Proposition. Spin lifts over a quadratic extension [fcap-003O]

Keep the quadratic-space hypotheses of Theorem [fcap-003N]. Let \(d\in K^\times \) be nonsquare, put \(L=K(\sqrt d)\), and let \(\sigma \) be its nontrivial \(K\)-automorphism. The square-class map has kernel \[\ker j=\{1,[d]_K\}.\] Consequently, for \(g\in SO(V,Q)\), \[g_L\text { has a Spin lift} \quad \Longleftrightarrow \quad \nu _Q(g)\in \{1,[d]_K\}.\] The trivial class already gives a lift over \(K\). If \(\nu _Q(g)=[d]_K\), the two lifts over \(L\) are exchanged by \(\sigma \); neither is the scalar extension of a \(K\)-defined Spin element.

For an explicit rational example, take \(K=\mathbb Q\) and \[Q(ae+bf)=a^2+2b^2,\qquad g=-\operatorname {Id}=r_e r_f.\] Here \(e^2=1\), \(f^2=2\), and \(ef=-fe\). Thus \(x=ef\) satisfies \[\operatorname {rev}(x)x=feef=2,\qquad x^2=efef=-2,\] and \(\nu _Q(g)=[2]_{\mathbb Q}\) is nontrivial. Over \(L=\mathbb Q(\sqrt 2)\), however, \[s=\frac {ef}{\sqrt 2},\qquad N_{\mathrm {rev}}(s)=1,\qquad s^2=-1.\] Its vector action is still \(-\operatorname {Id}\). The two Spin lifts \(s,-s\) are interchanged by conjugating \(\sqrt 2\); the rational orthogonal transformation acquires lifts without either lift becoming rational.