_ρ B_{σ\nu} + \partial_σ B_{\nu ρ} [/tex]

    And [itex] B^{μ \nu} [/itex] is an antisymmetric tensor.

    What are the global symmetries and what are the local symmetries?

    p.s how many degrees of freedom does it have?

    Thank you!
  2. jcsd
  3. Has [itex]A_{\mu}[/itex] anything to do with the [itex]B_{\mu \nu}[/itex]?

    And what does it have dofs?
    The Action is a (real) scalar quantity, so it has 1 dof.

    if [itex]A_{\mu}[/itex] is a massive bosonic field, it should have 3 dofs.
    and about [itex]B^{\mu \nu}[/itex] just by being an antisymmetric tensor (in Lorentz repr it is a 4x4 in your case matrix) will have:
    [itex] \frac{D^{2}}{2}-D = \frac{D(D-1)}{2} [/itex]
    free parameters. So for D=4, you have 6 dofs...
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  4. Thanks ChrisVer,
    [itex]A^\mu[/itex] has nothing to do with [itex]B_{\mu \nu}[/itex]
    I meant the number dof of the thoery.
    [itex]H_{\nu ρσ}[/itex] is antisymmetric, so it has only [itex]\binom{4}{3}=4[/itex] dof, doesn't it? thus in total it's 3X4=12 dof, isn't it?

    And more important for me is to know the action symmetries, both the global and the local ones.

  5. For the symmetries you should apply the Noether's procedure ...
    A global symmetry which I can see before hand is the Lorentz Symmetry (since you don't have any free indices flowing around)
  6. Also I don't think you need the dofs of the strength field tensor anywhere, do you?
    It gives the kinetic term of your field [itex]B_{\mu \nu}[/itex]
    I am not sure though about the dofs now...you might be right.
  7. For the H you were right.
    [itex]H[/itex] is a p=3-form, and a general p-form in n dimensions has:
    [itex]\frac{n!}{(n-p)!p!}[/itex] ind. components.
  8. You're probably right, it's the 6 dof of B that matters.
    But apparently B has a gauge symmetry, so only 3 dof left.
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