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In the basis (e0,e1,e2,e4∣e3,e5,e6,e7)(e_0, e_1, e_2, e_4 \mid e_3, e_5, e_6, e_7)(e0​,e1​,e2​,e4​∣e3​,e5​,e6​,e7​) the flow matrix is block diagonal

Proved
OctonionD8.flow_block_diag

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

characteristic-polynomialfano-planelinear-algebraoctonions

For all real c,sc, sc,s, reorder the basis of R8\mathbb{R}^8R8 as (e0,e1,e2,e4∣e3,e5,e6,e7)(e_0, e_1, e_2, e_4 \mid e_3, e_5, e_6, e_7)(e0​,e1​,e2​,e4​∣e3​,e5​,e6​,e7​). In this basis the flow matrix M=Re1+Lce1+se2M = R_{e_1} + L_{c e_1 + s e_2}M=Re1​​+Lce1​+se2​​ is block diagonal:

M=(A00B),M = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix},M=(A0​0B​),

with AAA and BBB the explicit 4×44\times44×4 blocks. So span(e0,e1,e2,e4)\mathrm{span}(e_0, e_1, e_2, e_4)span(e0​,e1​,e2​,e4​) and span(e3,e5,e6,e7)\mathrm{span}(e_3, e_5, e_6, e_7)span(e3​,e5​,e6​,e7​) are invariant subspaces of the flow.

Preamble
import Mathlib
import Definitions.Def_OctonionD8_blocks
Formal statement
namespace OctonionD8

open Polynomial

theorem flow_block_diag (c s : ℝ) :
    Matrix.reindex blockEquiv blockEquiv (flowMat c s) =
      Matrix.fromBlocks (blockA c s) 0 0 (blockB c s) := 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 ; 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

Statement. For all real numbers c,s∈Rc, s \in \mathbb{R}c,s∈R (no other hypotheses: in particular no relation such as c2+s2=1c^2+s^2=1c2+s2=1 is assumed, and c,sc,sc,s may be 000), the 8×88\times 88×8 real matrix Fc,sF_{c,s}Fc,s​ defined below, after its rows and columns are simultaneously permuted by the index bijection β\betaβ below, equals the block-diagonal matrix

(Ac,s00Bc,s),\begin{pmatrix} A_{c,s} & 0 \\ 0 & B_{c,s} \end{pmatrix},(Ac,s​0​0Bc,s​​),

with the explicit 4×44\times 44×4 blocks Ac,sA_{c,s}Ac,s​, Bc,sB_{c,s}Bc,s​ given below and 4×44\times 44×4 zero off-diagonal blocks.

The algebra. Let e0,…,e7e_0,\dots,e_7e0​,…,e7​ be the standard basis of R8\mathbb{R}^8R8 (indices 0,…,70,\dots,70,…,7). A bilinear product on R8\mathbb{R}^8R8 is defined by p⋅q=∑i,j,kpi qj T(i,j,k) ekp\cdot q = \sum_{i,j,k} p_i\, q_j\, T(i,j,k)\, e_kp⋅q=∑i,j,k​pi​qj​T(i,j,k)ek​, where the integer structure constants T(i,j,k)T(i,j,k)T(i,j,k) (so eiej=∑kT(i,j,k)eke_i e_j = \sum_k T(i,j,k) e_kei​ej​=∑k​T(i,j,k)ek​) are:

  • e0ej=eje_0 e_j = e_je0​ej​=ej​ for all jjj; eie0=eie_i e_0 = e_iei​e0​=ei​ for all iii (the first rule wins when i=0i=0i=0, giving e0e0=e0e_0e_0=e_0e0​e0​=e0​);
  • eiei=−e0e_i e_i = -e_0ei​ei​=−e0​ for i∈{1,…,7}i \in \{1,\dots,7\}i∈{1,…,7};
  • for distinct i,j∈{1,…,7}i,j \in\{1,\dots,7\}i,j∈{1,…,7}: the e0e_0e0​-coefficient is 000; and for k∈{1,…,7}k\in\{1,\dots,7\}k∈{1,…,7}, relabel i↦π(i)=(i+6) mod 7=i−1∈Z/7i\mapsto \pi(i) = (i+6) \bmod 7 = i-1 \in \mathbb{Z}/7i↦π(i)=(i+6)mod7=i−1∈Z/7. The Fano lines are the sets {l, l+1, l+3}⊂Z/7\{l,\,l+1,\,l+3\}\subset\mathbb{Z}/7{l,l+1,l+3}⊂Z/7, l∈Z/7l\in\mathbb{Z}/7l∈Z/7. Then T(i,j,k)=+1T(i,j,k)=+1T(i,j,k)=+1 if {π(i),π(j),π(k)}\{\pi(i),\pi(j),\pi(k)\}{π(i),π(j),π(k)} equals some line {l,l+1,l+3}\{l,l+1,l+3\}{l,l+1,l+3} and (π(i),π(j))(\pi(i),\pi(j))(π(i),π(j)) 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); T(i,j,k)=−1T(i,j,k)=-1T(i,j,k)=−1 if {π(i),π(j),π(k)}\{\pi(i),\pi(j),\pi(k)\}{π(i),π(j),π(k)} is a line but the ordering is not of that form; T(i,j,k)=0T(i,j,k)=0T(i,j,k)=0 otherwise (in particular when k∈{i,j}k\in\{i,j\}k∈{i,j}, since then the set has only 2 elements).

