Introduction
The classification of fusion categories is a structural problem at the meeting point of representation theory, quantum topology, conformal field theory, subfactor theory, and the mathematical theory of anyons. The Grothendieck ring makes the fusion rules discrete and computable, but it does not determine an associator: the pentagon equations impose additional, highly restrictive constraints. Consequently, a fusion ring may satisfy every evident numerical condition and still admit no categorification. Exact analyses of small and structured families are therefore valuable both for the classification program and for producing reliable algebraic models for topological and physical constructions.
Tambara–Yamagami categories are canonical nonpointed examples in which all but one simple object are invertible and the remaining object has Frobenius–Perron dimension ∣ G ∣ \sqrt{|G|} ∣ G ∣ [43 ] . They are explicit enough for detailed calculations, while already displaying irrational dimensions, nontrivial associativity, and arithmetic constraints. Near-integral fusion categories, formed by adjoining one simple object to a proper maximal-rank fusion subcategory [28 ] , provide a natural next layer of complexity. The near-Tambara–Yamagami case is especially well suited to a complete study: the Tambara–Yamagami subcategory supplies strong symmetry, whereas the extra simple object creates a sharp test of categorifiability.
Following [28 ] , we write C = C ( D , κ ) \mathcal C=\mathcal C(\mathcal D,\kappa) C = C ( D , κ ) D \mathcal D D G G G D \mathcal D D t t t ρ \rho ρ C ∖ D \mathcal C\setminus\mathcal D C ∖ D
∣ G ∣ = n 2 , FPdim ( t ) = n , ρ ⊗ ρ ≅ ⨁ g ∈ G g ⊕ n t ⊕ κ ρ , |G|=n^2,\qquad \operatorname{FPdim}(t)=n,\qquad
\rho\otimes \rho\cong \bigoplus_{g\in G}g\oplus n t\oplus\kappa \rho, ∣ G ∣ = n 2 , FPdim ( t ) = n , ρ ⊗ ρ ≅ g ∈ G ⨁ g ⊕ n t ⊕ κ ρ , so that, for d = FPdim ( ρ ) d=\operatorname{FPdim}(\rho) d = FPdim ( ρ )
d 2 = κ d + 2 n 2 . d^2=\kappa d+2n^2. d 2 = κ d + 2 n 2 . Our first result reduces the numerical possibilities to rank six. The following table records the candidates produced by the formal-codegree and indicator arguments; the irrational candidate κ = 8 \kappa=8 κ = 8 3.24 .
Theorem 1.1. Let C = C ( D , κ ) \mathcal C=\mathcal C(\mathcal D,\kappa) C = C ( D , κ ) C \mathbb C C D \mathcal D D n = 2 , G ≅ C 2 × C 2 , rank ( C ) = 6 , n=2,\qquad G\cong C_2\times C_2,\qquad \operatorname{rank}(\mathcal C)=6, n = 2 , G ≅ C 2 × C 2 , rank ( C ) = 6 , κ d FPdim ( C ) 0 2 2 16 8 4 + 2 6 48 + 16 6 2 4 24 7 8 72 \begin{array}{c|c|c}
\kappa&d&\operatorname{FPdim}(\mathcal C)\\\hline
0&2\sqrt2&16\\
8&4+2\sqrt6&48+16\sqrt6\\
2&4&24\\
7&8&72
\end{array} κ 0 8 2 7 d 2 2 4 + 2 6 4 8 FPdim ( C ) 16 48 + 16 6 24 72 κ = 8 \kappa=8 κ = 8 3.24 .
The irrational analysis is carried out in § 3 . Formal codegrees first force n = 2 n=2 n = 2 κ \kappa κ κ = 8 \kappa=8 κ = 8 [24 ] .
§ 4 treats the integral case independently. An integral parametrization gives two branches. The first branch leads to a Tannakian subcategory and a finite group with a normal subgroup whose nonidentity elements are conjugate; the second leads to a sharply doubly transitive group. The first branch uses the braided Tambara–Yamagami restriction and elementary arithmetic; the second uses the order bound on G G G n = 2 n=2 n = 2 κ = 7 \kappa=7 κ = 7 κ = 2 \kappa=2 κ = 2
The numerical result does not by itself classify associators. We therefore state the categorical conclusions separately. The parameter 7 7 7 Rep ( PSU ( 3 , 2 ) ) \operatorname{Rep}(\operatorname{PSU}(3,2)) Rep ( PSU ( 3 , 2 )) κ = 2 \kappa=2 κ = 2 § 4.2 constructs an explicit unitary group-theoretical fusion category with the fusion ring R 2 R_2 R 2
For κ ∈ Z ≥ 0 \kappa\in\mathbb Z_{\geq0} κ ∈ Z ≥ 0 R κ R_\kappa R κ G ∪ { t , ρ } G\cup\{t,\rho\} G ∪ { t , ρ } G = C 2 × C 2 G=C_2\times C_2 G = C 2 × C 2 G G G
g t = t g = t , g ρ = ρ g = ρ ( g ∈ G ) , t 2 = ∑ g ∈ G g , t ρ = ρ t = 2 ρ , ρ 2 = ∑ g ∈ G g + 2 t + κ ρ . \begin{equation}
\begin{gathered}
gt=tg=t,\qquad g\rho=\rho g=\rho\quad(g\in G),\qquad
t^2=\sum_{g\in G}g,\\
t\rho=\rho t=2\rho,\qquad
\rho^2=\sum_{g\in G}g+2t+\kappa\rho.
\end{gathered} \tag*{(1.1)}
\end{equation} g t = t g = t , g ρ = ρ g = ρ ( g ∈ G ) , t 2 = g ∈ G ∑ g , tρ = ρt = 2 ρ , ρ 2 = g ∈ G ∑ g + 2 t + κ ρ . (1.1) The following theorem classifies the categorifiable based rings in the family and records the available categorical conclusions.
Theorem 1.2. Let C \mathcal C C C \mathbb C C D \mathcal D D D \mathcal D D rank ( D ) = 5 , G ≅ C 2 × C 2 , K 0 ( C ) ≅ R κ for κ ∈ { 0 , 2 , 7 } . \operatorname{rank}(\mathcal D)=5,\qquad G\cong C_2\times C_2,\qquad
K_0(\mathcal C)\cong R_\kappa\quad\text{for }\kappa\in\{0,2,7\}. rank ( D ) = 5 , G ≅ C 2 × C 2 , K 0 ( C ) ≅ R κ for κ ∈ { 0 , 2 , 7 } .
For κ = 0 \kappa=0 κ = 0 Rep ( H 8 ) \operatorname{Rep}(H_8) Rep ( H 8 ) [24 Section 7 and Appendix C] ; see Theorem 3.13 for the precise scope of this assertion.
For κ = 2 \kappa=2 κ = 2 4.12 provides an explicit unitary group-theoretical categorification.
For κ = 7 \kappa=7 κ = 7 Rep ( PSU ( 3 , 2 ) ) \operatorname{Rep}(\operatorname{PSU}(3,2)) Rep ( PSU ( 3 , 2 )) 2.1 .
The dimensions are κ FPdim ( t ) FPdim ( ρ ) FPdim ( C ) 0 2 2 2 16 2 2 4 24 7 2 8 72. \begin{array}{c|c|c|c}
\kappa&\operatorname{FPdim}(t)&\operatorname{FPdim}(\rho)&\operatorname{FPdim}(\mathcal C)\\\hline
0&2&2\sqrt2&16\\
2&2&4&24\\
7&2&8&72.
\end{array} κ 0 2 7 FPdim ( t ) 2 2 2 FPdim ( ρ ) 2 2 4 8 FPdim ( C ) 16 24 72. κ = 2 , 7 \kappa=2,7 κ = 2 , 7
Preliminaries
Pseudo-unitary fusion categories
Throughout, let C \mathcal C C C \mathbb{C} C
Irr ( C ) = { X 1 = 1 , X 2 , … , X n } \operatorname{Irr}(\mathcal C)=\{X_1=\mathbf 1,X_2,\ldots,X_n\} Irr ( C ) = { X 1 = 1 , X 2 , … , X n } for the isomorphism classes of simple objects, with 1 \mathbf 1 1 Z \mathbb Z Z K 0 ( C ) K_0(\mathcal C) K 0 ( C )
For simple objects X , Y X,Y X , Y
X ⊗ Y ≅ ⨁ Z ∈ Irr ( C ) N X Y Z Z . X\otimes Y\cong
\bigoplus_{Z\in\operatorname{Irr}(\mathcal C)}N_{XY}^{Z} Z. X ⊗ Y ≅ Z ∈ Irr ( C ) ⨁ N X Y Z Z . The nonnegative integer N X Y Z N_{XY}^{Z} N X Y Z Z Z Z X ⊗ Y X\otimes Y X ⊗ Y C \mathcal C C X , Y , Z X,Y,Z X , Y , Z Irr ( C ) \operatorname{Irr}(\mathcal C) Irr ( C )
N X Y Z = N Z Y ∗ X = N X ∗ Z Y = N Y ∗ X ∗ Z ∗ , X , Y , Z ∈ Irr ( C ) . N_{XY}^{Z}
=N_{Z\,Y^*}^{X}
=N_{X^*\,Z}^{Y}
=N_{Y^*\,X^*}^{Z^*},
\qquad X,Y,Z\in\operatorname{Irr}(\mathcal C). N X Y Z = N Z Y ∗ X = N X ∗ Z Y = N Y ∗ X ∗ Z ∗ , X , Y , Z ∈ Irr ( C ) . For each X ∈ Irr ( C ) X\in\operatorname{Irr}(\mathcal C) X ∈ Irr ( C ) N X N_X N X
( N X ) Y Z = N X Y Z , Y , Z ∈ Irr ( C ) . (N_X)_{Y}^{\,Z}=N_{XY}^{Z},
\qquad Y,Z\in\operatorname{Irr}(\mathcal C). ( N X ) Y Z = N X Y Z , Y , Z ∈ Irr ( C ) . Its Perron–Frobenius eigenvalue is positive and is, by definition, the Frobenius–Perron dimension of X X X FPdim ( X ) \operatorname{FPdim}(X) FPdim ( X ) [33 Theorem 8.6] , shows that these numbers extend uniquely to a ring homomorphism FPdim : K 0 ( C ) ⟶ R \operatorname{FPdim}:K_0(\mathcal C)\longrightarrow\mathbb R FPdim : K 0 ( C ) ⟶ R C \mathcal C C
FPdim ( C ) = ∑ X ∈ Irr ( C ) FPdim ( X ) 2 . \operatorname{FPdim}(\mathcal C)
=\sum_{X\in\operatorname{Irr}(\mathcal C)}\operatorname{FPdim}(X)^2. FPdim ( C ) = X ∈ Irr ( C ) ∑ FPdim ( X ) 2 . A pivotal structure on C \mathcal C C i : Id C ⟹ ( − ) ∗ ∗ i:\operatorname{Id}_{\mathcal C}\Longrightarrow(-)^{**} i : Id C ⟹ ( − ) ∗∗ ( − ) ∗ (-)^* ( − ) ∗ dim ( X ) \dim(X) dim ( X )
dim ( C ) = ∑ X ∈ Irr ( C ) dim ( X ) dim ( X ∗ ) \dim(\mathcal C)=\sum_{X\in\operatorname{Irr}(\mathcal C)}\dim(X)\dim(X^*) dim ( C ) = X ∈ Irr ( C ) ∑ dim ( X ) dim ( X ∗ ) by [33 Proposition 2.9] . A pivotal structure is spherical when dim ( X ) = dim ( X ∗ ) \dim(X)=\dim(X^*) dim ( X ) = dim ( X ∗ ) X X X ∑ X dim ( X ) 2 \sum_X\dim(X)^2 ∑ X dim ( X ) 2 dim ( X ) = FPdim ( X ) > 0 \dim(X)=\operatorname{FPdim}(X)>0 dim ( X ) = FPdim ( X ) > 0 X X X [33 Proposition 8.23] . We use this positive spherical structure whenever working with a pseudo-unitary category. Unitarity is a stronger property and will be established separately where it is needed.
The Drinfeld center
We will also use the Drinfeld center Z ( C ) \mathcal Z(\mathcal C) Z ( C ) C \mathcal C C ( Z , c Z , − ) (Z,c_{Z,-}) ( Z , c Z , − ) Z Z Z C \mathcal C C c Z , X : Z ⊗ X → X ⊗ Z c_{Z,X}:Z\otimes X\to X\otimes Z c Z , X : Z ⊗ X → X ⊗ Z X ∈ C X\in\mathcal C X ∈ C
A principal tool that we will use is induction-restriction to the Drinfeld center. There is a forgetful functor F : Z ( C ) → C F:\mathcal Z(\mathcal C)\to\mathcal C F : Z ( C ) → C ( Z , c Z , − ) (Z,c_{Z,-}) ( Z , c Z , − ) Z Z Z I : C → Z ( C ) I:\mathcal C\to\mathcal Z(\mathcal C) I : C → Z ( C ) X ∈ C X\in\mathcal C X ∈ C Y ∈ Z ( C ) Y\in\mathcal Z(\mathcal C) Y ∈ Z ( C ) Hom Z ( C ) ( Y , I ( X ) ) ≅ Hom C ( F ( Y ) , X ) \operatorname{Hom}_{\mathcal Z(\mathcal C)}(Y,I(X)) \cong \operatorname{Hom}_{\mathcal C}(F(Y),X) Hom Z ( C ) ( Y , I ( X )) ≅ Hom C ( F ( Y ) , X )
The underlying object of an induced object admits the standard description [33 Proposition 5.4] :
Because the forgetful functor preserves Frobenius–Perron dimensions, this formula gives
FPdim ( I ( X ) ) = FPdim ( X ) FPdim ( C ) . \begin{equation}
\begin{aligned}
\operatorname{FPdim}(I(X))=\operatorname{FPdim}(X)\,\operatorname{FPdim}(\mathcal C).
\end{aligned} \tag*{(2.2)}
\end{equation} FPdim ( I ( X )) = FPdim ( X ) FPdim ( C ) . (2.2) Throughout, we write [ A , B ] = dim C Hom ( A , B ) [A,B]=\dim_{\mathbb C}\operatorname{Hom}(A,B) [ A , B ] = dim C Hom ( A , B ) A , B A,B A , B C \mathcal C C [ A , B ] [A,B] [ A , B ] B B B A A A B B B [ I ( X ) , Z ] [I(X),Z] [ I ( X ) , Z ] m X ( Z ) m_X(Z) m X ( Z )
[ Y , I ( X ) ] = [ F ( Y ) , X ] , [ I ( X ) , I ( Y ) ] = ∑ Z m X ( Z ) m Y ( Z ) . \begin{equation}
\begin{split}
[Y,I(X)]&=[F(Y),X],\\
[I(X),I(Y)]&=\sum_Zm_X(Z)m_Y(Z).
\end{split} \tag*{(2.3)}
\end{equation} [ Y , I ( X )] [ I ( X ) , I ( Y )] = [ F ( Y ) , X ] , = Z ∑ m X ( Z ) m Y ( Z ) . (2.3) A twist of a braided category C \mathcal C C θ : id C → id C \theta:\operatorname{id}_{\mathcal C}\to \operatorname{id}_{\mathcal C} θ : id C → id C θ 1 = id 1 \theta_1=\operatorname{id}_1 θ 1 = id 1 θ X ⊗ Y = ( θ X ⊗ θ Y ) c Y , X c X , Y \theta_{X\otimes Y}=(\theta_X\otimes \theta_Y)c_{Y,X}c_{X,Y} θ X ⊗ Y = ( θ X ⊗ θ Y ) c Y , X c X , Y X X X C \mathcal C C θ X \theta_X θ X C \mathbb{C} C θ \theta θ θ X \theta_X θ X X ∈ Irr ( C ) X\in \operatorname{Irr}(\mathcal C) X ∈ Irr ( C ) [44 ] .
When using spherical traces in the center, put Δ = dim ( C ) \Delta=\dim(\mathcal C) Δ = dim ( C ) Δ = FPdim ( C ) \Delta=\operatorname{FPdim}(\mathcal C) Δ = FPdim ( C ) Z ( C ) \mathcal Z(\mathcal C) Z ( C )
θ I ( 1 ) = id , Tr ( θ I ( X ) j ) = ν j ( X ) Δ ( j ≥ 1 ) , \begin{equation}
\theta_{I(\mathbf 1)}=\operatorname{id},\qquad
\operatorname{Tr}(\theta_{I(X)}^j)=\nu_j(X)\Delta\quad(j\geq1), \tag*{(2.4)}
\end{equation} θ I ( 1 ) = id , Tr ( θ I ( X ) j ) = ν j ( X ) Δ ( j ≥ 1 ) , (2.4) where ν j ( X ) \nu_j(X) ν j ( X ) j j j X X X [39 Proposition 2.4 and Theorem 2.5] and [38 Theorem 4.1] . For simple X ≠ 1 X\ne \mathbf 1 X = 1 ν 1 ( X ) = 0 \nu_1(X)=0 ν 1 ( X ) = 0 ν 2 ( X ) \nu_2(X) ν 2 ( X ) 0 0 0 X X X ± 1 \pm1 ± 1 Z ( C ) \mathcal Z(\mathcal C) Z ( C )
We consistently write S \mathsf S S S S S Z ( C ) \mathcal Z(\mathcal C) Z ( C ) S 1 , Z = dim Z \mathsf S_{\mathbf 1,Z}=\dim Z S 1 , Z = dim Z S = S / Δ S=\mathsf S/\Delta S = S /Δ
S V , W = 1 θ V θ W ∑ Z [ V ⊗ W , Z ] θ Z dim Z . \begin{equation}
\mathsf S_{V,W}=\frac{1}{\theta_V\theta_W}
\sum_Z[V\otimes W,Z]\theta_Z\dim Z. \tag*{(2.5)}
\end{equation} S V , W = θ V θ W 1 Z ∑ [ V ⊗ W , Z ] θ Z dim Z . (2.5) When K 0 ( C ) K_0(\mathcal C) K 0 ( C ) χ \chi χ Z χ Z_\chi Z χ I ( 1 ) I(\mathbf 1) I ( 1 ) f χ = ∑ X ∈ Irr ( C ) χ ( X ) χ ( X ∗ ) f_\chi=\sum_{X\in\operatorname{Irr}(\mathcal C)}\chi(X)\chi(X^*) f χ = ∑ X ∈ Irr ( C ) χ ( X ) χ ( X ∗ )
dim Z χ = Δ f χ , θ Z χ = 1 , S Z χ , W = dim ( Z χ ) χ ( F ( W ) ) , S Z χ , W = χ ( F ( W ) ) f χ . \begin{equation}
\dim Z_\chi=\frac{\Delta}{f_\chi},\qquad
\theta_{Z_\chi}=1,\qquad
\mathsf S_{Z_\chi,W}=\dim(Z_\chi)\chi(F(W)),\qquad
S_{Z_\chi,W}=\frac{\chi(F(W))}{f_\chi}. \tag*{(2.6)}
\end{equation} dim Z χ = f χ Δ , θ Z χ = 1 , S Z χ , W = dim ( Z χ ) χ ( F ( W )) , S Z χ , W = f χ χ ( F ( W )) . (2.6) These are the formal-codegree correspondence and the pullback character identity in [39 Section 2.3, Example 2.9, and Theorem 2.13] . The normalization in (2.6) will also be used for the Galois action on S S S
Near-Tambara–Yamagami categories
Inspired by near-group fusion categories, the authors [28 ] refer to fusion categories C = C ( D , κ ) \mathcal{C}=\mathcal{C}(\mathcal{D},\kappa) C = C ( D , κ ) D \mathcal{D} D D \mathcal D D C \mathcal C C
Example 2.1. The category Rep ( PSU ( 3 , 2 ) ) \operatorname{Rep}(\operatorname{PSU}(3,2)) Rep ( PSU ( 3 , 2 )) [28 Example 3.7 and Figure 4] for its character table. Write G 7 = PSU ( 3 , 2 ) G_7=\operatorname{PSU}(3,2) G 7 = PSU ( 3 , 2 ) G 7 G_7 G 7
1 , λ 1 , λ 2 , λ 3 , t , ρ , \mathbf{1},\quad \lambda_1,\quad \lambda_2,\quad \lambda_3,\quad
t,\quad \rho, 1 , λ 1 , λ 2 , λ 3 , t , ρ , with dimensions
dim ( 1 ) = dim ( λ i ) = 1 , dim ( t ) = 2 , dim ( ρ ) = 8. \dim(\mathbf{1})=\dim(\lambda_i)=1,\qquad
\dim(t)=2,\qquad
\dim(\rho)=8. dim ( 1 ) = dim ( λ i ) = 1 , dim ( t ) = 2 , dim ( ρ ) = 8. The fusion rules of Rep ( G 7 ) \operatorname{Rep}(G_7) Rep ( G 7 )
λ i ⊗ λ j ≅ ( λ i λ j ) , λ i ⊗ t ≅ t , t ⊗ t ≅ 1 ⊕ λ 1 ⊕ λ 2 ⊕ λ 3 , λ i ⊗ ρ ≅ ρ , t ⊗ ρ ≅ 2 ρ , ρ ⊗ ρ ≅ 1 ⊕ λ 1 ⊕ λ 2 ⊕ λ 3 ⊕ 2 t ⊕ 7 ρ . \begin{align}
&\lambda_i\otimes\lambda_j
\cong (\lambda_i\lambda_j), \lambda_i\otimes t\cong t,
t\otimes t\cong
\mathbf{1}\oplus\lambda_1\oplus\lambda_2\oplus\lambda_3, \tag*{(2.7)} \\
\lambda_i\otimes \rho
&\cong \rho, t\otimes \rho\cong 2 \rho, \rho\otimes \rho
\cong \mathbf{1}\oplus\lambda_1\oplus\lambda_2\oplus\lambda_3
\oplus 2 t\oplus 7 \rho. \tag*{(2.8)}
\end{align} λ i ⊗ ρ λ i ⊗ λ j ≅ ( λ i λ j ) , λ i ⊗ t ≅ t , t ⊗ t ≅ 1 ⊕ λ 1 ⊕ λ 2 ⊕ λ 3 , ≅ ρ , t ⊗ ρ ≅ 2 ρ , ρ ⊗ ρ ≅ 1 ⊕ λ 1 ⊕ λ 2 ⊕ λ 3 ⊕ 2 t ⊕ 7 ρ . (2.7) (2.8) Here λ i λ j \lambda_i\lambda_j λ i λ j i , j ∈ { 1 , 2 , 3 } i,j\in\{1,2,3\} i , j ∈ { 1 , 2 , 3 }
λ 1 ⊗ λ 1 ≅ λ 2 ⊗ λ 2 ≅ λ 3 ⊗ λ 3 ≅ 1 , λ 1 ⊗ λ 2 ≅ λ 3 . \lambda_1\otimes \lambda_1\cong \lambda_2\otimes \lambda_2\cong \lambda_3\otimes \lambda_3\cong \mathbf{1},
\quad
\lambda_1\otimes\lambda_2\cong \lambda_3. λ 1 ⊗ λ 1 ≅ λ 2 ⊗ λ 2 ≅ λ 3 ⊗ λ 3 ≅ 1 , λ 1 ⊗ λ 2 ≅ λ 3 . Throughout this subsection, C = C ( D , κ ) \mathcal{C}=\mathcal{C}(\mathcal{D},\kappa) C = C ( D , κ ) ρ \rho ρ D \mathcal{D} D X ⊗ ρ ≅ ρ ⊗ X ≅ FPdim ( X ) ρ X\otimes\rho\cong \rho\otimes X\cong \operatorname{FPdim}(X) \rho X ⊗ ρ ≅ ρ ⊗ X ≅ FPdim ( X ) ρ X ∈ D X\in\mathcal D X ∈ D D \mathcal D D [43 ] , the group G G G D \mathcal D D
g ⊗ t ≅ t ⊗ g ≅ t , g ⊗ ρ ≅ ρ ⊗ g ≅ ρ ( g ∈ G ) , ∣ G ∣ = n 2 , t ⊗ t ≅ ⨁ g ∈ G g , t ⊗ ρ ≅ ρ ⊗ t ≅ n ρ , ρ ⊗ ρ ≅ ⨁ g ∈ G g ⊕ n t ⊕ κ ρ . \begin{equation}
\begin{gathered}
g\otimes t\cong t\otimes g\cong t,\quad g\otimes\rho\cong \rho\otimes g\cong \rho\quad(g\in G),\qquad |G|=n^2,\\
t\otimes t\cong \bigoplus_{g\in G}g,\qquad t\otimes\rho\cong \rho\otimes t\cong n \rho,\qquad
\rho\otimes \rho\cong \bigoplus_{g\in G}g\oplus n t\oplus\kappa \rho.
