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Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

For normalized triplets $(E^{+},n,e{)}_L$ and $(E^{+},N,e{)}_R$, let the raising matrix have ones in entries $(1,2)$ and $(2,3)$. Its commutator with its adjoint is $\mathrm{diag}(1,0,-1)$. Neutral mixing in $n=c_{\alpha }\nu +s_{\alpha }N$ leaves this result unchanged because $c_{\alpha }^2+s_{\alpha }^2=1$. With current normalization using $1\mp {\gamma }^5=2P_{L,R}$, fermionic bilinear charge algebra gives the displayed proportionality to the leptonic electric charge $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$. Other current normalizations change the numerical prefactor, not closure. Equal-time algebra does not itself imply conservation of a fermionic weak charge in a symmetry-broken theory.