SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\psi : \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} \left(V _ {1}, \mathfrak {q} _ {1}\right) \hat {\otimes} \operatorname {C l} \left(V _ {2}, \mathfrak {q} _ {2}\right), \tag {125}
$$
$$
v _ {i _ {1}} \dots v _ {i _ {k}} \mapsto \left(\psi_ {1} \left(v _ {i _ {1}, 1}\right) + \psi_ {2} \left(v _ {i _ {1}, 2}\right)\right) \dots \left(\psi_ {1} \left(v _ {i _ {k}, 1}\right) + \psi_ {2} \left(v _ {i _ {k}, 2}\right)\right). \tag {126}
$$
One can see from this, by explicit calculation, that  $\psi$  also respects the  $\mathbb{Z} / 2\mathbb{Z}$ -grading. Furthermore, we see that  $\psi \circ \phi_l = \psi_l$  for  $l = 1, 2$ , and, also,  $\phi \circ \psi_l = \phi_l$  for  $l = 1, 2$ .
One easily sees that  $\phi$  and  $\psi$  are inverse to each other and the claim follows.
Theorem D.15 (The dimension of the Clifford algebra). Let  $(V, \mathfrak{q})$  be a finite dimensional quadratic vector space over  $\mathbb{F}$ ,  $\operatorname{char}(\mathbb{F}) \neq 2$ ,  $n \coloneqq \dim V < \infty$ . Then we have:
$$
\dim \operatorname {C l} (V, \mathfrak {q}) = 2 ^ {n}. \tag {127}
$$
Proof. Let  $e_1, \ldots, e_n$  be an orthogonal basis of  $(V, \mathfrak{q})$  and  $V_i := \operatorname{span}(e_i) \subseteq V$ ,  $\mathfrak{q}_i := \mathfrak{q}|_{V_i}$ . Then we get the orthogonal sum decomposition:
$$
(V, \mathfrak {q}) = \left(V _ {1}, \mathfrak {q} _ {1}\right) \oplus \dots \oplus \left(V _ {n}, \mathfrak {q} _ {n}\right), \tag {128}
$$
and thus by Proposition D.14 the isomorphism of  $\mathbb{Z} / 2\mathbb{Z}$ -graded algebras:
$$
\operatorname {C l} (V, \mathfrak {q}) \cong \operatorname {C l} \left(V _ {1}, \mathfrak {q} _ {1}\right) \hat {\otimes} \dots \hat {\otimes} \operatorname {C l} \left(V _ {n}, \mathfrak {q} _ {n}\right), \tag {129}
$$
and thus:
$$
\dim \operatorname {C l} (V, \mathfrak {q}) = \dim \operatorname {C l} \left(V _ {1}, \mathfrak {q} _ {1}\right) \dots \dim \operatorname {C l} \left(V _ {n}, \mathfrak {q} _ {n}\right), \tag {130}
$$
Since  $\dim V_{i} = 1$  and thus:
$$
\dim \operatorname {C l} \left(V _ {i}, \mathfrak {q} _ {i}\right) = \dim \left(\mathbb {F} \oplus V _ {i}\right) = 2, \tag {131}
$$
for  $i = 1, \dots, n$ , we get the claim:
$$
\dim \operatorname {C l} (V, \mathfrak {q}) = 2 ^ {n}. \tag {132}
$$
Corollary D.16 (Bases for the Clifford algebra). Let  $(V, \mathfrak{q})$  be a finite dimensional quadratic vector space over  $\mathbb{F}$ ,  $\operatorname{char}(\mathbb{F}) \neq 2$ ,  $n \coloneqq \dim V < \infty$ . Let  $e_1, \ldots, e_n$  be any basis of  $V$ , then  $(e_A)_{A \subseteq [n]}$  is a basis for  $\operatorname{Cl}(V, \mathfrak{q})$ , where we put for a subset  $A \subseteq [n] \coloneqq \{1, \ldots, n\}$ :
$$
e _ {A} := \prod_ {i \in A} ^ {<  } e _ {i}, \quad e _ {\emptyset} := 1 _ {\operatorname {C l} (V, \mathfrak {q})}. \tag {133}
$$
where the product is taken in increasing order of the indices  $i \in A$ .
