Deriving symmetries of a Gauge theory

Question

Hello, I don't know if this is a good place for exposing my problem but I'll try...

I have a gauge theory with action:

$S=\int\;dt L=\int d^4 x \;\epsilon^{\mu\nu\rho\sigma} B_{\mu\nu\;IJ} F_{\mu\nu}^{\;\;IJ} $

Where $B$ is an antisymmetric tensor of rank two and $F$ is the curvature of a connection $A$ i.e: $F=dA+A\wedge A$, $\mu,\nu...$ are space-time indices and $I,J...$ are Lie Algebra indices (internal indices) I would like to find its symmetries. So I rewrite the Lagrangian by splitting time and space indices $\{\mu,\nu...=0..3\}\equiv \{O; i,j,...=1..3\}$ I find:

$L = \int d^3 x\;(P^i_{\;IJ}\dot{A}_i+B_i^{\,IJ}\Pi^i_{\,IJ}+A_0^{\;IJ}\Pi_{IJ})$

Where $\dot{A}_i = \partial_0 A_i$, $P^i_{\;IJ} = 2\epsilon^{ijk}B_{jk\,IJ}$ is hence the conjugate momentum of $A_i^{\,IJ}$

$B_i^{\,IJ}$ and $A_0^{\;IJ}$ being Lagrange multipliers we obtain respectively two primary and two secondary constraints:

$\Phi_{IJ} = P^0_{\;IJ} \approx0$

$\Phi_{\;\;IJ}^{\mu\nu} = P^{\mu\nu}_{\;\;IJ} \approx0$

$\Pi^i_{\,IJ} = 2\epsilon^{ijk}F_{jk\,IJ} \approx0$

$\Pi_{IJ}=(D_i P^i)_{IJ} \approx0$

Where $P^0_{\;IJ}$ are the conjugate momentums of $A_0^{\,IJ}$ and $P^{\mu\nu}_{\;\;IJ}$ those of $B_{\mu\nu}^{\;\;IJ}$. Making these constraints constant in time produces no further constraints.

Whiche gives us a general constraint:

$\Phi = \int d^3 x \;(\epsilon^{IJ}P^0_{\,IJ}+\epsilon_{\mu\nu}^{IJ}\;P^{\mu\nu}_{\;\;IJ}+\eta^{IJ}\Pi_{IJ}+\eta_i^{IJ}\Pi^i_{\;IJ})$

Each quantity $F$ have thus a Gauge transformation $\delta F = \{F,\Phi\}$ where $\{...\}$ denotes the Poisson bracket.

Knowing that this theory have the following Gauge symmetry:

$\delta A = D\omega$

$\delta B = [B,\omega]$

Where $\omega$ is a 0-form, I would like to retrieve these transformations using the relation below. (where $\Phi$ is considered as the generator of the Gauge symmetry) but my problem is that I don't know how to proceed, I already did this with a Yang-Mills theory and it worked... but for this theory it seems to le intractable! Someone to guide me?

Answer 1

Unless for some reason you absolutely must work within the Hamiltonian approach, you can just directly look for the complete set of (infinitesimal Lie point) symmetries of the Euler--Lagrange equations or of the action itself. The procedure is standard and described in many good books. For instance, you can look into those by Olver (more math-y) or Stephani (somewhat closer to physics). Using the theory from these books you can also verify whether the transformation at the end of your question is indeed a symmetry.

Answer 2

Although this question is over a year old, some readers may be interested in the following comments.

1. The symmetries of the Lagrangian and the action are not identical.

2. I don't know why you cannot recover the gauge transformations from the constraints, perhaps it has to do with not including the $B_{0i}$ fields in the second expression for the Lagrangian?

3. The constraints give infinitesimal transformations leading to the usual gauge transformations, but there seem to be symmetries which are non-trivial extensions of these.

4. For example, for a gauge group SU(n), the transformations $$ A' = UAU^\dagger + \phi $$

$$ B' = UBU^\dagger + UAU^\dagger\wedge\phi + \phi\wedge UAU^\dagger + \phi\wedge\phi $$

where $U(x) \in$ SU(n), and $\phi = -dUU^\dagger$, leaves the action invariant.