SlowGuess/NeurIPS_2023_OCR huggingface dataset:
The latter implies that for  $e_A \in \mathfrak{Z}(\operatorname{Cl}(V, \mathfrak{q}))$  to hold,  $|A|$  necessarily needs to be even. Since in that case  $(-1)^{|A|-1} = -1$ , we necessarily need by the former case that for every  $i \in A$ ,  $\mathfrak{q}(e_i) = 0$ .
Now consider a linear combination  $x = \sum_{A\subseteq [n]}c_A\cdot e_A\in \mathrm{Cl}(V,\mathfrak{q})$  with  $c_{A}\in \mathbb{F}$ . Then abbreviate:
$$
\mathcal {Z} := \operatorname {s p a n} \left\{e _ {A} \mid A \subseteq [ n ], e _ {A} \in \mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})) \right\} \subseteq \mathfrak {Z} (\operatorname {C l} (V, \mathfrak {q})). \tag {308}
$$
We need to show that  $x \in \mathcal{Z}$ . We thus write:
$$
x = y + z, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \sum_ {A \subseteq [ n ]} c _ {A} \cdot e _ {A}, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad z := \sum_ {A \subseteq [ n ]} c _ {A} \cdot e _ {A} \in \mathcal {Z}.
$$
So with  $x,z\in \mathfrak{Z}(\operatorname {Cl}(V,\mathfrak{q}))$  also  $y\in \mathfrak{Z}(\operatorname {Cl}(V,\mathfrak{q}))$  and we are left to show that  $y = 0$
First note that, since  $e_{\emptyset} = 1 \in \mathcal{Z}$ , there is no  $e_{\emptyset}$ -component in  $y$ .
We now have for every  $i\in [n]$
$$
\begin{array}{l} 0 = y e _ {i} - e _ {i} y (310) \\ = \sum_ {\substack {A \subseteq [ n ] \\ e _ {A} \notin \mathcal {Z}}} c _ {A} \cdot \left(e _ {A} e _ {i} - e _ {i} e _ {A}\right) (311) \\ = \sum_ {\substack {A \subseteq [ n ] \\ e _ {A} \notin \mathcal {Z} \\ i \in A}} c _ {A} \cdot \left(e _ {A} e _ {i} - e _ {i} e _ {A}\right) + \sum_ {\substack {A \subseteq [ n ] \\ e _ {A} \notin \mathcal {Z} \\ i \notin A}} c _ {A} \cdot \left(e _ {A} e _ {i} - e _ {i} e _ {A}\right) (312) \\ = \sum_ {\substack {A \subseteq [ n ] \\ e _ {A} \notin \mathcal {Z} \\ i \in A}} \pm ((- 1) ^ {| A |} + 1) \mathfrak {q} (e _ {i}) c _ {A} \cdot e _ {A \setminus \{i \}} + \sum_ {\substack {A \subseteq [ n ] \\ e _ {A} \notin \mathcal {Z} \\ i \notin A}} \pm ((- 1) ^ {| A | + 1} + 1) c _ {A} \cdot e _ {A \cup \{i \}}. (313) \\ \end{array}
$$
Note that for  $A, B \subseteq [n]$  with  $A \neq B$  we always have:
$$
A \setminus \{i \} \neq B \setminus \{i \}, \quad \text {i f} i \in A, i \in B, \tag {314}
$$
$$
A \setminus \{i \} \neq B \dot {\cup} \{i \}, \quad \text {i f} i \in A, i \notin B, \tag {315}
$$
$$
A \dot {\cup} \{i \} \neq B \backslash \{i \}, \quad \text {i f} i \notin A, i \in B, \tag {316}
$$
$$
A \dot {\cup} \{i \} \neq B \dot {\cup} \{i \}, \quad \text {i f} i \notin A, i \notin B. \tag {317}
$$
So the above representation for  $ye_i - e_i y$  is already given in basis form. By their linear independence we then get that for every  $A \subseteq [n]$  with  $e_A \notin \mathcal{Z}$  and every  $i \in [n]$ :
$$
0 = \left((- 1) ^ {| A |} + 1\right) \mathfrak {q} \left(e _ {i}\right) c _ {A}, \quad \text {f o r} i \in A, \tag {318}
$$
$$
0 = \left((- 1) ^ {| A | + 1} + 1\right) c _ {A}, \quad \text {f o r} i \notin A. \tag {319}
$$
First consider the case  $e_{[n]} \notin \mathcal{Z}$ . By the previous result we then know that  $n$  is even and  $\mathfrak{q} \neq 0$ . So there exists  $e_i$  with  $\mathfrak{q}(e_i) \neq 0$ . So the above condition for  $i \in [n]$  then reads:
$$
0 = 2 \mathfrak {q} \left(e _ {i}\right) c _ {[ n ]}, \quad \text {f o r} i \in [ n ], \tag {320}
$$
which implies  $c_{[n]} = 0$  as  $\mathfrak{q}(e_i)\neq 0$ . So  $y$  does not have a  $e_{[n]}$ -component.