Proof. Since  $\operatorname{Cl}(V, \mathfrak{q}) = \operatorname{T}(V) / I(\mathfrak{q})$  and
$$
\mathrm {T} (V) = \operatorname {s p a n} \left\{e _ {i _ {1}} \otimes \dots \otimes e _ {i _ {m}} \mid m \geq 0, i _ {j} \in [ n ], j \in [ m ] \right\}, \tag {134}
$$
we see that:
$$
\operatorname {C l} (V, \mathfrak {q}) = \operatorname {s p a n} \left\{e _ {i _ {1}} \dots e _ {i _ {m}} \mid m \geq 0, i _ {j} \in [ n ], j \in [ m ] \right\}. \tag {135}
$$
By several applications of the fundamental identities D.3:
$$
e _ {i _ {k}} e _ {i _ {l}} = - e _ {i _ {l}} e _ {i _ {k}} + 2 \mathfrak {b} \left(e _ {i _ {k}}, e _ {i _ {l}}\right), \quad e _ {i _ {k}} e _ {i _ {k}} = \mathfrak {q} \left(e _ {i _ {k}}\right), \tag {136}
$$
we can turn products  $e_{i_1} \cdots e_{i_m}$  into sums of smaller products if some of the occurring indices agree, say  $i_k = i_l$ . Furthermore, we can also use those identities to turn the indices in increasing order. This then shows:
$$
\operatorname {C l} (V, \mathfrak {q}) = \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ] \right\}. \tag {137}
$$
Since  $\# \{A\subseteq [n]\} = 2^n$  and  $\dim \operatorname {Cl}(V,\mathfrak{q}) = 2^n$  by Theorem D.15 we see that  $\{e_A\mid A\subseteq [n]\}$  must already be a basis for  $\operatorname {Cl}(V,\mathfrak{q})$
D.5 Extending the Quadratic Form to the Clifford Algebra
We provide the extension of the quadratic form from the vector space to the Clifford algebra, which will lead to the construction of an orthogonal basis of the Clifford algebra.
Definition D.17 (The opposite algebra of an algebra). Let  $(\mathcal{A}, +, \cdot)$  be an algebra. The opposite algebra  $(\mathcal{A}^{\mathrm{op}}, +, \bullet)$  is defined to consist of the same underlying vector space  $(\mathcal{A}^{\mathrm{op}}, +) = (\mathcal{A}, +)$ , but where the multiplication is reversed in comparison to  $\mathcal{A}$ , i.e. for  $x, y \in \mathcal{A}^{\mathrm{op}}$  we have:
$$
x \cdot y := y \cdot x. \tag {138}
$$
Note that this really turns  $(\mathcal{A}^{\mathrm{op}}, + ,\bullet)$  into an algebra.
Definition D.18 (The main anti-involution of the Clifford algebra). Consider the following linear map:
$$
\beta : V \rightarrow \operatorname {C l} (V, \mathfrak {q}) ^ {\mathrm {o p}}, \quad v \mapsto v, \tag {139}
$$
which also satisfies  $v \cdot v = vv = \mathfrak{q}(v)$ . By the universal property of the Clifford algebra we get a unique extension to an algebra homomorphism:
$$
\begin{array}{l} \beta : \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}) ^ {\mathrm {o p}}, \quad \beta \left(c _ {0} + \sum_ {i \in I} c _ {i} \cdot v _ {i, 1} \dots v _ {i, k _ {i}}\right) (140) \\ = c _ {0} + \sum_ {i \in I} c _ {i} \cdot v _ {i, 1} \bullet \dots \bullet v _ {i, k _ {i}} (141) \\ = c _ {0} + \sum_ {i \in I} c _ {i} \cdot v _ {i, k _ {i}} \dots v _ {i, 1}, (142) \\ \end{array}
$$
for any finite sum representation with  $v_{i,j} \in V$  and  $c_i \in \mathbb{F}$ . We call  $\beta$  the main anti-involution of  $\operatorname{Cl}(V, \mathfrak{q})$ .
