Solution: the single-element spectral theorem for a Euclidean Jordan algebra #
Repeats the statement of SpectralChallenge.lean verbatim and discharges it from
EuclideanJordan.spectral_resolution_bilinear in EuclideanJordan/Spectral.lean.
The delegation is a single application because the library theorem is stated in exactly this
vocabulary: the product enters as a bundled bilinear map, so no Jordan-algebra instance has to
exist before the statement elaborates. The instances the proof needs (a non-unital commutative
ring on J, IsCommJordan, IsScalarTower ℝ J J, IsFormallyReal J) are built inside the
library from m, hcomm, hjordan and hfr, and none of them escapes into the statement.
The single-element spectral theorem for a Euclidean Jordan algebra.
Let J be a finite-dimensional real vector space and m : J →ₗ[ℝ] J →ₗ[ℝ] J a bilinear
product on it which is commutative (hcomm), satisfies the Jordan identity (hjordan), is
formally real (hfr : a vanishing sum of squares has vanishing summands), and has a unit e
(he). Then every x : J admits a spectral resolution: a finite family of idempotents
q i for m, pairwise orthogonal, summing to the unit, with x a real combination of them.
This is Theorem III.1.1 of J. Faraut and A. Koranyi, Analysis on Symmetric Cones, Oxford University Press (1994); the underlying classification is P. Jordan, J. von Neumann and E. Wigner, On an algebraic generalization of the quantum mechanical formalism, Ann. of Math. 35 (1934) 29-64.
Everything the statement mentions is Mathlib: bilinear maps, Finset.sum over Fin n, real
scalar multiplication. In particular "idempotent", "orthogonal" and "complete" are written out
inline as m (q i) (q i) = q i, i ≠ j → m (q i) (q j) = 0, and ∑ i, q i = e.
What is and is not assumed. No associativity, no power-associativity, no positivity of the
inner product against m (indeed no hypothesis at all connects m to ⟪·, ·⟫), no ordering, no
trace, no Peirce decomposition, no simplicity. Finite-dimensionality is essential: ℝ[X] under
polynomial multiplication satisfies every other hypothesis and has only the idempotents 0 and
1.