Is the $\mathbb{Z}$-grading of a Clifford algebra basis independent?

Question

Let $V$ be a finite dimensional vector space over a field $K$ of characteristic $\neq 2$, and let $q \colon V \to K$ be a quadratic form. One of the first things to show when learning the theory of Clifford algebras is that any orthogonal basis $\{ e_1, \dots, e_n \}$ of $V$ gives rise to a basis $\{ e_{i_1} \dots e_{i_k} : 1 \leq i_1 < i_2 < \dots < i_k \leq n \}$ of $\operatorname{C\ell}(V,q)$ as a $K$-vector space. In that way, we obtain a direct sum decomposition $$ \operatorname{C\ell}(V,q) = C_0 \oplus \dots \oplus C_n, $$ where each $C_k$ is generated by the $k$-products $e_{i_1} \dots e_{i_k}$. It is often regarded as a grading, although it is not a grading of a $K$-algebra in the strict sense.

Question 1: By looking at simple examples, it appears to me that this grading is independent of the orthogonal basis of $V$ chosen. Is this true?

Question 2: The Wikipedia article on Clifford algebras suggests that the map $\operatorname{C\ell}(V,q) \to K$ sending each element to its $C_0$-part is basis independent (of course this would be implied by a positive answer to the first question). Is this true? Is there some kind of basis independent formula?

Thank you in advance!

Answer 1

The answer to question 1 is yes. To see this, let $\beta=\{e_1,\ldots,e_n\}$ be an orthogonal basis for $V$ and $$Cl(V,q)=C_0\oplus\cdots\oplus C_n$$ be the vector space grading with respect to this basis.

Now, let $\gamma=\{v_1,\ldots,v_n\}$ be any orthogonal basis for $V$, and $A=(a_{ij})$ be the transition matrix $$v_j=\sum_ia_{ij}e_i.$$ It is enough to show that, for $i_1<\cdots<i_k$, $v_{i_1}\cdots v_{i_k}\subset C_k\backslash\{0\}$, since by dimension considerations $$\mathrm{span}\{v_{i_1}\cdots v_{i_k}\mid i_1<\cdots<i_k\}=C_k.$$ Indeed, one can show by induction that $$ v_{i_1}\cdots v_{i_k}=\sum_{r_1<r_2<\cdots<r_k}\left(\sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)a_{r_1i_{\sigma(1)}}a_{r_2i_{\sigma(2)}}\cdots a_{r_ki_{\sigma(k)}}\right)e_{r_1}e_{r_2}\cdots e_{r_k}. $$

I'll work out the $k=2$ case: Observe that, for $i<j$, $$v_iv_j=\left(\sum_k a_{ki}e_k\right)\left(\sum_\ell a_{\ell j}e_\ell\right) =\underbrace{\sum_k a_{ki}a_{kj}e_k^2}_{\in C_0} \;+\; \underbrace{\sum_{k<\ell}(a_{ki}a_{\ell j}-a_{kj}a_{\ell i})e_k e_\ell}_{\in C_2}.$$ There is no degree 1 component on the right hand side of this equation, while the degree 0 component is $$ \sum_k a_{k i}a_{k j} e_k^2=\sum_k a_{k i}a_{k j}q(e_k)=\langle v_i,v_j\rangle=0. $$ Since $v_iv_j\neq 0$, we have $v_iv_j\in C_2$.

EDIT: The proof is not too hard, but I was a bit lazy to typeset it yesterday.

Assume the formula for $v_{i_1}\cdots v_{i_k}$ holds by induction and rewrite $$v_{i_1}\cdots v_{i_k}=\sum_{\substack{r_1,\ldots,r_k\\r_i\neq r_j\;\forall i\neq j}}a_{r_1 i_1}a_{r_2 i_2}\cdots a_{r_k i_k}e_{r_1}e_{r_2}\cdots e_{r_k}.$$ Then, \begin{align} v_{i_1}\cdots v_{i_k}v_j=&\left(\sum_{\substack{r_1,\ldots,r_k\\r_i\neq r_j,\;\forall i\neq j}}a_{r_1 i_1}a_{r_2 i_2}\cdots a_{r_k i_k}e_{r_1}e_{r_2}\cdots e_{r_k}\right)\left(\sum_s a_{sj}e_s\right)\\ =&\sum_{t=0}^{k-1}(-1)^{k-t}\sum_s a_{s i_t}a_{sj}q(e_s)\sum_{\substack{r_1,\ldots,\widehat{r_t},\ldots,r_k\\r_i\neq r_j,\;\forall i\neq j\\r_i\neq s,\;\forall i\neq t}}a_{r_1 i_1}a_{r_2 i_2}\cdots \widehat{a_{r_t i_t}}\cdots a_{r_k i_k}e_{r_1}e_{r_2}\cdots \widehat{e_{r_t}}\cdots e_{r_k}\\ &+\sum_{\substack{r_1,\ldots,r_k,s\\r_i\neq r_j,\;\forall i\neq j\\r_i\neq s,\;\forall i}}a_{r_1 i_1}a_{r_2 i_2}\cdots a_{r_k i_k}a_{sj}e_{r_1}e_{r_2}\cdots e_{r_k}e_s \end{align} where the hat means omit that term from the product/summation. As in the degree 2 case $$ \sum_s a_{s i_t}a_{sj}q(e_s)=\langle v_{i_t}, v_j\rangle=0 $$ so, setting $j=i_{k+1}$ and $s=r_{k+1}$ \begin{align} v_{i_1}\cdots v_{i_k}v_{i_{k+1}}=&\sum_{\substack{r_1,\ldots,r_k,r_{k+1}\\r_i\neq r_j,\;\forall i\neq j}}a_{r_1 i_1}a_{r_2 i_2}\cdots a_{r_k i_k}a_{r_{k+1}j}e_{r_1}e_{r_2}\cdots e_{r_k}e_{r_{k+1}}\\ =&\sum_{r_1<r_2<\cdots<r_{k+1}}\left(\sum_{\sigma\in S_{k+1}}\mathrm{sgn}(\sigma)a_{r_1i_{\sigma(1)}}a_{r_2i_{\sigma(2)}}\cdots a_{r_{k+1}i_{\sigma(k+1)}}\right)e_{r_1}e_{r_2}\cdots e_{r_{k+1}}. \end{align}

Answer 2

This is an answer to the second question (can't believe I haven't seen this immediately). David's answer already implies that the map $\chi \colon \operatorname{C \ell}(V,q) \to K$ sending an element to its scalar part is well defined (i.e., independent of the orthogonal basis of $V$ chosen). But in fact, this map is even independent of the embedding $V \hookrightarrow \operatorname{C \ell}(V,q)$:

By choosing a standard basis of $\operatorname{C \ell}(V,q)$ given by an orthogonal basis of $V$, you can easily check that we have $\chi(x) = 2^{- \dim(V)} \cdot \operatorname{Tr}(x)$ for all $x \in \operatorname{C \ell}(V,q)$, where the trace is taken with respect to the standard action of $\operatorname{C \ell}(V,q)$ on itself by left multiplications (this is well defined since we are assuming $\operatorname{char}(K) \neq 2$).