Definition D.19 (The combined anti-involution of the Clifford algebra). The combined anti-involution or Clifford conjugation of  $\operatorname{Cl}(V, \mathfrak{q})$  is defined to be the  $\mathbb{F}$ -algebra homomorphism:
$$
\gamma : \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}) ^ {\mathrm {o p}}, \quad \gamma (x) := \beta (\alpha (x)). \tag {143}
$$
More explicitly, it is given by the formula:
$$
\gamma \left(c _ {0} + \sum_ {i \in I} c _ {i} \cdot v _ {i, 1} \dots v _ {i, k _ {i}}\right) = c _ {0} + \sum_ {i \in I} (- 1) ^ {k _ {i}} \cdot c _ {i} \cdot v _ {i, k _ {i}} \dots v _ {i, 1}, \tag {144}
$$
for any finite sum representation with  $v_{i,j} \in V$  and  $c_i \in \mathbb{F}$ .
Remark D.20. If  $e_1, \ldots, e_n \in V$  is an orthogonal basis for  $(V, \mathfrak{q})$ . Then we have for  $A \subseteq [n]$ :
$$
\alpha \left(e _ {A}\right) = (- 1) ^ {| A |} e _ {A}, \quad \beta \left(e _ {A}\right) = (- 1) ^ {\binom {| A |} {2}} e _ {A}, \quad \gamma \left(e _ {A}\right) = (- 1) ^ {\binom {| A | + 1} {2}} e _ {A}. \tag {145}
$$
Recall the definition of a trace of a linear map:
Definition D.21 (The trace of an endomorphism). Let  $\mathcal{Y}$  be a vector space over a field  $\mathbb{F}$  of dimension  $\dim \mathcal{Y} = m < \infty$  and  $\Phi : \mathcal{Y} \to \mathcal{Y}$  a vector space endomorphism<sup>14</sup>. Let  $B = \{b_{1},\ldots,b_{m}\}$  be a basis for  $\mathcal{Y}$  and  $B^{*} = \{b_{1}^{*},\ldots,b_{m}^{*}\}$  be the corresponding dual basis of  $\mathcal{Y}^{*}$ , defined via:  $b_{j}^{*}(b_{i}) := \delta_{i,j}$ . Let  $A = (a_{i,j})_{i=1,\ldots,m, j=1,\ldots,m}$  be the matrix representation of  $\Phi$  w.r.t.  $B$ :
$$
\forall j \in [ m ]. \quad \Phi (b _ {j}) = \sum_ {i = 1} ^ {m} a _ {i, j} b _ {i}. \tag {146}
$$
Then the trace of  $\Phi$  is defined via:
$$
\operatorname {T r} (\Phi) := \sum_ {j = 1} ^ {m} a _ {j, j} = \sum_ {j = 1} ^ {m} b _ {j} ^ {*} (\Phi (b _ {j})) \in \mathbb {F}. \tag {147}
$$
It is a well known fact that  $\mathrm{Tr}(\Phi)$  is not dependent on the initial choice of the basis  $B$ . Furthermore,  $\mathrm{Tr}$  is a well-defined  $\mathbb{F}$ -linear map (homomorphism of vector spaces):
$$
\operatorname {T r}: \operatorname {E n d} _ {\mathbb {F}} (\mathcal {Y}) \rightarrow \mathbb {F}, \quad \Phi \mapsto \operatorname {T r} (\Phi). \tag {148}
$$
We now want to define the projection of  $x \in \mathrm{Cl}(V, \mathfrak{q})$  onto its zero component  $x^{(0)} \in \mathbb{F}$  in a basis independent way.
Definition D.22 (The projection onto the zero component). We define the  $\mathbb{F}$ -linear map:
$$
\zeta : \operatorname {C l} (V, \mathfrak {q}) \rightarrow \mathbb {F}, \quad \zeta (x) := 2 ^ {- n} \operatorname {T r} (x), \tag {149}
$$
where  $n \coloneqq \dim V$  and  $\operatorname{Tr}(x) \coloneqq \operatorname{Tr}(L_x)$ , where  $L_x$  is the endomorphism of  $\mathrm{Cl}(V, \mathfrak{q})$  given by left multiplication with  $x$ :
$$
L _ {x}: \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad y \mapsto L _ {x} (y) := x y. \tag {150}
$$
We call  $\zeta$  the projection onto the zero component. We also often write for  $x\in \operatorname {Cl}(V,\mathfrak{q})$  ..
