The norm of a 3-vector: $|V|^2=-2\det V=\operatorname{tr}(V^2)$ is positive definite

Proved

SiegelFields.threeVector_norm_sq

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

lie-groupsmathematical-physicsspin

Throughout, $M_2(\mathbb C)$ is the space of complex $2\times2$ matrices, $V^\dagger$ the conjugate transpose, and a 3-vector is a matrix in $\mathcal V=\{V\in M_2(\mathbb C): V^\dagger=V,\ \operatorname{tr}V=0\}$ . For every $V\in\mathcal V$ :

$-2\det V=\operatorname{tr}(V^2)$ ;
$\operatorname{tr}(V^2)$ is a non-negative real number;
$\operatorname{tr}(V^2)=0$ if and only if $V=0$ .

|V|^2:=-2\det V=\operatorname{tr}(V^2)\ \ge 0,\qquad |V|^2=0\iff V=0.

This makes $|V|^2$ a positive definite quadratic form on $\mathcal V$ , the Euclidean norm of the 3-vector.

Preamble

import Mathlib
import Definitions.Def_SiegelFields_SpinDefs

open Matrix Complex

Formal statement

namespace SiegelFields
theorem threeVector_norm_sq (V : Matrix (Fin 2) (Fin 2) ℂ) (hV : IsThreeVector V) :
    -2 * V.det = trace (V * V) ∧ (trace (V * V)).im = 0 ∧ 0 ≤ (trace (V * V)).re ∧
      (trace (V * V) = 0 ↔ V = 0) := 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. 111

Read-back

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Let $V$ be a complex $2\times2$ matrix that equals its conjugate transpose and has trace $0$ . Then all of the following hold:

$-2\det V=\operatorname{tr}(VV)$ as complex numbers;
the imaginary part of $\operatorname{tr}(VV)$ is $0$ ;
the real part of $\operatorname{tr}(VV)$ is $\ge 0$ ;
$\operatorname{tr}(VV)=0$ (as a complex number) if and only if $V$ is the zero matrix.

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

The norm of a 3-vector: ∣V∣2=−2det⁡V=tr⁡(V2)|V|^2=-2\det V=\operatorname{tr}(V^2)∣V∣2=−2detV=tr(V2) is positive definite

The norm of a 3-vector: $|V|^2=-2\det V=\operatorname{tr}(V^2)$ is positive definite