# On the structure and representation theory of $q$-deformed Clifford algebras

Willie Aboumrad  The Institute for Computational and Mathematical Engineering
(ICME) at Stanford University<EMAIL_ADDRESS>https://web.stanford.edu/
willieab and  Travis Scrimshaw  Faculty of Science, Hokkaido University, 5
Chōme Kita 8 Jōnishi, Kita Ward, Sapporo, Hokkaidō 060-0808
<EMAIL_ADDRESS>https://tscrim.github.io/

###### Abstract.

We provide a generalized definition for the quantized Clifford algebra
introduced by Hayashi using another parameter $k$ that we call the twist. For
a field of characteristic not equal to $2$, we provide a basis for our
quantized Clifford algebra, show that it can be decomposed into rank $1$
components, and compute its center to show it is a classical Clifford algebra
over the group algebra of a product of cyclic groups of order $2k$. In
addition, we characterize the semisimplicity of our quantum Clifford algebra
in terms of the semisimplicity of a cyclic group of order $2k$ and give a
complete set of irreducible representations. We construct morphisms from
quantum groups and explain various relationships between the classical and
quantum Clifford algebras. By changing our generators, we provide a further
generalization to allow $k$ to be a half integer, where we recover certain
quantum Clifford algebras introduced by Fadeev, Reshetikhin, and Takhtajan as
a special case.

###### Key words and phrases:

quantum Clifford, basis, semisimple

###### 2010 Mathematics Subject Classification:

16G10, 16D60, 81R05, 15A66

T.S. was partially supported by Grant-in-Aid for JSPS Fellows 21F51028.

##  1\. Introduction

Clifford algebras (over $\mathbb{C}$) are a certain class of superalgebras
defined by a universal property involving a symmetric bilinear form
(equivalently a quadratic form) that appear naturally in the study of particle
systems to describe fermionic creation and annihilation operators. We build a
Clifford algebra $\operatorname{Cl}(V\oplus V^{*})$ by taking an $n$
dimensional vector space $V$ and its algebraic dual $V^{*}$ to be maximal
isotropic subspaces under the natural symmetric bilinear form defined by
$\psi_{i}^{*}(\psi_{j})=\delta_{ij}$ for dual bases $\\{\psi_{i}\\}_{i=1}^{n}$
and $\\{\psi_{i}^{*}\\}_{i=1}^{n}$ of $V$ and $V^{*}$ respectively. We can
also form the (s)pin group, which is the universal cover of the (special)
orthogonal group and is used to describe the symmetries of fermions, from the
Clifford algebra. By taking $n\to\infty$ in a suitable way, we arrive at the
(fermionic) Fock space that is used to define representations of the infinite
rank general linear Lie algebra $\mathfrak{gl}_{\infty}$ and the affine
general linear Lie algebra $\widehat{\mathfrak{gl}}_{n}$ through the boson-
fermion correspondence. More precisely, this comes from the spinor
representation of the Clifford algebra $\operatorname{Cl}(V\oplus V^{*})$ that
is constructed on the exterior algebra $\bigwedge V$, which is the unique (up
to isomorphism) simple module of a Clifford algebra. Taking $V$ to be the
natural $\mathfrak{gl}_{n}$ representation, the exterior power
$\bigwedge^{k}V$ is the irreducible fundamental representation
$V(\Lambda_{k})$ of $\mathfrak{gl}_{n}$. Given the relation with the spin
group, we also obtain representations of the special orthogonal Lie algebra
$\mathfrak{so}_{n}$, with a slight dependency on the parity of $n$. See, e.g.,
[FH91] for more details.

There is a $q$-deformation of the representation theory of a Lie algebra
$\mathfrak{g}$ through its corresponding (Drinfel’d–Jimbo) quantum group
$U_{q}(\mathfrak{g})$. We refer the reader to standard textbooks such as
[CP94, Jan96, Kas95] for more information. There are analogous
$q$-deformations of the exterior algebra that have been used to construct
$U_{q}(\mathfrak{g})$-representations, such as [DF94, Hay90, JMMO91, JMO00,
KMS95, Kwo14, LT96]. One interesting development is the relationship between
the global crystal basis, in the sense of Kashiwara [Kas90, Kas91], and Hecke
algebras at roots of unity first conjectured by Lascoux, Leclerc, and Thibon
[LLT96] and later proven by Ariki [Ari96]. Some additional recent developments
can be found in [Ger19].

In this paper, we focus on the construction of a $q$-deformation of the
Clifford algebra given by Hayashi [Hay90] that we call the quantum Clifford
algebra. In [Kwo14], Kwon uses the quantum Clifford algebra that is attributed
to Hayashi, but the presentation is slightly different. Indeed, the dimensions
of the two algebras are different, although this is done to account for the
difference in the quantum groups used in their constructions. In this paper,
we generalize these constructions to have a second parameter $k$ that we call
the twist; the case $k=1$ was used in [Kwo14] while the case $k=2$ in [Hay90].
Furthermore, since the Clifford algebra construction works over an arbitrary
field $\mathbf{k}$ of chararacteristic not equal to $2$, we work at this level
of generality.111Modifications can be made for the characteristic $2$ case,
but the situation is drastically different. We denote the rank $n$ and twist
$k$ quantum Clifford algebra as $\operatorname{Cl}_{q}(n,k)$, where
$q\in\mathbf{k}\setminus\\{0\\}$.

In order to work with these algebras, we first construct a basis (3.4)
analogous to the usual Clifford algebra basis through a series of reduction
rules to a standard form. This allows us to show the dimension of
$\operatorname{Cl}_{q}(n,k)$ is equal to $(8k)^{n}$ (3.5). This recovers the
remark at the end of [Hay90] due to M. Takeuchi about the dimension of
$\operatorname{Cl}_{q}(n,2)$ over $\mathbb{C}$. In Section 6, we expand this
remark to our level of generality provided we have all $2k$th roots of unity
in $\mathbf{k}$. Using our basis, we are then able to compute the center of
$\operatorname{Cl}_{q}(n,k)$ as the group algebra
$\mathbf{k}[\mathbb{Z}_{2k}^{n}]$ of a product of cyclic groups of order $2k$
(3.12). We also extend the morphisms from a quantum group (3.17) given in
[Hay90].

Since the quantum Clifford algebra is a superalgebra (that is, it has a
$\mathbb{Z}_{2}$-grading), we can decompose it into a signed tensor product of
rank $1$ Clifford algebras, so

$\operatorname{Cl}_{q}(n,k)\cong\operatorname{Cl}_{q}(1,k)\,\widehat{\otimes}\,\operatorname{Cl}_{q}(1,k)\,\widehat{\otimes}\,\cdots\,\widehat{\otimes}\,\operatorname{Cl}_{q}(1,k),$

analogous to the classical case. However, it will be useful to decompose this
as a usual tensor product of algebras (3.10),

$\operatorname{Cl}_{q}(n,k)\cong\operatorname{Cl}_{q}(1,k)\otimes\operatorname{Cl}_{q}(1,k)\otimes\cdots\otimes\operatorname{Cl}_{q}(1,k),$

which is also analogous to the classical case. Following the classical case,
we construct this isomorphism by introducing a quantum analog of the volume
element. The volume element of the usual Clifford algebra is an important
element as it can be used to prove Bott periodicity, as explained in [ML90,
Prop. 3.5, Thm. 3.7]. It also has important representation-theoretic
implications as it can be used to classify irreducible representations of any
Clifford algebra; see [ML90, Prop.5.9, 5.10]. In spin geometry, the volume
element gives any Dirac bundle a $\mathbb{Z}_{2}$-grading [ML90, Ch. 6], and
in relativistic quantum mechanics, it is used to define chirality for spinors
[TFV20].

While $\operatorname{Cl}_{q}(n,k)$ behaves much like its classical
counterpart, there are a number of important distinctions coming from the
existence of a nontrivial center. The first is that
$\operatorname{Cl}_{q}(n,k)$ is no longer a simple algebra, but its
semisimplicity is tied directly to the semisimplicity of its center (4.6).
This is a reflection of the fact that $\operatorname{Cl}_{q}(n,k)$ could be
seen as a classical Clifford algebra over its center. This also allows us to
construct all irreducible representations, which we can easily characterize
when we have all $2k$th roots of unity in $\mathbf{k}$ (4.1 and 4.5). Although
our given proof uses the explicit structure from our basis and the (unsigned)
tensor product decomposition. Moreover, our construction of the irreducible
representations uses the braided exterior algebra of [BZ05], which is
equivalent to the natural generalization of Hayashi’s construction [Hay90] on
the usual exterior algebra but is better behaved when taking tensor products
of quantum group representations.

Next, by observing the relations, we note that almost all relations can be
defined when $k\in\frac{1}{2}\mathbb{Z}_{>0}$. We extend to this case by
slightly changing our generating set in Section 5, which leads to analogous
results in the half twist cases. In particular, we show that the quantum
Clifford algebra can recover the usual Clifford algebra when $k=\frac{1}{2}$
(5.3). Furthermore, it becomes easy to see that this is the quantum Clifford
algebra construction given by Ding and Frenkel [DF94] based upon the FRT
construction [RTF89]. Thus, we show that our family of quantum Clifford
algebras encompasses the seemingly different constructions of [DF94] and
[Hay90].

Let us discuss an application of our results to quantized skew Howe duality,
which is a relationship between commutating actions of a quantum group and (a
coideal subalgebra of) a possibly different quantum group on a braided
exterior algebra [CKM14, LZZ11, QS19, ST19]. The skew
$U_{q}(\mathfrak{gl}_{n})\otimes U_{q}(\mathfrak{gl}_{m})$-duality [Abo22b,
Theorem 3.19] (resp. skew $U_{q}(\mathfrak{o}_{2n})\otimes
U_{q}^{\prime}(\mathfrak{so}_{m})$-duality [Abo22a, Theorem 3.19]) relies on
embeddings of $U_{q}(\mathfrak{gl}_{n})$ and $U_{q}(\mathfrak{gl}_{m})$ (resp.
$U_{q}(\mathfrak{so}_{2n})$ and $U_{q}^{\prime}(\mathfrak{so}_{m})$) into
$\operatorname{Cl}_{q}(nm,1)$. These embeddings leverage the structure theorem
3.10 in order to define quantum group actions on certain tensor products of
modules.

This paper is organized as follows. In Section 2, we recall some facts and
properties of the classical Clifford algebra. In Section 3, we prove a number
of structural properties of the quantum Clifford algebra. In Section 4, we
discuss the representation theory of the quantum Clifford algebra. In Section
5, we discuss the half twist cases of the quantum Clifford algebra. In Section
6, we generalize the remark of Takeuchi in [Hay90].

### Acknowledgements

The authors thank Daniel Bump and Jae-Hoon Kwon for useful discussions. The
first author also thanks Daniel Bump for his patient guidance and caring
support throughout the years. The second author thanks Stanford University for
its hospitality during his visit in May, 2022.

This work benefited from computations using SageMath [Sag22]. Some of the
results in this work were discovered by the first author as part of his
dissertation research. This work was partly supported by Osaka City University
Advanced Mathematical Institute (MEXT Joint Usage/Research Center on
Mathematics and Theoretical Physics JPMXP0619217849).

##  2\. Background: Classical Clifford algebras

In this section we define the classical Clifford algebra and recall some of
its basic properties.

