# Zoran Skoda VSVint

$S_0 \hat{y}_\alpha = S_0(x_\rho \phi(-\partial)^\rho_\alpha) = \phi(\partial)^\rho_\alpha x_\rho = x_\rho\phi(\partial)^\rho_\alpha + \partial_\rho\phi(\partial)^\rho_\alpha$

where we denoted $\partial_\alpha = \frac{\partial}{\partial(\partial^\alpha)}$. We seek for $V = V(\partial)$ such that $S(h) = V^{-1}S_0(h)V$. Then

$x_\rho\phi^\rho_\alpha = V^{-1}(x_\rho\phi(\partial)^\rho_\alpha + \partial_\rho\phi(\partial)^\rho_\alpha)V = V^{-1}x_\rho V\phi^\rho_\alpha + \partial_\rho\phi^\rho_\alpha$
$x_\alpha - (\partial_\rho \phi^\rho_\gamma)\phi^{-1\gamma}_\alpha = V^{-1} x_\alpha V$
$V^{-1} [V,x_\alpha] = (\partial_\rho\phi^\rho_\gamma)\phi^{-1\gamma}_\alpha$
$\partial_\alpha{ln|V|} = V^{-1}\partial_\alpha V = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma$
$R := ln|V|$
$\partial_\alpha R = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma$

If we set $F_\alpha = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma$ this is the system

$\partial_\alpha R = F_\alpha$

which is integrable if and only if $\partial_\beta F_\alpha = \partial_\alpha F_\beta$ that is

$\partial_\beta(\phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma) = \partial_\alpha(\phi^{-1\gamma}_\beta\partial_\rho\phi^\rho_\gamma)$

In the above derivation of the integrability condition this was derived under the assumption that $\hat{y}_\alpha = x_\rho\phi(-\partial)^\rho_\alpha$. That is true for the symmetric ordering and some other orderings. Thus the existance of twist $V$ for the antipode for such ones should be implied by the integrability condition, but it is not clear how to prove it.

If the solution exists we can write it as $V_0 V_1$ where $|V_1| = 1$ and $V_0$ is a normalization constant which therefore commutes. So without loss of generality we can set $V_0 = V(0) = 1$ what can be understood as the boundary condition $R(0) = 0$ for $R$ when solving the exact first order differential equation above.

Created on January 8, 2018 at 06:08:35. See the history of this page for a list of all contributions to it.