Irreducible graded Clifford modules and real, complex, and quaternionic types

In the quotient [F8], let $n=p+q>0$ in the real case, or $n>0$ in the complex case. A standard nondegenerate form has a generator $u$ with $u^2=1$ or $u^2=-1$, so $u$ is odd by [F9] and invertible. By [F1], irreducible graded $A$-modules correspond bijectively to simple $A^0$-modules; [F2] identifies $A^0$ with the lower-dimensional Clifford algebra stated in the theorem.

In standard complex dimension $n$, let $e_1,\dots ,e_n$ be the standard generators, so $e_j^2=1$ by [F8], and set $\mathrm{\Omega }=e_1\cdots e_n$ and $\omega =i^{\lfloor n\mathrm{/}2\rfloor }\mathrm{\Omega }$. Moving $e_j$ past the other $n-1$ orthogonal generators gives $\mathrm{\Omega }{e}_j=(-1{)}^{n-1}{e}_j\mathrm{\Omega }$ by [F7], while pairing the generators in ${\mathrm{\Omega }}^2$ gives ${\mathrm{\Omega }}^2=(-1{)}^{n(n-1)\mathrm{/}2}{\prod }_{j=1}^n{e}_j^2=(-1{)}^{n(n-1)\mathrm{/}2}$ by [F7] and [F8]. For $n=2m$ and for $n=2m+1$, the scalar factor has square $(-1{)}^m$, so ${\omega }^2=1$ in both cases. Its parity is $n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$ by [F9]; hence for even $n$ it is even and anticommutes with odd generators, while for odd $n$ it is odd and commutes with every generator.

By [F3], the even algebra $A^0$ is either a single matrix algebra $M_m(D)$ or a direct sum of two such algebras, with $D$ one of $\mathbb{R},\mathbb{C},\mathbb{H}$ in the real case and $D=\mathbb{C}$ in the complex case. Each $D$ is a division ring by [F5], so [F4] gives one simple column module $D^m$ in the single-factor case and exactly two simple column modules, each supported on one factor, in the split case.

For the column module $D^m$ (column vectors with left matrix action), every $M_m(D)$-linear endomorphism is right multiplication by a unique $d\in D$: commuting with $E_{11}$ forces the image of the first standard column to be $e_{1}d$, and commuting with $E_{i1}(a)$ gives the image $e_{i}ad$ of every column $e_{i}a$. Conversely right multiplication commutes with the left matrix action. Composition reverses the order of the right multipliers, so ${\mathrm{End}}_{M_m(D)}(D^m)\cong D^{\mathrm{o}\mathrm{p}}$. For $D=\mathbb{R},\mathbb{C}$ this is $D$, and for $D=\mathbb{H}$ quaternion conjugation gives $D^{\mathrm{o}\mathrm{p}}\cong D$ by [F5].

Combining steps 1.1, 2.1, and 3.1 yields the asserted graded-module counts and coefficient types, including both modules when the even algebra splits. If $n=0$, [F10] gives $A=F$ for $F=\mathbb{R}$ or $\mathbb{C}$, and there is no odd component. Each parity component of a graded module is then an $A$-submodule, so a graded-simple module must be nonzero in exactly one parity. Any nonzero vector in that component generates a nonzero submodule $Fv$, which by simplicity is the entire component; the map $F\to Fv$, $c\mapsto cv$, is injective and surjective, so the module is one-dimensional over $F$. The two parity placements are nonisomorphic as even graded modules.

If $n=2m>0$, [F2] and [F3] identify the even algebra with a split sum of two complex matrix algebras. The element $\omega$ is central in the even algebra; by [F6] it is not $1$ or $-1$, so $(1+\omega )\mathrm{/}2$ and $(1-\omega )\mathrm{/}2$ are the two nonzero factor idempotents. The simple degree-zero modules therefore have $\omega$-eigenvalue $+1$ or $-1$, and multiplication by any odd invertible generator switches that eigenvalue in degree $1$; this is precisely the two chiral parity choices. If $n=2m+1$, [F2] and [F3] give a simple even algebra and hence one graded-simple class by step 1.1. In the full algebra $\omega$ is odd central and not $\pm 1$ by [F6], so its idempotents $(1\pm \omega )\mathrm{/}2$ are nonzero and split the two ungraded matrix factors; the parity operator $P(x^0+x^1)=x^0-x^1$ satisfies $Pa{P}^{-1}=(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}a$ and exchanges these idempotents. Thus a nonzero graded module contains both ungraded factors, as stated.