Stückelberg mechanism and the axion

Question

In the so-called Stückelberg mechanism we have the BF term

$$ \sim \int m\; B_2\wedge F ~, $$

where the field $B_2$ is a 2-form and $F$ is the field strength arising from a $U(1)$.

The Stückelberg mechanism allows to introduce a scalar field $\eta$ (Axion) to rewrite the BF term in a local equivalent form according to

$$ \int_{4d} m\; B_2 \wedge F \rightarrow -\frac{1}{2}\int_{4d}(m\;A+d\eta)^2 \; $$ with $F=dA$.

Now, the 2-form $B_2$ is related (Hodge duals) to a $0$-form $\Phi$ in four dimensions.

$$ B_2 \sim -\Phi~. $$

Thus, it seems that we can get the following

$$ \int_{4d} m\; B_2 \wedge F \rightarrow -\frac{1}{2}\int_{4d}(m\;A+d\Phi)^2. $$

Question:

1. Why can we assume that the $\Phi$ corresponds to the axion $\eta$?

2. What happens to the minus? Shouldn't we get:

$$ \int_{4d} m\; B_2 \wedge F \rightarrow -\frac{1}{2}\int_{4d}(m\;A-d\Phi)^2. $$

The Stückelberg mechanism can be found e.g. here on page 63.

Answer 1

in addition to the comments above, I now checked the signs. Here is the maths. We start with the Lagrangian $$ \mathscr{L} = - \frac{1}{2}H_3 \wedge \star H_3 + m B_2 \wedge F \ ,$$ where I omitted the kinetic term $F \wedge \star F$, because it is not relevant for the dualisation.

Now I want to find the Lagrangian for the dual of $B_2$. For this, we need to add a Lagrange multiplier term:

$$ \mathscr{L} = - \frac{1}{2}H_3 \wedge \star H_3 + m B_2 \wedge F - \eta dH_3 \ .$$

Why minus and not plus? Because I want to get $|mA+d\eta|^2$ and not $|mA-d\eta|^2$. ;-) Here I use the notation $|F|^2 = F \wedge \star F$.

This extra term ensures that we have the Bianchi identity $dH_3=0$ (when integrating out $\eta$).

Since $F=dA$, we can now rewrite

$$ \mathscr{L} = - \frac{1}{2}H_3 \wedge \star H_3 - m H_3 \wedge A - H_3 \wedge d\eta \ ,$$

where I dropped all boundary terms $d(...)$.

When we vary with respect to $H_3$, we obtain: $$ 0 = \delta H_3 \wedge \left[- \star H_3 - (mA + d\eta) \right] \ , $$ i.e. we get the equation of motion $$ \star H_3 = - (mA + d\eta)$$ Since $H_3$ is a $3$-form in 4D with Lorentzian signature, we have $\star(\star H_3)=H_3$.

We can now go back to our Lagrangian (with the Lagrange multiplier) and substitute $\star H_3$ and $H_3$. I obtain $$ \mathscr{L} = - \frac{1}{2}|mA + d\eta|^2 \ ,$$ as desired.