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Finite 2-group background gauge consistency

Statement

Let $K$ be a finite ordered simplicial complex; let $G$ be a finite group, $A$ a finite abelian group written additively, $ρ : G \to Aut (A)$ a left action, and $β : G^{3} \to A$ a normalized group $3$ -cocycle with the convention that its associator is directed from $(g h) k$ to $g (h k)$ . Let $a_{i j} \in G$ be flat on every ordered triangle, let $B_{i j k} \in A$ be an $A$ -valued $2$ -cochain at the first vertex, and assume on every ordered tetrahedron that

ρ_{a_{i j}} B_{j k l} - B_{i k l} + B_{i j l} - B_{i j k} = β (a_{i j}, a_{j k}, a_{k l}) .

For arbitrary vertex labels $g_{i} \in G$ and edge labels $λ_{i j} \in A$ , put $a_{i j}^{'} = g_{i} a_{i j} g_{j}^{- 1}$ and

ζ_{i j k} = - β (g_{i}, a_{i j}, a_{j k}) + β (a_{i j}^{'}, g_{j}, a_{j k}) - β (a_{i j}^{'}, a_{j k}^{'}, g_{k}) .

B_{i j k}^{'} = ρ_{g_{i}} B_{i j k} + ζ_{i j k} + ρ_{a_{i j}^{'}} λ_{j k} - λ_{i k} + λ_{i j} .

Then $a^{'}$ is flat and $(a^{'}, B^{'})$ satisfies the same twisted tetrahedron equation. If $(g, λ)$ is followed by $(h, μ)$ , with $a_{i j}^{''} = h_{i} a_{i j}^{'} h_{j}^{- 1}$ , their composite is $(h g, ν)$ where

ν_{i j} = β (h_{i}, g_{i}, a_{i j}) + ρ_{h_{i}} λ_{i j} - β (h_{i}, a_{i j}^{'}, g_{j}) + μ_{i j} + β (a_{i j}^{''}, h_{j}, g_{j}) .

Each transformation is invertible: for a fixed source background, put $h_{i} = g_{i}^{- 1}$ and

μ_{i j} = - β (h_{i}, g_{i}, a_{i j}) - ρ_{h_{i}} λ_{i j} + β (h_{i}, a_{i j}^{'}, g_{j}) - β (a_{i j}, h_{j}, g_{j});

then its composite has $ν_{i j} = 0$ . Treating a gauge arrow as its induced map on background cochains, sequential maps compose associatively, and the displayed formula gives a representative for each composite. This is a mathematical consistency result and does not assert a physical anomaly theorem.

Statement references

Benini, Córdova, and Hsin, §2.3, PDF pp. 18–19

Baez and Lauda, §8.3, PDF pp. 54–59

Facts & Assumptions

Given: finite data $G, A, ρ, β, K$ , a flat edge cochain $a$ , a $2$ -cochain $B$ satisfying the displayed tetrahedron equation, and vertex/edge gauge parameters $g_{i}, λ_{i j}$ as in the Statement.

[F1]

Finite 2-group background cochains: a flat edge cochain satisfies $a_{i j} a_{j k} = a_{i k}$ , local coefficient transport from $j$ to $i$ is $ρ (a_{i j})$ , and the twisted differential on a $2$ -cochain is $ρ (a_{i j}) B_{j k l} - B_{i k l} + B_{i j l} - B_{i j k}$ .

[F2]

Finite 2-group background cochains: the skeletal $2$ -group tensor product is $(g, u) \otimes (h, v) = (g h, u + ρ_{g} v)$ , its associator has label $β (g, h, k)$ , and the normalized $3$ -cocycle identity is the associator pentagon equation.

Proof

1.1givenF1algebra

Flatness of $a$ gives $a_{i j}^{'} a_{j k}^{'} = g_{i} a_{i j} g_{j}^{- 1} g_{j} a_{j k} g_{k}^{- 1} = g_{i} a_{i k} g_{k}^{- 1} = a_{i k}^{'}$ , so $a^{'}$ is flat and $ρ (a_{i j}^{'})$ defines its local coefficient transport.

1.2F2givenconstructalgebra

For consecutive transformations $(g, λ)$ and $(h, μ)$ , start with the edge product $(h_{i} g_{i}) a_{i j}$ . The successive associator/edge-morphism labels along $(h_{i} g_{i}) a_{i j} \to h_{i} (g_{i} a_{i j}) \to h_{i} (a_{i j}^{'} g_{j}) \to (h_{i} a_{i j}^{'}) g_{j} \to (a_{i j}^{''} h_{j}) g_{j} \to a_{i j}^{''} (h_{j} g_{j})$ are $β (h_{i}, g_{i}, a_{i j})$ , $ρ_{h_{i}} λ_{i j}$ , $- β (h_{i}, a_{i j}^{'}, g_{j})$ , $μ_{i j}$ , and $β (a_{i j}^{''}, h_{j}, g_{j})$ , respectively; adding them gives exactly $ν_{i j}$ in the Statement.