Every algebra considered here is unital and associative unless otherwise
stated. We fix a field $\mathbf{k}$ of characteristic different from $2$ and
compute all tensor products over $\mathbf{k}$. We denote by $\zeta_{r}$ a
primitive $r$-th root of unity in $\mathbf{k}$ (assuming it exists); so
$\zeta_{r}=e^{2\pi\mathbf{i}/r}$ with $\mathbf{i}=\sqrt{-1}$ in $\mathbb{C}$.
We use $\mathbb{Z}_{r}=\mathbb{Z}/r\mathbb{Z}$ to denote the (cyclic) group of
integers modulo $r$ under $+$, and let $e_{j}$ denote the $j$th standard basis
vector in $\mathbb{Z}^{n}$. For any ring $R$, we let $R^{\times}$ denote its
group of units. We use $\\{A,B\\}$ (resp. $[A,B]_{q}$) to denote the algebra
anticommutator (resp. $q$-commutator) of $A$ and $B$:

$\\{A,B\\}=AB+BA,\qquad\qquad[A,B]_{q}=AB-q\,BA.$

######  Definition 2.1 ([GW09, Def. 6.1.1]).

Let $W$ be a vector space over $\mathbf{k}$ equipped with a symmetric bilinear
form $\beta$. A _Clifford algebra_ for $(W,\beta)$ is an associative
$\mathbf{k}$-algebra $\operatorname{Cl}(W,\beta)$ and a linear map
$\gamma\colon W\hookrightarrow\operatorname{Cl}(W,\beta)$ satisfying the
following conditions:

  1. (i)

$\\{\gamma(v),\gamma(w)\\}=\beta(v,w)$ for every $v,w\in W$.

  2. (ii)

$\gamma(W)$ generates $\operatorname{Cl}(W,\beta)$ as an algebra.

  3. (iii)

(Universal Property) Given any unital associative $\mathbf{k}$-algebra
$\mathcal{A}$ with a linear map $\varphi\colon W\to\mathcal{A}$ such that
$\\{\varphi(v),\varphi(w)\\}=\beta(v,w)$, there exists an algebra homomorphism
$\widetilde{\varphi}\colon\operatorname{Cl}(W,\beta)\to\mathcal{A}$ such that
$\varphi=\widetilde{\varphi}\circ\gamma$. In other words, there is an algebra
map $\widetilde{\varphi}$ making the following diagram commute.

${W}$${\mathcal{A}}$${Cl(W,\beta)}$$\scriptstyle{\gamma}$$\scriptstyle{\varphi}$$\scriptstyle{\widetilde{\varphi}}$

From the universal property, Clifford algebras are unique (up to isomorphism)
for the pair $(W,\beta)$, and so we call $\operatorname{Cl}(W,\beta)$ the
Clifford algebra of $(W,\beta)$. We can explicitly construct it as a quotient
of the tensor algebra $T(W)=\bigoplus_{m=0}^{\infty}W^{\otimes m}$ modulo the
ideal generated by elements of the form $v\otimes v-\beta(v,v)1_{T(W)}$.
Hence, the natural $\mathbb{Z}$-grading on $T(W)$ descends to a
$\mathbb{Z}_{2}$-grading, making $\operatorname{Cl}(W,\beta)$ a
_superalgebra_. The _exterior algebra_ is when $\beta=0$ and is denoted
$\bigwedge(W)=\operatorname{Cl}(W,0)$, and it is supercommutative.

In this work, we mostly consider Clifford algebras for spaces of the form
$V\oplus V^{*}$, where $V^{*}:=\operatorname{Hom}_{\mathbf{k}}(V,\mathbf{k})$
is the usual algebraic dual of $V$. We equip $V\oplus V^{*}$ with the
symmetric bilinear form $\beta$ resulting from the duality pairing between $V$
and $V^{*}$:

$\beta\big{(}(v,f),(w,h)\big{)}=f(w)+h(v),\quad v,w\in V,\,\,f,h\in V^{*}.$
(2.1)

Hence $V$ and $V^{*}$ are maximal isotropic subspaces with respect to $\beta$;
that is, they are Lagrangian subspaces of $V\oplus V^{*}$. For simplicity, we
write $\operatorname{Cl}(V\oplus V^{*})=\operatorname{Cl}(V\oplus
V^{*},\,\beta)$. We will work in a fixed basis $\\{\psi_{a}\\}_{a=1}^{n}$ for
$V$, and let $\\{\psi_{b}^{*}\\}_{b=1}^{n}$ denote the dual basis in $V^{*}$.
Then the $\psi_{a}$ and $\psi_{b}^{*}$ satisfy the _canonical anticommutation
relations_

$\displaystyle\begin{gathered}\psi_{a}\psi_{b}+\psi_{b}\psi_{a}=\psi_{a}^{*}\psi_{b}^{*}+\psi_{b}^{*}\psi_{a}^{*}=0,\\\
\psi_{a}\psi_{b}^{*}+\psi_{b}^{*}\psi_{a}=\delta_{ab}.\end{gathered}$ (2.4)

Recall that $\operatorname{Cl}(V\oplus V^{*})$ is a central simple algebra,
and therefore it has a unique irreducible representation called the _spinor
representation_ , up to isomorphism. See, e.g., [Bum04, Thm. 31.1], [Knu91,
Ch. IV], [Lam05, Ch. 5], or [O’M00, Ch. V] for details. The irreducible
$\operatorname{Cl}(V\oplus V^{*})$-module may be realized on the exterior
algebra $\bigwedge(V)$ by imposing that $\psi_{i}^{*}\cdot
1_{\bigwedge(V)}=0$; equivalently, this is $\operatorname{Cl}(V\oplus
V^{*})/J^{*}$, where $J^{*}$ is the left ideal generated by $V^{*}$.

We conclude this section by constructing an isomorphism

$\Gamma_{V}\colon\operatorname{Cl}(V\oplus V^{*})^{\otimes
m}\to\operatorname{Cl}\left(V\otimes W)\oplus(V\otimes W)^{*}\right),$

with $m=\dim W$ that we will generalize to the quantum setting in the sequel.
For each $r=1,\ldots,n$, define the _volume_ or _chirality element_ of the
subalgebra
$\operatorname{Cl}(\mathbf{k}^{r}\oplus(\mathbf{k}^{r})^{*})\subset\operatorname{Cl}(V\oplus
V^{*})$ by

$\displaystyle\overline{f}_{r}=[\psi_{1},\psi_{1}^{*}]\cdots[\psi_{r},\psi_{r}^{*}].$
(2.5)

This name comes from rewriting Equation 2.5 as

$\overline{f}_{r}=\epsilon_{1}\cdots\epsilon_{2r},$

where

$\epsilon_{2j-1}=\psi_{j}^{*}-\psi_{j}\quad\text{and}\quad\epsilon_{2j}=\psi_{j}^{*}+\psi_{j}$
(2.6)

(cf. [ML90]) since $\epsilon_{2j-1}\epsilon_{2j}=[\psi_{j},\psi_{j}^{*}]$. The
$\epsilon_{j}$ correspond to the (standard) orthonormal basis of
$\mathbf{k}^{2n}$ with the bilinear form defined by (2.1), which becomes a
diagonal bilinear form $\delta$ in this basis so that
$\operatorname{Cl}(V\oplus
V^{*})\cong\operatorname{Cl}(\mathbf{k}^{2n},\delta)$ (see Equation 6.2 and
Equation 6.3 for a precise example). Hence, $\overline{f}_{r}$ is the volume
element in the corresponding exterior algebra $\bigwedge(\mathbf{k}^{2r})$.
Contrast this with $\psi_{j},\psi_{j}^{*}$, which define an isotropic basis of
$\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}$ with the polarized bilinear form of
(2.1).

In this context we use the volume element to construct a superalgebra
isomorphism expressing a tensor product of Clifford algebras as another
Clifford algebra. This map is motivated by [Wen12, Lemma 1.2].

######  Proposition 2.2.

Suppose $V$ and $W$ are $\mathbf{k}$-vector spaces with $\dim V=n$ and $\dim
W=m$. Let $\phi_{a}$ denote either $\psi_{a}$ or $\psi_{a}^{*}$. There is a
superalgebra isomorphism $\Gamma_{V}\colon\operatorname{Cl}(V\oplus
V^{*})^{\otimes m}\to\operatorname{Cl}\left((V\otimes W)\oplus(V\otimes
W)^{*}\right)$ satisfying

$\displaystyle 1\otimes\cdots\otimes\phi_{a}\otimes\cdots\otimes
1\to\overline{f}_{(j-1)n}\phi_{a+(j-1)n},$

where on the left $\phi_{a}$ appears in the $j$th tensor factor of
$\operatorname{Cl}(V\oplus V^{*})^{\otimes m}$.

######  Remark 2.3.

As the Clifford algebra is a superalgebra, clearly $\operatorname{Cl}(V\oplus
W,\,\beta_{1}\oplus\beta_{2})\cong\operatorname{Cl}(V,\beta_{1})\,\widehat{\otimes}\,\operatorname{Cl}(W,\beta_{2})$,
with $\widehat{\otimes}$ denoting the graded (or signed) tensor product of
superalgebras. Furthermore, there is a natural superalgebra structure on
$\operatorname{Cl}(V\oplus V^{*})^{\otimes m}$ using the ordinary tensor
product, but the product structure is reflecting the fact that Clifford
algebras are not supercommutative. In particular, the volume elements
$\overline{f}_{r}$ introduce signs when appropriate.

##  3\. Quantized Clifford algebras

In this section we prove structural properties of the quantum Clifford
algebra. To begin, we provide a generalized definition of the quantized
Clifford algebra. The first main result in this section is an explicit basis
in 3.4 proven by giving a reduction algorithm. Our other main result is
computing the center in 3.12. This computation relies on the tensor product
factorization of 3.10, the quantum analog of 2.2, and the construction a
quantum volume element, which enjoys many of the same properties as its
classical counterpart.

######  Definition 3.1.

Choose $q\in\mathbf{k}^{\times}$ and let $n,k\in\mathbb{Z}_{>0}$. The _quantum
Clifford algebra_ $\operatorname{Cl}_{q}(n,k)$ of _rank_ $n$ and _twist_ $k$
is the associative $\mathbf{k}$-algebra generated by $\psi_{a}$,
$\psi_{a}^{*}$, $\omega_{a}$, and $\omega_{a}^{-1}$, for
$a\in\\{1,\dotsc,n\\}$, subject to the relations

$\displaystyle\omega_{a}\omega_{b}$ $\displaystyle=\omega_{b}\omega_{a},$
$\displaystyle\omega_{a}\omega_{a}^{-1}$ $\displaystyle=1,$
$\displaystyle\omega_{a}\psi_{b}$
$\displaystyle=q^{\delta_{ab}}\psi_{b}\omega_{a},$
$\displaystyle\omega_{a}\psi_{b}^{*}$
$\displaystyle=q^{-\delta_{ab}}\psi_{b}^{*}\omega_{a},$
$\displaystyle\psi_{a}\psi_{b}$ $\displaystyle+\psi_{b}\psi_{a}=0,$
$\displaystyle\psi_{a}^{*}\psi_{b}^{*}$
$\displaystyle+\psi_{b}^{*}\psi_{a}^{*}=0,$ (3.1)
$\displaystyle\psi_{a}\psi_{a}^{*}$
$\displaystyle+q^{k}\psi_{a}^{*}\psi_{a}=\omega_{a}^{-k},$
$\displaystyle\psi_{a}\psi_{a}^{*}$
$\displaystyle+q^{-k}\psi_{a}^{*}\psi_{a}=\omega_{a}^{k},$
$\displaystyle\psi_{a}\psi_{b}^{*}$ $\displaystyle+\psi_{b}^{*}\psi_{a}=0$
$\displaystyle\text{if }a\neq b.$

The parameter $n$ refers to the number of pairs of
$(\psi_{j},\psi_{j}^{\dagger})$ generators, so we have only defined a quantum
analog for the even dimensional Clifford algebra
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$. The
case $k=2$ was given by Hayashi [Hay90] and the $k=1$ version was utilized by
Kwon [Kwo14].