\end{gathered} \tag*{(2.9)}
\end{equation} g ⊗ t ≅ t ⊗ g ≅ t , g ⊗ ρ ≅ ρ ⊗ g ≅ ρ ( g ∈ G ) , ∣ G ∣ = n 2 , t ⊗ t ≅ g ∈ G ⨁ g , t ⊗ ρ ≅ ρ ⊗ t ≅ n ρ , ρ ⊗ ρ ≅ g ∈ G ⨁ g ⊕ n t ⊕ κ ρ . (2.9) Here t t t D \mathcal D D n = FPdim ( t ) ∈ Z ≥ 2 n=\operatorname{FPdim}(t)\in\mathbb Z_{\geq2} n = FPdim ( t ) ∈ Z ≥ 2 κ ∈ Z ≥ 0 \kappa\in\mathbb Z_{\geq0} κ ∈ Z ≥ 0 t t t ρ \rho ρ κ \kappa κ ρ \rho ρ ρ 2 \rho^2 ρ 2 max { n , κ } \max\{n,\kappa\} max { n , κ } max { 2 , κ } \max\{2,\kappa\} max { 2 , κ } R 0 R_0 R 0 R 2 R_2 R 2
N = 2 n 2 , d = d + = κ + κ 2 + 8 n 2 2 , d − = κ − κ 2 + 8 n 2 2 . \begin{equation}
N=2n^2,\qquad d=d_+=\frac{\kappa+\sqrt{\kappa^2+8n^2}}2,
\qquad d_-=\frac{\kappa-\sqrt{\kappa^2+8n^2}}2. \tag*{(2.10)}
\end{equation} N = 2 n 2 , d = d + = 2 κ + κ 2 + 8 n 2 , d − = 2 κ − κ 2 + 8 n 2 . (2.10) Then d d − = − N dd_-=-N d d − = − N d 2 = κ d + N d^2=\kappa d+N d 2 = κ d + N Δ : = FPdim ( C ) = N + d 2 = 4 n 2 + κ d \Delta:=\operatorname{FPdim}(\mathcal C)=N+d^2=4n^2+\kappa d Δ := FPdim ( C ) = N + d 2 = 4 n 2 + κ d d d d d ∉ Z d\notin\mathbb Z d ∈ / Z d ∉ Q d\notin\mathbb Q d ∈ / Q C \mathcal C C
By [28 Lemma 4.1] , a near-integral category is Galois conjugate to a pseudo-unitary category. Galois conjugation preserves fusion rules. We therefore use a pseudo-unitary conjugate whenever positivity of categorical dimensions is needed. Integral categories are already pseudo-unitary by [33 Proposition 8.24] , and all pseudo-unitary categories below carry their canonical positive spherical structures.
By ((2.1) ), we have
F ( I ( g ) ) ≅ n 2 g ⊕ ⨁ h ∈ G 2 h ⊕ n t ⊕ κ ρ , F ( I ( t ) ) ≅ ⨁ h ∈ G n h ⊕ 3 n 2 t ⊕ n κ ρ , F ( I ( ρ ) ) ≅ ⨁ h ∈ G κ h ⊕ n κ t ⊕ ( 4 n 2 + κ 2 ) ρ . \begin{equation}
\begin{split}
F(I(g))&\cong n^2 g\oplus\bigoplus_{h\in G}2 h\oplus n t\oplus\kappa \rho,\\
F(I(t))&\cong \bigoplus_{h\in G}n h\oplus 3n^2 t\oplus n\kappa \rho,\\
F(I(\rho))&\cong \bigoplus_{h\in G}\kappa h\oplus n\kappa t\oplus (4n^2+\kappa^2) \rho.
\end{split} \tag*{(2.11)}
\end{equation} F ( I ( g )) F ( I ( t )) F ( I ( ρ )) ≅ n 2 g ⊕ h ∈ G ⨁ 2 h ⊕ n t ⊕ κ ρ , ≅ h ∈ G ⨁ nh ⊕ 3 n 2 t ⊕ nκ ρ , ≅ h ∈ G ⨁ κh ⊕ nκ t ⊕ ( 4 n 2 + κ 2 ) ρ . (2.11) For a tensor functor, its image means the full fusion subcategory generated by the simple constituents of its values. A tensor functor L : A → B L:\mathcal A\to\mathcal B L : A → B B \mathcal B B L ( X ) L(X) L ( X ) X ∈ A X\in\mathcal A X ∈ A
Lemma 2.2. The characters of K 0 ( C ) K_0(\mathcal C) K 0 ( C ) g ∈ G t ρ f χ + 1 n d N + d 2 χ − 1 n d − N + d − 2 η 1 − n 0 N η ψ ψ ( g ) 0 0 n 2 \begin{array}{c|ccc|c}
&g\in G&t&\rho&f\\\hline
\chi_+&1&n&d&N+d^2\\
\chi_-&1&n&d_-&N+d_-^2\\
\eta&1&-n&0&N\\
\eta_\psi&\psi(g)&0&0&n^2
\end{array} χ + χ − η η ψ g ∈ G 1 1 1 ψ ( g ) t n n − n 0 ρ d d − 0 0 f N + d 2 N + d − 2 N n 2 ψ \psi ψ G G G I ( 1 ) I(\mathbf 1) I ( 1 ) 1 , d 2 N = 1 + κ N d , 2 + κ N d , 4 + 2 κ N d ,
1,\qquad \frac{d^2}{N}=1+\frac\kappa N d,\qquad
2+\frac\kappa N d,\qquad 4+\frac{2\kappa}N d,
1 , N d 2 = 1 + N κ d , 2 + N κ d , 4 + N 2 κ d , n 2 − 1 n^2-1 n 2 − 1 1 \mathbf 1 1
Proof. If a character is nontrivial on G G G g ⊗ t ≅ t g\otimes t\cong t g ⊗ t ≅ t g ⊗ ρ ≅ ρ g\otimes\rho\cong \rho g ⊗ ρ ≅ ρ t , ρ t,\rho t , ρ t t t ± n \pm n ± n − n -n − n ρ \rho ρ n n n d , d − d,d_- d , d − n 2 + 2 n^2+2 n 2 + 2 f χ = ∑ X χ ( X ) χ ( X ∗ ) f_\chi=\sum_X\chi(X)\chi(X^*) f χ = ∑ X χ ( X ) χ ( X ∗ ) [39 Theorem 2.13] assigns a summand of dimension Δ / f χ \Delta/f_\chi Δ/ f χ d d − = − N dd_-=-N d d − = − N
Lemma 2.3. Every g ∈ G g\in G g ∈ G ord ( g ) ≤ n \operatorname{ord}(g)\leq n ord ( g ) ≤ n n = 2 n=2 n = 2 G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2
Proof. The rank-one module category add ( ρ ) \operatorname{add}(\rho) add ( ρ ) Vec \operatorname{Vec} Vec X ∈ D X\in\mathcal D X ∈ D Φ ( X ) = Hom C ( ρ , X ⊗ ρ ) \Phi(X)=\operatorname{Hom}_{\mathcal C}(\rho,X\otimes\rho) Φ ( X ) = Hom C ( ρ , X ⊗ ρ )
Φ ( X ) ⊗ Φ ( Y ) → ∼ Φ ( X ⊗ Y ) , C → ∼ Φ ( 1 ) . \Phi(X)\otimes\Phi(Y)\xrightarrow{\ \sim\ }\Phi(X\otimes Y),
\qquad \mathbb C\xrightarrow{\ \sim\ }\Phi(\mathbf 1). Φ ( X ) ⊗ Φ ( Y ) ∼ Φ ( X ⊗ Y ) , C ∼ Φ ( 1 ) . On elementary tensors the first map sends f ⊗ g f\otimes g f ⊗ g ( id X ⊗ g ) ∘ f (\operatorname{id}_X\otimes g)\circ f ( id X ⊗ g ) ∘ f Φ \Phi Φ Φ ( t ∗ ) ≅ Φ ( t ) ∗ \Phi(t^*)\cong\Phi(t)^* Φ ( t ∗ ) ≅ Φ ( t ) ∗
For the order estimate, let
Φ : D ⟶ Vec , Φ ( X ) = Hom C ( ρ , X ⊗ ρ ) , \Phi:\mathcal D\longrightarrow\operatorname{Vec},\qquad
\Phi(X)=\operatorname{Hom}_{\mathcal C}(\rho,X\otimes\rho), Φ : D ⟶ Vec , Φ ( X ) = Hom C ( ρ , X ⊗ ρ ) , denote the functor just described. We use only its values and tensor structure in the argument below.
For g ∈ G g\in G g ∈ G g ⊗ ρ ≅ ρ g\otimes\rho\cong\rho g ⊗ ρ ≅ ρ
Φ ( g ) = Hom C ( ρ , g ⊗ ρ ) ≅ End C ( ρ ) ≅ C . \Phi(g)=\operatorname{Hom}_{\mathcal C}(\rho,g\otimes\rho)
\cong\operatorname{End}_{\mathcal C}(\rho)\cong\mathbb C. Φ ( g ) = Hom C ( ρ , g ⊗ ρ ) ≅ End C ( ρ ) ≅ C . The first isomorphism uses any chosen isomorphism g ⊗ ρ ≅ ρ g\otimes\rho\cong\rho g ⊗ ρ ≅ ρ Φ ( g ) \Phi(g) Φ ( g ) V = Φ ( t ) V=\Phi(t) V = Φ ( t )
V ≅ Hom C ( ρ , n ρ ) ≅ C n , dim C V = n . V\cong\operatorname{Hom}_{\mathcal C}(\rho,n \rho)\cong\mathbb C^n,
\qquad\dim_{\mathbb C}V=n. V ≅ Hom C ( ρ , n ρ ) ≅ C n , dim C V = n . Choose a nonzero vector v g ∈ Φ ( g ) v_g\in\Phi(g) v g ∈ Φ ( g ) ψ g : g ⊗ t → t \psi_g:g\otimes t\to t ψ g : g ⊗ t → t g ⊗ t ≅ t g\otimes t\cong t g ⊗ t ≅ t Φ ( g ) ⊗ V → Φ ( g ⊗ t ) \Phi(g)\otimes V\to\Phi(g\otimes t) Φ ( g ) ⊗ V → Φ ( g ⊗ t ) Φ ( ψ g ) \Phi(\psi_g) Φ ( ψ g ) U g ∈ End C ( V ) U_g\in\operatorname{End}_{\mathbb C}(V) U g ∈ End C ( V ) ψ g \psi_g ψ g t t t v 1 v_\mathbf 1 v 1 ψ 1 \psi_\mathbf 1 ψ 1 U 1 = id V U_\mathbf 1=\operatorname{id}_V U 1 = id V
For g , h ∈ G g,h\in G g , h ∈ G v g v_g v g v h v_h v h v g h v_{gh} v g h Φ ( g h ) \Phi(gh) Φ ( g h ) c ( g , h ) ∈ C × c(g,h)\in\mathbb C^\times c ( g , h ) ∈ C × ( g ⊗ h ) ⊗ t (g\otimes h)\otimes t ( g ⊗ h ) ⊗ t t t t ψ g , ψ h \psi_g,\psi_h ψ g , ψ h ψ g h \psi_{gh} ψ g h Φ \Phi Φ
U g U h = α ( g , h ) U g h , α ( g , h ) ∈ C × . U_gU_h=\alpha(g,h)U_{gh},\qquad \alpha(g,h)\in\mathbb C^\times. U g U h = α ( g , h ) U g h , α ( g , h ) ∈ C × . Associativity implies the cocycle identity for α \alpha α U g U_g U g
Rigidity gives Hom D ( g ⊗ t , t ) ≅ Hom D ( g , t ⊗ t ∗ ) \operatorname{Hom}_{\mathcal D}(g\otimes t,t)\cong
\operatorname{Hom}_{\mathcal D}(g,t\otimes t^*) Hom D ( g ⊗ t , t ) ≅ Hom D ( g , t ⊗ t ∗ ) t ⊗ t ∗ ≅ ⨁ h ∈ G h t\otimes t^*\cong\bigoplus_{h\in G}h t ⊗ t ∗ ≅ ⨁ h ∈ G h ψ g \psi_g ψ g g g g ⨁ g ∈ G g → t ⊗ t ∗ \bigoplus_{g\in G}g\to t\otimes t^* ⨁ g ∈ G g → t ⊗ t ∗ Φ \Phi Φ Φ ( t ∗ ) ≅ V ∗ \Phi(t^*)\cong V^* Φ ( t ∗ ) ≅ V ∗
⨁ g ∈ G Φ ( g ) ≅ V ⊗ V ∗ ≅ End C ( V ) . \bigoplus_{g\in G}\Phi(g)\cong V\otimes V^*\cong\operatorname{End}_{\mathbb C}(V). g ∈ G ⨁ Φ ( g ) ≅ V ⊗ V ∗ ≅ End C ( V ) . Under the standard identification V ⊗ V ∗ ≅ End ( V ) V\otimes V^*\cong\operatorname{End}(V) V ⊗ V ∗ ≅ End ( V ) v g v_g v g U g U_g U g End C ( V ) = ⨁ g ∈ G C U g \operatorname{End}_{\mathbb C}(V)=\bigoplus_{g\in G}\mathbb CU_g End C ( V ) = ⨁ g ∈ G C U g U g U_g U g
If ℓ = ord ( g ) \ell=\operatorname{ord}(g) ℓ = ord ( g ) U g j = c j U g j U_g^j=c_jU_{g^j} U g j = c j U g j c j ∈ C × c_j\in\mathbb C^\times c j ∈ C × id V , U g , … , U g ℓ − 1 \operatorname{id}_V,U_g,\ldots,U_g^{\ell-1} id V , U g , … , U g ℓ − 1 n + 1 n+1 n + 1 n × n n\times n n × n U g U_g U g ℓ ≤ n \ell\leq n ℓ ≤ n n = 2 n=2 n = 2 ∣ G ∣ = 4 |G|=4 ∣ G ∣ = 4 G G G G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2
The irrational near-Tambara–Yamagami categories
Throughout this section, d ∉ Z d\notin\mathbb Z d ∈ / Z § 2.3 is in force. For arguments depending only on the fusion rules we work in a pseudo-unitary Galois conjugate. In Theorem 3.13 , which concerns tensor equivalence, we return to the original category.
Proposition 3.1. One has n = 2 n=2 n = 2 κ = 8 k \kappa=8k κ = 8 k k ∈ Z ≥ 0 k\in\mathbb Z_{\geq0} k ∈ Z ≥ 0
Proof. The second summand in (2.12) forgets to an object of dimension a + b d a+bd a + b d a , b ∈ Z ≥ 0 a,b\in\mathbb Z_{\geq0} a , b ∈ Z ≥ 0 a = 1 a=1 a = 1 b = κ / N b=\kappa/N b = κ / N κ = N k \kappa=Nk κ = N k k ≥ 0 k\geq0 k ≥ 0 I ( 1 ) I(\mathbf 1) I ( 1 ) 1 , 1 , 2 , 4 1,1,2,4 1 , 1 , 2 , 4 1 \mathbf 1 1 n ≥ 4 n\geq4 n ≥ 4 t t t [ F ( I ( 1 ) ) , t ] = n [F(I(\mathbf 1)),t]=n [ F ( I ( 1 )) , t ] = n n ≤ 3 n\leq3 n ≤ 3
If n = 3 n=3 n = 3 2 + k d 2+kd 2 + k d 1 ⊕ f ⊕ k ρ \mathbf 1\oplus f\oplus k \rho 1 ⊕ f ⊕ k ρ f ≠ 1 f\ne \mathbf 1 f = 1 [41 Corollary 5.8] , the fusion category C \mathcal C C [41 Theorem 4.10(6)] , the right adjoint I I I F F F
I ( 1 ) ≅ I ( 1 ) ∗ . I(\mathbf 1)\cong I(\mathbf 1)^*. I ( 1 ) ≅ I ( 1 ) ∗ . Since the summand of dimension 2 + k d 2+kd 2 + k d I ( 1 ) I(\mathbf 1) I ( 1 ) f = f − 1 f=f^{-1} f = f − 1 n = 2 n=2 n = 2
The positive irrational parameter bound
The case where k k k 0 0 0 k ≥ 1 k\geq1 k ≥ 1
By Proposition 3.1 , we may write κ = 8 k \kappa=8k κ = 8 k k ≥ 1 k\geq1 k ≥ 1
r = 16 k 2 + 8 , q = 2 + k d , d = 4 k + r , d 2 = 8 k d + 8 , Δ = 8 q . \begin{equation}
r=\sqrt{16k^2+8},\qquad q=2+kd,\qquad
d=4k+r,\quad d^2=8kd+8,\quad\Delta=8q. \tag*{(3.1)}
\end{equation} r = 16 k 2 + 8 , q = 2 + k d , d = 4 k + r , d 2 = 8 k d + 8 , Δ = 8 q . (3.1) Lemma 3.2. For x ∈ G x\in G x ∈ G F ( I ( x ) ) ≅ 4 x ⊕ ⨁ y ∈ G 2 y ⊕ 2 t ⊕ 8 k ρ , F ( I ( t ) ) ≅ ⨁ y ∈ G 2 y ⊕ 12 t ⊕ 16 k ρ , F ( I ( ρ ) ) ≅ ⨁ y ∈ G 8 k y ⊕ 16 k t ⊕ ( 16 + 64 k 2 ) ρ . \begin{aligned}
F(I(x))&\cong 4 x\oplus\bigoplus_{y\in G}2 y\oplus 2 t\oplus 8k \rho,\\
F(I(t))&\cong \bigoplus_{y\in G}2 y\oplus 12 t\oplus 16k \rho,\\
F(I(\rho))&\cong \bigoplus_{y\in G}8k y\oplus 16k t\oplus (16+64k^2) \rho.
\end{aligned} F ( I ( x )) F ( I ( t )) F ( I ( ρ )) ≅ 4 x ⊕ y ∈ G ⨁ 2 y ⊕ 2 t ⊕ 8 k ρ , ≅ y ∈ G ⨁ 2 y ⊕ 12 t ⊕ 16 k ρ , ≅ y ∈ G ⨁ 8 k y ⊕ 16 k t ⊕ ( 16 + 64 k 2 ) ρ . G = { 1 , f , g , h } G=\{\mathbf 1,f,g,h\} G = { 1 , f , g , h } M X , Y = [ I ( X ) , I ( Y ) ] M_{X,Y}=[I(X),I(Y)] M X , Y = [ I ( X ) , I ( Y )] 1 , f , g , h , t , ρ \mathbf 1,f,g,h,t,\rho 1 , f , g , h , t , ρ M = ( 6 2 2 2 2 8 k 2 6 2 2 2 8 k 2 2 6 2 2 8 k 2 2 2 6 2 8 k 2 2 2 2 12 16 k 8 k 8 k 8 k 8 k 16 k 16 + 64 k 2 ) .
M=
\begin{pmatrix}
6&2&2&2&2&8k\\
2&6&2&2&2&8k\\
2&2&6&2&2&8k\\
2&2&2&6&2&8k\\
2&2&2&2&12&16k\\
8k&8k&8k&8k&16k&16+64k^2
\end{pmatrix}.
M = 6 2 2 2 2 8 k 2 6 2 2 2 8 k 2 2 6 2 2 8 k 2 2 2 6 2 8 k 2 2 2 2 12 16 k 8 k 8 k 8 k 8 k 16 k 16 + 64 k 2 .
Proof. Apply (2.3) to (2.11) and M X , Y = [ I ( X ) , I ( Y ) ] = [ F I ( X ) , Y ] M_{X,Y}=[I(X),I(Y)]=[FI(X),Y] M X , Y = [ I ( X ) , I ( Y )] = [ F I ( X ) , Y ]
Lemma 3.3. After relabeling f , g , h f,g,h f , g , h I ( 1 ) I(\mathbf 1) I ( 1 ) 1 , A , B , C 0 , C g , C h \mathbf 1,A,B,C_0,C_g,C_h 1 , A , B , C 0 , C g , C h Z dim Z F ( Z ) 1 1 1 A 1 + k d 1 ⊕ k ρ B q 1 ⊕ f ⊕ k ρ C 0 2 q 1 ⊕ f ⊕ g ⊕ h ⊕ 2 k ρ C g 2 q 1 ⊕ g ⊕ t ⊕ 2 k ρ C h 2 q 1 ⊕ h ⊕ t ⊕ 2 k ρ
\begin{array}{c|c|l}
Z&\dim Z&F(Z)\\ \hline
\mathbf 1&1&\mathbf 1\\
A&1+kd&\mathbf 1\oplus k \rho\\
B&q&\mathbf 1\oplus f\oplus k \rho\\
C_0&2q&\mathbf 1\oplus f\oplus g\oplus h\oplus 2k \rho\\
C_g&2q&\mathbf 1\oplus g\oplus t\oplus 2k \rho\\
C_h&2q&\mathbf 1\oplus h\oplus t\oplus 2k \rho
\end{array}
Z 1 A B C 0 C g C h dim Z 1 1 + k d q 2 q 2 q 2 q F ( Z ) 1 1 ⊕ k ρ 1 ⊕ f ⊕ k ρ 1 ⊕ f ⊕ g ⊕ h ⊕ 2 k ρ 1 ⊕ g ⊕ t ⊕ 2 k ρ 1 ⊕ h ⊕ t ⊕ 2 k ρ
Proof. The ring K 0 ( C ) K_0(\mathcal C) K 0 ( C ) χ + , χ − \chi_+,\chi_- χ + , χ − 1 1 1 G G G − 2 -2 − 2 t t t 0 0 0 ρ \rho ρ G G G t , ρ t,\rho t , ρ
8 + d 2 , 8 + d − 2 , 8 , 4 , 4 , 4. 8+d^2,\quad 8+d_-^2,\quad 8,\quad 4,\quad 4,\quad 4. 8 + d 2 , 8 + d − 2 , 8 , 4 , 4 , 4. The formal-codegree theorem [39 Theorem 2.13] gives the stated dimensions and multiplicities. In particular, every listed forgetful image contains 1 \mathbf 1 1 F ( A ) ≅ 1 ⊕ k ρ F(A)\cong \mathbf 1\oplus k \rho F ( A ) ≅ 1 ⊕ k ρ F ( B ) ≅ 1 ⊕ f ⊕ k ρ F(B)\cong \mathbf 1\oplus f\oplus k \rho F ( B ) ≅ 1 ⊕ f ⊕ k ρ f ≠ 1 f\ne \mathbf 1 f = 1
Each of the three objects of dimension 2 q = 4 + 2 k d 2q=4+2kd 2 q = 4 + 2 k d t t t 1 \mathbf 1 1 t t t F ( I ( 1 ) ) F(I(\mathbf 1)) F ( I ( 1 )) 2 2 2
1 ⊕ x ⊕ t ⊕ 2 k ρ , 1 ⊕ y ⊕ t ⊕ 2 k ρ , x , y ∈ { f , g , h } . \mathbf 1\oplus x\oplus t\oplus 2k \rho,\qquad \mathbf 1\oplus y\oplus t\oplus 2k \rho,
\qquad x,y\in\{f,g,h\}. 1 ⊕ x ⊕ t ⊕ 2 k ρ , 1 ⊕ y ⊕ t ⊕ 2 k ρ , x , y ∈ { f , g , h } . Each nonidentity element of G G G 2 2 2 F ( I ( 1 ) ) F(I(\mathbf 1)) F ( I ( 1 )) x = y = f x=y=f x = y = f g , h g,h g , h x , y x,y x , y { x , y } = { g , h } , { f , g } , { g , g } \{x,y\}=\{g,h\},\{f,g\},\{g,g\} { x , y } = { g , h } , { f , g } , { g , g }
For { x , y } = { g , g } \{x,y\}=\{g,g\} { x , y } = { g , g } 1 ⊕ f ⊕ 2 h ⊕ 2 k ρ \mathbf 1\oplus f\oplus 2 h\oplus 2k \rho 1 ⊕ f ⊕ 2 h ⊕ 2 k ρ f , g , h , t f,g,h,t f , g , h , t
( 4 2 0 2 2 4 2 0 0 2 2 2 2 0 2 10 ) . \begin{pmatrix}
4&2&0&2\\
2&4&2&0\\
0&2&2&2\\
2&0&2&10
\end{pmatrix}. 4 2 0 2 2 4 2 0 0 2 2 2 2 0 2 10 . Its determinant is − 16 -16 − 16
For { x , y } = { f , g } \{x,y\}=\{f,g\} { x , y } = { f , g } 1 ⊕ g ⊕ 2 h ⊕ 2 k ρ \mathbf 1\oplus g\oplus 2 h\oplus 2k \rho 1 ⊕ g ⊕ 2 h ⊕ 2 k ρ
( 4 2 2 1 2 4 0 1 2 0 2 2 1 1 2 10 ) . \begin{pmatrix}
4&2&2&1\\
2&4&0&1\\
2&0&2&2\\
1&1&2&10
\end{pmatrix}. 4 2 2 1 2 4 0 1 2 0 2 2 1 1 2 10 . These are inner products of nonnegative integer columns. The h h h 1 1 1 g g g f f f 2 2 2 2 2 2 4 4 4 2 2 2 g g g f f f 1 1 1 h h h ⟨ f , t ⟩ ≥ ⟨ h , t ⟩ = 2 \langle f,t\rangle\geq\langle h,t\rangle=2 ⟨ f , t ⟩ ≥ ⟨ h , t ⟩ = 2 ⟨ f , t ⟩ = 1 \langle f,t\rangle=1 ⟨ f , t ⟩ = 1
Only { x , y } = { g , h } \{x,y\}=\{g,h\} { x , y } = { g , h } F ( I ( 1 ) ) F(I(\mathbf 1)) F ( I ( 1 )) (3.3) . ◻
Proposition 3.4. Every simple object of Z ( C ) \mathcal Z(\mathcal C) Z ( C ) 1 1 1 − 1 -1 − 1 I ( 1 ) I(\mathbf 1) I ( 1 ) 1 1 1 P 1 , P 2 P_1,P_2 P 1 , P 2 F ( P i ) ≅ f ⊕ b i ρ , b 1 , b 2 ∈ Z ≥ 0 , b 1 + b 2 = k .