2.1F1step 1.1algebra

For an edge cochain $λ$ , expanding $δ_{a^{'}}^{2} λ$ on $[i j k l]$ gives terms $ρ (a_{i j}^{'}) ρ (a_{j k}^{'}) λ_{k l} - ρ (a_{i j}^{'}) λ_{j l} + ρ (a_{i j}^{'}) λ_{j k} - ρ (a_{i k}^{'}) λ_{k l} + λ_{i l} - λ_{i k} + ρ (a_{i j}^{'}) λ_{j l} - λ_{i l} + λ_{i j} - ρ (a_{i j}^{'}) λ_{j k} + λ_{i k} - λ_{i j}$ ; the first and fourth cancel because $a^{'}$ is flat and $ρ$ is an action, and every other term cancels with its displayed opposite. Hence the $1$ -form shift contributes zero to the transformed tetrahedron equation.

2.2F2step 1.1algebra

Write $u = a_{01}^{'}$ , $v = a_{12}^{'}$ , $w = a_{23}^{'}$ , $x = a_{01}$ , $y = a_{12}$ , and $z = a_{23}$ . Expanding $δ_{a^{'}} ζ$ using the three displayed terms in each $ζ_{i j k}$ and applying the five-term cocycle identity to the four ordered quadruples $(g_{0}, x, y, z)$ , $(u, g_{1}, y, z)$ , $(u, v, g_{2}, z)$ , and $(u, v, w, g_{3})$ yields $(δ_{a^{'}} ζ)_{0123} + ρ_{g_{0}} β (x, y, z) - β (u, v, w) = D (g_{0}, x, y, z) - D (u, g_{1}, y, z) + D (u, v, g_{2}, z) - D (u, v, w, g_{3}) = 0$ , where $D (s, t, r, q) : = ρ_{s} β (t, r, q) - β (s t, r, q) + β (s, t r, q) - β (s, t, r q) + β (s, t, r)$ . The cancellation uses flatness and $a_{i j}^{'} g_{j} = g_{i} a_{i j}$ to identify the adjacent products. Hence $δ_{a^{'}} ζ = β (a_{01}^{'}, a_{12}^{'}, a_{23}^{'}) - ρ_{g_{0}} β (a_{01}, a_{12}, a_{23})$ on each tetrahedron.

2.3F2step 1.2algebra

Let $τ_{i j} = β (h_{i}, g_{i}, a_{i j}) - β (h_{i}, a_{i j}^{'}, g_{j}) + β (a_{i j}^{''}, h_{j}, g_{j})$ be the $β$ part of $ν$ . The sequential and direct face formulas agree exactly when $ρ_{h_{i}} ζ_{i j k}^{g} + ζ_{i j k}^{h} (a^{'}) - ζ_{i j k}^{h g} (a) = (δ_{a^{''}} τ)_{i j k}$ , where $(δ_{a^{''}} τ)_{i j k} = ρ_{a_{i j}^{''}} τ_{j k} - τ_{i k} + τ_{i j}$ . For a fixed ordered triangle $i < j < k$ , write $x = a_{i j}$ , $y = a_{j k}$ , $b_{r s} = a_{r s}^{'}$ , and $c_{r s} = a_{r s}^{''}$ . For the cocycle expression $D$ , direct expansion gives the exact formal identity $ρ_{h_{i}} ζ_{i j k}^{g} + ζ_{i j k}^{h} (a^{'}) - ζ_{i j k}^{h g} (a) - (δ_{c} τ)_{i j k} = - D (h_{i}, g_{i}, x, y) + D (h_{i}, b_{i j}, g_{j}, y) - D (h_{i}, b_{i j}, b_{j k}, g_{k}) - D (c_{i j}, h_{j}, g_{j}, y) + D (c_{i j}, h_{j}, b_{j k}, g_{k}) - D (c_{i j}, c_{j k}, h_{k}, g_{k})$ . To verify it, substitute $D (s, t, r, q) = ρ_{s} β (t, r, q) - β (s t, r, q) + β (s, t r, q) - β (s, t, r q) + β (s, t, r)$ in these six terms, use $b_{i j} g_{j} = g_{i} a_{i j}$ and $c_{i j} h_{j} = h_{i} b_{i j}$ to identify mixed products, and use $a_{i k} = a_{i j} a_{j k}$ , $b_{i k} = b_{i j} b_{j k}$ , $c_{i k} = c_{i j} c_{j k}$ to identify long edges. The remaining terms are precisely the three $ζ$ expressions and the three terms of $- δ_{c} τ$ on the left; every other beta term cancels with its opposite. All six $D$ terms vanish by [F2]. In full, $B_{s e q} = ρ_{h g} B + ρ_{h} ζ^{g} + ζ^{h} + δ_{a^{''}} (ρ_{h} λ + μ)$ and $B_{d i r e c t} = ρ_{h g} B + ζ^{h g} + δ_{a^{''}} (ρ_{h} λ + μ + τ)$ , so the displayed identity proves equality on every triangle.

