Example. A planar Spin curve records the half-angle [fcap-0038]

Let \(Q(e_1)=Q(e_2)=1\) with \(B_Q(e_1,e_2)=0\), and let \(J(e_1)=-e_2\), \(J(e_2)=e_1\). The normalization in Proposition [fcap-0037] gives \[b(J)=\frac 12e_1e_2, \qquad g(\theta )=\exp \left (\frac \theta 2e_1e_2\right ) =\cos \frac \theta 2+\sin \frac \theta 2e_1e_2.\] Since \((e_1e_2)^2=-1\), direct multiplication gives \[g(\theta )e_1g(\theta )^{-1} =\cos \theta \,e_1-\sin \theta \,e_2.\] Thus differentiating at zero gives \(J\) with the displayed sign, while \[g(2\pi )=-1,\qquad g(4\pi )=1.\] The Spin parameter therefore has twice the period of its vector action. This is the rank-two curve used in the compact-cover calculation below.

[meinrenken2013clifford, Example 3.3, p. 54] computes the same Clifford exponential and planar action. His convention has \(B_M(e_i,e_i)=1\), hence the same generator squares; the preceding polar-form translation fixes the coefficient \(1/2\).