F(P_i)\cong f\oplus b_i \rho,\qquad b_1,b_2\in\mathbb Z_{\geq0},\qquad b_1+b_2=k.
F ( P i ) ≅ f ⊕ b i ρ , b 1 , b 2 ∈ Z ≥ 0 , b 1 + b 2 = k .
Proof. The part of I ( g ) I(g) I ( g ) I ( 1 ) I(\mathbf 1) I ( 1 ) C 0 ⊕ C g C_0\oplus C_g C 0 ⊕ C g 4 q = Δ / 2 4q=\Delta/2 4 q = Δ/2 1 1 1 (2.4) , twist trace − 4 q -4q − 4 q − 1 -1 − 1 I ( h ) I(h) I ( h ) Vec G \operatorname{Vec}_G Vec G ν 2 ( f ) = 1 \nu_2(f)=1 ν 2 ( f ) = 1 Tr ( θ I ( f ) 2 ) = Δ = dim I ( f ) \operatorname{Tr}(\theta^2_{I(f)})=\Delta=\dim I(f) Tr ( θ I ( f ) 2 ) = Δ = dim I ( f ) I ( f ) I(f) I ( f ) ± 1 \pm1 ± 1 Δ / 2 = 4 q \Delta/2=4q Δ/2 = 4 q I ( 1 ) I(\mathbf 1) I ( 1 ) B ⊕ C 0 B\oplus C_0 B ⊕ C 0 3 q 3q 3 q q q q
Subtracting the rows in (3.3) from (3.2) leaves, on f , g , h , t f,g,h,t f , g , h , t
R = ( 4 1 1 2 1 4 1 1 1 1 4 1 2 1 1 10 ) . \begin{equation}
R=
\begin{pmatrix}
4&1&1&2\\
1&4&1&1\\
1&1&4&1\\
2&1&1&10
\end{pmatrix}. \tag*{(3.5)}
\end{equation} R = 4 1 1 2 1 4 1 1 1 1 4 1 2 1 1 10 . (3.5) Every entry in each of the residual f , g , h f,g,h f , g , h 0 0 0 1 1 1 2 2 2 4 4 4 1 1 1
Here and in what follows, “contains” refers to the underlying object after applying the forgetful functor F F F I ( f ) I(f) I ( f ) Z Z Z I ( f ) I(f) I ( f ) g g g F ( Z ) F(Z) F ( Z )
[ I ( g ) , Z ] = [ F ( Z ) , g ] > 0. [I(g),Z]=[F(Z),g]>0. [ I ( g ) , Z ] = [ F ( Z ) , g ] > 0. Thus Z Z Z I ( g ) I(g) I ( g ) I ( g ) I(g) I ( g ) − 1 -1 − 1 h h h f f f
F ( Z ) ≅ f ⊕ u Z t ⊕ b Z ρ . F(Z)\cong f\oplus u_Z t\oplus b_Z \rho . F ( Z ) ≅ f ⊕ u Z t ⊕ b Z ρ . The total dimension of all residual positive-twist summands is 2 + k d 2+kd 2 + k d 1 + 2 u Z 1+2u_Z 1 + 2 u Z 2 2 2 u Z = 0 u_Z=0 u Z = 0 d d d (3.4) . ◻
Notation 3.5. For each Z ∈ Irr ( Z ( C ) ) Z\in\operatorname{Irr}(\mathcal Z(\mathcal C)) Z ∈ Irr ( Z ( C ))
F ( Z ) ≅ ⨁ x ∈ G m x ( Z ) x ⊕ u Z t ⊕ b Z ρ , s Z = ∑ x ∈ G m x ( Z ) , a Z = s Z + 2 u Z . F(Z)\cong \bigoplus_{x\in G}m_x(Z) x\oplus u_Z t\oplus b_Z \rho,\qquad
s_Z=\sum_{x\in G}m_x(Z),\qquad a_Z=s_Z+2u_Z. F ( Z ) ≅ x ∈ G ⨁ m x ( Z ) x ⊕ u Z t ⊕ b Z ρ , s Z = x ∈ G ∑ m x ( Z ) , a Z = s Z + 2 u Z . Thus dim Z = a Z + b Z d \dim Z=a_Z+b_Zd dim Z = a Z + b Z d Irr ( Z ( C ) ) \operatorname{Irr}(\mathcal Z(\mathcal C)) Irr ( Z ( C ))
U = { Z : s Z > 0 } , V = { Z : s Z = 0 , u Z > 0 } , W = { Z : s Z = u Z = 0 } . \mathscr U=\{Z:s_Z>0\},\quad
\mathscr V=\{Z:s_Z=0,\ u_Z>0\},\quad
\mathscr W=\{Z:s_Z=u_Z=0\}. U = { Z : s Z > 0 } , V = { Z : s Z = 0 , u Z > 0 } , W = { Z : s Z = u Z = 0 } . The summands in W \mathscr W W b Z ρ b_Z \rho b Z ρ b Z > 0 b_Z>0 b Z > 0 S \mathscr S S I ( X ) ∣ S I(X)|_{\mathscr S} I ( X ) ∣ S I ( X ) I(X) I ( X )
Lemma 3.6. Set T = ∑ Z ∈ U u Z 2 T=\sum_{Z\in\mathscr U}u_Z^2 T = ∑ Z ∈ U u Z 2 T ∈ { 4 , 8 } T\in\{4,8\} T ∈ { 4 , 8 } D \mathcal D D U \mathscr U U D -parts T I f ⊕ g ⊕ t , f ⊕ h ⊕ t , g ⊕ h , g , g , h , h 4 I I f ⊕ g ⊕ h , f ⊕ 2 t , g ⊕ t , h ⊕ t , g , g , h , h 8 I I I f ⊕ g ⊕ h ⊕ t , f ⊕ t , g , g , g , h , h , h 4
\begin{array}{c|l|c}
&\mathcal D\text{-parts}&T\\ \hline
\mathrm{I}& f\oplus g\oplus t,\ f\oplus h\oplus t,\ g\oplus h,\ g,\ g,\ h,\ h&4\\
\mathrm{II}& f\oplus g\oplus h,\ f\oplus 2 t,\ g\oplus t,\ h\oplus t,\ g,\ g,\ h,\ h&8\\
\mathrm{III}& f\oplus g\oplus h\oplus t,\ f\oplus t,\ g,\ g,\ g,\ h,\ h,\ h&4
\end{array}
I II III D -parts f ⊕ g ⊕ t , f ⊕ h ⊕ t , g ⊕ h , g , g , h , h f ⊕ g ⊕ h , f ⊕ 2 t , g ⊕ t , h ⊕ t , g , g , h , h f ⊕ g ⊕ h ⊕ t , f ⊕ t , g , g , g , h , h , h T 4 8 4 ρ \rho ρ
Proof. Remove the two rows in (3.4) from (3.5) . The remaining invertible columns have 2 , 4 , 4 2,4,4 2 , 4 , 4 1 1 1 1 1 1 f , g , h f,g,h f , g , h f , g f,g f , g f , h f,h f , h g , h g,h g , h
In the latter case the two rows containing f f f t t t 2 2 2 1 1 1 g t gt g t h t ht h t 1 1 1 1 1 1 t t t 0 0 0
In the former case let α \alpha α t t t g t gt g t h t ht h t α ∈ { 0 , 1 } \alpha\in\{0,1\} α ∈ { 0 , 1 } f f f t t t 2 − α 2-\alpha 2 − α α = 0 \alpha=0 α = 0 g g g h h h t t t 1 1 1 α = 1 \alpha=1 α = 1 0 0 0
The rows C g , C h C_g,C_h C g , C h I ( 1 ) I(\mathbf 1) I ( 1 ) 2 2 2 T T T P 1 , P 2 P_1,P_2 P 1 , P 2 0 0 0 2 , 6 , 2 2,6,2 2 , 6 , 2
Lemma 3.7. Choose a cyclotomic field containing the modular data of Z ( C ) \mathcal Z(\mathcal C) Z ( C ) σ \sigma σ Q ( d ) \mathbb Q(d) Q ( d ) π = σ ^ \pi=\widehat{\sigma} π = σ a Z > 0 a_Z>0 a Z > 0 m x ( π ( Z ) ) = m x ( Z ) , u π ( Z ) = u Z , b π ( Z ) = k a Z − b Z .
m_x(\pi(Z))=m_x(Z),\quad u_{\pi(Z)}=u_Z,\quad
b_{\pi(Z)}=ka_Z-b_Z.
m x ( π ( Z )) = m x ( Z ) , u π ( Z ) = u Z , b π ( Z ) = k a Z − b Z . U \mathscr U U V \mathscr V V π \pi π θ π ( Z ) = σ 2 ( θ Z )
\theta_{\pi(Z)}=\sigma^2(\theta_Z)
θ π ( Z ) = σ 2 ( θ Z ) Z Z Z θ π ( Z ) = θ Z \theta_{\pi(Z)}=\theta_Z θ π ( Z ) = θ Z θ Z 8 = 1 \theta_Z^8=1 θ Z 8 = 1
Proof. Use the normalization S = S / Δ S=\mathsf S/\Delta S = S /Δ § 2 . Galois symmetry gives signs ϵ Z ∈ { ± 1 } \epsilon_Z\in\{\pm1\} ϵ Z ∈ { ± 1 }
σ ( S X , Z ) = ϵ Z S X , π ( Z ) . \begin{equation}
\sigma(S_{X,Z})=\epsilon_ZS_{X,\pi(Z)}. \tag*{(3.9)}
\end{equation} σ ( S X , Z ) = ϵ Z S X , π ( Z ) . (3.9) The twist identity (3.8) holds for the canonical modular representation of a Drinfeld center [27 Section 4.1 and Lemma 4.2] . One has σ ( d ) = 8 k − d \sigma(d)=8k-d σ ( d ) = 8 k − d Δ / σ ( Δ ) = 1 + k d \Delta/\sigma(\Delta)=1+kd Δ/ σ ( Δ ) = 1 + k d (3.9) therefore gives
ϵ Z dim π ( Z ) = ( 1 + k d ) σ ( a Z + b Z d ) = a Z + ( k a Z − b Z ) d . \epsilon_Z\dim\pi(Z)
=(1+kd)\sigma(a_Z+b_Zd)=a_Z+(ka_Z-b_Z)d. ϵ Z dim π ( Z ) = ( 1 + k d ) σ ( a Z + b Z d ) = a Z + ( k a Z − b Z ) d . If a Z > 0 a_Z>0 a Z > 0 + 1 +1 + 1 1 , d 1,d 1 , d a π ( Z ) = − a Z < 0 a_{\pi(Z)}=-a_Z<0 a π ( Z ) = − a Z < 0 a π ( Z ) = a Z a_{\pi(Z)}=a_Z a π ( Z ) = a Z b π ( Z ) = k a Z − b Z b_{\pi(Z)}=ka_Z-b_Z b π ( Z ) = k a Z − b Z
To recover the full D \mathcal D D A ψ A_\psi A ψ I ( 1 ) I(\mathbf 1) I ( 1 ) [39 Section 2.3 and Theorem 2.13] :
S A ψ , Z = ψ ( F ( Z ) ) f ψ . S_{A_\psi,Z}=\frac{\psi(F(Z))}{f_\psi}. S A ψ , Z = f ψ ψ ( F ( Z )) . For the character associated with B B B S B , Z = ( s Z − 2 u Z ) / 8 S_{B,Z}=(s_Z-2u_Z)/8 S B , Z = ( s Z − 2 u Z ) /8 η : G → { ± 1 } \eta:G\to\{\pm1\} η : G → { ± 1 } S A η , Z = 1 4 ∑ x ∈ G η ( x ) m x ( Z ) S_{A_\eta,Z}=\frac14\sum_{x\in G}\eta(x)m_x(Z) S A η , Z = 4 1 ∑ x ∈ G η ( x ) m x ( Z ) ϵ Z = 1 \epsilon_Z=1 ϵ Z = 1 (3.9) shows that they are unchanged under π \pi π a Z = s Z + 2 u Z a_Z=s_Z+2u_Z a Z = s Z + 2 u Z s Z , u Z s_Z,u_Z s Z , u Z G G G m x ( Z ) m_x(Z) m x ( Z ) (3.7) . Finally every automorphism of the field of eighth roots acts by an odd power, whose square is 1 1 1 8 8 8
Proposition 3.8. With Notation 3.5 , ∑ U u Z 2 = 4 , ∑ V u Z 2 = 8 , ∑ V u Z b Z = 8 k .
\sum_{\mathscr U}u_Z^2=4,\qquad
\sum_{\mathscr V}u_Z^2=8,\qquad
\sum_{\mathscr V}u_Zb_Z=8k.
∑ U u Z 2 = 4 , ∑ V u Z 2 = 8 , ∑ V u Z b Z = 8 k . ν 2 ( t ) = 1 , dim I ( t ) ∣ V = Δ , Tr ( θ I ( t ) ∣ V ) = Tr ( θ I ( t ) ∣ V 2 ) = 0.
\nu_2(t)=1,\qquad
\dim I(t)|_{\mathscr V}=\Delta,\qquad
\operatorname{Tr}(\theta_{I(t)|_{\mathscr V}})=
\operatorname{Tr}(\theta^2_{I(t)|_{\mathscr V}})=0.
ν 2 ( t ) = 1 , dim I ( t ) ∣ V = Δ , Tr ( θ I ( t ) ∣ V ) = Tr ( θ I ( t ) ∣ V 2 ) = 0. U ∪ V \mathscr U\cup\mathscr V U ∪ V 8 8 8
Proof. The matrix (3.2) gives ∑ U s Z u Z = 8 \sum_\mathscr Us_Zu_Z=8 ∑ U s Z u Z = 8 π \pi π U \mathscr U U
2 ∑ U u Z b Z = k ∑ U u Z a Z = k ( 8 + 2 T ) . 2\sum_\mathscr Uu_Zb_Z
=k\sum_\mathscr Uu_Za_Z=k(8+2T). 2 U ∑ u Z b Z = k U ∑ u Z a Z = k ( 8 + 2 T ) . Consequently,
dim I ( t ) ∣ U = ∑ U u Z ( a Z + b Z d ) = 8 + 2 T + ( 4 k + k T ) d = ( 4 + T ) q . \begin{equation}
\dim I(t)|_\mathscr U
=\sum_\mathscr Uu_Z(a_Z+b_Zd)
=8+2T+(4k+kT)d=(4+T)q. \tag*{(3.12)}
\end{equation} dim I ( t ) ∣ U = U ∑ u Z ( a Z + b Z d ) = 8 + 2 T + ( 4 k + k T ) d = ( 4 + T ) q . (3.12) The positive-twist part of I ( t ) ∣ U I(t)|_\mathscr U I ( t ) ∣ U C g , C h C_g,C_h C g , C h 4 q 4q 4 q dim I ( t ) = 16 q \dim I(t)=16q dim I ( t ) = 16 q
dim I ( t ) ∣ V = ( 12 − T ) q , Tr ( θ I ( t ) ∣ V ) = ( T − 4 ) q . \begin{equation}
\dim I(t)|_\mathscr V=(12-T)q,\qquad
\operatorname{Tr}(\theta_{I(t)|_\mathscr V})=(T-4)q. \tag*{(3.13)}
\end{equation} dim I ( t ) ∣ V = ( 12 − T ) q , Tr ( θ I ( t ) ∣ V ) = ( T − 4 ) q . (3.13) If T = 8 T=8 T = 8 (3.13) both equal 4 q 4q 4 q I ( t ) ∣ V I(t)|_\mathscr V I ( t ) ∣ V 1 1 1 I ( t ) ∣ U I(t)|_\mathscr U I ( t ) ∣ U 1 1 1 Tr ( θ I ( t ) 2 ) = 2 Δ \operatorname{Tr}(\theta^2_{I(t)})=2\Delta Tr ( θ I ( t ) 2 ) = 2Δ ν 2 ( t ) ∈ { ± 1 } \nu_2(t)\in\{\pm1\} ν 2 ( t ) ∈ { ± 1 } 3.6 therefore gives T = 4 T=4 T = 4
Now both parts of I ( t ) I(t) I ( t ) Δ \Delta Δ V \mathscr V V 0 0 0 U \mathscr U U Δ \Delta Δ ν 2 ( t ) = − 1 \nu_2(t)=-1 ν 2 ( t ) = − 1 − 2 Δ -2\Delta − 2Δ Δ \Delta Δ ν 2 ( t ) = 1 \nu_2(t)=1 ν 2 ( t ) = 1 0 0 0 ( t , t ) (t,t) ( t , t ) ( t , ρ ) (t,\rho) ( t , ρ ) (3.2) now give ∑ V u Z 2 = 12 − 4 = 8 \sum_\mathscr Vu_Z^2=12-4=8 ∑ V u Z 2 = 12 − 4 = 8 ∑ V u Z b Z = 16 k − ( 4 k + 4 k ) = 8 k \sum_\mathscr Vu_Zb_Z=16k-(4k+4k)=8k ∑ V u Z b Z = 16 k − ( 4 k + 4 k ) = 8 k
The pivotal structure restricts to the canonical positive structure on D \mathcal D D [40 Theorem 3.4 and the formula immediately following it] gives
ν 4 ( t ) = 1 2 ∑ x ∈ G χ ( x , x ) − 1 ∈ { 0 , 2 } , \nu_4(t)=\frac12\sum_{x\in G}\chi(x,x)^{-1}\in\{0,2\}, ν 4 ( t ) = 2 1 x ∈ G ∑ χ ( x , x ) − 1 ∈ { 0 , 2 } , where χ \chi χ D \mathcal D D G G G 2 2 2 x ↦ χ ( x , x ) x\mapsto\chi(x,x) x ↦ χ ( x , x ) { ± 1 } \{\pm1\} { ± 1 } 0 0 0 4 4 4 U \mathscr U U I ( t ) I(t) I ( t ) Δ \Delta Δ
Tr ( θ I ( t ) ∣ V 4 ) = ( ν 4 ( t ) − 1 ) Δ = ± Δ . \operatorname{Tr}(\theta^4_{I(t)|_\mathscr V})=(\nu_4(t)-1)\Delta=\pm\Delta. Tr ( θ I ( t ) ∣ V 4 ) = ( ν 4 ( t ) − 1 ) Δ = ± Δ. As dim I ( t ) ∣ V = Δ \dim I(t)|_\mathscr V=\Delta dim I ( t ) ∣ V = Δ θ Z 8 = 1 \theta_Z^8=1 θ Z 8 = 1 Z ∈ V Z\in\mathscr V Z ∈ V U \mathscr U U 3.4 . ◻
Lemma 3.9. Let Q = ∑ Z ∈ U b Z 2 Q=\sum_{Z\in\mathscr U}b_Z^2 Q = ∑ Z ∈ U b Z 2 ∑ U a Z 2 = 96 , ∑ U a Z b Z = 48 k , Q ≥ 49 2 k 2 .
\sum_\mathscr Ua_Z^2=96,\qquad
\sum_\mathscr Ua_Zb_Z=48k,\qquad
Q\geq\frac{49}{2}k^2.
∑ U a Z 2 = 96 , ∑ U a Z b Z = 48 k , Q ≥ 2 49 k 2 .
Proof. The invertible block of (3.2) gives ∑ U s Z 2 = 4 ⋅ 6 + 12 ⋅ 2 = 48 \sum_\mathscr Us_Z^2=4\cdot6+12\cdot2=48 ∑ U s Z 2 = 4 ⋅ 6 + 12 ⋅ 2 = 48 ∑ U s Z u Z = 8 \sum_\mathscr Us_Zu_Z=8 ∑ U s Z u Z = 8 ∑ U u Z 2 = 4 \sum_\mathscr Uu_Z^2=4 ∑ U u Z 2 = 4 3.8 . Therefore
∑ U a Z 2 = ∑ U ( s Z + 2 u Z ) 2 = 48 + 32 + 16 = 96. \sum_\mathscr Ua_Z^2
=\sum_\mathscr U(s_Z+2u_Z)^2=48+32+16=96. U ∑ a Z 2 = U ∑ ( s Z + 2 u Z ) 2 = 48 + 32 + 16 = 96. Lemma 3.7 implies 2 ∑ U a Z b Z = k ∑ U a Z 2 = 96 k 2\sum_\mathscr Ua_Zb_Z=k\sum_\mathscr Ua_Z^2=96k 2 ∑ U a Z b Z = k ∑ U a Z 2 = 96 k
Q = 24 k 2 + ∑ U ( b Z − k a Z 2 ) 2 . Q=24k^2+\sum_\mathscr U\left(b_Z-\frac{k a_Z}{2}\right)^2. Q = 24 k 2 + U ∑ ( b Z − 2 k a Z ) 2 . The objects 1 , A \mathbf 1,A 1 , A ( a Z , b Z ) = ( 1 , 0 ) , ( 1 , k ) (a_Z,b_Z)=(1,0),(1,k) ( a Z , b Z ) = ( 1 , 0 ) , ( 1 , k ) k 2 / 2 k^2/2 k 2 /2
Proposition 3.10. Put p = b 1 2 + b 2 2 p=b_1^2+b_2^2 p = b 1 2 + b 2 2 b 1 , b 2 b_1,b_2 b 1 , b 2 (3.4) , and let ν = ν 2 ( ρ ) ∈ { ± 1 } \nu=\nu_2(\rho)\in\{\pm1\} ν = ν 2 ( ρ ) ∈ { ± 1 } ζ 1 , … , ζ L \zeta_1,\ldots,\zeta_L ζ 1 , … , ζ L L = 16 + 56 k 2 − Q ≤ 16 + 63 2 k 2 , ∑ i = 1 L ζ i = Q − 24 k 2 − 2 p − k r , ∑ i = 1 L ζ i 2 = 24 k 2 − Q + ( 2 ν − 6 k ) r . \begin{aligned}
L&=16+56k^2-Q\leq16+\frac{63}{2}k^2,\\
\sum_{i=1}^{L}\zeta_i
&=Q-24k^2-2p-kr,\\
\sum_{i=1}^{L}\zeta_i^2
&=24k^2-Q+(2\nu-6k)r.\end{aligned} L i = 1 ∑ L ζ i i = 1 ∑ L ζ i 2 = 16 + 56 k 2 − Q ≤ 16 + 2 63 k 2 , = Q − 24 k 2 − 2 p − k r , = 24 k 2 − Q + ( 2 ν − 6 k ) r .
