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For an adjoint vacuum expectation value proportional to $\mathrm{diag}(1,1,-2)$, the centralizer is $(SU(2)\times U(1))/{\mathbb{Z}}_2$. Four gauge bosons remain massless and four acquire a common mass. This differs from Fundamental-Higgs breaking of SU(3) to SU(2).

The unbroken local gauge symmetry associated with electric charge in the Standard Model. Its gauge boson is the massless photon. The neutral Higgs field vacuum preserves this subgroup while the W and Z acquire masses.

Sheldon Lee Glashow's 1979 lecture, Towards a Unified Theory: Threads in a Tapestry, recalls his 1961 $SU(2)\times U(1)$ construction: a photon, a massive neutral intermediary and a mixing angle. It also stresses its unresolved mass problem: symmetry-breaking masses had been inserted, rather than generated by the Higgs mechanism. Establishing the gauge structure and explaining the gauge boson masses were complementary steps.^[ref]

Steven Weinberg's A Model of Leptons (1967) builds a spontaneously broken gauge theory for the electron-type leptons, using a scalar doublet to give masses while retaining a massless photon. It derives neutral and charged couplings and discusses neutrino-electron scattering. Its closing question about renormalizability is explicitly unresolved; the paper is neither a proof of that property nor a complete treatment of quarks.^[ref]

Abdus Salam's 1979 lecture, Gauge Unification of Fundamental Forces, describes his and Steven Weinberg's completion of the spontaneously broken $SU(2)\times U(1)$ formulation for leptons. He separates that construction from the subsequent establishment of renormalizability. Its role is to connect the mass mechanism with predictive neutral and charged couplings, not to identify a photographed event. The Gargamelle evidence subsequently tested the predicted neutral current.^[ref]

A four-dimensional $\mathcal{N}=4$ supersymmetric gauge theory has one supersymmetric vector multiplet containing a gauge boson, four Weyl spinors, and six real scalar fields in the adjoint group representation. Its massless helicity multiplicities are $1,4,6,4,1$ from helicity $1$ to $-1$.

With canonically normalized gauge field kinetic terms, a gauge-boson mass matrix is the real symmetric matrix of coefficients in the displayed quadratic mass term. Its eigenvalues are squared physical gauge boson masses. For a complex scalar field vacuum ${\Phi }_0$ in the Higgs mechanism, $(M^2{)}_{ab}=2\mathrm{Re}[(g_a{T}_a{\Phi }_0{)}^{†}(g_b{T}_b{\Phi }_0)]$. Thus for real $x_a$, $x^T{M}^{2}x=2\Vert {\sum }_a{x}_a{g}_a{T}_a{\Phi }_0{\Vert }^2\ge 0$: it is a Gram matrix and a positive semidefinite matrix. An unbroken generator annihilates the vacuum expectation value and lies in its null space. Diagonalizing the matrix identifies the massive and massless field combinations.

For generators normalized by $\mathrm{Tr}(t^a{t}^b)={\delta }^{ab}/2$, a constant adjoint vacuum expectation value ${\Phi }_0$ gives this matrix. A generator commuting with ${\Phi }_0$ corresponds to a massless gauge boson.

Canonical gauge interactions preserve the tree-level supertrace mass sum rule at a vacuum with vanishing D-terms. Put $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. The extra real-scalar squared-mass trace is $2C$, the extra Weyl spinor squared-mass trace from gaugino mixing is $4C$, and the gauge boson squared-mass trace is $2C$. Their spin-weighted supertrace is $2C-2(4C)+3(2C)=0$. The scalar contribution comes from differentiating $\frac{1}{2}{\sum }_a{D}_a^2$; the fermionic one counts both blocks of the symmetric mixing matrix with entries $\sqrt{2}{g}_a{T}_{a}v$; the vector one follows from the covariant scalar kinetic term. Any zero-mass Goldstone bosons in the unreduced scalar Hessian do not change its trace.