The defining relations imply that

$(q\omega_{a})^{k}-(q\omega_{a})^{-k}=(q^{k}-q^{-k})\psi_{a}\psi_{a}^{*},\qquad\qquad\omega_{a}^{-k}-\omega_{a}^{k}=(q^{k}-q^{-k})\psi_{a}^{*}\psi_{a}.$
(3.2)

When $q^{2k}\neq 1$, we obtain an alternative presentation of
$\operatorname{Cl}_{q}(n,k)$ using Equation 3.2 by replacing the $q^{\pm
k}$-commutor relations for $\psi_{a}$ and $\psi_{a}^{*}$ with

$\psi_{a}\psi_{a}^{*}=\frac{(q\omega_{a})^{k}-(q\omega_{a})^{-k}}{q^{k}-q^{-k}},\qquad\qquad\psi_{a}^{*}\psi_{a}=-\frac{\omega_{a}^{k}-\omega_{a}^{-k}}{q^{k}-q^{-k}}.$
(3.3)

Notice there is a $\mathbb{Z}^{n}$-grading on $\operatorname{Cl}_{q}(n,k)$
defined by

$\deg(\psi_{a})=e_{a},\qquad\qquad\deg(\psi_{a}^{*})=-e_{a},\qquad\qquad\deg(\omega_{a})=0.$
(3.4)

This naturally yields a $\mathbb{Z}$-grading (by the linear map $e_{a}\mapsto
1\in\mathbb{Z}$ for all $a$) and a $\mathbb{Z}_{2}$-grading (by $e_{a}\mapsto
1\in\mathbb{Z}_{2}$), so $\operatorname{Cl}_{q}(n,k)$ is a superalgebra.

###  3.1. Basis

We construct a series of reduction rules by manipulating the defining
relations. We then obtain bases for $\operatorname{Cl}_{q}(n,k)$ using these
reduction rules.

######  Lemma 3.2.

The following identities hold in $\operatorname{Cl}_{q}(n,k)$.

$\displaystyle\psi_{a}\omega_{a}^{2k}$ $\displaystyle=q^{-2k}\psi_{a},$ (3.5)
$\displaystyle\psi_{a}^{*}\omega_{a}^{2k}$ $\displaystyle=\psi_{a}^{*},$ (3.6)
$\displaystyle\omega_{a}^{-k}$
$\displaystyle=(q^{2k}+1)\omega_{a}^{k}-q^{2k}\omega_{a}^{3k}.$ (3.7)

###### Proof.

The defining relations imply $\psi_{a}^{2}=(\psi_{a}^{*})^{2}=0$. In addition,
they imply

$\psi_{a}\omega_{a}^{-k}=\psi_{a}(\psi_{a}\psi_{a}^{*}+q^{k}\psi_{a}^{*}\psi_{a})=q^{k}\psi_{a}\psi_{a}^{*}\psi_{a}=q^{k}\psi_{a}(q^{k}\omega^{k}-q^{k}\psi_{a}\psi_{a}^{*})=q^{2k}\psi_{a}\omega_{a}^{k}.$

Multiplying both sides on the right by $q^{-2k}\omega^{k}$ yields Equation
3.5. Similarly, we calculate that

$\psi_{a}^{*}\omega_{a}^{k}=\psi_{a}^{*}(\psi_{a}\psi_{a}^{*}+q^{-k}\psi_{a}^{*}\psi_{a})=\psi_{a}^{*}\psi_{a}\psi_{a}^{*}=\psi_{a}^{*}(\omega_{a}^{-k}-q^{k}\psi_{a}^{*}\psi_{a})=\psi_{a}^{*}\omega_{a}^{-k}.$

This implies Equation 3.6. Applying Equations 3.2 and 3.5 yields

$\displaystyle\omega_{a}^{-k}-\omega_{a}^{k}$
$\displaystyle=(q^{k}-q^{-k})\psi_{a}^{*}\psi_{a}=q^{2k}(q^{k}-q^{-k})\psi_{a}^{*}\psi_{a}\omega_{a}^{2k}=q^{2k}(\omega_{a}^{-k}-\omega_{a}^{k})\omega_{a}^{2k}$
$\displaystyle=q^{2k}\omega_{a}^{k}-q^{2k}\omega_{a}^{3k}.$

Solving for $\omega_{a}^{-k}$ yields Equation 3.7. ∎

We can use Equation 3.7 to express $\omega_{a}^{-1}$ as a polynomial in
$\mathbf{k}[\omega_{a}]$, since it is equivalent to

$\omega_{a}^{4k}-(1+q^{-2k})\omega_{a}^{2k}+q^{-2k}=0.$ (3.8)

Thus, the quantum Clifford algebra is generated by the $3n$ generators
$\psi_{a},\psi_{a}^{*},\omega_{a}$, for $a=1,\ldots,n$. Notice that if
$q^{2k}=1$, Equation 3.8 factors as

$(\omega_{a}^{2k}-1)^{2}=0.$ (3.9)

######  Lemma 3.3.

The following relations hold in $\operatorname{Cl}_{q}(n,k)$:

$\displaystyle\psi_{a}^{*}\psi_{a}$
$\displaystyle=q^{k}\omega_{a}^{k}-q^{k}\psi_{a}\psi_{a}^{*},$
$\displaystyle\psi_{a}\psi_{a}^{*}\psi_{a}$
$\displaystyle=q^{k}\psi_{a}\omega_{a}^{k},$
$\displaystyle\psi_{a}\psi_{a}^{*}\psi_{a}\psi_{a}^{*}$
$\displaystyle=\psi_{a}\psi_{a}^{*}\omega_{a}^{k},$
$\displaystyle\psi_{a}^{*}\psi_{a}\psi_{a}^{*}$
$\displaystyle=\psi_{a}^{*}\omega_{a}^{k},$ $\displaystyle\omega_{a}^{2k}$
$\displaystyle=(1-q^{-2k})\psi_{a}\psi_{a}^{*}\omega_{a}^{k}+q^{-2k}.$

###### Proof.

We obtain the last relation as follows:

$\omega_{a}^{2k}=\psi_{a}\psi_{a}^{*}\omega_{a}^{k}+q^{-k}\psi_{a}^{*}\psi_{a}\omega_{a}^{k}=\psi_{a}\psi_{a}^{*}\omega_{a}^{k}+q^{-k}(q^{-k}\omega_{a}^{-k}-q^{-k}\psi_{a}\psi_{a}^{*})\omega_{a}^{k}=(1-q^{-2k})\psi_{a}\psi_{a}^{*}\omega_{a}^{k}+q^{-2k}.$

The remaining identities follow immediately from the defining relations (3.1).
∎

We construct a basis of $\operatorname{Cl}_{q}(n,k)$ by considering a normal
ordering of the generators analogous to the standard basis for the classical
Clifford algebra.

######  Theorem 3.4.

Consider the set $\mathcal{B}$ of monomials

$\prod_{a=1}^{n}\psi_{a}^{p_{a}}(\psi_{a}^{*})^{d_{a}}\omega_{a}^{v_{a}},$

where $p_{a},d_{a}\in\\{0,1\\}$ and $v_{a}\in\\{0,1,\dotsc,2k-1\\}$. Then
$\mathcal{B}$ is a basis of $\operatorname{Cl}_{q}(n,k)$.

###### Proof.

Clearly $\mathcal{B}$ contains $1$ and the generators of
$\operatorname{Cl}_{q}(n,k)$. Thus, we need to show that $\mathcal{B}$ is
linearly independent and closed under multiplication. Together with Lemma 3.3,
the $q$\- and skew-commuting relations show that $\mathcal{B}$ is closed under
multiplication. Indeed, these are a set of rewriting rules to reduce any
product of generators into a linear combination of elements in $\mathcal{B}$.
Furthermore, note that we cannot apply the rewriting rules to any element of
$\mathcal{B}$ (after having chosen an order on the elements
$\psi_{a},\psi_{a}^{*},\omega_{a}$). Hence, we just need to show $\mathcal{B}$
is linearly independent.

To show the linear independence, we need to show that the result of the
rewriting rules is unique and that they undo the application of any of the
defining relations. The latter property is clear (the only non-trivial one is
the $q$-commutor relations between $\psi_{a}$ and $\psi_{a}^{*}$). The
uniqueness is a straightforward argument using the diamond lemma. ∎

Observe that the linear independence of the monomials in $\mathcal{B}$ does
not depend on the particular order of the factors we chose in 3.4; other
orderings are possible and in fact they all differ by a power of $q$ and a
sign with the exception of $\psi_{a}^{*}\psi_{a}$. For the replacement
$\psi_{a}\psi_{a}^{*}\mapsto\psi_{a}^{*}\psi_{a}$, we simply use the
alternative reduction rule
$\psi_{a}\psi_{a}^{*}=\omega_{a}^{k}-q^{-k}\psi_{a}^{*}\psi_{a}$.

We immediately obtain the dimension of $\operatorname{Cl}_{q}(n,k)$ from 3.4.

######  Corollary 3.5.

We have

$\dim\operatorname{Cl}_{q}(n,k)=(8k)^{n}.$

Now notice that when $q^{2k}=1$, Equation 3.9 implies that
$\omega_{a}^{2k}=1$. We can also see this from the defining relations as

$\psi_{a}\psi_{a}^{*}+q^{k}\psi_{a}^{*}\psi_{a}=\omega_{a}^{-k}=\omega_{a}^{k}$

since $q^{k}=q^{-k}=\pm 1$. Conversely, when $q^{2k}\neq 1$, we can construct
another basis that demonstrates we have a commutative algebra of dimension
$(4k)^{n}$ generated by $\omega_{1},\ldots,\omega_{n}$.

######  Theorem 3.6.

Suppose $q^{2k}\neq 1$. Consider the set $\mathcal{B}^{\prime}$ of monomials

$\prod_{a=1}^{n}\psi_{a}^{p_{a}}(\psi_{a}^{*})^{d_{a}}\omega_{a}^{v_{a}},$

with $p_{a},d_{a}\in\\{0,1\\}$ and $v_{a}\in\\{0,1,\dotsc,4k-1\\}$, satisfying
the following conditions for all $a$:

  1. (1)

$p_{a}+d_{a}<2$,

  2. (2)

$v_{a}\leq k$ if $p_{a}+d_{a}=1$.

Then $\mathcal{B}^{\prime}$ is a basis of $\operatorname{Cl}_{q}(n,k)$.

We provide two proofs of 3.6. The first uses a series of reduction rules. It
is analogous to the proof of 3.4. The second uses an explicit invertible
change-of-basis map.

