Clifford Classification, Modules, and Graded Tensor Products

In the tensor-length parity grading of [F2], each $r_v=j(v{)}^2-Q(v)1$ is even, since its two terms have tensor lengths $2$ and $0$. Let $I$ be the subspace of finite sums of elements $a{r}_{v}b$ with $a,b\in T_k(V)$ and $v\in V$. It is a two-sided ideal containing every $r_v$, and every two-sided ideal containing all $r_v$ contains $I$; hence [F1] gives $I=J_Q$.

Write $a=a_0+a_1$ and $b=b_0+b_1$ according to tensor-length parity. Then $a{r}_{v}b={\sum }_{i,j\in \{0,1\}}{a}_i{r}_v{b}_j$, and each summand has parity $i+j(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and belongs to $I$. Thus every homogeneous component of every generator $a{r}_{v}b$ lies in $I$; by linearity the same holds for every element of $I=J_Q$. Therefore $J_Q$ is a graded ideal.

Since $J_Q$ is graded, its quotient decomposes as the direct sum of the images of the two parity components of $T_k(V)$, and the quotient multiplication preserves their parities. The tensor-algebra unit has even length, while each $j(v)$ has odd length; their quotient images therefore satisfy $1\in {\mathrm{C}\mathrm{l}}^0(V,Q)$ and $\bar{v}\in {\mathrm{C}\mathrm{l}}^1(V,Q)$. This is the claimed canonical grading, including when $Q$ is degenerate.

Since $V=0$, every tensor power $V^{\otimes m}$ with $m\ge 1$ is zero. By [F2], $T_k(0)=k$ and the canonical inclusion $j:V\to T_k(V)$ has only the value $j(0)=0$.

The only defining relation in [F1] is $j(0{)}^2-Q(0)1=0$. The zero ideal already contains this relation, so it is the smallest such two-sided ideal; hence the quotient is $T_k(0)\mathrm{/}0=k$ as a unital $k$-algebra.

Conjugation fixes $1$ and negates $i,j,k$. For an imaginary basis unit $u$, $\overline{u^2}=-1=(-u)(-u)$; for distinct imaginary units $u,v$, $uv$ is an imaginary basis unit up to sign and $\bar{v}\bar{u}=vu=-uv=\overline{uv}$. Bilinearity proves $\overline{hg}=\bar{g}\bar{h}$ for all $h,g$, and the coordinate formula gives $\overline{\stackrel{ˉ}{h}}=h$.

Write $h=a+u$ with $u=bi+cj+dk$. The relations in [F1] give $u^2=-(b^2+c^2+d^2)$ because each mixed term is a scalar multiple of $ij+ji$, $ik+ki$, or $jk+kj$. Since $a$ is central, $(a+u)(a-u)=(a-u)(a+u)=a^2-u^2=N(h)1$. Thus $h\bar{h}=\bar{h}h=N(h)1$.

If $h\mathrm{\ne }0$, at least one of $a,b,c,d$ is nonzero, so [F2] gives $N(h)>0$. Therefore $N(h{)}^{-1}$ exists in $\mathbb{R}$, and step 1.2 shows both $h(\bar{h}\mathrm{/}N(h))=1$ and $(\bar{h}\mathrm{/}N(h))h=1$. For $h=0$ the norm is $0$ and no inverse is asserted.

Write $u^{-1}=x_0+x_1$ with $x_i\in A^i$. From $1=u{u}^{-1}=u{x}_0+u{x}_1$, the odd component $u{x}_0$ and even component $u{x}_1$ must be $0$ and $1$, respectively. Since $u$ is invertible, $x_0=0$ and $u^{-1}=x_1\in A^1$. Consequently $A^1=u{A}^0$: inclusion follows from [F1], and for $a\in A^1$, $u^{-1}a\in A^0$ gives $a=u(u^{-1}a)$.

For any graded left $A$-module $M$, multiplication by $u$ maps $M^0$ bijectively onto $M^1$: it maps into $M^1$ by [F1], and its inverse $u^{-1}$ maps $M^1$ into $M^0$. Thus every homogeneous component of $M$ is generated from $M^0$ by the $A$-action.

Define $\mu :A{\otimes }_{A^0}{M}^0\to M$ by $\mu (a\otimes m)=am$. The multiplication map is balanced over $A^0$, so [F3] makes $\mu$ well defined; associativity makes it $A$-linear. Its even part $A^0{\otimes }_{A^0}{M}^0\to M^0$ is an isomorphism with inverse $m\mapsto 1\otimes m$, by the balancing relation. By step 1.1, the map $M^0\to A^1{\otimes }_{A^0}{M}^0$, $m\mapsto u\otimes m$, is an isomorphism: its inverse is $a_1\otimes m\mapsto (u^{-1}{a}_1)m$, which is well defined because $u^{-1}{a}_1\in A^0$ and respects balancing. Under this identification, the odd part of $\mu$ is $m\mapsto um$, bijective by step 2.1. Hence $\mu$ is a graded $A$-module isomorphism.

