No beta-invertible S_3 symmetry when 3 divides N

Abstract

When 3 divides N, no copy of the symmetric group S_3 inside the anyon permutation symmetries of the Z_N x Z_N SymTFT has all of its non-identity elements beta-invertible, as conjectured by D.-C. Lu, Z. Sun and Z. Zhang (arXiv:2406.12151, JHEP 11 (2025) 081). The paper infers from this, through the Etingof-Nikshych-Ostrik classification of G-extensions, that there are no S_3-ality defects for such N; that inference is not formalized here.

Definition 1.1 (The quadratic form of the SymTFT).

$$\forall v\in \mathrm{Fin}\left(2\right)+\mathrm{Fin}\left(2\right)\to \mathrm{ZMod}\left(N\right),Q\left(v\right)=v\left(\mathrm{inl}\left(0\right)\right)\cdot v\left(\mathrm{inr}\left(0\right)\right)+v\left(\mathrm{inl}\left(1\right)\right)\cdot v\left(\mathrm{inr}\left(1\right)\right)$$

Formalization. D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.Q (✓ std3).

Citation. Da-Chuan Lu; Zhengdi Sun; Zipei Zhang (2024). Exploring G-ality defects in 2-dim QFTs. DOI: 10.1007/JHEP11(2025)081. URL: https://arxiv.org/abs/2406.12151v3.

Commentary.

On the anyons (a, a’) of the Z_N x Z_N SymTFT, with a in A = (Z/N)^2 the left summand and a’ in the dual group identified with (Z/N)^2, Q is the dot product a . a’ modulo N; the self-statistics of the anyon is exp(2 pi i Q / N).

Definition 1.2 (The conjecture for N divisible by 3).

$$claim\iff (\forall N\in \mathbb{N},3\mid N\Rightarrow (\neg (\exists rho\in \mathrm{MonoidHom}\left(\mathrm{Perm}\left(\mathrm{Fin}\left(3\right)\right),\mathrm{GL}\left(\mathrm{Fin}\left(2\right)+\mathrm{Fin}\left(2\right),\mathrm{ZMod}\left(N\right)\right)\right),(\forall g\in \mathrm{Perm}\left(\mathrm{Fin}\left(3\right)\right),\forall v\in \mathrm{Fin}\left(2\right)+\mathrm{Fin}\left(2\right)\to \mathrm{ZMod}\left(N\right),Q\left(rho\left(g\right)v\right)=Q\left(v\right))\wedge (\forall g\in \mathrm{Perm}\left(\mathrm{Fin}\left(3\right)\right),g\ne 1\Rightarrow (\mathrm{IsUnit}\left(\det \left(\left({\mathrm{toBlocks}}_{12}\right)\left(rho\left(g\right)\right)\right)\right))))))$$

Formalization. D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.claim (✓ std3).

Citation. Da-Chuan Lu; Zhengdi Sun; Zipei Zhang (2024). Exploring G-ality defects in 2-dim QFTs. DOI: 10.1007/JHEP11(2025)081. URL: https://arxiv.org/abs/2406.12151v3.

Commentary.

For every N divisible by 3 there is no group homomorphism rho from S_3, the permutations of three letters, to the invertible 4 x 4 matrices over Z/N such that every rho(g) preserves Q and, for every g other than the identity, the upper-right 2 x 2 block beta of rho(g), the component from the dual group to A, has a unit determinant. A subgroup isomorphic to S_3 whose non-identity elements are all beta-invertible is exactly such a homomorphism, which is then injective.

Theorem 1.3 (Proof of the conjecture).

$$claim$$

Proof. Machine-checked in Lean as D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.result (✓ std3). ∎

Resolves. Problems/lu-sun-zhang-2024-beta-invertible-s3-obstruction (proved) by D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.result.

Source. Repository-derived.

Acknowledgement. Da-Chuan Lu; Zhengdi Sun; Zipei Zhang (2024). Exploring G-ality defects in 2-dim QFTs. DOI: 10.1007/JHEP11(2025)081. URL: https://arxiv.org/abs/2406.12151v3.

Commentary.

Reduce rho modulo 3. The reduced matrices still preserve Q, since every vector over Z/3 lifts to Z/N, and still have invertible beta blocks at g other than the identity, since the determinant reduces to a unit. Over Z/3 let S_g = delta_g beta_g^{-1}, where delta_g is the lower-right block. The matrix rho(g) sends (0, x) to (beta_g x, delta_g x), and Q vanishes at (0, x), so y . S_g y = 0 for every y; hence S_g has zero diagonal and opposite off-diagonal entries and is determined by its entry S_g(0, 1). If g and h are different non-identity permutations with S_g = S_h, then rho(h) sends (0, z) to rho(g)(0, x) for z = beta_h^{-1} beta_g x, so rho(h^{-1} g) sends (0, x) to (0, z) for every x and its beta block vanishes, although h^{-1} g is not the identity. So g -> S_g(0, 1) maps the five non-identity permutations injectively into Z/3, which has three elements, a contradiction. The same argument, which is not formalized here, bounds the order of every beta-invertible group by p + 1 for any prime p dividing N.

References

• Truth anchor: D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.Q
• Truth anchor: D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.claim
• Truth anchor: D5/S3/Quantum/Algebra/BetaInvertibleSThreeObstruction.result