At tree level and energies much greater than a gauge-vector mass, an amplitude with longitudinal external gauge bosons equals the corresponding amplitude with their would-be Goldstone bosons, up to conventional external-state phases and relative $O(m_V/E)$ corrections for a nonzero leading amplitude. In a renormalizable gauge, the broken-symmetry identities relate $k_{\mu }{\mathcal{M}}^{\mu }$ to $m_V{\mathcal{M}}_{\chi }$; inserting ${\epsilon }_L^{\mu }=k^{\mu }/m_V+O(m_V/E)$ explains the result. Loop-level versions include field-normalization factors. This is different from an equivalence theorem for field redefinitions.

In the Higgs mechanism, gauge fields associated with broken generators acquire longitudinal polarizations and masses by absorbing Goldstone modes. Gauge fields of the unbroken subgroup remain massless.

The phrase “absorbing” describes a rearrangement of degrees of freedom, not a collision in which a gauge boson swallows another particle. The gauge symmetry remains a redundancy of the description. A chosen vacuum expectation value leaves a smaller subgroup manifest, while gauge-invariant observables describe the physical masses.

A one-field calculation isolates the mechanism. In an Abelian Higgs model, write a complex scalar field locally as $\Phi =(v+H)e^{i\theta }/\sqrt{2}$, use $D_{\mu }={\partial }_{\mu }-ie{A}_{\mu }$, and define $B_{\mu }=A_{\mu }-{\partial }_{\mu }\theta /e$. ThenThe second term contains a Proca field mass $m_B=ev$ and its interactions with the radial Higgs boson. A local gauge transformation removes $\theta$ from the scalar field but leaves its information in the longitudinal vector polarization. The electroweak symmetry breaking calculation applies this idea to three broken directions.

In their 1964 paper, François Englert and Robert Brout showed that gauge bosons associated with generators that rotate the chosen vacuum acquire masses while gauge invariance is maintained. Their discussion included both scalar and fermionic models.^[ref]

Independently, Peter Higgs exhibited a massive vector and a surviving massive scalar field by linearizing a simple gauge theory. His paper explicitly distinguishes this classical calculation from a proof for the full quantized theory. These 1964 mechanisms supplied an ingredient for the later electroweak interaction, rather than the complete Standard Model.^[ref]

The following scalar toy isolates one step in renormalization: removing a local divergence while retaining a finite dependence on a physical mass. For $M,\mu >0$ and a sharp cutoff regularization in Euclidean space with momentum bound $\Lambda$, define the dimensionless Feynman integralThis is a zero-external-momentum scalar bubble integral stripped of coupling and combinatorial factors. Using the three-sphere area $2{\pi }^2$ and $t=k_E^2$ reduces the integral toThe antiderivative follows by writing $t=(t+M^2)-M^2$. At large $\Lambda$ the divergent part is $\log {\Lambda }^2/(16{\pi }^2)$, with the logarithm's dimensionful argument understood only within the displayed ratios. It has no external-momentum dependence. In a four-scalar-field vertex this structure can therefore be removed by a local operator proportional to ${\varphi }^4$, rather than a momentum-dependent nonlocal interaction.

Choose the fixed reference mass $\mu$ and subtract $I_{\Lambda }(\mu )$. If the loop contribution to a chosen four-point vertex coefficient is $c{\lambda }^2{I}_{\Lambda }(M)$, choose the additive counterterm in that same coefficient as $\delta {\lambda }_{\Lambda }=-c{\lambda }^2{I}_{\Lambda }(\mu )$, where $c$ includes its stated diagram factors. This defines a finite subtraction prescription:The renormalization condition for this toy loop contribution is $I_R(\mu ;\mu )=0$. The reference mass stays fixed when $M$ is varied; finite dependence on $M$ survives. A different finite subtraction changes the renormalization scheme, and the defined renormalized coupling must change consistently.

For $M=2\mu$, the limiting result is $16{\pi }^2{I}_R=-\log 4\approx -1.38629436$. At $\Lambda =10\mu$, the exact regulated difference is about $-1.32846343$; at $\Lambda =100\mu$, it is about $-1.38569459$. Expanding the exact expression gives the leading residual $2(M^2-{\mu }^2)/{\Lambda }^2$ in $16{\pi }^2[I_{\Lambda }(M)-I_{\Lambda }(\mu )]$. This explicitly separates finite-cutoff error from the chosen finite renormalization condition.

