What symmetries are in the following action:

1. Jun 25, 2014

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!

2. Jun 25, 2014

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...

3. Jun 25, 2014

bagherihan

Thanks ChrisVer,
$A^\mu$ has nothing to do with $B_{\mu \nu}$
I meant the number dof of the thoery.
$H_{\nu ρσ}$ 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.

4. Jun 25, 2014

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)

5. Jun 25, 2014

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.

6. Jun 25, 2014

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.

7. Jun 26, 2014

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.