Yang–Baxter Operators and Quantum Braid Representations

Let $k$ be a field and let $\mathcal{C}$ be a $k$-linear category (k-linear categories and k-linear functors). Since composition in $\mathcal{C}$ is $k$-bilinear, ${\mathrm{End}}_{\mathcal{C}}(X)$ is a $k$-algebra with unit ${\mathrm{id}}_X$ for every object $X$.

An object $X$ of $\mathcal{C}$ is absolutely simple when the $k$-algebra ${\mathrm{End}}_{\mathcal{C}}(X)$ is one-dimensional over $k$, that is,

${\mathrm{End}}_{\mathcal{C}}(X)=k\cdot {\mathrm{id}}_X,{\mathrm{id}}_X\mathrm{\ne }0.$

Equivalently, the evaluation map

$k⟶{\mathrm{End}}_{\mathcal{C}}(X),\lambda ⟼\lambda {\mathrm{id}}_X,$

is an isomorphism of $k$-algebras. These two formulations agree: the evaluation map is $k$-linear and multiplicative with $1\mapsto {\mathrm{id}}_X$, and it is injective because $\lambda {\mathrm{id}}_X=0$ with ${\mathrm{id}}_X\mathrm{\ne }0$ forces $\lambda =0$; surjectivity is exactly ${\mathrm{End}}_{\mathcal{C}}(X)=k\cdot {\mathrm{id}}_X$. In particular a nonzero endomorphism space $k\cdot {\mathrm{id}}_X$ has dimension one, so an absolutely simple object is not a zero object; if $\mathcal{C}$ has a zero object then $X\mathrm{\ne }0$ automatically.

In a $k$-linear abelian category, a simple object with $\mathrm{End}(X)=k$ satisfies this condition (Simple object). The converse need not hold: the representation $k\stackrel{1}{\to }k$ of the two-vertex quiver has only scalar endomorphisms but has the proper nonzero subrepresentation $0\to k$. Thus the definition records the scalar endomorphism condition without asserting simplicity in an arbitrary abelian category. Every simple object of a locally finite $k$-linear abelian category over an algebraically closed field $k$ is absolutely simple. Indeed, a nonzero endomorphism of a simple object has zero kernel and full image, hence is an isomorphism; its endomorphism algebra $D$ is therefore a division algebra. Local finiteness makes $D$ finite-dimensional (Locally finite k-linear abelian categories). For $a\in D$, a polynomial over $k$ annihilates $a$ by linear dependence of its powers; it splits into linear factors, and a division algebra has no zero divisors, so one factor $a-\lambda {\mathrm{id}}_X$ vanishes. Thus $D=k$. This definition fixes no semisimplicity and no algebraic-closedness hypothesis; it records them only as the standard situation in which absolute simplicity is automatic.

If $X$ is absolutely simple, every automorphism of $X$ is a scalar $\lambda {\mathrm{id}}_X$ with $\lambda \in k^{\times }$: it is an invertible endomorphism, hence by absolute simplicity equals some $\lambda {\mathrm{id}}_X$, and $\lambda$ is invertible with inverse ${\lambda }^{-1}$ because $\lambda {\mathrm{id}}_X$ is invertible.

For $n\ge 2$ let $B_n$ be the braid group of the Artin presentation (The braid group by Artin presentation), with generators ${\sigma }_1,\dots ,{\sigma }_{n-1}$. The assignment

${\sigma }_i⟼1(1\le i\le n-1)$

has equal values on the two sides of every defining relator of that presentation: a braid relator ${\sigma }_i{\sigma }_{i+1}{\sigma }_i={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}$ has three letters on each side, and a distant-commutativity relator ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i$ has two letters on each side. Hence Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group gives a unique homomorphism

$w_n\text{\hspace{0.17em}⁣}:B_n⟶\mathbb{Z},w_n({\sigma }_i^{\pm 1})=\pm 1.$

It is the exponent sum, or writhe, of a braid. If $\beta ={\sigma }_{i_1}^{{ϵ}_1}\cdots {\sigma }_{i_k}^{{ϵ}_k}$ is an Artin word for $\beta$ with ${ϵ}_l\in \{\pm 1\}$, then $w_n(\beta )={\sum }_l{ϵ}_l$: the number of positive letters minus the number of negative letters. For $n=0,1$ set $w_n=0$ on the trivial group $B_n$.

