Manuscript IX-Free@1 · IX-Free.tex · section · 33b20e96

Overlap diagnostics

F5–F6: actual maps and the remaining overlap gates › Overlap diagnostics · lines 1216–1249

IX-Free@1#eq:tensorintersection, IX-Free@1#eq:newoverlap, IX-Free@1#eq:oldoverlap

For vector subspaces $U, U^{'}, V, V^{'}$ , all contained in fixed ambient spaces, algebraic tensor products satisfy

(U \otimes V) \cap (U^{'} \otimes V^{'}) = (U \cap U^{'}) \otimes (V \cap V^{'}) .

One proof is to extend bases of the intersections to bases of the sums and compare the resulting tensor coordinates.

Field-degree bigrading shows that a left degree-four block and a right degree-four block have zero intersection: their argument degrees are $(4, 2)$ and $(2, 4)$ . The mixed degree-four/degree-two intersections are also zero.

The potentially nontrivial left/right degree-two intersection is

(L_{2} \otimes B) \cap (B \otimes R_{2}) = (L_{2} \cap B) \otimes (B \cap R_{2}) .

No vanishing of the two factors is proved here.

For each new block $U \otimes V$ , its old-product overlap is

(U \otimes V) \cap D_{old} = (U \cap P_{mc}) \otimes (V \cap P_{mc}) .

Thus the old-product obligation is a concrete value-preservation test, not a convergence assertion.

Displayed formulas in this passage

IX-Free@1#eq:tensorintersection · line 1220
$(U \otimes V) \cap (U^{'} \otimes V^{'}) = (U \cap U^{'}) \otimes (V \cap V^{'}) .$
IX-Free@1#eq:newoverlap · line 1234
$(L_{2} \otimes B) \cap (B \otimes R_{2}) = (L_{2} \cap B) \otimes (B \cap R_{2}) .$
IX-Free@1#eq:oldoverlap · line 1242
$(U \otimes V) \cap D_{old} = (U \cap P_{mc}) \otimes (V \cap P_{mc}) .$

Overlap diagnostics

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