$$
x ^ {(0)} := \zeta (x). \tag {151}
$$
The name is justified by following property:
Lemma D.23. Let  $e_1, \ldots, e_n$  be a fixed orthogonal basis of  $(V, \mathfrak{q})$ . Then we know that  $(e_A)_{A \subseteq [n]}$  is a basis for  $\operatorname{Cl}(V, \mathfrak{q})$ . So we can write every  $x \in \operatorname{Cl}(V, \mathfrak{q})$  as:
$$
x = \sum_ {A \subseteq [ n ]} x _ {A} \cdot e _ {A}, \tag {152}
$$
with  $x_A \in \mathbb{F}$ ,  $A \subseteq [n]$ . The claim is now that we have:
$$
\zeta (x) \stackrel {!} {=} x _ {\emptyset}. \tag {153}
$$
Proof. By the linearity of the trace we only need to investigate  $\operatorname{Tr}(e_A)$  for  $A \subseteq [n]$ . For  $A = \emptyset$ , we have  $e_{\emptyset} = 1$  and we get:
$$
\operatorname {T r} (1) = \sum_ {B \subseteq [ n ]} e _ {B} ^ {*} \left(1 \cdot e _ {B}\right) = \sum_ {B \subseteq [ n ]} 1 = 2 ^ {n}, \tag {154}
$$
$^{14}$ A map from a mathematical object space to itself.
which shows:  $\zeta(1) = 2^{-n} \operatorname{Tr}(1) = 1$ . Now, consider  $A \subseteq [n]$  with  $A \neq \emptyset$ . Let  $\triangleleft$  denote the symmetric difference of two sets. Further, we can write  $\pm$  to refrain from distinguishing between signs, which will not affect the result.
$$
\begin{array}{l} \operatorname {T r} \left(e _ {A}\right) = \sum_ {B \subseteq [ n ]} e _ {B} ^ {*} \left(e _ {A} e _ {B}\right) (155) \\ = \sum_ {B \subseteq [ n ]} \pm \prod_ {i \in A \cap B} \mathfrak {q} \left(e _ {i}\right) \cdot e _ {B} ^ {*} \left(e _ {A \bigtriangleup B}\right) (156) \\ = \sum_ {B \subseteq [ n ]} \pm \prod_ {i \in A \cap B} \mathfrak {q} \left(e _ {i}\right) \cdot \delta_ {B, A \Delta B} (157) \\ = \sum_ {B \subseteq [ n ]} \pm \prod_ {i \in A \cap B} \mathfrak {q} \left(e _ {i}\right) \cdot \delta_ {\emptyset , A} (158) \\ = 0, (159) \\ \end{array}
$$
where the third equality follows from the fact that  $B = A \triangle B$  holds if and only if  $A = \emptyset$ , regardless of  $B$ . However,  $A = \emptyset$  was ruled out by assumption, so then in the last equality we always have  $\delta_{\emptyset, A} = 0$ . So for  $A \neq \emptyset$  we have:  $\zeta(e_A) = 2^{-n} \operatorname{Tr}(e_A) = 0$ . Altogether we get:
$$
\zeta \left(e _ {A}\right) = \delta_ {A, \emptyset} = \left\{ \begin{array}{l l} 1, & \text {i f} A = \emptyset , \\ 0, & \text {e l s e .} \end{array} \right. \tag {160}
$$
With this and linearity we get:
$$
\zeta (x) = \zeta \left(\sum_ {A \subseteq [ n ]} x _ {A} \cdot e _ {A}\right) = \sum_ {A \subseteq [ n ]} x _ {A} \cdot \zeta \left(e _ {A}\right) = \sum_ {A \subseteq [ n ]} x _ {A} \cdot \delta_ {A, \emptyset} = x _ {\emptyset}. \tag {161}
$$
This shows the claim.
