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Casimir connections, braid monodromy and chiral WZW interfaces — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Two-insertion KZ monodromy is a Casimir exponential

Example

Let $g = {s l}_{2} (C)$ , let $V = C^{2}$ be its standard representation with its standard Hermitian inner product, and use the Killing-form Casimir tensor $Ω_{12}$ on $V \otimes V$ . For two insertions set $w = z_{1} - z_{2} \neq 0$ . At the basepoint $w_{*} = 1$ , take $z_{2} = 0$ and $z_{1} = w$ ; along a chosen logarithm branch, a horizontal section with value $v_{*}$ at $w_{*}$ is $s (w) = \exp ((Ω_{12} / κ) \log (w / w_{*})) v_{*}$ . The positive loop $w (t) = w_{*} e^{2 π i t}$ , equivalently $z_{1} (t) = e^{2 π i t}$ and $z_{2} (t) = 0$ , has horizontal transport

$P_{γ} = \exp (2 π i Ω_{12} / κ) .$

For the library's first-then-second fundamental-group product, the associated left monodromy representation uses $P_{γ}^{- 1}$ ; equivalently, $P_{γ}$ is the right monodromy action by horizontal transport. This transport need not be unitary: with $κ = i$ , it has eigenvalue $e^{π / 4}$ .

Example references

Infinitesimal braid identities imply KZ flatness and braid monodromy

Casimir KZ connection on ordered configurations

The Killing form of a semisimple Lie algebra

Diagram and convention

The diagram uses $w = z_{1} - z_{2}$ as the relative coordinate; the arrow indicates increasing $t$ in $w (t) = w_{*} e^{2 π i t}$ , and the removed origin is the collision $z_{1} = z_{2}$ . It records the positive loop only; the proof below derives transport from the differential equation and does not rely on the picture.

diagram failed to render:
\begin{tikzpicture}[>=stealth,thick]
  \draw[->] (-1.35,0) -- (1.45,0) node[right] {$\operatorname{Re}w$};
  \draw[->] (0,-1.35) -- (0,1.45) node[above] {$\operatorname{Im}w$};
  \draw (0,0) circle (1);
  \fill (0,0) circle (0.045) node[below right] {$0$};
  \fill (1,0) circle (0.045) node[below right] {$w_*$};
  \draw[->] (0.707,0.707) arc[start angle=45,end angle=90,radius=1];
  \node at (-0.25,1.12) {$w(t)$};
\end{tikzpicture}

Facts & Assumptions

Given: Two insertions, $g = {s l}_{2} (C)$ , the standard two-dimensional representation and Hermitian inner product, $w_{*} = z_{1} - z_{2} \neq 0$ , a selected logarithm branch near $w_{*}$ , and $κ \neq 0$ .

[F1]

Casimir KZ connection on ordered configurations: for two insertions, the connection is $\nabla = d - κ^{- 1} Ω_{12} d w / w$ on $C^{\times}$ .

[F2]

Infinitesimal braid identities imply KZ flatness and braid monodromy: this connection is flat; horizontal transport is the local-system monodromy, with the left-representation convention obtained by inverting path transport when loops multiply first-then-second.

[F3]

The Killing form of a semisimple Lie algebra: the form is $B (x, y) = tr ({ad}_{x} {ad}_{y})$ .

Verification

Given: [F1]–[F3] and the specified ${s l}_{2}$ representation.

1.1F1givenalgebra

The horizontal-section equation $\nabla s = 0$ is $d s / d w = (Ω_{12} / (κ w)) s$ . On a simply connected branch domain, differentiate the matrix exponential $s (w) = \exp ((Ω_{12} / κ) \log (w / w_{*})) v_{*}$ ; because its exponent is a scalar multiple of the fixed matrix $Ω_{12}$ , differentiation gives $s^{'} (w) = Ω_{12} s (w) / (κ w)$ and $s (w_{*}) = v_{*}$ .

1.2F3givenalgebra

In the defining two-dimensional representation take $H = d i a g (1, - 1)$ , $E = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ , and $F = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array})$ . Their brackets are $[H, E] = 2 E$ , $[H, F] = - 2 F$ , and $[E, F] = H$ . Computing the traces in [F3] gives $B (H, H) = 8$ , $B (E, F) = B (F, E) = 4$ , and all other pairings among these basis elements zero; hence the dual basis is $H / 8, F / 4, E / 4$ .

2.1F1step 1.1algebra

The loop $w (t) = w_{*} e^{2 π i t}$ changes the continued logarithm by $2 π i$ . Substitution into the solution from step 1.1 gives $P_{γ} = \exp (2 π i Ω_{12} / κ)$ .

3.1F2step 2.1algebra

With the first-then-second group product, uniqueness of horizontal transport gives $P_{α * β} = P_{β} P_{α}$ . Therefore $ρ ([γ]) : = P_{γ}^{- 1}$ satisfies $ρ ([α] [β]) = (P_{β} P_{α})^{- 1} = P_{α}^{- 1} P_{β}^{- 1}$ , so the left representation convention agrees with [F2] while the right action uses $P_{γ}$ .

4.1F1step 2.1step 1.2algebra∎

Thus $Ω_{12} = \frac{1}{8} H \otimes H + \frac{1}{4} (E \otimes F + F \otimes E)$ . On $e_{1} \otimes e_{1}$ it acts as $\frac{1}{8} (e_{1} \otimes e_{1})$ , so for $κ = i$ the positive-loop transport in step 2.1 has eigenvalue $\exp (2 π i (1 / 8) / i) = e^{π / 4}$ . A unitary operator on a finite-dimensional Hermitian space has every eigenvalue of modulus one, whereas $∣ e^{π / 4} ∣ > 1$ ; thus this KZ monodromy is not unitary. Its inverse has eigenvalue $e^{- π / 4}$ and is also nonunitary.

5 · Examples, experiments and counterexamples

None yet.