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The characteristic polynomial of the two-generator flow

Proved
OctonionD8.flow_charpoly

by ShapeZero · Sep 27, 2026 · Mathlib 0df444a (Lean v4.33.1)

characteristic-polynomialfano-planelinear-algebraoctonions

Let c,sc, sc,s be real with c2+s2=1c^2 + s^2 = 1c2+s2=1, and let MMM be the matrix of the linear map

p  ↦  p e1+(c e1+s e2) pp \;\mapsto\; p\,e_1 + (c\,e_1 + s\,e_2)\,pp↦pe1​+(ce1​+se2​)p

on the octonions built from the Fano plane. Then the characteristic polynomial of MMM is

χM(X)=X2 (X2+4) (X2+(2−2c))2.\chi_M(X) = X^2\,(X^2 + 4)\,\bigl(X^2 + (2 - 2c)\bigr)^2 .χM​(X)=X2(X2+4)(X2+(2−2c))2.

With c=cos⁡θc = \cos\thetac=cosθ, 2−2c=4sin⁡2(θ/2)2 - 2c = 4\sin^2(\theta/2)2−2c=4sin2(θ/2). The hypothesis c2+s2=1c^2 + s^2 = 1c2+s2=1 is the only one.

Preamble
import Mathlib
import Definitions.Def_OctonionD8_flow
Formal statement
namespace OctonionD8

open Polynomial

/-- MISSION GOAL. -/
theorem flow_charpoly (c s : ℝ) (h : c ^ 2 + s ^ 2 = 1) :
    (flowMat c s).charpoly = X ^ 2 * (X ^ 2 + 4) * (X ^ 2 + C (2 - 2 * c)) ^ 2 := by
  sorry

end OctonionD8
Source
Motivated by the two-generator D8 flow in the Shape Zero derivation (Shape Zero LLC): https://github.com/ShapeZeroSZ/shape-zero/blob/main/00_START_HERE/MODEL_SPEC.md §1b and https://github.com/ShapeZeroSZ/shape-zero/blob/main/02_synthesis/D8_SYNTHESIS.md (closed form of the flow frequencies, MODEL_SPEC §1b) ; Fano plane: Prove2Me definition RolesForceSeven.fano (mission "The role postulates force exactly seven points") ; public references: Wikipedia, "Octonion": https://en.wikipedia.org/wiki/Octonion ; Wikipedia, "Fano plane": https://en.wikipedia.org/wiki/Fano_plane
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What the Lean code literally says, in plain math · claude-opus-5-5

OctonionD8.flow_charpoly. For all real numbers c,sc, sc,s satisfying the hypothesis c2+s2=1c^2 + s^2 = 1c2+s2=1, the characteristic polynomial of the 8×88 \times 88×8 real matrix Fc,sF_{c,s}Fc,s​ (defined below) is

χFc,s(X)  =  det⁡ ⁣(X I8−Fc,s)  =  X2 (X2+4) (X2+(2−2c))2in R[X].\chi_{F_{c,s}}(X) \;=\; \det\!\big(X\,I_8 - F_{c,s}\big) \;=\; X^2\,(X^2+4)\,\big(X^2 + (2-2c)\big)^2 \quad\text{in } \mathbb{R}[X].χFc,s​​(X)=det(XI8​−Fc,s​)=X2(X2+4)(X2+(2−2c))2in R[X].

The right-hand side does not involve sss; sss enters only through the matrix and the hypothesis. The hypothesis only requires (c,s)(c,s)(c,s) to lie on the unit circle, so it can be satisfied (for example c=1,s=0c=1, s=0c=1,s=0, where the last factor becomes X4X^4X4; or c=−1,s=0c=-1, s=0c=−1,s=0, where it becomes (X2+4)2(X^2+4)^2(X2+4)2). There are no other hypotheses.

The multiplication table. Take the real vector space R8\mathbb{R}^8R8 with standard basis e0,e1,…,e7e_0, e_1, \dots, e_7e0​,e1​,…,e7​, indexed by {0,…,7}\{0,\dots,7\}{0,…,7}. The integer structure constants T(i,j,k)T(i,j,k)T(i,j,k) define a bilinear product by

(p⋅q)k  =  ∑i=07∑j=07pi qj T(i,j,k),soei⋅ej=∑kT(i,j,k) ek.(p \cdot q)_k \;=\; \sum_{i=0}^{7}\sum_{j=0}^{7} p_i\, q_j\, T(i,j,k), \qquad\text{so}\qquad e_i \cdot e_j = \sum_k T(i,j,k)\, e_k .(p⋅q)k​=i=0∑7​j=0∑7​pi​qj​T(i,j,k),soei​⋅ej​=k∑​T(i,j,k)ek​.

