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A Pin generator reflects one coordinate and has scalar kernel

Example

Let $V = R^{2}$ with orthonormal basis $e_{1}, e_{2}$ , positive quadratic form, and Clifford relations $e_{i}^{2} = 1$ , $e_{1} e_{2} = - e_{2} e_{1}$ . The Pin element $e_{1}$ acts by $(x, y) \mapsto (- x, y)$ under the twisted adjoint. The kernel of the twisted action restricted to Pin is exactly ${1, - 1}$ .

Facts & Assumptions

Given: The stated Clifford relations and the Pin twisted action.

[F1]

An anisotropic vector acts by its hyperplane reflection and the unrestricted twisted kernel is scalar (Twisted Clifford action, reflections, and scalar kernel).

[F2]

The Pin map is onto the orthogonal group with kernel ${1, - 1}$ (Real Pin and full-even Spin images, kernels, and definite compactness).

Verification

technique · direct

1.1givenF1algebra

From the polarized relation, $e_{1} e_{2} + e_{2} e_{1} = 0$ . Thus $ρ_{e_{1}} (e_{1}) = - e_{1} e_{1} e_{1} = - e_{1}$ and $ρ_{e_{1}} (e_{2}) = - e_{1} e_{2} e_{1} = e_{2}$ . Linearity gives $(x, y) \mapsto (- x, y)$ , the reflection in $e_{1}^{⊥}$ .

2.1F1F2algebra∎

By [F1], every Pin element acting trivially is a nonzero real scalar $λ$ . Its norm is $N (λ) = λ^{2}$ ; Pin requires this norm to be $\pm 1$ , so positivity gives $λ^{2} = 1$ and $λ = \pm 1$ . Both scalars act trivially, agreeing with [F2].

Depends on

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Sources

Jean Gallier, Clifford Algebras, Clifford Groups, and a Generalization of the Quaternions: The Pin and Spin Groups, §1.3 (standard reference, not scraped)
Eckhard Meinrenken, Lie groups and Clifford algebras, Chapter 2 §4.1 (standard reference, not scraped)