Universal Clifford Algebras and Their PBW Basis

For each $r\ge 1$, the displayed sum is multilinear in $(w_1,\dots ,w_r)$, so [F3] defines a linear map on $V^{\otimes r}$. Together with ${\iota }_v(1)=0$, these maps define ${\delta }_v:T_k(V)\to T_k(V)$. On a concatenated pair of pure tensors of degrees $p,q$, the first $p$ deletion terms give ${\delta }_v(x)y$, while each remaining deletion has the extra sign $(-1{)}^p$ and gives $(-1{)}^{p}x{\delta }_v(y)$. Thus ${\delta }_v(xy)={\delta }_v(x)y+(-1{)}^{p}x{\delta }_v(y)$; bilinearity extends this to homogeneous tensors.

Left multiplication gives ${ϵ}_v{ϵ}_w+{ϵ}_w{ϵ}_v=0$: in $\mathrm{\Lambda }V$, $(v+w{)}^2=v^2=w^2=0$, so expansion gives $vw+wv=0$.

Let $S$ be the $k$-linear span of the terms $xj(u{)}^{2}y$ for $x,y\in T_k(V)$ and $u\in V$. This span contains all the squares, and multiplying its terms on the left or right only changes $x$ or $y$, so it is a two-sided ideal by [F4]. Every two-sided ideal containing all the squares contains $S$, hence $S=J_{\wedge }$ by [F2]. For $u\in V$, ${\delta }_v(j(u)j(u))=B(v,u)j(u)-j(u)B(v,u)=0$. If $x$ is homogeneous of degree $p$, then for homogeneous $y$ the derivation rule gives ${\delta }_v(xj(u{)}^{2}y)={\delta }_v(x)j(u{)}^{2}y+(-1{)}^{p}x{\delta }_v(j(u{)}^2)y+(-1{)}^{p+2}xj(u{)}^2{\delta }_v(y)$. Each outer term lies in $J_{\wedge }$ by [F4], and the middle term is zero. By linearity, ${\delta }_v(J_{\wedge })\subseteq J_{\wedge }$, so ${\delta }_v$ descends to ${\iota }_v$ on $\mathrm{\Lambda }V$. Since $J_{\wedge }$ is a two-sided ideal, left multiplication by $j(v)$ also descends to ${ϵ}_v$.

On a decomposable wedge, every term of ${\iota }_v{\iota }_w+{\iota }_w{\iota }_v$ deletes two distinct positions. For fixed $i<j$ and fixed choice of which contraction deletes each factor, deleting position $i$ before $j$ changes the second deletion's sign from $(-1{)}^{j-1}$ to $(-1{)}^{j-2}$; the two orders therefore have opposite signs and identical scalar coefficient. Pairing all such terms gives zero. If $r<2$, both compositions vanish.

The contraction formula separates the term deleting the first factor and the terms deleting the remaining factors, giving ${\iota }_v(w\wedge \omega )=B(v,w)\omega -w\wedge {\iota }_v(\omega )$ on decomposable $\omega$ and hence on all $\omega$. Thus ${\iota }_v{ϵ}_w+{ϵ}_w{\iota }_v=B(v,w)I$.

Adding the three anticommutator identities yields $c(v)c(w)+c(w)c(v)=(B(v,w)+B(w,v))I=2B(v,w)I$. Taking $w=v$ and cancelling $2$ in $k$ gives $c(v{)}^2=Q(v)I$. Linearity of ${ϵ}_v$ and of the contraction formula in $v$ makes $c:V\to {\mathrm{End}}_k(\mathrm{\Lambda }V)$ linear.

Applying [F1] to $v+w$, $v$, and $w$, then subtracting the last two defining equalities from the first, gives $\bar{v}\bar{w}+\bar{w}\bar{v}=(Q(v+w)-Q(v)-Q(w))1$.

Bilinearity and symmetry give $Q(v+w)-Q(v)-Q(w)=B(v,w)+B(w,v)=2B(v,w)$, proving the relation.

Since $\psi$ is a homomorphism, its kernel is closed under addition and under left and right multiplication by elements of $T_k(V)$. It is therefore a two-sided ideal in the sense of [F2] containing each defining relation, and minimality in [F1] gives $J_Q\subseteq \ker \psi$.

Define $\bar{\psi }(t+J_Q)=\psi (t)$. If $t+J_Q=t^{\mathrm{\prime }}+J_Q$, then $t-t^{\mathrm{\prime }}\in J_Q\subseteq \ker \psi$, so $\psi (t)=\psi (t^{\mathrm{\prime }})$ and the definition is well defined.

The induced map preserves addition, multiplication, scalar multiplication, and the unit because $\psi$ does and the quotient operations are induced from $T_k(V)$. Since $q_Q$ is surjective, any map factoring $\psi$ agrees with $\bar{\psi }$ on every coset; thus the factor is unique.