The homomorphisms $w_n$ assemble to a function $w\text{\hspace{0.17em}⁣}:{⨆}_{n\ge 0}{B}_n\to \mathbb{Z}$. Let ${\iota }_n\text{\hspace{0.17em}⁣}:B_n\to B_{n+1}$, ${\iota }_n({\sigma }_i)={\sigma }_i$, be the standard inclusion, which is well defined because the defining relators of $B_n$ are among those of $B_{n+1}$. Then

$w_{n+1}({\iota }_n(\beta ))=w_n(\beta ),w_{n+1}({\iota }_n(\beta ){\sigma }_n^{\pm 1})=w_n(\beta )\pm 1,$

for all $n\ge 1$ and $\beta \in B_n$: the first identity holds because both sides are additive over an Artin word for $\beta$ and agree on generators, and the second adds the single letter ${\sigma }_n^{\pm 1}$. Consequently $w$ is invariant under conjugation,

$w(\gamma \beta {\gamma }^{-1})=w(\beta )$

for all braids $\gamma ,\beta$ for which the product is defined, since $w(\gamma \beta {\gamma }^{-1})=w(\gamma )+w(\beta )-w(\gamma )$.

The exponent sum is invariant under conjugation, while under stabilization it shifts by $\pm 1$; it is not a complete invariant of braids. For instance in $B_3$ the braids ${\sigma }_1{\sigma }_2$ and ${\sigma }_2{\sigma }_1$ have equal exponent sum $2$ and are distinct: their images under ${\sigma }_i\mapsto (ii+1)$ are the two different three-cycles. This generator assignment respects the Artin relations by direct permutation multiplication, so it extends by Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group.

Work in the smooth category and fix the slab

$W:=\mathbb{R}\times [0,1]\times [0,1],$

with coordinates $(x,y,t)$ and height function $t$. Its bottom face is $\mathbb{R}\times [0,1]\times \{0\}$, its top face is $\mathbb{R}\times [0,1]\times \{1\}$, and its side faces are $\mathbb{R}\times \{0\}\times [0,1]$ and $\mathbb{R}\times \{1\}\times [0,1]$.

The framed oriented tangle category $\mathcal{T}$ is the strict monoidal category whose objects are the finite sequences $\epsilon =({\epsilon }_1,\dots ,{\epsilon }_m)$ with ${\epsilon }_j\in \{+1,-1\}$, the empty sequence included. For a sequence of length $m$, prescribe boundary points $(j,\frac{1}{2},0)$ on the bottom and $(j,\frac{1}{2},1)$ on the top, $1\le j\le m$, with vertical strand collars and fixed normal vector $(0,1,0)$ on those collars. A morphism $\epsilon \to {\epsilon }^{\mathrm{\prime }}$ is a boundary-relative isotopy class of compact oriented framed $1$-manifolds properly embedded in $W$ and disjoint from the side faces, meeting the bottom face in $m$ points and the top face in $m^{\mathrm{\prime }}$ points at those prescribed positions, equal to the vertical framed collars near both faces, and with the signs ${\epsilon }_1,\dots ,{\epsilon }_m$ at the bottom and ${\epsilon }_1^{\mathrm{\prime }},\dots ,{\epsilon }_{m^{\mathrm{\prime }}}^{\mathrm{\prime }}$ at the top, ordered by the $x$-coordinate. A sign $+1$ means that the oriented tangent points in the increasing $t$ direction; $-1$ means decreasing $t$, at either face. A framing is a homotopy class, relative to the fixed collars, of nonvanishing normal vector fields. Isotopies are ambient isotopies of $W$, constant on the side faces and on the fixed top and bottom collars, that carry the framing class of one tangle to that of the other, fix every boundary point and its framing, and preserve strand orientation. This is the boundary-relative analogue of Oriented links in the three-sphere and ambient isotopy with diagrams in the slab in place of links in the sphere; the oriented crossing signs and kink conventions are those of Oriented Reidemeister moves.