A sharp cutoff regularization is convenient for this scalar demonstration, but shifts of a divergent loop momentum can change its integration domain. A gauge theory calculation cannot assume that it preserves the Ward identities. The toy contains no interacting gauge bosons, ghost loops or overlapping ultraviolet subdivergences, so its successful subtraction does not prove Non-Abelian gauge theory renormalizability.

A physical polarization of a massive spin-one gauge boson whose spatial polarization points along its spatial momentum. Together with two transverse polarizations it supplies the three physical spin states. In the Higgs mechanism, a broken-direction Goldstone boson supplies this extra state as the gauge boson acquires mass.

Consider one massive gauge field with $m>0$ after Higgs mechanism. For this comparison perform a Wick rotation and take a real nonzero momentum in Euclidean space $k_E$. The transverse and longitudinal orthogonal projectors areHere “longitudinal” means parallel to the off-shell four-momentum, not the physical longitudinal polarization of an on-shell massive gauge boson. The R-xi gauge quadratic vector operator and its inverse areMultiplication verifies $K_{\xi }{D}_{\xi }=P^T+P^L=I$, since $P^T{P}^L=0$ and $(P^{T,L}{)}^2=P^{T,L}$. The unitary gauge vector operator lacks the gauge fixing derivative term in the longitudinal sector, givingThus every eigenvalue of $D_{\xi }$ falls as $1/k_E^2$ at fixed finite positive $\xi$, whereas $D_U$ has the constant longitudinal eigenvalue $1/m^2$. Unitary gauge consequently obscures ordinary power counting; this is not a failure of the physical theory.

For a worked comparison choose $|k_E|=10m$. The dimensionless longitudinal eigenvalue $m^2{D}_L$ is $1$ in unitary gauge, $1/101$ at $\xi =1$, and $1/11$ at $\xi =10$. Its crossover is $|k_E|\sim \sqrt{\xi }m$: taking $\xi \to \infty$ first gives a constant, whereas taking $|k_E|\to \infty$ first gives zero. The two limits do not commute. Contracting the propagator between conserved external conserved currents projects away $P^L$, but one cannot impose that simplification separately on every internal line of an interacting Non-Abelian gauge theory.

In the MSSM, each fermion generation has chiral superfields $Q:(3,2,1/6)$, $U^c:(\bar{3},1,-2/3)$, $D^c:(\bar{3},1,1/3)$, $L:(1,2,-1/2)$, and $E^c:(1,1,1)$. This uses hypercharge normalized by $Q_{\mathrm{electric}}=T^3+Y$ and only left-handed Weyl spinors. Two Higgs chiral doublets have charges $\pm 1/2$. The three vector superfields transform as $(8,1,0)$, $(1,3,0)$ and $(1,1,0)$, supplying the gauge bosons and gauginos.

The Standard Model uses the local gauge group $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$. Its Higgs field is a complex Higgs doublet of hypercharge $Y=1/2$. A bare mass term for a weak gauge boson, or for a charged fermion joining inequivalent left- and right-handed gauge representations, would not respect the unbroken electroweak construction. Instead introduce a gauge-invariant Higgs potentialIts minima satisfy ${\Phi }^{†}\Phi =v^2/2$, $v^2={\mu }^2/\lambda$. Choosing a representative $⟨\Phi ⟩=(0,v{)}^T/\sqrt{2}$ leaves $Q=T_3+Y$ unbroken. In this description electroweak symmetry breaking is $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_{\mathrm{em}}$; the underlying local gauge symmetry remains a redundancy of the theory. A vacuum representative is not a physical explicit violation of gauge invariance.

In unitary gauge, write $\Phi =(0,v+H{)}^T/\sqrt{2}$. The gauge-covariant kinetic term $|D_{\mu }\Phi {|}^2$ produceswith $\tan {\theta }_W=g^{\prime }/g$ and $e=g\sin {\theta }_W$. Three angular scalar directions supply the longitudinal gauge-boson polarizations of $W^{+},W^{-},Z$, while one real radial Higgs boson survives with $m_H^2=2\lambda v^2$. Counting physical degrees of freedom checks the Higgs mechanism: the electroweak gauge sector and scalar doublet have $4\times 2+4=12$ before the rearrangement and $3\times 3+2+1=12$ afterwards.

