Non-abelian Goldstone bosons

Suppose we have a theory with a non-abelian symmetry group $G$ that is spontaneously broken to the subgroup $H\subset G$--this is a global symmetry, $not$ a gauge redundancy. Let $X^a$ be the generators of the unbroken subgroup $H$ and $Y^m$ be the remaining generators. For each broken generator $Y^m$, there is a corresponding Goldstone $\pi^m(x)$. My question is the following: if we integrate out all of the massive modes so that we are just left with a low-energy theory of the Goldstones corresponding to $G/H$, what do infinitesimal transformations of $\pi^a(x)$ look like under the $X^a$ and $Y^m$ generators?

For abelian symmetries we expect the Goldstones to have shift symmetries under broken generators, but that is clearly not possible for a generic non-abelian symmetry group. I expect something much more complicated happens but I can't quite see what.

• Related. Note a σ vev transforms the πs linearly under the 3 vector (isospin) charges but, quite clearly, contrary to your unwarranted expectation, shifts them under the 3 axial charges. – Cosmas Zachos Aug 5 '18 at 0:22

The action of the broken generators remains a shift. Specifically, let $\pi^a$ be coordinates for the vacuum manifold $G/H$ so that $\pi^a = 0$ at the vacuum. Assuming these coordinates are appropriately chosen, the lowest-order behavior of broken generators is a shift, which means that under the transformation $e^{i \alpha_a T^a}$ for broken generators $T^a$, $$\pi^a \to \pi^a + \alpha_b v^b + O(\alpha^2)$$ and for unbroken generators the symmetry is realized linearly as usual.
A classic example of this is chiral perturbation theory and the general method to find the coordinates $\pi^a$ is called the CCWZ construction.