Proof. By Lemma 3.7 and Proposition 3.8 , the Galois reflection preserves twists on U ∪ V \mathscr U\cup\mathscr V U ∪ V V \mathscr V V a Z = 2 u Z a_Z=2u_Z a Z = 2 u Z ζ \zeta ζ
∑ Z ∈ V θ Z = ζ u Z b Z = k ∑ Z ∈ V θ Z = ζ u Z 2 . \sum_{\substack{Z\in\mathscr V\\\theta_Z=\zeta}}u_Zb_Z
=k\sum_{\substack{Z\in\mathscr V\\\theta_Z=\zeta}}u_Z^2. Z ∈ V θ Z = ζ ∑ u Z b Z = k Z ∈ V θ Z = ζ ∑ u Z 2 . For j = 1 , 2 j=1,2 j = 1 , 2 (3.11) therefore give
0 = ∑ V u Z ( 2 u Z + b Z d ) θ Z j = ( 2 + k d ) ∑ V u Z 2 θ Z j . 0=\sum_\mathscr Vu_Z(2u_Z+b_Zd)\theta_Z^j
=(2+kd)\sum_\mathscr Vu_Z^2\theta_Z^j. 0 = V ∑ u Z ( 2 u Z + b Z d ) θ Z j = ( 2 + k d ) V ∑ u Z 2 θ Z j . It follows that
∑ V u Z 2 θ Z j = ∑ V u Z b Z θ Z j = 0 ( j = 1 , 2 ) . \begin{equation}
\sum_\mathscr Vu_Z^2\theta_Z^j
=\sum_\mathscr Vu_Zb_Z\theta_Z^j=0
\qquad (j=1,2). \tag*{(3.16)}
\end{equation} V ∑ u Z 2 θ Z j = V ∑ u Z b Z θ Z j = 0 ( j = 1 , 2 ) . (3.16) For Z ∈ V Z\in\mathscr V Z ∈ V δ Z = b Z − k u Z ∈ Z \delta_Z=b_Z-ku_Z\in\mathbb Z δ Z = b Z − k u Z ∈ Z
1 d Tr ( θ I ( ρ ) ∣ V j ) = ∑ V b Z 2 θ Z j = ∑ V δ Z 2 θ Z j ( j = 1 , 2 ) , \begin{equation}
\frac1d\operatorname{Tr}(\theta^j_{I(\rho)|_\mathscr V})
=\sum_\mathscr Vb_Z^2\theta_Z^j
=\sum_\mathscr V\delta_Z^2\theta_Z^j
\qquad (j=1,2), \tag*{(3.17)}
\end{equation} d 1 Tr ( θ I ( ρ ) ∣ V j ) = V ∑ b Z 2 θ Z j = V ∑ δ Z 2 θ Z j ( j = 1 , 2 ) , (3.17) where both equalities use (3.16) .
Take b Z 2 b_Z^2 b Z 2 θ Z \theta_Z θ Z Z ∈ W Z\in\mathscr W Z ∈ W δ Z 2 \delta_Z^2 δ Z 2 Z ∈ V Z\in\mathscr V Z ∈ V ρ ρ \rho\rho ρρ (3.2) and (3.10) , its cardinality is
L = ∑ W b Z 2 + ∑ V ( b Z − k u Z ) 2 = 16 + 64 k 2 − Q − 2 k ( 8 k ) + k 2 ( 8 ) = 16 + 56 k 2 − Q . \begin{align*}
L&=\sum_\mathscr Wb_Z^2+\sum_\mathscr V(b_Z-ku_Z)^2\\
&=16+64k^2-Q-2k(8k)+k^2(8)
=16+56k^2-Q.
\end{align*} L = W ∑ b Z 2 + V ∑ ( b Z − k u Z ) 2 = 16 + 64 k 2 − Q − 2 k ( 8 k ) + k 2 ( 8 ) = 16 + 56 k 2 − Q . Lemma 3.9 proves the inequality in (3.15) .
The positive-twist objects of U \mathscr U U 1 , A , B , C 0 , C g , C h , P 1 , P 2 \mathbf 1,A,B,C_0,C_g,C_h,P_1,P_2 1 , A , B , C 0 , C g , C h , P 1 , P 2 a Z 2 a_Z^2 a Z 2 1 + 1 + 4 + 16 + 16 + 16 + 1 + 1 = 56 1+1+4+16+16+16+1+1=56 1 + 1 + 4 + 16 + 16 + 16 + 1 + 1 = 56 ∑ U θ Z a Z 2 = 56 − ( 96 − 56 ) = 16 \sum_\mathscr U\theta_Za_Z^2=56-(96-56)=16 ∑ U θ Z a Z 2 = 56 − ( 96 − 56 ) = 16
∑ U θ Z a Z b Z = k 2 ∑ U θ Z a Z 2 = 8 k . \sum_\mathscr U\theta_Za_Zb_Z
=\frac{k}{2}\sum_\mathscr U\theta_Za_Z^2=8k. U ∑ θ Z a Z b Z = 2 k U ∑ θ Z a Z 2 = 8 k . The positive-twist b Z 2 b_Z^2 b Z 2 14 k 2 + p 14k^2+p 14 k 2 + p ∑ U θ Z b Z 2 = 28 k 2 + 2 p − Q \sum_\mathscr U\theta_Zb_Z^2=28k^2+2p-Q ∑ U θ Z b Z 2 = 28 k 2 + 2 p − Q I ( ρ ) I(\rho) I ( ρ )
0 = 8 k + ( 28 k 2 + 2 p − Q ) d + d ∑ i = 1 L ζ i . 0=8k+(28k^2+2p-Q)d+d\sum_{i=1}^L\zeta_i. 0 = 8 k + ( 28 k 2 + 2 p − Q ) d + d i = 1 ∑ L ζ i . Since 8 / d = d − 8 k 8/d=d-8k 8/ d = d − 8 k d = 4 k + r d=4k+r d = 4 k + r eq:first-sum .
For the second trace, every twist in U \mathscr U U 1 1 1 3.9 and (3.17) give
ν Δ = 48 k + Q d + d ∑ i = 1 L ζ i 2 . \nu\Delta=48k+Qd+d\sum_{i=1}^L\zeta_i^2. ν Δ = 48 k + Q d + d i = 1 ∑ L ζ i 2 . Now Δ / d = 2 r \Delta/d=2r Δ/ d = 2 r 48 k / d = 6 k r − 24 k 2 48k/d=6kr-24k^2 48 k / d = 6 k r − 24 k 2 eq:second-sum . ◻
For a positive integer M M M
v 2 ( M ) = max { r ∈ Z ≥ 0 : 2 r ∣ M } . v_2(M)=\max\{r\in\mathbb Z_{\geq0}:2^r\mid M\}. v 2 ( M ) = max { r ∈ Z ≥ 0 : 2 r ∣ M } . Thus v 2 ( M ) v_2(M) v 2 ( M ) 2 2 2 M M M
Lemma 3.11. Let c > 1 c>1 c > 1 ζ 1 , … , ζ L \zeta_1,\ldots,\zeta_L ζ 1 , … , ζ L ∑ i = 1 L ζ i = A + B c , ∑ i = 1 L ζ i 2 = A ′ + B ′ c , A , B , A ′ , B ′ ∈ Z . \sum_{i=1}^L\zeta_i=A+B\sqrt c,\qquad
\sum_{i=1}^L\zeta_i^2=A'+B'\sqrt c,
\qquad A,B,A',B'\in\mathbb Z. ∑ i = 1 L ζ i = A + B c , ∑ i = 1 L ζ i 2 = A ′ + B ′ c , A , B , A ′ , B ′ ∈ Z . L ≥ φ ( 2 c ) ( ∣ B ∣ + ∣ B ′ ∣ ) .
L\geq\varphi(2c)(|B|+|B'|).
L ≥ φ ( 2 c ) ( ∣ B ∣ + ∣ B ′ ∣ ) .
Proof. Let K = Q ( c ) K=\mathbb Q(\sqrt c) K = Q ( c ) E E E K K K H = Gal ( E / K ) H=\operatorname{Gal}(E/K) H = Gal ( E / K ) h = ∣ H ∣ h=|H| h = ∣ H ∣ K K K 4 c 4c 4 c c c c c = 2 c 0 c=2c_0 c = 2 c 0 c 0 c_0 c 0 4 c = 8 c 0 4c=8c_0 4 c = 8 c 0
v 2 ( 4 c ) = v 2 ( 8 c 0 ) = 3. v_2(4c)=v_2(8c_0)=3. v 2 ( 4 c ) = v 2 ( 8 c 0 ) = 3. For a root ζ \zeta ζ n n n T ( ζ ) = ∑ τ ∈ H τ ( ζ ) ∈ K \mathcal T(\zeta)=\sum_{\tau\in H}\tau(\zeta)\in K T ( ζ ) = ∑ τ ∈ H τ ( ζ ) ∈ K K ⊈ Q ( ζ ) K\not\subseteq\mathbb Q(\zeta) K ⊆ Q ( ζ ) K ∩ Q ( ζ ) = Q K\cap\mathbb Q(\zeta)=\mathbb Q K ∩ Q ( ζ ) = Q H H H Q ( ζ ) \mathbb Q(\zeta) Q ( ζ ) T ( ζ ) \mathcal T(\zeta) T ( ζ ) K ⊆ Q ( ζ ) K\subseteq\mathbb Q(\zeta) K ⊆ Q ( ζ ) 4 c ∣ n 4c\mid n 4 c ∣ n v 2 ( n ) ≥ 4 v_2(n)\geq4 v 2 ( n ) ≥ 4 ζ ↦ ζ 1 + n / 2 = − ζ \zeta\mapsto\zeta^{1+n/2}=-\zeta ζ ↦ ζ 1 + n /2 = − ζ K K K 4 c ∣ n / 2 4c\mid n/2 4 c ∣ n /2 H H H T ( ζ ) = 0 \mathcal T(\zeta)=0 T ( ζ ) = 0 c \sqrt c c T ( ζ ) \mathcal T(\zeta) T ( ζ ) v 2 ( n ) = 3 v_2(n)=3 v 2 ( n ) = 3
Let J 3 = { i : v 2 ( ord ζ i ) = 3 } J_3=\{i:v_2(\operatorname{ord}\zeta_i)=3\} J 3 = { i : v 2 ( ord ζ i ) = 3 } J 4 = { i : v 2 ( ord ζ i ) = 4 } J_4=\{i:v_2(\operatorname{ord}\zeta_i)=4\} J 4 = { i : v 2 ( ord ζ i ) = 4 } ord ( ζ i 2 ) = ord ( ζ i ) / gcd ( 2 , ord ζ i ) \operatorname{ord}(\zeta_i^2)=\operatorname{ord}(\zeta_i)/\gcd(2,\operatorname{ord}\zeta_i) ord ( ζ i 2 ) = ord ( ζ i ) / g cd( 2 , ord ζ i )
∑ i ∈ J 3 T ( ζ i ) = a + h B c , ∑ i ∈ J 4 T ( ζ i 2 ) = a ′ + h B ′ c \sum_{i\in J_3}\mathcal T(\zeta_i)=a+hB\sqrt c,\qquad
\sum_{i\in J_4}\mathcal T(\zeta_i^2)=a'+hB'\sqrt c i ∈ J 3 ∑ T ( ζ i ) = a + h B c , i ∈ J 4 ∑ T ( ζ i 2 ) = a ′ + h B ′ c for some integers a , a ′ a,a' a , a ′ K K K K K K Z [ c ] \mathbb Z[\sqrt c] Z [ c ]
By [36 Theorem 6.2] , writing an element a + b c a+b\sqrt c a + b c ∣ b ∣ φ ( 2 c ) |b|\varphi(2c) ∣ b ∣ φ ( 2 c ) h ∣ J 3 ∣ h|J_3| h ∣ J 3 ∣ h ∣ J 4 ∣ h|J_4| h ∣ J 4 ∣ ∣ J 3 ∣ ≥ ∣ B ∣ φ ( 2 c ) |J_3|\geq|B|\varphi(2c) ∣ J 3 ∣ ≥ ∣ B ∣ φ ( 2 c ) ∣ J 4 ∣ ≥ ∣ B ′ ∣ φ ( 2 c ) |J_4|\geq|B'|\varphi(2c) ∣ J 4 ∣ ≥ ∣ B ′ ∣ φ ( 2 c ) (3.18) follows. ◻
Theorem 3.12. If C = C ( D , κ ) \mathcal C=\mathcal C(\mathcal D,\kappa) C = C ( D , κ ) d + ∉ Z d_+\notin\mathbb Z d + ∈ / Z κ > 0 \kappa>0 κ > 0 n = 2 n=2 n = 2 κ = 8 \kappa=8 κ = 8
Proof. By Proposition 3.1 , κ = 8 k \kappa=8k κ = 8 k k ∈ Z ≥ 1 k\in\mathbb Z_{\geq1} k ∈ Z ≥ 1 n = 2 n=2 n = 2 r = m c r=m\sqrt c r = m c m ≥ 1 m\geq1 m ≥ 1 c c c r 2 = 8 ( 2 k 2 + 1 ) r^2=8(2k^2+1) r 2 = 8 ( 2 k 2 + 1 ) 2 2 2 3 3 3 c c c 3.11 to eq:first-sum –eq:second-sum . As ν ∈ { ± 1 } \nu\in\{\pm1\} ν ∈ { ± 1 } k ≥ 1 k\geq1 k ≥ 1
L ≥ m φ ( 2 c ) ( k + ∣ 2 ν − 6 k ∣ ) = ( 7 k − 2 ν ) m φ ( 2 c ) ≥ ( 7 k − 2 ) m φ ( 2 c ) . L\geq m\varphi(2c)\bigl(k+|2\nu-6k|\bigr)
=(7k-2\nu)m\varphi(2c)
\geq(7k-2)m\varphi(2c). L ≥ m φ ( 2 c ) ( k + ∣2 ν − 6 k ∣ ) = ( 7 k − 2 ν ) m φ ( 2 c ) ≥ ( 7 k − 2 ) m φ ( 2 c ) . Together with (3.15) , this gives
16 + 63 2 k 2 ≥ ( 7 k − 2 ) 16 k 2 + 8 φ ( 2 c ) c . \begin{equation}
16+\frac{63}{2}k^2
\geq(7k-2)\sqrt{16k^2+8}\,
\frac{\varphi(2c)}{\sqrt c}. \tag*{(3.19)}
\end{equation} 16 + 2 63 k 2 ≥ ( 7 k − 2 ) 16 k 2 + 8 c φ ( 2 c ) . (3.19) For even squarefree c c c
φ ( 2 c ) c = 2 ∏ ℓ ∣ c ℓ o d d p r i m e ℓ − 1 ℓ ≥ 2 . \frac{\varphi(2c)}{\sqrt c}
=\sqrt2\prod_{\substack{\ell\mid c\\\ell\ {\rm odd\ prime}}}
\frac{\ell-1}{\sqrt\ell}
\geq\sqrt2. c φ ( 2 c ) = 2 ℓ ∣ c ℓ odd prime ∏ ℓ ℓ − 1 ≥ 2 . Since 16 k 2 + 8 > 4 k \sqrt{16k^2+8}>4k 16 k 2 + 8 > 4 k 2 > 7 / 5 \sqrt2>7/5 2 > 7/5 (3.19) is greater than 28 5 k ( 7 k − 2 ) \frac{28}{5}k(7k-2) 5 28 k ( 7 k − 2 )
28 5 k ( 7 k − 2 ) − ( 16 + 63 2 k 2 ) = 77 k 2 − 112 k − 160 10 > 0 ( k ≥ 3 ) . \frac{28}{5}k(7k-2)-\left(16+\frac{63}{2}k^2\right)
=\frac{77k^2-112k-160}{10}>0
\qquad(k\geq3). 5 28 k ( 7 k − 2 ) − ( 16 + 2 63 k 2 ) = 10 77 k 2 − 112 k − 160 > 0 ( k ≥ 3 ) . The last quadratic is positive at 3 3 3 k ≤ 2 k\leq2 k ≤ 2
If k = 2 k=2 k = 2 r = 6 2 r=6\sqrt2 r = 6 2 m = 6 m=6 m = 6 c = 2 c=2 c = 2 φ ( 2 c ) = 2 \varphi(2c)=2 φ ( 2 c ) = 2 L ≥ 144 L\geq144 L ≥ 144 (3.15) gives L ≤ 142 L\leq142 L ≤ 142 k = 2 k=2 k = 2 k = 1 k=1 k = 1 κ = 8 \kappa=8 κ = 8
Together with the case κ = 0 \kappa=0 κ = 0 d + ∉ Z ⟹ ( n , κ ) ∈ { ( 2 , 0 ) , ( 2 , 8 ) } d_+\notin\mathbb Z\Longrightarrow (n,\kappa)\in\{(2,0),(2,8)\} d + ∈ / Z ⟹ ( n , κ ) ∈ {( 2 , 0 ) , ( 2 , 8 )}
The zero-parameter case
Theorem 3.13. Let C = C ( D , 0 ) \mathcal C=\mathcal C(\mathcal D,0) C = C ( D , 0 ) C \mathbb C C D \mathcal D D D ≃ Rep ( H 8 ) \mathcal D\simeq\operatorname{Rep}(H_8) D ≃ Rep ( H 8 ) H 8 H_8 H 8 C \mathcal C C [24 Section 7, Tables 3–4, and Appendix C.1] .
Proof. Since d = n 2 d=n\sqrt2 d = n 2 3.1 gives n = 2 n=2 n = 2 FPdim ( D ) = 8 \operatorname{FPdim}(\mathcal D)=8 FPdim ( D ) = 8 M = add ( ρ ) \mathcal M=\operatorname{add}(\rho) M = add ( ρ )
C = D ⊕ M , D ⊗ M , M ⊗ D ⊆ M , M ⊗ M ⊆ D . \mathcal C=\mathcal D\oplus\mathcal M,\qquad
\mathcal D\otimes\mathcal M,\ \mathcal M\otimes\mathcal D\subseteq\mathcal M,
\qquad \mathcal M\otimes\mathcal M\subseteq\mathcal D. C = D ⊕ M , D ⊗ M , M ⊗ D ⊆ M , M ⊗ M ⊆ D . Thus C \mathcal C C Z / 2 Z \mathbb Z/2\mathbb Z Z /2 Z D \mathcal D D 2.3 , the action on M \mathcal M M
Φ : D ⟶ End C ( M ) ≃ Vec , Φ ( X ) = Hom C ( ρ , X ⊗ ρ ) . \Phi:\mathcal D\longrightarrow\operatorname{End}_{\mathbb C}(\mathcal M)\simeq\operatorname{Vec},
\qquad \Phi(X)=\operatorname{Hom}_{\mathcal C}(\rho,X\otimes\rho). Φ : D ⟶ End C ( M ) ≃ Vec , Φ ( X ) = Hom C ( ρ , X ⊗ ρ ) . Here End C ( M ) \operatorname{End}_{\mathbb C}(\mathcal M) End C ( M ) H H H D ≃ Rep ( H ) \mathcal D\simeq\operatorname{Rep}(H) D ≃ Rep ( H ) dim H = 8 \dim H=8 dim H = 8
The odd component M \mathcal M M D \mathcal D D
D o p ≃ End D ( M ) , \mathcal D^{\mathrm{op}}\simeq\operatorname{End}_{\mathcal D}(\mathcal M), D op ≃ End D ( M ) , where o p \mathrm{op} op [32 Theorem 6.1 and Proposition 4.2] . The fiber-functor duality of [29 Example 7.12.26] , with the correction recorded in [30 ] , identifies this module endofunctor category with Rep ( H ∗ c o p ) \operatorname{Rep}(H^{*\mathrm{cop}}) Rep ( H ∗ cop ) Rep ( H ∗ ) \operatorname{Rep}(H^*) Rep ( H ∗ ) D \mathcal D D H H H H ∗ H^* H ∗ H H H [37 ] now gives H ≅ H 8 H\cong H_8 H ≅ H 8 D ≃ Rep ( H 8 ) \mathcal D\simeq\operatorname{Rep}(H_8) D ≃ Rep ( H 8 )
To apply the unitary calculation in [24 ] , observe directly from the fusion rules that
C a d = D , D a d = D p t , ( D p t ) a d = Vec . \mathcal C_{\mathrm{ad}}=\mathcal D,\qquad
\mathcal D_{\mathrm{ad}}=\mathcal D_{\mathrm{pt}},\qquad
(\mathcal D_{\mathrm{pt}})_{\mathrm{ad}}=\operatorname{Vec}. C ad = D , D ad = D pt , ( D pt ) ad = Vec . Thus C \mathcal C C [31 ] . By [34 Theorem 5.20] , both C \mathcal C C D \mathcal D D C \mathcal C C Rep ( H 8 ) \operatorname{Rep}(H_8) Rep ( H 8 ) [34 Definition 5.13(1)] . After choosing the identification of the neutral component with Rep ( H 8 ) \operatorname{Rep}(H_8) Rep ( H 8 ) [24 Section 7, Tables 3–4, and Appendix C.1] . That calculation lists eight solutions modulo changes of bases in the fusion spaces, with the neutral category fixed. Forgetting these choices gives the claimed membership in the eight-entry list. We do not need to decide here which entries remain distinct under tensor equivalences that relabel the neutral component. ◻
Nonexistence for parameter eight
From now on suppose, for contradiction, that C = C ( D , 8 ) \mathcal C=\mathcal C(\mathcal D,8) C = C ( D , 8 ) D \mathcal D D G G G t t t 8 8 8 [28 Lemma 4.1] , a Galois conjugate of C \mathcal C C C \mathcal C C
s = 6 , d = 4 + 2 s , D = 48 + 16 s = 8 ( d + 2 ) , ε = 1 + d . \begin{equation}
s=\sqrt6,\qquad d=4+2s,\qquad
D=48+16s=8(d+2),\qquad \varepsilon=1+d. \tag*{(3.20)}
\end{equation} s = 6 , d = 4 + 2 s , D = 48 + 16 s = 8 ( d + 2 ) , ε = 1 + d . (3.20) The identities
d 2 = 8 + 8 d , D d = 4 s , D ′ = 48 − 16 s , D D ′ = ε \begin{equation}
d^2=8+8d,\qquad \frac Dd=4s,\qquad
D'=48-16s,\qquad \frac D{D'}=\varepsilon \tag*{(3.21)}
\end{equation} d 2 = 8 + 8 d , d D = 4 s , D ′ = 48 − 16 s , D ′ D = ε (3.21) will be used repeatedly. The matrix (3.2) now is
M = ( 6 2 2 2 2 8 2 6 2 2 2 8 2 2 6 2 2 8 2 2 2 6 2 8 2 2 2 2 12 16 8 8 8 8 16 80 ) . \begin{equation}
M=\begin{pmatrix}
6&2&2&2&2&8\\
2&6&2&2&2&8\\
2&2&6&2&2&8\\
2&2&2&6&2&8\\
2&2&2&2&12&16\\
8&8&8&8&16&80
\end{pmatrix}. \tag*{(3.22)}
\end{equation} M = 6 2 2 2 2 8 2 6 2 2 2 8 2 2 6 2 2 8 2 2 2 6 2 8 2 2 2 2 12 16 8 8 8 8 16 80 . (3.22) Here D = Δ = dim ( C ) D=\Delta=\dim(\mathcal C) D = Δ = dim ( C ) S \mathsf S S S = S / D S=\mathsf S/D S = S / D (2.6) fixed in § 2 .
The Ng–Schauenburg indicator formula [38 Theorem 4.1] is
∑ Z ∈ Irr ( Z ( C ) ) m X ( Z ) dim ( Z ) θ Z n = D ν n ( X ) . \begin{equation}
\sum_{Z\in\operatorname{Irr}(\mathcal Z(\mathcal C))}m_X(Z)\dim(Z)\theta_Z^n=D\nu_n(X). \tag*{(3.23)}
\end{equation} Z ∈ Irr ( Z ( C )) ∑ m X ( Z ) dim ( Z ) θ Z n = D ν n ( X ) . (3.23) For a nontrivial simple X X X ν 1 ( X ) = 0 \nu_1(X)=0 ν 1 ( X ) = 0 C \mathcal C C ν 2 ( X ) ∈ { 1 , − 1 } \nu_2(X)\in\{1,-1\} ν 2 ( X ) ∈ { 1 , − 1 } Hom ( 1 , X ⊗ 4 ) \operatorname{Hom}(\mathbf 1,X^{\otimes4}) Hom ( 1 , X ⊗ 4 )
ν 4 ( X ) ∈ Z [ i ] . \begin{equation}
\nu_4(X)\in\mathbb Z[\mathrm i]. \tag*{(3.24)}
\end{equation} ν 4 ( X ) ∈ Z [ i ] . (3.24) The balancing identity is (2.5) .
