Documentation

Mathlib.LinearAlgebra.CliffordAlgebra.Inversion

Results about inverses in Clifford algebras #

This contains some basic results about the inversion of vectors, related to the fact that $ι(m)^{-1} = \frac{ι(m)}{Q(m)}$.

If the quadratic form of a vector is invertible, then so is that vector.

Equations
Instances For
    theorem CliffordAlgebra.invOf_ι {R : Type u_1} {M : Type u_2} [CommRing R] [AddCommGroup M] [Module R M] (Q : QuadraticForm R M) (m : M) [Invertible (Q m)] [Invertible ((CliffordAlgebra.ι Q) m)] :

    For a vector with invertible quadratic form, $v^{-1} = \frac{v}{Q(v)}$

    theorem CliffordAlgebra.isUnit_ι_of_isUnit {R : Type u_1} {M : Type u_2} [CommRing R] [AddCommGroup M] [Module R M] (Q : QuadraticForm R M) {m : M} (h : IsUnit (Q m)) :

    $aba^{-1}$ is a vector.

    $a^{-1}ba$ is a vector.

    Over a ring where 2 is invertible, Q m is invertible whenever ι Q m.

    Equations
    • One or more equations did not get rendered due to their size.
    Instances For
      theorem CliffordAlgebra.isUnit_of_isUnit_ι {R : Type u_1} {M : Type u_2} [CommRing R] [AddCommGroup M] [Module R M] (Q : QuadraticForm R M) [Invertible 2] {m : M} (h : IsUnit ((CliffordAlgebra.ι Q) m)) :
      IsUnit (Q m)
      @[simp]
      theorem CliffordAlgebra.isUnit_ι_iff {R : Type u_1} {M : Type u_2} [CommRing R] [AddCommGroup M] [Module R M] (Q : QuadraticForm R M) [Invertible 2] {m : M} :