Flatness of the KZ connection and braid relations

Put ${\alpha }_{ij}=d\log (z_i-z_j)$. Since $z_i-z_j$ is nonzero on ${\mathrm{Conf}}_q(\mathbb{C})$, in local coordinates ${\alpha }_{ij}=d(z_i-z_j)\mathrm{/}(z_i-z_j)$ and $d{\alpha }_{ij}=0$ by equality of mixed partial derivatives. Thus $d{\mathrm{\Omega }}_q=0$.

Write $a=t_{ij}$, $b=t_{kl}$ for two distinct chord generators. In the expansion of ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q$, the two ordered terms for these generators combine as $[a,b]{\alpha }_{ij}\wedge {\alpha }_{kl}$. If the pairs have four distinct indices, $[a,b]=0$ by [F1]; if they are the same pair, ${\alpha }_{ij}\wedge {\alpha }_{ij}=0$. Hence only pairs of generators supported on a common triple of strands can contribute.

Fix $i<j<k$ and put $a=t_{ij}$, $b=t_{ik}$, $c=t_{jk}$, $x={\alpha }_{ij}$, $y={\alpha }_{ik}$, and $z={\alpha }_{jk}$. The tangled 4T relation on these three strands is exactly the pair of infinitesimal braid relations $[a,b+c]=0$ and $[b,a+c]=0$. The contribution of this triple is $[a,b]x\wedge y+[a,c]x\wedge z+[b,c]y\wedge z$. The infinitesimal braid relations in [F1] give $[a,c]=-[a,b]$ and $[b,c]=[a,b]$, so this is $[a,b](x\wedge y-x\wedge z+y\wedge z)$. With $u=z_i-z_j$ and $v=z_j-z_k$, we have $x=du\mathrm{/}u$, $z=dv\mathrm{/}v$, and $y=d(u+v)\mathrm{/}(u+v)$. Direct expansion gives $x\wedge z+z\wedge y+y\wedge x=(1\mathrm{/}(uv)-1\mathrm{/}(v(u+v))-1\mathrm{/}(u(u+v)))du\wedge dv=0$, which is equivalent to $x\wedge y-x\wedge z+y\wedge z=0$. Thus every triple contribution vanishes.

The repeated-pair, disjoint-pair, and three-index cases exhaust the expansion of ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q$; steps 2.1–3.1 show that each contribution is zero. Together with step 1.1 this gives $d{\mathrm{\Omega }}_q=0$ and ${\mathrm{\Omega }}_q\wedge {\mathrm{\Omega }}_q=0$.

Fix a smooth path $\gamma$ and let $U(t)$ solve $\dot{U}=A_{t}U$, $U(0)=1$, where $A_t={\mathrm{\Omega }}_q(\dot{\gamma }(t))$. In each degree this is a finite linear ODE system, solved recursively by the iterated-integral formula. For a homotopy $\gamma (s,t)$ with fixed endpoints, set $A_s={\mathrm{\Omega }}_q({\mathrm{\partial }}_s\gamma )$ and $W=U^{-1}({\mathrm{\partial }}_{s}U-A_{s}U)$. Differentiating and using ${\mathrm{\partial }}_{t}U=A_{t}U$ gives ${\mathrm{\partial }}_{t}W=U^{-1}({\mathrm{\partial }}_s{A}_t-{\mathrm{\partial }}_t{A}_s+[A_t,A_s])U$. The coefficient in parentheses is the evaluation of the flatness expression from step 4.1, so it is zero. Since the initial endpoint is fixed, $W(s,0)=0$ and hence $W(s,t)=0$; at the fixed terminal endpoint $A_s=0$, so ${\mathrm{\partial }}_{s}U(s,1)=0$. The holonomy is therefore constant throughout the homotopy.

The KZ form is unchanged by relabeling the strands, since its sum runs over every unordered pair; endpoint strand permutations therefore act on holonomy by the same permutation of the $t_{ij}$. For $\mathrm{\mid }i-j\mathrm{\mid }\ge 2$, the paths for ${\sigma }_i$ and ${\sigma }_j$ lie in disjoint two-point disks, so exchanging their time intervals is a fixed-endpoint homotopy and step 5.1 gives ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i$. For three adjacent points, start at $(-1,0,1)$ and follow the configuration path $(-e^{\pi it},0,e^{\pi it})$, $0\le t\le 1$. Its endpoints agree as unordered configurations with the three points reversed, and its pairwise distances stay bounded below by $1$. At $t=1\mathrm{/}2$ the three real projections coincide. Small perturbations of the trajectories therefore remain collision-free and can separate the three projected crossing times in either of the two orders; both perturbed paths are homotopic to this rigid half-turn and give respectively ${\sigma }_i{\sigma }_{i+1}{\sigma }_i$ and ${\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}$. By step 5.1 their transports agree.