Order parameter

Remarks

If $\mu$ is invariant under $K$, then $\langle f \rangle_\mu = \rho(k) \langle f \rangle_\mu$ for all $k$, so $\langle f \rangle_\mu = 0$. Hence an ordered state breaks the symmetry $K$.

Example: for the Ising model on $\Z^d$ with zero field, $K = \Z/2$ acts by the global flip $\sigma \mapsto -\sigma$, $V = \R$, $\rho$ is the sign representation and $f(\sigma) = \sigma_0$. The flip sends $\sigma_0$ to $-\sigma_0$, so the equivariance holds. The order parameter is the magnetization $\langle \sigma_0 \rangle$; at low temperature in the plus state it equals the spontaneous magnetization, which is positive for $d \ge 2$.

Non-example: the energy observable $f(\sigma) = \sigma_0 \sigma_{e_1}$ is unchanged by the flip. The trivial representation fixes every vector, which fails the condition on $\rho$, and indeed $\langle \sigma_0 \sigma_{e_1} \rangle > 0$ at every temperature.

For the classical Heisenberg model, $K = SO(3)$ and the order parameter is a vector in $\R^3$.