Characters and the pointed subcategory
The characters of R 8 R_8 R 8 χ ± \chi_\pm χ ± η \eta η λ \lambda λ G G G
x ∈ G t ρ f χ χ + 1 2 4 + 2 s D χ − 1 2 4 − 2 s D ′ η 1 − 2 0 8 λ λ ( x ) 0 0 4. \begin{equation}
\begin{array}{c|ccc|c}
&x\in G&t&\rho& f_\chi\\ \hline
\chi_+&1&2&4+2s&D\\
\chi_-&1&2&4-2s&D'\\
\eta&1&-2&0&8\\
\lambda&\lambda(x)&0&0&4.
\end{array} \tag*{(3.25)}
\end{equation} χ + χ − η λ x ∈ G 1 1 1 λ ( x ) t 2 2 − 2 0 ρ 4 + 2 s 4 − 2 s 0 0 f χ D D ′ 8 4. (3.25) The table follows directly from (2.9) . Its six distinct rows exhaust the characters of the complexified rank-six ring.
Lemma 3.14. The pointed subcategory of C \mathcal C C Vec G \operatorname{Vec}_G Vec G x ∈ G x\in G x ∈ G ν 2 ( x ) = 1 \nu_2(x)=1 ν 2 ( x ) = 1 x ∈ G x\in G x ∈ G
Proof. The additive subcategory generated by ρ \rho ρ x ⊗ ρ ≅ ρ x\otimes\rho\cong \rho x ⊗ ρ ≅ ρ x ∈ G x\in G x ∈ G x ⊗ ρ ≅ ρ x\otimes\rho\cong\rho x ⊗ ρ ≅ ρ Vec G \operatorname{Vec}_G Vec G G G G 1 1 1
A tensor automorphism of Id C \operatorname{Id}_{\mathcal C} Id C a X a_X a X X X X a X a Y = a Z a_Xa_Y=a_Z a X a Y = a Z Z Z Z X ⊗ Y X\otimes Y X ⊗ Y ρ ⊗ ρ \rho\otimes \rho ρ ⊗ ρ ρ \rho ρ a ρ 2 = a ρ a_\rho^2=a_\rho a ρ 2 = a ρ a ρ = 1 a_\rho=1 a ρ = 1 a X = 1 a_X=1 a X = 1
Lemma 3.15. For x ∈ G ∖ { 1 } x\in G\setminus\{1\} x ∈ G ∖ { 1 } I ( x ) I(x) I ( x ) 1 1 1 − 1 -1 − 1 D / 2 D/2 D /2
Proof. By Lemma 3.14 and (3.23) ,
∑ Z m x ( Z ) dim Z θ Z 2 = D = ∑ Z m x ( Z ) dim Z . \sum_Zm_x(Z)\dim Z\,\theta_Z^2=D
=\sum_Zm_x(Z)\dim Z. Z ∑ m x ( Z ) dim Z θ Z 2 = D = Z ∑ m x ( Z ) dim Z . Every summand has a positive dimension and ∣ θ Z ∣ = 1 |\theta_Z|=1 ∣ θ Z ∣ = 1 θ Z 2 = 1 \theta_Z^2=1 θ Z 2 = 1 m x ( Z ) > 0 m_x(Z)>0 m x ( Z ) > 0 x x x
Call a simple central object low if its restriction contains 1 \mathbf 1 1 I ( 1 ) I(\mathbf 1) I ( 1 )
By (3.25) and (2.6) ,
I ( 1 ) ≅ 1 ⊕ A ⊕ B ⊕ ⨁ λ ≠ 1 C λ , \begin{equation}
I(\mathbf 1)\cong \mathbf 1\oplus A\oplus B\oplus\bigoplus_{\lambda\ne1}C_\lambda, \tag*{(3.26)}
\end{equation} I ( 1 ) ≅ 1 ⊕ A ⊕ B ⊕ λ = 1 ⨁ C λ , (3.26) where A , B , C λ A,B,C_\lambda A , B , C λ χ − , η , λ \chi_-,\eta,\lambda χ − , η , λ
dim A = 1 + d , dim B = d + 2 , dim C λ = 2 ( d + 2 ) . \begin{equation}
\dim A=1+d,\qquad \dim B=d+2,\qquad
\dim C_\lambda=2(d+2). \tag*{(3.27)}
\end{equation} dim A = 1 + d , dim B = d + 2 , dim C λ = 2 ( d + 2 ) . (3.27) All of them have twist 1 1 1
The coefficient of d d d ρ \rho ρ 1 \mathbf 1 1
F ( A ) ≅ 1 ⊕ ρ , F ( B ) ≅ 1 ⊕ g ⊕ ρ \begin{equation}
F(A)\cong \mathbf 1\oplus \rho,\qquad F(B)\cong \mathbf 1\oplus g\oplus \rho \tag*{(3.28)}
\end{equation} F ( A ) ≅ 1 ⊕ ρ , F ( B ) ≅ 1 ⊕ g ⊕ ρ (3.28) for some nontrivial g ∈ G g\in G g ∈ G g g g h , k = g h h,k=gh h , k = g h f , g , h f,g,h f , g , h § 3.1 ; this local relabeling is used throughout the R 8 R_8 R 8 F C λ FC_\lambda F C λ ρ \rho ρ 4 4 4 b λ = [ F C λ , t ] b_\lambda=[FC_\lambda,t] b λ = [ F C λ , t ] S \mathsf S S B , C λ B,C_\lambda B , C λ
2 ( d + 2 ) ( 1 + λ ( g ) ) = ( d + 2 ) ( 4 − 4 b λ ) . 2(d+2)(1+\lambda(g))=(d+2)(4-4b_\lambda). 2 ( d + 2 ) ( 1 + λ ( g )) = ( d + 2 ) ( 4 − 4 b λ ) . Thus
b λ = 1 − λ ( g ) 2 . \begin{equation}
b_\lambda=\frac{1-\lambda(g)}2. \tag*{(3.29)}
\end{equation} b λ = 2 1 − λ ( g ) . (3.29) Precisely one of the three objects, denoted C 0 C_0 C 0 b λ = 0 b_\lambda=0 b λ = 0 b λ = 1 b_\lambda=1 b λ = 1
By the first row of (3.22) , the low contributions to I ( g ) , I ( h ) I(g),I(h) I ( g ) , I ( h ) I ( k ) I(k) I ( k )
3 ( d + 2 ) , 4 ( d + 2 ) , 4 ( d + 2 ) , \begin{equation}
3(d+2),\qquad4(d+2),\qquad4(d+2), \tag*{(3.30)}
\end{equation} 3 ( d + 2 ) , 4 ( d + 2 ) , 4 ( d + 2 ) , (3.30) respectively. For g g g B B B C λ C_\lambda C λ h h h k k k C λ C_\lambda C λ 3.15 implies that every nonlow central simple whose restriction contains h h h k k k − 1 -1 − 1
Lemma 3.16. There is a central invertible P P P F ( P ) ≅ g F(P)\cong g F ( P ) ≅ g P ⊗ P ≅ 1 P\otimes P\cong \mathbf 1 P ⊗ P ≅ 1 θ P = 1 \theta_P=1 θ P = 1 C h , C k C_h,C_k C h , C k F ( C 0 ) ≅ 1 ⊕ g ⊕ h ⊕ k ⊕ 2 ρ , F ( C h ) ≅ 1 ⊕ h ⊕ t ⊕ 2 ρ , F ( C k ) ≅ 1 ⊕ k ⊕ t ⊕ 2 ρ . \begin{aligned}
F(C_0)&\cong \mathbf 1\oplus g\oplus h\oplus k\oplus 2 \rho,\\
F(C_h)&\cong \mathbf 1\oplus h\oplus t\oplus 2 \rho,\qquad
F(C_k)\cong \mathbf 1\oplus k\oplus t\oplus 2 \rho.\end{aligned} F ( C 0 ) F ( C h ) ≅ 1 ⊕ g ⊕ h ⊕ k ⊕ 2 ρ , ≅ 1 ⊕ h ⊕ t ⊕ 2 ρ , F ( C k ) ≅ 1 ⊕ k ⊕ t ⊕ 2 ρ . P ⊗ B ≅ B P\otimes B\cong B P ⊗ B ≅ B P ⊗ C 0 ≅ C 0 P\otimes C_0\cong C_0 P ⊗ C 0 ≅ C 0 I ( g ) I(g) I ( g ) B , C 0 , P , P ⊗ A , P ⊗ C h , P ⊗ C k , B,\ C_0,\ P,\ P\otimes A,\ P\otimes C_h,\ P\otimes C_k, B , C 0 , P , P ⊗ A , P ⊗ C h , P ⊗ C k , 1 1 1 − 1 -1 − 1
Proof. By (3.30) and Lemma 3.15 , the nonlow twist-1 1 1 I ( g ) I(g) I ( g ) d + 2 d+2 d + 2 h h h k k k r g ⊕ b t ⊕ c ρ r g\oplus b t\oplus c \rho r g ⊕ b t ⊕ c ρ r ≥ 1 r\ge1 r ≥ 1 b , c ≥ 0 b,c\ge0 b , c ≥ 0 I ( g ) I(g) I ( g ) r r r 1 , d 1,d 1 , d
∑ r c = 1 , ∑ ( r 2 + 2 r b ) = 2. \begin{equation}
\sum rc=1,\qquad \sum(r^2+2rb)=2. \tag*{(3.32)}
\end{equation} ∑ r c = 1 , ∑ ( r 2 + 2 r b ) = 2. (3.32) The first equation gives exactly one constituent with r = c = 1 r=c=1 r = c = 1 b b b 1 1 1 ( r , b , c ) = ( 1 , 0 , 0 ) (r,b,c)=(1,0,0) ( r , b , c ) = ( 1 , 0 , 0 ) P P P F P ≅ g FP\cong g F P ≅ g 1 1 1 3.14 implies P ⊗ P ≅ 1 P\otimes P\cong \mathbf 1 P ⊗ P ≅ 1 Z Z Z F ( Z ) ≅ g ⊕ ρ F(Z)\cong g\oplus\rho F ( Z ) ≅ g ⊕ ρ P ⊗ Z P\otimes Z P ⊗ Z 1 ⊕ ρ \mathbf 1\oplus\rho 1 ⊕ ρ I ( 1 ) I(\mathbf 1) I ( 1 ) I ( 1 ) I(\mathbf 1) I ( 1 ) 1 + d 1+d 1 + d A A A P ⊗ Z ≅ A P\otimes Z\cong A P ⊗ Z ≅ A P 2 ≅ 1 P^2\cong\mathbf 1 P 2 ≅ 1 Z ≅ P ⊗ A Z\cong P\otimes A Z ≅ P ⊗ A
The restriction of P ⊗ B P\otimes B P ⊗ B F B FB F B 1 \mathbf 1 1 B B B d + 2 d+2 d + 2 P ⊗ B ≅ B P\otimes B\cong B P ⊗ B ≅ B (2.6) and (2.5) give
θ P ⊗ C λ = λ ( g ) . \begin{equation}
\theta_{P\otimes C_\lambda}=\lambda(g). \tag*{(3.33)}
\end{equation} θ P ⊗ C λ = λ ( g ) . (3.33) If F ( C 0 ) F(C_0) F ( C 0 ) g g g P ⊗ C 0 P\otimes C_0 P ⊗ C 0 1 1 1 h h h k k k F ( C 0 ) F(C_0) F ( C 0 ) 4 4 4 1 \mathbf 1 1 t t t (3.30) . Thus P ⊗ C 0 P\otimes C_0 P ⊗ C 0 t t t P ⊗ C 0 ≅ C 0 P\otimes C_0\cong C_0 P ⊗ C 0 ≅ C 0 (3.29) . Invariance under g g g [ F ( C 0 ) , g ] = 1 [F(C_0),g]=1 [ F ( C 0 ) , g ] = 1 h , k h,k h , k 4 4 4 (3.31) .
The first row of (3.22) now forces the two remaining restrictions to be those in r8:eq:Chk . Their P P P
F ( P ⊗ C h ) ≅ g ⊕ k ⊕ t ⊕ 2 ρ , F ( P ⊗ C k ) ≅ g ⊕ h ⊕ t ⊕ 2 ρ \begin{equation}
F(P\otimes C_h)\cong g\oplus k\oplus t\oplus 2 \rho,\qquad
F(P\otimes C_k)\cong g\oplus h\oplus t\oplus 2 \rho \tag*{(3.34)}
\end{equation} F ( P ⊗ C h ) ≅ g ⊕ k ⊕ t ⊕ 2 ρ , F ( P ⊗ C k ) ≅ g ⊕ h ⊕ t ⊕ 2 ρ (3.34) and twist − 1 -1 − 1 (3.33) . These, together with B , C 0 , P , P ⊗ A B,C_0,P,P\otimes A B , C 0 , P , P ⊗ A g g g M g , g = 6 M_{g,g}=6 M g , g = 6 I ( g ) I(g) I ( g )
Table 1. All central simples whose restrictions meet G G G ε = 1 + d \varepsilon=1+d ε = 1 + d
Proposition 3.17. There is a central invertible object Q Q Q F Q ≅ h FQ\cong h F Q ≅ h Table 1 are exactly the central simples whose restrictions contain an invertible object of C \mathcal C C { 1 , P , Q , P ⊗ Q } ≅ C 2 2 \{\mathbf 1,P,Q,P\otimes Q\}\cong C_2^2 { 1 , P , Q , P ⊗ Q } ≅ C 2 2
Proof. Consider the ten objects consisting of the six in (3.26) and P , P ⊗ A , P ⊗ C h , P ⊗ C k P,P\otimes A,P\otimes C_h,P\otimes C_k P , P ⊗ A , P ⊗ C h , P ⊗ C k 3.16 exhausts both I ( 1 ) I(\mathbf 1) I ( 1 ) I ( g ) I(g) I ( g ) C h , C k , P ⊗ C h , P ⊗ C k C_h,C_k,P\otimes C_h,P\otimes C_k C h , C k , P ⊗ C h , P ⊗ C k M h , t = M k , t = 2 M_{h,t}=M_{k,t}=2 M h , t = M k , t = 2 h h h k k k
a h ⊕ b k ⊕ c ρ . a h\oplus b k\oplus c \rho. ah ⊕ bk ⊕ c ρ . Subtracting the contributions of the ten known objects from (3.22) gives
∑ a 2 = ∑ b 2 = 3 , ∑ a b = 1 , ∑ a c = ∑ b c = 2. \begin{equation}
\sum a^2=\sum b^2=3,\qquad
\sum ab=1,\qquad
\sum ac=\sum bc=2. \tag*{(3.35)}
\end{equation} ∑ a 2 = ∑ b 2 = 3 , ∑ ab = 1 , ∑ a c = ∑ b c = 2. (3.35) The first two square sums imply a , b ∈ { 0 , 1 } a,b\in\{0,1\} a , b ∈ { 0 , 1 } h , k h,k h , k h h h k k k z z z ρ \rho ρ z ∈ { 0 , 1 , 2 } z\in\{0,1,2\} z ∈ { 0 , 1 , 2 }
If z = 2 z=2 z = 2 h h h h h h 3.14 . Suppose z = 0 z=0 z = 0 h ⊕ k h\oplus k h ⊕ k E E E h , k h,k h , k E ∗ ≅ E E^*\cong E E ∗ ≅ E P ⊗ E ≅ E P\otimes E\cong E P ⊗ E ≅ E E ⊗ E E\otimes E E ⊗ E 2 ( 1 ⊕ g ) 2 (\mathbf 1\oplus g) 2 ( 1 ⊕ g ) I ( 1 ) I(\mathbf 1) I ( 1 ) I ( g ) I(g) I ( g ) 1 , P \mathbf 1,P 1 , P { 1 , g } \{\mathbf 1,g\} { 1 , g } E ⊗ E E\otimes E E ⊗ E dim ( E ⊗ E ) = 4 \dim(E\otimes E)=4 dim ( E ⊗ E ) = 4
Hence z = 1 z=1 z = 1 h h h h h h h ⊕ ρ h\oplus \rho h ⊕ ρ Q Q Q Q ⊗ A Q\otimes A Q ⊗ A P P P k k k Q ⊗ B Q\otimes B Q ⊗ B h ⊕ k ⊕ ρ h\oplus k\oplus \rho h ⊕ k ⊕ ρ − 1 -1 − 1 (3.30) . Their restrictions distinguish them and exhaust the possibilities in (3.35) . Lemma 3.14 gives Q ⊗ Q ≅ 1 Q\otimes Q\cong \mathbf 1 Q ⊗ Q ≅ 1 Z ( C ) \mathcal Z(\mathcal C) Z ( C )
Let R \mathcal R R Table 1 . By Proposition 3.17 , each has restriction
F Z ≅ b Z t ⊕ c Z ρ , b Z , c Z ∈ Z ≥ 0 . \begin{equation}
FZ\cong b_Z t\oplus c_Z \rho,\qquad b_Z,c_Z\in\mathbb Z_{\ge0}. \tag*{(3.36)}
\end{equation} F Z ≅ b Z t ⊕ c Z ρ , b Z , c Z ∈ Z ≥ 0 . (3.36) Subtracting the fifteen restriction vectors from (3.22) gives
∑ Z ∈ R b Z 2 = 8 , ∑ Z ∈ R b Z c Z = 8 , ∑ Z ∈ R c Z 2 = 54. \begin{equation}
\sum_{Z\in\mathcal R}b_Z^2=8,\qquad
\sum_{Z\in\mathcal R}b_Zc_Z=8,\qquad
\sum_{Z\in\mathcal R}c_Z^2=54. \tag*{(3.37)}
\end{equation} Z ∈ R ∑ b Z 2 = 8 , Z ∈ R ∑ b Z c Z = 8 , Z ∈ R ∑ c Z 2 = 54. (3.37) Dimension restrictions
Lemma 3.18. For every Z ∈ R Z\in\mathcal R Z ∈ R b t , b ( t ⊕ ρ ) , b ( t ⊕ 2 ρ ) , c ρ , b t,\qquad b (t\oplus\rho),\qquad b (t\oplus 2 \rho),\qquad c \rho, b t , b ( t ⊕ ρ ) , b ( t ⊕ 2 ρ ) , c ρ ,
Proof. If b Z = 0 b_Z=0 b Z = 0 b = b Z b=b_Z b = b Z c = c Z c=c_Z c = c Z b ∈ { 1 , 2 } b\in\{1,2\} b ∈ { 1 , 2 } (3.37) . Let σ \sigma σ Q ( s ) \mathbb Q(s) Q ( s ) S / D \mathsf S/D S / D σ ^ \widehat\sigma σ
dim σ ^ ( Z ) = D D ′ ∣ σ ( dim Z ) ∣ = ∣ ε σ ( 2 b + c d ) ∣ = ∣ 2 b + ( 2 b − c ) d ∣ . \begin{equation}
\dim\widehat\sigma(Z)=\frac D{D'}\,|\sigma(\dim Z)|
=\left|\varepsilon\sigma(2b+cd)\right|
=\left|2b+(2b-c)d\right|. \tag*{(3.38)}
\end{equation} dim σ ( Z ) = D ′ D ∣ σ ( dim Z ) ∣ = ∣ ε σ ( 2 b + c d ) ∣ = ∣ 2 b + ( 2 b − c ) d ∣ . (3.38) See [27 Section 4.1, equation (4.3)] ; the absolute value records the sign in the signed permutation, since all dimensions are positive. If c > 2 b c>2b c > 2 b 2 b + ( 2 b − c ) d < 0 2b+(2b-c)d<0 2 b + ( 2 b − c ) d < 0 d > 4 ≥ 2 b d>4\ge2b d > 4 ≥ 2 b − 2 b -2b − 2 b 1 , d 1,d 1 , d C \mathcal C C 0 ≤ c ≤ 2 b 0\le c\le2b 0 ≤ c ≤ 2 b
For every central simple Z Z Z D / dim Z D/\dim Z D / dim Z Z Z Z Z ( C ) \mathcal Z(\mathcal C) Z ( C )
f Z = FPdim ( Z ( C ) ) dim ( Z ) 2 = ( D dim Z ) 2 . f_Z=\frac{\operatorname{FPdim}(\mathcal Z(\mathcal C))}{\dim(Z)^2}=\left(\frac D{\dim Z}\right)^2. f Z = dim ( Z ) 2 FPdim ( Z ( C )) = ( dim Z D ) 2 . Formal codegrees are algebraic integers. Thus D / dim Z D/\dim Z D / dim Z X 2 − f Z = 0 X^2-f_Z=0 X 2 − f Z = 0 b = 2 b=2 b = 2 c = 1 , 3 c=1,3 c = 1 , 3
D 4 + d = 24 + 4 s 5 , D 4 + 3 d = 24 − 4 s 5 . \frac D{4+d}=\frac{24+4s}{5},\qquad
\frac D{4+3d}=\frac{24-4s}{5}. 4 + d D = 5 24 + 4 s , 4 + 3 d D = 5 24 − 4 s . Both have nonintegral field trace 48 / 5 48/5 48/5 ( b , c ) (b,c) ( b , c )
Lemma 3.19. There is at most one central simple U U U t t t F ( U ) ≅ t F(U)\cong t F ( U ) ≅ t b ( t ⊕ 2 ρ ) b (t\oplus 2 \rho) b ( t ⊕ 2 ρ ) U ⊗ A U\otimes A U ⊗ A F ( U ⊗ A ) ≅ t ⊕ 2 ρ , θ U ⊗ A = θ U . F(U\otimes A)\cong t\oplus 2 \rho,\qquad \theta_{U\otimes A}=\theta_U. F ( U ⊗ A ) ≅ t ⊕ 2 ρ , θ U ⊗ A = θ U . U U U
Proof. Suppose F U ≅ b t FU\cong b t F U ≅ b t U ⊗ U ∗ U\otimes U^* U ⊗ U ∗ G G G 3.17 , every simple constituent is one of 1 , P , Q , P ⊗ Q \mathbf 1,P,Q,P\otimes Q 1 , P , Q , P ⊗ Q L L L
[ U ⊗ U ∗ , L ] = dim Hom ( L ⊗ U , U ) ≤ 1. [U\otimes U^*,L]=\dim\operatorname{Hom}(L\otimes U,U)\le1. [ U ⊗ U ∗ , L ] = dim Hom ( L ⊗ U , U ) ≤ 1. Consequently 4 b 2 = dim ( U ⊗ U ∗ ) ≤ 4 4b^2=\dim(U\otimes U^*)\le4 4 b 2 = dim ( U ⊗ U ∗ ) ≤ 4 b = 1 b=1 b = 1
U ∗ ⊗ U ≅ U ⊗ U ∗ ≅ 1 ⊕ P ⊕ Q ⊕ ( P ⊗ Q ) . \begin{equation}
U^*\otimes U\cong U\otimes U^*\cong \mathbf 1\oplus P\oplus Q\oplus (P\otimes Q). \tag*{(3.39)}
\end{equation} U ∗ ⊗ U ≅ U ⊗ U ∗ ≅ 1 ⊕ P ⊕ Q ⊕ ( P ⊗ Q ) . (3.39) If distinct U , V U,V U , V t t t U ⊗ V ∗ U\otimes V^* U ⊗ V ∗ 1 \mathbf 1 1 dim ( U ⊗ V ∗ ) = 4 \dim(U\otimes V^*)=4 dim ( U ⊗ V ∗ ) = 4
The dimension transformation in (3.38) interchanges 2 b 2b 2 b 2 b + 2 b d 2b+2bd 2 b + 2 b d b = 1 , 2 b=1,2 b = 1 , 2 Table 1 . A simple of either dimension must therefore be residual, and irrationality identifies its restriction as b t b t b t b ( t ⊕ 2 ρ ) b (t\oplus 2 \rho) b ( t ⊕ 2 ρ ) U U U
To identify that object, reciprocity and (3.39) give
dim End ( U ⊗ A ) = ∑ L ∈ { 1 , P , Q , P ⊗ Q } dim Hom ( A , L ⊗ A ) = 1. \dim\operatorname{End}(U\otimes A)=\sum_{L\in\{\mathbf 1,P,Q,P\otimes Q\}}\dim\operatorname{Hom}(A,L\otimes A)=1. dim End ( U ⊗ A ) = L ∈ { 1 , P , Q , P ⊗ Q } ∑ dim Hom ( A , L ⊗ A ) = 1. Indeed, the four restrictions F ( L ⊗ A ) ≅ F L ⊕ ρ F(L\otimes A)\cong FL\oplus \rho F ( L ⊗ A ) ≅ F L ⊕ ρ U ⊗ A U\otimes A U ⊗ A F ( U ⊗ A ) ≅ t ⊕ 2 ρ F(U\otimes A)\cong t\oplus 2 \rho F ( U ⊗ A ) ≅ t ⊕ 2 ρ (2.6) , for the character χ − \chi_- χ − A A A S A , U = 2 ε = dim ( U ⊗ A ) \mathsf S_{A,U}=2\varepsilon=\dim(U\otimes A) S A , U = 2 ε = dim ( U ⊗ A ) U ⊗ A U\otimes A U ⊗ A θ A = 1 \theta_A=1 θ A = 1 (2.5) implies θ U ⊗ A = θ U \theta_{U\otimes A}=\theta_U θ U ⊗ A = θ U
A multiset of forty-six twists
Let x ∈ { 0 , 1 } x\in\{0,1\} x ∈ { 0 , 1 } U U U V V V W W W
F V ≅ b V ( t ⊕ ρ ) , F W ≅ c W ρ . FV\cong b_V (t\oplus\rho),\qquad FW\cong c_W \rho. F V ≅ b V ( t ⊕ ρ ) , F W ≅ c W ρ . Equations (3.37) and Lemma 3.19 become
∑ V b V 2 = 8 − 2 x , ∑ W c W 2 = 46 − 2 x . \begin{equation}
\sum_Vb_V^2=8-2x,\qquad
\sum_Wc_W^2=46-2x. \tag*{(3.40)}
\end{equation} V ∑ b V 2 = 8 − 2 x , W ∑ c W 2 = 46 − 2 x . (3.40) Define a multiset Λ \Lambda Λ c W 2 c_W^2 c W 2 θ W \theta_W θ W W W W x = 1 x=1 x = 1 θ U \theta_U θ U
∣ Λ ∣ = 46. \begin{equation}
|\Lambda|=46. \tag*{(3.41)}
\end{equation} ∣Λ∣ = 46. (3.41) Put p n = ∑ ζ ∈ Λ ζ n p_n=\sum_{\zeta\in\Lambda}\zeta^n p n = ∑ ζ ∈ Λ ζ n
Proposition 3.20. One has ν 2 ( t ) = 1 \nu_2(t)=1 ν 2 ( t ) = 1 ϵ ρ = ν 2 ( ρ ) ∈ { 1 , − 1 } \epsilon_{\rho}=\nu_2(\rho)\in\{1,-1\} ϵ ρ = ν 2 ( ρ ) ∈ { 1 , − 1 } p 1 = − 2 s , p 2 = − 2 + ( 4 ϵ ρ − 12 ) s , p 4 = − 2 + 4 s ( ν 4 ( ρ ) − ν 4 ( t ) − 2 ) . \begin{aligned}
p_1&=-2s,\\
p_2&=-2+(4\epsilon_{\rho}-12)s,\\
p_4&=-2+4s\bigl(\nu_4(\rho)-\nu_4(t)-2\bigr).
