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.