lemmaunknowninferred_dep

Repeated Taylor degrees force determinant decay

Bounds ∣det⁡T∣|\det T| by e−c∑na2+…e^{-c\sum n_a^2+\dots}: repeated Taylor indices shrink it
supplies the quadratic collision saving
Lemma·det:collision·§3.3 Repeated Taylor degrees force a quadratic saving

Statement (paper)

Choose one translated test row from each row expansion in Lemma , and remove its scalar factor. Let TT be the resulting square matrix, and let nan_a be the number of its rows with index aa. With c=log⁡24,Ehol=100Kw0+log⁡2v0+log⁡(200K)w∗, c=\frac{\log2}{4},\qquad E_{\rm hol}=\frac{100K}{w_0}+\frac{\log2}{v_0} +\frac{\log(200K)}{w_*}, there is a quantity ρH→0\rho_H\to0, independent of all these row choices and of the selected columns, such that ∣det⁡T∣≤exp⁡{−c∑ana2+MH(Ehol+ρH)}. |\det T| \le\exp\left\{-c\sum_a n_a^2+MH(E_{\rm hol}+\rho_H)\right\}.

In plain English

After picking one translated test row from each row expansion and stripping its scalar factor, the resulting square matrix TT satisfies ∣det⁡T∣≤exp⁡{−c∑ana2+MH(Ehol+ρH)}|\det T|\le\exp\{-c\sum_a n_a^2+MH(E_{\rm hol}+\rho_H)\}, where nan_a counts rows with index aa, c=log⁡24c=\frac{\log 2}{4}, and ρH→0\rho_H\to0 does not depend on the row choices or the selected columns.

Why it's here

This is the quadratic saving: when many rows share the same index aa, they are forced to be nearly dependent, so the determinant shrinks like e−c na2e^{-c\,n_a^2} for each index. It acts on the entire-function test rows produced by the translation lemma, and the determinant comparison and the final contradiction use it to turn index collisions into a determinant bound.

Sources

PaperOpen
Section
§3.3 Repeated Taylor degrees force a quadratic saving
Page
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Label
det:collision
Line
sections/determinant.tex:267
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    critique: The collision lemma explicitly chooses rows from the expansion in Lemma 3.2 (sections/determinant.tex:267–270).