claimroutineinferred_dep

Kernel factorization extends to k columns

Kernel factorization holds on kk columns with the same scs_{\rm c} 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 ee with residual basis Be\mathcal B_e and translation kernels Ke,vK_{e,v} (forward or inverse) for v∈Bev\in\mathcal B_e, on kk columns one has (CVCU−1)⊗k=∏v∈BeKe,v⊗k\bigl(\mathcal C_V\mathcal C_U^{-1}\bigr)^{\otimes k}=\prod_{v\in\mathcal B_e}K_{e,v}^{\otimes k}. Thus each residual basis vector contributes one factor acting on all kk columns, and there are scs_{\rm c} such factors over the network.

In plain English

On kk columns, the phase-frame difference (CVCU−1)⊗k(\mathcal C_V\mathcal C_U^{-1})^{\otimes k} of each edge equals the product over the residual basis Be\mathcal B_e of the tensor powers Ke,v⊗kK_{e,v}^{\otimes k} of the translation kernels (forward or inverse). Each basis vector thus gives one factor acting on all kk columns, and there are scs_{\rm c} such factors over the network.

Why it's here

This follows from the single-column factorization because tensor powers respect products, (AB)⊗k=A⊗kB⊗k(AB)^{\otimes k}=A^{\otimes k}B^{\otimes k}, so the factor count stays scs_{\rm c} regardless of kk. It lets the later telescoping argument, which yields C⊗mC^{\otimes m} 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, (AB)⊗k=A⊗kB⊗k(AB)^{\otimes k}=A^{\otimes k}B^{\otimes k}.

Sources

PaperOpen
Section
§3.5 The binary phase interface
Page
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Label
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Line
sections/03-motifs.tex:616
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    outline: "Kernel factorization extends to k columns" rests on claim-prop-complex-motif-interface-1