Composition and tensor product. The composition of $\epsilon \to {\epsilon }^{\mathrm{\prime }}$ and ${\epsilon }^{\mathrm{\prime }}\to {\epsilon }^{\mathrm{\prime }\mathrm{\prime }}$ is stacking: put the first tangle in $\mathbb{R}\times [0,1]\times [0,\frac{1}{2}]$, the second in $\mathbb{R}\times [0,1]\times [\frac{1}{2},1]$, and glue along the common boundary. The tensor product is horizontal juxtaposition, placing the first factor to the left of the second and normalizing the ordered endpoints to the prescribed positions for the concatenated sequence. Use horizontal embeddings with disjoint image strips and fixed vertical collars, followed by an order-preserving horizontal adjustment near the boundary. Different such adjustments are isotopic relative to the prescribed framed collars: interpolate the increasing horizontal coordinate maps, whose derivatives remain positive, and extend through the collars. Thus the result is independent of those adjustments. The vertical collars make stacking smooth; different height rescalings and collar lengths are related by increasing height reparametrizations relative to the boundary. These isotopies prove associativity, the identity laws, strict associativity of juxtaposition on isotopy classes, and interchange. The empty sequence is the strict tensor unit. Hence these operations define a strict monoidal category.

Blackboard framing and elementary tangles. When a tangle is drawn in the interior picture plane $y=\frac{1}{2}$ with its bands parallel to that plane, its blackboard framing is represented by the normal to the band surface, equal to $(0,1,0)$ away from small crossing neighborhoods and on the fixed collars. A diagram is a projection to the picture plane; at crossings the bands are separated in the $y$ direction, rather than literally contained in that plane. All elementary tangles below carry the blackboard framing relative to the plane of the picture. The elementary morphisms of $\mathcal{T}$ are:

1. for each pair of signs $(\epsilon ,{\epsilon }^{\mathrm{\prime }})$ the positive crossing $X_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{+}\text{\hspace{0.17em}⁣}:(\epsilon ,{\epsilon }^{\mathrm{\prime }})\to ({\epsilon }^{\mathrm{\prime }},\epsilon )$ and the negative crossing $X_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{-}\text{\hspace{0.17em}⁣}:(\epsilon ,{\epsilon }^{\mathrm{\prime }})\to ({\epsilon }^{\mathrm{\prime }},\epsilon )$, the two blackboard-framed crossings of the two adjacent strands. The superscript fixes the over/under geometry, independently of their orientations: $X^{+}$ has the geometry of the positive Artin crossing on two $+$ strands, and $X^{-}$ has the inverse geometry. For mixed endpoint signs its oriented crossing sign is reversed. Following the two strands, the crossing sends the left bottom endpoint to the right top endpoint and conversely, which is why the sign sequence is reversed in the target. The two crossings are mutually inverse, $X_{{\epsilon }^{\mathrm{\prime }},\epsilon }^{+}{X}_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{-}={\mathrm{id}}_{(\epsilon ,{\epsilon }^{\mathrm{\prime }})}$ and $X_{{\epsilon }^{\mathrm{\prime }},\epsilon }^{-}{X}_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{+}={\mathrm{id}}_{(\epsilon ,{\epsilon }^{\mathrm{\prime }})}$;
2. for each sign $\epsilon$ the blackboard-framed cup ${\cup }_{\epsilon }\text{\hspace{0.17em}⁣}:\mathrm{\varnothing }\to (\epsilon ,-\epsilon )$ and cap ${\cap }_{\epsilon }\text{\hspace{0.17em}⁣}:(-\epsilon ,\epsilon )\to \mathrm{\varnothing }$, single arcs with both endpoints on the top face respectively the bottom face, oriented by the sign; these are not invertible;
3. for each sign $\epsilon$ the positive full twist ${\phi }_{\epsilon }^{+}\text{\hspace{0.17em}⁣}:(\epsilon )\to (\epsilon )$ and the negative full twist ${\phi }_{\epsilon }^{-}\text{\hspace{0.17em}⁣}:(\epsilon )\to (\epsilon )$, the blackboard-framed bands carrying one positive respectively negative full twist of the band; they are mutually inverse, ${\phi }_{\epsilon }^{+}{\phi }_{\epsilon }^{-}={\phi }_{\epsilon }^{-}{\phi }_{\epsilon }^{+}=\mathrm{id}$.

