Remark. The graph half-turn fixes the quadratic center [fdip-0045]

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The alternating-cycle half-turn \(P=(02)(13)\) acts on the rational configuration space, preserving the counting pairing, the action and the opposite-vertex splitting. Its determinant is \(+1\). For every rational \(t>0\), its induced Clifford automorphism fixes the even center pointwise. This action is different from the nontrivial involution of that quadratic algebra.

More generally, let \(q\) be a nondegenerate rational quadratic form of positive even rank \(n\), and let \(T\) be a \(q\)-isometry. Choose a rational orthogonal basis \(e_1,\ldots ,e_n\) and put \(w=\iota _q(e_1)\cdots \iota _q(e_n)\). Then \[\begin {aligned} Z(\operatorname {Cl}^+(q))&=\mathbb Q\cdot 1\oplus \mathbb Q\cdot w,\\ \operatorname {Cl}(T)(w)&=(\det T)w. \end {aligned}\] Thus determinant \(+1\) fixes the center pointwise, while determinant \(-1\) induces its nontrivial involution.

For the center description, use the orthogonal Clifford monomial basis. A basis monomial in the even algebra commutes with \(\iota _q(e_i)\iota _q(e_j)\), for \(i\ne j\), precisely when its index set contains both indices or neither. Every generator square is nonzero, so these pairs are invertible and their conjugations act diagonally on that basis. A central linear combination can therefore involve only the empty and full index sets, since their membership indicators must agree for every pair. Conversely both \(1\) and \(w\) commute with every even word. They are independent because \(n>0\). This gives the stated center and is the general even-rank center description in TauCeti's CliffordAlgebra.mem_center_even_iff_exists_eq_add_smul_volume, under its nonempty even nonisotropic orthogonal spanning-list hypotheses over a field of characteristic different from two.

For the action, consider the alternating multilinear map \[A(v_1,\ldots ,v_n)=\frac 1{n!} \sum _{\sigma \in S_n}\operatorname {sgn}(\sigma ) \iota _q(v_{\sigma (1)})\cdots \iota _q(v_{\sigma (n)}).\] On an orthogonal list anticommutation makes this its ordered Clifford product. Both \((e_i)\) and \((Te_i)\) are orthogonal, so alternation gives \[\begin {aligned} \operatorname {Cl}(T)(w)&=A(Te_1,\ldots ,Te_n)\\ &=(\det T)A(e_1,\ldots ,e_n)=(\det T)w. \end {aligned}\] Nondegeneracy gives \((\det T)^2=1\). The center is quadratic étale since \(w^2\ne 0\); sending \(w\) to \(-w\) is its unique nontrivial involution. This determinant-action argument is separate from the cited formal center description.

To check the assertion in the actual family, let \(h_1,\ldots ,h_4\) be the columns of the matrix \(H\) in Example [fdip-000U]. They are a rational orthogonal basis for \(q_t\); their nonzero squares are \(4,12,4(2t+1),4(2t+3)\). Direct application of the half-turn gives \[\begin {aligned} Ph_1&=h_1,& Ph_2&=-h_2,\\ Ph_3&=-h_3,& Ph_4&=h_4. \end {aligned}\] The four-dimensional center calculation identifies the center with the span of \(1\) and \[w=\iota _{q_t}(h_1)\iota _{q_t}(h_2) \iota _{q_t}(h_3)\iota _{q_t}(h_4).\] Its image under \(P\) is \((+1)(-1)(-1)(+1)w=w\). Scalars are fixed as well, proving the assertion.

On the same center, the canonical nontrivial algebra involution sends \(w\mapsto -w\). Its square is the identity and it is nontrivial since \(1,w\) are linearly independent. For a field center it is quadratic Galois conjugation; for \(\mathbb Q\times \mathbb Q\) it exchanges the factors. Thus the two-element graph group and the two-element group of center automorphisms have different specified actions: the former acts trivially on the center, whereas the latter acts nontrivially. The center character in Definition [fdip-0049] records the arithmetic Galois action, not this supplied graph symmetry.