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Two-insertion KZ monodromy is a Casimir exponential
Example
Let , let be its standard representation with its standard Hermitian inner product, and use the Killing-form Casimir tensor on . For two insertions set . At the basepoint , take and ; along a chosen logarithm branch, a horizontal section with value at is . The positive loop , equivalently and , has horizontal transport
For the library's first-then-second fundamental-group product, the associated left monodromy representation uses ; equivalently, is the right monodromy action by horizontal transport. This transport need not be unitary: with , it has eigenvalue .
Example references
Diagram and convention
The diagram uses as the relative coordinate; the arrow indicates increasing in , and the removed origin is the collision . 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, , the standard two-dimensional representation and Hermitian inner product, , a selected logarithm branch near , and .
Casimir KZ connection on ordered configurations: for two insertions, the connection is on .
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.
The Killing form of a semisimple Lie algebra: the form is .
Verification
Given: [F1]–[F3] and the specified representation.
The horizontal-section equation is . On a simply connected branch domain, differentiate the matrix exponential ; because its exponent is a scalar multiple of the fixed matrix , differentiation gives and .
In the defining two-dimensional representation take , , and . Their brackets are , , and . Computing the traces in [F3] gives , , and all other pairings among these basis elements zero; hence the dual basis is .
The loop changes the continued logarithm by . Substitution into the solution from step 1.1 gives .
With the first-then-second group product, uniqueness of horizontal transport gives . Therefore satisfies , so the left representation convention agrees with [F2] while the right action uses .
Thus . On it acts as , so for the positive-loop transport in step 2.1 has eigenvalue . A unitary operator on a finite-dimensional Hermitian space has every eigenvalue of modulus one, whereas ; thus this KZ monodromy is not unitary. Its inverse has eigenvalue and is also nonunitary.