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Topological Responses of the Standard Model Gauge Group

T0 review · 2 major / 5 minor · reviewed 2026-08-10 · deepseek-v4-flash

Pith's one-line read This paper claims that a single fractional topological response $\sigma_n(q,k)$ — the coefficient of $\int \sigma_n\, B_m\, dA_{X_n}$ — uniquely fixes which of the four global gauge groups $(SU(3)\times SU(2)\times U(1))/\mathbb{Z}_q$ the…

desk verdict A solid generalization of Hsin–Gomis with sound number theory, but the X_n symmetry is only a global symmetry if you stop gauging hypercharge, so the abstract's 'measurable' claim overreaches. read the letter →

arxiv 2412.21196 v2 pith:UVE7OGVX submitted 2024-12-30 hep-th cond-mat.str-elhep-ph

classification hep-thcond-mat.str-elhep-ph
keywords StandardModelglobalgaugegrouptopologicalresponsesymmetryfractionalizationhigher-formbaryon-minus-leptoncobordisminvariantsfractionalquantumHall
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

The Standard Model's force carriers are described by the Lie algebra $su(3)\times su(2)\times u(1)$, but that same algebra admits four different global gauge groups, $G_{\mathrm{SM}_q}=SU(3)\times SU(2)\times U(1)/\mathbb{Z}_q$ with $q=1,2,3,6$, and current experiments do not tell which one is realized. The paper introduces a family of global symmetries $X_n\equiv n(B-L)+(1-\frac{n}{3})\tilde{Y}$, combinations of baryon-minus-lepton number and electroweak hypercharge treated as a global symmetry, and computes the fractional topological response $\sigma_n(q,k)$ they induce through symmetry fractionalization. The central claim is that for any fixed $n\ge 7$ outside the four exceptional values $10,12,15,30$, the response $\sigma_n(q,k)=\frac{q(1-n)\gcd(2,n)}{2n}+\frac{kq}{6}\bmod 1$ takes a distinct value for each allowed pair $(q,k)$, so one measurement would identify both the global gauge group and the fractionalization class. This matters because it turns an abstract ambiguity about the global structure of the Standard Model into a concrete, quantum-Hall-like measurable quantity.

What carries the argument

The load-bearing objects are the new $U(1)_{X_n}$ symmetries with integer charges $q_{X_n}=(2-n,-3,1,-4+n,6-n,n,3-n)$ on the Standard Model fields, and two structural constraints that tie them to the SM gauge bundle: the gauge-bundle constraint relating the hypercharge flux $da_{\tilde{Y}}$ to the $X_n$ flux and the Stiefel-Whitney classes, and the fractionalization constraint $f_k^*\tilde{B}_e=\frac{k}{6/q}dA_{X_n}$ coming from the class $k\in H^2(BG_{[0]},\mathbb{Z}_{6/q})=\mathbb{Z}_{6/q}$, where symmetry fractionalization is the data specifying how the 0-form symmetry acts on the 1-form symmetry's charged line operators. These feed the mixed anomaly $\int_{M^5}-\frac{1}{2\pi}\tilde{B}_e\, dB_m$, whose inflow produces the fractional SPT term $\int_{M^4}\frac{k}{6/q}A_{X_n}\, dB_m$. The identity that carries the argument is $\sigma_n(q,k)=\frac{q(1-n)\gcd(2,n)}{2n}+\frac{kq}{6}\bmod 1$; its two terms come respectively from the fractional part of the magnetic 2-current $J_m^{(2)}=q\star\frac{da_{\tilde{Y}}}{2\pi}$ and from the anomaly-cancelling fractionalization term.

What would settle it

A finite modular computation settles the uniqueness claim: enumerate all pairs $(q,k)$ with $q\in\{1,2,3,6\}$ and $k\in\{0,\dots,6/q-1\}$; for any $n\ge 7$ outside $\{10,12,15,30\}$, two distinct pairs with equal $\sigma_n(q,k)\bmod 1$ would refute the 'if and only if' statement. A measurement of the coefficient of $\int B_m\, dA_{X_n}$ on a $\mathrm{Spin}^c$ (odd $n$) or $\mathrm{Spin}$ (even $n$) 4-manifold that disagrees with the formula would refute the derivation.

Watch

Extended reading notes

Core claim

The paper establishes that the symmetry-enriched Standard Model carries a fractional topological response whose coefficient is a function of two discrete labels that experiments have not yet fixed: $q$, the order of the quotient in $G_{\mathrm{SM}_q}$, and $k$, the electric symmetry-fractionalization class in $\mathbb{Z}_{6/q}$. Using the new $U(1)_{X_n}$ symmetry, $X_n\equiv n(B-L)+(1-\frac{n}{N_c})\tilde{Y}$ with $N_c=3$, the paper derives $\sigma_n(q,k)=\frac{q(1-n)\gcd(2,n)}{2n}+\frac{kq}{6}\bmod 1$ as the coefficient of the 4d response $\int \frac{\sigma_n}{2\pi} B_m\, dA_{X_n}$. The derivation runs through two constraints: the gauge-bundle constraint $\frac{da_{\tilde{Y}}}{2\pi}=\frac{1}{\mathrm{lcm}(2,n)}\frac{dA_{X_n}}{2\pi}-\frac{\gcd(2,n)}{2}w_2(TM)+\frac{1}{q}w_2^{(q)}\bmod 1$, and the fractionalization constraint $f_k^*\tilde{B}_e=\frac{k}{6/q}dA_{X_n}$, which cancels the mixed anomaly between the electric and magnetic 1-form symmetries. For a fixed $n$ with $n\ge 7$ and $n\notin\{10,12,15,30\}$, the map $(q,k)\mapsto \sigma_n$ is injective modulo 1, and the same injectivity is achieved by measuring pairs such as $(n_1,n_2)=(2,3),(2,5),(3,4),(3,5),(4,5)$. The result is independent of the number of fermion families and of whether right-handed neutrinos are present.

