we should consider the tensor product decomposition:

$$\frac{1}{2} \otimes \frac{1}{2} \otimes \frac{1}{2}
= \frac{1}{2}  \oplus \frac{1}{2}  \oplus
\frac{3}{2}
\label{11xh11}$$

The particular way of writing $H_B$ given in Eq. (\[11xh10\]) suggests to compose first $\vec{S}_1$ and $\vec{S}_3$ and then, the resulting spin with $\vec{S}_2$. The result of this compositions is given as follows:

$$| \frac{3}{2},  \frac{3}{2} \rangle = | \uparrow
\uparrow \uparrow \rangle,  \ \ E_B = J/2       \label{11xh12a}$$

$$| \frac{3}{2},  \frac{1}{2} \rangle =
\frac{1}{\sqrt{3}}  ( | \uparrow \downarrow \uparrow \rangle  +  |
\downarrow \uparrow \uparrow \rangle  +
   | \uparrow \uparrow \downarrow \rangle  ),     \ \ E_B =
J/2             \label{11xh12b}$$

$$| \frac{1}{2},  \frac{1}{2} \rangle_1 =
\frac{1}{\sqrt{2}}
 (| \uparrow \uparrow \uparrow \rangle - | \downarrow \uparrow
\uparrow \rangle),  \ \ \ E_B =
0
\label{11xh12c}$$

$$| \frac{1}{2},  \frac{1}{2} \rangle_0 =
\frac{1}{\sqrt{6}}  ( 2 | \uparrow \downarrow \uparrow \rangle  -  |
\downarrow \uparrow \uparrow \rangle  -
   | \uparrow \uparrow \downarrow \rangle  ),     \ \ E_B =
-J             \label{11xh12d}$$

Hence for $\epsilon =0$ we could choose the spin $1/2$ irrep. with basis vectors $|\frac{1}{2}, \frac{1}{2} \rangle_0$ and $|\frac{1}{2}, -\frac{1}{2} \rangle_0$ in order to define the intertwiner operator $T_0$.

However, if $\epsilon \neq 0$ the states (\[11xh12a\]) -(\[11xh12d\]) are not eigenstates of (\[11xh10\]). The full 