2.4F1F2step 1.2algebra

For a fixed source background, set $h_{i} = g_{i}^{- 1}$ and choose $μ_{i j}$ by the displayed reverse formula. Substitution into the composite law cancels its five terms and gives $h_{i} g_{i} = e$ , $ν_{i j} = 0$ , and $a^{''} = a$ . The identity transformation fixes $B$ because $β$ is normalized and $ρ_{e} = {id}_{A}$ .

3.1givenF1F2step 2.1step 2.2algebra

The coefficient transport identity $ρ (a_{i j}^{'}) ρ (g_{j}) = ρ (g_{i}) ρ (a_{i j})$ gives $(δ_{a^{'}} (ρ_{g} B))_{i j k l} = ρ_{g_{i}} (δ_{a} B)_{i j k l}$ . Combining this with $B^{'} = ρ_{g} B + ζ + δ_{a^{'}} λ$ , steps 2.1–2.2, and $δ_{a} B = a^{*} β$ gives $δ_{a^{'}} B^{'} = ρ_{g_{i}} β (a_{i j}, a_{j k}, a_{k l}) + β (a_{i j}^{'}, a_{j k}^{'}, a_{k l}^{'}) - ρ_{g_{i}} β (a_{i j}, a_{j k}, a_{k l}) = a^{' *} β$ .

4.1F1F2step 1.2step 2.4algebra∎

To check the opposite composite on the target $(b, C) = (a^{'}, B^{'})$ , first apply $(h, μ)$ and then $(g, λ)$ . Its $0$ -form labels are $g_{i} h_{i} = e$ . For a fixed edge $i j$ , write $x = a_{i j}$ , $u = g_{i}$ , $v = g_{j}$ , and $b = b_{i j} = u x v^{- 1}$ ; substituting the reverse formula for $μ_{i j}$ in the composite law gives $ν_{i j}^{o p} = β (u, u^{- 1}, b) - ρ_{u} β (u^{- 1}, u, x) + ρ_{u} β (u^{- 1}, b, v) - ρ_{u} β (x, v^{- 1}, v) - β (u, x, v^{- 1}) + β (b, v, v^{- 1})$ . Set $σ_{i} = β (g_{i}, g_{i}^{- 1}, g_{i})$ and $σ_{j} = β (g_{j}, g_{j}^{- 1}, g_{j})$ . The identities $D = 0$ on $(u, u^{- 1}, u, x)$ , $(u, u^{- 1}, b, v)$ , $(u, x, v^{- 1}, v)$ and $(b, v, v^{- 1}, v)$ , with normalization of $β$ , give respectively $ρ_{u} β (u^{- 1}, u, x) = β (u, u^{- 1}, u x) - σ_{i}$ , $ρ_{u} β (u^{- 1}, b, v) = - β (u, x v^{- 1}, v) + β (u, u^{- 1}, u x) - β (u, u^{- 1}, b)$ , $ρ_{u} β (x, v^{- 1}, v) = β (u x, v^{- 1}, v) - β (u, x v^{- 1}, v) - β (u, x, v^{- 1})$ , and $ρ_{b} σ_{j} = β (b v, v^{- 1}, v) - β (b, v, v^{- 1})$ . Substituting the first two identities into the first three terms of $ν_{i j}^{o p}$ gives $σ_{i} - β (u, x v^{- 1}, v)$ ; the third identity then combines all remaining terms involving $u$ into $- β (u x, v^{- 1}, v)$ , and the fourth gives $ν_{i j}^{o p} = σ_{i} - ρ_{b} σ_{j} = - (δ_{b} σ)_{i j}$ , since $b v = u x$ . The composite has identity vertex labels, so normalization gives $ζ^{e} = 0$ and its $B$ -shift is $δ_{b} ν^{o p} = - δ_{b}^{2} σ = 0$ : explicitly, on $i < j < k$ , flatness of $b$ and the action law give $ρ_{b_{i j}} (ρ_{b_{j k}} σ_{k} - σ_{j}) - (ρ_{b_{i k}} σ_{k} - σ_{i}) + (ρ_{b_{i j}} σ_{j} - σ_{i}) = 0$ . Thus the opposite composite also induces the identity on background cochains. Each gauge arrow therefore has a two-sided inverse, while sequential induced maps compose associatively as ordinary function composition.

Depends on

Finite 2-group background cochains