Example. A mixed-sign isometry has no norm-one Spin lift [fcap-002E]

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On \(\mathbb R^2\), let \(Q(a,b)=a^2-b^2\), \(e_+=(1,0)\), and \(e_-=(0,1)\). The corresponding reflections are \[r_{e_+}=\operatorname {diag}(-1,1),\qquad r_{e_-}=\operatorname {diag}(1,-1),\] so \(g=r_{e_+}r_{e_-}=-I\) lies in \(SO(Q)\). Its spinor norm is \[\nu _Q(g)=[Q(e_+)Q(e_-)]=[-1]\ne [1].\] With the two norm conventions of Remark [fcap-0028], the even Clifford lift \(x=\widehat e_+\widehat e_-\) has \[N_{\mathrm {rev}}(x)=N_*(x)=-1.\] For every \(c\in \mathbb R^\times \), the scalar multiple \(cx\) has norm \(-c^2\), which cannot equal \(1\). More generally, any two Lipschitz lifts of \(g\) differ by a scalar by Proposition [fcap-002D], so no lift can be normalized into the norm-one Spin carrier. Equivalently, the image theorem in Theorem [fcap-0023] excludes a Spin lift because \(\nu _Q(g)\) is nontrivial.

This is the mixed-sign counterpart to the successful square normalizations in Lemma [fcap-000B] and Theorem [fcap-000C]. Meinrenken notes that over \(\mathbb R\) one sometimes uses the weaker carrier condition \(N(x)=\pm 1\); under that different convention this \(x\) passes the norm test [meinrenken2013clifford, Remark 3.2, p. 52]. TauCeti's Spin group uses the norm-one carrier, so the two conventions must not be identified.