= Solution
Since the <Weak hypercharge>[hypercharge] is zero, <Electric charge>[electric neutrality] means $T^3v=0$. Choose the real-triplet <Vacuum state>[vacuum] $v_i=v\delta_{i3}$, with $v^2=m^2/(4\lambda)$. With <Canonical field normalization>[canonical] real-scalar kinetic density $-\tfrac12(D_\mu\Phi)^T D^\mu\Phi$, insert this <Vacuum state>[vacuum] into the <covariant derivative>. The <Weak hypercharge>[hypercharge] field $B_\mu$ has no coupling to this <Scalar field>[scalar], while only $T^1v,T^2v$ are nonzero. Thus
$$
\boxed{M_V^2=g^2v^2\operatorname{diag}(1,1,0,0)\quad\text{in }(W^1,W^2,W^3,B).}
$$
The <Electric charge>[charged] combinations $W^\pm=(W^1\mp iW^2)/\sqrt2$ have
$$
\boxed{m_{W^\pm}^2=g^2v^2=\frac{g^2m^2}{4\lambda},\qquad m_{W^3}=m_B=0.}
$$
Both $T^3$ and the independent <Weak hypercharge>[hypercharge] generator annihilate the <Vacuum state>[vacuum]. The connected <Stabilizer>[unbroken subgroup] is therefore
$$
\boxed{U(1)_{T^3}\times U(1)_Y.}
$$
The electromagnetic generator $Q=T^3+Y/2$ belongs to it, but it is only one linear combination of two surviving neutral generators. This model has two massless <Gauge field>[neutral gauge fields], rather than the single massless <photon> and massive neutral vector of the usual scalar-doublet electroweak model. A neutral-field rotation does not create a massive <eigenvalue> from this zero block.
The two transverse triplet components are eaten, leaving one real, electrically neutral radial <Scalar field>[scalar] $H=\phi_3-v$. The radial <eigenvalue> of the <Hessian matrix>[Hessian] found in part i is
$$
\boxed{m_H^2=8\lambda v^2=2m^2,\qquad m_H=\sqrt2\,m.}
$$
Indeed the radial <Scalar potential>[potential] is a constant plus $4\lambda v^2H^2+4\lambda vH^3+\lambda H^4$, fixing the factor of two between its quadratic coefficient and its <Mass>[mass squared]. The physical degree-of-freedom count is preserved: three real <Scalar fields>[scalars] plus four massless vectors give $3+4\times2=11$, while one <Scalar field>[scalar], two massive vectors and two massless vectors give $1+2\times3+2\times2=11$.
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