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LeanPool.CompositionAlgebrasOverRings.QuadraticForm.PlaneSplitting

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 #

Main results #

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 #

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.

def QuadraticMap.IsHyperbolicPair.planeEmbedding {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] (u v : M) :
R × R →ₗ[R] M

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.

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    @[simp]
    theorem QuadraticMap.IsHyperbolicPair.planeEmbedding_apply {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {u v : M} (x : R × R) :
    (planeEmbedding u v) x = x.1 • u + x.2 • v
    theorem QuadraticMap.IsHyperbolicPair.apply_planeEmbedding {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {Q : QuadraticForm R M} {u v : M} (h : IsHyperbolicPair Q u v) (x : R × R) :

    The embedding carries the hyperbolic plane to Q.

    theorem QuadraticMap.IsHyperbolicPair.polar_planeEmbedding_right {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {Q : QuadraticForm R M} {u v : M} (h : IsHyperbolicPair Q u v) (x : R × R) :
    polar (⇑Q) ((planeEmbedding u v) x) v = x.1

    Pairing a u + b v with v recovers a.

    theorem QuadraticMap.IsHyperbolicPair.polar_planeEmbedding_left {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {Q : QuadraticForm R M} {u v : M} (h : IsHyperbolicPair Q u v) (x : R × R) :
    polar (⇑Q) ((planeEmbedding u v) x) u = x.2

    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.

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      noncomputable def QuadraticMap.IsHyperbolicPair.planeEquiv {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {Q : QuadraticForm R M} {u v : M} (h : IsHyperbolicPair Q u v) :
      (R × R) ≃ₗ[R] ↥(planeSubmodule u v)

      The plane of a hyperbolic pair, as a module, is R × R.

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        @[simp]
        theorem QuadraticMap.IsHyperbolicPair.planeEquiv_coe {R : Type u} [CommRing R] {M : Type v} [AddCommGroup M] [Module R M] {Q : QuadraticForm R M} {u v : M} (h : IsHyperbolicPair Q u v) (x : R × R) :
        ↑(h.planeEquiv x) = x.1 • u + x.2 • v

        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).

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          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).

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          • One or more equations did not get rendered due to their size.
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