What symmetries are in the following action:

[bagherihan]

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

And B^{μ \nu} 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!

 

[ChrisVer]

Has A_{\mu} anything to do with the B_{\mu \nu}?

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

if A_{\mu} is a massive bosonic field, it should have 3 dofs.
and about B^{\mu \nu} just by being an antisymmetric tensor (in Lorentz repr it is a 4x4 in your case matrix) will have:
\frac{D^{2}}{2}-D = \frac{D(D-1)}{2}
free parameters. So for D=4, you have 6 dofs...

 

[bagherihan]

Thanks ChrisVer,
A^\mu has nothing to do with B_{\mu \nu}
I meant the number dof of the theory.
H_{\nu ρσ} is antisymmetric, so it has only \binom{4}{3}=4 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.

thanks.

 

[ChrisVer]

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)

 

[ChrisVer]

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 B_{\mu \nu}
I am not sure though about the dofs now...you might be right.

 

[ChrisVer]

For the H you were right.
H is a p=3-form, and a general p-form in n dimensions has:
\frac{n!}{(n-p)!p!} ind. components.

 

[bagherihan]

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.