- For what concern the full potential, it's clear that the $\Lambda$ coupling breaks the $SO(3)$ custodial symmetry. In fact, using only the neutral component of T and opening the Higgs we have:
$$V=m^2_T(t^0)^2-\Lambda\frac{t^0}{2}(\phi_{1b}^2-\phi_{2b}^2+2i\phi_{1b}\phi_{2b})-\Lambda\frac{t^0}{2}(\phi_{1b}^2-\phi_{2b}^2-2i\phi_{1b}\phi_{2b})-\frac{m^2}{2}(\phi_{1a}^2+\phi_{2a}^2+\phi_{1b}^2+\phi_{2b}^2)+\lambda(\phi_{1a}^2+\phi_{2a}^2+\phi_{1b}^2+\phi_{2b}^2)$$
so, I would say that $\Delta\rho\sim\Lambda/v$.
Performing the first derivative I find:
\begin{equation}
V_T=2m^2_TTr[T_{ab}]-\Lambda\epsilon^{ac}\epsilon^{bd}H_cH_d=0\end{equation}
\begin{equation}
V_H=-2\Lambda T_{ab}\epsilon^{ac}\epsilon^{bd}-m^2H^\dagger_c+2\lambda|H|^2H^\dagger_c=0.
\end{equation}
Now I have a doubt: for the vev of the triplet T, I can choose whatever "gauge" I want or is there a more privileged choice? I would say that, according to CP invariance, only an em neutral component can acquire a vev, so i would let $t^0$ acquires a vev: $$<T_{ab}>=v_0\delta_{a1}\delta_{b1} \, . $$
So I'm not sure if this choice should be an input in the eqs. and see if it minimizes the potential or rather an output.
Anyway, substituting that in the first eq. I find the condition for the Higgs vev (could be seen as a consistency check of my choice?):
$$V_T=2m^2_Tv_0-\Lambda (<H_2>)^2=0$$
which, denoting $v$ the Higgs vev, implies:
$$\frac{m^2_T}{\Lambda}=\frac{v^2}{v_0} \, .$$
- After the usual manipulations for the kinetic term of $T$ in the unitary gauge, I find:
$$L_{mass}=\frac{v_0^2}{4}\biggl[\bigl(gW_\mu^3-2g'B\mu)^2+g^2W_\mu^+W_\mu^-\biggr]$$
doubt: I haven't done any calculation yet, but how is it possible to find a mass term for the $Z$ if I have already a precise constraint (the one from Higgs sector) for the $(B,W^3)$ rotation into $(A,Z)$?