SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Note that the positional index  $t$  can occur before or after the positional index  $s$ . Depending of its position the elements  $e_{i_t}$  will appear before or after the element  $e_{i_s}$  in the product. We look at both cases separately and suppress the dots in between for readability.
First consider  $t > s$ :
$$
\begin{array}{l} \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} \sum_ {\substack {i _ {1}, i _ {2},.., i _ {t},.., i _ {m} \\ \forall k \neq l. i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \cdot a _ {i _ {t}, j _ {s}} \cdot a _ {i _ {t}, j _ {t}} \cdot a _ {i _ {m}, j _ {m}} a _ {i _ {t}, j} \cdot e _ {i _ {1}} \cdot e _ {i _ {s}} \cdot e _ {i _ {t}} \cdot e _ {i _ {m}} \cdot (-1) ^ {m - s} \mathfrak {q} \left(e _ {i _ {t}}\right) (219) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} \sum_ {\substack {i _ {1},.., i _ {s},.., i _ {t},.., i _ {m} \\ \forall k \neq l. i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \cdot a _ {i _ {t}, j _ {s}} \cdot a _ {i _ {t}, j _ {t}} \cdot a _ {i _ {m}, j _ {m}} a _ {i _ {t}, j} \cdot (-1) ^ {t - 2} e _ {i _ {t}} \cdot e _ {i _ {1}} \cdot e _ {i _ {s}} \cdot e _ {i _ {t}} \cdot e _ {i _ {m}} \cdot (-1) ^ {m - s} \mathfrak {q} \left(e _ {i _ {t}}\right) (220) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} \sum_ {i _ {t} = 1} ^ {n} a _ {i _ {t}, j _ {s}} \cdot a _ {i _ {t}, j _ {t}} \cdot a _ {i _ {t}, j} (- 1) ^ {m - s} \mathfrak {q} \left(e _ {i _ {t}}\right) (- 1) ^ {t} e _ {i _ {t}} \cdot \pi_ {s, t} (i _ {t}) (221) \\ \end{array}
$$
with  $\pi_{s,t}(i)\coloneqq \sum_{\substack{i_1,\ldots ,i_s,\ldots ,j_t,\ldots ,i_m\\ \forall k\neq l,i_k\neq i_l\\ \forall k,i_k\neq i}}a_{i_1,j_1}\cdot \underline{a}_{i_\tau ,j_s}\cdot \underline{a}_{i_\tau ,j_t}\cdot a_{i_m,j_m}\cdot e_{i_1}\cdot \underline{e}_{i_s}\cdot \underline{e}_{j_t}\cdot e_{i_m}$  (222)
$$
\begin{array}{l} = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} \sum_ {i = 1} ^ {n} (- 1) ^ {m + s + t} a _ {i, j _ {s}} \cdot a _ {i, j _ {t}} \cdot a _ {i, j} \cdot \mathfrak {q} (e _ {i}) \cdot e _ {i} \cdot \pi_ {s, t} (i) (223) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} y (s, t), (224) \\ \end{array}
$$
with  $y(s,t)\coloneqq \sum_{i = 1}^{n}(-1)^{m + s + t}a_{i,j_s}\cdot a_{i,j_t}\cdot a_{i,j}\cdot \mathfrak{q}(e_i)\cdot e_i\cdot \pi_{s,t}(i).$  (225)
It is important to note that for all  $s \neq t$  we have:
$$
y (s, t) = y (t, s). \tag {226}
$$
Now consider  $t < s$ :
$$
\begin{array}{l} \sum_{s = 1}^{m}\sum_{\substack{t = 1\\ t <   s}}^{m}\sum_{\substack{i_{1},\ldots ,i_{t},\ldots ,i_{s},\ldots ,i_{m}\\ \forall k\neq l. i_{k}\neq i_{l}}}a_{i_{1},j_{1}}\cdot a_{i_{t},j_{t}}\cdot a_{i_{t},j_{s}}\cdot a_{i_{m},j_{m}}a_{i_{t},j}\cdot e_{i_{1}}\cdot e_{i_{t}}\cdot e_{i_{s}}\cdot e_{i_{m}}\cdot (-1)^{m - s}\mathfrak{q}(e_{i_{t}}) (227) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t <   s}} ^ {m} \sum_ {\substack {i _ {1},.., i _ {t},.., i _ {s},.., i _ {m} \\ \forall k \neq l. i _ {k} \neq i _ {l}}} a _ {i _ {1}, j _ {1}} \cdot a _ {i _ {t}, j _ {t}} \cdot a _ {i _ {t}, j _ {s}} \cdot a _ {i _ {m}, j _ {m}} a _ {i _ {t}, j} \cdot (-1) ^ {t - 1} e _ {i _ {t}} \cdot e _ {i _ {1}} \cdot e _ {i _ {t}} \cdot e _ {i _ {s}} \cdot e _ {i _ {m}} \cdot (-1) ^ {m - s} \mathfrak {q} \left(e _ {i _ {t}}\right) (228) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t <   s}} ^ {m} \sum_ {i _ {t} = 1} ^ {n} a _ {i _ {t}, j _ {s}} \cdot a _ {i _ {t}, j _ {t}} \cdot a _ {i _ {t}, j} (- 1) ^ {m - s} \mathfrak {q} \left(e _ {i _ {t}}\right) (- 1) ^ {t - 1} e _ {i _ {t}} \cdot \pi_ {s, t} \left(i _ {t}\right) (229) \\ = \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t <   s}} ^ {m} \sum_ {i = 1} ^ {n} (- 1) ^ {m + s + t + 1} a _ {i, j _ {s}} \cdot a _ {i, j _ {t}} \cdot a _ {i, j} \cdot \mathfrak {q} (e _ {i}) \cdot e _ {i} \cdot \pi_ {s, t} (i) (230) \\ = - \sum_ {s = 1} ^ {m} \sum_ {t = 1 \atop t <   s} ^ {m} y (s, t) (231) \\ = - \sum_ {t = 1} ^ {m} \sum_ {\substack {s = 1 \\ s <   t}} ^ {m} y (t, s) (232) \\ = - \sum_ {t = 1} ^ {m} \sum_ {\substack {s = 1 \\ s <   t}} ^ {m} y (s, t) (233) \\ = - \sum_ {s = 1} ^ {m} \sum_ {\substack {t = 1 \\ t > s}} ^ {m} y (s, t). (234) \\ \end{array}
$$
In total we see that both terms appear with a different sign and thus cancel out. This shows the claim.
Corollary D.28. Let  $e_1, \ldots, e_n$  and  $b_1, \ldots, b_n$  be two orthogonal bases of  $(V, \mathfrak{q})$  with basis transition matrix  $C = (c_{i,j})_{i \in [n], j \in [n]}$ :
$$
\forall j \in [ n ]. \quad b _ {j} = \sum_ {i \in [ n ]} c _ {i, j} \cdot e _ {i}. \tag {235}
$$
Then  $C$  is invertible and we have the following matrix relations:
$$
\operatorname {d i a g} \left(\mathfrak {q} \left(b _ {1}\right), \dots , \mathfrak {q} \left(b _ {n}\right)\right) = C ^ {\top} \operatorname {d i a g} \left(\mathfrak {q} \left(e _ {1}\right), \dots , \mathfrak {q} \left(e _ {n}\right)\right) C. \tag {236}
$$
Furthermore, for every subset  $J \subseteq [n]$  we have the formula:
$$
b _ {J} = \sum_ {\substack {I \subseteq [ n ] \\ | I | = | J |}} \det C _ {I, J} \cdot e _ {I}, \tag{237}
$$
with the submatrix:  $C_{I,J} = (c_{i,j})_{i\in I,j\in J}$
Proof. For  $j, l \in [n]$  we have:
$$
\begin{array}{l} \operatorname {d i a g} \left(\mathfrak {q} \left(b _ {1}\right), \dots , \mathfrak {q} \left(b _ {n}\right)\right) _ {j, l} = \mathfrak {q} \left(b _ {j}\right) \cdot \delta_ {j, l} (238) \\ = \mathfrak {b} \left(b _ {j}, b _ {l}\right) (239) \\ = \sum_ {i = 1} ^ {n} \sum_ {k = 1} ^ {n} c _ {i, j} \cdot c _ {k, l} \cdot \mathfrak {b} \left(e _ {i}, e _ {k}\right) (240) \\ = \sum_ {i = 1} ^ {n} \sum_ {k = 1} ^ {n} c _ {i, j} \cdot c _ {k, l} \cdot \mathfrak {q} \left(e _ {i}\right) \cdot \delta_ {i, k} (241) \\ = \left(C ^ {\top} \operatorname {d i a g} (\mathfrak {q} (e _ {1}), \dots , \mathfrak {q} (e _ {n})) C\right) _ {j, l}. (242) \\ \end{array}
$$
This shows the matrix identity:
$$
\operatorname {d i a g} \left(\mathfrak {q} \left(b _ {1}\right), \dots , \mathfrak {q} \left(b _ {n}\right)\right) = C ^ {\top} \operatorname {d i a g} \left(\mathfrak {q} \left(e _ {1}\right), \dots , \mathfrak {q} \left(e _ {n}\right)\right) C. \tag {243}
$$
Furthermore, we have the following identities:
