tend toward the same value $d\Omega/dt$ in accordance with the fact that the Hubble constant is the same in any direction. Since the equations (\[betapm\]) and the expression for $N$ lead to: $$\label{isoC1}
\frac{d\beta_\pm}{dt}=-\frac{p_\pm e^{3\Omega}}{12\pi R_0^3}$$ a stable isotropic state needs $p_\pm e^{3\Omega}\rightarrow 0$. We now look for the conditions allowing this limit.\
Let us assume that isotropy leads to a static Universe, i.e. $\Omega$ tends toward a constant. Then $p_\pm$ must tend toward zero such that $p_\pm e^{3\Omega}$ vanishes. However, from (\[lapse\]) and (\[ppp\]-\[pmp\]), it comes that $dp_\pm/dt\propto \partial V/\partial \beta_\pm e^{3\Omega}$. Hence for the Bianchi $I$, $VI_0$ and $VIII$ models, when $\beta_+$ and $\Omega$ tend toward some constants, the conjugate momentum diverges with the proper time $t$ since $\dot{p}_\pm$ tends toward a non vanishing constant and $p_+e^{3\Omega}\rightarrow \infty$. Thus from the reasonable assumption that isotropy happens when $t$ diverges, we deduce that isotropisation can not lead to a static Universe for these three models (it would be deeply unnatural that a static Universe ends for a finite value of $t$). For the $VII_0$ and $IX$ models, the demonstration is not so simple since then, when $\beta_\pm\rightarrow 0$ and $\Omega\rightarrow const$, $\partial V/\partial \beta_\pm\rightarrow 0$ and $\dot{p}_\pm$ vanishes. We 