Proposition. The discriminant center of a four-dimensional Spin group [fcap-0043]
✍️source

    Let \(K\) be a field of characteristic different from two, and let \(q=\langle a,b,c,d\rangle \) with \(a,b,c,d\in K^\times \). Thus \(q(x)=ax_1^2+bx_2^2+cx_3^2+dx_4^2\). The center of its even Clifford algebra determines whether the quaternion description of its Spin group has two factors over \(K\) or one factor over a quadratic field. Put \[\begin {aligned} \delta &=abcd,\\ E&=K[t]/(t^2-\delta ),\\ D&=(-ab,-ac\mid K), \end {aligned}\] using the quaternion algebra notation. The class of \(\delta \) in \(K^\times /(K^\times )^2\) is the signed discriminant in dimension four; changing the diagonalizing basis changes the determinant by a square. Lam's center theorem identifies the corresponding quadratic algebra; the explicit quaternion and group identifications below follow by calculation [lam2005quadratic, V.2, thms. 2.2--2.5, pp. 109--111].

    Extend quaternion conjugation \(\overline i=-i\), \(\overline j=-j\) \(E\)-linearly to \(D_E=D\otimes _K E\). Write \(SL_1(D_E)=\{x\in D_E:x\overline x=1\}\) for its norm-one group. Indeed, for \(x=x_0+x_1i+x_2j+x_3ij\), with \(x_r\in E\), \[x\overline x=\overline x x =x_0^2+abx_1^2+acx_2^2+a^2bcx_3^2\in E.\] This extends the reduced norm to the displayed quaternion algebra, and \(x^{-1}=\overline x\) when the norm is one. There are algebra and group isomorphisms \[\begin {aligned} \mathcal {C}\kern -2pt\ell ^+(q)&\simeq D_E,\\ Z(\mathcal {C}\kern -2pt\ell ^+(q))&\simeq E,\\ \operatorname {Spin}(q)&\simeq SL_1(D_E), \end {aligned}\] where the algebra isomorphism takes grade reversion to quaternion conjugation. If \(\delta \) is a square in \(K^\times \), this gives \[\operatorname {Spin}(q)\simeq SL_1(D)\times SL_1(D).\] If \(\delta \) is nonsquare, it gives \[\operatorname {Spin}(q)\simeq SL_1(D\otimes _K K(\sqrt \delta )).\] All groups here consist of points over the indicated fields.

    For a concrete comparison, take \(q_d=\langle 1,-1,1,-d\rangle \). Then \(\delta =d\) and \(D=(1,-1\mid K)\). The matrices \[i\longmapsto \begin {pmatrix}1&0\\0&-1\end {pmatrix},\qquad j\longmapsto \begin {pmatrix}0&1\\-1&0\end {pmatrix}\] have squares \(1,-1\), anticommute, and, together with their product and the identity, form a basis of \(M_2(K)\). They therefore identify \(D\) with \(M_2(K)\). Conjugation becomes matrix adjugation and the norm becomes the determinant, giving \(\operatorname {Spin}(q_d)\simeq SL_2(E)\). In particular, over \(K=\mathbb Q\), the members \(q_1\) and \(q_2\) satisfy \[\begin {aligned} \operatorname {Spin}(q_1)(\mathbb Q) &\simeq SL_2(\mathbb Q)\times SL_2(\mathbb Q),\\ \operatorname {Spin}(q_2)(\mathbb Q) &\simeq SL_2(\mathbb Q(\sqrt 2)). \end {aligned}\] These identities apply the proposition to the two centers \(\mathbb Q\times \mathbb Q\) and \(\mathbb Q(\sqrt 2)\).

    Proof. 