Similarly, for  $A \subseteq [n]$  with  $A \neq [n]$  and  $e_A \notin \mathcal{Z}$  and  $|A|$  odd there exists  $i \notin A$  and the above condition reads:
$$
0 = 2 c _ {A}, \quad \text {f o r} i \notin A, \tag {321}
$$
which implies  $c_A = 0$ . So  $y$  does not have any  $e_A$ -components with odd  $|A|$ .
Now let  $A \subseteq [n]$  with  $A \neq [n]$  and  $e_A \notin \mathcal{Z}$  and  $|A|$  even. Then by our previous analysis we know that there exists  $i \in A$  with  $\mathfrak{q}(e_i) \neq 0$ . Otherwise  $e_A \in \mathcal{Z}$ . So the above condition reads:
$$
0 = 2 \mathfrak {q} \left(e _ {i}\right) c _ {A}, \quad \text {f o r} i \in A, \tag {322}
$$
which implies  $c_A = 0$  as  $\mathfrak{q}(e_i) \neq 0$ . This shows that  $y$  does not have any  $e_A$ -component with even  $|A|$ .
Overall, this shows that  $y = 0$  and thus the claim.
D.9 The Twisted Center of the Clifford Algebra
Notation D.48. 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  $(e_A)_{A \subseteq [n]}$  forms a basis for  $\operatorname{Cl}(V, \mathfrak{q})$ .
Definition D.49 (The twisted center of a  $\mathbb{Z}/2\mathbb{Z}$ -graded algebra). We define the twisted center of a  $\mathbb{Z}/2\mathbb{Z}$ -graded algebra  $\mathcal{A}$  as the following subset:
$$
\mathcal {R} (\mathcal {A}) := \left\{y \in \mathcal {A} \mid \forall x \in \mathcal {A}. y x ^ {[ 0 ]} + (y ^ {[ 0 ]} - y ^ {[ 1 ]}) x ^ {[ 1 ]} = x y \right\}. \tag {323}
$$
Theorem D.50. We have the following identification of the twisted center with the radical subalgebra of the Clifford algebra  $\operatorname{Cl}(V, \mathfrak{q})$  and the set:
$$
\mathfrak {R} (\operatorname {C l} (V, \mathfrak {q})) = \bigwedge (\mathcal {R}) = \left\{y \in \operatorname {C l} (V, \mathfrak {q}) \mid \forall v \in V. \alpha (y) v = v y \right\}. \tag {324}
$$
Proof. Let  $y \in \bigwedge(\mathcal{R})$  then by Lemma D.42 we get:
$$
y x ^ {[ 0 ]} + \alpha (y) x ^ {[ 1 ]} = x ^ {[ 0 ]} y + x ^ {[ 1 ]} y = x y. \tag {325}
$$
This shows that:
$$
y \in \Re (\operatorname {C l} (V, \mathfrak {q})), \tag {326}
$$
$$
\bigwedge (\mathcal {R}) \subseteq \Re (\operatorname {C l} (V, \mathfrak {q})). \tag {327}
$$
Note that the following inclusion is clear as  $V \subseteq \operatorname{Cl}(V, \mathfrak{q})$ :
$$
\Re (\operatorname {C l} (V, \mathfrak {q})) \subseteq \{y \in \operatorname {C l} (V, \mathfrak {q}) | \forall v \in V. \alpha (y) v = v y \}. \tag {328}
$$
For the final inclusion, let  $y = \sum_{B \subseteq [n]} c_B \cdot e_B \in \operatorname{Cl}(V, \mathfrak{q})$  such that for all  $v \in V$  we have  $\alpha(y)v = vy$ . Then for all orthogonal basis vectors  $e_i$  we get the requirement:
$$
e _ {i} y = \alpha (y) e _ {i}, \tag {329}
