The spinor kernel is open

Let Q be a nondegenerate quadratic form on a finite-dimensional space over a locally compact nontrivially normed field K in which 2 is invertible, such as ℝ or ℚ_p, and suppose that the squares are open in Kˣ. Then the kernel of the spinor norm θ : O(Q) → Kˣ ⧸ (Kˣ)² is open in the orthogonal group, and so are the kernel of its restriction to SO(Q) and the image of the Spin group in SO(Q), which is that kernel.

The spinor norm is the Clifford norm N x = reverse x * x of a Lipschitz lift read modulo squares. So the kernel of θ is the image, under the vector representation, of the Lipschitz elements whose Clifford norm is a square. That set is open because the Clifford norm is continuous and the squares are open, and its image is open because the vector representation is an open map.

Equivalently, both spinor-norm homomorphisms are continuous when their square-class codomain has the quotient topology. The openness of the squares cannot be dropped, and it does not follow from the topology on V: over ℝ the squares are the positive reals, while over ℚ_p it is the local-field fact that the units deep enough in the unit filtration are squares. Discreteness of the square-class group alone would only show that a continuous map into it is locally constant.

Nor does the module topology supply openness on its own. Over a field with the indiscrete topology, every module topology is indiscrete and so is O(Q), whose only open subgroup is therefore O(Q) itself. The spinor kernel is then not open as soon as Q takes a nonsquare value, for instance for Q x = a * x ^ 2 with a a nonsquare, where O(Q) = {1, -1} and θ(-1) = [a]. In that example the squares are not open in Kˣ either.

An open subgroup contains the identity component. So when the Spin group is connected, or more generally generated by its identity component and the kernel of the Spin action, its image in SO(Q) and the spinor kernel of SO(Q) are exactly the identity component of SO(Q).

Main results

• CliffordAlgebra.isOpen_ker_orthogonalSpinorNorm: the kernel of the spinor norm on O(Q) is open.
• CliffordAlgebra.isOpen_ker_spinorNorm: the kernel of the spinor norm on SO(Q) is open.
• CliffordAlgebra.isOpen_range_spinToSpecialOrthogonal: the image of the Spin group in SO(Q) is open.
• CliffordAlgebra.continuous_orthogonalSpinorNorm, CliffordAlgebra.continuous_spinorNorm: the spinor norms on O(Q) and SO(Q) are continuous.
• CliffordAlgebra.range_spinToSpecialOrthogonal_eq_connectedComponentOfOne, CliffordAlgebra.ker_spinorNorm_eq_connectedComponentOfOne: if the Spin group is generated by its identity component and the kernel of the Spin action, then its image in SO(Q) and the spinor kernel of SO(Q) are the identity component of SO(Q).
• CliffordAlgebra.not_isOpen_ker_orthogonalSpinorNorm_of_indiscreteTopology: over a field with the indiscrete topology, the spinor kernel of O(Q) is not open once Q takes a nonsquare value.

References

• O. T. O'Meara, Introduction to Quadratic Forms, §55.

theorem CliffordAlgebra.isOpen_ker_orthogonalSpinorNorm {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) : IsOpen ↑(orthogonalSpinorNorm Q hQ).ker

The spinor kernel of O(Q) is open. For a nondegenerate quadratic form on a finite-dimensional space over a locally compact nontrivially normed field in which 2 is invertible and the squares are open in the units, the kernel of the spinor norm is an open subgroup of the orthogonal group.

theorem CliffordAlgebra.isOpen_ker_spinorNorm {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) : IsOpen ↑(spinorNorm Q hQ).ker

The spinor kernel of SO(Q) is open. Under the hypotheses of isOpen_ker_orthogonalSpinorNorm, the kernel of the spinor norm on the special orthogonal group is open.

theorem CliffordAlgebra.isOpen_range_spinToSpecialOrthogonal {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) : IsOpen ↑(spinToSpecialOrthogonal Q).range

The image of Spin is open in SO(Q). Under the hypotheses of isOpen_ker_orthogonalSpinorNorm, the image of the Spin group in the special orthogonal group is open, being the kernel of the spinor norm.

theorem CliffordAlgebra.continuous_orthogonalSpinorNorm {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) : Continuous ⇑(orthogonalSpinorNorm Q hQ)

The spinor norm on O(Q) is continuous. If Q is nondegenerate and the squares are open in Kˣ, then the spinor norm O(Q) → Kˣ ⧸ (Kˣ)² is continuous, where the square-class group carries its quotient topology from Kˣ.

theorem CliffordAlgebra.continuous_spinorNorm {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) : Continuous ⇑(spinorNorm Q hQ)

The spinor norm on SO(Q) is continuous. If Q is nondegenerate and the squares are open in Kˣ, then the spinor norm SO(Q) → Kˣ ⧸ (Kˣ)² is continuous, where the square-class group carries its quotient topology from Kˣ.

theorem CliffordAlgebra.range_spinToSpecialOrthogonal_eq_connectedComponentOfOne {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) (hgen : Subgroup.connectedComponentOfOne ↥(spinGroup Q) ⊔ (spinToSpecialOrthogonal Q).ker = ⊤) : (spinToSpecialOrthogonal Q).range = Subgroup.connectedComponentOfOne ↥(TauCeti.QuadraticMap.specialOrthogonalGroup Q)

The image of Spin is the identity component of SO(Q) under the hypotheses of isOpen_ker_orthogonalSpinorNorm, as soon as the Spin group is generated by its identity component together with the kernel of the Spin action, for instance when it is connected.

theorem CliffordAlgebra.ker_spinorNorm_eq_connectedComponentOfOne {K : Type u_1} {V : Type u_2} [NontriviallyNormedField K] [LocallyCompactSpace K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) (hsq : IsOpen ↑(Subgroup.square Kˣ)) (hgen : Subgroup.connectedComponentOfOne ↥(spinGroup Q) ⊔ (spinToSpecialOrthogonal Q).ker = ⊤) : (spinorNorm Q hQ).ker = Subgroup.connectedComponentOfOne ↥(TauCeti.QuadraticMap.specialOrthogonalGroup Q)

The spinor kernel of SO(Q) is its identity component under the hypotheses of range_spinToSpecialOrthogonal_eq_connectedComponentOfOne.

theorem CliffordAlgebra.not_isOpen_ker_orthogonalSpinorNorm_of_indiscreteTopology {K : Type u_3} {V : Type u_4} [Field K] [TopologicalSpace K] [IndiscreteTopology K] [Invertible 2] [AddCommGroup V] [Module K V] [FiniteDimensional K V] (Q : QuadraticForm K V) (hQ : QuadraticMap.Nondegenerate) {v : V} (hv : ¬IsSquare (Q v)) : ¬IsOpen ↑(orthogonalSpinorNorm Q hQ).ker

Over a field with the indiscrete topology, the kernel of the spinor norm on O(Q) is not open as soon as Q takes a nonsquare value Q v: the orthogonal group is indiscrete, and the kernel is a nonempty subgroup missing the reflection in v, whose spinor norm is [Q v].