The rules for TTT, tried in order, are:

  • e0⋅ej=eje_0 \cdot e_j = e_je0​⋅ej​=ej​ and ei⋅e0=eie_i \cdot e_0 = e_iei​⋅e0​=ei​, so e0e_0e0​ is a two-sided unit;
  • for i∈{1,…,7}i \in \{1,\dots,7\}i∈{1,…,7}, ei⋅ei=−e0e_i \cdot e_i = -e_0ei​⋅ei​=−e0​;
  • for distinct i,j∈{1,…,7}i, j \in \{1,\dots,7\}i,j∈{1,…,7}, the product has no e0e_0e0​-component. It is ±ek\pm e_k±ek​, where kkk is the unique third index such that {i−1,j−1,k−1}\{i-1, j-1, k-1\}{i−1,j−1,k−1} is a line of the Fano plane on Z/7\mathbb{Z}/7Z/7. The lines are {l,l+1,l+3}\{l, l+1, l+3\}{l,l+1,l+3} for l∈Z/7l \in \mathbb{Z}/7l∈Z/7, and index i∈{1,…,7}i \in \{1,\dots,7\}i∈{1,…,7} corresponds to the point i−1 mod 7i-1 \bmod 7i−1mod7. The sign is +++ when the ordered pair of points (i−1,j−1)(i-1, j-1)(i−1,j−1) is one of (l,l+1)(l, l+1)(l,l+1), (l+1,l+3)(l+1, l+3)(l+1,l+3), (l+3,l)(l+3, l)(l+3,l) for that line, and −-− otherwise. For distinct i,ji, ji,j a unique line always exists, so the final "else 000" case never gives the whole product.

In index terms, the positive cyclic triples eaeb=ece_a e_b = e_cea​eb​=ec​ (with also ebec=eae_b e_c = e_aeb​ec​=ea​ and ecea=ebe_c e_a = e_bec​ea​=eb​) are

(1,2,4), (2,3,5), (3,4,6), (4,5,7), (5,6,1), (6,7,2), (7,1,3),(1,2,4),\ (2,3,5),\ (3,4,6),\ (4,5,7),\ (5,6,1),\ (6,7,2),\ (7,1,3),(1,2,4), (2,3,5), (3,4,6), (4,5,7), (5,6,1), (6,7,2), (7,1,3),

and reversing the order flips the sign. For example, e1e2=e4e_1 e_2 = e_4e1​e2​=e4​, e1e3=e7e_1 e_3 = e_7e1​e3​=e7​, e1e4=−e2e_1 e_4 = -e_2e1​e4​=−e2​, e1e5=e6e_1 e_5 = e_6e1​e5​=e6​, e1e6=−e5e_1 e_6 = -e_5e1​e6​=−e5​, e1e7=−e3e_1 e_7 = -e_3e1​e7​=−e3​, e2e3=e5e_2 e_3 = e_5e2​e3​=e5​, e2e5=−e3e_2 e_5 = -e_3e2​e5​=−e3​, e2e6=e7e_2 e_6 = e_7e2​e6​=e7​, e2e7=−e6e_2 e_7 = -e_6e2​e7​=−e6​. This is a real octonion multiplication table with unit e0e_0e0​.

The matrix. Let RaR_aRa​ be the matrix whose entry in row kkk, column jjj is (ej⋅a)k(e_j \cdot a)_k(ej​⋅a)k​. So column jjj holds the coordinates of ej⋅ae_j \cdot aej​⋅a, and RaR_aRa​ is the matrix of right multiplication x↦x⋅ax \mapsto x\cdot ax↦x⋅a acting on column vectors in the standard basis. Likewise LbL_bLb​ has row-kkk, column-jjj entry (b⋅ej)k(b\cdot e_j)_k(b⋅ej​)k​ and is the matrix of left multiplication x↦b⋅xx \mapsto b\cdot xx↦b⋅x. Then

Fc,s  =  Re1+L c e1+s e2,F_{c,s} \;=\; R_{e_1} + L_{\,c\,e_1 + s\,e_2},Fc,s​=Re1​​+Lce1​+se2​​,

which is the matrix, in the basis e0,…,e7e_0,\dots,e_7e0​,…,e7​ acting on column vectors, of the linear map

x  ⟼  x⋅e1  +  (c e1+s e2)⋅x.x \;\longmapsto\; x\cdot e_1 \;+\; (c\,e_1 + s\,e_2)\cdot x .x⟼x⋅e1​+(ce1​+se2​)⋅x.

Written out, with rows and columns indexed 0,…,70,\dots,70,…,7:

Fc,s=(0−1−c−s000001+c000s000s0001−c00000000−s01−c0−sc−100000000s001−c000000c−10−s000c−100s0).F_{c,s} = \begin{pmatrix} 0 & -1-c & -s & 0 & 0 & 0 & 0 & 0\\ 1+c & 0 & 0 & 0 & s & 0 & 0 & 0\\ s & 0 & 0 & 0 & 1-c & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -s & 0 & 1-c\\ 0 & -s & c-1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & s & 0 & 0 & 1-c & 0\\ 0 & 0 & 0 & 0 & 0 & c-1 & 0 & -s\\ 0 & 0 & 0 & c-1 & 0 & 0 & s & 0 \end{pmatrix}.Fc,s​=​01+cs00000​−1−c000−s000​−s000c−1000​00000s0c−1​0s1−c00000​000−s00c−10​000001−c0s​0001−c00−s0​​.

Here the characteristic polynomial is Mathlib's det⁡(XI−F)\det(X I - F)det(XI−F), which is monic of degree 8, and 2−2c2-2c2−2c appears as a constant polynomial.

Human review
  • Endorsed by Shuze Chen · Sep 27, 2026

    Confirmed by the moderator at approval.

  • Endorsed by ShapeZero · Sep 27, 2026

    Confirmed by the mission captain (proposal self-audit).

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