$$
b _ {J} = b _ {j _ {1}} \dots b _ {j _ {m}} \tag {244}
$$
$$
\begin{array}{l} = \left(\sum_ {i _ {1} \in [ n ]} c _ {i _ {1}, j _ {1}} \cdot e _ {i _ {1}}\right) \dots \left(\sum_ {i _ {m} \in [ n ]} c _ {i _ {m}, j _ {m}} \cdot e _ {i _ {m}}\right) (245) \\ = \sum_ {i _ {1} \in [ n ], \dots , i _ {m} \in [ n ]} \left(c _ {i _ {1}, j _ {1}} \dots c _ {i _ {m}, j _ {m}}\right) \cdot \left(e _ {i _ {1}} \dots e _ {i _ {m}}\right) (246) \\ \end{array}
$$
$$
\begin{array}{l} \stackrel {D. 2 7} {=} \sum_ {\substack {i _ {1} \in [ n ], \dots , i _ {m} \in [ n ] \\ | \{i _ {1}, \dots , i _ {m} \} | = m}} \left(c _ {i _ {1}, j _ {1}} \dots c _ {i _ {m}, j _ {m}}\right) \cdot \left(e _ {i _ {1}} \dots e _ {i _ {m}}\right) (247) \\ = \sum_ {\substack {i _ {1} \in [ n ], \dots , i _ {m} \in [ n ] \\ | \{i _ {1}, \dots , i _ {m} \} | = m}} \operatorname {sgn} \left(i _ {1}, \dots , i _ {m}\right) \cdot \left(c _ {i _ {1}, j _ {1}} \dots c _ {i _ {m}, j _ {m}}\right) \cdot e _ {\left\{i _ {1}, \dots , i _ {m} \right\}} (248) \\ = \sum_ {\substack {i _ {1}, \dots , i _ {m} \in [ n ] \\ i _ {1} <   \dots <   i _ {m}}} \left(\sum_ {\sigma \in \mathrm {S} _ {m}} \operatorname {sgn} (\sigma) \cdot c _ {i _ {\sigma (1)}, j _ {1}} \dots c _ {i _ {\sigma (m)}, j _ {m}}\right) \cdot e _ {\{i _ {1}, \dots , i _ {m} \}} (249) \\ = \sum_ {\substack {I \subseteq [ n ] \\ | I | = m}} \det C _ {I, J} \cdot e _ {I}. (250) \\ \end{array}
$$
Notation D.29. For  $m \notin \{0, \dots, n\}$  it is sometimes convenient to put:
$$
\operatorname {C l} ^ {(m)} (V, \mathfrak {q}) := 0. \tag {251}
$$
Corollary D.30. We have the following orthogonal sum decomposition (w.r.t.  $\bar{\mathfrak{q}}$ ) of the Clifford algebra  $\mathrm{Cl}(V, \mathfrak{q})$  into its  $\mathbb{F}$ -vector spaces of multivector components:
$$
\operatorname {C l} (V, \mathfrak {q}) = \bigoplus_ {m = 0} ^ {n} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}), \tag {252}
$$
which is independent of the choice of orthogonal basis of  $(V,\mathfrak{q})$ . Also note that for all  $m = 0,\dots ,n$
$$
\dim \operatorname {C l} ^ {(m)} (V, \mathfrak {q}) = \binom {n} {m}. \tag {253}
$$
Definition D.31. We call an element  $x \in \operatorname{Cl}^{(m)}(V, \mathfrak{q})$  an  $m$ -multivector or an element of  $\operatorname{Cl}(V, \mathfrak{q})$  of pure grade  $m$ . For  $x \in \operatorname{Cl}(V, \mathfrak{q})$  we have a decomposition:
$$
x = x ^ {(0)} + x ^ {(1)} + \dots + x ^ {(n)}, \tag {254}
$$
with  $x^{(m)}\in \mathrm{Cl}^{(m)}(V,\mathfrak{q})$ $m = 0,\dots ,n$  . We call  $x^{(m)}$  the grade-  $m$  -component of  $x$
Remark D.32. Note that the multivector grading of  $\operatorname{Cl}(V, \mathfrak{q})$  is only a grading of  $\mathbb{F}$ -vector spaces, but not of  $\mathbb{F}$ -algebras. The reason is that multiplication can make the grade drop. For instance, for  $v \in V = \operatorname{Cl}^{(1)}(V, \mathfrak{q})$  we have  $vv = \mathfrak{q}(v) \in \operatorname{Cl}^{(0)}(V, \mathfrak{q})$ , while a grading for algebras would require that  $vv \in \operatorname{Cl}^{(2)}(V, \mathfrak{q})$ , which is here not the case.