All isotopies are ambient isotopies of the slab fixing the boundary points and their framings, so isotopic framed colored tangles represent the same morphism of $\mathcal{T}$.

Let $\mathcal{C}$ be a monoidal category and let $X$ be an object of $\mathcal{C}$. Write $X^{\otimes 3}$ for the triple tensor product and write $1_X^{\otimes k}$ for the $k$-fold tensor power of ${\mathrm{id}}_X$. The equation below is interpreted in a strict model of $\mathcal{C}$: fix a monoidal equivalence from $\mathcal{C}$ to a strict monoidal category, as in Mac Lane strictification, with tensor constraint $J_{A,B}:E(A)\otimes E(B)\to E(A\otimes B)$. Put $X^{\mathrm{\prime }}=E(X)$ and $R^{\mathrm{\prime }}=J_{X,X}^{-1}E(R)J_{X,X}$, and read the displayed equation with $X^{\mathrm{\prime }},R^{\mathrm{\prime }}$ in that strict category. When $\mathcal{C}$ is strict the display below is literal.

A Yang–Baxter operator on $X$ is an invertible morphism $R\text{\hspace{0.17em}⁣}:X\otimes X\to X\otimes X$ such that

$(R\otimes 1_X)(1_X\otimes R)(R\otimes 1_X)=(1_X\otimes R)(R\otimes 1_X)(1_X\otimes R)\mathrm{.}$

Both sides are endomorphisms of $X^{\otimes 3}$ in the strict model, so the display is a well-formed equality of morphisms of the strict model; transported back along the equivalence it is a well-formed statement about $X$. It is the Yang–Baxter equation, and an invertible solution $R$ is also called an $R$-matrix on $X$ in the categorical sense. Invertibility is part of the data: a solution of the cubic equation that is not invertible is not a Yang–Baxter operator in this sense.

In a strict braided monoidal category the braiding gives the basic example $R=c_{X,X}$ on any object $X$ (Braided monoidal category): the Yang–Baxter equation for the braiding is In a strict braided monoidal category the braiding satisfies the Yang-Baxter equation, and $c_{X,X}$ is invertible with inverse $c_{X,X}^{-1}$ because each component of a braiding is an isomorphism. In a general braided monoidal category the same example is read in a strict model via the braided strictification of the category.

Let $\mathcal{C}$ be a monoidal category, let $X\in \mathcal{C}$, let $R$ be a Yang–Baxter operator on $X$ (Yang–Baxter operators on an object), and let $n\ge 2$. Write $X^{\otimes n}$ for the left-nested tensor power: $X^{\otimes 0}:=\mathbf{1}$, $X^{\otimes 1}:=X$ and $X^{\otimes (k+1)}:=X^{\otimes k}\otimes X$ for $k\ge 1$; write $1_X^{\otimes k}$ for the $k$-fold tensor power of ${\mathrm{id}}_X$.

