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Hexagon in F/R coordinates

Statement

Let $C$ be a unitary braided fusion category with associator $α$ and braiding $c$ . For all objects $X, Y, Z$ , its two hexagons are

$α_{Y, Z, X} \circ c_{X, Y \otimes Z} \circ α_{X, Y, Z} = ({id}_{Y} \otimes c_{X, Z}) \circ α_{Y, X, Z} \circ (c_{X, Y} \otimes {id}_{Z}),$

$α_{Z, X, Y}^{- 1} \circ c_{X \otimes Y, Z} \circ α_{X, Y, Z}^{- 1} = (c_{X, Z} \otimes {id}_{Y}) \circ α_{X, Z, Y}^{- 1} \circ ({id}_{X} \otimes c_{Y, Z}) .$

For simple input charges and a fixed simple total charge, the two route matrices in each equation are equal in fusion-tree coordinates. Each entry is the sum over all intermediate simple charges and multiplicity indices of the products of the corresponding F- and R-block entries.

Facts & Assumptions

Given: a braided fusion category, simple input charges, a simple total charge, and the orthonormal fusion-tree bases.

[F1]

Braiding: the braiding is a natural isomorphism and satisfies the two hexagon equations displayed in the Statement.

[F2]

F-symbol as a recoupling matrix: each associator change between tree bases is represented by the corresponding F block, with all multiplicity indices.

[F3]

R-symbol as a channel exchange map: each braiding exchange on a channel is represented by the corresponding typed R block.

[F4]

Fusion-tree basis and dimension recursion: each intermediate tree space has a complete basis indexed by all admissible simple labels and vertex multiplicity indices.

Proof

1.1F2F3F4construct

On a fixed fusion-tree basis, the matrix of each associator factor is the appropriate F block by [F2], and the matrix of each channel exchange is the appropriate R block by [F3]. An identity factor preserves all untouched labels and multiplicity indices. If a braiding input is a tensor product rather than a simple charge, decompose it into its finite simple summands; its matrix is the resulting finite block matrix, with the summand and multiplicity indices kept. Thus every factor in either hexagon composite has a specified finite matrix.

2.1F1F2F3F4step 1.1algebra

For the first hexagon, let $M_{1}, M_{2}, M_{3}$ be the matrices of $α_{X, Y, Z}$ , $c_{X, Y \otimes Z}$ , and $α_{Y, Z, X}$ in order along its left route, and let $N_{1}, N_{2}, N_{3}$ be those of $c_{X, Y} \otimes {id}_{Z}$ , $α_{Y, X, Z}$ , and ${id}_{Y} \otimes c_{X, Z}$ along its right route. For fixed endpoint indices $x, z$ , the entries are respectively $\sum_{u, v} (M_{3})_{z, v} (M_{2})_{v, u} (M_{1})_{u, x}$ and $\sum_{u^{'}, v^{'}} (N_{3})_{z, v^{'}} (N_{2})_{v^{'}, u^{'}} (N_{1})_{u^{'}, x}$ . Every intermediate index is the full tuple of simple charges and vertex multiplicity indices for that tree. The two structural maps are equal by the first hexagon [F1], so their coordinate entries and these finite sums are equal.

2.2F1F2F3F4step 1.1algebra

For the second hexagon, let $P_{1}, P_{2}, P_{3}$ be the matrices of $α_{X, Y, Z}^{- 1}$ , $c_{X \otimes Y, Z}$ , and $α_{Z, X, Y}^{- 1}$ along its left route; let $Q_{1}, Q_{2}, Q_{3}$ be those of ${id}_{X} \otimes c_{Y, Z}$ , $α_{X, Z, Y}^{- 1}$ , and $c_{X, Z} \otimes {id}_{Y}$ along its right route. The entries at fixed endpoints are $\sum_{u, v} (P_{3})_{z, v} (P_{2})_{v, u} (P_{1})_{u, x}$ and $\sum_{u^{'}, v^{'}} (Q_{3})_{z, v^{'}} (Q_{2})_{v^{'}, u^{'}} (Q_{1})_{u^{'}, x}$ . The indices again range over every simple intermediate charge and multiplicity coordinate. The second root hexagon [F1] makes the structural maps equal, so the two finite sums agree.

3.1step 2.1step 2.2∎

The two matrix equalities in steps 2.1 and 2.2 are the coordinate forms of the first and second hexagons, respectively, and each includes all admissible intermediate labels and multiplicities.

Hexagon diagrams

The first diagram shows the two routes from $(X \otimes Y) \otimes Z$ to $Y \otimes (Z \otimes X)$ ; the second shows the routes from $X \otimes (Y \otimes Z)$ to $(Z \otimes X) \otimes Y$ . The arrow labels are exactly the maps in the two displayed equations. They illustrate the equations; the proof above uses the typed composites and their matrix coefficients.

diagram failed to render:
((X\otimes Y)\otimes Z)
  \arrow[r,"\alpha_{X,Y,Z}"]
  \arrow[d,"c_{X,Y}\otimes\operatorname{id}_Z"']
& X\otimes(Y\otimes Z)
  \arrow[r,"c_{X,Y\otimes Z}"]
& (Y\otimes Z)\otimes X
  \arrow[d,"\alpha_{Y,Z,X}"] \\
(Y\otimes X)\otimes Z
  \arrow[r,"\alpha_{Y,X,Z}"']
& Y\otimes(X\otimes Z)
  \arrow[r,"\operatorname{id}_Y\otimes c_{X,Z}"']
& Y\otimes(Z\otimes X)

diagram failed to render:
X\otimes(Y\otimes Z)
  \arrow[r,"\operatorname{id}_X\otimes c_{Y,Z}"]
  \arrow[d,"\alpha_{X,Y,Z}^{-1}"']
& X\otimes(Z\otimes Y)
  \arrow[r,"\alpha_{X,Z,Y}^{-1}"]
& (X\otimes Z)\otimes Y
  \arrow[d,"c_{X,Z}\otimes\operatorname{id}_Y"] \\
(X\otimes Y)\otimes Z
  \arrow[r,"c_{X\otimes Y,Z}"']
& Z\otimes(X\otimes Y)
  \arrow[r,"\alpha_{Z,X,Y}^{-1}"']
& (Z\otimes X)\otimes Y

Depends on

Used by

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Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, author final text (standard reference, not scraped)
David Penneys, Unitary Fusion Categories, Math 8800 lecture notes (standard reference, not scraped)