Remark D.33 (Parity grading and multivector filtration in terms of multivector grading).
1. We clearly have:
$$
\operatorname {C l} ^ {[ 0 ]} (V, \mathfrak {q}) = \bigoplus_ {\substack {m = 0, \dots , n \\ m \text{even}}} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}), \quad \operatorname {C l} ^ {[ 1 ]} (V, \mathfrak {q}) = \bigoplus_ {\substack {m = 1, \dots , n \\ m \text{odd}}} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}). \tag{255}
$$
2. It is also clear that for general  $m$  we have:
$$
\operatorname {C l} ^ {(\leq m)} (V, \mathfrak {q}) = \bigoplus_ {l = 0} ^ {m} \operatorname {C l} ^ {(m)} (V, \mathfrak {q}) \subseteq \operatorname {C l} (V, \mathfrak {q}). \tag {256}
$$
A simpler proof of Theorem D.27 can be obtained if we assume that  $\mathrm{char}(\mathbb{F}) = 0$ . We would then argue as follows.
Definition D.34 (Antisymmetrization). For  $m \in \mathbb{N}$  and  $x_1, \ldots, x_m \in \operatorname{Cl}(V, \mathfrak{q})$  we define their antisymmetrization as:
$$
[ x _ {1}; \dots ; x _ {m} ] := \frac {1}{m !} \sum_ {\sigma \in S _ {m}} \operatorname {s g n} (\sigma) \cdot x _ {\sigma (1)} \dots x _ {\sigma (m)}, \tag {257}
$$
where  $\mathrm{S}_m$  denotes the group of all permutations of  $[m]$ . Note that, due to the division by  $m!$  we need that  $\operatorname{char}(\mathbb{F}) = 0$  if we want to accommodate arbitrary  $m \in \mathbb{N}$ .
Lemma D.35. Let  $e_1, \ldots, e_n$  be an orthogonal basis of  $(V, \mathfrak{q})$  and  $x_1, \ldots, x_m \in \mathrm{Cl}(V, \mathfrak{q})$ . Then we have:
1.  $[x_1; \ldots; x_m]$  is linear in each of its arguments (if the other arguments are fixed).
2.  $[x_1; \ldots; x_k; \ldots; x_l; \ldots; x_m] = -[x_1; \ldots; x_l; \ldots; x_k; \ldots; x_m]$ .
3.  $[x_1; \ldots; x_k; \ldots; x_l; \ldots; x_m] = 0$  if  $x_k = x_l$ .
4.  $[e_{i_1};\ldots ;e_{i_m}] = e_{i_1}\dots e_{i_m}$  if  $|\{i_1,\ldots ,i_m\} | = m$  (i.e.if all indices are different).
Definition D.36 (Multivector grading - alternative, basis independent definition). For  $m = 0,\dots ,n$  we (re-)define:
$$
\operatorname {C l} ^ {(m)} (V, \mathfrak {q}) := \operatorname {s p a n} \left\{\left[ v _ {1}; \dots ; v _ {m} \right] \mid v _ {1}, \dots , v _ {m} \in V \right\}. \tag {258}
$$
Theorem D.37. Let  $e_1, \ldots, e_n$  be an orthogonal basis of  $(V, \mathfrak{q})$ ,  $\operatorname{char}(\mathbb{F}) = 0$ . Then for every  $m = 0, \ldots, n$  we have the equality:
$$
\operatorname {C l} ^ {(m)} (V, \mathfrak {q}) = \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ], | A | = m \right\}. \tag {259}
$$
Note that the rhs is seemingly dependent of the choice of the orthogonal basis while the lhs is defined in a basis independent way.
