How to count on-shell DoF of a gauge theory?

[ismaili]

Suppose I have a gauge potential $$A_{\mu\nu}$$, which is totally antisymmetric, if, say, the theory is in 6 dimensions, so that there are $$6\times5/2 = 15$$ degrees of freedom.

For the action $$S = \int d^6x F_{\mu\nu\rho}F^{\mu\nu\rho}$$, where
$$F_{\mu\nu\rho}\equiv \partial_\mu A_{\nu\rho} + \partial_{\nu}A_{\rho\mu} + \partial_{\rho}A_{\mu\nu}$$, we would have the following equation of motion
$$\partial_\lambda F^{\mu\nu\lambda} = 0$$

The question is, how to count the on-shell degrees of freedom of the gauge potential? or, before solving the equations of motion, how to know the number of independent equations?
Naively the number would be 15, but it turns out to be 9.

Is there any ideas? Thanks in advance.

Sincerely

 

[azntoon]

The answer is that the number of independent equations is actually 9. This is because the antisymmetry of the gauge potential implies that 6 of the equations are actually redundant. To see this, consider the equation \partial_\lambda F^{\mu\nu\lambda} = 0 for a fixed $\mu$. Permuting the indices $\nu$ and $\lambda$, we obtain \partial_\lambda F^{\mu\nu\lambda} = -\partial_\nu F^{\mu\nu\lambda} If we now add these equations together, we get 0 = \partial_\lambda F^{\mu\nu\lambda} + \partial_\nu F^{\mu\nu\lambda} = 0 which is obviously redundant. Thus, by subtracting out the 6 redundant equations, we obtain an equation of motion with only 9 independent degrees of freedom.