Work first in a strict model ${\mathcal{C}}^{\mathrm{\prime }}$ of $\mathcal{C}$, obtained from a monoidal equivalence as in Mac Lane strictification, and let $R^{\mathrm{\prime }}\text{\hspace{0.17em}⁣}:X^{\mathrm{\prime }}\otimes X^{\mathrm{\prime }}\to X^{\mathrm{\prime }}\otimes X^{\mathrm{\prime }}$ be the transported Yang–Baxter operator. The local Yang–Baxter operators are the automorphisms of the $n$-fold tensor power

$R_i:=1_X^{\otimes (i-1)}\otimes R\otimes 1_X^{\otimes (n-i-1)}\in {\mathrm{Aut}}_{\mathcal{C}}(X^{\otimes n}),1\le i\le n-1,$

where in the strict model the display is literal and each $R_i$ is the endomorphism of $X^{\mathrm{\prime }\otimes n}$ acting as $R^{\mathrm{\prime }}$ on the $i$-th and $(i+1)$-st tensor factors and as the identity elsewhere.

Outside the strict model the same operators are obtained by conjugating the displayed word with canonical associativity isomorphisms: the composite

$X^{\otimes n}\stackrel{\kappa }{\to }{X}^{\otimes (i-1)}\otimes (X\otimes X)\otimes X^{\otimes (n-i-1)}\stackrel{1^{\otimes (i-1)}\otimes R\otimes 1^{\otimes (n-i-1)}}{\to }{X}^{\otimes (i-1)}\otimes (X\otimes X)\otimes X^{\otimes (n-i-1)}\stackrel{{\kappa }^{-1}}{\to }{X}^{\otimes n},$

where $\kappa$ is any canonical morphism between the two parenthesised tensor words built from associators and unitors, is independent of the chosen $\kappa$ by Mac Lane coherence in canonical-map form (Mac Lane coherence in canonical-map form), and it is this composite that defines $R_i$ in $\mathcal{C}$.

Each $R_i$ is invertible, with inverse $1_X^{\otimes (i-1)}\otimes R^{-1}\otimes 1_X^{\otimes (n-i-1)}$ (respectively its bracket-corrected conjugate): the displayed inverse is a two-sided inverse of the displayed word in the strict model, and conjugation by $\kappa$ preserves the inverse relation.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Every morphism of the framed oriented tangle category $\mathcal{T}$ of The framed oriented tangle category is a finite composite of tensor products of identities, crossings, cups, caps and full twists.

A complete presentation on this elementary family is Turaev's reduced presentation (3.1.a), (3.2.a)--(3.2.h), together with the expressions (2.5.d)--(2.5.e) and (3.1.b)--(3.1.c) defining its additional orientation variants. Thus two elementary words represent the same framed tangle exactly when these relations and the strict monoidal axioms relate them.

Here the source's $X^{\delta }$ denotes our $X_{+,+}^{\delta }$, its $Z^{\delta }$ denotes our $X_{+,-}^{-\delta }$, and its $Y^{\delta }$ denotes our $X_{-,+}^{-\delta }$; its $T^{\delta }$ denotes our $X_{-,-}^{\delta }$. The source's positive cup and cap are ${\cup }_{+}$ and ${\cap }_{+}$, and its $\phi ,{\phi }^{\mathrm{\prime }}$ are ${\phi }_{+}^{+},{\phi }_{+}^{-}$. The reduced generators are therefore $X_{+,+}^{\delta }$, $X_{+,-}^{-\delta }$, ${\phi }_{+}^{\pm 1}$, ${\cup }_{+}$ and ${\cap }_{+}$. The other cups, caps and crossings are the expressions above; a twist on a negative strand is obtained by bending the corresponding twisted positive strand with a cup and cap. Turaev's convention that a positive strand points downwards is transported to ours by reversing all strand orientations; this preserves composition, tensor product and framed isotopy. Crossing superscripts in his mixed-orientation pictures are oriented signs, explaining the minus signs in the dictionary.