Load-bearing premise

The prediction that $\sigma_n$ is measurable rests on $U(1)_{X_n}$ being an exact global symmetry of the Standard Model, which requires treating $B-L$ as exact and the hypercharge $U(1)$ as restricted to global transformations; if neutrino masses, quantum gravity, or other higher-energy physics break $X_n$, the response vanishes or changes and the uniqueness argument loses its observable target.

Editorial extensions

If this is right

  • A measurement of $\sigma_n$ at any admissible $n$ distinguishes all four candidate gauge groups $q=1,2,3,6$, turning the global-structure ambiguity of the Standard Model into a single experimental question.
  • The same measurement also fixes the fractionalization label $k$, selecting one of the symmetry-enriched variants $\mathrm{SM}_{(q,k)}$ from the kinematical possibilities.
  • The earlier $n=3$ response cannot separate $\mathrm{SM}_{(1,2)}$, $\mathrm{SM}_{(2,2)}$, $\mathrm{SM}_{(3,0)}$, and $\mathrm{SM}_{(6,0)}$; the $X_n$ family removes these degeneracies for $n\ge 7$ outside the four exceptional values.
  • The response is independent of the fermion family number and of the presence of right-handed neutrinos, so the distinguishing power survives Standard Model extensions that preserve $X_n$.
  • Because the response is fractional and quantized modulo 1, it behaves like a fractional quantum Hall conductance and is insensitive to continuous deformations.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • Beyond the paper's explicit list, one could classify all finite sets of $n$ values whose combined responses $\{\sigma_n\}$ uniquely identify $\mathrm{SM}_{(q,k)}$; the listed pairs are instances, not a complete classification.
  • If $X_n$ is broken by neutrino masses, quantum gravity, or other ultraviolet physics, the predicted response would vanish or shift, so a positive measurement of $\sigma_n$ would simultaneously certify that $B-L$ is an exact low-energy global symmetry.
  • Because the derivation uses only symmetry and topology, a condensed-matter or lattice system with the same 0-form and 1-form symmetry data could emulate the response, making the abstract Standard Model distinction potentially testable in a tabletop experiment.
  • The odd-$n$ versus even-$n$ distinction ($\mathrm{Spin}^c$ versus $\mathrm{Spin}$ manifolds) suggests the response depends on spacetime topology; extending the computation to manifolds with boundary could yield edge or defect observables that are easier to measure.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

2 major / 5 minor

Summary. The paper studies the global form of the Standard Model gauge group, GSM_q = (SU(3)×SU(2)×U(1))/Z_q with q=1,2,3,6, which is not fixed by the local algebra. For each integer n≥1, the authors introduce a new U(1)_X_n symmetry, X_n = n(B−L)+(1−n/3)Y~, and consider the symmetry-enriched Standard Model with 0-form and 1-form symmetries. They derive a fractional topological response σ_n(q,k) = q(1−n)gcd(2,n)/(2n) + kq/6 mod 1, where k∈Z_{6/q} is a symmetry fractionalization label. The first term comes from a gauge bundle constraint for the U(1)_Y~ and U(1)_X_n fields, while the second comes from canceling a mixed 1-form anomaly. The paper then proves, by explicit case checks and divisibility arguments, that for a fixed n, σ_n uniquely fixes (q,k) if and only if n≥7 and n∉{10,12,15,30}, and that certain pairs (n1,n2) also distinguish all SM_{(q,k)} variants. The result is presented as extending the n=3 case of Hsin–Gomis and as yielding 'measurable topological responses' that illuminate the global structure of the SM gauge group.

Significance. If the construction is accepted, the paper provides a clean and internally consistent derivation of a family of fractional SPT responses that depend nontrivially on the global gauge group parameter q and fractionalization class k. The derivation has no fitted continuous parameters: the first contribution follows from the gauge bundle constraint quoted in Eqs. (49)/(55), the second from the mixed anomaly and the fractionalization class, and the remaining number-theoretic uniqueness claim is supported by explicit manual checks in Sec. V and Table II. The introduction of the X_n family is a natural generalization of the B−L and Wilczek–Zee symmetries and the result, conditional on the symmetry assumption, is a falsifiable prediction. The main weakness is the physical status of U(1)_X_n as an exact global symmetry of the actual Standard Model, on which the central distinguishing claim hinges.

