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A Q-exact deformation with a nonzero boundary term

Physical scope

A finite-dimensional graded integral on a compact manifold with boundary, included only to exhibit the boundary obstruction; it is not an S4 field theory, empirical result, or continuum claim.

Example

Let M=[0,1]×(R/2πZ) be the cylinder as a smooth manifold with boundary, with dimensionless product coordinates x,y; dx,dy denote the global product and angular one-forms. Orient M by the volume form dx∧dy and define the graded integral I(F)=∫MF[2].

The complex differential-form algebra is Ω∙(M;C)=Ω0⊕Ω1⊕Ω2, where C∞(M;C) is the space of functions smooth up to x=0,1 and periodic in y, Ω0=C∞(M;C), Ω1={a dx+b dy:a,b∈C∞(M;C)}, Ω2={c dx∧dy:c∈C∞(M;C)}, and Ωk=0 for k>2. The product is the exterior product, determined by dx∧dx=dy∧dy=0 and dy∧dx=−dx∧dy. The exterior derivative is the degree-one linear map d defined by df=(∂xf)dx+(∂yf)dy, d(a dx+b dy)=(∂xb−∂ya)dx∧dy, and dα=0 for α∈Ω2; these formulas give the graded Leibniz rule and d2=0.

Take the odd derivation Q=d, dimensionless action S=0, even insertion O=1, and odd deformation V=x dy. For every dimensionless real t, Z(t)=I(Oe−S−tQV)=−2πt. The change occurs because I(Q(Ve−S−tQV))=2π≠0: this cycle violates the zero-boundary Q-invariance condition and therefore is not an instance of the finite-dimensional Ward identity.

Facts & Assumptions

Given: The compact oriented cylinder M, its boundary circles, the graded integral of top-degree two-forms, Q=d, S=0, O=1, and V=x dy as specified in the Example.

[A1]

The integration cycle is the whole cylinder with its two boundary circles included; no boundary condition cancels their flux.

Verification

1.1givenA1algebra

The coefficient product rule in the displayed coordinate formulas gives d(fg)=df g+f dg for f,g∈Ω0 and d(fα)=df∧α+f dα, d(αf)=dα f−α∧df for α∈Ω1; for α,β∈Ω1, both sides of the graded-Leibniz identity have degree three and vanish because Ω3=0. Thus d is an odd derivation. For f∈Ω0, d2f=(∂x∂yf−∂y∂xf)dx∧dy=0 by equality of mixed partial derivatives, and d2=0 on Ω1⊕Ω2 because dΩ2=0. Thus Q=d has Q2=0. The differential gives QV=d(x dy)=dx∧dy. Since (dx∧dy)2=0, the finite exterior-degree expansion is e−tQV=1−t dx∧dy. Also QS=QO=0.

2.1givenstep 1.1algebra

The degree-two part of Oe−S−tQV is −t dx∧dy, so Z(t)=−t∫01dx∫02πdy=−2πt. In particular, Z′(t)=−2π.

3.1givenA1step 1.1algebra∎

Since Qe−tQV=−(Q2V)t e−tQV=0, the graded Leibniz rule gives Q(Ve−tQV)=(QV)e−tQV=dx∧dy. Its integral is 2π, the nonzero flux from the boundary circles. Thus the integral fails the Q-invariance condition exactly at the boundary term.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.