\end{aligned} p 1 p 2 p 4 = − 2 s , = − 2 + ( 4 ϵ ρ − 12 ) s , = − 2 + 4 s ( ν 4 ( ρ ) − ν 4 ( t ) − 2 ) .
Proof. In the difference between the two indicator sums (3.23) for ρ \rho ρ t t t V V V W W W c W 2 d θ W n c_W^2d\theta_W^n c W 2 d θ W n U U U − 2 θ U n -2\theta_U^n − 2 θ U n U ⊗ A U\otimes A U ⊗ A ( 2 + 2 d ) θ U n (2+2d)\theta_U^n ( 2 + 2 d ) θ U n 3.19 ; their sum is 2 d θ U n 2d\theta_U^n 2 d θ U n d p n dp_n d p n
Let K n K_n K n Table 1 to this difference. The table gives
K n = ( 12 + 7 d ) ( 1 + ( − 1 ) n ) + 8 + 4 d = { 4 ( d + 2 ) , n odd , 18 d + 32 , n even . \begin{equation}
K_n=(12+7d)(1+(-1)^n)+8+4d
=\begin{cases}
4(d+2),&n\text{ odd},\\
18d+32,&n\text{ even}.
\end{cases} \tag*{(3.43)}
\end{equation} K n = ( 12 + 7 d ) ( 1 + ( − 1 ) n ) + 8 + 4 d = { 4 ( d + 2 ) , 18 d + 32 , n odd , n even . (3.43) Therefore
p n = D ( ν n ( ρ ) − ν n ( t ) ) − K n d . \begin{equation}
p_n=\frac{D(\nu_n(\rho)-\nu_n(t))-K_n}{d}. \tag*{(3.44)}
\end{equation} p n = d D ( ν n ( ρ ) − ν n ( t )) − K n . (3.44) The first indicators of t , ρ t,\rho t , ρ p 1 = − 4 ( d + 2 ) / d = − 2 s p_1=-4(d+2)/d=-2s p 1 = − 4 ( d + 2 ) / d = − 2 s
Exactly four objects in Table 1 contain t t t 2 ( d + 2 ) 2(d+2) 2 ( d + 2 ) 1 , 1 , − 1 , − 1 1,1,-1,-1 1 , 1 , − 1 , − 1 I ( t ) I(t) I ( t ) 8 ( d + 2 ) = D 8(d+2)=D 8 ( d + 2 ) = D I ( t ) I(t) I ( t ) D D D dim I ( t ) = 2 D \dim I(t)=2D dim I ( t ) = 2 D ν 2 ( t ) = − 1 \nu_2(t)=-1 ν 2 ( t ) = − 1 − 2 D -2D − 2 D ν 2 ( t ) = 1 \nu_2(t)=1 ν 2 ( t ) = 1
Finally D / d = 4 s D/d=4s D / d = 4 s ( 18 d + 32 ) / d = 2 + 8 s (18d+32)/d=2+8s ( 18 d + 32 ) / d = 2 + 8 s (3.44) for n = 2 , 4 n=2,4 n = 2 , 4 r8:eq:p2 and r8:eq:p4 . ◻
Let K = Q ( 6 ) K=\mathbb Q(\sqrt6) K = Q ( 6 ) L = Q ( ζ N ) L=\mathbb Q(\zeta_N) L = Q ( ζ N ) 24 ∣ N 24\mid N 24 ∣ N N N N 2 , 3 2,3 2 , 3 ζ m = exp ( 2 π i / m ) \zeta_m=\exp(2\pi\mathrm i/m) ζ m = exp ( 2 π i / m )
P = 1 [ L : K ] Tr L / K , P ( z ) = a ( z ) + b ( z ) 6 , a ( z ) , b ( z ) ∈ Q . \begin{equation}
P=\frac1{[L:K]}\operatorname{Tr}_{L/K},\qquad
P(z)=a(z)+b(z)\sqrt6,\qquad a(z),b(z)\in\mathbb Q. \tag*{(3.45)}
\end{equation} P = [ L : K ] 1 Tr L / K , P ( z ) = a ( z ) + b ( z ) 6 , a ( z ) , b ( z ) ∈ Q . (3.45) The functions a , b a,b a , b Q \mathbb Q Q L L L
Lemma 3.21. If ζ ∈ L \zeta\in L ζ ∈ L m m m m ∣ b ( ζ ) ∣ ∣ b ( ζ 2 ) ∣ a ( ζ 2 ) + a ( ζ 4 ) 1 , 2 0 0 2 3 , 6 , 8 0 0 − 1 24 1 / 4 0 1 / 2 48 0 1 / 4 0 all other m 0 0 0.
\begin{array}{c|ccc}
m&|b(\zeta)|&|b(\zeta^2)|&a(\zeta^2)+a(\zeta^4)\\ \hline
1,2&0&0&2\\
3,6,8&0&0&-1\\
24&1/4&0&1/2\\
48&0&1/4&0\\
\text{all other }m&0&0&0.
\end{array}
m 1 , 2 3 , 6 , 8 24 48 all other m ∣ b ( ζ ) ∣ 0 0 1/4 0 0 ∣ b ( ζ 2 ) ∣ 0 0 0 1/4 0 a ( ζ 2 ) + a ( ζ 4 ) 2 − 1 1/2 0 0.
Proof. The conductor of K K K 24 24 24 24 ∤ m 24\nmid m 24 ∤ m K ∩ Q ( ζ m ) = Q K\cap\mathbb Q(\zeta_m)=\mathbb Q K ∩ Q ( ζ m ) = Q P ( ζ ) P(\zeta) P ( ζ ) 24 ∣ m 24\mid m 24 ∣ m m > 24 m>24 m > 24 p ∈ { 2 , 3 } p\in\{2,3\} p ∈ { 2 , 3 } 24 ∣ m / p 24\mid m/p 24 ∣ m / p m m m Q ( ζ m / p ) \mathbb Q(\zeta_{m/p}) Q ( ζ m / p ) p p p P ( ζ ) = 0 P(\zeta)=0 P ( ζ ) = 0
For m = 24 m=24 m = 24 K K K { 1 , 5 , 19 , 23 } ⊂ ( Z / 24 Z ) × \{1,5,19,23\}\subset(\mathbb Z/24\mathbb Z)^\times { 1 , 5 , 19 , 23 } ⊂ ( Z /24 Z ) ×
ζ 24 + ζ 24 5 + ζ 24 19 + ζ 24 23 = 6 . \zeta_{24}+\zeta_{24}^{5}+\zeta_{24}^{19}+\zeta_{24}^{23}=\sqrt6. ζ 24 + ζ 24 5 + ζ 24 19 + ζ 24 23 = 6 . The sum over the other coset is − 6 -\sqrt6 − 6 24 24 24 6 / 4 \sqrt6/4 6 /4 − 6 / 4 -\sqrt6/4 − 6 /4 ζ 2 \zeta^2 ζ 2 ζ 2 \zeta^2 ζ 2 24 24 24 m = 48 m=48 m = 48
Since a ( z ) a(z) a ( z ) L L L Q \mathbb Q Q
a ( ζ j ) = μ ( e j ) φ ( e j ) , e j = m gcd ( m , j ) . \begin{equation}
a(\zeta^j)=\frac{\mu(e_j)}{\varphi(e_j)},\qquad
e_j=\frac m{\gcd(m,j)}. \tag*{(3.47)}
\end{equation} a ( ζ j ) = φ ( e j ) μ ( e j ) , e j = g cd( m , j ) m . (3.47) Here μ \mu μ φ \varphi φ m = 2 u 3 v m=2^u3^v m = 2 u 3 v j = 2 , 4 j=2,4 j = 2 , 4 (3.47) gives the final column of (3.46) . In particular, if v ≥ 2 v\ge2 v ≥ 2 v ≤ 1 v\le1 v ≤ 1 u ≤ 3 u\le3 u ≤ 3 u u u
Theorem 3.22. Let Λ \Lambda Λ 2 , 3 2,3 2 , 3 p j = ∑ ζ ∈ Λ ζ j p_j=\sum_{\zeta\in\Lambda}\zeta^j p j = ∑ ζ ∈ Λ ζ j L L L ζ 24 \zeta_{24} ζ 24 a , b a,b a , b (3.45) . Then ∣ Λ ∣ ≥ 6 ∣ b ( p 1 ) ∣ + 4 ∣ b ( p 2 ) ∣ − a ( p 2 + p 4 ) .
|\Lambda|\ge 6|b(p_1)|+4|b(p_2)|-a(p_2+p_4).
∣Λ∣ ≥ 6∣ b ( p 1 ) ∣ + 4∣ b ( p 2 ) ∣ − a ( p 2 + p 4 ) .
Proof. Each row of (3.46) gives the pointwise inequality
1 + a ( ζ 2 ) + a ( ζ 4 ) ≥ 6 ∣ b ( ζ ) ∣ + 4 ∣ b ( ζ 2 ) ∣ . 1+a(\zeta^2)+a(\zeta^4)
\ge6|b(\zeta)|+4|b(\zeta^2)|. 1 + a ( ζ 2 ) + a ( ζ 4 ) ≥ 6∣ b ( ζ ) ∣ + 4∣ b ( ζ 2 ) ∣. Sum over the multiset and apply the triangle inequality separately to the two sums of b b b a , b a,b a , b (3.48) . ◻
Example 3.23. The conditions
p 1 = − 2 6 , p 2 = − 2 − 8 6 , p 4 = − 2 \begin{equation}
p_1=-2\sqrt6,\qquad p_2=-2-8\sqrt6,\qquad p_4=-2 \tag*{(3.49)}
\end{equation} p 1 = − 2 6 , p 2 = − 2 − 8 6 , p 4 = − 2 (3.49) are attained by a multiset of size 48 48 48
E 24 = { ζ 24 r : r ∈ { 7 , 11 , 13 , 17 } } , E 48 = { ζ 48 r : 0 ≤ r < 48 , r m o d 24 ∈ { 7 , 11 , 13 , 17 } } , E_{24}=\{\zeta_{24}^r:r\in\{7,11,13,17\}\},\qquad
E_{48}=\{\zeta_{48}^r:0\le r<48,\ r\bmod24\in\{7,11,13,17\}\}, E 24 = { ζ 24 r : r ∈ { 7 , 11 , 13 , 17 }} , E 48 = { ζ 48 r : 0 ≤ r < 48 , r mod 24 ∈ { 7 , 11 , 13 , 17 }} , and let P m \mathcal P_m P m m m m
Λ ∗ = 2 E 24 ⊔ 4 E 48 ⊔ P 8 ⊔ P 3 ⊔ P 6 . \Lambda_*=2E_{24}\sqcup4E_{48}\sqcup\mathcal P_8
\sqcup\mathcal P_3\sqcup\mathcal P_6. Λ ∗ = 2 E 24 ⊔ 4 E 48 ⊔ P 8 ⊔ P 3 ⊔ P 6 . The contributions to ( p 1 , p 2 , p 4 ) (p_1,p_2,p_4) ( p 1 , p 2 , p 4 )
( − 2 6 , 0 , 4 ) , ( 0 , − 8 6 , 0 ) , ( 0 , 0 , − 4 ) , ( − 1 , − 1 , − 1 ) , ( 1 , − 1 , − 1 ) . (-2\sqrt6,0,4),\quad (0,-8\sqrt6,0),\quad
(0,0,-4),\quad(-1,-1,-1),\quad(1,-1,-1). ( − 2 6 , 0 , 4 ) , ( 0 , − 8 6 , 0 ) , ( 0 , 0 , − 4 ) , ( − 1 , − 1 , − 1 ) , ( 1 , − 1 , − 1 ) . They sum to (3.49) , and the cardinality is 8 + 32 + 4 + 2 + 2 = 48 8+32+4+2+2=48 8 + 32 + 4 + 2 + 2 = 48
Theorem 3.24. The near-Tambara–Yamagami fusion category C = C ( D , κ ) \mathcal C=\mathcal C(\mathcal D,\kappa) C = C ( D , κ ) n = 2 n=2 n = 2 κ = 8 \kappa=8 κ = 8
Proof. Work with the hypothetical pseudo-unitary category and the multiset Λ \Lambda Λ 3.20 .
The norm of the global dimension is
N K / Q ( D ) = ( 48 + 16 6 ) ( 48 − 16 6 ) = 768 = 2 8 ⋅ 3. N_{K/\mathbb Q}(D)=(48+16\sqrt6)(48-16\sqrt6)=768=2^8\cdot3. N K / Q ( D ) = ( 48 + 16 6 ) ( 48 − 16 6 ) = 768 = 2 8 ⋅ 3. Let N 0 N_0 N 0 C \mathcal C C [23 Theorem 3.9] identifies the prime ideals dividing the global dimension with those dividing N 0 N_0 N 0 N 0 N_0 N 0 2 2 2 3 3 3 N 0 N_0 N 0 Z ( C ) \mathcal Z(\mathcal C) Z ( C ) [38 Theorem 5.5] , so all roots in Λ \Lambda Λ 3.22 , taking N = lcm ( 24 , N 0 ) N=\operatorname{lcm}(24,N_0) N = lcm ( 24 , N 0 )
Equations (3.42) and r8:eq:p2 give
b ( p 1 ) = − 2 , b ( p 2 ) = 4 ϵ ρ − 12 , a ( p 2 ) = − 2 , ϵ ρ = ν 2 ( ρ ) ∈ { 1 , − 1 } . b(p_1)=-2,\qquad b(p_2)=4\epsilon_{\rho}-12,\qquad a(p_2)=-2,
\qquad \epsilon_{\rho}=\nu_2(\rho)\in\{1,-1\}. b ( p 1 ) = − 2 , b ( p 2 ) = 4 ϵ ρ − 12 , a ( p 2 ) = − 2 , ϵ ρ = ν 2 ( ρ ) ∈ { 1 , − 1 } . In particular ∣ b ( p 2 ) ∣ ≥ 8 |b(p_2)|\ge8 ∣ b ( p 2 ) ∣ ≥ 8 (3.24) , ν 4 ( ρ ) − ν 4 ( t ) ∈ Z [ i ] \nu_4(\rho)-\nu_4(t)\in\mathbb Z[\mathrm i] ν 4 ( ρ ) − ν 4 ( t ) ∈ Z [ i ] K K K K K K r8:eq:p4 , the term multiplied by 6 \sqrt6 6 a ( p 4 ) = − 2 a(p_4)=-2 a ( p 4 ) = − 2
a ( p 2 + p 4 ) = − 4. a(p_2+p_4)=-4. a ( p 2 + p 4 ) = − 4. The mixed-moment inequality now requires
46 = ∣ Λ ∣ ≥ 6 ⋅ 2 + 4 ⋅ 8 + 4 = 48 , 46=|\Lambda|\ge6\cdot2+4\cdot8+4=48, 46 = ∣Λ∣ ≥ 6 ⋅ 2 + 4 ⋅ 8 + 4 = 48 , a contradiction. This proves the theorem. ◻
The rational near-Tambara–Yamagami categories
Assume in this section that d = d + ∈ Z d=d_+\in\mathbb Z d = d + ∈ Z C \mathcal C C n = 2 n=2 n = 2
Integral parameter restrictions
Lemma 4.1. There is a positive integer m m m d − = − m , d m = 2 n 2 , d > m , κ = d − m . d_-= -m,\qquad dm=2n^2,\qquad d>m,\qquad \kappa=d-m. d − = − m , d m = 2 n 2 , d > m , κ = d − m . a , m ≥ 1 a,m\geq1 a , m ≥ 1 n = a m , d = 2 a 2 m , κ = m ( 2 a 2 − 1 ) .
n=am,\qquad d=2a^2m,\qquad \kappa=m(2a^2-1).
n = am , d = 2 a 2 m , κ = m ( 2 a 2 − 1 ) .
Proof. The equation d ( d − κ ) = 2 n 2 d(d-\kappa)=2n^2 d ( d − κ ) = 2 n 2 d − = − m d_-=-m d − = − m m m m d > m d>m d > m κ > 0 \kappa>0 κ > 0 κ = 0 \kappa=0 κ = 0 d = n 2 d=n\sqrt2 d = n 2 d − d_- d −
N + d 2 N + d − 2 = − d d − = d 2 N . \frac{N+d^2}{N+d_-^2}=-\frac d{d_-}=\frac{d^2}{N}. N + d − 2 N + d 2 = − d − d = N d 2 . This rational algebraic integer is an integer. Thus ( d / n ) 2 = 2 r (d/n)^2=2r ( d / n ) 2 = 2 r r ∈ Z > 0 r\in\mathbb Z_{>0} r ∈ Z > 0 d / n d/n d / n d / n = 2 a d/n=2a d / n = 2 a a ≥ 1 a\geq1 a ≥ 1 d = 2 a n d=2an d = 2 an d m = 2 n 2 dm=2n^2 d m = 2 n 2 n = a m n=am n = am d = 2 a 2 m d=2a^2m d = 2 a 2 m κ = d − m = m ( 2 a 2 − 1 ) \kappa=d-m=m(2a^2-1) κ = d − m = m ( 2 a 2 − 1 )
Lemma 4.2. With the notation of (4.1) , write the multiplicity-free decomposition I ( 1 ) ≅ 1 ⊕ A ⊕ B ⊕ ⨁ i = 1 n 2 − 1 Y i . I(\mathbf 1)\cong \mathbf 1\oplus A\oplus B\oplus
\bigoplus_{i=1}^{n^2-1}Y_i . I ( 1 ) ≅ 1 ⊕ A ⊕ B ⊕ ⨁ i = 1 n 2 − 1 Y i . 1 , A , B , Y 1 , … , Y n 2 − 1 \mathbf 1,A,B,Y_1,\ldots,Y_{n^2-1} 1 , A , B , Y 1 , … , Y n 2 − 1 Z ( C ) \mathcal Z(\mathcal C) Z ( C ) FPdim ( A ) = 2 a 2 , FPdim ( B ) = 2 a 2 + 1 , FPdim ( Y i ) = 4 a 2 + 2. \operatorname{FPdim}(A)=2a^2,\qquad
\operatorname{FPdim}(B)=2a^2+1,\qquad
\operatorname{FPdim}(Y_i)=4a^2+2. FPdim ( A ) = 2 a 2 , FPdim ( B ) = 2 a 2 + 1 , FPdim ( Y i ) = 4 a 2 + 2. Z Z Z [ F ( Z ) , 1 ] = 1 [F(Z),\mathbf 1]=1 [ F ( Z ) , 1 ] = 1
Proof. This is Lemma 2.2 after substituting d = 2 a 2 m d=2a^2m d = 2 a 2 m m = − d − m=-d_- m = − d − K 0 ( C ) K_0(\mathcal C) K 0 ( C )
Proposition 4.3. In (4.1) , m ∈ { 1 , 2 } m\in\{1,2\} m ∈ { 1 , 2 }
Proof. The largest dimension of a summand of I ( 1 ) I(\mathbf 1) I ( 1 ) 4 a 2 + 2 4a^2+2 4 a 2 + 2 4.2 . Since κ > 0 \kappa>0 κ > 0 (2.11) implies that some summand Z Z Z [ F ( Z ) , ρ ] > 0 [F(Z),\rho]>0 [ F ( Z ) , ρ ] > 0 1 \mathbf 1 1
FPdim ( Z ) ≥ 1 + d = 1 + 2 a 2 m . \operatorname{FPdim}(Z)\geq 1+d=1+2a^2m. FPdim ( Z ) ≥ 1 + d = 1 + 2 a 2 m . If m ≥ 3 m\geq3 m ≥ 3 1 + 2 a 2 m > 4 a 2 + 2 1+2a^2m>4a^2+2 1 + 2 a 2 m > 4 a 2 + 2 a ≥ 1 a\geq1 a ≥ 1 m ≤ 2 m\leq2 m ≤ 2
The branch m = 1 m=1 m = 1
Here n = a n=a n = a d = 2 n 2 d=2n^2 d = 2 n 2 κ = 2 n 2 − 1 \kappa=2n^2-1 κ = 2 n 2 − 1 N = 2 n 2 = d N=2n^2=d N = 2 n 2 = d I ( 1 ) I(\mathbf 1) I ( 1 ) 1 , N , N + 1 , 2 N + 2 1,N,N+1,2N+2 1 , N , N + 1 , 2 N + 2
Lemma 4.4. Let A , B , Y i A,B,Y_i A , B , Y i 4.2 . Thus FPdim ( A ) = N , FPdim ( B ) = N + 1 , FPdim ( Y i ) = 2 N + 2. \operatorname{FPdim}(A)=N,\qquad \operatorname{FPdim}(B)=N+1,\qquad
\operatorname{FPdim}(Y_i)=2N+2 . FPdim ( A ) = N , FPdim ( B ) = N + 1 , FPdim ( Y i ) = 2 N + 2. G G G Y i Y_i Y i g ∈ G ∖ { 1 } g\in G\setminus\{\mathbf 1\} g ∈ G ∖ { 1 } X g X_g X g F ( A ) ≅ ⨁ g ∈ G g ⊕ n t , F ( B ) ≅ 1 ⊕ ρ , F ( X g ) ≅ 1 ⊕ g ⊕ 2 ρ ( g ∈ G ∖ { 1 } ) . F(A)\cong \bigoplus_{g\in G}g\oplus n t,\qquad F(B)\cong \mathbf 1\oplus \rho,\qquad
F(X_g)\cong \mathbf 1\oplus g\oplus 2 \rho\quad(g\in G\setminus\{\mathbf 1\}). F ( A ) ≅ ⨁ g ∈ G g ⊕ n t , F ( B ) ≅ 1 ⊕ ρ , F ( X g ) ≅ 1 ⊕ g ⊕ 2 ρ ( g ∈ G ∖ { 1 }) .