major comments (2)
  1. [Sec. II, Eq. (5)] The construction of U(1)_X_n as an exact global symmetry of the Standard Model is not fully justified. Since hypercharge Y~ is a local gauge symmetry, a spacetime-independent Y~ transformation is a gauge redundancy on gauge-invariant states; consequently, on gauge-invariant operators the X_n generator coincides with n times the B−L generator up to a field-dependent gauge transformation. The paper states in Sec. II that one can 'restrict the Y~ gauge transformation as a global symmetry transformation,' but this does not address whether the background field A_X_n is an independent probe or can be absorbed by a shift of the dynamical hypercharge field. If the reduction is complete, the n-dependence of σ_n may be only a normalization convention for the B−L current rather than new discriminating physical content. The authors should either provide a gauge-invariant definition of the A_X_n coupling that cannot be removed by a field redefinition, or explicitly restrict the central claim to the formal symmetry-enriched variants SM_{(q,k)} rather than to the actual Standard Model.
  2. [Sec. II and Sec. V] The exactness of X_n is not stable under standard beyond-the-SM operators. The Weinberg operator (l_L H)^2 carries X_n charge −2n, and a right-handed neutrino Majorana mass ν_R ν_R carries charge +2n, so any neutrino-mass-generating extension of the SM breaks X_n. Since Eq. (3) allows the presence of right-handed neutrinos, the statement that X_n is an exact global symmetry is conditional on excluding these operators. The paper should state this condition explicitly and discuss whether σ_n remains well-defined in the presence of small X_n-breaking perturbations. Moreover, Sec. V lists 'Experimental Measurement of Topological Responses' as an open future direction, so the abstract's phrase 'measurable topological responses' is stronger than what is established; a concrete measurement protocol, or an explicit weakening of the claim to a formal SPT classification, is needed.
minor comments (5)
  1. [Table I caption] The caption contains a typo: 'inteer series' should read 'integer series.'
  2. [Eq. (38) vs Eq. (46)] In Eq. (38), σ_n denotes only the first contribution q(1−n)gcd(2,n)/(2n), while Eq. (46) defines the full σ_n including the fractionalization term; the paper should use different symbols (e.g., σ_n^{(0)} and σ_n) or explicitly state the convention change.
  3. [Sec. V] The proof of the 'if and only if' condition for n≥7 is written out only for n up to 30; while the divisibility argument is standard, the paper would be clearer if it stated the general divisibility step explicitly (n | 3(q'−q) for odd n, and m | 3(q−q') for even n) before the case checks.
  4. [Sec. V, Table II] The claim that pairs such as (n1,n2)=(2,5),(3,4),(3,5),(4,5) 'all such pairs can discern SM_{(q,k)}' is asserted without a proof or a table of the combined distinguishability; a short argument or a supplementary table would make this claim verifiable.
  5. [Appendix A.2, A.6] There are typographical artifacts in the text: 'Eilenberg?MacLane' appears in Appendix A.2 and 'Here?s' in Appendix A.6; these should be corrected to 'Eilenberg–MacLane' and 'Here's'.

Circularity Check

0 steps flagged · score 1.0 of 10

No significant circularity: the central σ_n formula is derived from gauge-bundle constraints and anomaly inflow, with the fractionalization label k imported transparently from [24]; the uniqueness proof is self-contained arithmetic.

full rationale

The derivation chain is self-contained. The first contribution to σ_n is obtained by substituting the independently derived gauge bundle constraints (Eqs. (34), (47), (53)) into the magnetic 1-form current J_m^(2) = q ⋆ da_Ỹ/2π, giving σ_n = q(1−n)gcd(2,n)/(2n) mod 1 (Eqs. (38)-(39)); no parameter is fitted and no target response is assumed. The second contribution, kq/6, follows from the standard symmetry-fractionalization constraint f_k^* B̃_e = k/(6/q) dA_Xn and anomaly inflow from the mixed anomaly ∫ −1/(2π) B̃_e dB_m (Eqs. (14)-(16), (45)-(46)); k is an external label from Hsin-Gomis [arXiv:2411.18160], and the paper explicitly identifies this term as already obtained from that fractionalization class, so it is an input-consequence relation rather than a circular reduction. The paper also re-derives the mixed anomaly in Sec. IV.A and computes the relevant cohomology in App. A.4, so the self-citations to [18,19] are supporting rather than load-bearing. The 'uniquely determines q and k iff n≥7 and n≠10,12,15,30' claim is proven by explicit modular arithmetic in Sec. V, not imported from a self-cited theorem. The only weakness is physical rather than circular: Sec. V lists experimental measurement of σ_n as an open future direction, so the abstract's 'measurable topological responses' is stronger than what the paper demonstrates, but this does not amount to the derivation being equivalent to its inputs.

Assumptions & free parameters 0 free parameters · 5 assumptions · 1 invented entities

No continuous parameters are fitted. The main external inputs are the SM matter content, the 1-form symmetry data from prior self-cited papers, and the gauge bundle constraints from [37,38]. The only invented entity is the X_n symmetry family, which is a proposed probe rather than a new fundamental particle.