$$
which always holds if  $\mathfrak{q}(e_i) = 0$ , and is only a condition for  $\mathfrak{q}(e_i)\neq 0$ . For such  $e_i$  we get:
$$
\begin{array}{l} e _ {i} \left(\sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \notin B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \notin B}} c _ {B} \cdot e _ {B}\right) (330) \\ = e _ {i} y (331) \\ = \alpha (y) e _ {i} (332) \\ = \alpha \left(\sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \notin B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \notin B}} c _ {B} \cdot e _ {B}\right) e _ {i} (333) \\ = \left(\sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \notin B}} c _ {B} \cdot e _ {B} - \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \in B}} c _ {B} \cdot e _ {B} - \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \notin B}} c _ {B} \cdot e _ {B}\right) e _ {i} (334) \\ = e _ {i} \left(- \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {even} \\ i \notin B}} c _ {B} \cdot e _ {B} - \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \in B}} c _ {B} \cdot e _ {B} + \sum_ {\substack {B \subseteq [ n ] \\ | B | \text {odd} \\ i \notin B}} c _ {B} \cdot e _ {B}\right). (335) \\ \end{array}
$$
Since  $e_i$  with  $\mathfrak{q}(e_i) \neq 0$  is invertible, we can cancel  $e_i$  on both sides and make use of the linear independence of  $(e_B)_{B \subseteq [n]}$  to get that:
$$
c _ {B} = 0 \quad \text {i f} \quad i \in B. \tag {336}
$$
Since for given  $B$  this can be concluded from every  $e_i$  with  $\mathfrak{q}(e_i) \neq 0$  we can only have  $c_B \neq 0$  if  $\mathfrak{q}(e_j) = 0$  for all  $j \in B$ . Note that elements  $e_j$  with  $\mathfrak{q}(e_j) = 0$  that are part of an orthogonal basis satisfy  $e_j \in \mathcal{R}$ . This shows that:
$$
y = \sum_{\substack{B\subseteq [n]\\ \forall j\in B.\mathfrak{q}(e_{j}) = 0}}c_{B}\cdot e_{B}\in \operatorname {span}\left\{e_{A}\mid A\subseteq [n],\forall i\in A.e_{i}\in \mathcal{R}\right\} = \bigwedge (\mathcal{R}). \tag{337}
$$
This shows the remaining inclusion:
$$
\{y \in \operatorname {C l} (V, \mathfrak {q}) \mid \forall v \in V. \alpha (y) v = v y \} \subseteq \bigwedge (\mathcal {R}). \tag {338}
$$
This shows the equality of all three sets.
E The Clifford Group and its Clifford Algebra Representations
We saw that Cartan-Dieudonné (Theorem C.13) generates the orthogonal group of a (non-degenerate) quadratic space by composing reflections. Considering this, we seek in the following a group representation that acts on the entire Clifford algebra, but reduces to a reflection when restricted to  $V$ . Further, we ensure that the action is an algebra homomorphism and will therefore respect the geometric product.
