Ivancevic_Applied-Diff-Geom

466 Applied Differential Geometry: A Modern Introductionbut we deform the co–sectorwhere∆ˆδ ξ = e − i 2 hθρσ ˆ∂ρ⊗ ˆ∂ σ(ˆδ ξ ⊗ 1 + 1 ⊗ ˆδ ξ )e i 2 hθρσ ˆ∂ρ⊗ ˆ∂ σ,(3.285)[ ˆ∂ ρ , ˆδ ξ ] = ˆδ (∂ρξ).The deformed coproduct (3.285) reduces to the undeformed one (3.282) inthe limit θ → 0. Antipode and counit remain undeformedS(ˆδ ξ ) = −ˆδ ξ ε(ˆδ ξ ) = 0.We have to check whether the above deformation is a good one in thesense that it leads to a consistent action on Â. First we need a differentialoperator acting on fields in  which represents the algebra (3.284). Let usconsider the differential operatorˆX ξ :=∞∑n=01n! (− i 2 )n θ ρ 1 σ1 · · · θ ρ n σn ( ˆ∂ ρ1 · · · ˆ∂ ρnˆξµ ) ˆ∂µ ˆ∂σ1 · · · ˆ∂ σn . (3.286)This is to be understood like that: A vector–field ξ = ξ µ ∂ µ is determined byits coefficient functions ξ µ . Before we have seen that there is a vectorspaceisomorphism W from the space of commutative to the space of noncommutativefunctions which is given by the symmetric ordering prescription. Theimage of a commutative function f under the isomorphism W is denotedby ˆfW : f ↦→ W (f) = ˆf.In (3.286) ˆξ µ is therefore to be interpreted as the image of ξ µ with respectto W . Then indeed we haveTo see this we use result (3.276) to rewrite ( ˆX ξˆφ) :( ˆX ξˆφ) = ∞ ∑∞∑n=0n=0[ ˆX ξ , ˆX η ] = ˆX ξ×η . (3.287)1n! (− i 2 )n θ ρ 1 σ1 · · · θ ρ n σn ( ˆ∂ ρ1 · · · ˆ∂ ρnˆξµ )( ˆ∂µ ˆ∂σ1 · · · ˆ∂ σnˆφ) = (3.288)1n! (− i 2 )n θ ρ 1 σ1 · · · θ ρ n σn ( ˆ∂ ρ1 · · · ˆ∂ ρnˆξµ )( ˆ∂σ1 · · · ˆ∂ σn̂∂µ φ) = ̂ ξ µ (∂ µ φ) =From (3.288) it follows( ˆX ξ ( ˆX ηˆφ)) − ( ˆXη ( ˆX ξˆφ)) = ̂ ([ξ, η]φ) = ( ˆXξ×ηˆφ),̂ (ξφ).