In section 19.6, page 212-213 in Weinberg's Quantum Theory of Fields Volume II, Weinberg shows that on can always eliminate the Goldstone Bosons from a field $\psi(x)$, by use of a space-dependent transform $\gamma(x)$ in the symmetry group $G$. To do this, he considers the quantity $V_\psi(g) = \psi_m g_{mn} \psi_n^*$ (implicit summation), where I have denoted the vacuum of the theory with $\psi^*$ and $g \in G$. He then argues that for any point $x$, $V_{\psi(x)}$ has a maximum $g = \gamma(x)$ as $G$ is compact, and that therefore the variation at this point is zero, $$ \delta V_{\psi(x)}(\gamma(x)) = \psi_n(x) \gamma_{nm}(x) t^\alpha_{mk}\psi_k^* = 0, $$ where $t_{mn}^\alpha$ is a generator of $G$. So far so good, in my opinion. However, next he does a small trick, writing $$ \psi_n(x) \gamma_{nm}(x) = [ \gamma^{-1}(x)]_{mn}\psi_n(x) = \tilde \psi_m(x). $$ Why can he do this? I am guessing it is some sort of property of the representation of $G$, but that is not my strongest area. Can anyone identity the property, why it is true, and where to read about it?