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A Q-exact deformation with a nonzero boundary term
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 be the cylinder as a smooth manifold with boundary, with dimensionless product coordinates ; denote the global product and angular one-forms. Orient by the volume form and define the graded integral .
The complex differential-form algebra is , where is the space of functions smooth up to and periodic in , , , , and for . The product is the exterior product, determined by and . The exterior derivative is the degree-one linear map defined by , , and for ; these formulas give the graded Leibniz rule and .
Take the odd derivation , dimensionless action , even insertion , and odd deformation . For every dimensionless real , . The change occurs because : 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 , its boundary circles, the graded integral of top-degree two-forms, , , , and as specified in the Example.
The integration cycle is the whole cylinder with its two boundary circles included; no boundary condition cancels their flux.
Verification
The coefficient product rule in the displayed coordinate formulas gives for and , for ; for , both sides of the graded-Leibniz identity have degree three and vanish because . Thus is an odd derivation. For , by equality of mixed partial derivatives, and on because . Thus has . The differential gives . Since , the finite exterior-degree expansion is . Also .
The degree-two part of is , so . In particular, .
Since , the graded Leibniz rule gives . Its integral is , 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.