Infinitesimal braid identities imply KZ flatness and braid monodromy

If $\{i,j\}$ and $\{k,l\}$ are disjoint pairs, then ${\mathrm{\Omega }}_{ij}$ and ${\mathrm{\Omega }}_{kl}$ act on disjoint tensor factors. Each factorwise operator commutes with every operator on the other factors, so $[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{kl}]=0$.

For three distinct indices $i,j,k$, write $\mathrm{\Omega }={\sum }_a{x}_a\otimes x^a$. Apply [F2] with $x=x_b$, place its two tensor factors in slots $i,j$, apply ${\rho }_i\otimes {\rho }_j$, multiply on slot $k$ by ${\rho }_k(x^b)$, and sum over $b$. The first resulting sum is $-[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{ik}]$ and the second is $-[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{jk}]$, because $[x_b,x_a]=-[x_a,x_b]$ and $[x_b,x^a]=-[x^a,x_b]$. Hence $[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{ik}+{\mathrm{\Omega }}_{jk}]=0$.

Set $u=z_i-z_j$ and $v=z_j-z_k$, so $z_i-z_k=u+v$. Then ${\eta }_{ij}=du\mathrm{/}u$, ${\eta }_{jk}=dv\mathrm{/}v$, and ${\eta }_{ik}=(du+dv)\mathrm{/}(u+v)$. Direct expansion gives ${\eta }_{ij}\wedge {\eta }_{jk}+{\eta }_{jk}\wedge {\eta }_{ik}+{\eta }_{ik}\wedge {\eta }_{ij}=(1\mathrm{/}(uv)-1\mathrm{/}(v(u+v))-1\mathrm{/}(u(u+v)))du\wedge dv=0$, since the numerator after taking a common denominator is $(u+v)-u-v=0$.

Put $B_0:={\sum }_{i<j}{\mathrm{\Omega }}_{ij}{\eta }_{ij}$, so $\mathrm{\nabla }=d-{\kappa }^{-1}{B}_0$. Each ${\eta }_{ij}$ is closed because $d(dw\mathrm{/}w)=0$ on $w\mathrm{\ne }0$, hence $d{B}_0=0$, and the curvature is $R^{\mathrm{\nabla }}={\kappa }^{-2}{B}_0\wedge B_0$. Expanding the wedge product groups terms by disjoint pairs and by triples. The disjoint terms vanish by step 1.1. For $i<j<k$, the triple contribution is $C_{ijk}({\eta }_{ij}\wedge {\eta }_{ik}-{\eta }_{ij}\wedge {\eta }_{jk}+{\eta }_{ik}\wedge {\eta }_{jk})$, where $C_{ijk}:=[{\mathrm{\Omega }}_{ij},{\mathrm{\Omega }}_{ik}]$: step 1.2 gives the other two commutators as $-C_{ijk}$ and $C_{ijk}$. This form is the negative of the zero Arnold sum in step 1.3, so every triple contribution vanishes and $R^{\mathrm{\nabla }}=0$.

By [F3], every point has a neighborhood with a parallel frame. A parallel section has constant coordinates in that frame. On an overlap, if one parallel frame is the other times a smooth matrix $G(x)$, the connection equation gives $dG=0$, so $G$ is constant on each connected component of the overlap. These charts therefore define a locally constant sheaf of horizontal sections; define its transport along a continuous path by subdividing the path into chart intervals and retaining the same frame coordinates. For a piecewise smooth path this agrees with differential parallel transport by [F4] and uniqueness in [F5].

Suppose $V_i=V_0$ and ${\rho }_i={\rho }_0$ for every $i$. The coordinate-permutation action from [F7] lifts to the same permutation of tensor factors. Since ${\eta }_{ij}$ is symmetric in $i,j$, and the factor permutation sends ${\mathrm{\Omega }}_{ij}$ to the correspondingly relabeled $\mathrm{\Omega }$, it preserves the connection. The action is free on $F_n(\mathbb{C})$ by [F7]. The usual $\mathbb{C}$ is a connected Hausdorff real 2-manifold, so [F10] makes the quotient map a regular covering.