Proof. We can write every  $v_{k} \in V$  as a linear combination of its basis vectors:
$$
v _ {k} = \sum_ {j _ {k} \in [ n ]} c _ {k, j _ {k}} \cdot e _ {j _ {k}}. \tag {260}
$$
With this we get:
$$
\begin{array}{l} [ v _ {1}; \dots ; v _ {m} ] = \sum_ {j _ {1}, \dots , j _ {m} \in [ n ]} \prod_ {k \in [ m ]} c _ {k, j _ {k}} \cdot \left[ e _ {j _ {1}}; \dots ; e _ {j _ {m}} \right] (261) \\ = \sum_ {\substack {j _ {1}, \dots , j _ {m} \in [ n ] \\ | \{j _ {1}, \dots , j _ {m} \} | = m}} \prod_ {k \in [ m ]} c _ {k, j _ {k}} \cdot \left[ e _ {j _ {1}}; \dots ; e _ {j _ {m}} \right] (262) \\ = \sum_ {\substack {j _ {1}, \dots , j _ {m} \in [ n ] \\ \left| \left\{j _ {1}, \dots , j _ {m} \right\} \right| = m}} \prod_ {k \in [ m ]} c _ {k, j _ {k}} \cdot e _ {j _ {1}} \dots e _ {j _ {m}} (263) \\ = \sum_ {\substack {j _ {1}, \dots , j _ {m} \in [ n ] \\ \left| \left\{j _ {1}, \dots , j _ {m} \right\} \right| = m}} \pm \prod_ {k \in [ m ]} c _ {k, j _ {k}} \cdot e _ {\left\{j _ {1}, \dots , j _ {m} \right\}} (264) \\ \in \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ], | A | = m \right\}. (265) \\ \end{array}
$$
This shows the inclusion:
$$
\operatorname {C l} ^ {(m)} (V, \mathfrak {q}) \subseteq \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ], | A | = m \right\}. \tag {266}
$$
The reverse inclusion is also clear as:
$$
e _ {A} = \left[ e _ {j _ {1}}; \dots ; e _ {j _ {m}} \right] \in \mathrm {C l} ^ {(m)} (V, \mathfrak {q}), \tag {267}
$$
where  $A = \{j_1,\dots ,j_m\}$  and  $|A| = m$  . This shows the equality of both sets.
D.7 The Radical Subalgebra of the Clifford Algebra
Again, let  $(V,\mathfrak{q})$  be a quadratic vector space of finite dimensions  $\dim V = n < \infty$  over a field  $\mathbb{F}$  of  $\operatorname {char}(\mathbb{F})\neq 2$ . Let  $\mathfrak{b}$  the corresponding bilinear form of  $\mathfrak{q}$ .
Notation D.38. We denote the group of the invertible elements of  $\operatorname{Cl}(V, \mathfrak{q})$ :
$$
\operatorname {C l} ^ {\times} (V, \mathfrak {q}) := \{x \in \operatorname {C l} (V, \mathfrak {q}) \mid \exists y \in \operatorname {C l} (V, \mathfrak {q}). x y = y x = 1 \}. \tag {268}
$$
Let  $\mathcal{R} \subseteq V$  be  $V$ 's radical subspace. Recall:
$$
\mathcal {R} := \{f \in V | \forall v \in V. \mathfrak {b} (f, v) = 0 \}. \tag {269}
$$
Definition D.39 (The radical subalgebra). We define the radical subalgebra of  $\operatorname{Cl}(V, \mathfrak{q})$  to be:
$$
\bigwedge (\mathcal {R}) := \operatorname {s p a n} \left\{1, f _ {1} \dots f _ {k} \mid k \in \mathbb {N} _ {0}, f _ {l} \in \mathcal {R}, l = 1, \dots , k \right\} \subseteq \operatorname {C l} (V, \mathfrak {q}). \tag {270}
$$
Note that  $\mathfrak{q}|_{\mathcal{R}} = 0$  and that  $\bigwedge (\mathcal{R})$  coincides with  $\operatorname {Cl}(\mathcal{R},\mathfrak{q}|_{\mathcal{R}})$
Notation D.40. We make the following further abbreviations:
$$
\bigwedge^ {[ i ]} (\mathcal {R}) := \bigwedge (\mathcal {R}) \cap \operatorname {C l} ^ {[ i ]} (V, \mathfrak {q}), \tag {271}
$$
$$
\bigwedge^ {(\geq 1)} (\mathcal {R}) := \operatorname {s p a n} \left\{f _ {1} \dots f _ {k} \mid k \geq 1, f _ {l} \in \mathcal {R}, l = 1, \dots , k \right\}, \tag {272}
$$
$$
\bigwedge^ {\times} (\mathcal {R}) := \mathbb {F} ^ {\times} + \bigwedge^ {(\geq 1)} (\mathcal {R}), \tag {273}
$$
$$
\bigwedge^ {[ \times ]} (\mathcal {R}) := \mathbb {F} ^ {\times} + \operatorname {s p a n} \left\{f _ {1} \dots f _ {k} \mid k \geq 2 \text {e v e n}, f _ {l} \in \mathcal {R}, l = 1, \dots , k \right\}, \tag {274}
$$
$$
\bigwedge^ {*} (\mathcal {R}) := 1 + \bigwedge^ {(\geq 1)} (\mathcal {R}), \tag {275}
$$
$$
\bigwedge^ {[ * ]} (\mathcal {R}) := 1 + \operatorname {s p a n} \left\{f _ {1} \dots f _ {k} \mid k \geq 2 \text {e v e n}, f _ {l} \in \mathcal {R}, l = 1, \dots , k \right\}. \tag {276}
$$
Here,  $\mathbb{F}^{\times}$  denotes the set of invertible elements of  $\mathbb{F}$ .
