propositionroutine

One-stage copy step reduces density and active vertices

Copy stage keeps entries ≤ f/qf/q, at most doubles active vertices, C+Pr=PrDC^+P^r=P^rD
one-stage step iterated to build the hierarchy
Proposition·prop:hier-copy-step·§4.4 Copies reduce density while controlling active vertices

Statement (paper)

Choose the detector rank JJ large enough that J≥2Cπ(D0−1)k∗(k∗+1)δ0, J\geq\frac{2C_\pi(D_0-1)k_*(k_*+1)}{\delta_0}, and then choose R∗R_* as in Equation . The matrix C+C^+ is strictly forward and substochastic, and C+(xi,yj)≤f/q,∣A(C+)∣≤2∣A(C)∣. C^+(x_i,y_j)\leq f/q, \qquad |\mathcal A(C^+)|\leq2|\mathcal A(C)|. If PrP^r is the matrix from V+V^+ to VV defined by Pr(xi,x)=ri(x),Pr(xi,z)=0(z≠x), P^r(x_i,x)=r_i(x),\qquad P^r(x_i,z)=0\quad(z\ne x), then C+Pr=PrD. C^+P^r=P^rD.

In plain English

With the detector rank JJ and R∗R_* chosen as specified, the copy matrix C+C^+ is strictly forward and substochastic, has entries at most f/qf/q, and has at most twice as many active vertices as CC. It also satisfies the one-step intertwining C+Pr=PrDC^+P^r=P^rD with the projection PrP^r from V+V^+ to VV.

Why it's here

This packages one copy stage: the column test (Lemma 4.3) and the pilot bounds (Lemma 4.2) keep heavy columns out of C+C^+, which gives the density cap f/qf/q and the active-vertex bound. The residual DD (Lemma 4.4) is what C+C^+ reproduces under projection, so later stages can compare the copied walk to the original through PrP^r and control the reward identity and shallow target that follow.
Classified routine:
This one-stage packaging follows directly from the column test, pilot bounds and residual lemmas: the density cap, active-vertex bound and intertwining identity C^+P^r = P^rD come from bookkeeping on the copy construction rather than from a new idea.

Sources

PaperOpen
Section
§4.4 Copies reduce density while controlling active vertices
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Label
prop:hier-copy-step
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sections/hierarchy.tex:522
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