shall make some general considerations concerning Hamiltonians which commute with a symmetry group ${\cal G}$. Notice that for the AF Heisenberg model ${\cal G}$ is nothing but the rotation group $SU(2)$. Let us call $g$ an element of the group ${\cal G}$ and $\Pi _{\cal H} (g)$ a representation of $g$ acting on the Hilbert space ${\cal H}$. We say that ${\cal G}$ is a symmetry group of the Hamiltonian $H$ if

$$[H ,  \Pi _{\cal H} (g) ] = 0 \ \ \ \ \  \forall g
\in    {\cal G}             \label{11xh1}$$

Similarly we want the effective Hamiltonian $H'$ to be invariant under the action of ${\cal G}$ acting now on the Hilbert space ${\cal H'}$,

$$[H' ,  \Pi _{\cal H'} (g) ] = 0 \ \ \ \ \  \forall g
\in    {\cal G}             \label{11xh2}$$

For this to be the case, the RG-transformation must preserve the symmetry of the original hamiltonian. This can be simply achieved if one choses $T$ as the intertwiner or Clebchs-Gordan operator. Indeed we may recall that if a representation say $\Pi _{V_3}$ is contained in the tensor product $\Pi _{V_1} \otimes \Pi _{V_2}$ then one can define the CG-operator as,

$$C^{12}_3: \ \ \ 