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The SU(5) grand unified theory embeds the strong and electroweak groups in a single simple gauge group. Its 24 gauge generators contain the eight gluons, three weak generators and one hypercharge generator, plus twelve additional generators connecting color and weak directions. A common gauge coupling applies at the unification scale with the properly normalized hypercharge generator; low-energy couplings differ through their renormalization group evolution.

Using left-chiral fields throughout, one Standard Model generation fits intoThe dimensions are $3+2=5$ and $6+3+1=10$, so no right-handed neutrino is required in these multiplets. A singlet neutrino can be added. A suitable adjoint scalar vacuum breaks SU(5) to the Standard Model subgroup; a fundamental scalar contains an electroweak doublet and a colored triplet. The extra heavy vector bosons can mediate transitions between quarks and leptons, hence proton decay in the minimal model. This is a proposed unification, not an experimentally established identification of new particles.

The hierarchy problem concerns keeping an elementary weak-scale scalar light in the presence of much heavier physical scales. A heavy state coupled to it generically induces a threshold correction of order $\delta m_H^2\sim c{M}^2/(16{\pi }^2)$, with $c$ a product of couplings. Without protection, a weak-scale result then needs cancellation against a much larger parameter. In SU(5), this is sharpened by doublet-triplet splitting: the electroweak doublet must stay light while its colored partner must be sufficiently heavy to avoid rapid proton decay. Choosing a small tree-level doublet mass is not by itself a mechanism protecting it from subsequent heavy-particle corrections. Supersymmetry and other proposals change this protection problem, but the minimal SU(5) embedding alone does not solve it.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

The SU(5) grand unified theory has simply connected space $G=SU(5)$ and a connected Standard Model subgroup. More precisely that subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$; the product notation suppresses a finite central quotient. The long exact sequence of homotopy groups of a fibration $H\to G\to G/H$ givesThus stable magnetic monopoles form, whereas this transition does not require walls or strings. The monopole core and mass scales in an SU(5) transition are set by the massive gauge and scalar fields:For order-one couplings, the core size is of order ${\eta }^{-1}$ and the mass is of order $\eta$ up to the often substantial factor $4\pi /g$. These heavy objects are nonrelativistic soon after formation and can survive as stable relics.

Use the canonically normalized Abelian coupling $g_1=-\sqrt{5/3}{g}^{\prime }$, with $g_2=g$ and $g_3=g_s$, and write ${\alpha }_i=g_i^2/(4\pi )$. The sign in $g_1$ is a generator convention; its square is what enters running and matching. Let ${\alpha }_5=g_5^2/(4\pi )$ and $L_X=\log (M_X^2/{\mu }^2)$. To leading order,Use exactly the common-flavor coefficients stipulated for this model:The same $n_f$ cancels from their differences. In particular $b_3-b_2=11/3$ and $b_2-b_1=22/3$, so eliminating the unified coupling givesAt the low reference scale, $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. With $w={\sin }^2{\theta }_W$ this meansThe last difference equation becomes $3(1-w)/(5\alpha )-w/\alpha =2(w/\alpha -1/{\alpha }_s)$. Solving givesSubstitute this into the first difference equation to obtainAll low-scale couplings on the right must be evaluated at the same specified $\mu$. A dimensionful $M_X$ cannot be fixed by dimensionless coupling values without such a reference scale. The result is independent of the common $n_f$ in these differences, although ${\alpha }_5$ is not. An inferred $M_X>\mu$ requires ${\alpha }^{-1}>8/(3{\alpha }_s)$, and perturbative matching requires a positive small ${\alpha }_5$.

This is the one-loop SU(5) prediction with common flavor coefficients. It is the prediction of the supplied simplified coefficient model, not the full matter-and-Higgs calculation for a realistic SU(5) grand unified theory. Actual representations, intermediate thresholds and higher-loop terms change the running and require the corresponding matching calculation.

The block-diagonal subgroupis locally $SU(3)\times SU(2)\times U(1)$, with only a finite central quotient distinguishing the global groups. The hypercharge direction in the fundamental representation iswhich is traceless and commutes with the $SU(3)$ and $SU(2)$ blocks.

The canonically normalized $SU(5)$ generator must satisfy $\mathrm{Tr}(T_Y^2)=1/2$. Since $\mathrm{Tr}(Y^2)=5/6$,The unified gauge covariant derivative contains $g_5{T}_Y{B}_{\mu }$, whereas the Standard Model convention contains $g_{Y}Y{B}_{\mu }$. Hence $g_Y=\sqrt{3/5}{g}_5$. The non-Abelian generators already have the canonical normalization, so gauge coupling unification in the SU(5) grand unified theory predicts

Use a Hermitian gauge field $A_{\mu }=A_{\mu }^A{T}^A$ and $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. With fundamental covariant derivative ${\partial }_{\mu }-ig{A}_{\mu }$, the adjoint derivative isIt transforms as $D_{\mu }\Phi \mapsto U(D_{\mu }\Phi )U^{†}$, provided $A_{\mu }^{\prime }=U{A}_{\mu }{U}^{†}-(i/g)({\partial }_{\mu }U)U^{†}$.

The infinitesimal parameter in $U=1+iϵ+O({ϵ}^2)$ must be Hermitian and traceless, from unitarity and unit determinant. It is not generally antisymmetric as the hint states. The valid generator $ϵ=\mathrm{diag}(1,-1,0,0,0)$ is a concrete counterexample to that assertion.

Choose canonical adjoint kinetic normalization ${\mathcal{L}}_{\Phi }=\mathrm{tr}(D_{\mu }\Phi D^{\mu }\Phi )$, which gives $\frac{1}{2}\partial {\Phi }^A\partial {\Phi }^A$. For the diagonal vacuum expectation value $D=v\mathrm{diag}(2,2,2,-3,-3)$, the matrix unit $E_{ij}$ obeys $[D,E_{ij}]=(d_i-d_j)E_{ij}$. Thus the mass of the gauge boson associated with either normalized Hermitian generator joining $i,j$ is $g|d_i-d_j|$.

Within the three-dimensional and two-dimensional blocks, or for diagonal generators, the commutator vanishes. There are therefore $8+3+1=12$ massless gauge bosons. The $3\times 2$ interblock pairs supply two real generators each, hence 12 equally massive gauge bosons:The absolute mass formula assumes the stated kinetic normalization, which the question leaves implicit. Writing instead $\frac{1}{2}\mathrm{tr}(D\Phi {)}^2$ without rescaling $\Phi$ would multiply these squared masses by $1/2$.

This is the gauge group of the Standard Model, up to its global quotient, arising from one adjoint Higgs mechanism in a simple unified group. The unbroken abelian generator can be chosen as $Y=\mathrm{diag}(-1/3,-1/3,-1/3,1/2,1/2)$, so $D=-6vY$. This is the characteristic first breaking step of an SU(5) grand unified theory, rather than electroweak breaking all the way to electromagnetism.

An SU(5) grand unified theory embeds $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ into $SU(5)$. Canonical normalization of its hypercharge generator predicts $g_s^2=g_w^2=(5/3)g_Y^2$ at the unification scale.