Proof. By Lemma 4.2 , every image contains 1 \mathbf 1 1 A , B , Y i A,B,Y_i A , B , Y i ρ \rho ρ F ( I ( 1 ) ) F(I(\mathbf 1)) F ( I ( 1 )) κ = N − 1 \kappa=N-1 κ = N − 1 0 , 1 , 2 0,1,2 0 , 1 , 2 ρ \rho ρ A , B , Y i A,B,Y_i A , B , Y i
0 + 1 + 2 ( n 2 − 1 ) = N − 1 , 0+1+2(n^2-1)=N-1, 0 + 1 + 2 ( n 2 − 1 ) = N − 1 , which is the total coefficient. Thus all three bounds are attained. The remaining dimension gives F ( B ) ≅ 1 ⊕ ρ F(B)\cong \mathbf 1\oplus \rho F ( B ) ≅ 1 ⊕ ρ Y i Y_i Y i 1 ⊕ u i ⊕ 2 ρ \mathbf 1\oplus u_i\oplus 2 \rho 1 ⊕ u i ⊕ 2 ρ u i ∈ G u_i\in G u i ∈ G t t t n n n t t t F ( A ) F(A) F ( A )
F ( A ) ≅ ⨁ g ∈ G a g g ⊕ n t . F(A)\cong \bigoplus_{g\in G}a_g g\oplus n t . F ( A ) ≅ g ∈ G ⨁ a g g ⊕ n t . Since FPdim ( A ) = N = 2 n 2 \operatorname{FPdim}(A)=N=2n^2 FPdim ( A ) = N = 2 n 2 ∑ g ∈ G a g = n 2 \sum_{g\in G}a_g=n^2 ∑ g ∈ G a g = n 2 g g g t t t
n = [ F ( I ( g ) ) , t ] = ∑ Z [ F ( Z ) , g ] [ F ( Z ) , t ] ≥ a g n , n=[F(I(g)),t]
=\sum_Z[F(Z),g][F(Z),t]\geq a_g n, n = [ F ( I ( g )) , t ] = Z ∑ [ F ( Z ) , g ] [ F ( Z ) , t ] ≥ a g n , so a g ≤ 1 a_g\leq1 a g ≤ 1 g g g a g = 1 a_g=1 a g = 1 g ∈ G g\in G g ∈ G g g g F ( I ( 1 ) ) F(I(\mathbf 1)) F ( I ( 1 )) F ( A ) F(A) F ( A ) Y i Y_i Y i G G G X g X_g X g
We use the convention for the image of a tensor functor from § 2.3 .
Lemma 4.5. Let L : U → V L:\mathcal U\to\mathcal V L : U → V R R R FPdim ( U ) FPdim ( V ) = FPdim ( R ( 1 ) ) = ∑ X ∈ Irr ( U ) [ L ( X ) , 1 ] FPdim ( X ) .
\frac{\operatorname{FPdim}(\mathcal U)}{\operatorname{FPdim}(\mathcal V)}
=\operatorname{FPdim}(R(\mathbf 1))=\sum_{X\in\operatorname{Irr}(\mathcal U)}[L(X),\mathbf 1]\operatorname{FPdim}(X).
FPdim ( V ) FPdim ( U ) = FPdim ( R ( 1 )) = ∑ X ∈ Irr ( U ) [ L ( X ) , 1 ] FPdim ( X ) . R ( 1 ) ≅ 1 R(\mathbf 1)\cong \mathbf 1 R ( 1 ) ≅ 1 L L L U ⊂ Z ( C ) \mathcal U\subset\mathcal Z(\mathcal C) U ⊂ Z ( C ) L = F ∣ U L=F|_{\mathcal U} L = F ∣ U R ( 1 ) R(\mathbf 1) R ( 1 ) I ( 1 ) I(\mathbf 1) I ( 1 ) U \mathcal U U
Proof. The image of the canonical regular element of K 0 ( U ) K_0(\mathcal U) K 0 ( U ) FPdim ( U ) / FPdim ( V ) \operatorname{FPdim}(\mathcal U)/\operatorname{FPdim}(\mathcal V) FPdim ( U ) / FPdim ( V ) K 0 ( V ) K_0(\mathcal V) K 0 ( V ) [29 Proposition 3.3.13] . Taking the coefficient of 1 \mathbf 1 1 (4.2) . If R ( 1 ) ≅ 1 R(\mathbf 1)\cong \mathbf 1 R ( 1 ) ≅ 1 Hom ( L ( X ) , L ( Y ) ) \operatorname{Hom}(L(X),L(Y)) Hom ( L ( X ) , L ( Y )) Hom ( X , Y ) \operatorname{Hom}(X,Y) Hom ( X , Y ) [ R ( 1 ) , Z ] = [ F ( Z ) , 1 ] = [ I ( 1 ) , Z ] [R(\mathbf 1),Z]=[F(Z),\mathbf 1]=[I(\mathbf 1),Z] [ R ( 1 ) , Z ] = [ F ( Z ) , 1 ] = [ I ( 1 ) , Z ]
Proposition 4.6. If m = 1 m=1 m = 1 n = 2 n=2 n = 2 G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2 d = 8 d=8 d = 8 κ = 7 \kappa=7 κ = 7
Proof. Let C p t \mathcal C_{\mathrm{pt}} C pt C \mathcal C C P = F − 1 ( C p t ) ⊂ Z ( C ) \mathcal P=F^{-1}(\mathcal C_{\mathrm{pt}})\subset\mathcal Z(\mathcal C) P = F − 1 ( C pt ) ⊂ Z ( C ) 4.4 , the only summand of I ( 1 ) I(\mathbf 1) I ( 1 ) 1 \mathbf 1 1 F ∣ P F|_{\mathcal P} F ∣ P 1 \mathbf 1 1 1 \mathbf 1 1 4.5 makes F ∣ P F|_{\mathcal P} F ∣ P P \mathcal P P P \mathcal P P
The unique dimension-N N N A A A I ( 1 ) I(\mathbf 1) I ( 1 ) D \mathcal D D F ( A ) 2 ≅ N F ( A ) F(A)^2\cong N F(A) F ( A ) 2 ≅ N F ( A ) I ( 1 ) I(\mathbf 1) I ( 1 ) 1 \mathbf 1 1 A A A A 2 A^2 A 2 ρ \rho ρ F F F [ F ( A 2 ) , 1 ] = N [F(A^2),\mathbf 1]=N [ F ( A 2 ) , 1 ] = N
A 2 ≅ 1 ⊕ ( N − 1 ) A ⊕ W , F ( W ) ≅ ⨁ g ≠ 1 g ⊕ n t . \begin{equation}
A^2\cong\mathbf 1\oplus(N-1)A\oplus W,
\qquad F(W)\cong\bigoplus_{g\ne\mathbf 1}g\oplus nt. \tag*{(4.3)}
\end{equation} A 2 ≅ 1 ⊕ ( N − 1 ) A ⊕ W , F ( W ) ≅ g = 1 ⨁ g ⊕ n t . (4.3) For a central simple Y Y Y a g ( Y ) = [ F ( Y ) , g ] a_g(Y)=[F(Y),g] a g ( Y ) = [ F ( Y ) , g ] b ( Y ) = [ F ( Y ) , t ] b(Y)=[F(Y),t] b ( Y ) = [ F ( Y ) , t ]
n = [ I ( g ) , I ( t ) ] = ∑ Y a g ( Y ) b ( Y ) ( g ∈ G ) . n=[I(g),I(t)]=\sum_Y a_g(Y)b(Y)\qquad(g\in G). n = [ I ( g ) , I ( t )] = Y ∑ a g ( Y ) b ( Y ) ( g ∈ G ) . The summand A A A 1 ⋅ n = n 1\cdot n=n 1 ⋅ n = n
a g ( Y ) b ( Y ) = 0 ( g ∈ G , Y ≠ A ) . \begin{equation}
a_g(Y)b(Y)=0\qquad(g\in G,\;Y\ne A). \tag*{(4.4)}
\end{equation} a g ( Y ) b ( Y ) = 0 ( g ∈ G , Y = A ) . (4.4) Every simple constituent of W W W F F F G G G t t t P \mathcal P P F ( W ) F(W) F ( W ) 1 \mathbf 1 1 g ~ \widetilde g g g ∈ G g\in G g ∈ G
Let T T T W W W F ( T ) ≅ b t F(T)\cong bt F ( T ) ≅ b t b > 0 b>0 b > 0 T ⊗ T ∗ T\otimes T^* T ⊗ T ∗ P \mathcal P P
F ( T ⊗ T ∗ ) ≅ ⨁ g ∈ G b 2 g , [ T ⊗ T ∗ , g ~ ] = [ T , g ~ ⊗ T ] ≤ 1. F(T\otimes T^*)\cong\bigoplus_{g\in G}b^2g,
\qquad
[T\otimes T^*,\widetilde g]
=[T,\widetilde g\otimes T]\le1. F ( T ⊗ T ∗ ) ≅ g ∈ G ⨁ b 2 g , [ T ⊗ T ∗ , g ] = [ T , g ⊗ T ] ≤ 1. Uniqueness of the lifts therefore forces b = 1 b=1 b = 1 T ′ T' T ′ F ( T ′ ) ≅ t F(T')\cong t F ( T ′ ) ≅ t T ⊗ T ′ ∗ T\otimes T'^* T ⊗ T ′ ∗ P \mathcal P P 1 \mathbf 1 1 [ T , T ′ ] = [ T ⊗ T ′ ∗ , 1 ] = 1 [T,T']=[T\otimes T'^*,\mathbf 1]=1 [ T , T ′ ] = [ T ⊗ T ′ ∗ , 1 ] = 1 T ≅ T ′ T\cong T' T ≅ T ′ T ≅ T ∗ T\cong T^* T ≅ T ∗ g ~ T ≅ T \widetilde gT\cong T g T ≅ T
T 2 ≅ ⨁ g ∈ G g ~ , g ~ h ~ ≅ g h ~ . T^2\cong\bigoplus_{g\in G}\widetilde g,
\qquad
\widetilde g\widetilde h\cong\widetilde{gh}. T 2 ≅ g ∈ G ⨁ g , g h ≅ g h . The coefficient of t t t F ( W ) F(W) F ( W ) [ W , T ] = n [W,T]=n [ W , T ] = n
[ T A , A ] = [ A 2 , T ] = n , T A ≅ n A . [T A,A]=[A^2,T]=n,\qquad TA\cong nA. [ T A , A ] = [ A 2 , T ] = n , T A ≅ n A . Thus the subcategory on { g ~ : g ∈ G } ∪ { T } \{\widetilde g:g\in G\}\cup\{T\} { g : g ∈ G } ∪ { T } [42 Theorem 1.2(1)] , G G G ∣ G ∣ = n 2 |G|=n^2 ∣ G ∣ = n 2 n ≥ 2 n\ge2 n ≥ 2 n = 2 s n=2^s n = 2 s s ≥ 1 s\ge1 s ≥ 1
We next establish the closure and twists needed for the Tannakian argument. The simple g ~ A \widetilde gA g A N N N F ( g ~ A ) F(\widetilde gA) F ( g A ) 1 \mathbf 1 1 I ( 1 ) I(\mathbf 1) I ( 1 ) N N N A A A g ~ A ≅ A \widetilde gA\cong A g A ≅ A (4.3) becomes
A 2 ≅ 1 ⊕ ( N − 1 ) A ⊕ ⨁ g ≠ 1 g ~ ⊕ n T . \begin{equation}
A^2\cong\mathbf 1\oplus(N-1)A\oplus
\bigoplus_{g\ne\mathbf 1}\widetilde g\oplus nT. \tag*{(4.5)}
\end{equation} A 2 ≅ 1 ⊕ ( N − 1 ) A ⊕ g = 1 ⨁ g ⊕ n T . (4.5) Together with the preceding product formulas and self-duality of A , T A,T A , T
{ g ~ : g ∈ G } ∪ { T , A } \{\widetilde g:g\in G\}\cup\{T,A\} { g : g ∈ G } ∪ { T , A } is the complete simple-object list of a fusion subcategory E ⊂ Z ( C ) \mathcal E\subset\mathcal Z(\mathcal C) E ⊂ Z ( C )
The character associated with A A A χ − \chi_- χ − χ − ( g ) = 1 \chi_-(g)=1 χ − ( g ) = 1 χ − ( t ) = n \chi_-(t)=n χ − ( t ) = n χ − ( ρ ) = − 1 \chi_-(\rho)=-1 χ − ( ρ ) = − 1 (2.6) gives
S A , A = dim ( A ) χ − ( F ( A ) ) = N 2 . \mathsf S_{A,A}
=\dim(A)\chi_-(F(A))=N^2. S A , A = dim ( A ) χ − ( F ( A )) = N 2 . As A ⊂ I ( 1 ) A\subset I(\mathbf 1) A ⊂ I ( 1 ) θ A = 1 \theta_A=1 θ A = 1 (4.5) consequently gives
N 2 = 1 + ( N − 1 ) N + ∑ g ≠ 1 θ g ~ + n 2 θ T , N^2=1+(N-1)N+
\sum_{g\ne\mathbf 1}\theta_{\widetilde g}+n^2\theta_T, N 2 = 1 + ( N − 1 ) N + g = 1 ∑ θ g + n 2 θ T , or
∑ g ≠ 1 θ g ~ + n 2 θ T = N − 1 = ( n 2 − 1 ) + n 2 . \sum_{g\ne\mathbf 1}\theta_{\widetilde g}+n^2\theta_T
=N-1=(n^2-1)+n^2. g = 1 ∑ θ g + n 2 θ T = N − 1 = ( n 2 − 1 ) + n 2 . All twists have absolute value one. Equality with the sum of the positive weights forces θ g ~ = θ T = 1 \theta_{\widetilde g}= \theta_T=1 θ g = θ T = 1 E \mathcal E E E \mathcal E E [29 Theorem 9.9.22] makes E \mathcal E E
Write E ≃ Rep ( K ) \mathcal E\simeq\operatorname{Rep}(K) E ≃ Rep ( K )
∣ K ∣ = ∣ G ∣ + n 2 + N 2 = N ( N + 1 ) . |K|=|G|+n^2+N^2=N(N+1). ∣ K ∣ = ∣ G ∣ + n 2 + N 2 = N ( N + 1 ) . Its Tambara–Yamagami subcategory is Rep ( K / V ) \operatorname{Rep}(K/V) Rep ( K / V ) V ⊲ K V\lhd K V ⊲ K ∣ K / V ∣ = N |K/V|=N ∣ K / V ∣ = N ∣ V ∣ = N + 1 |V|=N+1 ∣ V ∣ = N + 1 K / V K/V K / V A A A N N N
Res V K C [ K ] ≅ N C [ V ] . \operatorname{Res}^K_V\mathbb C[K]\cong N\mathbb C[V]. Res V K C [ K ] ≅ N C [ V ] . The inflated part of C [ K ] \mathbb C[K] C [ K ] N 1 V N\mathbf 1_V N 1 V A A A N N N
Res V K A ⊕ 1 V ≅ C [ V ] . \operatorname{Res}^K_VA\oplus\mathbf 1_V\cong\mathbb C[V]. Res V K A ⊕ 1 V ≅ C [ V ] . Consequently χ A ( v ) = − 1 \chi_A(v)=-1 χ A ( v ) = − 1 v ∈ V ∖ { 1 } v\in V\setminus\{1\} v ∈ V ∖ { 1 } K / V K/V K / V V V V T T T n n n K K K V ∖ { 1 } V\setminus\{1\} V ∖ { 1 } K K K
All its elements have a common prime order p p p V V V p p p Z ( V ) Z(V) Z ( V ) V V V K K K V ∖ { 1 } V\setminus\{1\} V ∖ { 1 } Z ( V ) = V Z(V)=V Z ( V ) = V V V V p p p V ≅ ( C p ) r V\cong(C_p)^r V ≅ ( C p ) r N + 1 = p r N+1=p^r N + 1 = p r
We obtain 2 2 s + 1 + 1 = p r 2^{2s+1}+1=p^r 2 2 s + 1 + 1 = p r p = 3 p=3 p = 3 r = 2 u r=2u r = 2 u
( 3 u − 1 ) ( 3 u + 1 ) = 2 2 s + 1 . (3^u-1)(3^u+1)=2^{2s+1}. ( 3 u − 1 ) ( 3 u + 1 ) = 2 2 s + 1 . The two factors are positive powers of two differing by two, so they are 2 2 2 4 4 4 u = s = 1 u=s=1 u = s = 1 n = 2 n=2 n = 2 G G G G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2 d = N = 8 d=N=8 d = N = 8 κ = N − 1 = 7 \kappa=N-1=7 κ = N − 1 = 7
The branch m = 2 m=2 m = 2
Write n = 2 a n=2a n = 2 a q = n 2 / 2 = 2 a 2 q=n^2/2=2a^2 q = n 2 /2 = 2 a 2
d = 2 q , κ = 2 q − 2 , FPdim ( D ) = 4 q , Δ = 4 q ( q + 1 ) . \begin{equation}
d=2q,\qquad \kappa=2q-2,\qquad
\operatorname{FPdim}(\mathcal D)=4q,\qquad \Delta=4q(q+1). \tag*{(4.6)}
\end{equation} d = 2 q , κ = 2 q − 2 , FPdim ( D ) = 4 q , Δ = 4 q ( q + 1 ) . (4.6) Lemma 4.7. Let A , B , Y i A,B,Y_i A , B , Y i 4.2 . In the m = 2 m=2 m = 2 FPdim ( A ) = q , FPdim ( B ) = q + 1 , FPdim ( Y i ) = 2 q + 2. \operatorname{FPdim}(A)=q,\qquad \operatorname{FPdim}(B)=q+1,\qquad
\operatorname{FPdim}(Y_i)=2q+2. FPdim ( A ) = q , FPdim ( B ) = q + 1 , FPdim ( Y i ) = 2 q + 2. 2 q − 1 2q-1 2 q − 1 Y i Y_i Y i 2 q + 2 2q+2 2 q + 2 ρ \rho ρ X 0 X_0 X 0 F ( X 0 ) F(X_0) F ( X 0 ) t t t
Proof. The summands of dimensions q q q q + 1 q+1 q + 1 ρ \rho ρ 2 q + 2 2q+2 2 q + 2 ρ \rho ρ ρ \rho ρ 2 q − 2 2q-2 2 q − 2 X 0 X_0 X 0 ρ \rho ρ F ( X ) ≅ 1 ⊕ g ⊕ ρ F(X)\cong \mathbf 1\oplus g\oplus \rho F ( X ) ≅ 1 ⊕ g ⊕ ρ g ≠ 1 g\ne \mathbf 1 g = 1 q = 2 a 2 q=2a^2 q = 2 a 2 n = 2 a n=2a n = 2 a
[ F ( A ) , t ] ≤ a − 1 , [ F ( B ) , t ] ≤ a . [F(A),t]\le a-1,\qquad [F(B),t]\le a. [ F ( A ) , t ] ≤ a − 1 , [ F ( B ) , t ] ≤ a . Since [ F ( I ( 1 ) ) , t ] = 2 a [F(I(\mathbf 1)),t]=2a [ F ( I ( 1 )) , t ] = 2 a [ F ( X 0 ) , t ] ≥ 1 [F(X_0),t]\ge1 [ F ( X 0 ) , t ] ≥ 1
Proposition 4.8. Let E 2 \mathcal E_2 E 2 Z ( C ) \mathcal Z(\mathcal C) Z ( C ) D \mathcal D D E 2 = F − 1 ( D ) . \mathcal E_2=F^{-1}(\mathcal D). E 2 = F − 1 ( D ) . Z ( C ) \mathcal Z(\mathcal C) Z ( C ) E 2 ′ = { Z ∈ Z ( C ) : c W , Z ∘ c Z , W = id Z ⊗ W for every W ∈ E 2 } , \mathcal E_2'
=\left\{Z\in\mathcal Z(\mathcal C):
c_{W,Z}\circ c_{Z,W}=\operatorname{id}_{Z\otimes W}
\text{ for every }W\in\mathcal E_2\right\}, E 2 ′ = { Z ∈ Z ( C ) : c W , Z ∘ c Z , W = id Z ⊗ W for every W ∈ E 2 } , c c c Z ( C ) \mathcal Z(\mathcal C) Z ( C ) J = E 2 ′ \mathcal J=\mathcal E_2' J = E 2 ′ FPdim ( E 2 ) = 4 Δ , FPdim ( J ) = Δ / 4 = q ( q + 1 ) , \operatorname{FPdim}(\mathcal E_2)=4\Delta,\qquad
\operatorname{FPdim}(\mathcal J)=\Delta/4=q(q+1), FPdim ( E 2 ) = 4Δ , FPdim ( J ) = Δ/4 = q ( q + 1 ) , J \mathcal J J J ≃ Rep ( K ) , ∣ K ∣ = q ( q + 1 ) , F ( J ) ≃ Rep ( H ) , ∣ H ∣ = q , \mathcal J\simeq\operatorname{Rep}(K),\quad |K|=q(q+1),\qquad
F(\mathcal J)\simeq\operatorname{Rep}(H),\quad |H|=q, J ≃ Rep ( K ) , ∣ K ∣ = q ( q + 1 ) , F ( J ) ≃ Rep ( H ) , ∣ H ∣ = q , H ≤ K H\le K H ≤ K [ K : H ] = q + 1 [K:H]=q+1 [ K : H ] = q + 1 H ^ \widehat H H G G G F ∣ J F|_{\mathcal J} F ∣ J Res H K \operatorname{Res}^K_H Res H K Fun ( K / H ) ≅ 1 ⊕ A \operatorname{Fun}(K/H)\cong\mathbf 1\oplus A Fun ( K / H ) ≅ 1 ⊕ A
Proof. The image of F ∣ E 2 F|_{\mathcal E_2} F ∣ E 2 t t t 4.7 and hence is all of D \mathcal D D I ( 1 ) I(\mathbf 1) I ( 1 ) E 2 \mathcal E_2 E 2 1 ⊕ A ⊕ B ⊕ X 0 \mathbf 1\oplus A\oplus B\oplus X_0 1 ⊕ A ⊕ B ⊕ X 0 4 ( q + 1 ) 4(q+1) 4 ( q + 1 ) 4.5 gives FPdim ( E 2 ) = 4 q ⋅ 4 ( q + 1 ) = 4 Δ \operatorname{FPdim}(\mathcal E_2)=4q\cdot4(q+1)=4\Delta FPdim ( E 2 ) = 4 q ⋅ 4 ( q + 1 ) = 4Δ Z ( C ) \mathcal Z(\mathcal C) Z ( C ) [29 Corollary 8.21.7] gives FPdim ( J ) = Δ 2 / ( 4 Δ ) = Δ / 4 \operatorname{FPdim}(\mathcal J)=\Delta^2/(4\Delta)=\Delta/4 FPdim ( J ) = Δ 2 / ( 4Δ ) = Δ/4
If F ( J ) F(\mathcal J) F ( J ) ρ \rho ρ C \mathcal C C 4.5 would be 1 / 4 1/4 1/4 t t t D \mathcal D D ( q + 1 ) / 4 (q+1)/4 ( q + 1 ) /4 q q q F ( J ) F(\mathcal J) F ( J ) J ⊆ E 2 \mathcal J\subseteq\mathcal E_2 J ⊆ E 2 J ′ = E 2 \mathcal J'=\mathcal E_2 J ′ = E 2 J ⊆ J ′ \mathcal J\subseteq\mathcal J' J ⊆ J ′ J \mathcal J J
If J \mathcal J J [29 Corollary 9.9.32] . Its centralizer would then have dimension 8 Δ 8\Delta 8Δ E 2 \mathcal E_2 E 2 FPdim ( E 2 ) = 4 Δ \operatorname{FPdim}(\mathcal E_2)=4\Delta FPdim ( E 2 ) = 4Δ E 2 \mathcal E_2 E 2 ρ \rho ρ C \mathcal C C 4.5 is therefore 8 8 8 1 , A , B , X 0 \mathbf 1,A,B,X_0 1 , A , B , X 0 4 ( q + 1 ) ≥ 12 4(q+1)\ge12 4 ( q + 1 ) ≥ 12 J ≃ Rep ( K ) \mathcal J\simeq\operatorname{Rep}(K) J ≃ Rep ( K )
It remains to identify the right-adjoint unit image exactly. Let η B \eta_B η B K 0 ( C ) K_0(\mathcal C) K 0 ( C ) B B B
η B ( g ) = 1 ( g ∈ G ) , η B ( t ) = − n , η B ( ρ ) = 0 , FPdim ( B ) = q + 1. \eta_B(g)=1\quad(g\in G),\qquad
\eta_B(t)=-n,\qquad \eta_B(\rho)=0,
\qquad \operatorname{FPdim}(B)=q+1. η B ( g ) = 1 ( g ∈ G ) , η B ( t ) = − n , η B ( ρ ) = 0 , FPdim ( B ) = q + 1. Equation (2.6) gives, for every central simple Z Z Z
S B , Z = ( q + 1 ) η B ( F Z ) , \mathsf S_{B,Z}=(q+1)\eta_B(FZ), S B , Z = ( q + 1 ) η B ( F Z ) , with the normalization of (2.6) . Since F ( X 0 ) F(X_0) F ( X 0 ) t t t ρ \rho ρ F ( X 0 ) ≅ ⨁ g a g g ⊕ γ t F(X_0)\cong \bigoplus_g a_g g\oplus \gamma t F ( X 0 ) ≅ ⨁ g a g g ⊕ γ t γ ≥ 1 \gamma\ge1 γ ≥ 1 B B B J \mathcal J J B B B X 0 X_0 X 0
( q + 1 ) η B ( F X 0 ) = S B , X 0 = FPdim ( B ) FPdim ( X 0 ) . (q+1)\eta_B(FX_0)=\mathsf S_{B,X_0}
=\operatorname{FPdim}(B)\operatorname{FPdim}(X_0). ( q + 1 ) η B ( F X 0 ) = S B , X 0 = FPdim ( B ) FPdim ( X 0 ) . But
η B ( F X 0 ) = ∑ g a g − n γ < ∑ g a g + n γ = FPdim ( X 0 ) , \eta_B(FX_0)=\sum_g a_g-n\gamma
<\sum_g a_g+n\gamma=\operatorname{FPdim}(X_0), η B ( F X 0 ) = g ∑ a g − nγ < g ∑ a g + nγ = FPdim ( X 0 ) , a contradiction. Thus B ∉ J B\notin\mathcal J B ∈ / J X 0 X_0 X 0 J \mathcal J J t t t Y i Y_i Y i J \mathcal J J ρ \rho ρ I ( 1 ) I(\mathbf 1) I ( 1 ) J \mathcal J J 1 ⊕ A \mathbf 1\oplus A 1 ⊕ A 1 \mathbf 1 1 q ( q + 1 ) > ∣ G ∣ = 2 q q(q+1)>|G|=2q q ( q + 1 ) > ∣ G ∣ = 2 q C \mathcal C C ∣ G ∣ |G| ∣ G ∣ 1 ⊕ A \mathbf 1\oplus A 1 ⊕ A 4.5 gives FPdim ( F ( J ) ) = q \operatorname{FPdim}(F(\mathcal J))=q FPdim ( F ( J )) = q
Set B = F ( J ) \mathcal B=F(\mathcal J) B = F ( J ) L = F ∣ J : J → B L=F|_{\mathcal J}:\mathcal J\to
\mathcal B L = F ∣ J : J → B B \mathcal B B L L L [25 Lemma 3.5] , its right-adjoint unit object A 0 = 1 ⊕ A A_0=\mathbf 1\oplus A A 0 = 1 ⊕ A J ≃ Rep ( K ) \mathcal J\simeq\operatorname{Rep}(K) J ≃ Rep ( K )
B ≃ Rep ( K ) A 0 . \mathcal B\simeq\operatorname{Rep}(K)_{A_0}. B ≃ Rep ( K ) A 0 . A finite-dimensional commutative separable algebra in Rep ( K ) \operatorname{Rep}(K) Rep ( K ) K K K H ≤ K H\le K H ≤ K
A 0 ≅ Fun ( K / H ) , [ K : H ] = dim ( A 0 ) = q + 1. A_0\cong\operatorname{Fun}(K/H),\qquad [K:H]=\dim(A_0)=q+1. A 0 ≅ Fun ( K / H ) , [ K : H ] = dim ( A 0 ) = q + 1. Taking the fiber of an equivariant module at the coset H H H Rep ( K ) A 0 ≃ Rep ( H ) \operatorname{Rep}(K)_{A_0}\simeq\operatorname{Rep}(H) Rep ( K ) A 0 ≃ Rep ( H ) Res H K \operatorname{Res}^K_H Res H K L L L B \mathcal B B H H H ∣ H ∣ = ∣ K ∣ / ( q + 1 ) = q |H|=|K|/(q+1)=q ∣ H ∣ = ∣ K ∣/ ( q + 1 ) = q C p t \mathcal C_{\mathrm{pt}} C pt H ^ \widehat H H H H H G G G
Proposition 4.9. If m = 2 m=2 m = 2 n = 2 n=2 n = 2 G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2 d = 4 d=4 d = 4 κ = 2 \kappa=2 κ = 2
Proof. By Proposition 4.8 , K K K Ω = K / H \Omega=K/H Ω = K / H ∣ Ω ∣ = q + 1 |\Omega|=q+1 ∣Ω∣ = q + 1 ∣ H ∣ = q |H|=q ∣ H ∣ = q
C [ Ω ] ≅ 1 ⊕ A . \mathbb C[\Omega]\cong\mathbf 1\oplus A. C [ Ω ] ≅ 1 ⊕ A . Thus dim End K ( C [ Ω ] ) = 2 \dim\operatorname{End}_K(\mathbb C[\Omega])=2 dim End K ( C [ Ω ]) = 2 K K K Ω × Ω \Omega\times\Omega Ω × Ω
∣ K ∣ = q ( q + 1 ) = ∣ Ω ∣ ( ∣ Ω ∣ − 1 ) . |K|=q(q+1)=|\Omega|(|\Omega|-1). ∣ K ∣ = q ( q + 1 ) = ∣Ω∣ ( ∣Ω∣ − 1 ) . The stabilizer of an ordered pair of distinct points is therefore trivial. In particular the action is faithful and sharply doubly transitive.
