Fermionic statictics in $SU(2)$ slave-boson representation One of the $SU(2)$ slave-boson decompositions has been introduced by X.-G. Wen and P. A. Lee in PRL, 76, 503 (1996). (A generic recipe for constructing the SU(2) slave-particle framework has been discussed Here.) It has been suggested that the electronic annihilation operators can be expressed as 
$$
\big(c_{\uparrow}\quad c_{\downarrow} \big)= \frac{1}{\sqrt{2}} \big( b^{\dagger}_{1} \quad b^{\dagger}_{2} \big)
\begin{pmatrix}
f_{\uparrow} & f_{\downarrow} \\
f_{\downarrow}^{\dagger} & -f^{\dagger}_{\uparrow}
\end{pmatrix} ,
$$
where $c_{\sigma}$ is electronic annihilation operator with spin $\sigma$, $b~(b^{\dagger})$ is a slave-boson annihilation (creation) operator, and $f^{\dagger}_{i}(f_{i})$ creates (annihilates) a fermion with spin $i$. 
It has been shown that the above representation preserves the $SU(2)$ symmetry if slave-particles satisfy
$$
b^{\dagger}_{1}b_{1} -b^{\dagger}_{2}b_{2}+\sum\limits_{\sigma \in \{\uparrow, \downarrow\}} f^{\dagger}_{\sigma}f_{\sigma}=1.
$$ 
To assess the anti-commutation relation of electrons I have followed the common approach in slave-particle methods, namely employing the constraint to enforce the Fermionic statistics.
$$
\begin{align}
&\qquad \qquad \qquad \{ c_{\sigma}, c^{\dagger}_{\sigma} \}=1,\\
&2c^{\dagger}_{\sigma} c_{\sigma}= f^{\dagger}_{\sigma} b_{1} b^{\dagger}_{1} f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}b_{2}b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}}
+\sigma f^{\dagger}_{\sigma} b_{1} b_{2}^{\dagger}f_{\overline{\sigma}}^{\dagger}
+\sigma f_{\overline{\sigma}} b_{2} b_{1}^{\dagger}f_{\sigma},\\
&2c_{\sigma}c^{\dagger}_{\sigma}= b^{\dagger}_{1} f_{\sigma} f^{\dagger}_{\sigma} b_{1} +\sigma^{2} b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}} b_{2}
+\sigma b^{\dagger}_{1}f_{\sigma} f_{\overline{\sigma}}b_{2}
+\sigma b_{2}^{\dagger}f_{\overline{\sigma}}^{\dagger} f_{\sigma}^{\dagger}b_{1}
\end{align}
$$ 
It is straight forward to show that the summation of last two terms of $c^{\dagger}_{\sigma}c_{\sigma}$ and $c_{\sigma} c^{\dagger}_{\sigma}$ are zero. Following the same strategy, I can show that
$$
\begin{align}
2c^{\dagger}_{\sigma} c_{\sigma} &=
 f^{\dagger}_{\sigma} b_{1} b^{\dagger}_{1} f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}b_{2}b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}} \\
&= f^{\dagger}_{\sigma} (1+ b^{\dagger}_{1}b_{1}  ) f_{\sigma} +
\sigma^{2} f_{\overline{\sigma}} (1+ b^{\dagger}_{2} b_{2} ) f^{\dagger}_{\overline{\sigma}}
\\
&=
f^{\dagger}_{\sigma}f_{\sigma} +f^{\dagger}_{\sigma} b^{\dagger}_{1}b_{1}f_{\sigma} 
+ \sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}}
+ \sigma^{2}f_{\overline{\sigma}}b^{\dagger}_{2} b_{2}f^{\dagger}_{\overline{\sigma}}
\\
&=
f^{\dagger}_{\sigma}f_{\sigma} +
b^{\dagger}_{1} ( 1- f_{\sigma}f^{\dagger}_{\sigma} )b_{1}
+ \sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}}
+ \sigma^{2}
b^{\dagger}_{2} (1- f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}}) )b_{2}
\\
&=
f^{\dagger}_{\sigma}f_{\sigma} +
\sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}}
+b^{\dagger}_{1}  b_{1} +b^{\dagger}_{2}b_{2}
-b^{\dagger}_{1}  f_{\sigma}f^{\dagger}_{\sigma} b_{1}
-\sigma^{2}
b^{\dagger}_{2}f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}}b_{2}.
\end{align} 
$$
By setting $\sigma^{2}=1$, and cancelling the last two terms of $2c^{\dagger}_{\sigma}c_{\sigma}$ by the first two terms of $2 c_{\sigma} c^{\dagger}_{\sigma}$, I would obtain
$$
\begin{align}
2c^{\dagger}_{\sigma}c_{\sigma} + 2c_{\sigma} c^{\dagger}_{\sigma}
&= f^{\dagger}_{\sigma}f_{\sigma} +
f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}}
+b^{\dagger}_{1}  b_{1} +b^{\dagger}_{2}b_{2} \\
&=
f^{\dagger}_{\sigma}f_{\sigma} +
1- f^{\dagger}_{\overline{\sigma}}f_{\overline{\sigma}}
+b^{\dagger}_{1}  b_{1} +b^{\dagger}_{2}b_{2}
\end{align}
$$
which is not the same as the constraint. Where did I make a mistake?
 A: After trying it a couple of times, I get the same as you at the equation for $2c^{\dagger}_{\sigma}c_{\sigma}$, but then I don't understand where the last two terms went. Basically, I get
$$
\begin{array}{ccl}
\{c_{\sigma},c^{\dagger}_{\sigma}\} & = & f^{\dagger}_{\sigma}f_{\sigma} b_1b_1^{\dagger} + f_{\bar{\sigma}}f^{\dagger}_{\bar{\sigma}} b_2b_2^{\dagger} \\
 & = & f^{\dagger}_{\sigma} f_{\sigma} + 1 - f^{\dagger}_{\bar{\sigma}} f_{\bar{\sigma}} +b^{\dagger}_2 b_2 + b^{\dagger}_1 b_1 f^{\dagger}_{\sigma} f_{\sigma} - b^{\dagger}_2 b_2 f^{\dagger}_{\bar{\sigma}} f_{\bar{\sigma}}.
\end{array}
$$
By summing on sigma, the $ff$ terms disappear and I get
$$
\begin{array}
a\sum_{\sigma} \{c_{\sigma}, c^{\dagger}_{\sigma}\} & = & 2 + 2 b^{\dagger}_2 b_2 + (b^{\dagger}_1 b_1 - b^{\dagger}_2 b_2) \sum_{\sigma} f^{\dagger}_{\sigma} f_{\sigma} \\
& = & 2 + b^{\dagger}_1b_1 + b^{\dagger}_2 b_2 + (b^{\dagger}_2b_2 - b^{\dagger}_1b_1)\left(\sum_{\sigma}f^{\dagger}_{\sigma}f_{\sigma}+1\right)
\end{array}
$$
which has some similar features to the condition for slave-particles, but it's not quite it! Good luck with that, I'll keep an eye on it.