Definition D.24 (The bilinear form on the Clifford algebra). For our quadratic  $\mathbb{F}$ -vector space  $(V, \mathfrak{q})$  with corresponding bilinear form  $\mathfrak{b}$  and corresponding Clifford algebra  $\operatorname{Cl}(V, \mathfrak{q})$  we now define the following  $\mathbb{F}$ -bilinear form on  $\operatorname{Cl}(V, \mathfrak{q})$ :
$$
\bar {\mathfrak {b}}: \operatorname {C l} (V, \mathfrak {q}) \times \operatorname {C l} (V, \mathfrak {q}) \rightarrow \mathbb {F}, \quad \bar {\mathfrak {b}} (x, y) := \zeta (\beta (x) y). \tag {162}
$$
We also define the corresponding quadratic form on  $\operatorname{Cl}(V, \mathfrak{q})$  via:
$$
\bar {\mathfrak {q}}: \operatorname {C l} (V, \mathfrak {q}) \rightarrow \mathbb {F}, \quad \bar {\mathfrak {q}} (x) := \bar {\mathbf {b}} (x, x) = \zeta (\beta (x) x). \tag {163}
$$
We will see below that  $\bar{\mathfrak{b}}$  and  $\bar{\mathfrak{q}}$  will agree with  $\mathfrak{b}$  and  $\mathfrak{q}$ , resp., when they are restricted to  $V$ . From that point on we will denote  $\bar{\mathfrak{b}}$  just by  $\mathfrak{b}$ , and  $\bar{\mathfrak{q}}$  with  $\mathfrak{q}$ , resp., without (much) ambiguity.
Lemma D.25. For  $v, w \in V$  we have:
$$
\bar {\mathfrak {b}} (v, w) = \mathfrak {b} (v, w). \tag {164}
$$
Proof. We pick an orthogonal basis  $e_1, \ldots, e_n$  for  $V$  and write:
$$
v = \sum_ {i = 1} ^ {n} a _ {i} \cdot e _ {i}, \quad w = \sum_ {j = 1} ^ {n} c _ {j} \cdot e _ {j}. \tag {165}
$$
We then get by linearity:
$$
\begin{array}{l} \bar {\mathfrak {h}} (v, w) = \sum_ {i = 1} ^ {n} \sum_ {j = 1} ^ {n} a _ {i} c _ {j} \cdot \bar {\mathfrak {h}} \left(e _ {i}, e _ {j}\right) (166) \\ = \sum_ {i = 1} ^ {n} \sum_ {j = 1} ^ {n} a _ {i} c _ {j} \cdot \zeta \left(\beta \left(e _ {i}\right) e _ {j}\right) (167) \\ = \sum_ {i = 1} ^ {n} \sum_ {j = 1} ^ {n} a _ {i} c _ {j} \cdot \zeta \left(e _ {i} e _ {j}\right) (168) \\ = \sum_ {i \neq j} a _ {i} c _ {j} \cdot \underbrace {\zeta \left(e _ {i} e _ {j}\right)} _ {= 0} + \sum_ {i = j} a _ {i} c _ {j} \cdot \zeta \left(e _ {i} e _ {j}\right) (169) \\ = \sum_ {i = 1} ^ {n} a _ {i} c _ {i} \cdot \mathfrak {q} \left(e _ {i}\right) \cdot \underbrace {\zeta (1)} _ {= 1} (170) \\ = \sum_ {i = 1} ^ {n} \sum_ {j = 1} ^ {n} a _ {i} c _ {j} \cdot \overbrace {\mathfrak {b} \left(e _ {i} , e _ {j}\right)} ^ {\mathfrak {q} \left(e _ {i}\right) \cdot \delta_ {i, j} =} (171) \\ = \mathfrak {b} \left(\sum_ {i = 1} ^ {n} a _ {i} \cdot e _ {i}, \sum_ {j = 1} ^ {n} c _ {j} \cdot e _ {j}\right) (172) \\ = \mathfrak {b} (v, w). (173) \\ \end{array}
$$
Theorem D.26. Let  $e_1, \ldots, e_n$  be an orthogonal basis for  $(V, \mathfrak{q})$  then  $(e_A)_{A \subseteq [n]}$  is an orthogonal basis for  $\operatorname{Cl}(V, \mathfrak{q})$  w.r.t. the induced bilinear form  $\bar{\mathfrak{b}}$ . Furthermore, for  $x, y \in \operatorname{Cl}(V, \mathfrak{q})$  of the form:
$$