assumptions (5)
  • domain assumption SM fermion content: 3 families of 15 Weyl fermions with the charge assignments in Table I.
    The entire analysis relies on these charge representations, including the new X_n charges q_{X_n} = (2−n, −3, 1, −4+n, 6−n, n, 3−n).
  • domain assumption B−L is an exact global symmetry of the SM and can be combined with the gauged hypercharge U(1)_{Y~} restricted to global transformations to form the new U(1)_{X_n}.
    Sec. II justifies X_n by this combination; if B−L is broken by higher-energy physics, the probe is not a symmetry of the actual theory.
  • domain assumption The 1-form symmetries of the gauged SM are Z_{6/q} electric and U(1) magnetic with mixed anomaly −1/(2π) B̃_e dB_m.
    Taken from prior work [18,19,27-29]; central to the fractionalization second term in σ_n.
  • domain assumption The gauge bundle constraints (49) and (55) from Benini-Hsin-Seiberg and Cheng-Hsin-Jian [37,38] apply to the quotient groups considered here.
    Used to derive the first term q(1−n)gcd(2,n)/(2n) of σ_n.
  • domain assumption For odd n, G[0] = Spin^c = Spin ×_{Z2} U(1)_{X_n}/Z_n, imposing w2(TM) = dA_{X_n}/2π mod 2.
    This identification sets the spacetime-internal structure and enters the gauge bundle constraint (34).
invented entities (1)
  • U(1)_{X_n} global symmetry
    purpose: A family of baryon-minus-lepton-like probes with integer charges; the topological response σ_n is defined by coupling its background field A_{X_n}.
    New linear combination X_n = n(B−L) + (1 − n/3) Y~; its predicted response is internal to the paper, no external evidence yet.

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Pith. "Pith review of Topological Responses of the Standard Model Gauge Group." pith.science (2026). https://pith.science/paper/UVE7OGVX

@misc{pith2026241221196,
  author       = {Pith},
  title        = {Pith review of: Topological Responses of the Standard Model Gauge Group},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/UVE7OGVX}},
  note         = {Machine review of arXiv:2412.21196}
}
abstract

The local Lie algebra of the Standard Model (SM) is $su(3)\times su(2) \times u(1)$, yet its global gauge group, $G_{{\rm SM}_{\rm q}}=$SU(3)$\times$SU(2)$\times$U(1)/$\mathbb{Z}_{\rm q}$, q$=1,2,3,6$ remains undetermined. Building on previous work on 4d anomalies and 5d cobordism invariants, we classify lower-dimensional invertible field theories (iFTs) or symmetry-protected topological states (SPTs) in 4d, 3d, 2d, and 1d. While the integer SPTs are hard to detect, the fractional SPTs produce measurable topological responses. In particular, the symmetry fractionalization labeled by $k\in\mathbb{Z}_{6/{\rm q}}$ in [arXiv:2411.18160] introduces the symmetry-enriched SM variants, denoted as SM$_{({\rm q},k)}$. We further introduce a new integer $n$ series of baryon-minus-lepton $({\bf B}-{\bf L})$-like U(1) symmetries, $X_n \equiv n (\mathbf{B}-\mathbf{L}) + (1-\frac{n}{N_c})\tilde{Y}$ with electroweak hypercharge $\tilde{Y}$, $n\ge1$, $N_c=3$, where the charge $q_{X_n} = q_{\tilde{Y}} \mod n$. Analyzing the symmetry-enriched SM with 0-form and 1-form symmetries $(G_{[0]}, G_{[1]})$, symmetry-twist group homomorphism $\rho$, and symmetry frationalization obstruction $[\beta]$, their spacetime-internal gauge bundle constraints, and their mixed anomalies, we derive the fractional topological response $\sigma_n({\rm q},k)=\frac{{\rm q}(1-n)\gcd(2,n)}{2n}+\frac{k{\rm q}}{6}\mod1.$ Our $\sigma_n$ response requires more general Spin$^c$ manifolds for odd $n$ and Spin manifolds for even $n$. For a given $n$ (with $n\ge 7$ and $n\ne 10,12,15,30$), $\sigma_n$ uniquely fixes the gauge group parameter q and fractionalization label $k$. Moreover, using pairs such as $(n_1,n_2)=(2,3),(2,5),(3,4),(3,5),(4,5)$, etc. uniquely distinguishes SM$_{({\rm q},k)}$. Our results illuminate the global structure of the SM gauge group via measurable topological responses.

Figures

Figures reproduced from arXiv: 2412.21196 by the authors.

Figure 1
Figure 1. FIG. 1. The flow chart of the Standard Model (SM)’s symmetry fractionalization [PITH_FULL_IMAGE:figures/full_fig_p009_1.png] view at source ↗
Figure 2
Figure 2. FIG. 2. Serre spectral sequence ( [PITH_FULL_IMAGE:figures/full_fig_p019_2.png] view at source ↗

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Reference graph

Works this paper leans on

127 extracted references · 46 canonical work pages · cited by 3 Pith papers

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    = Ze 6/q,[1] × U(1)m [1]. The ρ : π1(BG[0]) → Aut(G[1]) is trivial, such that [ β] = ([ βe], [βm]) contains the trivial electric [ βe] ∈ [BG[0], B3Ge [1]]ρ = 0 twisted homotopy class (thus the zero H 3 ρ(BG[0], Ge [1]) obstruction class) and a nontrivial magnetic [ βm] ∈ [BG[0], B3Gm [1]]ρ = H 4 ρ(BG[0], Z) = Z2 obstruction class. Thus, the symmetry fract...

  2. [2]

    and Gm [1], the symmetry fractionalization ˜Be ∼ k 6/q dAXn demands a 4d fractional SPTs 1 2π k 6/q AXn dBm. In addition, the spacetime-internal structure imposes the gauge bundle constraint between U(1) ˜Y flux, U(1) Xn flux, and Stiefel- Whitney classes of spacetime and SM: da ˜Y 2π = 1 lcm(2,n) dAXn 2π − gcd(2,n) 2 w2(T M) + 1 q w(q) 2 mod 1. Without t...

