A hyperbolic pair splits off a hyperbolic plane #
If (u, v) is a hyperbolic pair of (M, q) — q u = q v = 0 and q(u, v) = 1 — then the
submodule R u ⊕ R v they span is free of rank two, the restriction of q to it is the
hyperbolic plane (a, b) ↦ ab, and that restriction is regular. So the plane splits off
orthogonally:
q ≅ ⟨plane⟩ ⊥ q|_{(Ru ⊕ Rv)ᗮ}.
Nothing here needs M finitely generated or projective, q regular, or 2 invertible. The
regularity that the orthogonal splitting of QuadraticForm/OrthogonalComplement.lean consumes is
regularity of the restricted form, and that is supplied by the pair itself: the plane is
hyperbolic, and hyperbolic forms on finitely generated projective — here free of rank two —
modules are regular.
This is the missing half of GPR 11.17: QuadraticForm/Isotropic.lean turns an isotropic vector
into a hyperbolic pair (isGPRIsotropic_iff_exists_isHyperbolicPair), and this file turns the pair
into a splitting.
Main definitions #
QuadraticMap.IsHyperbolicPair.planeEmbedding: the map(a, b) ↦ a u + b v.QuadraticMap.IsHyperbolicPair.planeSubmodule: its range,R u ⊕ R v.
Main results #
QuadraticMap.IsHyperbolicPair.isGPRRegular_comp_planeSubmodule_subtype: the restriction ofqtoR u ⊕ R vis regular.QuadraticMap.IsHyperbolicPair.isometryEquivPlaneProd:q ≅ plane ⊥ q|_{(Ru ⊕ Rv)ᗮ}.QuadraticForm.isGPRRegular_plane: the hyperbolic plane is regular over any commutative ring.
Implementation notes #
The embedding is injective because the pair recovers coordinates: q(a u + b v, v) = a and
q(a u + b v, u) = b, using q(u, u) = 2 q u = 0 and q(v, v) = 0. This is why no hypothesis on
M is needed — the splitting is entirely local to the plane.
References #
- [GPR] Garibaldi, Petersson, Racine, Albert Algebras over Commutative Rings, 11.11, 11.17.
- R. Baeza, Quadratic Forms over Semilocal Rings, Lecture Notes in Math. 655, Springer 1978, I, (3.3).
The hyperbolic plane is regular over any commutative ring: (a, b) ↦ ab is the pairing of
R against R, so QuadraticForm.isGPRRegular_of_basis_pairing applies to the one-element basis
of R. No hypothesis on 2.
The hyperbolic plane is non-singular.
The map R × R → M, (a, b) ↦ a u + b v, whose image is the plane spanned by a hyperbolic
pair. Defined for any two vectors; it is the hypothesis IsHyperbolicPair that makes it injective
and an isometry from QuadraticForm.plane.
Equations
- QuadraticMap.IsHyperbolicPair.planeEmbedding u v = LinearMap.toSpanSingleton R M u ∘ₗ LinearMap.fst R R R + LinearMap.toSpanSingleton R M v ∘ₗ LinearMap.snd R R R
Instances For
The embedding carries the hyperbolic plane to Q.
Pairing a u + b v with v recovers a.
Pairing a u + b v with u recovers b.
The submodule R u ⊕ R v spanned by a hyperbolic pair, presented as the range of
QuadraticMap.IsHyperbolicPair.planeEmbedding.
Equations
Instances For
The plane of a hyperbolic pair, as a module, is R × R.
Equations
Instances For
The restriction of Q to the plane of a hyperbolic pair is the hyperbolic plane, read
through QuadraticMap.IsHyperbolicPair.planeEquiv.
The hyperbolic plane, as a quadratic module, is the plane of the pair inside (M, Q).
Equations
- h.planeIsometryEquiv = { toLinearEquiv := h.planeEquiv, map_app' := ⋯ }
Instances For
The restriction of Q to the plane of a hyperbolic pair is regular. This is what lets the
plane split off orthogonally, and it holds over any commutative ring with no hypothesis on 2.
A hyperbolic pair splits off a hyperbolic plane: Q ≅ plane ⊥ Q|_{(Ru ⊕ Rv)ᗮ}.
The orthogonal complement is taken for the polar form of Q
(QuadraticForm.polarOrthogonal).
Equations
- One or more equations did not get rendered due to their size.