is due to the impossibility to eliminate textures and form domains, because this would require activated processes with $\Delta E_p>0$ given by Eq. (\[activation\]). Quenching to a finite temperature those processes are no longer forbidden and we expect textures to start being removed after a characteristic time $\tau _p ^{cross}(T_f)$. In order to estimate the crossover time let us consider again the situation of Fig. \[figenergy\]. The activated process described by the thin arrow, where the spins with $n=1$ are rotated to $n=2$, is a first action towards the removal of the texture, but the texture is not disappeared yet. The second action is the rotation of spins from $n=2$ to $n=3$, indicated by a bold arrow in the figure [@nota1]. This requires an energy E\_p\^[(2)]{}= J\[(2/p) +(4/p) - (6/p)-1\]. \[activation2\] Then a third action is required, where spins with $n=3$ are rotated to $n=4$, and so on, until, after $p-1$ steps all the spins in the region considered have $n=p$. It is easy to generalize Eqs. (\[activation\],\[activation2\]) to the generic $m$-th action: E\_p\^[(m)]{}= J\[(2/p) +(2m/p) - (2(m+1)/p)-1\]. \[activationm\] Let us consider $\Delta E_p^{(2)}$. This quantity is positive for $p>6$. For $p=5,6$, therefore, the second action is not an activated 