If $L\subseteq M$ is a graded submodule, then $L^1=u{L}^0$ by invertibility of $u$ and the grading. Conversely, for any $A^0$-submodule $N\subseteq M^0$, the subspace $N\oplus uN$ is graded and is stable under $A$: for $a\in A^0$, $u^{-1}au\in A^0$, so $a(un)=u(u^{-1}au)n\in uN$; for $b\in A^1$, step 1.1 writes $b=u{a}_0$ with $a_0\in A^0$, whence $bN\subseteq uN$ and $b(uN)\subseteq N$ because $u{a}_{0}u\in A^0$. Thus it is an $A$-submodule. By the definition of graded-simple in [F2], $M$ has a proper nonzero graded submodule exactly when $M^0$ has a proper nonzero $A^0$-submodule.

For any left $A^0$-module $N$, the induced module $A{\otimes }_{A^0}N$ splits as $A^0{\otimes }_{A^0}N\oplus A^1{\otimes }_{A^0}N$: the map $(a_0+a_1)\otimes n\mapsto (a_0\otimes n,a_1\otimes n)$ is well defined by balancing and has inverse given by addition. This is a grading because left multiplication respects the parity of $A$, while the balancing relations use only even elements. Its degree-zero part is isomorphic to $N$ by $a_0\otimes n\mapsto a_{0}n$, with inverse $n\mapsto 1\otimes n$. If $N$ is simple, step 3.2 makes the induced module graded-simple. Conversely, step 3.1 identifies any graded $M$ with the module induced from $M^0$, and step 3.2 says $M$ is graded-simple exactly when $M^0$ is simple. Restriction to degree zero and induction are therefore inverse on isomorphism classes: even $A$-isomorphisms restrict to $A^0$-isomorphisms, and $A^0$-isomorphisms induce even $A$-isomorphisms.

Define $\varphi :\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{C}\to \mathbb{C}\oplus \mathbb{C}$ by $\varphi (z\otimes w)=(zw,z\bar{w})$. The formula is real-bilinear in $(z,w)$, so [F2] extends it linearly. Since $\mathbb{C}$ is commutative, componentwise multiplication gives $\varphi ((z\otimes w)(z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}))=\varphi (z\otimes w)\varphi (z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})$. The images of $1\otimes 1$, $i\otimes 1$, $1\otimes i$, and $i\otimes i$ are $(1,1)$, $(i,i)$, $(i,-i)$, and $(-1,1)$. If a real linear combination with coefficients $a,b,c,d$ vanishes, its real coordinates give $a-d=a+d=0$ and its imaginary coordinates give $b+c=b-c=0$, so all coefficients vanish. They form a basis of the four-dimensional target by [F3], hence $\varphi$ is an isomorphism.

Regard $\mathbb{H}$ as a two-dimensional left $\mathbb{C}$-vector space with basis $1,j$, so $k=ij$ and $jz=\bar{z}j$ for $z\in \mathbb{C}$, using the real basis of $\mathbb{H}$ from [F1] and the field $\mathbb{C}$ from [F3]. Define $\psi :\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{H}\to {\mathrm{End}}_{\mathbb{C}}(\mathbb{H})$ by $\psi (y\otimes z)(x)=yx\bar{z}$. The formula is real-bilinear and therefore defines a linear map by [F2]. It is multiplicative because $\psi (y\otimes z)\psi (y^{\mathrm{\prime }}\otimes z^{\mathrm{\prime }})(x)=y{y}^{\mathrm{\prime }}x\overline{z{z}^{\mathrm{\prime }}}$. In the basis $1,j$, the maps $\psi (1\otimes z)$ for $z=1,i,j,k$ have matrices $I$, $\left(\begin{array}{cc}-i& 0\\ 0& i\end{array}\right)$, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, and $\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)$. Their diagonal entries force the coefficients of the first two matrices in a linear relation to vanish, and their upper-right and lower-left entries force the other two coefficients to vanish; hence they are linearly independent over $\mathbb{C}$. They form a basis of the four-dimensional complex vector space ${\mathrm{End}}_{\mathbb{C}}(\mathbb{H})$, so $\psi$ is an algebra isomorphism with $M_2(\mathbb{C})$.

