Example. A norm-one even unit can leave the vector space in dimension six [fcap-003I]

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Let \(K\) be a field of characteristic different from two, and let \((V,Q)\) be a finite-dimensional nondegenerate quadratic space containing six orthogonal vectors \(e_1,\ldots ,e_6\) with \(Q(e_i)\ne 0\). Put \[\omega =e_1\cdots e_6,\qquad \delta =\prod _{i=1}^6 Q(e_i).\] Now \(\omega \) is even, \(\operatorname {rev}(\omega )=-\omega \), and \(\omega ^2=-\delta 1\). It anticommutes with each of the six displayed vectors. If \(a,b\in K\) satisfy \(a^2+\delta b^2=1\), then \[x=a+b\omega ,\qquad x^{-1}=\operatorname {rev}(x)=a-b\omega \] is an even unit of reverse norm one. Its action on a displayed basis vector is \[xe_i x^{-1}=(a^2-\delta b^2)e_i+2ab\omega e_i.\]

For \(ab\ne 0\), the last term is a nonzero degree-five monomial, up to a scalar. Extend the six vectors to an orthogonal basis; independence of the Clifford monomials shows that the displayed conjugate is not a vector. Since \(x\) is even, ordinary and twisted conjugation agree. Thus \(x\) is not Lipschitz and cannot belong to Spin, despite satisfying the norm equation CliffordAlgebra.notMem_lipschitzGroup_of_mem_evenUnitaryGroup_of_coe_eq. Only the selected six vectors are used: in higher dimension \(\omega \) commutes with an additional orthogonal vector, which does not affect this counterexample.

For a concrete compact example, take \(K=\mathbb R\) and \(Q(v)=v_1^2+\cdots +v_6^2\). With \(a=b=1/\sqrt 2\), one obtains \[x=\frac {1+\omega }{\sqrt 2},\qquad \operatorname {rev}(x)x=1,\qquad xe_1x^{-1}=\omega e_1=-e_2e_3e_4e_5e_6.\] Porteous [porteous1995clifford, Chapter 16, p. 146] exhibits the unnormalized element \(1+\omega \) of norm two.

Gallier [gallier2014clifford, Section 1.4, closing Remark, p. 31] uses the normalized element above.

Their compact generators square to \(-1\). With that choice the same six-vector product still squares to \(-1\), but \(\omega e_1=+e_2e_3e_4e_5e_6\). Both signs give the same obstruction. The calculation here uses the positive-square convention of Example [fcap-002H].

Over any infinite \(K\) as above, the conic always supplies \(a,b\ne 0\). Indeed, choose \(t\ne 0\) with \(1+\delta t^2\ne 0\) and \(1-\delta t^2\ne 0\), and set \[a=\frac {1-\delta t^2}{1+\delta t^2},\qquad b=\frac {2t}{1+\delta t^2}.\] Only finitely many values of \(t\) are excluded, and substitution gives \(a^2+\delta b^2=1\). Consequently the Spin image is a proper subgroup of the even-unitary group for every nondegenerate form of dimension at least six over an infinite field of characteristic different from two CliffordAlgebra.range_spinGroup_toUnits_ne_evenUnitaryGroup_of_six_le_finrank. This field extension follows from the displayed calculation; the cited book passages concern real spaces. Over a finite field the conditional six-vector construction remains valid, but this infinitude argument makes no assertion that a suitable conic point exists.