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A conjugate representation reverses the cubic anomaly polynomial

Statement

Let $H = S U (3) \times U (1)$ , let $E = C^{3}$ have representation $ρ_{E} (g, z) = z g$ , and let $E^{*}$ be its dual representation. For a smooth background connection $A_{E}$ on a smooth manifold $X$ , use the induced dual connection $A_{E^{*}} = - A_{E}^{T}$ and curvatures $F_{E} = d A_{E} + A_{E} \land A_{E}$ and $F_{E^{*}} = d A_{E^{*}} + A_{E^{*}} \land A_{E^{*}}$ . With the degree-six polynomial $I_{6}$ defined above, one has

$I_{6} (E^{*}, A_{E^{*}}) = - I_{6} (E, A_{E}) .$

For the canonical Chern–Simons transgression

$Q_{5} (E, A_{E}) : = 3 \frac{i}{24 π^{2}} \int_{0}^{1} {tr}_{E} ⁣ (A_{E} \land F_{E, t} \land F_{E, t}) d t, F_{E, t} : = t d A_{E} + t^{2} A_{E} \land A_{E},$

the dual transgression satisfies $Q_{5} (E^{*}, A_{E^{*}}) = - Q_{5} (E, A_{E})$ . Thus any common linear descent of this conjugate pair has zero total cubic polynomial and zero total first consistent descent representative.

Statement references

Bilal, §§7.1–7.2 and 9.4

Alvarez, §§2.2–2.3

Facts & Assumptions

Given: The group $H$ , its defining representation $E$ , the dual representation $E^{*}$ , a smooth connection $A_{E}$ , its induced dual connection, and the cubic polynomial normalization.

[F1]

Degree-six cubic trace polynomial for a chiral representation: $I_{6} (ρ, A_{ρ}) = \frac{i}{24 π^{2}} {tr}_{V_{ρ}} (F_{ρ} \land F_{ρ} \land F_{ρ})$ , and the dual connection is represented by $A_{E^{*}} = - A_{E}^{T}$ .

Proof

1.1givenF1algebra

Transposing the curvature formula gives $F_{E^{*}} = - F_{E}^{T}$ : for the one-form $A_{E}$ , graded transposition gives $(A_{E} \land A_{E})^{T} = - A_{E}^{T} \land A_{E}^{T}$ , so $d (- A_{E}^{T}) + (- A_{E}^{T}) \land (- A_{E}^{T}) = - (d A_{E} + A_{E} \land A_{E})^{T}$ .

2.1F1step 1.1algebra

Since each entry of $F_{E}$ is a two-form, reversing the order of two curvature factors contributes $(- 1)^{2 \cdot 2} = 1$ . Hence $tr ((F_{E}^{T})^{3}) = tr ((F_{E}^{3})^{T}) = tr (F_{E}^{3})$ . The three minus signs in $F_{E^{*}}^{3} = (- F_{E}^{T})^{3}$ therefore give $I_{6} (E^{*}, A_{E^{*}}) = - I_{6} (E, A_{E})$ by [F1].

3.1F1step 1.1algebra∎

For every $t \in [0, 1]$ , the same transposition calculation applied to $A_{E, t} = t A_{E}$ gives $F_{E^{*}, t} = - F_{E, t}^{T}$ . Thus the integrand defining $Q_{5} (E^{*}, A_{E^{*}})$ is $- tr (A_{E}^{T} \land F_{E, t}^{T} \land F_{E, t}^{T})$ . The trace of a transpose reverses the order of the three factors; cyclically moving the degree-one factor past the two degree-two factors contributes sign $(- 1)^{1 (2 + 2)} = 1$ . The integrand is therefore $- tr (A_{E} \land F_{E, t} \land F_{E, t})$ . Integration over $t$ proves the transgression identity and the stated cancellation under a common linear descent.

Depends on

Degree-six cubic trace polynomial for a chiral representation