Example. A null-frame Spin transvection and a radical shear [fcap-002N]

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Let \(K\) be a field in which \(2\) is invertible and take the three-dimensional quadratic space \[Q(au+bv+cz)=ab+c^2.\] Thus \(u,v\) are null, \(B_Q(u,v)=1\), \(Q(z)=1\), and \(z\) is orthogonal to their hyperbolic plane. For \(w=tz\), the preceding lift and action become \[\begin {gathered} L_t=1+t\widehat z\widehat u =\widehat z\,\widehat {(z+tu)},\\ E_t(u)=u,\qquad E_t(z)=z-2tu,\qquad E_t(v)=v+tz-t^2u. \end {gathered}\] Both factors in the displayed product have quadratic value \(1\). Hence the pair criterion puts \(L_t\) directly in the actual Spin group, including at \(t=0\) CliffordAlgebra.ι_mul_ι_mem_spinGroup_of_norm_mul_norm_eq_one. In coordinates \((a,c,b)\) the transformation is \[E_t= \begin {pmatrix} 1&-2t&-t^2\\ 0&1&t\\ 0&0&1 \end {pmatrix}.\] Substitution gives \(Q(E_t x)=Q(x)\), while multiplication gives \(E_sE_t=E_{s+t}\) and \(L_sL_t=L_{s+t}\). At \(t=1/2\), \[z\longmapsto z-u,\qquad v\longmapsto v+\frac 12z-\frac 14u,\] and composing this map with itself gives \(z\mapsto z-2u\) and \(v\mapsto v+z-u\). For \(t\ne 0\), the vector action has \((E_t-1)^2v=-2t^2u\) and \((E_t-1)^3=0\), although the Clifford increment satisfies \((L_t-1)^2=0\). Over \(\mathbb R\), this is the null frame inside the split model of Example [fcap-002I]; over every stated field it is a concrete element of the spinor-norm kernel in Theorem [fcap-0023].

There is also a precise radical comparison. On the degenerate real form \[Q_{\mathrm {PGA}}(ar+xe+yf)=x^2+y^2\] the line \(\mathbb Rr\) is radical. With \(u=r\) and \(w=te\), the same ring-level transvection formula gives \[\begin {gathered} L_t=1+t\widehat e\widehat r =\widehat e\,\widehat {(e+tr)},\\ E_t(r)=r,\qquad E_t(e)=e-2tr,\qquad E_t(f)=f. \end {gathered}\] The nondegenerate packaged Spin-transvection theorem does not apply here. Nevertheless this particular lift still belongs to the vector-generated Spin carrier, because both displayed factors have quadratic value \(1\); the pair criterion itself needs no nondegeneracy. The quadratic action term vanishes because \(B_Q(r,-)=0\), and now \((E_t-1)^2=0\). This retains the original degenerate Clifford algebra rather than replacing it by the radical quotient of Remark [fcap-002C]. It does not identify the shear with a particular PGA point translation, motor parameter, dual encoding, or orientation convention.