Example. The split four-dimensional Spin action is left and inverse-right multiplication [fcap-003Y]
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For $Q_{2,2}(v)=v_0^2+v_1^2-v_2^2-v_3^2$, the even Clifford algebra is a pair of real matrix algebras. We construct its coordinates to specify which Spin factor acts on each side of a vector. Set $A=e_0e_1$, $B=e_0e_2$, $C=e_1e_2$, and $I=e_0e_1e_2e_3$. The subalgebra $H_0=\langle 1,A,B,C\rangle _{\mathbb R}$ has $A^2=-1$, $B^2=C^2=1$, and $AB=-C$. Define \[J=\begin {pmatrix}0&1\\-1&0\end {pmatrix},\quad S=\begin {pmatrix}0&1\\1&0\end {pmatrix},\quad D=\begin {pmatrix}-1&0\\0&1\end {pmatrix}.\] The assignments $\rho (A)=-J$, $\rho (B)=D$, $\rho (C)=S$ satisfy these relations and give a basis of $M_2(\mathbb R)$ together with the identity. They define an algebra equivalence $\rho $. The volume element $I$ is central in the even algebra and $I^2=1$. The eight even basis monomials give $\mathcal {C}\kern -2pt\ell ^+(Q_{2,2})=H_0\oplus I H_0$. Hence the map \[\Phi (u+Iw)=\bigl (\rho (u)-\rho (w),\ S(\rho (u)+\rho (w))S\bigr )\] is an algebra equivalence to $M_2(\mathbb R)\times M_2(\mathbb R)$. For a pair $(P,R)$, its inverse is recovered from \[\rho (u)=\tfrac 12(P+SRS),\qquad \rho (w)=\tfrac 12(SRS-P).\]

The reversion fixes $I$ and negates $A,B,C$. On $H_0$, this is matrix adjugation, since $\rho (A), \rho (B),\rho (C)$ are traceless. Conjugation by $S$ commutes with adjugation, so \[\Phi (\operatorname {rev}(x))= \bigl (\operatorname {adj}(\Phi _1(x)), \operatorname {adj}(\Phi _2(x))\bigr ).\] Consequently $\operatorname {rev}(x)x=1$ if and only if both component determinants are one. The carrier lemma Lemma [fcap-002J] therefore restricts $\Phi $, with the displayed inverse, to the actual group \[\operatorname {Spin}(Q_{2,2})\simeq SL_2(\mathbb R)\times SL_2(\mathbb R).\] Porteous's norm-one subgroup is denoted $\operatorname {Spin}^+$; his split-four entry is a pair of $Sp_2(\mathbb R)$ factors [porteous1995clifford, Chapter 17, Theorem 17.9 and Table 14, p. 159]. In degree two, $M^{\mathsf T}JM=(\det M)J$, so these are precisely $SL_2(\mathbb R)$. The norm-one restriction is essential to that comparison.

Our pair map has the following six bivector images: \[\begin {aligned} e_0e_1&\longmapsto (-J,J),& e_0e_2&\longmapsto (D,-D),\\ e_0e_3&\longmapsto (-S,S),& e_1e_2&\longmapsto (S,S),\\ e_1e_3&\longmapsto (D,D),& e_2e_3&\longmapsto (-J,-J). \end {aligned}\] In particular $\Phi (I)=(-1,1)$. These are the specializations of TauCeti's bilinear-generator formula TauCeti.realCliffordTwoTwoEvenEquivMatrixProd_ι. Since the bivectors generate the even algebra, this identifies $\Phi $ with its concrete even-algebra map and hence identifies the restricted group equivalence TauCeti.realSpinTwoTwoEquivSpecialLinearProd.

Now encode vectors by \[X(v)=\Phi _1(ve_3)S= \begin {pmatrix}-v_0-v_2&-v_1-v_3\\v_1-v_3&-v_0+v_2\end {pmatrix}.\] The inverse sends $\begin {pmatrix}a&b\\c&d\end {pmatrix}$ to $(-(a+d)/2,(c-b)/2,(d-a)/2,-(b+c)/2)$. A determinant expansion gives $\det X(v)=Q_{2,2}(v)$. Thus $X$ identifies the whole vector space with the determinant quadratic space of real two-by-two matrices TauCeti.realCliffordTwoTwoVectorEquivMatrix_apply TauCeti.realCliffordTwoTwoVectorEquivMatrix_det. Writing $\Phi (s)=(P,R)$ for a Spin element, the action is \[X(svs^{-1})=P X(v)R^{-1}.\]

For the proof, $e_3^2=-1$, and conjugation $\gamma (x)=-e_3xe_3$ fixes $H_0$ and negates $I$. Therefore \[\Phi (\gamma (x))= \bigl (S\Phi _2(x)S,\ S\Phi _1(x)S\bigr ).\] The identities \[e_0e_3=IC,\qquad e_1e_3=-IB,\qquad e_2e_3=-IA\] give the displayed formula for $X$. Apply $\Phi _1$ to $(svs^{-1})e_3=s(ve_3)\gamma (s^{-1})$, then multiply on the right by $S$. Since $S^2=1$, the result is $P X(v)R^{-1}$. This proves TauCeti's exact coordinate action TauCeti.realSpinTwoTwoEquivSpecialLinearProd_action.

The diagram records the two ways to calculate the same action:

$\text{[math]}$ For example, take $P=\operatorname {diag}(a,a^{-1})$, $R=1$, where $a\ne 0$. The two rows of $X(v)$ are multiplied by $a$ and $a^{-1}$. Thus the null cone is exactly the matrices of rank at most one, and its rank-one part is preserved. For $P=R$, the matrix $X(e_0)=-1$ is fixed; this diagonal subgroup fixes $e_0$.

As in the compact action of Example [fcap-003K], multiplication by a fixed vector makes an odd vector accessible to the even algebra, and conjugation by that vector interchanges its two factors. Here the factors are real matrices, with the explicit $S$-conjugations above. This is the even-algebra action transport of [porteous1995clifford, Chapter 16, Proposition 16.9, pp. 143--144] in the stated axes. The construction gives the group and its vector action; its orthogonal image retains the spinor-norm qualification of Theorem [fcap-0023].