Lemma D.41. 1. For every  $h\in \bigwedge^{(\geq 1)}(\mathcal{R})$  there exists a  $k\geq 0$  such that:
$$
h ^ {k + 1} = 0. \tag {277}
$$
In particular, no  $h \in \bigwedge^{(\geq 1)}(\mathcal{R})$  is ever invertible.
2. Every  $y \in \bigwedge^{\times}(\mathcal{R}) = \bigwedge (\mathcal{R}) \setminus \bigwedge^{(\geq 1)}(\mathcal{R})$  is invertible. Its inverse is given by:
$$
y ^ {- 1} = c ^ {- 1} \left(1 - h + h ^ {2} - \dots + (- 1) ^ {k} h ^ {k}\right), \tag {278}
$$
where we write:  $y = c\cdot (1 + h)$  with  $c\in \mathbb{F}^{\times}$ ,  $h\in \bigwedge^{(\geq 1)}(\mathcal{R})$ , and  $k$  is such that  $h^{k + 1} = 0$ .
3. In particular, we get:
$$
\bigwedge^ {\times} (\mathcal {R}) = \bigwedge (\mathcal {R}) \cap \mathrm {C l} ^ {\times} (V, \mathfrak {q}). \tag {279}
$$
Proof. Items 1 and 3 are clear. For item 2 we refer to Example E.21.
Lemma D.42 (Twisted commutation relationships). the anticommutation relationship:
1. For every  $f \in \mathcal{R}$  and  $v \in V$  we have
$$
f v = - v f + 2 \underbrace {\mathfrak {b} (f , v)} _ {= 0} = - v f. \tag {280}
$$
2. For every  $f \in \mathcal{R}$  and  $x \in \mathrm{Cl}(V, \mathfrak{q})$  we get the following twisted commutation relationship:
$$
\alpha (x) f = \left(x ^ {[ 0 ]} - x ^ {[ 1 ]}\right) f = f \left(x ^ {[ 0 ]} + x ^ {[ 1 ]}\right) = f x. \tag {281}
$$
3. For every  $y \in \bigwedge(\mathcal{R})$  and every  $x \in \mathrm{Cl}^{[0]}(V, \mathfrak{q})$  we get:
$$
x y = y x. \tag {282}
$$
4. For every  $y \in \bigwedge(\mathcal{R})$  and  $v \in V$  we get:
$$
\alpha (y) v = \left(y ^ {[ 0 ]} - y ^ {[ 1 ]}\right) v = v \left(y ^ {[ 0 ]} + y ^ {[ 1 ]}\right) = v y. \tag {283}
$$
5. For every  $y \in \bigwedge^{[0]}(\mathcal{R})$  (of even parity) and every  $x \in \operatorname{Cl}(V, \mathfrak{q})$  we get:
$$
y x = x y. \tag {284}
$$
Remark D.43. A direct consequence from Lemma D.42 is that the even parity parts:  $\bigwedge^{[0]}(\mathcal{R})$ ,  $\bigwedge^{[\times ]}(\mathcal{R})$  and  $\bigwedge^{[*]}(\mathcal{R})$  all lie in the center of  $\operatorname {Cl}(V,\mathfrak{q})$ , which we denote by  $\Im (\operatorname {Cl}(V,\mathfrak{q}))$ :
$$
\bigwedge^ {[ * ]} (\mathcal {R}) \subseteq \bigwedge^ {[ \times ]} (\mathcal {R}) \subseteq \bigwedge^ {[ 0 ]} (\mathcal {R}) \subseteq \Im (\operatorname {C l} (V, \mathfrak {q})), \tag {285}
$$
i.e. every  $y \in \bigwedge^{[0]}(\mathcal{R})$  (of even parity) commutes with every  $x \in \operatorname{Cl}(V, \mathfrak{q})$ . For more detail we refer to Theorem D.47.
In the following we will study the center of the Clifford algebra  $\mathfrak{Z}(\operatorname {Cl}(V,\mathfrak{q}))$  more carefully.
D.8 The Center of the Clifford Algebra
In the following final subsections, we study additional properties of the Clifford algebra that will aid us in studying group representations and actions on the algebra in the upcoming sections.
Definition D.44 (The center of an algebra). The center of an algebra  $\mathcal{A}$  is defined to be:
$$
\mathfrak {3} (\mathcal {A}) := \{z \in \mathcal {A} | \forall x \in \mathcal {A}. x z = z x \}. \tag {286}
$$
Lemma D.45. Let  $e_1, \ldots, e_n$  be an orthogonal basis of  $(V, \mathfrak{q})$ . For  $A \subseteq [n] := \{1, \ldots, n\}$  let  $e_A := \prod_{i \in A}^{<} e_i$  be the product in  $\operatorname{Cl}(V, \mathfrak{q})$  in increasing index order,  $e_\emptyset := 1$ . Then we get for two subsets  $A, B \subseteq [n]$ :
$$
e _ {A} e _ {B} = (- 1) ^ {| A | \cdot | B | - | A \cap B |} \cdot e _ {B} e _ {A}. \tag {287}
$$
In particular, for  $j \notin A$  we get:
$$
e _ {A} e _ {j} = (- 1) ^ {| A |} \cdot e _ {j} e _ {A}, \tag {288}
$$
$$
e _ {A} e _ {j} - e _ {j} e _ {A} = \left((- 1) ^ {| A |} - 1\right) \cdot e _ {j} e _ {A} = (- 1) ^ {t} \left((- 1) ^ {| A | + 1} + 1\right) e _ {A \dot {\cup} \{j \}}, \tag {289}
$$
where  $t$  is the position of  $j$  in the ordered set  $A \dot{\cup} \{j\}$ .
For  $i\in A$  we get:
$$
e _ {A} e _ {i} = (- 1) ^ {| A | - 1} \cdot e _ {i} e _ {A} = (- 1) ^ {| A | - s} \mathfrak {q} (e _ {i}) e _ {A \backslash \{i \}}, \tag {290}
$$
$$
e _ {A} e _ {i} - e _ {i} e _ {A} = (- 1) ^ {s} \left((- 1) ^ {| A |} + 1\right) \mathfrak {q} \left(e _ {i}\right) e _ {A \backslash \{i \}}, \tag {291}
$$
where  $s$  is the position of  $i$  in the ordered set  $A$ .
Proof. Let  $B \coloneqq \{j_1, \ldots, j_{|B|}\} \subseteq [n]$ .
$$
\begin{array}{l} e _ {A} e _ {B} = \prod_ {i \in A} ^ {<  } e _ {i} \prod_ {j \in B} ^ {<  } e _ {j} (292) \\ = (- 1) ^ {| A | - \mathbb {1} [ j _ {1} \in A ]} e _ {j _ {1}} \prod_ {i \in A} ^ {<  } e _ {i} \prod_ {j \in B \backslash j _ {1}} ^ {<  } e _ {j} (293) \\ = (- 1) ^ {| B | | A | - \sum_ {j \in B} 1 [ j \in A ]} e _ {B} e _ {A} (294) \\ = (- 1) ^ {| B | | A | - | A \cap B |} e _ {B} e _ {A} (295) \\ \end{array}
$$
For the other two identities, we similarly make use of the fundamental Clifford identity.
Lemma D.46. Let  $(V,\mathfrak{q})$  be a quadratic space with  $\dim V = n < \infty$ . Let  $e_1, \ldots, e_n$  be an orthogonal basis for  $(V,\mathfrak{q})$ . For  $x \in \operatorname{Cl}(V,\mathfrak{q})$  we have the equivalence:
$$
x \in \mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})) \quad \Longleftrightarrow \quad \forall i \in [ n ]. \quad x e _ {i} = e _ {i} x. \tag {296}
$$
Proof. This is clear as  $\operatorname{Cl}(V, \mathfrak{q})$  is generated by  $e_{k_1} \cdots e_{k_l}$ .
