Zoran Skoda

S 0y^ α=S 0(x ρϕ() α ρ)=ϕ() α ρx ρ=x ρϕ() α ρ+ ρϕ() α ρ 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 α=( α)\partial_\alpha = \frac{\partial}{\partial(\partial^\alpha)}. We seek for V=V()V = V(\partial) such that S(h)=V 1S 0(h)VS(h) = V^{-1}S_0(h)V. Then

x ρϕ α ρ=V 1(x ρϕ() α ρ+ ρϕ() α ρ)V=V 1x ρVϕ α ρ+ ρϕ α ρ 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 α( ρϕ γ ρ)ϕ α 1γ=V 1x αV x_\alpha - (\partial_\rho \phi^\rho_\gamma)\phi^{-1\gamma}_\alpha = V^{-1} x_\alpha V
V 1[V,x α]=( ρϕ γ ρ)ϕ α 1γ V^{-1} [V,x_\alpha] = (\partial_\rho\phi^\rho_\gamma)\phi^{-1\gamma}_\alpha
αln|V|=V 1 αV=ϕ α 1γ ρϕ γ ρ \partial_\alpha{ln|V|} = V^{-1}\partial_\alpha V = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma
R:=ln|V| R := ln|V|
αR=ϕ α 1γ ρϕ γ ρ \partial_\alpha R = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma

If we set F α=ϕ α 1γ ρϕ γ ρF_\alpha = \phi^{-1\gamma}_\alpha\partial_\rho\phi^\rho_\gamma this is the system

αR=F α \partial_\alpha R = F_\alpha

which is integrable if and only if βF α= αF β\partial_\beta F_\alpha = \partial_\alpha F_\beta that is

β(ϕ α 1γ ρϕ γ ρ)= α(ϕ β 1γ ρϕ γ ρ) \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 y^ α=x ρϕ() α ρ\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 VV 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 0V 1V_0 V_1 where |V 1|=1|V_1| = 1 and V 0V_0 is a normalization constant which therefore commutes. So without loss of generality we can set V 0=V(0)=1V_0 = V(0) = 1 what can be understood as the boundary condition R(0)=0R(0) = 0 for RR 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.