Stückelberg mechanism and the axion 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:

*

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


*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.
 A: 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.