E.1 Adjusting the Twisted Conjugation
Recall the notation  $\mathrm{Cl}^{\times}(V,\mathfrak{q})\coloneqq \{x\in \mathrm{Cl}(V,\mathfrak{q})|\exists y\in \mathrm{Cl}(V,\mathfrak{q}).xy = yx = 1\}$
Motivation E.1 (Generalizing reflection operations). For  $v, w \in V$  with  $\mathfrak{q}(w) \neq 0$  the reflection of  $v$  onto the hyperplane that is normal to  $w$  is given by the following formula, which we then simplify:
$$
\begin{array}{l} r _ {w} (v) = v - 2 \frac {\mathfrak {b} (w , v)}{\mathfrak {b} (w , w)} w (339) \\ = w w v / \mathfrak {q} (w) - 2 \frac {\mathfrak {b} (w , v)}{\mathfrak {q} (w)} w (340) \\ = - w (- w v + 2 \mathfrak {b} (w, v)) / \mathfrak {q} (w) (341) \\ = - w v w / \mathfrak {q} (w) (342) \\ = - w v w ^ {- 1}. (343) \\ \end{array}
$$
So we have  $r_w(v) = -vww^{-1}$  for  $v, w \in V$  with  $\mathfrak{q}(w) \neq 0$ . We would like to generalize this to an operation  $\rho$  for  $w$  from a subgroup of  $\Gamma \subseteq \mathrm{Cl}^\times(V, \mathfrak{q})$  (as large as possible) onto all elements  $x \in \mathrm{Cl}(V, \mathfrak{q})$ . More explicitly, we want for all  $w_1, w_2 \in \Gamma$  and  $x_1, x_2 \in \mathrm{Cl}(V, \mathfrak{q})$ ,  $v, w \in V$ ,  $\mathfrak{q}(w) \neq 0$ :
$$
\rho (w) (v) = - w v w ^ {- 1} = r _ {w} (v), \tag {344}
$$
$$
\left(\rho \left(w _ {2}\right) \circ \rho \left(w _ {1}\right)\right) \left(x _ {1}\right) = \rho \left(w _ {2} w _ {1}\right) \left(x _ {1}\right), \tag {345}
$$
$$
\rho \left(w _ {1}\right) \left(x _ {1} + x _ {2}\right) = \rho \left(w _ {1}\right) \left(x _ {1}\right) + \rho \left(w _ {1}\right) \left(x _ {2}\right), \tag {346}
$$
$$
\rho \left(w _ {1}\right) \left(x _ {1} x _ {2}\right) = \rho \left(w _ {1}\right) \left(x _ {1}\right) \rho \left(w _ {1}\right) \left(x _ {2}\right). \tag {347}
$$
The second condition makes sure that  $\operatorname{Cl}(V, \mathfrak{q})$  will be a group representation of  $\Gamma$ , i.e. we get a group homomorphism:
$$
\rho : \Gamma \rightarrow \operatorname {A u t} (\operatorname {C l} (V, \mathfrak {q})) \tag {348}
$$
where  $\operatorname{Aut}(\operatorname{Cl}(V, \mathfrak{q}))$  denotes the set of automorphisms  $\operatorname{Cl}(V, \mathfrak{q}) \to \operatorname{Cl}(V, \mathfrak{q})$ .
In the literature, the following versions of reflection operations were studied:
$$
\rho_ {0} (w): x \mapsto w x w ^ {- 1}, \quad \rho_ {1} (w): x \mapsto \alpha (w) x w ^ {- 1}, \quad \rho_ {2} (w): x \mapsto w x \alpha (w) ^ {- 1}. \tag {349}
$$
Since  $\rho_0(w)$  is missing the minus sign it only generalizes compositions of reflections for elements  $w = w_{1}\dots w_{k},w_{l}\in V,\mathfrak{q}(v_{l})\neq 0$  , of even parity  $k$  . The map  $\rho_{1}(w)$  , on the other hand, takes the minus sign into account and generalizes also to such elements  $w = w_{1}\dots w_{k}$  of odd party  $k = \mathrm{prt}(w)$  as then:  $\alpha (w) = (-1)^{\mathrm{prt}(w)}w$  . However, in contrast to  $\rho_0(w)$  , which is an algebra homomorphism for all  $w\in \operatorname {Cl}^{\times}(V,\mathfrak{q})$  , the map  $\rho_{1}(w)$  is not multiplicative in  $x$  , as can be seen with  $v_{1},v_{2}\in V$  and  $w\in V$  with  $\mathfrak{q}(w)\neq 0$  ..
$$
\begin{array}{l} \rho_ {1} (w) \left(v _ {1} v _ {2}\right) = \alpha (w) v _ {1} v _ {2} w ^ {- 1} (350) \\ = (- 1) ^ {\operatorname {p r t} (w)} w v _ {1} w ^ {- 1} (- 1) ^ {\operatorname {p r t} (w)} (- 1) ^ {\operatorname {p r t} (w)} w v _ {2} w ^ {- 1} (351) \\ = (- 1) ^ {\operatorname {p r t} (w)} \rho_ {1} (w) \left(v _ {1}\right) \rho_ {1} (w) \left(v _ {2}\right) (352) \\ \neq \rho_ {1} (w) \left(v _ {1}\right) \rho_ {1} (w) \left(v _ {2}\right). (353) \\ \end{array}
$$
The lack of multiplicativity means that reflection and taking geometric product does not commute.