Concretely, in terms of basis indices, eiej=eke_ie_j=e_kei​ej​=ek​ for (i,j,k)(i,j,k)(i,j,k) any cyclic rotation of

(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 ejei=−eke_je_i=-e_kej​ei​=−ek​ for these, all other products of distinct imaginary units having zero coefficient.

The matrix Fc,sF_{c,s}Fc,s​. For a,b∈R8a,b\in\mathbb{R}^8a,b∈R8, RaR_aRa​ is the matrix with entry (k,j)(k,j)(k,j) equal to the kkk-th coordinate of ej⋅ae_j\cdot aej​⋅a, i.e. the matrix (acting on column vectors, column jjj = image of eje_jej​) of right multiplication x↦x⋅ax\mapsto x\cdot ax↦x⋅a in the standard basis; LbL_bLb​ is likewise 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​​,

the matrix of the linear map x↦x e1+(c e1+s e2) xx \mapsto x\,e_1 + (c\,e_1+s\,e_2)\,xx↦xe1​+(ce1​+se2​)x, with (Fc,s)k,j(F_{c,s})_{k,j}(Fc,s​)k,j​ = the eke_kek​-coefficient of the image of eje_jej​.

The bijection and reindexing. β:{0,…,7}→{0,1,2,3}⊔{0,1,2,3}\beta:\{0,\dots,7\}\to\{0,1,2,3\}\sqcup\{0,1,2,3\}β:{0,…,7}→{0,1,2,3}⊔{0,1,2,3} sends 0,1,2,40,1,2,40,1,2,4 to the first copy's 0,1,2,30,1,2,30,1,2,3 and 3,5,6,73,5,6,73,5,6,7 to the second copy's 0,1,2,30,1,2,30,1,2,3 (so β−1\beta^{-1}β−1 of first-copy iii is pip_ipi​ with (p0,p1,p2,p3)=(0,1,2,4)(p_0,p_1,p_2,p_3)=(0,1,2,4)(p0​,p1​,p2​,p3​)=(0,1,2,4), and of second-copy iii is qiq_iqi​ with (q0,q1,q2,q3)=(3,5,6,7)(q_0,q_1,q_2,q_3)=(3,5,6,7)(q0​,q1​,q2​,q3​)=(3,5,6,7)). The reindexed matrix has entry Fc,s(β−1(x),β−1(y))F_{c,s}(\beta^{-1}(x),\beta^{-1}(y))Fc,s​(β−1(x),β−1(y)) at position (x,y)(x,y)(x,y), the same bijection being used for rows and columns. Hence the statement is equivalent to: for all i,j∈{0,1,2,3}i,j\in\{0,1,2,3\}i,j∈{0,1,2,3},

(Fc,s)pi,pj=(Ac,s)ij,(Fc,s)qi,qj=(Bc,s)ij,(Fc,s)pi,qj=(Fc,s)qi,pj=0,(F_{c,s})_{p_i,p_j} = (A_{c,s})_{ij},\qquad (F_{c,s})_{q_i,q_j} = (B_{c,s})_{ij},\qquad (F_{c,s})_{p_i,q_j} = (F_{c,s})_{q_i,p_j} = 0,(Fc,s​)pi​,pj​​=(Ac,s​)ij​,(Fc,s​)qi​,qj​​=(Bc,s​)ij​,(Fc,s​)pi​,qj​​=(Fc,s​)qi​,pj​​=0,

i.e. the coordinate subspaces span⁡(e0,e1,e2,e4)\operatorname{span}(e_0,e_1,e_2,e_4)span(e0​,e1​,e2​,e4​) and span⁡(e3,e5,e6,e7)\operatorname{span}(e_3,e_5,e_6,e_7)span(e3​,e5​,e6​,e7​) are each mapped into themselves, with the stated matrices in those ordered bases.

The blocks.

Ac,s=(0−c−1−s0c+100ss001−c0−sc−10),Bc,s=(0−s01−cs01−c00c−10−sc−10s0).A_{c,s}=\begin{pmatrix} 0 & -c-1 & -s & 0\\ c+1 & 0 & 0 & s\\ s & 0 & 0 & 1-c\\ 0 & -s & c-1 & 0\end{pmatrix},\qquad B_{c,s}=\begin{pmatrix} 0 & -s & 0 & 1-c\\ s & 0 & 1-c & 0\\ 0 & c-1 & 0 & -s\\ c-1 & 0 & s & 0\end{pmatrix}.Ac,s​=​0c+1s0​−c−100−s​−s00c−1​0s1−c0​​,Bc,s​=​0s0c−1​−s0c−10​01−c0s​1−c0−s0​​.

The equation is an equality of real 8×88\times 88×8 matrices indexed by {0,1,2,3}⊔{0,1,2,3}\{0,1,2,3\}\sqcup\{0,1,2,3\}{0,1,2,3}⊔{0,1,2,3}, required to hold identically in ccc and sss.

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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