The reduced relations are Yang--Baxter, the two zig-zags, inverse crossings and twists, curl slide, the bent-crossing inverse relation (3.2.g), and the twist-square relation (3.2.h). For example, the latter has the well-typed form

$({\phi }_{+}^{+}{)}^2=({\cap }_{+}\otimes 1_{+})(1_{-}\otimes X_{+,+}^{+})(X_{+,-}^{-}\otimes 1_{+})({\cup }_{+}\otimes 1_{+})\mathrm{.}$

Let $\mathcal{C}$ be a monoidal category, let $X\in \mathcal{C}$, let $R$ be a Yang–Baxter operator on $X$, let $n\ge 2$, and let $R_1,\dots ,R_{n-1}$ be the local operators of Local Yang–Baxter operators on tensor powers, so that $R_i\in {\mathrm{Aut}}_{\mathcal{C}}(X^{\otimes n})$ is the local operator acting on the $i$-th and $(i+1)$-st tensor factors. Then

$R_i{R}_j=R_j{R}_i\text{whenever }\mathrm{\mid }i-j\mathrm{\mid }>1,$

$R_i{R}_{i+1}{R}_i=R_{i+1}{R}_i{R}_{i+1}\text{for }1\le i\le n-2.$

In the non-strict model the identities are those of the bracket-corrected local operators defined in Local Yang–Baxter operators on tensor powers.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $\mathcal{C}$ be a ribbon category (Twist and ribbon structure) with chosen left duals (Left dual and right dual object), braiding $c$ (Braiding) and twist $\theta$, and let $X\in \mathcal{C}$. Write $X_{+1}:=X$ and $X_{-1}:=X^{\vee }$. In a strict model of $\mathcal{C}$ (Every braided monoidal category is monoidally equivalent to a strict braided one) there is a unique strict monoidal functor

$F_X\text{\hspace{0.17em}⁣}:\mathcal{T}⟶\mathcal{C}$

from the framed oriented tangle category $\mathcal{T}$ with

$F_X(+)=X,F_X(-)=X^{\vee },$

$F_X\left(X_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{+}\right)=c_{X_{\epsilon },X_{{\epsilon }^{\mathrm{\prime }}}},F_X\left(X_{\epsilon ,{\epsilon }^{\mathrm{\prime }}}^{-}\right)=(c_{X_{{\epsilon }^{\mathrm{\prime }}},X_{\epsilon }}{)}^{-1},F_X({\cup }_{+1})={\mathrm{coev}}_X,F_X({\cap }_{+1})={\mathrm{ev}}_X,F_X({\phi }_{+1}^{\pm 1})={\theta }_X^{\pm 1},$

where the cup and cap are the coevaluation and evaluation maps attached to the chosen left duals and the signs are the blackboard conventions of The framed oriented tangle category. The remaining elementary tangles of $\mathcal{T}$, those carrying a negative sign at a cup, cap or twist, receive the values forced by these data through the relations. Explicitly put $j_X=u_X{\theta }_X:X\to X^{\vee \vee }$, with $u$ as in A braided rigid category has a Drinfeld morphism. Then

$F_X({\cup }_{-})=(1_{X^{\vee }}\otimes j_X^{-1}){\mathrm{coev}}_{X^{\vee }},F_X({\cap }_{-})={\mathrm{ev}}_{X^{\vee }}(j_X\otimes 1_{X^{\vee }}),F_X({\phi }_{-}^{\pm 1})={\theta }_{X^{\vee }}^{\pm 1}\mathrm{.}$

These are the right coevaluation and evaluation on $X$ induced by the ribbon structure; they need no identification $X^{\vee \vee }=X$ on objects. For a general ribbon category the same assignment determines a strong monoidal functor, unique up to the canonical coherence isomorphisms of the strictification. Isotopic framed tangles are equal morphisms of $\mathcal{T}$, so $F_X$ is an invariant of framed tangles. Uniqueness needs no choice.

