# 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.