Define $\theta :\mathbb{H}{\otimes }_{\mathbb{R}}\mathbb{H}\to {\mathrm{End}}_{\mathbb{R}}(\mathbb{H})$ by $\theta (a\otimes b)(x)=ax\bar{b}$. The real-bilinearity gives a well-defined linear map by [F2]. It is multiplicative since $\theta (a\otimes b)\theta (c\otimes d)(x)=acx\overline{bd}$. Set $T_{u,v}(x)=ux\bar{v}$ for basis units $u,v\in \{1,i,j,k\}$. Direct multiplication using [F1] gives the trace matrix $(\mathrm{Tr}{T}_{u,v}{)}_{u,v=1,i,j,k}=\left(\begin{array}{cccc}4& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$, with rows and columns ordered $1,i,j,k$. Equip $\mathbb{H}$ with the real inner product making $1,i,j,k$ orthonormal. Each $T_{a,b}$ is a signed permutation of this basis, hence orthogonal with $T_{a,b}^T=T_{\bar{a},\bar{b}}$. The composition formula and trace matrix give $⟨T_{a,b},T_{c,d}{⟩}_{\mathrm{H}\mathrm{S}}=\mathrm{Tr}(T_{a,b}^T{T}_{c,d})=4{\delta }_{a,c}{\delta }_{b,d}$. The sixteen nonzero images are linearly independent; since ${\mathrm{End}}_{\mathbb{R}}(\mathbb{H})$ has dimension $16$, $\theta$ is an isomorphism with $M_4(\mathbb{R})$.

Choose real bases of $D$ and $E$; these exist for $\mathbb{R},\mathbb{C},\mathbb{H}$ by their displayed coordinate descriptions. The corresponding matrix units with basis coefficients form bases of $M_m(D)$ and $M_n(E)$, and [F4] makes their elementary tensors a basis of the tensor product. They multiply by the two Kronecker-delta rules in their matrix indices and by $(d\otimes e)(d^{\mathrm{\prime }}\otimes e^{\mathrm{\prime }})=d{d}^{\mathrm{\prime }}\otimes e{e}^{\mathrm{\prime }}$ in their coefficients. Sending such a basis tensor $E_{ij}(d)\otimes E_{kl}(e)$ to the matrix unit indexed by row $(i,k)$ and column $(j,l)$ with coefficient $d\otimes e$ sends this basis bijectively to the corresponding basis of $M_{mn}(D{\otimes }_{\mathbb{R}}E)$ and is multiplicative. This proves the stated matrix identity.

Suppose first that $p>0$. Let $e_1$ be a positive generator, let $e_2,\dots ,e_p$ be the other positive generators, and let $f_1,\dots ,f_q$ be the negative generators. Set $u_i=e_1{e}_{i+1}$ for $1\le i<p$ and $v_j=e_1{f}_j$ for $1\le j\le q$. These elements are even by [F1]. Using $e_1^2=1$ and the orthogonal anticommutation in [F3], each $u_i^2=-1$, each $v_j^2=+1$, and every two distinct elements in the combined list anticommute. Thus the $v_j$ are the $q$ positive generators and the $u_i$ the $p-1$ negative generators for the defining relations of ${\mathrm{C}\mathrm{l}}_{q,p-1}$; [F4] gives an algebra homomorphism ${\mathrm{C}\mathrm{l}}_{q,p-1}\to {\mathrm{C}\mathrm{l}}_{p,q}^0$.

If $p=0$, choose a negative generator $f_1$ and set $u_j=f_1{f}_{j+1}$ for $1\le j<q$. Each $u_j$ is even by [F1], squares to $-1$, and anticommutes with the other $u_l$ by [F2]–[F3]. For distinct remaining generators $f_i,f_j$, the identity $(f_1{f}_i)(f_1{f}_j)=f_i{f}_j$ shows that these pair products generate every even ordered monomial; when $q=1$ there are no $u_j$ and the only even monomial is the unit. Thus [F4] gives a surjection ${\mathrm{C}\mathrm{l}}_{0,q-1}\to {\mathrm{C}\mathrm{l}}_{0,q}^0$.

The image contains every product of two distinct standard generators: $e_1{e}_i$ and $e_1{f}_j$ are generators of the image, while for $i,j>1$, $(e_1{e}_i)(e_1{e}_j)=-e_i{e}_j$, and the same identity holds for a pair of negative generators or one positive and one negative generator. Every even ordered monomial is a product of such pairs, so the homomorphism is surjective; if $p=1,q=0$, the only even monomial is the unit and the scalar map is already surjective.

Both algebras in step 1.2 have dimension $2^{q-1}$ by [F5], so the surjection is an isomorphism. This includes $q=1$, when [F7] identifies the source with $\mathbb{R}$ and the even part of ${\mathrm{C}\mathrm{l}}_{0,1}$ is its scalar subalgebra.