Let $\mathcal{C}$ be a monoidal category, let $X\in \mathcal{C}$, and let $R$ be a Yang–Baxter operator on $X$. For every $n\ge 2$ there is a unique group homomorphism

${\rho }_n\text{\hspace{0.17em}⁣}:B_n⟶{\mathrm{Aut}}_{\mathcal{C}}(X^{\otimes n}),{\rho }_n({\sigma }_i)=R_i,$

where $B_n$ is the braid group of The braid group by Artin presentation and $R_1,\dots ,R_{n-1}$ are the local operators of Local Yang–Baxter operators on tensor powers. The family $({\rho }_n{)}_{n\ge 2}$ is compatible with the standard inclusions ${\iota }_n\text{\hspace{0.17em}⁣}:B_n\to B_{n+1}$, ${\iota }_n({\sigma }_i)={\sigma }_i$, in the sense that

${\rho }_{n+1}({\iota }_n(\beta ))={\rho }_n(\beta )\otimes 1_X\text{for all }\beta \in B_n\mathrm{.}$

Let $\mathcal{C}$ be a braided monoidal category with braiding $c$ and let $X\in \mathcal{C}$. For every $n\ge 2$ there is a canonical group homomorphism

${\rho }_n\text{\hspace{0.17em}⁣}:B_n⟶{\mathrm{Aut}}_{\mathcal{C}}(X^{\otimes n})$

characterized by the property that ${\rho }_n({\sigma }_i)$ is the local braiding of the $i$-th and $(i+1)$-st tensor factors: in a strict model ${\rho }_n({\sigma }_i)=1_X^{\otimes (i-1)}\otimes c_{X,X}\otimes 1_X^{\otimes (n-i-1)}$, while in general $c_{X,X}$ is transported by the associativity isomorphisms and the result is independent of those choices by braided coherence (Braided coherence is controlled by underlying braids). The actions are compatible with the standard inclusions ${\iota }_n\text{\hspace{0.17em}⁣}:B_n\to B_{n+1}$ for $n\ge 1$:

${\rho }_{n+1}({\iota }_n(\beta ))={\rho }_n(\beta )\otimes 1_X\mathrm{.}$

For $n=0,1$, define ${\rho }_n$ to be the unique homomorphism from the trivial group $B_n$ to $\mathrm{Aut}(X^{\otimes n})$, with $X^{\otimes 0}=\mathbf{1}$. There are no local generators in these cases. At $n=0$, let ${\lambda }_X:\mathbf{1}\otimes X\to X$ be the left unitor. Compatibility means

${\rho }_1({\iota }_0(e))={\lambda }_X({\rho }_0(e)\otimes 1_X){\lambda }_X^{-1}=1_X\mathrm{.}$

For $n=1$ the preceding compatibility formula is literal because both sides are the identity of $X\otimes X$.

The statement holds for the braiding of Braided monoidal category with no choice principle.

Let $\mathcal{C}$ be a monoidal category, let $X\in \mathcal{C}$, and let $R$ be a Yang–Baxter operator on $X$ with $R^2=1_{X\otimes X}$. Then for every $n\ge 2$ the homomorphism ${\rho }_n\text{\hspace{0.17em}⁣}:B_n\to \mathrm{Aut}(X^{\otimes n})$ of A Yang–Baxter operator gives braid-group representations factors through the canonical surjection ${\pi }_n\text{\hspace{0.17em}⁣}:B_n\to S_n$ of The braid group surjects onto the symmetric group: there is a unique homomorphism ${\psi }_n\text{\hspace{0.17em}⁣}:S_n\to \mathrm{Aut}(X^{\otimes n})$ with ${\rho }_n={\psi }_n\circ {\pi }_n$.

Conversely, if ${\rho }_n$ factors through ${\pi }_n$ for some $n\ge 2$, then $R_i^2=1_{X^{\otimes n}}$ for every local operator $R_i$, because ${\sigma }_i^2$ lies in the kernel of ${\pi }_n$; and since ${\rho }_2({\sigma }_1)=R$, if ${\rho }_2$ factors through ${\pi }_2$ then $R^2=1_{X\otimes X}$. Consequently an involutive Yang–Baxter operator is exactly one whose two-strand braid action factors through $S_2$, and an involutive Yang–Baxter operator has its braid actions factoring through $S_n$ for every $n$.