Paper III visual guide · Visual 16 of 24 · MODEL F / PACKET BLOCKS FROM INPUT TO OUTPUT

Resolve the entire obstruction

Traditional paper: Theorem 7.5 (p. 17), Corollary 7.6 (p. 18)

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Original plate in the visual-guide PDF · Original atlas source

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Use three coordinate packets with projectors \(P_0,P_1,P_2\), and set

\[ H=I+2E_{01}+3E_{12},\qquad {\operatorname{Ob}}=H-I. \]

Here \(H\) is invertible, so it can be the loop product of a reversible connection on three charts. Its entries outside the diagonal encode two effects that mix packets.

Every input-output block of an obstruction
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

The board retains all nine embedded blocks. Its rows are output packets and columns are input packets; two off-diagonal addresses carry the nonzero maps.

Drawing labels and data 37 labels

Read the original labels as selectable text. Tables preserve grid relationships; other short labels are grouped compactly.

  • \(H\)
Drawing grid 1 · rows top to bottom, columns left to right
Row / columnColumn 1Column 2Column 3
Row 1\(1\)\(2\)\(0\)
Row 2\(0\)\(1\)\(3\)
Row 3\(0\)\(0\)\(1\)
  • \({\operatorname{Ob}}=H-I\)
Drawing grid 2 · rows top to bottom, columns left to right
Row / columnColumn 1Column 2Column 3
Row 1\(0\)\(2\)\(0\)
Row 2\(0\)\(0\)\(3\)
Row 3\(0\)\(0\)\(0\)
Drawing grid 3 · rows top to bottom, columns left to right
Row / column\(\nu=0\)\(\nu=1\)\(\nu=2\)
\(\mu=0\)\(0\)\(2E_{01}\)\(0\)
\(\mu=1\)\(0\)\(0\)\(3E_{12}\)
\(\mu=2\)\(0\)\(0\)\(0\)

\({\operatorname{Ob}}(4,5,6)^\mathsf T=(10,18,0)^\mathsf T\)

  • sum all nine blocks

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Evidence note from the source edition

The board on the right contains all nine actual embedded blocks \(P_\mu{\operatorname{Ob}} P_\nu\). Row labels are output packets and column labels are input packets. Only the addresses \((0,1)\) and \((1,2)\) carry nonzero maps.

Completeness on both sides supplies the reconstruction:

\[ {\operatorname{Ob}}=(P_0+P_1+P_2){\operatorname{Ob}}(P_0+P_1+P_2) =\sum_{\mu,\nu}P_\mu{\operatorname{Ob}} P_\nu. \]

For \(x=(4,5,6)^\mathsf T\), the input in packet \(1\) contributes \(10e_0\) and the input in packet \(2\) contributes \(18e_1\). Their sum is the full displacement \({\operatorname{Ob}} x\).

A block becomes a witness when it acts

The nonzero block \(2E_{01}\) has the explicit source vector \(e_1\), supported entirely in packet \(1\), and \(P_0{\operatorname{Ob}} e_1=2e_0\ne0\). Similarly, \(P_1{\operatorname{Ob}} e_2=3e_1\ne0\). The general theorem supplies such a witness classically when a nonzero block is known. The example displays it concretely.

Read this as

An obstruction packet is not just a colored matrix cell. It consists of a specific label for a pair of packets and an exact nonvanishing statement about the corresponding map.

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