Lemma. The volume term does not obstruct Spin in dimension five [fcap-003H]
Lemma. The volume term does not obstruct Spin in dimension five [fcap-003H]
Let \(K\) be a field of characteristic different from two, and let
\((V,Q)\) be a nondegenerate quadratic space of dimension five over \(K\).
For \(x\in \mathcal {C}\kern -2pt\ell ^{+}(Q)\),
\[x\text { is the Clifford value of a Spin element}
\quad \Longleftrightarrow \quad \operatorname {rev}(x)x=1.\]
The norm condition implies Lipschitz membership in this dimension
CliffordAlgebra.evenUnitaryGroup_le_lipschitzGroup_of_finrank_eq_five.
The point is to remove the possible degree-five term left by the
dimension-four argument of Lemma [fcap-002J].
Choose an orthogonal basis \(e_1,\ldots ,e_5\), suppress the generator map, and put \(\omega =e_1\cdots e_5\). Nondegeneracy gives \[\omega ^2=\delta 1,\qquad \delta =\prod _{i=1}^5 Q(e_i)\ne 0.\] The element \(\omega \) is central and invertible. Suppose \(\operatorname {rev}(x)x=1\). Finite dimensionality makes the left inverse a two-sided inverse. For any \(v\in V\), the element \(y=xvx^{-1}\) is odd and satisfies \(\operatorname {rev}(y)=y\). In the orthogonal monomial basis, reversal has signs \(+1,-1,+1\) on degrees one, three and five. Hence \[y=w+c\omega \qquad (w\in V,\ c\in K).\]
Conjugation preserves the scalar square of a vector. Since \(\omega \) is central, \[Q(v)1=y^2=Q(w)1+2c\omega w+c^2\delta 1.\] Each \(\omega e_i\) is a nonzero scalar multiple of the product of the other four basis vectors. Thus \(\omega w\) has pure degree four in this basis, whereas the other terms in the equation are scalar. Independence of the Clifford basis gives \(2c\omega w=0\). As two and \(\omega \) are invertible, either \(c=0\) or \(w=0\). In the latter case \(y=c\omega \) is central, so conjugating back gives \(v=x^{-1}yx=y=c\omega \). The degree-one and degree-five subspaces have zero intersection; therefore \(v=0\) and \(c=0\). In both cases \(y\in V\).
Thus \(x\) normalizes the vector space. The Clifford-group
characterization makes it Lipschitz; together with evenness and reverse
norm one, this is the Spin criterion of Remark [fcap-0028]. The reverse
implication is that same criterion. In fact the carrier equality holds in
every dimension from one through five
CliffordAlgebra.range_spinGroup_toUnits_eq_evenUnitaryGroup.
The lower bound matters: in dimension zero the vector-generated Lipschitz
group is trivial, whereas the even-unitary group contains both \(1\) and
\(-1\). This carrier equality alone supplies no symplectic matrix model;
the spinor pairing and matrix inverse for complex dimension five remain
the separate argument of Proposition [fcap-0033].
Porteous gives the real dimension-five proof by separating the vector, trivector and volume terms [porteous1995clifford, Chapter 16, Proposition 16.15, p. 147]. His relation is \(v^2=-q(v)\), so here \(Q=-q\); his conjugation is \(\operatorname {rev}\circ \alpha \), equal to reversal on even elements. His indefinite Spin carrier permits norm \(\pm 1\); we select norm \(+1\). The basis calculation above extends the norm-one argument from real spaces to the stated fields: orthogonality, nonzero basis squares and invertibility of two replace any need for real unit-length normalization. Gallier records the compact boundary at five and the failure at six [gallier2014clifford, Section 1.4, closing Remark, p. 31]; the next example computes that failure.