  3. [3]

    Systematic Notations and Definitions 17

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    Classifying Space 17

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    Higher Classifying Space 18

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    Second and Third Integral Cohomology of BSpin c and B(Spin × U(1)) 18

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    ( G[0], G[1], ρ: π1(BG[0]) → Aut(G[1]), [β] ∈ [BG[0], B3G[1]]ρ) and k 19 b

    Symmetry-Enriched Standard Model SM (q,k): (G[0], G[1], ρ, [β]) and symmetry fractionalization k 19 a. ( G[0], G[1], ρ: π1(BG[0]) → Aut(G[1]), [β] ∈ [BG[0], B3G[1]]ρ) and k 19 b. Explain the Symmetry fractionalization k ∈ [BG[0], B2G[1]]ρ when [β] = 0 via a lifting diagram 21

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    Cobordism group classification 22 References 23 I. INTRODUCTION AND SUMMAR Y The Standard Model (SM) of particle physics [1–4], while remarkably successful for describing various subatomic phenomena, harbors a nuanced structure when scrutinized through the lens of global topology [5], generalized sym- metry [6], and nonperturbative quantum regularization ...

Show all 127 references
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    Moreover the 0d SM gauge charge fermionic particle (3) explicitly breaks the Ge

    and Gm [1], which are determined from the center of gauge group Z(GSMq ), and the Pontryagin dual of the homotopy group π1(GSMq )∨ = Hom(π1(GSMq ), U(1)) = Hom( Z, U(1)) = U(1). Moreover the 0d SM gauge charge fermionic particle (3) explicitly breaks the Ge

  2. [10]

    from Z(GSMq ) to a subgroup [27–29] that acts trivially on all particle representations: Z(GSMq ) π1(GSMq )∨ 1-form e sym Ge

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    (8) Thus, starting from the ungauged spacetime-internal symmetry eq

    1-form m sym Gm [1] GSMq ≡ SU(3)×SU(2)×U(1) ˜Y Zq Z6/q × U(1) U(1) Ze 6/q,[1] U(1)m [1] , q = 1, 2, 3, 6. (8) Thus, starting from the ungauged spacetime-internal symmetry eq. (6), by dynamically gauging GSMq , we have to remove GSMq and include Ge

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    (6), so the gauged SM has the following spacetime-internal symmetry: G′U(1)Xn SMq ≡ Spin ×ZF 2 U(1)Xn Zn × Ze 6/q,[1] × U(1)m

    into eq. (6), so the gauged SM has the following spacetime-internal symmetry: G′U(1)Xn SMq ≡ Spin ×ZF 2 U(1)Xn Zn × Ze 6/q,[1] × U(1)m

  5. [13]

    for odd n, G′U(1)Xn SMq ≡ Spin × U(1)Xn Zn × Ze 6/q,[1] × U(1)m

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    (9) N¨ aively, we obtainG[1] = Ge

    for even n. (9) N¨ aively, we obtainG[1] = Ge

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    However, there is a possibility of symmetry- fractionalization between 0-symmetry and 1-symmetry in the SM recently pointed out by Ref

    [18, 19, 27–29]. However, there is a possibility of symmetry- fractionalization between 0-symmetry and 1-symmetry in the SM recently pointed out by Ref. [24]. Follow the general procedure of the symmetry fractionalization (e.g. [30–32]), we will compute ( G[0], G[1], ρ,[β]) ex...

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    The obstruction of the symmetry-fractionalization is determined by the homotopy class [β] = ([βe], [βm]) ∈ [BG[0], B3G[1]]ρ = ([BG[0], B3Ge [1]]ρ, [BG[0], B3Gm [1]]ρ) = (0, Z2)

    × Gm [1]) = ( ( Spin ×ZF 2 U(1)Xn Zn = Spinc, n ∈ Zodd Spin × U(1)Xn Zn = Spin × U(1), n ∈ Zeven , Ze 6/q,[1] × U(1)m [1]) (10) The symmetry twist is a group homomorphism ρ (from the first homotopy group of classifying space B G[0] to the automorphism group of G[1]) which is t...

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    = Ze 6/q,[1] and the 0-form symmetry G[0]: ˜Be ∼ k 6/q dAXn . (14) Here ˜Be = 2π 6/q Be, with Be ∈ H2(B2Z 6 q , Z 6 q ) = Z 6 q is the background gauge field of the electric 1-form Ze 6/q,[1] symmetry of the SM, while AXn ∈ Ω1(B( U(1)Xn Zlcm(2,n) ), R) is the background gauge ...

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    = Ze 6/q,[1] × U(1)m

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    of the Standard Model, Z M 5 − 1 2π ˜BedBm = Z M 5 − 1 6/q BedBm, (15) for q = 1 , 2, 3, 6, where Bm ∈ Ω2(B2U(1), R) is the background gauge field of the magnetic 1-form U(1) m

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    symmetry of the SM, and M 5 is a 5-manifold with the boundary ∂M 5 = M 4, see Sec. IV A. Due to the mixed anomaly − 1 2π ˜BedBm, the symmetry fractionalization ˜Be ∼ k 6/q dAXn demands a 4d fractional SPTs Z M 4 1 2π k 6/q AXn dBm, (16) thanks to the anomaly inflow from the 5d...

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    Line Operators in the Standard Model,

    David Tong, “Line Operators in the Standard Model,” JHEP 07, 104 (2017), arXiv:1705.01853 [hep-th]

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    IV B and Sec

    One contribution q(1−n)gcd(2,n) 2n of the SM’s σn response is obtained from the fractional part of ⋆J (2) m due to the coupling Bm ⋆ J(2) m (see footnote 4) where J (2) m is the 2-form current of the magnetic 1-form symmetry U(1) m [1] without turning on the background gauge f...