To fix this, it makes sense to first restrict  $\rho_{1}(w)$  to  $V$ , where it coincides with  $r_w$  and also still is multiplicative in  $w$ , and then study under which conditions on  $w$  it extends to an algebra homomorphism  $\operatorname{Cl}(V, \mathfrak{q}) \to \operatorname{Cl}(V, \mathfrak{q})$ .
More formally, by the universal property of the Clifford algebra, the obstruction for:
$$
\rho (w): V \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \rho (w) (v) := \alpha (w) v w ^ {- 1} = w \eta (w) v w ^ {- 1}, \quad \eta (w) := w ^ {- 1} \alpha (w), \tag {354}
$$
with general invertible  $w \in \mathrm{Cl}^{\times}(V, \mathfrak{q})$ , to extend to an  $\mathbb{F}$ -algebra homomorphism:
$$
\rho (w): \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \tag {355}
$$
is the following:
$$
\begin{array}{l} \forall v \in V. \quad \mathfrak {q} (v) \stackrel {!} {=} (\rho (w) (v)) ^ {2} (356) \\ = \rho (w) (v) \rho (w) (v) (357) \\ = (w \eta (w) v w ^ {- 1}) (w \eta (w) v w ^ {- 1}) (358) \\ = w \eta (w) v \eta (w) v w ^ {- 1}, (359) \\ \end{array}
$$
which reduces to:
$$
\forall v \in V. \quad \mathbf {q} (v) \stackrel {!} {=} \eta (w) v \eta (w) v. \tag {360}
$$
The latter is, for instance, satisfied if  $\eta(w)$  commutes with every  $v \in V$  and  $1 = \eta(w)^2$ . In particular, the above requirement is satisfied for all  $w \in \mathrm{Cl}^{\times}(V, \mathfrak{q})$  with  $\eta(w) \in \{\pm 1\}$ , which is equivalent to  $\alpha(w) = \pm w$ , which means that  $w$  is a homogeneous element of  $\mathrm{Cl}(V, \mathfrak{q})$  in the parity grading. This discussion motivates the following definitions and analysis.
Notation E.2 (The coboundary of  $\alpha$ ). The coboundary  $\eta$  of  $\alpha$  is defined on  $w \in \mathrm{Cl}^{\times}(V, \mathfrak{q})$  as follows:
$$
\eta : \mathrm {C l} ^ {\times} (V, \mathfrak {q}) \rightarrow \mathrm {C l} ^ {\times} (V, \mathfrak {q}), \quad \eta (w) := w ^ {- 1} \alpha (w). \tag {361}
$$
Remark E.3. 1.  $\eta$  is a crossed group homomorphism (aka 1-cocycle), i.e. for  $w_{1}, w_{2} \in \mathrm{Cl}^{\times}(V, \mathfrak{q})$  we have:
$$
\eta \left(w _ {1} w _ {2}\right) = \eta \left(w _ {1}\right) ^ {w _ {2}} \eta \left(w _ {2}\right), \quad \text {w i t h} \quad \eta \left(w _ {1}\right) ^ {w _ {2}} := w _ {2} ^ {- 1} \eta \left(w _ {1}\right) w _ {2}. \tag {362}
$$
2. For  $w\in \mathrm{Cl}^{\times}(V,\mathfrak{q})$  we have:
$$
\alpha (\eta (w)) = \alpha (w) ^ {- 1} w = \eta (w) ^ {- 1}. \tag {363}
$$
3. For  $w \in \mathrm{Cl}^\times(V, \mathfrak{q})$  we have that  $w$  is an homogeneous element in  $\operatorname{Cl}(V, \mathfrak{q})$ , in the sense of parity, if and only if  $\eta(w) \in \{\pm 1\}$ .