x = \sum_ {A \subseteq [ n ]} x _ {A} \cdot e _ {A}, \quad y = \sum_ {A \subseteq [ n ]} y _ {A} \cdot e _ {A}, \tag {174}
$$
with  $x_A, y_A \in \mathbb{F}$  we get:
$$
\bar {\mathfrak {b}} (x, y) = \sum_ {A \subseteq [ n ]} x _ {A} \cdot y _ {A} \cdot \prod_ {i \in A} \mathfrak {q} \left(e _ {i}\right), \quad \bar {\mathfrak {q}} (x) = \sum_ {A \subseteq [ n ]} x _ {A} ^ {2} \cdot \prod_ {i \in A} \mathfrak {q} \left(e _ {i}\right). \tag {175}
$$
Note that:  $\bar{\mathfrak{q}} (e_{\emptyset}) = \bar{\mathfrak{q}} (1) = 1$
Proof. We already know that  $(e_A)_{A\subseteq [n]}$  is a basis for  $\operatorname {Cl}(V,\mathfrak{q})$  . So we only need to check the orthogonality condition. First note that for  $e_C = e_{i_1}\dots e_{i_r}$  we get:
$$
\bar {\mathfrak {q}} \left(e _ {C}\right) = \zeta \left(\beta \left(e _ {C}\right) e _ {C}\right) = \zeta \left(e _ {i _ {r}} \dots e _ {i _ {1}} \cdot e _ {i _ {1}} \dots e _ {i _ {r}}\right) = \mathfrak {q} \left(e _ {i _ {1}}\right) \dots \mathfrak {q} \left(e _ {i _ {r}}\right). \tag {176}
$$
Now let  $A, B \subseteq [n]$  with  $A \neq B$ , i.e. with  $A \triangle B \neq \emptyset$ . We then get:
$$
\bar {\mathfrak {b}} \left(e _ {A}, e _ {B}\right) = \zeta \left(\beta \left(e _ {A}\right) e _ {B}\right) = \pm \prod_ {i \in A \cap B} \mathfrak {q} \left(e _ {i}\right) \cdot \zeta \left(e _ {A \bigtriangleup B}\right) = 0. \tag {177}
$$
This shows that  $(e_A)_{A\subseteq [n]}$  is an orthogonal basis for  $\operatorname {Cl}(V,\mathfrak{q})$  . For  $x,y\in \operatorname {Cl}(V,\mathfrak{q})$  from above we get:
$$
\begin{array}{l} \bar {\mathfrak {b}} (x, y) = \bar {\mathfrak {b}} \left(\sum_ {A \subseteq [ n ]} x _ {A} \cdot e _ {A}, \sum_ {B \subseteq [ n ]} y _ {B} \cdot e _ {B}\right) (178) \\ = \sum_ {A \subseteq [ n ]} \sum_ {B \subseteq [ n ]} x _ {A} \cdot y _ {B} \cdot \bar {\mathfrak {b}} \left(e _ {A}, e _ {B}\right) (179) \\ = \sum_ {A \subseteq [ n ]} \sum_ {B \subseteq [ n ]} x _ {A} \cdot y _ {B} \cdot \bar {\mathfrak {q}} (e _ {A}) \cdot \delta_ {A, B} (180) \\ = \sum_ {A \subseteq [ n ]} x _ {A} \cdot y _ {A} \cdot \bar {\mathfrak {q}} (e _ {A}) (181) \\ = \sum_ {A \subseteq [ n ]} x _ {A} \cdot y _ {A} \cdot \prod_ {i \in A} \mathfrak {q} \left(e _ {i}\right). (182) \\ \end{array}
$$
D.6 The Multivector Grading
Now that we have an orthogonal basis for the algebra, we show that the Clifford algebra allows a vector space grading that is independent of the chosen orthogonal basis.
Let  $(V, \mathfrak{q})$  be a quadratic space over a field  $\mathbb{F}$  with  $\mathrm{char}(\mathbb{F}) \neq 2$  and  $\dim V = n < \infty$ .
In the following we present a technical proof that works for fields  $\mathbb{F}$  with  $\mathrm{char}(\mathbb{F})\neq 2$ . An alternative, simpler and more structured proof, but for the more restrictive case of  $\mathrm{char}(\mathbb{F}) = 0$ , can be found in Theorem D.37 later.
