Paper IV visual guide · Visual 5 of 26 · DIFFERENTIAL SYSTEM

One signed edge determines the cycle

Traditional monograph: Theorem 5.3 (p. 8), Theorem 6.4 (p. 11), Corollary 6.5 (p. 12), Corollary 6.6 (p. 12).

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Let \(J_3e_0=e_1\), \(J_3e_1=e_2\), and \(J_3e_2=-e_0\). A complete turn changes sign, so \(J_3^3=-I\). The signed sections are the coordinates of \(e^{zJ_3}e_0\).

The signed shift matrix and its exponential are shown beside their cycle, initial-value, and determinant identities.
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

Read the two matrices by row and column in the ordered basis e₀, e₁, e₂. Columns describe inputs and rows describe outputs. The minus sign in the top-right corner closes the signed cycle. In the exponential matrix, each upper-triangular minus sign records a wrap across that edge. The derivative rules underneath use the abbreviated section names explained in the plate.

Drawing labels and data 24 labels

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\(J_3\)
Row / columnColumn 0Column 1Column 2
Row 0\(0\)\(0\)\(-1\)
Row 1\(1\)\(0\)\(0\)
Row 2\(0\)\(1\)\(0\)
\(e^{zJ_3}\)
Row / columnColumn 0Column 1Column 2
Row 0\(D_{0}\)\(- D_{2}\)\(- D_{1}\)
Row 1\(D_{1}\)\(D_{0}\)\(- D_{2}\)
Row 2\(D_{2}\)\(D_{1}\)\(D_{0}\)

Identities below the matrices

  • \(J_3^3=-I\)

\(\mathbf D_3(0)=e_0\)

  • \(\det e^{zJ_3}=1\)
\[D_0\mapsto-D_2,\qquad D_1\mapsto D_0,\qquad D_2\mapsto D_1\]

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Here \(D_j\) inside the matrix means \(D_{3,j}(z)\). For example,

\[ D_{3,0}=1-\frac{z^3}{3!}+\frac{z^6}{6!}-\cdots, \qquad D_{3,1}=z-\frac{z^4}{4!}+\cdots. \]

Rows are output coordinates and columns are input coordinates. The upper-triangular minus signs record crossing the signed edge while moving from the column to the row. They are not arbitrary orientation choices.

Two checks connect the matrix and the scalar equation

The initial derivative matrix is the identity. Each initial jet \((y(0),y'(0),y''(0))\) therefore determines a unique linear combination. Also,

\[ \frac13{\operatorname{Tr}}(e^{zJ_3})=D_3(z),\qquad \det(XI-J_3)=X^3+1. \]

The roots of \(-1\) appearing later in the exponential sum are already the spectrum of this concrete matrix.

Read this as

The functions, their initial data, the addition law, and the exponent polygon come from one signed shift.

Evidence note from the source edition

The exact tests check every matrix-entry Taylor coefficient through degree 14 for \(N=1,\ldots,5\). The displayed determinant-one law is for \(N\ge2\). \(N=1\) is a separate boundary case.

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