Note: I have a related question here. I was advised to split my questions into a series of posts.
I am trying to understand the mixed 't Hooft anomaly between the electric and magnetic one-form symmetries in 4d Maxwell. SE answers I've already read include 1, 3, 4 and 5, but I wasn't able to get a complete picture.
In this post I will ask about what happens when we start trying to gauge the electric one-form symmetry in 4d Maxwell on a manifold without boundary - i.e. for now I will neglect the magnetic symmetry.
To review: the Maxwell action $$ S_0 = \int \frac{1}{2g^2} F\wedge \star F $$ has a conserved current $$ J_e = \frac{2}{g^2} \star F \; \; (\text{from the equation of motion for } A). $$
Now we set about trying to gauge the electric symmetry. This involves introducing a background field $B_e$ coupled to the current $J_e$. The minimal coupling is $$ S_e = \int \frac{1}{2g^2} F\wedge \star F - \frac{2}{g^2} \star F \wedge B_e $$
This changes the equation of motion to $d\star F = 2 d \star B$, which usually signals an anomaly. But I believe there should be no anomaly at this stage.
We can apparently express the action as $$ S_e' = \int \frac{1}{2g^2} (F - B_e)\wedge \star (F - B_e) $$
Why are we allowed to do this? I get $$ S_e' = S_e + \int \frac{1}{2g^2} B_e \wedge \star B_e + \frac{1}{g^2} \star F \wedge B_e, $$ This answer says we are 'allowed to add local terms' quadratic in the background field, like $B\wedge \star B$ - why?
If we do gauge it, the term is no longer a background term. Why are we still allowed to equate the two expressions?
There's also an extra $\star F \wedge B_e$ because of a relative factor of $4$ between the original terms. What do we do with this? Why did we even define the current with that particular coefficient?
