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Definition

Braided monoidal category

A monoidal category with natural exchanges of tensor factors satisfying both hexagon axioms.

A braided monoidal category is a with a

βX,Y:X⊗Y⟶Y⊗X\beta_{X,Y}:X\otimes Y\longrightarrow Y\otimes X

satisfying both hexagon axioms below. The associator is α\alpha, and compositions act from right to left:

αY,Z,XβX,Y⊗ZαX,Y,Z=(1Y⊗βX,Z)αY,X,Z(βX,Y⊗1Z),\alpha_{Y,Z,X}\beta_{X,Y\otimes Z}\alpha_{X,Y,Z} =(1_Y\otimes\beta_{X,Z})\alpha_{Y,X,Z}(\beta_{X,Y}\otimes1_Z),
αZ,X,Y−1βX⊗Y,ZαX,Y,Z−1=(βX,Z⊗1Y)αX,Z,Y−1(1X⊗βY,Z).\alpha^{-1}_{Z,X,Y}\beta_{X\otimes Y,Z}\alpha^{-1}_{X,Y,Z} =(\beta_{X,Z}\otimes1_Y)\alpha^{-1}_{X,Z,Y}(1_X\otimes\beta_{Y,Z}).

The maps β\beta are the braiding. Unlike a symmetry, a braiding is not required to satisfy βY,XβX,Y=1\beta_{Y,X}\beta_{X,Y}=1.

The two hexagons

The first says that moving XX past a pair Y,ZY,Z agrees with moving it past each factor in turn:

Commutative diagram& X\otimes(Y\otimes Z) \arrow[r, "\beta_{X,Y\otimes Z}"] & (Y\otimes Z)\otimes X \arrow[dr, "\alpha_{Y,Z,X}"] & \\ (X\otimes Y)\otimes Z \arrow[ur, "\alpha_{X,Y,Z}"] \arrow[dr, "\beta_{X,Y}\otimes1_Z"'] & & & Y\otimes(Z\otimes X) \\ & (Y\otimes X)\otimes Z \arrow[r, "\alpha_{Y,X,Z}"'] & Y\otimes(X\otimes Z) \arrow[ur, "1_Y\otimes\beta_{X,Z}"'] &

The second moves a pair past ZZ; it is an independent requirement for a general braiding:

Commutative diagram& (X\otimes Y)\otimes Z \arrow[r, "\beta_{X\otimes Y,Z}"] & Z\otimes(X\otimes Y) \arrow[dr, "\alpha^{-1}_{Z,X,Y}"] & \\ X\otimes(Y\otimes Z) \arrow[ur, "\alpha^{-1}_{X,Y,Z}"] \arrow[dr, "1_X\otimes\beta_{Y,Z}"'] & & & (Z\otimes X)\otimes Y \\ & X\otimes(Z\otimes Y) \arrow[r, "\alpha^{-1}_{X,Z,Y}"'] & (X\otimes Z)\otimes Y \arrow[ur, "\beta_{X,Z}\otimes1_Y"'] &
Braided versus symmetric

Imposing βY,XβX,Y=1\beta_{Y,X}\beta_{X,Y}=1 gives a . A need not have any braiding.

For an explicit nonsymmetric braiding, take finite-dimensional Z\mathbb Z-graded vector spaces over kk, choose q∈k×q\in k^\times with q2≠1q^2\ne1, and put β(vm⊗wn)=qmnwn⊗vm\beta(v_m\otimes w_n)=q^{mn}w_n\otimes v_m. The hexagons reduce to qm(n+p)=qmnqmpq^{m(n+p)}=q^{mn}q^{mp} and its analogue in the first variable. Two exchanges on degree-one vectors multiply by q2q^2, so this braiding is not symmetric.

References

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories (2015), Definitions 8.1.1–8.1.2 and §8.2; author-hosted text.