In Griffiths' Introduction to Elementary Particles, it is mentioned p. 179 that the $\pi^0$ is a singlet under $SU(2)$ isospin. But it is also part of the $\pi^-,\pi^0,\pi^+$ isospin triplet. How can it be both?
Don't particles of a given $SU(2)$ multiplet mix under a corresponding transformation?
In a travesty of overlapping historical notations, we have both strong isospin, under which the neutron and proton are the two projections of the nucleon and are raised and lowered by the pion, and weak isospin, in which the left-handed parts of the $(u,d)$, $(e,\nu_e)$, $(c,s)$, $(\mu, \nu_\mu)$, $(t,b)$, and $(\tau,\nu_\tau)$ doublets are raised and lowered by the $W^\pm$ bosons. The six right-handed quarks and the six right-handed leptons are weak isospin singlets.
It is possible in principle (I think) for a particle like the $\pi^0$ to be the neutral member of a strong isospin triplet, but a singlet under weak isospin.
However, that doesn't seem to be the case for the $\pi^0$. The "lowering operator" vertex $\pi^\pm \leftrightarrow W^\pm\pi^0$ actually does exist, as evidenced by the existence of the decay mode $\pi^+\to \pi^0 e^+ \nu_e$, which has branching ratio $10^{-8}$. You have a fresh comment suggesting that your textbook makes this claim incorrectly.
$\pi^0$ is a singlet under $SU(2)$ isospin because if you interchange the u by d (as a rotation in 2- dimensional flavor space) then you will not be able to mix up $\pi^0$ with any other pion, and hence it is a singlet. It is in this sense you should understand it.
The pi zero is not a singlet, but part of an Ispin triplet. The eta meson and the omega meson are Ispin SU(2) singlets.