Theorem D.47 (The center of the Clifford algebra). Let  $(V, \mathfrak{q})$  be a quadratic space with  $\dim V = n < \infty$ ,  $\operatorname{char} \mathbb{F} \neq 2$ , and let  $\mathcal{R} \subseteq V$  be the radical subspace of  $(V, \mathfrak{q})$ . Then for the center of  $\operatorname{Cl}(V, \mathfrak{q})$  we have the following cases:
1. If  $n$  is odd then:
$$
\mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})) = \bigwedge^ {[ 0 ]} (\mathcal {R}) \oplus \operatorname {C l} ^ {(n)} (V, \mathfrak {q}). \tag {297}
$$
2. If  $n$  is even then:
$$
\Im (\operatorname {C l} (V, \mathfrak {q})) = \bigwedge^ {[ 0 ]} (\mathcal {R}). \tag {298}
$$
In all cases we have:
$$
\bigwedge^ {[ 0 ]} (\mathcal {R}) \subseteq \Im (\operatorname {C l} (V, \mathfrak {q})). \tag {299}
$$
Proof. Let  $e_1, \ldots, e_n$  be an orthogonal basis for  $(V, \mathfrak{q})$ . The statement can then equivalently be expressed as:
1. If  $n$  is odd or  $\mathfrak{q} = 0$  (on all vectors), then:
$$
\begin{array}{l} \mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})) = \operatorname {s p a n} \left\{1, e _ {A}, e _ {[ n ]} \mid | A | \text {e v e n}, \forall i \in A. \mathfrak {q} \left(e _ {i}\right) = 0 \right\} (300) \\ = \bigwedge^ {[ 0 ]} (\mathcal {R}) + \operatorname {C l} ^ {(n)} (V, \mathfrak {q}). (301) \\ \end{array}
$$
2. If  $n$  is even and  $\mathfrak{q} \neq 0$  (on some vector), then:
$$
\begin{array}{l} \mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})) = \operatorname {s p a n} \{1, e _ {A} \mid | A | \text {e v e n}, \forall i \in A. \mathfrak {q} \left(e _ {i}\right) = 0 \} (302) \\ = \bigwedge^ {[ 0 ]} (\mathcal {R}). (303) \\ \end{array}
$$
Note that if  $n$  is even and  $\mathfrak{q} = 0$  then both points would give the same answer, as then  $V = \mathcal{R}$  and thus:
$$
\operatorname {C l} ^ {(n)} (V, \mathfrak {q}) = \bigwedge^ {(n)} (\mathcal {R}) \subseteq \bigwedge^ {[ 0 ]} (\mathcal {R}). \tag {304}
$$
We first only consider basis elements  $e_A$  with  $A \subseteq [n]$  and check when we have  $e_A \in 3(\operatorname{Cl}(V, \mathfrak{q}))$ .
We now consider three cases:
1.  $A = \emptyset$ . We always have:  $e_{\emptyset} = 1 \in \mathfrak{Z}(\operatorname{Cl}(V, \mathfrak{q}))$ .
2.  $A = [n]$ . For every  $i \in [n]$  we have:
$$
e _ {[ n ]} e _ {i} = (- 1) ^ {n - 1} \cdot e _ {i} e _ {[ n ]} = \pm \mathfrak {q} \left(e _ {i}\right) \cdot e _ {[ n ] \backslash \{i \}}. \tag {305}
$$
So  $e_{[n]}\in \mathfrak{Z}(\mathrm{Cl}(V,\mathfrak{q}))$  iff either  $n$  is odd or  $\mathfrak{q} = 0$  (on all  $e_i$ ).
3.  $\emptyset \neq A\subsetneq [n]$  . Note that for  $i\in A$  we have:
$$
e _ {A} e _ {i} = (- 1) ^ {| A | - 1} \cdot e _ {i} e _ {A} = \pm \mathfrak {q} (e _ {i}) \cdot e _ {A \backslash \{i \}}, \tag {306}
$$
while for  $j\notin A$  we have:
$$
e _ {A} e _ {j} = (- 1) ^ {| A |} \cdot e _ {j} e _ {A}. \tag {307}
$$