Theorem D.27 (The multivector grading of the Clifford algebra). Let  $e_1, \ldots, e_n$  be an orthogonal basis of  $(V, \mathfrak{q})$ . Then for every  $m = 0, \ldots, n$  we define the following sub-vector space of  $\operatorname{Cl}(V, \mathfrak{q})$ :
$$
\begin{array}{l} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}) := \operatorname {s p a n} \left\{e _ {i _ {1}} \dots e _ {i _ {m}} \mid 1 \leq i _ {1} <   \dots <   i _ {m} \leq n \right\} (183) \\ = \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ], | A | = m \right\}, (184) \\ \end{array}
$$
where  $\operatorname{Cl}^{(0)}(V, \mathfrak{q}) \coloneqq \mathbb{F}$ .
Then the sub-vector spaces  $\mathrm{Cl}^{(m)}(V,\mathfrak{q})$ ,  $m = 0,\ldots ,n$ , are independent of the choice of the orthogonal basis, i.e. if  $b_{1},\ldots ,b_{n}$  is another orthogonal basis of  $(V,\mathfrak{q})$ , then:
$$
\operatorname {C l} ^ {(m)} (V, \mathfrak {q}) = \operatorname {s p a n} \left\{b _ {i _ {1}} \dots b _ {i _ {m}} \mid 1 \leq i _ {1} <   \dots <   i _ {m} \leq n \right\}. \tag {185}
$$
Proof. First note that by the orthogonality and the fundamental relation of the Clifford algebra we have for all  $i \neq j$ :
$$
e _ {i} e _ {j} = - e _ {j} e _ {i}, \quad b _ {i} b _ {j} = - b _ {j} b _ {i}. \tag {186}
$$
We now abbreviate:
$$
B ^ {(m)} := \operatorname {s p a n} \left\{b _ {j _ {1}} \dots b _ {j _ {m}} \mid 1 \leq j _ {1} <   \dots <   j _ {m} \leq n \right\}, \tag {187}
$$
and note that:
$$
\begin{array}{l} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}) = \operatorname {s p a n} \left\{e _ {i _ {1}} \dots e _ {i _ {m}} \mid 1 \leq i _ {1} <   \dots <   i _ {m} \leq n \right\} (188) \\ = \operatorname {s p a n} \left\{e _ {i _ {1}} \dots e _ {i _ {m}} \mid 1 \leq i _ {1}, \dots , i _ {m} \leq n, \forall s \neq t. i _ {s} \neq i _ {t} \right\}. (189) \\ \end{array}
$$
We want to show that for  $1 \leq j_1 < \dots < j_m \leq n$  we have that:
$$
b _ {j _ {1}} \dots b _ {j _ {m}} \in \operatorname {C l} ^ {(m)} (V, \mathfrak {q}). \tag {190}
$$
Since we have two bases we can write an orthogonal change of basis:
$$
b _ {j} = \sum_ {i = 1} ^ {n} a _ {i, j} e _ {i} \in V = \operatorname {C l} ^ {(1)} (V, \mathfrak {q}). \tag {191}
$$
Using this, we can now write the above product as the sum of two terms:
$$
\begin{array}{l} b _ {j _ {1}} \dots b _ {j _ {m}} = \sum_ {i _ {1}, \dots , i _ {m}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} \cdot e _ {i _ {1}} \dots e _ {i _ {m}} (192) \\ = \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall s \neq t. i _ {s} \neq i _ {t}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} \cdot e _ {i _ {1}} \dots e _ {i _ {m}} + \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \exists s \neq t. i _ {s} = i _ {t}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} \cdot e _ {i _ {1}} \dots e _ {i _ {m}}. (193) \\ \end{array}
$$
Our claim is equivalent to the vanishing of the second term. Note that the above equation for  $b_{j}$  already shows the claim for  $m = 1$ . The case  $m = 0$  is trivial.
