Example. A rational Spin obstruction vanishes over the 7-adic field [fcap-004B]
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On the rational hyperbolic plane with basis $u,v$, put $q(au+bv)=ab$. Its unhalved polar form has matrix $\left (\begin {smallmatrix}0&1\\1&0\end {smallmatrix}\right )$, so it is nondegenerate over both $\mathbb {Q}$ and $\mathbb {Q}_7$. We construct a rational special orthogonal transformation with no rational Spin lift, then exhibit its lift over $\mathbb {Q}_7$. This specializes the scalar-extension criterion of Theorem [fcap-003N].

For $c\ne 0$, write $r_c=r_{u+cv}$. The reflection formula $r_z(w)=w-B_q(w,z)z/q(z)$ gives, in the ordered basis $u,v$, \[\begin {gathered} r_c=\begin {pmatrix}0&-c^{-1}\\-c&0\end {pmatrix},\\ g=r_2r_1=\begin {pmatrix}1/2&0\\0&2\end {pmatrix}. \end {gathered}\] Thus $g$ preserves $q$ and has determinant one. Since $q(u+2v)=2$ and $q(u+v)=1$, its spinor norm is \[\nu _q(g)=[2]_{\mathbb {Q}}\] by Proposition [fcap-002D]. This class is nontrivial: if $(a/b)^2=2$ for coprime integers $a,b$ with $b\ne 0$, then $a^2=2b^2$ first makes $a$ even and then makes $b$ even, a contradiction. The image/kernel theorem Theorem [fcap-0023] therefore rules out a rational Spin lift of $g$.

In $\mathbb {Z}_7$, take $f(T)=T^2-2$. We have $f(3)=7$ and $f'(3)=6$, a unit. Hensel's lemma [lam2005quadratic, VI.2, Theorem 2.18, p. 161] gives \[\begin {gathered} \alpha \in \mathbb {Z}_7,\qquad \alpha ^2=2,\\ \alpha \equiv 3\pmod {7}. \end {gathered}\] Indeed, in the theorem use the ideal $7\mathbb {Z}_7$ and $e=f'(3)=6$: since $e$ is a unit, $f(3)\in e^2(7\mathbb {Z}_7)$. In Mathlib's norm formulation hensels_lemma, the same input satisfies $|f(3)|_7=1/7<1=|f'(3)|_7^2$. In particular $\alpha \ne 0$ and $[2]_{\mathbb {Q}_7}=1$.

Identify $\mathbb {Q}_7\otimes _{\mathbb {Q}}V$ with the hyperbolic plane over $\mathbb {Q}_7$, and denote the extended form and transformation by $q_7,g_7$. Put $\mathcal {S}_K=K^{\times }/(K^{\times })^2$. The square in Theorem [fcap-003N] becomes

$\text{[math]}$ Here $j([a]_{\mathbb {Q}})=[a]_{\mathbb {Q}_7}$. Its value on $\nu _q(g)$ is $j([2]_{\mathbb {Q}})=1$, so $g_7$ has a Spin lift. The following calculation supplies that lift explicitly.

Suppress the Clifford inclusion symbols and use $u^2=v^2=0$, $uv+vu=1$ in $\mathcal {C}\kern -2pt\ell (q_7)$. Set $p=uv$ and $x=(u+2v)(u+v)$. Then \[\begin {gathered} p^2=u(vu)v,\\ u(vu)v=u(1-uv)v=p,\\ x=uv+2vu=2-p,\\ \operatorname {rev}(x)=(u+v)(u+2v),\\ \operatorname {rev}(x)=1+p,\\ \operatorname {rev}(x)x=(1+p)(2-p),\\ (1+p)(2-p)=2. \end {gathered}\] Define \[s=\frac {x}{\alpha } =\left (\frac {u+2v}{\alpha }\right )(u+v).\] Both displayed vector factors have quadratic value one. They are Clifford units in the vector-generated Lipschitz carrier; their even product has reverse norm $2/\alpha ^2=1$. On an even element the star and reverse norms agree by Remark [fcap-0028]. Thus $s$ lies in the actual $\operatorname {Spin}(q_7)$, by the product-square construction Theorem [fcap-000C]. Rescaling a nonisotropic vector leaves its reflection unchanged, so the Spin action of $s$ is \[r_{(u+2v)/\alpha }\,r_{u+v}=r_2r_1=g_7.\] This is the normalized-pair construction CliffordAlgebra.spinReflectionPair and its action CliffordAlgebra.spinToOrthogonal_spinReflectionPair: the constructor assumes both quadratic values are one, and the action identity also assumes that two is invertible. These declarations certify the general pair construction; the arithmetic specialization above is proved here.

The extension $\mathbb {Q}\subset \mathbb {Q}_7$ has killed this particular square-class obstruction. The quadratic-extension example Proposition [fcap-003O] exhibits the same class becoming trivial over $\mathbb {Q}(\sqrt {2})$ for a different quadratic form. Here the Hensel root and the displayed Clifford product give a lift at the single completion $\mathbb {Q}_7$.