# Field Strength and Source terms

This question is related to my recent unanswered question, but it was too complicated so please let me make this new question at first.

First, I consider a field strength which is expressed as \begin{align} F = dA \end{align} with $$A$$ ($$A$$ is some differential form). It is just a generalization of usual Maxwell theory.

The Bianchi identity is clearly \begin{align} dF = 0. \end{align}

Now consider the action is given as

\begin{align} S = \int F \wedge \star F. \end{align}

Then EOM for $$F$$ is

\begin{align} d\star F = 0, \end{align}

It's OK. Next, I consider an additional source term for $$S$$ s.t. \begin{align} S = \int F \wedge \star F - q \int A \wedge \delta. \end{align} Here $$\delta$$ is just a Poincaré dual, which depends on where $$A$$ lives.

Then EOM becomes

\begin{align} d\star F = q \delta, \end{align} It's OK again.

Then, please consider the case $$F$$ is self-dual i.e. $$F = \star F$$.

Now $$dF=0$$ from Bianchi, which contradicts the EOM unless $$q=0$$.

I know when we write $$F=dA$$ automatically source terms like magnetic monopole vanishes, but how can I solve this contradiction?

I think I should re-define $$F$$ to match its EOM, but the action with source term don't tell me to do so I think.

If $$F$$ is self dual \begin{align} F = \star F \end{align} The action \begin{align} S = \int F \wedge \star F = \int F \wedge F \end{align} turns into a topological Pontryagin term, where EOM does NOT apply.
• Sorry, I am not familiar with a topological Pontryagin term, but in string theory (IIB SUGRA) there is a self-dual 5-form $\tilde{F}_5$ and several textbooks say if we consider EOM for $\tilde{F}_5$ we can get Bianchi identity and above my question is a sketch of that discussion. Please tell me what does "where EOM does NOT apply" mean. We cannot get EOM from this action? Feb 24 '20 at 18:18
• @Keyspire, my answer is based on 4D. What is your action for the 5-form $\tilde{F}_5$ and what's the dimension, and is the gauge group Abelian? Feb 24 '20 at 18:21
• oh I thought my question does not depend on dimension too much but it's not... I consider 10D action in fact. and gauge group is Abelian I think, hence $F_5\wedge \star F_5$ is actually zero. This my old question (physics.stackexchange.com/questions/532675/…) has some detail. Feb 24 '20 at 18:43