      Choose orthogonal Clifford generators with \(e_1^2=a,e_2^2=b,e_3^2=c,e_4^2=d\). Distinct generators anticommute. Set \(I=e_1e_2\), \(J=e_1e_3\), and \(w=e_1e_2e_3e_4\). Then \[\begin {aligned} I^2&=-ab,&J^2&=-ac,\\ IJ&=-JI,&w^2&=abcd. \end {aligned}\] The element \(w\) commutes with every even word. Thus \(i\mapsto I\), \(j\mapsto J\), \(t\mapsto w\) define an algebra homomorphism \(D\otimes _K E\longrightarrow \mathcal {C}\kern -2pt\ell ^+(q)\). The eight elements \[1,I,J,IJ,w,Iw,Jw,IJw\] map to nonzero scalar multiples of the eight distinct even Clifford basis monomials. Explicitly, \[\begin {aligned} IJ&=-ae_2e_3,&Iw&=-abe_3e_4,\\ Jw&=ace_2e_4,&IJw&=abce_1e_4. \end {aligned}\] They form a basis, proving bijectivity. Writing an element of \(D_E\) in the basis \(1,i,j,ij\), commutation with \(i\) forces its \(j,ij\) coefficients to vanish, and commutation with \(j\) forces its \(i\) coefficient to vanish. Here \(2,ab,ac\) are units, even when \(E\) is a product. The center is therefore exactly \(E\). Reversion negates \(I,J,IJ\) and fixes \(w\), so it is \(\overline {(\,\cdot \,)}\) on \(D_E\).

      It remains to identify the actual vector-generated Spin group with all even elements of reverse norm one. The real four-dimensional argument of Lemma [fcap-002J] works over the stated field as follows. Suppose \(x\in \mathcal {C}\kern -2pt\ell ^+(q)\) and \(\operatorname {rev}(x)x=1\). Finite-dimensionality makes this left inverse a two-sided inverse. For every vector \(v\), \(xvx^{-1}\) is odd and fixed by reversion. In the orthogonal Clifford basis, reversal is \(+1\) on degree one and \(-1\) on degree three. Since two is invertible, \(xvx^{-1}\) is a vector. Its square is \(q(v)\), and conjugation is injective, hence induces an orthogonal automorphism \(g\) of the four-dimensional vector space.

      By the Clifford group exact sequence, lift \(g\) to a vector-generated Clifford unit \(y\). For the grading involution \(\alpha \), equality of the twisted actions of \(y\) and the even element \(x\) implies that \(z=y^{-1}x\) satisfies \(\alpha (z)v=vz\) for every vector \(v\). The graded-centrality theorem used in Lemma [fcap-002J] gives \(z=c1\) with \(c\in K^\times \). Every such scalar belongs to the generated Clifford group: for \(q(v)\ne 0\), the two nonisotropic vectors \((c/q(v))v\) and \(v\) have Clifford product \(c1\). Thus \(x=yz\) is in that group. It is even with reverse norm one, so the carrier criterion Remark [fcap-0028] places it in Spin. Conversely, every Spin element is even with reverse norm one by that same criterion. Under the algebra isomorphism this is exactly \(SL_1(D_E)\).

      If \(\delta =r^2\) with \(r\in K^\times \), evaluation at \(r,-r\) identifies \(E\) with \(K\times K\). The corresponding central idempotents are \((1+w/r)/2\) and \((1-w/r)/2\). Hence \(D_E\simeq D\times D\), with componentwise conjugation and norm, giving the displayed product of norm-one groups. If \(\delta \) is nonsquare, \(t^2-\delta \) is irreducible, so \(E=K(\sqrt \delta )\) and the second description follows.

      The split-center group equivalence has a formal counterpart in CliffordAlgebra.spinGroupEquivSplitCenterCorners, given a center isomorphism with \(K\times K\). The field-center existence theorem CliffordAlgebra.exists_spinGroupEquivQuaternionUnitaryOverCenter_of_finrank_eq_four gives a quaternion norm-one model over the actual even center. Both assume nondegeneracy and dimension four over a field with two invertible. The explicit parameters \(-ab,-ac\) and the matrices for \(q_d\) are supplied by the calculation above.