The Standard Model gauge group: charges, centers, and four global forms

The Lie algebra $\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ and the usual matter multiplets do not by themselves select one global gauge group. Their central action has a common kernel, and any subgroup of that kernel can be divided out while preserving those local representations.

Open the product gauge-group workbench and load the Standard Model example. It records the central subgroup by explicit coordinates and displays each field’s phase. This page derives the result, explains the convention choices, and identifies the physical information that remains outside the calculation.

1. Fix the circle before entering hypercharges

Use the compact covering group

$$\widetilde G=SU(3)_c\times SU(2)_L\times U(1),$$

with a circle parameter $\vartheta$ of period $2\pi$. The charge entered in the workbench is

$$q=6Y,\qquad e^{i\vartheta}\mapsto e^{iq\vartheta}.$$

Thus $q$ is an integer, while the conventional hypercharge is $Y=q/6$. Electric charge is normalized by $Q_{\mathrm{em}}=T_3+Y$.

The table below uses left-handed Weyl fermions throughout. The fields $u^c,d^c,e^c$ are the left-handed charge conjugates of the usual right-handed fields. Conjugation changes both the nonabelian representation and the sign of the abelian charge. The scalar $H$ is included with $Y=1/2$.

Multiplet	$SU(3)\times SU(2)$ representation	Dynkin labels	$q=6Y$
$Q$	$(\mathbf3,\mathbf2)$	$(1,0);(1)$	$1$
$u^c$	$(\overline{\mathbf3},\mathbf1)$	$(0,1);(0)$	$-4$
$d^c$	$(\overline{\mathbf3},\mathbf1)$	$(0,1);(0)$	$2$
$L$	$(\mathbf1,\mathbf2)$	$(0,0);(1)$	$-3$
$e^c$	$(\mathbf1,\mathbf1)$	$(0,0);(0)$	$6$
$H$	$(\mathbf1,\mathbf2)$	$(0,0);(1)$	$3$

Tong’s gauge-theory notes, §2.6.3, give the matter charges in an integer normalization; the table here rewrites the right-handed fields as left-handed conjugates and orders color before weak isospin.

Additional generations repeat the same central constraints. Adjoint gauge bosons have neutral center action. A sterile neutrino with labels $(0,0);(0)$ and $q=0$ also leaves the kernel unchanged.

Rescaling every integer charge while keeping the circle period fixed changes the representation of the compact group and can introduce an extra kernel. A charge convention must therefore state both the generator and its period; it is more than a cosmetic relabeling of a coupling.

2. Derive the embedded $\mathbb Z_6$

Let

$$\omega_3=e^{2\pi i/3}, \qquad \gamma= \left( \omega_3\mathbf1_3,\, -\mathbf1_2,\, e^{2\pi i/6} \right).$$

Write $t_3\in\mathbb Z_3$ for triality and $t_2\in\mathbb Z_2$ for the weak center charge. The action of $\gamma$ on a multiplet is

$$\rho(\gamma)= \exp\!\left[ \frac{2\pi i}{6}(2t_3+3t_2+q) \right]\mathbf1.$$

Every row in the table satisfies

$$2t_3+3t_2+q=0\pmod6.$$

For example, $Q$ gives $2+3+1=6$, while $u^c$ gives $4-4=0$. The antifundamental has triality $t_3=2$, equivalently $-1$ modulo $3$; treating a color triplet and an antitriplet as the same dimension label would lose the sign information.

To show that this is the entire common central kernel, rather than merely a subgroup of it, start with

$$c=(\omega_3^a\mathbf1_3,(-1)^b\mathbf1_2,e^{2\pi i\theta}).$$

The $e^c$ multiplet requires $6\theta\in\mathbb Z$, so set $\theta=m/6$. The lepton doublet then requires $b=m\pmod2$. The $d^c$ multiplet requires $2a+m=0\pmod3$, equivalently $a=m\pmod3$. Hence

$$c=\gamma^m,\qquad K=\langle\gamma\rangle\cong\mathbb Z_6.$$

The other rows obey the same condition. This direct derivation is also reproduced by the workbench’s exact rational enumeration. The embedded generator and quotient choices are discussed in Tong, Line Operators in the Standard Model, §3, pp. 7–9.

3. Four compatible quotient choices

The compatible connected gauge groups, with this covering group and charge convention fixed, are

$$G_d=\widetilde G/\Gamma_d, \qquad \Gamma_d=\langle\gamma^{6/d}\rangle, \qquad d\in\{1,2,3,6\}.$$

Order $d$	Embedded generator of $\Gamma_d$	Condition on a general multiplet
$1$	Identity	No additional central condition
$2$	$(\mathbf1_3,-\mathbf1_2,-1)$	$t_2+q=0\pmod2$
$3$	$(\omega_3^2\mathbf1_3,\mathbf1_2,e^{2\pi i/3})$	$2t_3+q=0\pmod3$
$6$	$\gamma$	$2t_3+3t_2+q=0\pmod6$

All usual Standard Model multiplets descend to all four groups. The kernel calculation therefore permits four choices; it does not select one as the observed global group.

