Manuscript XIV-I@1 · XIV-I.tex · section · 29df501d
Both internal placements of K 5
I0: copied inputs and the exact predecessor stop › Both internal placements of K 5 · lines 808–908
XIV-I@1#prop:pole, XIV-I@1#eq:pole, XIV-I@1#sec:amplitudespace
Partial fractions followed by the same distributional limit give (eq:partialfraction) when .
The cone formula has physical homogeneity degree in dimension four. Its Fourier transform consequently has degree as well. Alternatively the same identity follows by positive dilation in (eq:MR).
The damping in this proof computes the Fourier transform of an existing kernel. It is not a regulator for a graph product, and no restoring limit of the quantum coefficient is used.
uniform pole chart on either characteristic sheet
For , define
Take a conic neighborhood of that nonzero sheet satisfying
The opposite factor is then nonzero.
Let
On the stated conic neighborhood,
The coefficient is smooth and homogeneous of degree , and is smooth and homogeneous of degree there.
Off , the complete expression equals . Its singular and regular pieces in (eq:pole) are both retained.
Temporal inverse Fourier transformation gives exactly
Thus the pole and noncharacteristic multiplier reconstruct the same retarded kernel, whose separated-cone coefficient is the one in (eq:R).
Proof. For , isolate the first term of (eq:partialfraction); for , isolate the second. The opposite denominator is smooth on the selected chart, giving (eq:pole). The homogeneities are immediate.
Away from the pole, both boundary values become their ordinary reciprocals. Combining the two fractions gives .
The elementary Fourier identity
follows by Fourier transforming the right-hand side with positive damping. Applying it to (eq:partialfraction) gives . Its spatial inverse transform is the fixed cone kernel (eq:R), by Lemma lem:multiplier and Fourier injectivity.
The part is fixed by the original retarded support. It is not an adjustable internal counterterm. Replacing it would change and hence the frozen pair.
This pole chart is not a characteristic prescription for . Only one pre-existing internal line is being represented.
2b: a finite distribution-valued amplitude space
coordinates for the two placements
Displayed formulas in this passage
XIV-I@1#eq:pole · line 844
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Nearby in this section
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5
- section: Both internal placements of K 5