We now prove the claim for  $m = 2$  by hand before doing induction after. Recall that  $j_{1}\neq j_{2}$  ..
$$
\begin{array}{l} b _ {j _ {1}} b _ {j _ {2}} = \sum_ {\substack {i _ {1}, i _ {2} \\ i _ {1} \neq i _ {2}}} a _ {i _ {1}, j _ {1}} a _ {i _ {2}, j _ {2}} \cdot e _ {i _ {1}} e _ {i _ {2}} + \sum_ {i = 1} ^ {n} a _ {i, j _ {1}} a _ {i, j _ {2}} \cdot e _ {i} e _ {i} (194) \\ = \sum_ {\substack {i _ {1}, i _ {2} \\ i _ {1} \neq i _ {2}}} a _ {i _ {1}, j _ {1}} a _ {i _ {2}, j _ {2}} \cdot e _ {i _ {1}} e _ {i _ {2}} + \sum_ {i = 1} ^ {n} a _ {i, j _ {1}} \underbrace {a _ {i , j _ {2}} \cdot \mathfrak {q} (e _ {i})} _ {= \mathfrak {b} (e _ {i}, b _ {j _ {2}})} (195) \\ = \sum_ {\substack {i _ {1}, i _ {2} \\ i _ {1} \neq i _ {2}}} a _ {i _ {1}, j _ {1}} a _ {i _ {2}, j _ {2}} \cdot e _ {i _ {1}} e _ {i _ {2}} + \underbrace {\mathfrak {b} \left(b _ {j _ {1}} , b _ {j _ {2}}\right)} _ {= 0} (196) \\ = \sum_ {\substack {i _ {1}, i _ {2} \\ i _ {1} \neq i _ {2}}} a _ {i _ {1}, j _ {1}} a _ {i _ {2}, j _ {2}} \cdot e _ {i _ {1}} e _ {i _ {2}} (197) \\ \in \operatorname {C l} ^ {(2)} (V, \mathfrak {q}). (198) \\ \end{array}
$$
This shows the claim for  $m = 2$ .
By way of induction we now assume that we have shown the claim until some  $m \geq 2$ , i.e. we have:
$$
b _ {j _ {1}} \dots b _ {j _ {m}} = \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l. i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} e _ {i _ {1}} \dots e _ {i _ {m}} \in \operatorname {C l} ^ {(m)} (V, \mathfrak {q}). \tag{199}
$$
Now consider another  $b_{j}$  with  $j \coloneqq j_{m + 1} \neq j_{k}, k = 1, \ldots, m$ . We then get:
$$
\begin{array}{l} b _ {j _ {1}} \dots b _ {j _ {m}} b _ {j} = \left(\sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l, i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} e _ {i _ {1}} \dots e _ {i _ {m}}\right) \left(\sum_ {i = 1} ^ {n} a _ {i, j} e _ {i}\right) (200) \\ = \sum_ {i = 1} ^ {n} \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l, i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} a _ {i, j} e _ {i _ {1}} \dots e _ {i _ {m}} e _ {i} (201) \\ = \sum_ {i = 1} ^ {n} \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l, i _ {k} \neq i _ {l} \\ i \notin \{i _ {1}, \dots , i _ {m} \}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} a _ {i, j} e _ {i _ {1}} \dots e _ {i _ {m}} e _ {i} (202) \\ + \sum_ {i = 1} ^ {n} \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l, i _ {k} \neq i _ {l} \\ i \in \{i _ {1}, \dots , i _ {m} \}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} a _ {i, j} e _ {i _ {1}} \dots e _ {i _ {m}} e _ {i} (203) \\ = \sum_ {\substack {i _ {1}, \dots , i _ {m}, i _ {m + 1} \\ \forall k \neq l. i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} a _ {i _ {m + 1}, j _ {m + 1}} e _ {i _ {1}} \dots e _ {i _ {m}} e _ {i _ {m + 1}} (204) \\ + \sum_ {i = 1} ^ {n} \sum_ {s = 1} ^ {m} \sum_ {\substack {i _ {1}, \dots , i _ {m} \\ \forall k \neq l, i _ {k} \neq i _ {l} \\ i _ {s} = i}} a _ {i _ {1}, j _ {1}} \dots a _ {i _ {m}, j _ {m}} a _ {i, j} e _ {i _ {1}} \dots e _ {i _ {m}} e _ {i}. (205) \\ \end{array}
$$
We have to show that the last term vanishes. The last term can be written as: