comes that $\dot{\beta}_+-\sqrt{3}\dot{\beta}_-=p_0H^{-1}$. Thus, isotropisation needs a diverging Hamiltonian but for the zero measure case with $p_0=0$. If now we consider (\[eq5\]) near $(y,w)=(0,0)$, we deduce that $w$ behaves as $e^{-2\Omega}$ and vanishes when $\Omega\rightarrow +\infty$. Introducing this expression in (\[v4\]), we derive that $H$ should asymptotically behave as $e^{2(\beta_++\sqrt{3}\beta_-)}$ and thus should tend toward a constant value when Universe approaches isotropy. This contradict the fact that the Hamiltonian have to diverge and thus the set of points $(y,w)=(0,0)$ is not compatible with isotropisation.\
The second set of points is not real and such that $x_\pm\not =0$. Hence, it does not correspond to an isotropic equilibrium state and we discard it. The third set of points is such that $x_\pm=0$ and respects the constraint. It writes $(x_{+},x_{-},y,z,w)=(0,0,\pm \sqrt{3-\ell^2}(6\sqrt{2})^{-1},\ell/6,0)$ and is real if $\ell^2<3$, implying that isotropisation is not possible if this last quantity diverges[^2]. Indeed since asymptotically $z$ behaves as $\ell/6$, it means that an isotropic stable state needs $\ell$ to tend to a constant value with no limit cycle otherwise $\dot{z}\not\rightarrow 0$ when $\Omega\rightarrow -\infty$. This is the only set of points representing an isotropic stable state and the only one we will consider below. The fourth set of 