Definition E.4 (The group of homogeneous invertible elements). With the introduced notation we can define the group of all invertible elements of  $\operatorname{Cl}(V, \mathfrak{q})$  that are also homogeneous (in the sense of parity) as:
$$
\begin{array}{l} \operatorname {C l} ^ {[ \times ]} (V, \mathfrak {q}) := \left(\operatorname {C l} ^ {\times} (V, \mathfrak {q}) \cap \operatorname {C l} ^ {[ 0 ]} (V, \mathfrak {q})\right) \cup \left(\operatorname {C l} ^ {\times} (V, \mathfrak {q}) \cap \operatorname {C l} ^ {[ 1 ]} (V, \mathfrak {q})\right) (364) \\ = \left\{w \in \operatorname {C l} ^ {\times} (V, \mathfrak {q}) \mid \eta (w) \in \{\pm 1 \} \right\}. (365) \\ \end{array}
$$
Notation E.5 (The main involution - revisited). We now make the following abbreviations:
$$
\alpha^ {0} := \operatorname {i d}: \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \alpha^ {0} (x) := x ^ {[ 0 ]} + x ^ {[ 1 ]} = x, \tag {366}
$$
$$
\alpha^ {1} := \alpha : \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \alpha^ {1} (x) := x ^ {[ 0 ]} - x ^ {[ 1 ]}. \tag {367}
$$
For  $w\in \mathrm{Cl}(V,\mathfrak{q})$  we then have:
$$
\alpha^ {\operatorname {p r t} (w)}: \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \alpha^ {\operatorname {p r t} (w)} (x) = x ^ {[ 0 ]} + (- 1) ^ {\operatorname {p r t} (w)} x ^ {[ 1 ]}. \tag {368}
$$
Note that  $\alpha^{\mathrm{prt}(w)}$  is an  $\mathbb{F}$ -algebra involution of  $\operatorname{Cl}(V, \mathfrak{q})$  that preserves the parity grading of  $\operatorname{Cl}(V, \mathfrak{q})$ . We also need the following slight variation  $\alpha^w$ , which in many, but not all cases, coincides with  $\alpha^{\mathrm{prt}(w)}$ . For  $w \in \operatorname{Cl}^\times(V, \mathfrak{q})$  we define the  $w$ -twisted map:
$$
\alpha^ {w}: \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \alpha^ {w} (x) := x ^ {[ 0 ]} + \eta (w) x ^ {[ 1 ]}. \tag {369}
$$
Remark E.6. Note that for  $w \in \mathrm{Cl}^{[\times]}(V, \mathfrak{q})$  we have that  $\eta(w) = (-1)^{\mathrm{prt}(w)}$  and thus  $\alpha^w = \alpha^{\mathrm{prt}(w)}$ , in which case  $\alpha^w$  is an  $\mathbb{F}$ -algebra involution of  $\operatorname{Cl}(V, \mathfrak{q})$  that preserves the parity grading of  $\operatorname{Cl}(V, \mathfrak{q})$ .
Definition E.7 (Adjusted twisted conjugation). For  $w \in \mathrm{Cl}^{\times}(V, \mathfrak{q})$  we define the twisted conjugation:
$$
\begin{array}{l} \rho (w): \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad \rho (w) (x) := w x ^ {[ 0 ]} w ^ {- 1} + \alpha (w) x ^ {[ 1 ]} w ^ {- 1}. (370) \\ = w \left(x ^ {[ 0 ]} + \eta (w) x ^ {[ 1 ]}\right) w ^ {- 1} (371) \\ = w \alpha^ {w} (x) w ^ {- 1}. (372) \\ \end{array}
$$
We now want to re-investigate the action of  $\rho$  on  $\operatorname{Cl}(V, \mathfrak{q})$ .
Lemma E.8. For every  $w\in \mathrm{Cl}^{[\times ]}(V,\mathfrak{q})$  the map:
$$
\rho (w): \operatorname {C l} (V, \mathfrak {q}) \rightarrow \operatorname {C l} (V, \mathfrak {q}), \quad x \mapsto \rho (w) (x) = w \left(x ^ {[ 0 ]} + \eta (w) x ^ {[ 1 ]}\right) w ^ {- 1}, \tag {373}
$$
is an  $\mathbb{F}$ -algebra automorphism that preserves the parity grading of  $\operatorname{Cl}(V, \mathfrak{q})$ . Its inverse is given by  $\rho(w^{-1})$ .
Proof. First note that  $\alpha^w$  agrees with the involution  $\alpha^{\mathrm{prt}(w)}$  for  $w\in \operatorname {Cl}^{[\times ]}(V,\mathfrak{q})$ . Since  $\alpha^{\mathrm{prt}(w)}$  is an  $\mathbb{F}$ -algebra automorphism that preserves the parity grading of  $\operatorname {Cl}(V,\mathfrak{q})$  so is  $\alpha^w$  for  $w\in \operatorname {Cl}^{[\times ]}(V,\mathfrak{q})$ .