  15. [24]

    We derive the fractional part of σn, which consists of two parts:

    restricted to w2(T M) = 0). We derive the fractional part of σn, which consists of two parts:

  16. [25]

    Another contribution kq 6 of the SM’s σn response is already obtained earlier from the symmetry fractionalization of the 0-form symmetry G[0] and electric 1-form symmetry Ge

  17. [26]

    (15) (see Sec

    = Ze 6/q,[1] of the gauged SM with the fraction- alization class k ∈ Z6/q, obtained by canceling the mixed anomaly of eq. (15) (see Sec. IV D and Sec. IV E). Our topological response result of the SM is the combination of the two contributions σn = σn(q, k) = q(1 − n)gcd(2, n)...

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    The Xn charges of the quarks and leptons fermions, and Higgs boson, ( ¯dR, lL, qL, ¯uR, ¯eR, ¯νj,R, ϕH ), are qXn = (−n + 2p, −3p, p, n− 4p, −n + 6p, n, −n + 3p)

    Since the Xn charge of qL is n/3 + λ ∈ Z, let n/3 + λ ≡ p ∈ Z. The Xn charges of the quarks and leptons fermions, and Higgs boson, ( ¯dR, lL, qL, ¯uR, ¯eR, ¯νj,R, ϕH ), are qXn = (−n + 2p, −3p, p, n− 4p, −n + 6p, n, −n + 3p)

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    Since at least one of the Xn charges of the quarks and leptons fermions, and Higgs boson, ( ¯dR, lL, qL, ¯uR, ¯eR, ¯νj,R, ϕH ), is ±1, then p = ±1 or n = ±1 or n − 2p = ±1 or n − 3p = ±1 or n − 4p = ±1 or n − 6p = ±1. We find that the special solution p = 1 has the advantage t...

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    (27) known as a symmetry twist such that the 0-form symmetry’s codimension-1 topological defect acts on the 1-form symmetry’s charged line as ρg[0] (g[1]) as in [30]

    The homomorphism ρ : π1(BG[0]) → Aut(G[1]). (27) known as a symmetry twist such that the 0-form symmetry’s codimension-1 topological defect acts on the 1-form symmetry’s charged line as ρg[0] (g[1]) as in [30]. It turns out that our SM (9) here has a trivial ρ = 0

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    Obstruction to symmetry fractionalization is given by [ β], the twisted homotopy class [β] ∈ [BG[0], B3G[1]]ρ (28) in the SM such that the electric’s obstruction [βe] ∈ [BG[0], B3Ge [1]]ρ = 0 (29) is always trivial, while magnetic obstruction [βm] ∈ [BG[0], B3Gm [1]]ρ = Z2 (30...

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    Either [ βm] ̸= 0, or [ βm] = 0 but still km ∈ [BG[0], B2Gm [1]]ρ = 0; (32) for both scenarios, there is no magnetic symmetry fractionalization km = 0

    Because [ βe] = 0, then the electric symmetry fractionalization ke ∈ [BG[0], B2Ge [1]]ρ = Z6/q (31) can be defined. Either [ βm] ̸= 0, or [ βm] = 0 but still km ∈ [BG[0], B2Gm [1]]ρ = 0; (32) for both scenarios, there is no magnetic symmetry fractionalization km = 0. Since we ...

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    (36) Let Bm be the background gauge field for the magnetic 1-form symmetry U(1) m [1]

    is generated by a 2-form current J (2) m = q ⋆ da ˜Y 2π . (36) Let Bm be the background gauge field for the magnetic 1-form symmetry U(1) m [1]. We consider the topological response3 Z M 4 σn 2π BmdAXn (37) of the Standard Model, this term comes from the Standard Model Lagrang...

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    We first consider the first way, then J (2) m = σn ⋆ dAXn 2π

    of the Standard Model. We first consider the first way, then J (2) m = σn ⋆ dAXn 2π . (38) By (34), (35), (36), and (38), we have σn = q(1 − n)gcd(2, n) 2n mod 1. (39) Now we consider the second way. We consider the Standard Model Lagrangian. The Standard Model gauge group is ...

  25. [34]

    magnetic 1-form symmetry. However, this term is not gauge invariant under the gauge transformations qa ˜Y → qa ˜Y + λ, ˜Be → ˜Be + dλ, Bm → Bm + dµ (41) where λ = 2π 6/q C and C is a Z6/q-valued 1-cochain with the normalization H dλ ∈ 2π 6/q Z, µ is a R-valued 1-form with the ...

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    This anomaly appears as an SPTs or cobordism invariant in Table IV and Table VI

    1-form symmetry of the Standard Model is R M 5 − 1 2π ˜BedBm for q = 1 , 2, 3, 6 where ˜Be = 2π 6/q Be. This anomaly appears as an SPTs or cobordism invariant in Table IV and Table VI. In Sec. IV D and Sec. IV E, we derive the constraint f ∗ k ˜Be = k 6/q dAXn (45) for the sym...

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    Here, k ∈ Z6/q = H2(BG[0], Ge [1]) is represented by the map fk : BG[0] → B2Ge [1]

    = Ze 6/q,[1] symmetry of the gauged Standard Model. Here, k ∈ Z6/q = H2(BG[0], Ge [1]) is represented by the map fk : BG[0] → B2Ge [1]. To cancel the anomaly R M 5 − 1 2π ˜BedBm, we need to add R M 4 1 2π k 6/q BmdAXn in the Standard Model Lagrangian. The total topological res...

  28. [37]

    Given an n in the row, we consider q = 1 , 2, 3, 6 in the column, and the σn(q, k∈ Z6/q) values in the {..} running from {k = 0,

    magnetic 1-symmetry. Given an n in the row, we consider q = 1 , 2, 3, 6 in the column, and the σn(q, k∈ Z6/q) values in the {..} running from {k = 0, . . . ,6/q − 1} so k ∈ Z6/q. Here n = 3 does not discern all the four q cases [24]. For a fixed n specified by Xn, the σn can u...

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    (65) We discuss for which values of n, σn can distinguish q = 1 , 2, 3, 6

    magnetic 1-form symmetry, and σn = q(1 − n)gcd(2, n) 2n + kq 6 mod 1. (65) We discuss for which values of n, σn can distinguish q = 1 , 2, 3, 6. For odd n, if σn(q, k) = σn(q′, k′), then (q′ − q)(n − 1) 2n = k′q′ − kq 6 mod 1 (66) or equivalently (q′ − q)3(n − 1) n = k′q′ − kq...

  30. [39]

    σ3(1, 2) = σ3(2, 2) = σ3(3, 0) = σ3(6, 0) = 0, so the topological response cannot distinguish SM (1,2), SM (2,2), SM(3,0), and SM (6,0)

  31. [40]

    σ3(1, 0) = σ3(2, 1) = 2 3 , so the topological response cannot distinguish SM (1,0) and SM(2,1)

  32. [41]

    σ3(1, 4) = σ3(2, 0) = 1 3 , so the topological response cannot distinguish SM (1,4) and SM(2,0)

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    In summary, our introduction of U(1) Xn symmetry in eq

    σ3(1, 5) = σ3(3, 1) = 1 2 , so the topological response cannot distinguish SM (1,5) and SM(3,1). In summary, our introduction of U(1) Xn symmetry in eq. (5) for generic n in Table II extends the results of [24], enabling a complete distinction of the SM variants SM (q,k) compl...

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    Investigating Potential Symmetry Extensions: There is a possible symmetry extension between the 1- form electric Ge

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    = Ze 6 q ,[1] symmetry and 1-form magnetic Gm

  36. [45]

    However, in this article, we write the 1-form symmetry of the SM as the direct product G[1] = Ge

    symmetry. However, in this article, we write the 1-form symmetry of the SM as the direct product G[1] = Ge

  37. [46]

    We need to explore if this symmetry extension affects our results in this article

    × Gm [1]. We need to explore if this symmetry extension affects our results in this article

  38. [47]

    It will be beneficial to explore the topological response associated with those symmetries

    Baryon plus lepton and other discrete symmetries: There are discrete ( B + L) symmetries [19, 39, 40] and possibly other symmetries survived in the SM. It will be beneficial to explore the topological response associated with those symmetries

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    Such empirical investigations will validate the theoretical predictions presented in this work and enhance our understanding of the SM’s topological characteristics

    Experimental Measurement of T opological Responses: It is crucial to design and conduct experiments aimed at measuring the topological response σn, analogous to the quantum Hall conductance. Such empirical investigations will validate the theoretical predictions presented in t...

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    Throughout this article, all fields of a p-form symmetry G (defined as a differential form or a cohomology class) are pulled back to M via the classifying map M → Bp+1G

    Systematic Notations and Definitions A p-form symmetry G on a manifold M is characterized by a map M → Bp+1G [6]. Throughout this article, all fields of a p-form symmetry G (defined as a differential form or a cohomology class) are pulled back to M via the classifying map M → ...

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    AXn ∈ Ω1(B( U(1)Xn Zlcm(2,n) ), R) is the real-valued connection 1-form field for the U(1)Xn Zlcm(2,n) symmetry, A = AXn for odd n,

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    c1 = dA 2π ∈ H2(B( U(1)Xn Zlcm(2,n) ), Z) = Z is the first Chern class of the U(1)Xn Zlcm(2,n) bundle where d is the differential operator,

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    Be ∈ H2(B2Z 6 q , Z 6 q ) = Z 6 q is the 2-form field for the Ze 6/q,[1] electric 1-form symmetry, while ˜Be = 2π 6/q Be

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    P (Be,q=3) is the Pontryagin square of Be,q=3 with P (Be,q=3) = B2 e,q=3 mod 2,

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    Bm ∈ Ω2(B2U(1), R) is the real-valued connection 2-form field for the U(1) m

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    magnetic 1-form symmetry, and dBm 2π ∈ H3(B2U(1), Z) = Z is the generator of the cohomology group where d is the differential operator,

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    wi = wi(T M) is the i-th Stiefel-Whitney class of the tangent bundle,

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    p1 is the first Pontryagin class of the tangent bundle,

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    CS T M 3 is the gravitational Chern-Simons 3-form for the first Pontryagin class p1 of the tangent bundle,

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    w(q) 2 is the obstruction class to lifting a SU(3)×SU(2) Zq bundle to a SU(3) × SU(2) bundle,

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    ˜w(n) 2 is the obstruction class to lifting a SU(3)×SU(2)×U(1) ˜Y Zq Zn bundle to a SU(3)×SU(2)×U(1) ˜Y Zq bundle,

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    c1,Xn is the first Chern class of the U(1) Xn bundle,

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    a ˜Y ∈ Ω1(BU(1) ˜Y , R) is the real-valued connection 1-form field for the U(1) ˜Y symmetry,

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    We denote G1 × G2 Nshared ≡ G1 ×Nshared G2

    c1, ˜Y is the first Chern class of the U(1) ˜Y bundle. We denote G1 × G2 Nshared ≡ G1 ×Nshared G2. The Nshared is the shared common normal subgroup symmetry between G1 and G2

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    Classifying Space The classifying space B G of a topological group G is a space such that:

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    {G-bundles over X} ∼= [X, BG]

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    The classifying space B G is constructed as follows:

    The space B G has a principal G-bundle E G → BG, called the universal G-bundle, such that every G-bundle over any X is a pullback of E G → BG. The classifying space B G is constructed as follows:

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    E G: A contractible space on which G acts freely (e.g., G acting on itself by left translation)

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    Here are some examples of the classifying space:

    B G = EG/G: The quotient of E G by this free G-action. Here are some examples of the classifying space:

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    B G = CP∞, the infinite complex projective space

    For G = U(1) (the circle group): E G is the infinite-dimensional sphere S∞ with U(1)-action. B G = CP∞, the infinite complex projective space

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    For G = Zn (finite cyclic group): B G is the Eilenberg?MacLane space K(Zn, 1). 18

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    It is used to classify n-form symmetries or other higher categorical structures: B nG is the classifying space for ( n − 1)-form G-gauge fields or G-symmetries

    Higher Classifying Space A higher classifying space BnG generalizes BG to higher homotopy contexts. It is used to classify n-form symmetries or other higher categorical structures: B nG is the classifying space for ( n − 1)-form G-gauge fields or G-symmetries. The higher class...

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    B 2U(1): The classifying space for 1-form U(1) gauge fields

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    B nZ: The Eilenberg?MacLane space K(Z, n), classifying n-dimensional cohomology with Z-coefficients. The Eilenberg?MacLane space is related to cohomology as follows: Maps from a space X to K(G, n) correspond to elements of the cohomology group H n(X, G): [X, K(G, n)] ∼= Hn(X, ...

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    The short exact sequence of groups 1 → U(1) → Spinc → SO → 1 (A1) induces a fibration BU(1) → BSpinc → BSO

    Second and Third Integral Cohomology of BSpinc and B(Spin × U(1)) The group Spin c is defined as Spin×U(1) Z2 . The short exact sequence of groups 1 → U(1) → Spinc → SO → 1 (A1) induces a fibration BU(1) → BSpinc → BSO. (A2) We have the following Serre spectral sequence Hp(BSO...

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    (G[0], G[1], ρ: π1(BG[0]) → Aut(G[1]), [β] ∈ [BG[0], B3G[1]]ρ) and k We study the Symmetry-Enriched SM labeled by the following data (G[0], G[1], ρ,[β])

    Symmetry-Enriched Standard Model SM (q,k): (G[0], G[1], ρ, [β]) and symmetry fractionalization k a. (G[0], G[1], ρ: π1(BG[0]) → Aut(G[1]), [β] ∈ [BG[0], B3G[1]]ρ) and k We study the Symmetry-Enriched SM labeled by the following data (G[0], G[1], ρ,[β]). (A6) In [32] notation, ...

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    = Ze 6/q,[1] × U(1)m [1]. (A8)

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    However, in the definition of twisted cohomology H ∗ ρ(X, A), ρ is a homomorphism π1(X) → Aut(A); so in our case X = BG[0], ρ is a homomorphism ρ : π1(BG[0]) → Aut(G[1])

    Homomorphism ρ : π1(BG[0]) → Aut(G[1]) as a twist— In [32], their ρ is a homomorphism G → Aut(A). However, in the definition of twisted cohomology H ∗ ρ(X, A), ρ is a homomorphism π1(X) → Aut(A); so in our case X = BG[0], ρ is a homomorphism ρ : π1(BG[0]) → Aut(G[1]). (A9) Onl...

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    Regarding ρ: (a) When ρ is nontrivial, the Postnikov class [ β] is a twisted homotopy class or cohomology class

    Postnikov class[β] as a twisted homotopy class [β] ∈ [BG[0], B3G[1]]ρ, (A10) 20 which is equivalent to ( if and only ifG[1] is discrete) a degree-3 twisted cohomology class [30, 32], H3 ρ(BG[0], G[1]). Regarding ρ: (a) When ρ is nontrivial, the Postnikov class [ β] is a twiste...

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    By the results in App

    = B3Z6/q = K(Z6/q, 3) as the Eilenberg-MacLane space, the Postnikov class [βe] is a degree-3 twisted cohomology class H3 ρ(BG[0], Z6/q), because the degree- l cohomology class with coefficients in discrete A is a homotopy class of maps to the Eilenberg-MacLane space, K(A, l), ...

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    The magnetic Postnikov class [βm] ∈ [BG[0], B3Gm [1]]ρ = H4 ρ(BG[0], Z) = Z2 (A12) may be a nontrivial obstruction class

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    (A13) Only when the earlier Postnikov class [β] is trivial in eq

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Reviewed August 10, 2026 · model on record in the stance chip above.