b_0(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_1+t_3, 
\quad
&b_1(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_1t_2t_3, \\[2mm]
c_1(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_2+t_4,
\quad
&c_2(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_2t_3t_4, \\[2mm]
d_1(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_1t_2+t_1t_4+t_3t_4, 
\quad
&d_2(\mathbf{x}_{2,1,2,1}(\mathbf{t})) = t_1t_2t_3t_4.
\end{array}$$ It follows that, $$\label{eqnotCA}
t_4 = \frac{c_2}{a_1},\qquad
t_3 = \frac{\left|
\begin{matrix}
b_0& b_1\\
1 & a_1
\end{matrix}
\right|}{a_1},\qquad
t_2 = \frac{\left|
\begin{matrix}
c_1& c_2\\
1 & a_1
\end{matrix}
\right|}{a_1},\qquad 
t_1= \frac{b_1}{a_1}.$$ In fact, these formulas, although similar in spirit, are *not* the Chamber Ansatz formulas. Indeed the regular functions $a_1$, $b_1$, $c_2$, $b_0a_1-b_1$, and $c_1a_1-c_2$, are *not* cluster variables for the cluster algebra structure on $\C[N^w]$. Compare Example \[example6.2\] below.

\[Example3.7\] For the sake of comparison with Example \[Example3.6\], let us describe the groups $N$, $N(w)$, and $N'(w)$, for the same $w=s_2s_1s_2s_1$ and $\g = \slchap_2$. First, we have $$N = \left\{\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix}
\mid
a, b, c, d \in \C[[z]],\ ad-bc=1,\ 
 a(0)=d(0)=1,\ c(0)=0 
\right\},$$ where $\C[[z]]$ is the ring of formal power series in $z$. Next, $$\begin{array}{lcl}
N(w) &=&  \left\{\begin{pmatrix}
1 & b_0+b_1z+b_2z^2+b_3z^3 \\
0 & 1 
\end{pmatrix}
\mid
b_0, b_1, b_2, b_3 \in \C
\right\},\\  [5mm]
N'(w) &=&  \left\{\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix} \in N
\mid
b \in z^4\C[[z]]
\right\}.
\end{array}$$ Finally, for $g\in 