Furthermore, the conjugation  $\rho_0(w):x\mapsto wxx^{-1}$  is an  $\mathbb{F}$ -algebra automorphism, which preserves the parity grading of  $\operatorname {Cl}(V,\mathfrak{q})$  if  $w\in \mathrm{Cl}^{[\times ]}(V,\mathfrak{q})$ . To see this, note that  $0 = \operatorname {prt}(1) = \operatorname {prt}(ww^{-1}) = \operatorname {prt}(w) + \operatorname {prt}(w^{-1})$ . As such,  $\operatorname {prt}(w) = \operatorname {prt}(w^{-1})$ . Then,  $\operatorname {prt}(wxw^{-1}) = \operatorname {prt}(w) + \operatorname {prt}(x) + \operatorname {prt}(w^{-1}) = 2\operatorname {prt}(w) + \operatorname {prt}(x) = \operatorname {prt}(x)$ . Here, we use the fact that the Clifford algebra is  $\mathbb{Z} / 2\mathbb{Z}$ -graded.
So, their composition  $\rho(w) = \rho_0(w) \circ \alpha^w$  is also an  $\mathbb{F}$ -algebra automorphism that preserves the parity grading  $\operatorname{Cl}(V, \mathfrak{q})$  for  $w \in \operatorname{Cl}^{[\times]}(V, \mathfrak{q})$ .
As a direct corollary we get:
Corollary E.9. Let  $F(T_{0,1},\ldots ,T_{1,\ell})\in \mathbb{F}[T_{0,1},\ldots ,T_{1,\ell}]$  be a polynomial in  $2\ell$  variables with coefficients in  $\mathbb{F}$ . Let  $x_{1},\dots,x_{\ell}\in \mathrm{Cl}(V,\mathfrak{q})$  be  $\ell$  elements of the Clifford algebra and  $w\in \operatorname{Cl}^{[\times ]}(V,\mathfrak{q})$  be a homogeneous invertible element of  $\operatorname {Cl}(V,\mathfrak{q})$ . Then we have the following equivariance property:
$$
\rho (w) \left(F \left(x _ {1} ^ {[ 0 ]}, \dots , x _ {l} ^ {[ i ]}, \dots , x _ {\ell} ^ {[ 1 ]}\right)\right) = F (\rho (w) \left(x _ {1}\right) ^ {[ 0 ]}, \dots , \rho (w) \left(x _ {l}\right) ^ {[ i ]}, \dots , \rho (w) \left(x _ {\ell}\right) ^ {[ 1 ]}). \tag {374}
$$
Proof. This directly follows from Lemma E.8.  $\square$
Furthermore, we get the following result:
Theorem E.10. The map:
$$
\rho : \operatorname {C l} ^ {[ \times ]} (V, \mathfrak {q}) \rightarrow \operatorname {A u t} _ {\mathbf {A l g}, \operatorname {p r t}} (\operatorname {C l} (V, \mathfrak {q})) , \quad w \mapsto \rho (w), \tag {375}
$$
is a well-defined group homomorphism from the group of all homogeneous invertible elements of  $\operatorname{Cl}(V, \mathfrak{q})$  to the group of  $\mathbb{F}$ -algebra automorphisms of  $\operatorname{Cl}(V, \mathfrak{q})$  that preserve the parity grading of  $\operatorname{Cl}(V, \mathfrak{q})$ . In particular,  $\operatorname{Cl}(V, \mathfrak{q})$ ,  $\operatorname{Cl}^{[0]}(V, \mathfrak{q})$ ,  $\operatorname{Cl}^{[1]}(V, \mathfrak{q})$  are group representations of  $\operatorname{Cl}^{[\times]}(V, \mathfrak{q})$  via  $\rho$ .
Proof. By the previous Lemma E.8 we already know that  $\rho$  is a well-defined map. We only need to check if it is a group homomorphism. Let  $w_{1}, w_{2} \in \mathrm{Cl}^{[\times]}(V, \mathfrak{q})$  and  $x \in \mathrm{Cl}(V, \mathfrak{q})$ , then we get: