PBW basis and exterior associated graded for a Clifford algebra

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 $n=0$.