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ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Two-insertion KZ monodromy is a Casimir exponential

Example

Let g=sl2(C), let V=C2 be its standard representation with its standard Hermitian inner product, and use the Killing-form Casimir tensor Ω12 on V⊗V. For two insertions set w=z1−z2≠0. At the basepoint w∗=1, take z2=0 and z1=w; along a chosen logarithm branch, a horizontal section with value v∗ at w∗ is s(w)=exp⁡((Ω12/κ)log⁡(w/w∗))v∗. The positive loop w(t)=w∗e2πit, equivalently z1(t)=e2πit and z2(t)=0, has horizontal transport

Pγ=exp⁡(2πiΩ12/κ).

For the library's first-then-second fundamental-group product, the associated left monodromy representation uses Pγ−1; equivalently, Pγ is the right monodromy action by horizontal transport. This transport need not be unitary: with κ=i, it has eigenvalue eπ/4.

Diagram and convention

The diagram uses w=z1−z2 as the relative coordinate; the arrow indicates increasing t in w(t)=w∗e2πit, and the removed origin is the collision z1=z2. 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=sl2(C), the standard two-dimensional representation and Hermitian inner product, w∗=z1−z2≠0, a selected logarithm branch near w∗, and κ≠0.

[F1]

Casimir KZ connection on ordered configurations: for two insertions, the connection is ∇=d−κ−1Ω12 dw/w on C×.

[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⁡xad⁡y).

Verification

Given: [F1]–[F3] and the specified sl2 representation.

1.1F1givenalgebra

The horizontal-section equation ∇s=0 is ds/dw=(Ω12/(κw))s. On a simply connected branch domain, differentiate the matrix exponential s(w)=exp⁡((Ω12/κ)log⁡(w/w∗))v∗; because its exponent is a scalar multiple of the fixed matrix Ω12, differentiation gives s′(w)=Ω12s(w)/(κw) and s(w∗)=v∗.

1.2F3givenalgebra

In the defining two-dimensional representation take H=diag(1,−1), E=(0100), and F=(0010). Their brackets are [H,E]=2E, [H,F]=−2F, 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∗e2πit changes the continued logarithm by 2πi. Substitution into the solution from step 1.1 gives Pγ=exp⁡(2πiΩ12/κ).

3.1F2step 2.1algebra

With the first-then-second group product, uniqueness of horizontal transport gives Pα∗β=PβPα. Therefore ρ([γ]):=Pγ−1 satisfies ρ([α][β])=(PβPα)−1=Pα−1Pβ−1, so the left representation convention agrees with [F2] while the right action uses Pγ.

4.1F1step 2.1step 1.2algebra∎

Thus Ω12=18H⊗H+14(E⊗F+F⊗E). On e1⊗e1 it acts as 18(e1⊗e1), so for κ=i the positive-loop transport in step 2.1 has eigenvalue exp⁡(2πi(1/8)/i)=eπ/4. A unitary operator on a finite-dimensional Hermitian space has every eigenvalue of modulus one, whereas ∣eπ/4∣>1; thus this KZ monodromy is not unitary. Its inverse has eigenvalue e−π/4 and is also nonunitary.

Depends on

Dependency tree · 2 levels

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