Real Spin modules from the even Clifford algebra

Choose a pairwise orthogonal anisotropic basis $e_1,\dots ,e_n$ by [F1], and put $u_i=e_i\mathrm{/}\sqrt{\mathrm{\mid }Q(e_i)\mathrm{\mid }}$, so each $u_i$ is normalized. By PBW in this ordered basis, every even basis monomial $e_{i_1}\cdots e_{i_{2k}}$ is a nonzero real scalar multiple of the even normalized-vector product $u_{i_1}\cdots u_{i_{2k}}$. That product belongs to full-even Spin, so the real span of full-even Spin contains a basis of ${\mathrm{C}\mathrm{l}}^0$. Conversely every full-even Spin element belongs to ${\mathrm{C}\mathrm{l}}^0$. Therefore ${\mathrm{span}}_{\mathbb{R}}({\mathrm{Spin}}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}})={\mathrm{C}\mathrm{l}}^0$. A real subspace is Spin-invariant exactly when it is invariant under this linear span, so a given ${\mathrm{C}\mathrm{l}}^0$-module and its Spin restriction have the same invariant subspaces and commutants. Moreover any Spin intertwiner between two such restrictions intertwines the spanning algebra; hence their module-isomorphism and Spin-isomorphism classes agree, and an extending algebra action is uniquely determined.

For a division algebra $D$ and $r\ge 1$, the standard column module $D^r$ is simple over $M_r(D)$: from any nonzero column, matrix units move a nonzero coordinate to any position and left multiplication by $D$ produces arbitrary coordinates. Conversely, if $M$ is a finite-dimensional simple left $M_r(D)$-module, some diagonal matrix unit $E_{ii}$ acts nontrivially; matrix units transfer a nonzero vector to $E_{11}M$. This is a left $D$-module. If it had dimension greater than one, choose a proper nonzero left $D$-subspace $W\subset E_{11}M$ and form $N={\sum }_{i=1}^r{E}_{i1}W$. Matrix multiplication shows $N$ is an $M_r(D)$-submodule, and $E_{11}N=W$; hence $N$ is nonzero and proper, contradicting simplicity of $M$. Thus $E_{11}M\cong D$, and $D^r\to M$, $(d_i)\mapsto {\sum }_i{E}_{i1}(d_i)v$, is a nonzero module map; simplicity of $D^r$ makes it an isomorphism.

An endomorphism of $D^r$ commuting with $M_r(D)$ must commute with every $E_{ij}$ and every diagonal scalar matrix; it is therefore the same right multiplication by some $d\in D$ in every coordinate. The commutant is $D^{\mathrm{o}\mathrm{p}}$. For a product of matrix algebras, the central identity idempotents split every module; in a simple module exactly one factor acts nontrivially, so each factor gives one simple type and distinct factors give inequivalent modules.

Applying [L1] to identify the even algebra and [A1] to classify its real matrix factors, the matrix-algebra result of steps 1.2–2.1 gives exactly the column modules, dimensions, and commutant types stated. By step 1.1 these are exactly the irreducible Clifford spinor restrictions to full-even Spin. This calculation does not identify the type of a full Clifford module without computing its even-subalgebra factors.

In dimensions $0$ and $1$, the even monomials are just scalars, so ${\mathrm{C}\mathrm{l}}^0\cong \mathbb{R}$. In dimension $2$, put $E=e_1{e}_2$ for an orthogonal basis normalized so that $Q(e_i)=\pm 1$. Such normalization is obtained by dividing each anisotropic basis vector by $\sqrt{\mathrm{\mid }Q(e_i)\mathrm{\mid }}$. Then $E^2=-Q(e_1)Q(e_2)$ and ${\mathrm{C}\mathrm{l}}^0=\mathbb{R}\oplus \mathbb{R}E$. If the signature is definite, $E^2=-1$ and this is $\mathbb{C}$; its standard real module has dimension $2$ and commutant $\mathbb{C}$. If the signature is $(1,1)$, $E^2=1$ and $\mathbb{R}[E]\mathrm{/}(E^2-1)\cong \mathbb{R}\oplus \mathbb{R}$ via $a+bE\mapsto (a+b,a-b)$, giving two one-dimensional real modules. Step 1.1 transfers these classifications to Spin.