By [F5], ${\mathrm{C}\mathrm{l}}_{p,q}^0$ has dimension $2^{p+q-1}$, equal to the dimension of ${\mathrm{C}\mathrm{l}}_{q,p-1}$. The surjective map of finite-dimensional vector spaces is therefore injective, proving the first isomorphism.

For the complex standard form, choose the first generator $e_1$ and set $u_j=e_1{e}_{j+1}$. The construction and checks in steps 1.1, 2.1, and 3.1 show that the even subalgebra is generated by $n-1$ pairwise anticommuting elements of square $-1$ and has the PBW dimension $2^{n-1}$; hence it is ${\mathrm{C}\mathrm{l}}_{0,n-1}(\mathbb{C})$. Multiplying each of these generators by $i\in \mathbb{C}$ changes its square to $+1$ and gives a surjective homomorphism from the standard complex Clifford algebra ${\mathrm{C}\mathrm{l}}_{n-1}(\mathbb{C})$; the equal PBW dimensions make it an isomorphism. The argument includes $n=1$, where the even subalgebra is $\mathbb{C}$.

For ${\mathrm{C}\mathrm{l}}_{1,0}$, send its positive generator to $(1,-1)\in \mathbb{R}\oplus \mathbb{R}$; every multiple $xe$ then squares to $x^2(1,1)$, so [F2] gives a homomorphism. Its image contains $(1,1)$ and $(1,-1)$, which span the target over $\mathbb{R}$, and [F3] gives dimension $2$ on the source, so the map is an isomorphism. For ${\mathrm{C}\mathrm{l}}_{0,1}$, send its negative generator to $i\in \mathbb{C}$; every multiple $xe$ squares to $-x^2$, and the images $1,i$ are a basis, so [F2]–[F3] give an isomorphism.

Let $A=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $B_{+}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, and $B_{-}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Direct multiplication gives $A^2=B_{+}^2=I$, $B_{-}^2=-I$, and $A{B}_{+}=-B_{+}A$, $A{B}_{-}=-B_{-}A$. Hence $(xA+y{B}_{+}{)}^2=(x^2+y^2)I$ and $(xA+y{B}_{-}{)}^2=(x^2-y^2)I$ for all real $x,y$, so [F2] gives the required homomorphisms from ${\mathrm{C}\mathrm{l}}_{2,0}$ and ${\mathrm{C}\mathrm{l}}_{1,1}$. In each case the images of $1$, the two generators, and their product are the four linearly independent matrices $I,A,B_{+},B_{-}$ in some order, so the maps are surjective; [F3] gives dimension $4$ on each source, making both maps isomorphisms.

Send the two negative generators of ${\mathrm{C}\mathrm{l}}_{0,2}$ to $i,j\in \mathbb{H}$. They square to $-1$ and anticommute, so every linear combination $xi+yj$ squares to $-(x^2+y^2)$ and [F2] gives a homomorphism. Its image contains $1,i,j,ij=k$, the basis in [F4]; [F3] gives dimension $4$ on the source, so this homomorphism is an isomorphism.

For $v\in V$ and $w\in W$, set $c(v,w)=\bar{v}\otimes 1+1\otimes \bar{w}$. Each summand is odd by [F5]. By [F1], $(\bar{v}\otimes 1)(1\otimes \bar{w})=\bar{v}\otimes \bar{w}$ and $(1\otimes \bar{w})(\bar{v}\otimes 1)=-\bar{v}\otimes \bar{w}$, so their cross terms cancel in the square. By [F2], the two squares are $Q(v)1$ and $R(w)1$, so $c(v,w{)}^2=(Q(v)+R(w))1$.

By [F3], the linear map $(v,w)\mapsto c(v,w)$ extends uniquely to a unital algebra homomorphism $\mathrm{\Phi }$ with the stated generator values. It preserves parity because every generator maps to an odd element by [F5].

The image of $\mathrm{\Phi }$ contains $\bar{v}\otimes 1$ by taking $(v,0)$ and $1\otimes \bar{w}$ by taking $(0,w)$. These elements generate both Clifford factors, and [F4] gives finite bases consisting of products of canonical generators; their elementary tensors span the graded tensor product by [F6]. Thus $\mathrm{\Phi }$ is surjective.

By [F4], the source has dimension $2^{\dim V+\dim W}$, and [F6] gives the same dimension for the tensor-product target. A surjective linear map between equal finite-dimensional vector spaces is injective, hence $\mathrm{\Phi }$ is an isomorphism. This includes either zero-dimensional factor and degenerate forms.