Why is $J/\psi$ not able to decay into $\pi^+\pi^-\pi^0$ (Ozi-rule)?

Question

There are already (very similar) questions like What is the precise statement of the OZI Rule? and Why does charmonium (and phi mesons) not decay via quark and antiquark annihilation? but, to be honest, I don't get it. Therefore, I would like to get an explanation on an important example:

$$ J/\psi \to \pi^+ \pi^- \pi^0 $$

I read that this is not allowed because you can cut it into halves by cutting only gluonic interactions.

1. Why are there only gluons involved?
2. Hence, isn't it sufficient when one of the charm-quarks changes to another quark ($ u $ or $ d $), leaving one straight line for the charm-quarks so you cannot cut the process? My third question: How the hell do I know that I have to draw the process like depicted? Above all, again two gluon exchanges after the first three. It looks like the $ \bar{d} $ of the $ \Pi^+ $ is the same like the $ d$ of the $ \Pi^0 $ . And in addition connected via a gluon with a $ u $-quark.. how do I know this has to be this way?

Answer 1

Take a look at this statement from Wikipedia (emphasis mine):

[The OZI rule] states that any strongly occurring process will be suppressed if, through only the removal of internal gluon lines, its Feynman diagram can be separated into two disconnected diagrams: one containing all of the initial-state particles and one containing all of the final-state particles.

So we're only looking at QCD processes. As for your questions:

1. A "strongly occurring process" means that only the strong interactions take place. So no electromagnetic interactions (= no photons) and no weak interactions (no $W^\pm$ or $Z^0$ bosons). This only leaves gluons, because they are the only gauge bosons in the theory of strong interactions = QCD.
2. Changing the quark flavor can only happen in weak processes; QCD preserves quark flavor.