Given $z=(z_1,\dots ,z_n)\in F_n(\mathbb{C})$, choose pairwise disjoint open discs $D_i\subset \mathbb{C}$ about $z_i$ and put $\tilde{U}:=D_1\times \cdots \times D_n$. This is an open connected subset of $F_n(\mathbb{C})$, and its $S_n$-translates are pairwise disjoint because the $D_i$ are disjoint. By [F10] these sheets evenly cover their open image $U$ in the quotient; write $p(z)=[z]$ for the orbit quotient, so $p{\mathrm{\mid }}_{\tilde{U}}:\tilde{U}\to U$ is a homeomorphism. These product charts give the quotient its smooth-manifold structure: on overlaps the change of ordered representative is locally one fixed coordinate permutation, hence smooth. The map $(p(z),v)\mapsto [(z,v)]$, for the unique $z\in \tilde{U}$ over $p(z)$, is a local trivialization of $(F_n(\mathbb{C})\times V_0^{\otimes n})\mathrm{/}{S}_n$ over $U$. On each connected component of an overlap, changing the chosen sheet acts on the fiber by one fixed ${\tau }_{\sigma }$, a linear isomorphism, so these are smooth vector-bundle transition maps. In these local trivializations the connection is obtained by restricting the KZ connection to the chosen sheet. If a different sheet is chosen, the connection agrees under the transition ${\tau }_{\sigma }$ because the KZ connection is equivariant. Thus the quotient is a vector bundle and its connection descends, with fiber identified with $V_0^{\otimes n}$ at $[q]$ by the representative $q$.

For an endpoint-fixed homotopy of paths, pull back the parallel-frame neighborhoods to the compact square. A finite subcover has a positive Lebesgue number: choose finitely many smaller metric balls covering the square whose doubled balls lie in cover members, and take the minimum radius $\delta$ of those smaller balls. Choose a rectangular grid whose cell diameters are less than $\delta$; every cell then maps into one parallel-frame neighborhood. Transport around each cell is the identity because its four edges use one constant frame. Multiplying the cell identities cancels every interior edge with its reverse, so the boundary paths have identical local-system transport. By concatenating the unique parallel sections on successive path segments, [F5] gives $P_{\alpha \ast \beta }=P_{\beta }{P}_{\alpha }$; reversing a parallel section gives $P_{\bar{\alpha }}=P_{\alpha }^{-1}$. The same composition and inverse rules hold for local-system transport by its subdivision definition.

Let $T_{\gamma }$ denote the local-system transport for a continuous path. Step 4.2 makes it depend only on the path-homotopy class. Under [F6], the first-then-second product satisfies $T_{\alpha \ast \beta }=T_{\beta }{T}_{\alpha }$, so $\rho ([\gamma ]):=T_{\gamma }^{-1}$ is a left representation because $\rho ([\alpha ][\beta ])=(T_{\beta }{T}_{\alpha }{)}^{-1}=T_{\alpha }^{-1}{T}_{\beta }^{-1}$. On piecewise smooth loops $T_{\gamma }=P_{\gamma }$ by step 3.1. The map $h(z)=z\mathrm{/}(1+\mathrm{\mid }z\mathrm{\mid })$ has inverse $h^{-1}(w)=w\mathrm{/}(1-\mathrm{\mid }w\mathrm{\mid })$ from $D^{\circ }$ to $\mathbb{C}$; its coordinatewise product is a homeomorphism of ordered configuration spaces and takes $q$ to $\tilde{q}$. It induces a based isomorphism on fundamental groups, and the inclusion $D^{\circ }\hookrightarrow D^2$ induces the based isomorphism in [F8]. Hence the ordered monodromy representation is based at $\tilde{q}$ when identified with $P{B}_n={\pi }_1(F_n(D^2),\tilde{q})$.

The descended connection is flat because its pullback is the flat connection of step 2.1. Applying the local-system monodromy construction of steps 3.1 and 5.1 on the quotient gives its left monodromy representation of ${\pi }_1(C_n(\mathbb{C}),[q])$. The coordinatewise homeomorphism $h$ is $S_n$-equivariant, so it descends to a homeomorphism of unordered configuration spaces and takes $[q]$ to $[\tilde{q}]$. The inclusion-induced isomorphism at $[\tilde{q}]$ in [F11] then identifies this based fundamental group with $B_n^{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{f}}={\pi }_1(C_n(D^2),[\tilde{q}])$. No identification with any quantum-group ribbon representation follows from this construction.