The Frobenius kernel theorem for finite sharply doubly transitive groups [26 Section 7.6] gives
K = V ⋊ H , ∣ V ∣ = q + 1 , K=V\rtimes H,\qquad |V|=q+1, K = V ⋊ H , ∣ V ∣ = q + 1 , where V V V H H H V ∖ { 1 } V\setminus\{1\} V ∖ { 1 } V V V p p p V V V p p p Z ( V ) Z(V) Z ( V ) Z ( V ) = V Z(V)=V Z ( V ) = V V ≅ ( C p ) r V\cong(C_p)^r V ≅ ( C p ) r q + 1 = p r q+1=p^r q + 1 = p r p ∤ ∣ H ∣ p\nmid|H| p ∤ ∣ H ∣
The group H H H 4.8 . Its representation on V ⊗ F p F ‾ p V\otimes_{\mathbb F_p}\overline{\mathbb F}_p V ⊗ F p F p h ∈ H h\in H h ∈ H h − I h-I h − I F ‾ p \overline{\mathbb F}_p F p F p \mathbb F_p F p F p \mathbb F_p F p h h h V ∖ { 0 } V\setminus\{0\} V ∖ { 0 } H H H F p \mathbb F_p F p H H H H ^ \widehat H H q q q H ^ ≤ G \widehat H\le G H ≤ G G G G q = n 2 / 2 q=n^2/2 q = n 2 /2
Lemma 2.3 gives ord ( g ) ≤ n \operatorname{ord}(g)\le n ord ( g ) ≤ n g ∈ G g\in G g ∈ G n 2 / 2 ≤ n n^2/2\le n n 2 /2 ≤ n n ≥ 2 n\ge2 n ≥ 2 n = 2 n=2 n = 2 G G G G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2 (4.6) gives d = 4 d=4 d = 4 κ = 2 \kappa=2 κ = 2
Theorem 4.10. If d + ∈ Z d_+\in\mathbb Z d + ∈ Z ( n , d + , κ ) = ( 2 , 4 , 2 ) or ( 2 , 8 , 7 ) , (n,d_+,\kappa)=(2,4,2)\quad\text{or}\quad(2,8,7), ( n , d + , κ ) = ( 2 , 4 , 2 ) or ( 2 , 8 , 7 ) , G ≅ C 2 × C 2 G\cong C_2\times C_2 G ≅ C 2 × C 2 rank ( C ) = 6 \operatorname{rank}(\mathcal C)=6 rank ( C ) = 6
Proof. Proposition 4.3 leaves m = 1 m=1 m = 1 m = 2 m=2 m = 2 4.6 and 4.9 give the two alternatives. ◻
Thus the integral case leaves R 2 R_2 R 2 R 7 R_7 R 7 (1.1) . Example 2.1 categorifies R 7 R_7 R 7 R 2 R_2 R 2
A group-theoretical categorification of R 2 R_2 R 2
Group-theoretical data
For a finite group L L L 3 3 3 ω \omega ω H H H 2 2 2 ψ \psi ψ δ ψ = ω ∣ H \delta\psi=\omega|_H δ ψ = ω ∣ H C ( L , ω , H , ψ ) \mathcal C(L,\omega,H,\psi) C ( L , ω , H , ψ ) H H H Vec L ω \operatorname{Vec}_L^\omega Vec L ω ( x , π ) (x,\pi) ( x , π ) x x x H x H HxH H x H π \pi π H x = H ∩ x H x − 1 H_x=H\cap xHx^{-1} H x = H ∩ x H x − 1
FPdim ( x , π ) = [ H : H x ] dim π . \begin{equation}
\operatorname{FPdim}(x,\pi)=[H:H_x]\dim\pi. \tag*{(4.7)}
\end{equation} FPdim ( x , π ) = [ H : H x ] dim π . (4.7) The double-coset support of tensor products is multiplicative, and the component supported on H H H Rep ( H ) \operatorname{Rep}(H) Rep ( H ) [35 Theorems 4.1 and 5.1] .
Set
L = PGL 2 ( F 3 ) ≅ S 4 . L=\operatorname{PGL}_2(\mathbb F_3)\cong S_4. L = PGL 2 ( F 3 ) ≅ S 4 . For x ∈ L x\in L x ∈ L s ( x ) ∈ GL 2 ( F 3 ) s(x)\in\operatorname{GL}_2(\mathbb F_3) s ( x ) ∈ GL 2 ( F 3 ) 1 1 1
s ( x ) s ( y ) = ( − I ) β ( x , y ) s ( x y ) , α ( x ) = { 0 , det s ( x ) = 1 , 1 , det s ( x ) = 2. \begin{equation}
s(x)s(y)=(-I)^{\beta(x,y)}s(xy),\qquad
\alpha(x)=
\begin{cases}
0,&\det s(x)=1,\\
1,&\det s(x)=2.
\end{cases} \tag*{(4.8)}
\end{equation} s ( x ) s ( y ) = ( − I ) β ( x , y ) s ( x y ) , α ( x ) = { 0 , 1 , det s ( x ) = 1 , det s ( x ) = 2. (4.8) Here β \beta β F 2 \mathbb F_2 F 2 2 2 2 { I , − I } \{I,-I\} { I , − I } α : L → F 2 \alpha:L\to\mathbb F_2 α : L → F 2 3 3 3
ω ( x , y , z ) = ( − 1 ) α ( x ) β ( y , z ) . \begin{equation}
\omega(x,y,z)=(-1)^{\alpha(x)\beta(y,z)}. \tag*{(4.9)}
\end{equation} ω ( x , y , z ) = ( − 1 ) α ( x ) β ( y , z ) . (4.9) Take H = { 1 , a , b , c } H=\{1,a,b,c\} H = { 1 , a , b , c }
s ( 1 ) = I , s ( a ) = ( 0 1 1 0 ) , s ( b ) = ( 0 1 2 0 ) , s ( c ) = ( 1 0 0 2 ) . \begin{equation}
s(1)=I,\qquad
s(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
s(b)=\begin{pmatrix}0&1\\2&0\end{pmatrix},\qquad
s(c)=\begin{pmatrix}1&0\\0&2\end{pmatrix}. \tag*{(4.10)}
\end{equation} s ( 1 ) = I , s ( a ) = ( 0 1 1 0 ) , s ( b ) = ( 0 2 1 0 ) , s ( c ) = ( 1 0 0 2 ) . (4.10) In the action on P 1 ( F 3 ) \mathbb P^1(\mathbb F_3) P 1 ( F 3 ) H ≅ C 2 2 H\cong C_2^2 H ≅ C 2 2 ψ ( x , y ) = i q ( x , y ) \psi(x,y)=\mathrm i^{q(x,y)} ψ ( x , y ) = i q ( x , y ) q ( 1 , x ) = q ( x , 1 ) = 0 q(1,x)=q(x,1)=0 q ( 1 , x ) = q ( x , 1 ) = 0
q a b c a 0 2 0 b 3 0 1 c 1 0 1 in Z / 4 Z . \begin{equation}
\begin{array}{c|rrr}
q&a&b&c\\ \hline
a&0&2&0\\
b&3&0&1\\
c&1&0&1
\end{array}
\qquad\text{in }\mathbb Z/4\mathbb Z. \tag*{(4.11)}
\end{equation} q a b c a 0 3 1 b 2 0 0 c 0 1 1 in Z /4 Z . (4.11) Lemma 4.11. The cochains (4.9) and (4.11) satisfy δ ψ = ω ∣ H \delta\psi=\omega|_H δ ψ = ω ∣ H
Proof. Write x = c v a u x=c^v a^u x = c v a u y = c s a r y=c^s a^r y = c s a r z = c w a p z=c^w a^p z = c w a p { 0 , 1 } \{0,1\} { 0 , 1 } (4.10) have s ( c ) s ( a ) = s ( b ) s(c)s(a)=s(b) s ( c ) s ( a ) = s ( b ) s ( a ) s ( c ) = − s ( c ) s ( a ) s(a)s(c)=-s(c)s(a) s ( a ) s ( c ) = − s ( c ) s ( a )
α ( x ) = u + v ( m o d 2 ) , β ( y , z ) = r w ( m o d 2 ) . \alpha(x)=u+v\pmod2,\qquad \beta(y,z)=rw\pmod2. α ( x ) = u + v ( mod 2 ) , β ( y , z ) = r w ( mod 2 ) . The table (4.11) is equivalently
q ( x , y ) = v ( r + s ) + 2 ( v r s + u r s + u v r ) ( m o d 4 ) . q(x,y)=v(r+s)+2(vrs+urs+uvr)\pmod4. q ( x , y ) = v ( r + s ) + 2 ( v r s + u r s + uv r ) ( mod 4 ) . Expanding products in H H H r ⊕ s = r + s − 2 r s r\mathbin{\oplus}s=r+s-2rs r ⊕ s = r + s − 2 r s
q ( y , z ) + q ( x , y z ) − q ( x y , z ) − q ( x , y ) ≡ 2 ( u + v ) r w ( m o d 4 ) . q(y,z)+q(x,yz)-q(xy,z)-q(x,y)
\equiv2(u+v)rw\pmod4. q ( y , z ) + q ( x , y z ) − q ( x y , z ) − q ( x , y ) ≡ 2 ( u + v ) r w ( mod 4 ) . This is precisely the exponent identity for δ ψ ( x , y , z ) = ω ( x , y , z ) \delta\psi(x,y,z)=\omega(x,y,z) δ ψ ( x , y , z ) = ω ( x , y , z )
The three double cosets
Choose
x = [ ( 1 1 1 2 ) ] , u = [ ( 0 1 1 1 ) ] . \begin{equation}
x=\left[\begin{pmatrix}1&1\\1&2\end{pmatrix}\right],\qquad
u=\left[\begin{pmatrix}0&1\\1&1\end{pmatrix}\right]. \tag*{(4.12)}
\end{equation} x = [ ( 1 1 1 2 ) ] , u = [ ( 0 1 1 1 ) ] . (4.12) Label the four projective points by ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) , ( 1 , 2 ) (1,0),(0,1),(1,1),(1,2) ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) , ( 1 , 2 ) a = ( 12 ) a=(12) a = ( 12 ) c = ( 34 ) c=(34) c = ( 34 ) x = ( 13 ) ( 24 ) x=(13)(24) x = ( 13 ) ( 24 ) u = ( 1234 ) u=(1234) u = ( 1234 ) H = ⟨ ( 12 ) , ( 34 ) ⟩ H=\langle(12),(34)\rangle H = ⟨( 12 ) , ( 34 )⟩ 8 8 8 x x x H H H u H u − 1 = ⟨ ( 23 ) , ( 14 ) ⟩ uHu^{-1}=\langle(23),(14)\rangle u H u − 1 = ⟨( 23 ) , ( 14 )⟩ H H H 1 , x , u 1,x,u 1 , x , u 4 , 4 , 16 4,4,16 4 , 4 , 16 L L L
L = H ⊔ x H ⊔ H u H , N L ( H ) = H ⊔ x H . \begin{equation}
L=H\sqcup xH\sqcup HuH,\qquad N_L(H)=H\sqcup xH. \tag*{(4.13)}
\end{equation} L = H ⊔ x H ⊔ H u H , N L ( H ) = H ⊔ x H . (4.13) For any y ∈ L y\in L y ∈ L H × H H\times H H × H H y H HyH H y H
∣ H y H ∣ = ∣ H ∣ 2 ∣ H ∩ y H y − 1 ∣ . |HyH|=\frac{|H|^2}{|H\cap yHy^{-1}|}. ∣ H y H ∣ = ∣ H ∩ y H y − 1 ∣ ∣ H ∣ 2 . For y = 1 y=1 y = 1 y = x y=x y = x H H H N L ( H ) N_L(H) N L ( H ) 4 4 4 y = u y=u y = u H H H 16 16 16 24 = ∣ L ∣ 24=|L| 24 = ∣ L ∣
For clarity, in the conventions of [35 Section 5.1] , the projective factor on H x H_x H x
ψ x ( h , k ) = ψ ( h , k ) ψ ( x − 1 k − 1 x , x − 1 h − 1 x ) ω ( h , k , x ) ω ( h , k x , x − 1 k − 1 x ) ω ( h k x , x − 1 k − 1 x , x − 1 h − 1 x ) . \begin{equation}
\psi^x(h,k)=\psi(h,k)\psi(x^{-1}k^{-1}x,x^{-1}h^{-1}x)
\frac{\omega(h,k,x)\omega(h,kx,x^{-1}k^{-1}x)}
{\omega(hkx,x^{-1}k^{-1}x,x^{-1}h^{-1}x)}. \tag*{(4.14)}
\end{equation} ψ x ( h , k ) = ψ ( h , k ) ψ ( x − 1 k − 1 x , x − 1 h − 1 x ) ω ( hk x , x − 1 k − 1 x , x − 1 h − 1 x ) ω ( h , k , x ) ω ( h , k x , x − 1 k − 1 x ) . (4.14) Substitution of (4.8) –(4.11) gives the exponent matrix for ψ x \psi^x ψ x 1 , a , b , c 1,a,b,c 1 , a , b , c
( 0 0 0 0 0 1 1 0 0 3 0 1 0 2 3 1 ) ( m o d 4 ) . \begin{equation}
\begin{pmatrix}
0&0&0&0\\
0&1&1&0\\
0&3&0&1\\
0&2&3&1
\end{pmatrix}\pmod4. \tag*{(4.15)}
\end{equation} 0 0 0 0 0 1 3 2 0 1 0 3 0 0 1 1 ( mod 4 ) . (4.15) For example, substituting ( h , k ) = ( a , a ) , ( a , b ) , ( b , a ) , ( b , c ) (h,k)=(a,a),(a,b),(b,a),(b,c) ( h , k ) = ( a , a ) , ( a , b ) , ( b , a ) , ( b , c ) (4.14) gives the exponents 1 , 1 , 3 , 1 1,1,3,1 1 , 1 , 3 , 1 ψ x ( h , k ) / ψ x ( k , h ) = − 1 \psi^x(h,k)/\psi^x(k,h)=-1 ψ x ( h , k ) / ψ x ( k , h ) = − 1 H H H
Z a l g ( C ψ x [ H ] ) = C , C ψ x [ H ] ≅ M 2 ( C ) . Z_{\mathrm{alg}}\bigl(\mathbb C_{\psi^x}[H]\bigr)=\mathbb C,\qquad
\mathbb C_{\psi^x}[H]\cong M_2(\mathbb C). Z alg ( C ψ x [ H ] ) = C , C ψ x [ H ] ≅ M 2 ( C ) . Theorem 4.12. The category C ( L , ω , H , ψ ) \mathcal C(L,\omega,H,\psi) C ( L , ω , H , ψ ) R 2 R_2 R 2
Proof. The component over H H H H ^ ≅ C 2 2 \widehat H\cong C_2^2 H ≅ C 2 2 (4.15) gives one simple object t t t 2 2 2 x H xH x H H u H HuH H u H ρ \rho ρ 4 4 4 (4.7) .
Both x H xH x H H u H HuH H u H N L ( H ) / H N_L(H)/H N L ( H ) / H N L ( H ) N_L(H) N L ( H ) L L L t t t ρ \rho ρ t t t ρ \rho ρ x H xH x H H H H g ∈ H ^ g\in\widehat H g ∈ H g ⊗ t ≅ t g\otimes t\cong t g ⊗ t ≅ t
[ t ⊗ t , g ] = [ t , g ⊗ t ] = 1 , [t\otimes t,g]=[t,g\otimes t]=1, [ t ⊗ t , g ] = [ t , g ⊗ t ] = 1 , while no other simple is supported on H H H t ⊗ t ≅ ⨁ g ∈ H ^ g t\otimes t\cong \bigoplus_{g\in\widehat H}g t ⊗ t ≅ ⨁ g ∈ H g N L ( H ) N_L(H) N L ( H ) H u H HuH H u H t ⊗ ρ t\otimes\rho t ⊗ ρ ρ ⊗ t \rho\otimes t ρ ⊗ t H u H HuH H u H ρ \rho ρ t ⊗ ρ ≅ ρ ⊗ t ≅ 2 ρ t\otimes\rho\cong \rho\otimes t\cong 2 \rho t ⊗ ρ ≅ ρ ⊗ t ≅ 2 ρ g ∈ H ^ g\in\widehat H g ∈ H [ ρ ⊗ ρ , g ] = [ ρ , g ⊗ ρ ] = 1 [\rho\otimes \rho,g]=[\rho,g\otimes\rho]=1 [ ρ ⊗ ρ , g ] = [ ρ , g ⊗ ρ ] = 1 [ ρ ⊗ ρ , t ] = [ ρ , t ⊗ ρ ] = 2 [\rho\otimes \rho,t]=[\rho,t\otimes\rho]=2 [ ρ ⊗ ρ , t ] = [ ρ , t ⊗ ρ ] = 2
ρ ⊗ ρ ≅ ⨁ g ∈ H ^ g ⊕ 2 t ⊕ m ρ \rho\otimes \rho\cong \bigoplus_{g\in\widehat H}g\oplus 2 t\oplus m \rho ρ ⊗ ρ ≅ g ∈ H ⨁ g ⊕ 2 t ⊕ m ρ for some m ≥ 0 m\ge0 m ≥ 0 16 = 4 + 4 + 4 m 16=4+4+4m 16 = 4 + 4 + 4 m m = 2 m=2 m = 2
The associator in Vec L ω \operatorname{Vec}_L^\omega Vec L ω H H H ψ \psi ψ
We can now prove the classification stated in the introduction.
Proof of Theorem 1.2 . If d + ∉ Z d_+\notin\mathbb Z d + ∈ / Z 3.1 and Theorem 3.12 leave n = 2 n=2 n = 2 κ ∈ { 0 , 8 } \kappa\in\{0,8\} κ ∈ { 0 , 8 } 3.24 excludes κ = 8 \kappa=8 κ = 8 3.13 treats κ = 0 \kappa=0 κ = 0 d + ∈ Z d_+\in\mathbb Z d + ∈ Z 4.10 gives exactly the two possibilities ( n , d + , κ ) = ( 2 , 4 , 2 ) , ( 2 , 8 , 7 ) (n,d_+,\kappa)=(2,4,2),(2,8,7) ( n , d + , κ ) = ( 2 , 4 , 2 ) , ( 2 , 8 , 7 ) 2.3 identifies G G G (2.9) gives K 0 ( C ) ≅ R κ K_0(\mathcal C)\cong R_\kappa K 0 ( C ) ≅ R κ 4.12 and Example 2.1 establish existence for R 2 R_2 R 2 R 7 R_7 R 7 FPdim ( C ) = 8 + d + 2 \operatorname{FPdim}(\mathcal C)=8+d_+^2 FPdim ( C ) = 8 + d + 2
Acknowledgments
We sincerely thank Andrew Schopieray for proposing this topic and helping us prove Proposition 3.1. Sébastien Palcoux is supported by the National Natural Science Foundation of China (NSFC), Grant no. 12471031. GPT-6 Astra was used as a research assistant. The authors take full responsibility for the final manuscript.
