Complex spinor modules and complete reducibility for Euclidean Spin

Choose a pairwise orthogonal anisotropic real basis $e_1,\dots ,e_n$ by [F1], and put $u_i=e_i\mathrm{/}\sqrt{\mathrm{\mid }Q(e_i)\mathrm{\mid }}$. The ordered monomials in the $e_i$ are a real PBW basis of the real Clifford algebra and a complex PBW basis after complexification. The natural scalar-extension map ${\mathrm{C}\mathrm{l}}_{\mathbb{R}}(V,Q){\otimes }_{\mathbb{R}}\mathbb{C}\to {\mathrm{C}\mathrm{l}}_{\mathbb{C}}(V{\otimes }_{\mathbb{R}}\mathbb{C},Q_{\mathbb{C}})$, induced by the common generator relations, matches these monomials, so is an isomorphism and embeds the real Clifford algebra into its complexification. Viewed as a complex basis, the $e_i$ have the same squares and remain pairwise orthogonal. Each even PBW basis monomial $e_{i_1}\cdots e_{i_{2k}}$ is the nonzero real scalar ${\prod }_j\sqrt{\mathrm{\mid }Q(e_{i_j})\mathrm{\mid }}$ times the product $u_{i_1}\cdots u_{i_{2k}}$. This product belongs to the real full-even Spin group by [F3] and hence to the complexified Clifford algebra. Thus the complex span of full-even Spin contains a basis of ${\mathrm{C}\mathrm{l}}^0$. Conversely every real Spin element is even, so its complex span lies in ${\mathrm{C}\mathrm{l}}^0$. Therefore these complex linear spans are equal, and a complex subspace is Spin-invariant exactly when it is a ${\mathrm{C}\mathrm{l}}^0$-submodule.

For a complex matrix algebra $M_r(\mathbb{C})$, every module is the sum of its diagonal matrix-unit subspaces $E_{ii}M$, and $E_{i1}$ identifies these subspaces with $E_{11}M$. Consequently $M\cong {\mathbb{C}}^r{\otimes }_{\mathbb{C}}{E}_{11}M$ with the algebra acting on the first factor. Its simple module is therefore the column ${\mathbb{C}}^r$, and each finite-dimensional module is a sum of columns. For a product of matrix algebras the central factor idempotents split the module, and simples supported on different factors are inequivalent. This is a complex-linear calculation, separate from the real commutant types.

If $n=2m\ge 2$, [A1] gives one full Clifford simple of complex dimension $2^m$, and [A1]–[A2] identify the even algebra with two matrix factors of size $2^{m-1}$. Set $\omega =i^m{u}_1\cdots u_{2m}$ using the normalized vectors of step 1.1. Since $Q(u_j)=1$, the Clifford relations give ${\omega }^2=(-1{)}^m(-1{)}^{m(2m-1)}=1$, and $\omega$ commutes with every even monomial. PBW makes $\omega$ nonscalar, so $p_{\pm }=(1\pm \omega )\mathrm{/}2$ are nonzero orthogonal central idempotents of ${\mathrm{C}\mathrm{l}}^0$. An odd unit vector anticommutes with $\omega$ and conjugates $p_{+}$ to $p_{-}$. The full Clifford module is faithful because ${\mathrm{C}\mathrm{l}}_{2m}(\mathbb{C})$ is a simple matrix algebra; hence both idempotents act nontrivially. Each factor therefore contributes at least one simple module of dimension $2^{m-1}$; their dimensions already sum to $2^m$, so each occurs exactly once. Step 1.1 makes these two summands the inequivalent half-spin modules.

If $n=2m+1$, [A1] gives two full Clifford simple modules of dimension $2^m$, and [A1]–[A2] identify a single even matrix factor of size $2^m$. Each full module restricts nontrivially to that even algebra, and its dimension equals the unique simple even-module dimension. Hence each restriction is the unique irreducible even module; by step 1.1 the two Spin restrictions are isomorphic irreducible modules. For $n=0$, ${\mathrm{C}\mathrm{l}}^0=\mathbb{C}$ and the spinor has dimension one. For $n=1$, ${\mathrm{C}\mathrm{l}}^0=\mathbb{C}$ while the full algebra has two one-dimensional simples, so both restrict to the same one-dimensional Spin module.

By [F2], $\mathrm{Spin}(n)$ is a compact subgroup of a finite-dimensional Hausdorff real algebra, hence is a compact Hausdorff topological group. For any continuous finite-dimensional complex representation in [F5], [F4] and [A3] give a direct sum of irreducible subrepresentations.