###### Proof: Reduction rules.

The reduction rules are given by 3.2, the defining skew-commuting and
$q$-commuting relations, and

$\displaystyle\psi_{a}\psi_{a}^{*}$
$\displaystyle=\frac{q^{k}\omega_{a}^{3k}-q^{-k}\omega_{a}^{k}}{q^{k}-q^{-k}}=\frac{\omega_{a}^{k}(q^{2k}\omega_{a}^{2k}-1)}{q^{2k}-1},$
(3.10a) $\displaystyle\psi_{a}^{*}\psi_{a}$
$\displaystyle=\frac{q^{2k}\omega_{a}^{k}(1-\omega_{a}^{2k})}{q^{k}-q^{-k}}=\frac{q^{3k}\omega_{a}^{k}(1-\omega_{a}^{2k})}{q^{2k}-1},$
(3.10b)

which are obtained from Equation 3.3 by removing $\omega_{a}^{-k}$ using
Equation 3.7. The remainder of the proof is similar to the proof of 3.4. ∎

###### Proof: Change-of-basis.

We will construct an invertible linear map from $\mathcal{B}$ from Theorem 3.4
to $\mathcal{B}^{\prime}$. Consider a monomial
$\prod_{a=1}^{n}\psi_{a}^{p_{a}}(\psi_{a}^{*})^{d_{a}}\omega_{a}^{v_{a}}$ in
$\mathcal{B}$. It belongs to $\mathcal{B}^{\prime}$ if and only if
$p_{a}+d_{a}<2$ for all $a$, so we need only consider the case when
$p_{a}+d_{a}=2$ for some $a$. Using Equation 3.10a, we can rewrite

$\psi_{a}\psi_{a}^{*}\omega_{a}^{v_{a}}=\frac{q^{2k}\omega_{a}^{3k+v_{a}}-\omega_{a}^{k+v_{a}}}{q^{2k}-1}.$

If $0\leq v_{a}<k$, this is an element of $\mathcal{B}^{\prime}$ with inverse
image

$\omega_{a}^{3k+v_{a}}=q^{-2k}\bigl{(}(q^{2k}-1)\psi_{a}\psi_{a}^{*}\omega_{a}^{v_{a}}+\omega_{a}^{k+v_{a}}\bigr{)}\in\mathcal{B}.$

Otherwise we have $k\leq v_{a}<2k$, and so we may apply Equation 3.8 to obtain

$\displaystyle\psi_{a}\psi_{a}^{*}\omega_{a}^{v_{a}}$
$\displaystyle=\frac{q^{2k}\omega_{a}^{v_{a}-k}\bigl{(}(1+q^{-2k})\omega_{a}^{2k}-q^{-2k}\bigr{)}-\omega_{a}^{k+v_{a}}}{q^{2k}-1}=\frac{\omega_{a}^{v_{a}+k}-q^{-2k}\omega_{a}^{v_{a}-k}}{q^{2k}-1}\in\mathcal{B}^{\prime}.$

In this case, the inverse maps to

$\omega_{a}^{2k+u_{a}}=(q^{2k}-1)\psi_{a}\psi_{a}^{*}\omega_{a}^{u_{a}+k}+q^{-2k}\omega_{a}^{u_{a}}\in\mathcal{B},$

where $u_{a}=v_{a}-k$ and $0\leq u_{a}<v_{k}$. ∎

We note that there is a commutative subalgebra $\mathcal{C}_{q}(n,k)$ of
dimension $(4k)^{n}$ in $\operatorname{Cl}_{q}(n,k)$ generated by
$\langle\omega_{a},\psi_{a}\psi_{a}^{*}\mid a=1,\dotsc,n\rangle$ regardless of
whether $q^{2k}=1$. 3.6 says that when $q^{2k}\neq 1$, the subalgebra
$\mathcal{C}_{q}(n,k)$ is isomorphic to

$\mathbf{k}[\omega_{1},\dotsc,\omega_{n}]/(\omega_{1}^{4k}-(1+q^{-2k})\omega_{1}^{2k}+q^{-2k},\dotsc,\omega_{n}^{4k}-(1+q^{-2k})\omega_{n}^{2k}+q^{-2k}).$
(3.11)

Contrast this with the case $q^{2k}=1$, where this commutative subalgebra is
isomorphic to

$\mathbf{k}[\omega_{1},\eta_{1},\dotsc,\omega_{n},\eta_{n}]/(\omega_{1}^{2k}-1,\eta_{1}^{2}-\eta_{1}\omega_{1}^{k},\dotsc,\omega_{n}^{2k}-1,\eta_{n}^{2}-\eta_{n}\omega_{n}^{k})$
(3.12)

under the map $\omega_{a}\mapsto\omega_{a}$ and
$\eta_{a}\mapsto\psi_{a}\psi_{a}^{*}$. We can compute a Gröbner basis for the
ideal in (3.12) with respect to degree reverse lexicographic order by taking
its generators and adding $\eta_{a}^{3}-\eta_{a}$ for all $a=1,\dotsc,n$.
While not immediate, it turns out the commutative algebras are in fact
isomorphic as we will see below.

The bases in 3.4 and 3.6 have been implemented in SageMath by the second
author.

###  3.2. Structure Theorem

We construct a quantum analog of the isomorphism $\Gamma_{V}$ defined in 2.2.
The (anti)commutators

$\\{\psi_{a},\psi_{a}^{*}\\}=(1-q^{k})\psi_{a}\psi_{a}^{*}+q^{k}\omega_{a}^{k},\qquad\qquad[\psi_{a},\psi_{a}^{*}]=(1+q^{k})\psi_{a}\psi_{a}^{*}-q^{k}\omega_{a}^{k},$
(3.13)

will play an important role in many calculations in this section. Clearly
$\\{\psi_{a},\psi_{a}^{*}\\}\neq 0,1$ for all $q\in\mathbf{k}^{\times}$. We
first show that the anticommutators are nontrivial central elements. This does
not have a classical analog as the classical anticommutation relations state
$\\{\psi_{a},\psi_{a}^{*}\\}=1$. Let
$Z_{q}(n,k):=Z(\operatorname{Cl}_{q}(n,k))$ denote the center of
$\operatorname{Cl}_{q}(n,k)$. The next result follows from a straightforward
computation, left as an exercise to the reader, using the reduction rules of
3.3.

######  Lemma 3.7.

For each $a=1,\ldots,n$, the anticommutator $\\{\psi_{a},\psi_{a}^{*}\\}$
belongs to $Z_{q}(n,k)$ and $\\{\psi_{a},\psi_{a}^{*}\\}^{2}=1$.

We need quantum analogs of the volume elements defined by Equation 2.5. We
proceed exactly as in the classical case. For each $r=1,\ldots,n$, the
_quantum volume_ or _chirality_ element $f_{r}$ of the subalgebra
$\operatorname{Cl}_{q}(r,k)\subseteq\operatorname{Cl}_{q}(n,k)$ is given by

$\displaystyle f_{r}=[\psi_{1},\psi_{1}^{*}]\cdots[\psi_{r},\psi_{r}^{*}].$
(3.14)

Like in the classical case, we can express each volume element as a product of
standardized coordinates:

$f_{r}=\epsilon_{1}\cdots\epsilon_{2r},$

with $\epsilon_{j}$ given by the same formulae as in Equation 2.6. The
standardized coordinates in $\operatorname{Cl}_{q}(n,k)$ are a quantum analog
of a basis of $\mathbf{k}^{2n}\cong\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}$
where the defining bilinear form is diagonal, but they do not satisfy
$\\{\epsilon_{a},\epsilon_{b}\\}=\delta_{ab}$. Nevertheless, the
anticommutators $\\{\epsilon_{a},\epsilon_{b}\\}$ are still central; this
follows from an easy calculation left as an exercise to the reader.

######  Proposition 3.8.

Let $\epsilon_{1},\ldots,\epsilon_{2n}$ denote the standardized coordinates of
$\operatorname{Cl}_{q}(n,k)$ defined using formulae as in Equation 2.6. Then

$\\{\epsilon_{a},\epsilon_{b}\\}=\epsilon_{a}\epsilon_{b}+\epsilon_{b}\epsilon_{a}=2\delta_{ab}\\{\psi_{\lceil
a/2\rceil},\psi_{\lceil a/2\rceil}^{*}\\}$

is central in $\operatorname{Cl}_{q}(n,k)$.

Next we prove a quantum analog of [ML90, Prop. 3.3], obtaining some relations
involving $f_{r}$. We note that many of the signs there vanish here because
$f_{r}$ is the product of an even number of factors.

######  Proposition 3.9.

Let $f_{r}$ denote a quantum volume element in $\operatorname{Cl}_{q}(n,k)$ as
defined in (3.14). Let $\phi_{a}$ denote either $\psi_{a}$ or $\psi_{a}^{*}$.

  * •

If $a>r$, then $\phi_{a}f_{r}=f_{r}\phi_{a}$ and
$\omega_{a}f_{r}=f_{r}\omega_{a}$.

  * •

If $a\leq r$, then $\phi_{a}f_{r}=-f_{r}\phi_{a}$.

  * •

For any $r,s\leq n$, we have $f_{r}f_{s}=f_{s}f_{r}$ and $f_{r}^{2}=1$.

###### Proof.

The anticommutation relations imply $\phi_{a}$ commutes with
$[\psi_{b},\psi_{b}^{*}]$ whenever $a\neq b$. In addition, combining Equation
3.13 with 3.3 implies

$\displaystyle[\psi_{a},\psi_{a}^{*}]\psi_{a}$
$\displaystyle=(1+q^{k})\psi_{a}\psi_{a}^{*}\psi_{a}-q^{2k}\psi_{a}\omega_{a}^{k}=q^{k}\psi_{a}\omega_{a}^{k}$
$\displaystyle=-\psi_{a}\left((1+q^{k})\psi_{a}\psi_{a}^{*}-q^{k}\omega^{k}\right)=-\psi_{a}[\psi_{a},\psi_{a}^{*}].$

A similar calculation shows that
$[\psi_{a},\psi_{a}^{*}]\psi_{a}^{*}=-\psi_{a}^{*}[\psi_{a},\psi_{a}^{*}]$.
This means $\phi_{a}f_{r}=f_{r}\phi_{a}$ if $a>r$ and
$\phi_{a}f_{r}=-f_{r}\phi_{a}$ if $a\leq r$. The remaining relations follow
from straightforward calculations. ∎

We are now ready to construct an isomorphism
$\Gamma_{q}\colon\operatorname{Cl}_{q}(n,k)^{\otimes
m}\to\operatorname{Cl}_{q}(nm,k)$, obtaining the quantum analog of 2.2. Much
like its classical counterpart
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$, the
quantum Clifford algebra $\operatorname{Cl}_{q}(n,k)$ is a superalgebra: it is
equipped with the $\mathbb{Z}_{2}$-grading associated to the
$\mathbb{Z}^{n}$-grading of (3.4). Therefore, the considerations in 2.3 also
apply here. In particular, there is a natural isomorphism
$\operatorname{Cl}_{q}(nm,k)\cong\operatorname{Cl}_{q}(n,k)\,\widehat{\otimes}\,\cdots\,\widehat{\otimes}\,\operatorname{Cl}_{q}(n,k)$.
However, using the ordinary tensor product here helps us factor quantum group
actions on tensor products of modules through $\operatorname{Cl}_{q}(nm,k)$,
e.g., as in [Abo22b, Abo22a]. As in the classical case, our isomorphism relies
on the volume elements $f_{r}$ to compensate that we are using the ordinary
tensor product instead of the superalgebra tensor product: the $f_{r}$
introduce signs where needed.

######  Theorem 3.10.

Let $\phi_{a}$ denote either $\psi_{a}$ or $\psi_{a}^{*}$. Then the map
$\Gamma_{q}\colon\operatorname{Cl}_{q}(n,k)^{\otimes
m}\to\operatorname{Cl}_{q}(nm,k)$ defined by

$\displaystyle 1\otimes\cdots\otimes\phi_{a}\otimes\cdots\otimes 1$
$\displaystyle\mapsto f_{(j-1)n}\phi_{a+(j-1)n}\text{ and}$ $\displaystyle
1\otimes\cdots\otimes\omega_{a}\otimes\cdots\otimes 1$
$\displaystyle\mapsto\omega_{a+(j-1)n},$

is an isomorphism of $\mathbb{Z}^{nm}$-graded algebras. The inverse
$\Gamma_{q}^{-1}\colon\operatorname{Cl}_{q}(nm,k)\to\operatorname{Cl}_{q}(n,k)^{\otimes
m}$ is defined by

$\displaystyle\phi_{a+(j-1)n}$ $\displaystyle\mapsto f_{n}\otimes\cdots\otimes
f_{n}\otimes\phi_{a}\otimes 1\otimes\cdots\otimes 1\text{ and}$
$\displaystyle\omega_{a+(j-1)n}$ $\displaystyle\mapsto
1\otimes\cdots\otimes\omega_{a}\otimes\cdots\otimes 1.$

In particular,
$\Gamma_{q}(1\otimes\cdots\otimes\phi_{a}\phi_{b}\otimes\cdots\otimes
1)=\phi_{a+(j-1)n}\phi_{b+(j-1)n}.$ In all cases $\phi_{a}$,
$\phi_{a}\phi_{b}$, and $\omega_{a}$ appear in the $j$th tensor factor of
$\operatorname{Cl}_{q}(n,k)^{\otimes m}$.

