$M^2=M\operatorname{tr}M-I\det M$ , hence $V^2=-I\det V=I\tfrac12|V|^2$

Proved

SiegelFields.sq_eq_trace_smul_sub_det

by Lucas · Sep 25, 2026 · Mathlib 0df444a (Lean v4.33.1)

lie-groupsmathematical-physicsspin

For every complex $2\times2$ matrix $M$ , $\;M^2=(\operatorname{tr}M)\,M-(\det M)\,I$ .
For every 3-vector $V$ (hermitian, traceless),

V^2=-(\det V)\,I=\tfrac12\operatorname{tr}(V^2)\,I=\tfrac12|V|^2 I .

This is the Cayley–Hamilton identity in dimension two, specialized to 3-vectors.

Preamble

import Mathlib
import Definitions.Def_SiegelFields_SpinDefs

open Matrix Complex

Formal statement

namespace SiegelFields
theorem sq_eq_trace_smul_sub_det :
    (∀ M : Matrix (Fin 2) (Fin 2) ℂ,
      M * M = trace M • M - M.det • (1 : Matrix (Fin 2) (Fin 2) ℂ)) ∧
    (∀ V : Matrix (Fin 2) (Fin 2) ℂ, IsThreeVector V →
      V * V = (-V.det) • (1 : Matrix (Fin 2) (Fin 2) ℂ) ∧
      V * V = ((1 / 2 : ℂ) * trace (V * V)) • (1 : Matrix (Fin 2) (Fin 2) ℂ)) := by sorry
end SiegelFields

Source

W. Siegel, *Fields*, arXiv:hep-th/9912205v3 (2005), https://arxiv.org/abs/hep-th/9912205, Chapter II (Spin), Section A (Two components), §IIA1 p. 112

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic), same agent as the drafter (non-blind)

Non-blind read-back. This read-back was written by the same agent that drafted the Lean statement, with full knowledge of the source text and of the intended meaning. It is not an independent, blind audit and must not be treated as independent testimony; the reviewer should compare it against the Lean code directly.

The statement is a conjunction of two universally quantified claims. (1) For every complex $2\times2$ matrix $M$ , $MM=(\operatorname{tr}M)M-(\det M)I$ . (2) For every complex $2\times2$ matrix $V$ that equals its conjugate transpose and has trace $0$ : $VV=(-\det V)\,I$ , and also $VV=\big(\tfrac12\operatorname{tr}(VV)\big)I$ , where $\tfrac12$ and the trace are complex numbers.

Human review

Endorsed by Shuze Chen · Sep 26, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Sep 26, 2026
Confirmed by the mission captain (proposal self-audit).

View graph

M2=Mtr⁡M−Idet⁡MM^2=M\operatorname{tr}M-I\det MM2=MtrM−IdetM, hence V2=−Idet⁡V=I12∣V∣2V^2=-I\det V=I\tfrac12|V|^2V2=−IdetV=I21​∣V∣2

$M^2=M\operatorname{tr}M-I\det M$ , hence $V^2=-I\det V=I\tfrac12|V|^2$