Rho meson decay to three neutral pions In my assignment, I am asked to show whether $\rho^0\to3\pi^0$ is allowed. From the particle data group, I cannot find the decay mode; hence, I am guessing it is not allowed. Based on the knowledge that $\rho^0\to2\pi^0$ is not allowed, I focus on parity. However, the action contains an angular momentum of 3 bodies. Then I get stuck. Can someone tell me whether this action is allowed? Or at least give me some hint. (I have not learned weak interaction and strong interaction yet.)  
 A: I confess I might be overthinking this, having missed something more obvious (C? -- violated by the weak interactions).
The G-parity of the ρ is +, so it prefers to decay to 2π s, since the π has G parity -. (It requires an antisymmetric state because of angular momentum, as you already see.)
But G-parity relies on isospin, which is broken somewhat in the strong decay, so there is also the 3π  mode at the $10^{-4}$ level, where all 3π s are different, instead of your 3π0 which has to be fully symmetric. 
You cannot first combine the two π0s to a vector by an L=1 as you notice, by their unavoidable symmetry. So you have to combine them to a symmetric S-wave spinless neutral dipion, and then combine that antisymmetrically with an L=1 to the remaining neutral π0 to net a vector ρ0.
However, this is impossible; ignoring normalizations,
$$
(12+21)3-3(12+21)= 123+213-312-321= (123-321)+(-312+213),
$$
but the terms in the first and second final parentheses are 1-3 antisymmetric, and 3-2 antisymmetric respectively, which is impossible for identical neutral pions. 
This argument fails, of course, (phew!),  for the extant decay mode π0π+π- which is allowed, although ferociously suppressed. 