###### Proof.

Let $\phi_{a,j}$ denote the tensor product
$1\otimes\cdots\otimes\phi_{a}\otimes\cdots\otimes 1$ with $\phi_{a}$
appearing in the $j$th tensor factor. First note that $\Gamma_{q}$ maps each
$\operatorname{Cl}_{q}(n,k)$ factor homomorphically into
$\operatorname{Cl}_{q}(nm,k)$. In particular, whenever $a\neq b$, 3.9 implies

$\displaystyle\Gamma_{q}(\phi_{a,j})\Gamma_{q}(\phi_{b,j})$
$\displaystyle=f_{(j-1)n}^{2}\phi_{a+(j-1)n}\phi_{b+(j-1)n}=-\Gamma_{q}(\phi_{b,j})\Gamma_{q}(\phi_{a,j}).$

In addition, for any $a,b$, 3.9 implies

$\displaystyle\Gamma_{q}(\omega_{a,j})\Gamma_{q}(\psi_{b,j})\Gamma_{q}(\omega_{a,j})^{-1}$
$\displaystyle=f_{(j-1)n}\omega_{a+(j-1)n}\psi_{b+(j-1)n}\omega_{a+(j-1)n}^{-1}=q^{\delta_{ab}}\Gamma_{q}(\psi_{b,j}),$

and

$\displaystyle\Gamma_{q}(\psi_{a,j})\Gamma_{q}(\psi_{a,j}^{*})+q^{\pm
k}\Gamma_{q}(\psi_{a,j}^{*})\Gamma_{q}(\psi_{a,j})=\big{\\{}\psi_{a+(j-1)n},\psi_{a+(j-1)n}^{*}\big{\\}}_{q^{\pm
k}}=w_{a+(j-1)n}^{\mp k}=\Gamma_{q}(\omega_{a,j})^{\mp k}.$

Next notice $\Gamma_{q}$ indeed defines an algebra map on
$\operatorname{Cl}_{q}(n,k)^{\otimes m}$. Without loss of generality, suppose
that $j<k$. 3.9 implies $f_{(j-1)n}$ commutes with $\phi_{b+(k-1)n}$ and
$\omega_{b+(k-1)n}$. Thus for any $a,b$,

$\displaystyle\Gamma_{q}(\phi_{a,j})\Gamma_{q}(\phi_{b,k})$
$\displaystyle=-f_{(j-1)n}f_{(k-1)n}\phi_{a+(j-1)n}\phi_{b+(k-1)n}=\Gamma_{q}(\phi_{b,k})\Gamma_{q}(\phi_{a,k})\text{
and}$ $\displaystyle\Gamma_{q}(\phi_{a,j})\Gamma_{q}(\omega_{b,k})$
$\displaystyle=\omega_{b+(k-1)n}f_{(j-1)n}\phi_{a+(j-1)n}=\Gamma_{q}(\omega_{b,k})\Gamma_{q}(\phi_{a,j}).$

We conclude by showing $\Gamma_{q}$ is an isomorphism with inverse
$\Gamma_{q}^{-1}$. It suffices to show that $\Gamma_{q}\circ\Gamma_{q}^{-1}$
and $\Gamma_{q}^{-1}\circ\Gamma_{q}$ coincide with appropriate identity maps
on a set of generators. To this end, notice that if the commutator appears in
the $j$th factor of
$1\otimes\cdots\otimes[\psi_{a}^{\phantom{\dagger}},\psi_{a}^{*}]\otimes\cdots\otimes
1$, then

$\displaystyle\Gamma_{q}(1\otimes\cdots\otimes[\psi_{a}^{\phantom{\dagger}},\psi_{a}^{*}]\otimes\cdots\otimes
1)=f_{(j-1)n}^{2}[\psi_{a+(j-1)n}^{\phantom{\dagger}},\psi_{a+(j-1)n}^{*}]=\big{[}\psi_{a+(j-1)n}^{\phantom{\dagger}},\psi_{a+(j-1)n}^{*}\big{]}$

because $\Gamma_{q}$ is an algebra map and $f_{r}^{2}=1$ by 3.9. Since
$f_{n}=[\psi_{1},\psi_{1}^{*}]\cdots[\psi_{n},\psi_{n}^{*}]$, we see that

$\displaystyle\Gamma_{q}\circ\Gamma_{q}^{-1}(\phi_{a+(j-1)n})$
$\displaystyle=\Gamma_{q}(f_{n}\otimes\cdots\otimes
f_{n}\otimes\phi_{a}\otimes 1\otimes\cdots\otimes 1)$
$\displaystyle=\prod_{k=1}^{j-1}\left(\prod_{a=1}^{n}\left[\psi_{a+(j-1)n}^{\phantom{\dagger}},\psi_{a+(j-1)n}^{*}\right]\right)\cdot
f_{(j-1)n}\phi_{a+(j-1)n}$ $\displaystyle=f_{(j-1)n}^{2}\phi_{a+(j-1)n}$
$\displaystyle=\phi_{a+(j-1)n},$

and similarly
$\Gamma_{q}\circ\Gamma_{q}^{-1}(\omega_{a+(j-1)n})=\Gamma_{q}(\omega_{a,j})=\omega_{a+(j-1)n}$.
Hence $\Gamma_{q}^{-1}$ is in fact a right inverse of $\Gamma_{q}$. Similar
calculations show $\Gamma_{q}^{-1}$ also inverts $\Gamma_{q}$ on the left. ∎

The following structure result is a special case of 3.10.

######  Corollary 3.11.

There is an isomorphism of $\mathbb{Z}^{n}$-graded algebras

$\operatorname{Cl}_{q}(n,k)\cong\operatorname{Cl}_{q}(1,k)^{\otimes n}.$

###  3.3. The center of a quantum Clifford algebra

As we saw in 3.7, the anticommutators $\\{\psi_{a},\psi_{a}^{*}\\}$ are
nontrivial central elements that are square roots of unity. We now construct
their $k$th roots in $\operatorname{Cl}_{q}(n,k)$ and show that these roots
generate the center $Z_{q}(n,k)$ of $\operatorname{Cl}_{q}(n,k)$.

######  Theorem 3.12.

We have

$Z_{q}(n,k)\cong\mathbf{k}[\mathbb{Z}_{2k}]\otimes\cdots\otimes\mathbf{k}[\mathbb{Z}_{2k}]=\mathbf{k}[\mathbb{Z}_{2k}^{n}].$

In particular, $Z_{q}(n,k)$ is generated by the order $2k$ elements

$z_{a}=q\omega_{a}-(q-1)\psi_{a}^{\phantom{\dagger}}\psi_{a}^{*}\omega_{a}^{k+1},$
(3.15)

for $a=1,\ldots,n$. Moreover, $\dim Z_{q}(n,k)=(2k)^{n}$ and each $z_{a}$
satisfies

$\psi_{a}^{\phantom{\dagger}}\psi_{a}^{*}+\psi_{a}^{*}\psi_{a}^{\phantom{\dagger}}=z_{a}^{k}.$
(3.16)

We first show that the elements $z_{a}$ are in the center, and we write their
powers explicitly in terms of our basis from 3.4.

######  Lemma 3.13.

The elements $z_{a}$ defined in 3.12 are central in
$\operatorname{Cl}_{q}(n,k)$ and they satisfy

$\displaystyle
z_{a}^{r}=q^{r}\omega_{a}^{r}-(q^{r}-1)\psi_{a}^{\phantom{\dagger}}\psi_{a}^{*}\omega_{a}^{k+r}.$
(3.17)

In particular, $z_{a}^{k}=\\{\psi_{a},\psi_{a}^{*}\\}$, so $z_{a}^{2k}=1$.

###### Proof.

That $z_{a}$ is central is a straightforward computation showing it commutes
with all of the generators of $\operatorname{Cl}_{q}(n,k)$ using the reduction
rules. Equation 3.17 is a straightforward calculation using induction on $r$.
That $z_{a}^{k}=\\{\psi_{a},\psi_{a}^{*}\\}$ is immediate from comparing
Equation 3.13 with Equation 3.17 with $r=k$. The last claim is 3.7. ∎

###### Proof of 3.12..

Using the decomposition of 3.11, it is sufficient to show the claim for
$\operatorname{Cl}_{q}(1,k)$; that is,
$Z_{q}(1,k)\cong\mathbf{k}[\mathbb{Z}_{2k}]$.

We will show the central elements $z_{1},\ldots,z_{1}^{2k}$ constitute a basis
of $Z_{q}(1,k)$. The elements $z_{1},\ldots,z_{1}^{2k}$ are linearly
independent because the elements

$1,\omega_{1},\ldots,\omega_{1}^{2k-1},\psi_{1}\psi_{1}^{*},\psi_{1}\psi_{1}^{*}\omega_{1},\ldots,\psi_{1}\psi_{1}^{*}\omega_{1}^{2k-1}$

are linearly independent by 3.4. Hence the subalgebra generated by $z_{1}$ is
central and isomorphic to $\mathbf{k}[\mathbb{Z}_{2k}]$ by 3.13. To show there
is nothing else in the center, consider an arbitrary central element $\zeta\in
Z_{q}(1,k)$. In what follows we omit the subscript $1$ on
$\operatorname{Cl}_{q}(1,k)$ generators for brevity. Using 3.4, we write

$\zeta=\sum_{j=0}^{2k-1}a_{j}\omega^{j}+\psi\sum_{j=0}^{2k-1}b_{j}\omega^{j}+\psi^{*}\sum_{j=0}^{2k-1}c_{j}\omega^{j}+\psi\psi^{*}\sum_{j=0}^{2k-1}d_{j}\omega^{j}$

for some constants $a_{j},b_{j},c_{j},d_{j}\in\mathbf{k}$. By direct
computation using the reduction rules, we have

$\displaystyle 0=[\zeta,\psi]$
$\displaystyle=\psi\sum_{j=0}^{2k-1}\left((q^{j}-1)a_{j}+q^{j}d_{j+k}\right)\omega^{j}+\sum_{j=0}^{2k-1}q^{j+k}c_{j}\omega^{j}-\psi\psi^{*}\sum_{j=0}^{2k-1}(q^{j+k}+1)c_{j}\omega^{j},$
$\displaystyle 0=[\zeta,\psi^{*}]$
$\displaystyle=\psi^{*}\sum_{j=0}^{2k-1}\left((q^{-j}-1)a_{j}+d_{j+k}\right)\omega^{j}-q^{k}\sum_{j=0}^{2k-1}b_{j}\omega^{j+k}+q^{k}\psi\psi^{*}\sum_{j=0}^{2k-1}(q^{-j-k}+1)b_{j}\omega^{j},$
$\displaystyle\sum_{j=0}^{2k-1}b_{j}\omega^{j+k}$
$\displaystyle=\sum_{j=0}^{k-1}(b_{j}\omega^{k}+b_{j+k}q^{-2k})\omega^{j}+(1-q^{-2k})\sum_{j=0}^{k-1}b_{j+k}\psi\psi^{*}\omega^{j+k},$

where we take the indices modulo $2k$. Therefore, by 3.2 and linear
independence from 3.4, we must have $b_{j}=c_{j}=0$ for all $j$. Furthermore,
we must have $d_{j}=(q^{-j}-1)a_{k+j}$. This means there are at most $2k$
degrees of freedom, which yields $\dim\left(Z_{q}(1,k)\right)\leq 2k$ and the
claim follows. ∎

We comment on two consequences of 3.12. The first is revealed by Equation
3.16. We may view $\operatorname{Cl}_{q}(n,k)$ as a classical Clifford algebra
$\operatorname{Cl}(R^{n}\oplus(R^{*})^{n})$, albeit with a slightly rescaled
pairing, over the group ring
$R=\mathbf{k}[z_{1},\ldots,z_{n}]\cong\mathbf{k}[\mathbb{Z}_{2k}^{n}]$, once
we relax the vector space requirement in 2.1 to admit more general (free)
modules over commutative rings. This interpretation elucidates the conclusion
of 3.12: the classical Clifford algebra is simple, so the only central
elements in $\operatorname{Cl}_{q}(n,k)$ can be the scalars
$\mathbf{k}[\mathbb{Z}_{2k}^{n}]$ in the base ring. This also yields that
$\mathcal{C}_{q}(n,k)$ (see (3.11) and (3.12)) are isomorphic for any value of
$q\in\mathbf{k}^{\times}$ as $\mathcal{C}_{q}(n,k)\cong R\oplus
R\langle\psi_{1}\psi_{1}^{*},\dotsc,\psi_{n}\psi_{n}^{*}\rangle$. The second
consequence is revealed by Equation 3.15: the center belongs to the degree $0$
component of $\operatorname{Cl}_{q}(n,k)$ under the $\mathbb{Z}^{n}$-grading
given in Equation 3.4.

###  3.4. Quantum Clifford (anti-)involutions

We define some (anti-)involutions on the quantum Clifford algebra
$\operatorname{Cl}_{q}(n,k)$ that are quantum analogs of (anti-)involutions of
the classical Clifford algebra. For example, we define quantum analogs of the
transpose and conjugation maps.

First recall $\operatorname{Cl}_{q}(n,k)$ is a superalgebra, and so there is
the _grade involution_ $\alpha$ on $\operatorname{Cl}_{q}(n,k)$ defined by

$\psi_{a}\mapsto-\psi_{a},\qquad\qquad\psi_{a}^{*}\mapsto-\psi_{a}^{*},\qquad\qquad\omega_{a}\mapsto\omega_{a},\qquad\qquad
q\mapsto q.$

When the twist $k$ is even, we could also define a variant
$\widetilde{\alpha}$ by having $\omega_{a}\mapsto-\omega_{a}$ instead.

Next, we define the _transpose_ anti-involution on
$\operatorname{Cl}_{q}(n,k)$ by

$\psi_{a}\mapsto\psi_{a},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a}^{*},\qquad\qquad\omega_{a}\mapsto
q^{-1}\omega_{a}^{-1},\qquad\qquad q\mapsto q.$

This is the quantum analog of the transpose on the classical Clifford algebra,
and we denote it by $x^{t}$.

######  Proposition 3.14.

The transpose is an anti-involution on $\operatorname{Cl}_{q}(n,k)$. Moreover,
the transpose commutes with the grade involution.

###### Proof.

A direct computation shows the transpose preserves the defining relations and
that both $(x^{t})^{t}=x$ and $\alpha(x^{t})=\alpha(x)^{t}$ for all generators
$x\in\operatorname{Cl}_{q}(n,k)$. ∎

With the transpose and grade involution in hand, we may define the _Clifford
conjugation_ on $\operatorname{Cl}_{q}(n,k)$: for any
$x\in\operatorname{Cl}_{q}(n,k)$, we set

$\overline{x}=\alpha(x^{t}).$

Next, we separate the anti-involution $\ast$ defined in [Hay90, Sec. 5]
depending on the value $q$ into the product of two separate anti-involutions
that do not depend on $q$. Hayashi used the map $\ast$ to make
$\operatorname{Cl}_{q}(n,k)$ into a $\ast$-algebra and to prove the unitarity
of his $\operatorname{Cl}_{q}(n,k)$-modules [Hay90]. The first is the anti-
involution that we call the _dagger_ map, which we define by

$\psi_{a}\mapsto\psi_{a}^{*},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a},\qquad\qquad\omega_{a}\mapsto\omega_{a},\qquad\qquad
q\mapsto q,$

and denote by $x^{\dagger}$ for $x\in\operatorname{Cl}_{q}(n,k)$. The second
anti-involution is the _duality_ map on $\operatorname{Cl}_{q}(n,k)$, which we
define by

$\psi_{a}\mapsto\psi_{a}^{*},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a},\qquad\qquad\omega_{a}\mapsto\omega_{a}^{-1},\qquad\qquad
q\mapsto q^{-1}$

and denote by $x^{\vee}$ for $x\in\operatorname{Cl}_{q}(n,k)$. Both the dagger
and duality maps are different quantum analogs of the natural duality anti-
involution $\ast$ on
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$
sending $\psi_{a}^{*}\leftrightarrow\psi_{a}$, although arguably the dagger
map is the most natural analog. We justify calling the latter map the duality
map by the relationship with the quantum group discussed below in Section 3.5.

######  Proposition 3.15.

The grade involution, transpose, dagger, and duality maps all commute.

###### Proof.

This is a direct computation on the generators, such as
$x^{\vee\dagger}=x^{\dagger\vee}$ for any generator
$x\in\operatorname{Cl}_{q}(n,k)$. ∎

Now by composing the duality and dagger maps,we obtain the involution of
$\operatorname{Cl}_{q}(n,k)$ given by

$\psi_{a}\mapsto\psi_{a},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a}^{*},\qquad\qquad\omega_{a}\mapsto\omega_{a}^{-1},\qquad\qquad
q\mapsto q^{-1}.$

This could be said to not have a classical analog as it acts as the identity
map on $\psi_{a}$ and $\psi_{a}^{*}$. The composition of the transpose and
duality maps is the involution defined by

$\psi_{a}\mapsto\psi_{a}^{*},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a},\qquad\qquad\omega_{a}\mapsto
q\omega_{a},\qquad\qquad q\mapsto q^{-1}.$

If $q^{2k}=1$, then we can construct another involution $\check{\kappa}$ by

$\psi_{a}\mapsto\psi_{a}^{*},\qquad\qquad\psi_{a}^{*}\mapsto\psi_{a},\qquad\qquad\omega_{a}\mapsto
q^{-1}\omega_{a},\qquad\qquad q\mapsto q^{-1}.$

These arise from the isomorphism of
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$
induced from
$\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\cong(\mathbf{k}^{n})^{*}\oplus\mathbf{k}^{n}$.

###  3.5. Quantum group morphisms

We provide the twist $k$ version of [Hay90, Thm. 3.2] through a similar direct
computation.

Let $U_{q}(\mathfrak{g},k)$ denote a twisted form of the Drinfel’d–Jimbo
[Dri85, Jim85] quantum group, defined as follows. Let $\mathfrak{g}$ denote a
semisimple Lie algebra and let $A$ denote its generalized Cartan matrix with
the diagonal matrix of root lengths $D=\mathrm{diag}(d_{1},\ldots,d_{r})$ such
that $DA$ is symmetric. Let $q\in\mathbf{k}$ be transcendental and write
$q_{i}=q^{d_{i}}$. Then $U_{q}(\mathfrak{g},k)$ is the associative
$\mathbf{k}$-algebra generated by the elements $E_{i},F_{i},K_{i}$, and
$K_{i}^{-1}$, for $i=1,\ldots,r$, subject to the relations:

$\displaystyle K_{i}K_{j}=K_{j}K_{i},\quad K_{i}K_{i}^{-1}=K_{i}^{-1}K_{i}=1,$
$\displaystyle K_{i}E_{j}K_{i}^{-1}=q_{i}^{a_{ij}}E_{j},\quad
K_{i}F_{j}K_{i}^{-1}=q_{i}^{-a_{ij}}F_{j},$ $\displaystyle
E_{i}F_{j}-F_{j}E_{i}=\delta_{ij}\frac{K_{i}^{k}-K_{i}^{-k}}{q_{i}^{k}-q_{i}^{-k}},$
(3.18)

together with the $q$-Serre relations for $i\neq j$,

$\sum_{m=0}^{1-a_{ij}}(-1)^{m}\begin{bmatrix}1-a_{ij}\\\
m\end{bmatrix}_{q_{i}^{k}}X_{i}^{k}X_{j}X_{i}^{1-a_{ij}-m}=0.$ (3.19)

In the last equality every $X$ is either an $E$ or an $F$. The $q$-integer is
given by $[n]_{q_{i}}=\frac{q_{i}^{n}-q_{i}^{-n}}{q_{i}-q_{i}^{-1}}$, the
$q$-factorial is $[n]_{q_{i}}!=[n]_{q_{i}}\cdots[1]_{q_{i}}$, and the
$q$-binomial coefficient is defined analogously. The following identity is
often useful for verifying $q$-Serre relations when $a_{ij}=-1$:

$\displaystyle\begin{split}\sum_{m=0}^{1-a_{ij}}(-1)^{m}\begin{bmatrix}1-a_{ij}\\\
m\end{bmatrix}_{q_{i}^{k}}X_{i}^{k}X_{j}X_{i}^{1-a_{ij}-m}&=X_{i}(X_{i}X_{j}-q_{i}^{k}X_{j}X_{i})-q_{i}^{-k}(X_{i}X_{j}-q_{i}^{k}X_{j}X_{i})X_{i}\\\
&=\bigl{[}X_{i},[X_{i},X_{j}]_{q_{i}^{k}}\bigr{]}_{q_{i}^{-k}}.\end{split}$
(3.20)

Note that $U_{q}(\mathfrak{g},k)$ differs from the standard
$U_{q}(\mathfrak{g})$ by the exponents on $K_{i}$ in Equation 3.18 and the
exponents on $q_{i}$ in Equations 3.18 and 3.19. In particular, we recover the
standard presentation when $k=1$. For more information on quantum groups, we
refer the reader to one of the standard textbooks such as [CP94, Kas95,
Jan96].

We need the following twist $k$ versions of [Hay90, Lemma 3.1].

######  Lemma 3.16.

Suppose $q^{2k}\neq 1$. Let $a,b,c$ be distinct elements of $\\{1,\ldots,n\\}$
and let $\phi_{a}$ denote either $\psi_{a}$ or $\psi_{a}^{*}$. The following
identities hold in $\operatorname{Cl}_{q}(n,k)$:

$\displaystyle[\psi_{a}\psi_{b}^{*},\psi_{b}\psi_{a}^{*}]$
$\displaystyle=\frac{(\omega_{a}\omega_{b}^{-1})^{k}-(\omega_{a}\omega_{b}^{-1})^{-k}}{q^{k}-q^{-k}},$
(3.21) $\displaystyle[\psi_{a}\psi_{b},\psi_{b}^{*}\psi_{a}^{*}]$
$\displaystyle=\frac{(q\omega_{a}\omega_{b})^{k}-(q\omega_{a}\omega_{b})^{-k}}{q^{k}-q^{-k}},$
(3.22) $\displaystyle[\phi_{a}\psi_{b},\psi_{b}^{*}\phi_{c}]_{q^{\pm k}}$
$\displaystyle=\omega_{b}^{\mp k}\phi_{a}\phi_{c},$ (3.23)
$\displaystyle\psi_{a}\psi_{a}^{*}\pm\psi_{a}^{*}\psi_{a}$
$\displaystyle=\frac{q^{k}\omega_{a}^{k}\pm\omega_{a}^{-k}}{q^{k}\pm 1}.$
(3.24)

###### Proof.

Using the defining relations, we see that

$[\psi_{a}\psi_{b}^{*},\psi_{b}\psi_{a}^{*}]=(\psi_{a}\psi_{a}^{*})(\psi_{b}^{*}\psi_{b})-(\psi_{a}^{*}\psi_{a})(\psi_{b}\psi_{b}^{*}).$