The conditions are useful when adding new matter. A hypothetical $(\mathbf3,\mathbf1)_0$ has nontrivial color-center action and permits only $d=1$ or $2$. A nonabelian singlet with integer charge $q=1$ permits only the unquotiented covering group in this fixed normalization. The workbench can test these changes while retaining the original proposed quotient, making an obstruction visible.

A quotient is still a well-defined Lie group when a proposed field fails to descend. The failure says that this group and that local field representation cannot be used together in the stated ordinary representation category. It is not a failure of the field’s Lie-algebra representation.

4. What changes in the physics?

The adjoint gauge algebra and the listed local multiplets can be the same for all four choices, while the global theory has different allowed electric and magnetic probes and different bundle data. A representation of the covering group can label an ordinary Wilson representation of the quotient only if it descends.

For instance, a color and weak singlet with charge $q$ satisfies the $G_d$ descent test precisely when $d\mid q$. This is a group-level constraint on representations. A full discussion of genuine line operators must additionally account for screening, magnetic charges, and mutual locality; the calculator does not perform that classification.

The distinction between a global group and a fully specified quantum gauge theory is developed in Aharony, Seiberg, and Tachikawa, §1.1 and §2. The Standard Model application, including magnetic/dyonic lines and theta-angle periodicities, is treated in Tong, §3.

Likewise, a successful descent check is not an anomaly test. It does not establish perturbative gauge-anomaly cancellation, cancellation of global anomalies, or consistency on a chosen spacetime with a particular spin or other tangential structure. Those questions require information beyond the center characters used here.

5. The connection with $SU(5)$

The usual block homomorphism can be written in the same $q=6Y$ circle convention as

$$\begin{aligned} \Phi:\ SU(3)\times SU(2)\times U(1)&\longrightarrow SU(5),\\ (A,B,z)&\longmapsto \begin{pmatrix} z^{-2}A&0\\ 0&z^3B \end{pmatrix}. \end{aligned}$$

Its determinant is one, and its kernel is exactly $\langle\gamma\rangle$. Consequently its image is

$$S(U(3)\times U(2)) \cong \frac{SU(3)\times SU(2)\times U(1)}{\mathbb Z_6}.$$

This is a statement about the actual embedded subgroup. If one instead starts from $G_d$ with $d<6$, the same homomorphism still has a residual kernel $\mathbb Z_6/\Gamma_d$ and is not injective.

The fundamental branches as

$$\mathbf5\downarrow =(\mathbf3,\mathbf1)_{-2}\oplus(\mathbf1,\mathbf2)_3.$$

Taking the exterior square with the displayed $\Phi$ gives

$$\left.\Lambda^2\mathbf5\right|_{\Phi} =(\overline{\mathbf3},\mathbf1)_{-4} \oplus(\mathbf3,\mathbf2)_1 \oplus(\mathbf1,\mathbf1)_6,$$

and these are precisely $u^c,Q,e^c$. Together with

$$\left.\overline{\mathbf5}\right|_{\Phi} =(\overline{\mathbf3},\mathbf1)_2 \oplus(\mathbf1,\mathbf2)_{-3},$$

one obtains the familiar left-handed $\mathbf{10}\oplus\overline{\mathbf5}$ matter assignment.

The charged block-branching workbench uses the general primitive convention $\operatorname{diag}(z^2A,z^{-3}B)$ for the $3+2$ split. Its integer charge is $X=-q=-6Y$. Reversing the circle parameter gives $\Phi$ above. The SU(5) guide carries this sign choice explicitly through every summand.

Identifying the low-energy group with this particular $SU(5)$ matrix subgroup selects $d=6$. That is additional ultraviolet structure. The local matter spectrum alone does not establish such a completion.

6. Why all four groups have $\pi_1\cong\mathbb Z$

For these specific subgroups, a power of $\gamma^{6/d}$ has trivial circle component only when it is the identity. The phase-zero subgroup is therefore trivial. The product-center method gives

$$\pi_1(G_d)\cong\mathbb Z \qquad(d=1,2,3,6).$$

This does not erase the global-form distinction. The exact sequence of the degree-$d$ cover is

$$0\longrightarrow\mathbb Z \xrightarrow{\ \times d\ } \mathbb Z \longrightarrow\mathbb Z_d \longrightarrow0.$$

The covering group’s once-winding loop maps to $d$ times a generator in the quotient. Thus the abstract fundamental group can remain the same while its relation to the fixed charge normalization changes. See the product-center reference for the universal-cover derivation and an example where a central quotient also produces torsion in $\pi_1$.

7. A reproducible workflow

Begin with the Standard Model preset and inspect the explicit $q=6Y$ charges. Compare the four displayed quotient orders, then add or modify a multiplet and recalculate. A failed descent remains visible, together with the exact generator phase responsible for it.

Preserve the JSON export when using a result in a calculation. It records the highest weights, charge normalization, subgroup generators, common kernel, topology, sources, and implementation version. A saved setup should also state additional physical choices when they matter; this workbench does not infer them from a familiar group name.

The examples and methods are source checked and automatically verified. Independent human mathematical and physics review is recorded separately from source checking and automated verification.