By [F4], $A$ has the central unital $k$-algebra structure required by [F1], so $f$ extends uniquely to a unital $k$-algebra homomorphism $\tilde{f}:T_k(V)\to A$ with $\tilde{f}\circ j=f$.

For every $v\in V$, $\tilde{f}(j(v{)}^2-Q(v)1)=f(v{)}^2-Q(v)1_A=0$. Thus [F2] gives a unique unital homomorphism $\bar{f}:\mathrm{C}\mathrm{l}(V,Q)\to A$ satisfying $\bar{f}\circ q_Q=\tilde{f}$.

Since $i_Q=q_Q\circ j$, step 2.1 gives $\bar{f}\circ i_Q=\tilde{f}\circ j=f$. Let $g:\mathrm{C}\mathrm{l}(V,Q)\to A$ be any other unital $k$-algebra homomorphism with $g\circ i_Q=f$. Then $g\circ q_Q:T_k(V)\to A$ and $\tilde{f}$ are unital algebra extensions of the same map $f$ along $j$, because $(g\circ q_Q)\circ j=g\circ i_Q=f=\tilde{f}\circ j$. Uniqueness in [F1] gives $g\circ q_Q=\tilde{f}=\bar{f}\circ q_Q$. Since $q_Q$ is surjective by [F3], $g=\bar{f}$.

By [F6], $c:V\to {\mathrm{End}}_k(\mathrm{\Lambda }V)$ is linear and satisfies the quadratic relation. With endomorphism multiplication given by composition, [F3] gives a representation $\rho :\mathrm{C}\mathrm{l}(V,Q)\to {\mathrm{End}}_k(\mathrm{\Lambda }V)$. For $I=(i_1<\cdots <i_r)$, the degree-$r$ part of $\rho ({\bar{e}}_{i_1}\cdots {\bar{e}}_{i_r})(1)$ is exactly $e_{i_1}\wedge \cdots \wedge e_{i_r}$, since it comes from choosing every exterior-multiplication term; any contraction lowers the resulting degree.

The defining relation [F1] gives ${\bar{e}}_i^2=Q(e_i)1$, while [F2] gives ${\bar{e}}_j{\bar{e}}_i=-{\bar{e}}_i{\bar{e}}_j+2B(e_i,e_j)1$ whenever $j>i$.

Use well-founded induction on the lexicographically ordered pair (word length, inversion count), where the inversion count of ${\bar{e}}_{i_1}\cdots {\bar{e}}_{i_r}$ is the number of pairs $a<b$ with $i_a>i_b$. Swapping an adjacent inversion lowers the inversion count in the same-length term and creates only a scalar correction of length two less; a sorted word with a repeated index has an adjacent square, which [F1] replaces by a scalar. Both resulting measures are smaller, so induction reduces every word to a linear combination of the displayed words and proves spanning.

If a linear relation among the spanning words were nonzero, choose its greatest word length $r$ with a nonzero coefficient and apply $\rho$ to $1$. Shorter words have exterior degree at most their word length, so the degree-$r$ component is a relation among distinct basis wedges from [F5]; their linear independence forces all coefficients of length $r$ to vanish, a contradiction. Descending over the finitely many lengths proves independence.

For each $m$, the relations in [F1] and [F2] never increase word length, so the words of length at most $m$ span $F_m$. Steps 2.1 and 2.2 show that the ordered square-free words of length at most $m$ are a basis of $F_m$.

The filtration is multiplicative because concatenating words of lengths at most $p$ and $q$ has length at most $p+q$. Send $v$ to the class of $\bar{v}$ in $F_1\mathrm{/}{F}_0$ and extend to $T_k(V)$ by [F7]. The square of each degree-one class is zero, since ${\bar{v}}^2=Q(v)1\in F_0\subseteq F_1$; hence this tensor-algebra map kills each exterior relation in [F4]. Its kernel is a two-sided ideal because the map preserves sums and products, so it contains the ideal generated by those relations and induces a graded algebra map $\mathrm{\Phi }:\mathrm{\Lambda }V\to {\mathrm{gr}}_F\mathrm{C}\mathrm{l}(V,Q)$.

In degree $m$, [F5] gives the basis of ${\mathrm{\Lambda }}^{m}V$ and step 3.1 gives the corresponding basis of $F_m\mathrm{/}{F}_{m-1}$. The map $\mathrm{\Phi }$ sends each basis wedge to its matching Clifford-word class, so it is a vector-space isomorphism in every degree and therefore a graded algebra isomorphism.

Steps 2.1 and 2.2 prove the ordered-word basis; its words are indexed by subsets of the $n$ basis vectors, and partitioning those subsets by size gives ${\sum }_{r=0}^n\left(\genfrac{}{}{0px}{}{n}{r}\right)=2^n$. Steps 4.1 and 5.1 prove the associated-graded assertion, including the boundary case $n=0$.