Applying Equation 3.3 then yields Equation 3.21:

$\displaystyle[\psi_{a}\psi_{b}^{*},\psi_{b}\psi_{a}^{*}]$
$\displaystyle=\frac{-1}{(q^{k}-q^{-k})^{2}}\big{(}\bigl{(}q\omega_{a})^{k}-(q\omega_{a})^{-k}\bigr{)}(\omega_{b}^{k}-\omega_{b}^{-k})-(\omega_{a}-\omega_{a}^{-1})\bigl{(}q\omega_{b})^{k}-(q\omega_{b})^{-k}\bigr{)}\bigr{)}$
$\displaystyle=\frac{(\omega_{a}\omega_{b}^{-1})^{k}-(\omega_{a}\omega_{b}^{-1})^{-k}}{q^{k}-q^{-k}}.$

Equation 3.22 follows from a similar calculation. To establish Equation 3.24,
we use Equation 3.3:

$\displaystyle\psi_{a}\psi_{a}^{*}\pm\psi_{a}^{*}\psi_{a}$
$\displaystyle=\frac{1}{q^{k}-q^{-k}}\big{(}(q\omega_{a})^{k}-(q\omega_{a})^{-k}\bigr{)}\mp(\omega_{a}^{k}-\omega_{a}^{-k})\big{)}$
$\displaystyle=\frac{q^{k}\mp
1}{q^{2k}-1}\big{(}q^{k}\omega_{a}\pm\omega_{a}^{-k}\big{)}$
$\displaystyle=\frac{q^{k}\omega_{a}^{k}\pm\omega_{a}^{-k}}{q^{k}\pm 1}.$

Equation 3.23 is similarly easy and we omit its proof. ∎

If $k$ is even or $q$ has a square root in $\mathbf{k}$, we may write Equation
3.24 as the more symmetric

$\displaystyle\psi_{a}\psi_{a}^{*}\pm\psi_{a}^{*}\psi_{a}$
$\displaystyle=\frac{(q^{\frac{1}{2}}\omega_{a})^{k}\pm(q^{\frac{1}{2}}\omega_{a})^{-k}}{q^{\frac{k}{2}}\pm
q^{-\frac{k}{2}}}.$ (3.25)

######  Proposition 3.17.

Suppose $q^{2k}\neq 1$. There is an algebra homomorphism $\Theta\colon
U_{q}(\mathfrak{g},k)\to\operatorname{Cl}_{q}(n,k)$ defined on generators as
follows.

  1. (1)

If $\mathfrak{g}=\mathfrak{sl}_{n}$ then

$\displaystyle E_{i}$
$\displaystyle\mapsto\psi_{i}^{\phantom{\dagger}}\psi_{i+1}^{*},$
$\displaystyle F_{i}$
$\displaystyle\mapsto\psi_{i+1}^{\phantom{\dagger}}\psi_{i}^{*},$
$\displaystyle K_{i}$
$\displaystyle\mapsto\omega_{i}^{\phantom{\dagger}}\omega_{i+1}^{-1},$
$\displaystyle i=1,\ldots,n-1.$

  2. (2)

If $\mathfrak{g}=\mathfrak{so}_{2n}$ then

$\displaystyle E_{i}$
$\displaystyle\mapsto\psi_{i}^{\phantom{\dagger}}\psi_{i+1}^{*},$
$\displaystyle F_{i}$
$\displaystyle\mapsto\psi_{i+1}^{\phantom{\dagger}}\psi_{i}^{*},$
$\displaystyle K_{i}$
$\displaystyle\mapsto\omega_{i}^{\phantom{\dagger}}\omega_{i+1}^{-1},$
$\displaystyle i=1,\ldots,n-1,$ $\displaystyle E_{n}$
$\displaystyle\mapsto\psi_{n-1}\psi_{n},$ $\displaystyle F_{n}$
$\displaystyle\mapsto\psi_{n}^{*}\psi_{n-1}^{*},$ $\displaystyle K_{n}$
$\displaystyle\mapsto q\omega_{n-1}\omega_{n}.$

  3. (3)

If $\mathfrak{g}=\mathfrak{so}_{2n+1}$ and $q^{\frac{1}{2}}\in\mathbf{k}$,
there is an algebra homomorphism
$U_{q^{\frac{1}{2}}}(\mathfrak{g},k)\mapsto\operatorname{Cl}_{q}(n,k)$ defined
on generators by

$\displaystyle E_{i}$
$\displaystyle\mapsto\psi_{i}^{\phantom{\dagger}}\psi_{i+1}^{*},$
$\displaystyle F_{i}$
$\displaystyle\mapsto\psi_{i+1}^{\phantom{\dagger}}\psi_{i}^{*},$
$\displaystyle K_{i}$
$\displaystyle\mapsto\omega_{i}^{\phantom{\dagger}}\omega_{i+1}^{-1},$
$\displaystyle i=1,\ldots,n-1,$ $\displaystyle E_{n}$
$\displaystyle\mapsto\psi_{n},$ $\displaystyle F_{n}$
$\displaystyle\mapsto\psi_{n}^{*},$ $\displaystyle K_{n}$
$\displaystyle\mapsto q^{\frac{1}{2}}\omega_{n}.$

###### Proof.

The claim follows from direct calculation. We must verify that the images
$\widetilde{E}_{i},\widetilde{F}_{i}$, and $\widetilde{K}_{i}$ of
$E_{i},F_{i}$, and $K_{i}$ satisfy the relations defining
$U_{q}(\mathfrak{g},k)$. For starters, if $i<n$ then

$\displaystyle[\widetilde{E}_{i},\widetilde{F}_{i}]=[\psi_{i}^{\phantom{\dagger}}\psi_{i+1}^{*},\psi_{i+1}^{\phantom{\dagger}}\psi_{i}^{*}]=\frac{(\omega_{i}\omega_{i+1}^{-1})^{k}-(\omega_{i}\omega_{i+1}^{-1})^{-t}}{q^{k}-q^{-k}}=\frac{\widetilde{K}_{i}^{k}-\widetilde{K}_{i}^{-k}}{q^{k}-q^{-k}}.$

If $i,j<n$ and $\lvert i-j\rvert>1$, the canonical anticommutation relations
imply $[E_{i},E_{j}]=0$. When $\lvert i-j\rvert=1$, each summand in the
$q$-Serre eq. 3.19 vanishes. Since
$\widetilde{F}_{i}=\psi_{i+1}^{\phantom{\dagger}}\psi_{i}^{*}=\widetilde{E}_{i}^{\vee}$,
this means the $\widetilde{F}_{i}$ also satisfy the $q$-Serre relations.

To prove the theorem, it remains to check that
$\widetilde{E}_{i},\widetilde{F}_{i}$, and $\widetilde{K}_{i}$ satisfy the
appropriate relations when $i=n$. For instance, Equation 3.25 implies

$[\widetilde{E}_{n},\widetilde{F}_{n}]=[\psi_{n},\psi_{n}^{*}]=\frac{(q^{\frac{1}{2}}\omega_{n})^{k}-(q^{\frac{1}{2}}\omega_{n})^{-k}}{q^{\frac{k}{2}}-q^{-\frac{k}{2}}}=\frac{\widetilde{K}_{i}^{k}-\widetilde{K}_{i}^{-k}}{q^{\frac{k}{2}}-q^{-\frac{k}{2}}}.$

We leave the remaining verifications as an exercise to the reader. ∎

The formulas in 3.17 capture an underlying correspondence between the simple
roots of $\mathfrak{g}$ and $\operatorname{Cl}_{q}(n,k)$. Using the grading
from Equation 3.4, notice that

$\deg(\Theta(E_{i}))=e_{i}-e_{i+1}\quad\text{and}\quad\deg(\Theta(F_{i}))=-(e_{i}-e_{i+1})$

whenever $i<n$. Moreover, if $\mathfrak{g}=\mathfrak{so}_{2n}$ we have

$\deg(\Theta(E_{n}))=e_{n-1}+e_{n}\quad\text{and}\quad\deg(\Theta(F_{n}))=-(e_{n-1}+e_{n}),$

and similarly if $\mathfrak{g}=\mathfrak{so}_{2n+1}$ then

$\deg(\Theta(E_{n}))=e_{n}\quad\text{and}\quad\deg(\Theta(F_{n}))=-e_{n}.$

To conclude, we compute the maps induced by the various (anti-)involutions
defined in Section 3.4. These become algebra and coalgebra
(anti-)automorphisms on the quantum group image. To begin, note that the
dagger maps acts on $\Theta\left(U_{q}(\mathfrak{g},k)\right)$ as the anti-
automorphism [Hay90, Eq. (5.6)]

$E_{i}^{\dagger}=F_{i},\qquad\qquad F_{i}^{\dagger}=E_{i},\qquad\qquad
K_{i}^{\dagger}=K_{i},\qquad\qquad q\mapsto q.$

This map becomes a coalgebra automorphism of the quantum group as indicated in
[Hay90, Sec. 5.3]. Next observe that the duality map induces the dual
representation: the action on the quantum group image in
$\operatorname{Cl}_{q}(n,k)$ is [Hay90, Eq. (5.8)]

$E_{i}^{\vee}=F_{i},\qquad\qquad F_{i}^{\vee}=E_{i},\qquad\qquad
K_{i}^{\vee}=K_{i}^{-1},\qquad\qquad q\mapsto q^{-1}.$

As noted by [Hay90, Sec. 5.3], this is an anti-(co)algebra automorphism of the
quantum group, which agrees with the fact that we need to also reverse the
order of a tensor product of representations. Finally, observe that the
transpose on $\operatorname{Cl}_{q}(n,k)$ becomes the star (anti-)involution
map (see, e.g., [Lus90, Kas91, Kas93]) on the quantum group image:

$E_{i}^{t}=E_{i},\qquad\qquad F_{i}^{t}=F_{i},\qquad\qquad
K_{i}^{t}=K_{i}^{-1},\qquad\qquad q\mapsto q,$

which is important in Kashiwara’s grand loop argument for crystal bases
[Kas91].

##  4\. Representation theory

In this section we discuss the representation-theoretic aspects of
$\operatorname{Cl}_{q}(n,k)$. In particular, 4.1 defines a collection of
irreducible $\operatorname{Cl}_{q}(n,k)$-actions on the braided exterior
algebra $\bigwedge\nolimits_{q}(\mathbf{k}^{n})$, which is defined below, and
4.6 proves that $\operatorname{Cl}_{q}(n,k)$ is semisimple exactly when its
center $Z_{q}(n,k)$ is semisimple. We note that only one of the
$\operatorname{Cl}_{q}(n,k)$-modules has a classical analog. In addition, we
discuss a representation-theoretic implication of the quantum volume elements
defined in Equation 3.14.

Following [BZ05], the _braided exterior algebra_ $\bigwedge\nolimits_{q}(V)$
is the associative algebra generated by a basis $\\{v_{1},\ldots,v_{n}\\}$ of
$V$ subject to the following relations:

$\displaystyle v_{j}^{2}$ $\displaystyle=0,$ $\displaystyle j$
$\displaystyle=1,\ldots,n$ (4.1a) $\displaystyle v_{j}v_{k}$
$\displaystyle=-qv_{k}v_{j},$ $\displaystyle j$ $\displaystyle<k.$ (4.1b)

The elements

$\displaystyle v(\ell)=\begin{cases}v_{1}^{\ell_{1}}v_{2}^{\ell_{2}}\cdots
v_{n}^{\ell_{n}}&\text{if }\ell_{j}\in\\{0,1\\},\\\
0&\text{otherwise},\end{cases}$ (4.2)

for $\ell\in\\{0,1\\}^{n}$, form a basis of $\bigwedge\nolimits_{q}(V)$. Both
the classical and braided exterior algebras can be understood as quotients of
a tensor algebra modulo an ideal generated by “symmetric” $2$-tensors, which
are “positive” eigenvectors of appropriate braiding operators. In the
classical case, the braiding operators are trivial flip maps that transpose
tensor factors. In the quantum case, braiding operators are induced by the
universal $R$-matrix of some quantum group [BZ05, Def. 2.1].