The Higgs potential also fixes the scalar self-interactions:The color gauge group is unbroken, so gluons remain massless. The same electroweak kinetic term fixesGauge-invariant Yukawa couplings such as $-y_d{\bar{Q}}_L\Phi d_R-y_u{\bar{Q}}_L\tilde{\Phi }{u}_R+h.c.$ give $m_f=y_{f}v/\sqrt{2}$ and the physical coupling $-m_{f}H\bar{f}f/v$. These relations predict stronger Higgs couplings to heavier massive particles; they do not determine the Yukawa parameters themselves. The minimal Standard Model does not supply observed neutrino masses. The surviving Higgs boson is neutral and has spin zero; its mass and self-coupling are tied to the curvature of the Higgs potential, not fixed by the gauge couplings alone. Higgs exchange also provides the Higgs cancellation in longitudinal vector scattering, maintaining the required high-energy behavior of weak amplitudes.

At a proton collider, important production mechanisms are gluon-fusion Higgs production through a heavy-quark loop, vector-boson fusion with two outgoing quark jets, associated $WH$ or $ZH$ production, and production with a top quark pair. At an electron–positron collider, Higgsstrahlung $e^{+}{e}^{-}\to ZH$ is a clean example. Tree-level decays include Higgs decay to a fermion pair and, when allowed, Higgs decay to two W bosons or Higgs decay to two Z bosons. Below an on-shell vector-pair threshold, one or both vectors can be virtual. The $gg$ and $\gamma \gamma$ channels arise through loops; neutral colorless $H$ has no corresponding direct tree-level gauge coupling. Rates and branching fractions depend on both couplings and kinematics, so “heaviest available particle” is a guide rather than a complete rule.

The Higgs boson discovery in 2012 supplied an experimental test. The ATLAS experiment and CMS experiment reconstructed two-photon and four-lepton invariant masses in Large Hadron Collider proton collisions. Electromagnetic calorimeters measure photon energies, while tracking and muon systems identify charged leptons; narrow reconstructed peaks can be distinguished from smooth backgrounds with control samples and a likelihood analysis. The ATLAS discovery paper reported a new resonance near $126.5\mathrm{GeV}$. An observed mass peak supports the existence of a new state; production channels, decay rates and angular distributions test whether it has the particular Higgs couplings and quantum numbers.^[ref]

Finally, the hierarchy problem asks why the electroweak scalar mass scale remains much smaller than a putative heavy physical scale without delicate cancellations. A cutoff estimate suggests corrections $\delta m_H^2\sim {\Lambda }^2/(16{\pi }^2)$ times couplings, but the quadratic cutoff term itself depends on the regularization convention. The physical concern persists when a heavy state of mass $M$ couples to the Higgs field and produces threshold corrections of order its coupling times $M^2/(16{\pi }^2)$. Matching a much smaller Higgs mass may then require cancellation between independent parameters. Supersymmetry, compositeness and other protective structures propose different resolutions; renormalizability of the Standard Model does not by itself resolve this sensitivity or establish that new particles must occur at one particular mass.

For eigenvalues $(v,v,-2v)$, the first two entries coincide. Consequently the $12$ generators $t^1,t^2$, as well as $t^3,t^8$, commute with the vacuum. The $13$ and $23$ eigenvalue differences both have magnitude $3|v|$, givingThere are four massless and four massive gauge bosons. The centralizer of a Hermitian matrix in SU(n) preserves the degenerate two-dimensional eigenspace, so the unbroken group isAt the Lie algebra level this is $\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, generated by $t^{1,2,3}$ and $t^8$. The finite quotient specifies the actual subgroup inside SU(3); simply writing $SU(2)\times U(1)$ suppresses that global identification. The four real massive fields combine into a complex SU(2) doublet and its conjugate. Unbroken SU(2) rotations relate its two components, explaining the equal masses of the $13$ and $23$ sectors in this SU(3) gauge theory with an adjoint Higgs field.

The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

Let $Y$ be the compact coordinate, identified modulo $2\pi R$. The closed string may wrap the circle:Single-valued momentum eigenfunctions $e^{ipY}$ require $p=n/R$, $n\in \mathbb{Z}$. These are the momentum and winding modes. Write the chiral zero-mode momenta asThe left-moving string oscillators will be tilded: $N_L=\tilde{N}$, $N_R=N$. This fixes the sign in compact-circle closed-string level matching.

