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A singular value decomposition of the charged-lepton Yukawa matrix uses independent unitary rotations of left- and right-handed leptons. Their gauge-covariant kinetic terms are unchanged because each rotation mixes identical Standard Model representations. After electroweak symmetry breaking, the nonnegative singular values give Dirac masses $m_i=y_{i}v/\sqrt{2}$ and diagonal Higgs boson couplings $m_i/v$. When the neutrinos are massless, rotating their left-handed fields with the charged partners preserves the diagonal weak charged current.

With the plus-sign convention $D_{\mu }={\partial }_{\mu }+ig{T}_3{W}_{\mu }^3+i{g}^{\prime }Y{B}_{\mu }$, let $\tan {\theta }_W=g^{\prime }/g$ and $A_{\mu }=\sin {\theta }_W{W}_{\mu }^3+\cos {\theta }_W{B}_{\mu }$. The coefficient of $A_{\mu }$ is $g\sin {\theta }_W{T}_3+g^{\prime }\cos {\theta }_{W}Y=e(T_3+Y)$, where $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The extra minus sign in the interaction Lagrangian comes from the $i$ in $\bar{\psi }i{\gamma }^{\mu }{D}_{\mu }\psi$. It reproduces the electric charges of every component of an Standard Model representation.

The Standard Model Higgs field is a complex $SU(2{)}_L$ doublet of hypercharge $1/2$. Its nonzero vacuum expectation value triggers the Higgs mechanism.

The Higgs field is a scalar field: its components carry internal weak isospin and hypercharge, but no spacetime spin. A complex Higgs doublet contains four real field components. The vacuum expectation value is a property of the lowest-energy state, not a gas of Higgs bosons. A Higgs boson is an excitation about that state.

Why introduce this field? A bare mass term for a gauge field generally fails to respect the local gauge symmetry. A bare Dirac mass term for a charged Standard Model fermion also joins left- and right-handed fields with different Standard Model representations. A Higgs doublet permits gauge-invariant interactions whose expansion about a nonzero vacuum expectation value supplies both masses and definite Higgs boson interactions. The field therefore links the mass problem to experimentally testable coupling relations, as illustrated by the Higgs boson coupling to Z bosons.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The renormalization-group beta function impliesIntegrating the one-loop running coupling equation givesHere the sign of ${\beta }_i$ is the sign convention in the question: a negative coefficient makes the inverse coupling increase towards high energy.

For gauge coupling unification, let $L=\log (M_{\mathrm{GUT}}/M_Z)/(2\pi )$ and let $a_G$ be the common inverse coupling at the unification scale. The normalized hypercharge coupling is $(5/3){\alpha }_1$, soSubtract the second relation from the first to determine $L$, then subtract it from the third. Provided $3{\beta }_1/5-{\beta }_2\ne 0$, this provesThe factors $3/5$ belong to the inverse normalized coupling; dropping them changes the prediction.

Use the Standard Model gauge group $SU(3{)}_C\times SU(2{)}_L\times U(1{)}_Y$ and convention $Q=T_3+Y$. One fermion family and one complex Higgs doublet have the following Standard Model representations. The left-handed quark and lepton doublets contain two Weyl spinors each; the right-handed fields are single Weyl species.The row $u_R$ denotes the right-handed up quark. Equivalently an all-left-handed table replaces the three right-handed fields by $u_R^c:(\overline{3},1{)}_{-2/3}$, $d_R^c:(\overline{3},1{)}_{1/3}$, and $e_R^c:(1,1{)}_1$; these are alternative descriptions, not extra fermions. The minimal model has no right-handed neutrino. For completeness, the family-independent gauge fields are the spin-one gluons $(8,1{)}_0$, weak gauge bosons $(1,3{)}_0$, and hypercharge gauge boson $(1,1{)}_0$; they are not counted once per family.

For the Standard Model one-loop gauge coefficients, include three fermion families and one Higgs doublet. With $T(\mathbf{N})=1/2$, the colour Dynkin index sum in one family isThe first factor two counts the two members of $Q_L$; the colour trace is already included in the Dynkin index. There is no coloured scalar. HenceFor $SU(2{)}_L$, the three colours of $Q_L$ and the lepton doublet give $3(1/2)+1/2=2$ per family. The Higgs contributes $T_2(\varphi )=1/2$. ThusFor $U(1{)}_Y$, sum squared hypercharges over every colour and doublet component. One family contributesThe two complex Higgs components contribute $2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction term, soThis ${\beta }_1$ is for the unnormalized coupling $g_Y$ used in the question. For $g_1^{\mathrm{GUT}}=\sqrt{5/3}{g}_Y$, the corresponding coefficient is $(3/5){\beta }_1=41/10$. This distinction is required for a consistent unification calculation.

For one generation, the Standard Model representation content under $SU(2{)}_L\times SU(3{)}_C$ iswith no right-handed neutrino. For $SU(3{)}_C$, three generations give $n_L=6$ from the two left-handed quark flavors and $n_R=6$ from the two right-handed quark flavors, while $n_s=0$. ThereforeFor $SU(2{)}_L$, each generation supplies three colored quark doublets and one lepton doublet, so $n_L=12$, $n_R=0$, and the single complex Higgs doublet gives $n_s=1$. Hence