claimroutineinferred_dep
Kernel factorization extends to k columns
Kernel factorization holds on columns with the same factors
extends the factorization to any column count
Claim·§3.5 The binary phase interface·inside propComplex network edges factor into translation kernels
Statement (paper)
For each edge with residual basis and translation kernels (forward or inverse) for , on columns one has . Thus each residual basis vector contributes one factor acting on all columns, and there are such factors over the network.
In plain English
On columns, the phase-frame difference of each edge equals the product over the residual basis of the tensor powers of the translation kernels (forward or inverse). Each basis vector thus gives one factor acting on all columns, and there are such factors over the network.
Why it's here
This follows from the single-column factorization because tensor powers respect products, , so the factor count stays regardless of . It lets the later telescoping argument, which yields on the roles, be carried out on any positive number of columns.
Classified routine:
Direct consequence of the single-column factorization (claim-prop-complex-motif-interface-1) via the multiplicativity of tensor powers, .
Uses
Used by
Sources
PaperOpen
- Section
- §3.5 The binary phase interface
- Page
- —
- Label
—- Line
sections/03-motifs.tex:616
Leannot aligned
Phase 2 aligns this node to a Lean declaration.
Friction
- infoinferred_depoutline: "Kernel factorization extends to k columns" rests on claim-prop-complex-motif-interface-1