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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.

Please help me.

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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.

You should be talking about its Chern class instead.

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  • $\begingroup$ 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? $\endgroup$
    – Keyspire
    Feb 24 '20 at 18:18
  • $\begingroup$ @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? $\endgroup$
    – MadMax
    Feb 24 '20 at 18:21
  • $\begingroup$ 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. $\endgroup$
    – Keyspire
    Feb 24 '20 at 18:43

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