Define the lower-dimensional mass by $M^2=-p_{\mathrm{noncompact}}^2$. The two closed-string physical-state Virasoro conditions, with bosonic string intercept one, areTheir average and difference derive the complete free bosonic string mass spectrum on the circle:The allowed states have nonnegative integer oscillator levels and obey the Virasoro constraints within each chiral sector. Compactification has not removed oscillator polarizations; it has added quantized compact momentum and allowed winding while changing the lower-dimensional interpretation of those polarizations.

At large $R$, the momentum and winding modes with $w=0$ have closely spaced momentum energies $|n|/R$, whereas nonzero winding costs $|w|R/{\alpha }^{\prime }$. At small $R$ their roles reverse. The precise equality is T-duality:It leaves $p_L$ unchanged and reverses $p_R$, and leaves the mass formula and closed-string level matching unchanged. On the string embedding map it reverses the right-moving part of $Y$. Therefore the small-radius spectrum is the large-radius spectrum of the dual theory, rather than a spectrum in which all charged states simply become heavy.

At a generic radius, the neutral level $(N_L,N_R)=(1,1)$ contains the lower-dimensional graviton, Kalb–Ramond field and dilaton, two gauge bosons from the mixed components of the metric and two-form, and the radius modulus of a string compactification. The two vector charges are the Cartan charges of $U(1{)}_L\times U(1{)}_R$.

Consider instead the following charged states:Every state satisfies compact-circle closed-string level matching. The single noncompact string oscillator supplies a vector polarization. At the massless point, the level-one Virasoro constraint imposes transverse polarization and its physical null string state identifies polarizations differing by the noncompact momentum. Thus these are genuine lower-dimensional vector states. Their squared mass isThus four additional charged massless spin-one particles appear atFor $(N_L,N_R)=(0,1)$ the charges have $p_R=0$, $p_L=\pm 2/\sqrt{{\alpha }^{\prime }}$; for $(1,0)$ they have $p_L=0$, $p_R=\pm 2/\sqrt{{\alpha }^{\prime }}$. This identifies the two pairs of charged roots.

The enhanced gauge group can be derived rather than just inferred from counting. At the self-dual circle, the chiral compact boson satisfies $Y_L(z)Y_L(u)\sim -({\alpha }^{\prime }/2)\log (z-u)$. The self-dual circle current algebra has currentsThe exponentials have conformal weight ${\alpha }^{\prime }{p}_L^2/4=1$. Free-boson contractions give the operator product expansionsThey form the level-one SU(2) Lie algebra, and there is an independent right-moving copy. The string vertex operators $J_L^a\bar{\partial }{X}^{\mu }$ and $\partial X^{\mu }{J}_R^a$ give the six massless vectors. ConsequentlyChanging $R$ gives the four charged vectors the mass found above, consistent with the radius deformation as a Higgs mechanism.

For completeness, the enhanced spectrum also has scalar fields. Using a compact rather than noncompact oscillator in the four singly excited charged families gives four scalars. At $(N_L,N_R)=(0,0)$, closed-string level matching requires $nw=0$; the mass equation then gives the four charges $(\pm 2,0),(0,\pm 2)$. These exceptional massless ground states of a bosonic circle give another four scalars. Together with the radius scalar they form nine scalars in the $(3,3)$ representation, in addition to the neutral dilaton. The ever-present neutral bosonic tachyon still has $M^2=-4/{\alpha }^{\prime }$. The enhancement establishes the additional massless vectors, not stability of the entire bosonic vacuum.

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.

In a supersymmetric Higgs mechanism, an internal gauge symmetry breaks while supersymmetry can remain unbroken. A massless vector multiplet absorbs a chiral multiplet and becomes a massive multiplet with spins $(1,\frac{1}{2},\frac{1}{2},0)$: the gauge boson acquires its longitudinal particle polarization from the Goldstone mode. In the super-Higgs mechanism, local supersymmetry itself breaks: the spin-$3/2$ gravitino absorbs the spin-$1/2$ goldstino, gaining two longitudinal polarizations and becoming massive. With only global supersymmetry breaking, the goldstino instead remains a physical massless fermion.