Recall that unlike its classical counterpart
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$, the
quantum Clifford algebra $\operatorname{Cl}_{q}(n,k)$ has a non-trivial
center, which is described in 3.12. This means there is some flexibility in
defining an action of $\operatorname{Cl}_{q}(n,k)$ on
$\bigwedge\nolimits_{q}(\mathbf{k}^{n})$ depending on which roots of unity are
in $\mathbf{k}$. There are $n$ linearly independent central elements $z_{a}$
such that $z_{a}^{2k}=1$, so the representations are naturally indexed by
$\mathbb{Z}_{2k}^{n}$. In other words, there is a unique irreducible
representation for each element in the discrete subgroup
$\mathbb{Z}_{2k}^{n}\subset Z_{q}(n,k)$. Recall that $\zeta_{r}$ denotes a
primitive $r$-th root of unity.

######  Proposition 4.1.

Suppose $\zeta_{2k}\in\mathbf{k}$. For each $p=(p_{1},\ldots,p_{n})$ in
$\mathbb{Z}_{2k}^{n}$, there is an irreducible $\operatorname{Cl}_{q}(n,k)$
representation $\pi_{p}$ on $\bigwedge\nolimits_{q}(\mathbf{k}^{n})$ defined
by

$\displaystyle\psi_{a}\rhd v(\ell)$
$\displaystyle=(-1)^{\ell_{1}+\cdots\ell_{a-1}}v(\ell+e_{a}),$ (4.3a)
$\displaystyle\psi_{a}^{*}\rhd v(\ell)$
$\displaystyle=(-1)^{\ell_{1}+\cdots\ell_{a-1}}\cdot\zeta_{2k}^{p_{a}}\cdot
v(\ell-e_{a}),$ (4.3b) $\displaystyle\omega_{a}\rhd v(\ell)$
$\displaystyle=\zeta_{2k}^{p_{a}}q^{\ell_{a}-1}v(\ell),$ (4.3c)

where $v(\ell\pm e_{a})=0$ if $\ell\pm e_{a}\notin\\{0,1\\}^{n}$. Moreover,
$\pi_{p}$ and $\pi_{p^{\prime}}$ are not isomorphic whenever $p\neq
p^{\prime}$.

###### Proof.

An easy calculation shows that $\pi_{p}$ indeed defines an action of
$\operatorname{Cl}_{q}(n,k)$. Each $\pi_{p}$ is irreducible because we can
obtain any basis vector $v(\ell)$ from any other by first applying a sequence
of lowering operators to obtain $v(0)$ (up to a scalar), and then using an
appropriate sequence of raising operators. Using Relations (4.3), we see that

$\displaystyle\pi_{p}(z_{a})v(\ell)=\zeta_{2k}^{p_{a}}\left(q^{\ell_{a}}-(q-1)\ell_{a}\right)v(\ell)=\zeta_{2k}^{p_{a}}v(\ell),$

where $z_{a}$ is the central element defined in 3.12. Hence $\pi_{p}$ and
$\pi_{p^{\prime}}$ are distinct representations whenever $p\neq p^{\prime}$. ∎

######  Remark 4.2.

We have defined the $\operatorname{Cl}_{q}(n,k)$-action directly in terms of
basis vectors without using Equation 4.1b. As such, there is no difference
between defining this on $\bigwedge\nolimits_{q}(V)$, $\bigwedge(V)$, or
$\mathbf{k}^{2n}$, as they are all isomorphic as $\mathbf{k}$-modules. Hence,
the formulae defining the $\operatorname{Cl}_{q}(n,2)$-module $\pi_{0}$ have
appeared previously in [Hay90, Prop. 2.1], where the representation was
defined on the usual exterior algebra $\bigwedge(V)$. Yet, the braided
exterior algebra is better behaved for taking tensor products of quantum group
representations; see, e.g., [Abo22b, Abo22a].

######  Remark 4.3.

The spinor representation considered in [Kwo14] is a “dual” version of the
spinor representation considered here: it is obtained by interchanging
$\ell_{a}\leftrightarrow 1-\ell_{a}$ for all $a$.

Amongst the $\operatorname{Cl}_{q}(n,k)$-modules described in 4.1, only
$\pi_{0}$ has a classical analog. To see this, compare Equation 4.3 with the
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$-action
on the exterior algebra. In particular, the operators $\pi_{0}(\psi_{a})$ and
$\pi_{0}(\psi_{a}^{*})$ satisfy the canonical anticommutation relations (2.4).

The action of classical Clifford algebra
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$ on
its spinor representation can be described in terms of inner and exterior
multiplication. However, in the quantum Clifford algebra, even under
$\pi_{0}$, the generators $\psi_{a}^{*}$ and $\psi_{a}$ no longer act by inner
and exterior multiplication because the algebra structure of
$\bigwedge\nolimits_{q}(\mathbf{k}^{n})$ is such that a product $v(\ell)$
acquires powers of $\pm q$ when re-arranging factors of $v_{k}$. For instance,
exterior multiplication of $v(e_{1})=v_{1}$ by $v_{2}$ results in
$-q^{-1}v(e_{1}+e_{2})=-q^{-1}v_{1}v_{2}$.

In order to obtain quantum analogs of the inner and exterior multiplication
operators that correctly account for these powers of $\pm q$, we must use the
$\omega_{k}$. Concretely, we let $\iota_{j}^{q}$ and $\varepsilon_{j}^{q}$
denote quantum analogs of the inner and exterior multiplication operators
defined on any $v(\ell)$ by

$\displaystyle\iota_{j}^{q}(v(\ell))$
$\displaystyle=(-q)^{\ell_{1}+\cdots+\ell_{j-1}}v(\ell-e_{j})$
$\displaystyle=\pi_{0}\bigg{(}\prod_{k<j}\omega_{k}\cdot
q^{j-1}\psi_{i}^{*}\bigg{)}v(\ell),$
$\displaystyle\varepsilon_{j}^{q}(v(\ell))$
$\displaystyle=(-q^{-1})^{\ell_{1}+\cdots+\ell_{j-1}}v(\ell+e_{j})$
$\displaystyle=\pi_{0}\bigg{(}\prod_{k<j}\omega_{k}^{-1}\cdot
q^{1-j}\psi_{j}\bigg{)}v(\ell).$

There is likely a bilinear form on the spinor representation $\pi_{0}$ of
$\operatorname{Cl}_{q}(n,k)$ that gives rise to the polarization as a
$U_{q}(\mathfrak{g},k)$-module [Kas91, Eq. (2.5.1)]. This is motivated by the
fact that the transpose on $\operatorname{Cl}_{q}(n,k)$ goes to the star
involution on $U_{q}(\mathfrak{g},k)$ and there is a symmetric bilinear form
on the classical Clifford algebra defined by $\left\langle
x,y\right\rangle=\langle x^{t}y\rangle_{0}$, where $\langle x\rangle_{0}$
denotes the degree $0$ component (i.e., constant term) of
$x\in\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$.

The $\operatorname{Cl}_{q}(n,k)$-modules defined in 4.1 are compatible with
the isomorphism $\Gamma_{q}\colon\operatorname{Cl}_{q}(n,k)^{\otimes
m}\to\operatorname{Cl}_{q}(nm,k)$ of 3.10 in the sense that the following
diagram commutes:

${\operatorname{Cl}_{q}(nm,k)}$${\operatorname{End}\left(\bigwedge\nolimits_{q}(V^{(nm)})\right)}$${\operatorname{Cl}_{q}(n,k)^{\otimes
m}}$${\operatorname{End}\left(\bigwedge\nolimits_{q}(\mathbf{k}^{n})^{\otimes
m}\right).}$ (4.4)

######  Proposition 4.4.

Let $T\colon\bigwedge\nolimits_{q}(\mathbf{k}^{n})^{\otimes
m}\to\bigwedge\nolimits_{q}(\mathbf{k}^{nm})$ denote the linear map defined by

$T\left(v(\ell^{(1)})\otimes\cdots\otimes
v(\ell^{(m)})\right)=v\left(\sum_{j=1}^{m}\sum_{i=1}^{n}\ell_{i}^{(j)}e_{i+(j-1)n}\right).$

For $p\in\mathbb{Z}_{2k}^{n}$, let $\pi_{p}^{(r)}$ denote the
$\operatorname{Cl}_{q}(r,k)$-module defined by 4.1. Then for any
$\varphi\in\operatorname{Cl}_{q}(n,k)^{\otimes m}$,

$\displaystyle\pi_{p}^{(nm)}\left(\Gamma_{q}(\varphi)\right)\circ
T=T\circ\left(\pi_{p}^{(n)}\right)^{\otimes m}(\varphi).$ (4.5)

###### Proof.

For simplicity of notation, we assume $p=0$ without loss of generality. It
suffices to verify Equation 4.5 for a set of
$\operatorname{Cl}_{q}(n,k)^{\otimes m}$ generators. To this end, let
$\phi_{a}$ denote either $\psi_{a}$ or $\psi_{a}^{*}$ and set
$\phi_{a,j}=1\otimes\cdots\otimes\phi_{a}\otimes\cdots\otimes 1$ as in the
proof of 3.10. Define $\omega_{a,j}$ analogously and notice 4.1 implies

$\displaystyle\pi_{0}\left(\Gamma_{q}(\omega_{a,j})\right)$ $\displaystyle
T\left(v(\ell^{(1)})\otimes\cdots\otimes
v(\ell^{(m)})\right)=\omega_{a+(j-1)n}v\left(\sum_{k=1}^{m}\sum_{i=1}^{n}\ell_{i}^{(k)}e_{i+(k-1)n}\right)$
$\displaystyle=T\left(v(\ell^{(1)})\otimes\cdots\otimes\left(q^{\ell_{a}^{(j)}-1}v(\ell^{(j)})\right)\otimes\cdots\otimes
v(\ell^{(m)})\right)$
$\displaystyle=T\left(\left(\pi_{0}^{(n)}\right)^{\otimes
m}\left(\omega_{a,j}\right)v(\ell^{(1)})\otimes\cdots\otimes
v(\ell^{(m)})\right).$

To conclude, we take $\varphi=\phi_{a,j}$. We will need the action of the
volume element $f_{r}$ on $\bigwedge\nolimits_{q}(\mathbf{k}^{n})$. Equation
3.24 implies that each (non-central) commutator
$[\psi_{a}^{\dagger},\psi_{a}]$ acts diagonally:

$\displaystyle\pi_{p}\left([\psi_{a}^{\dagger},\psi_{a}]\right)v(\ell)$
$\displaystyle=\frac{q^{k(1-\ell_{a})}-q^{k\ell_{a}}}{q^{k}-1}\,v(\ell)=(-1)^{\ell_{a}}v(\ell).$

Together with 4.1 and Equation 3.14, the last calculation implies the desired
relation:

$\displaystyle\pi_{0}\left(\Gamma_{q}(\phi_{a,j})\right)$ $\displaystyle
T\left(v(\ell^{(1)})\otimes\cdots\otimes
v(\ell^{(m)})\right)=f_{(j-1)n}\phi_{a+(j-1)n}v\left(\sum_{k=1}^{m}\sum_{i=1}^{n}\ell_{i}^{(k)}e_{i+(k-1)n}\right)$
$\displaystyle=(-1)^{\sum_{i=1}^{a-1}\ell_{i}^{(j)}}v\left(\sum_{k=1}^{m}\sum_{i=1}^{n}\ell_{i}^{(k)}e_{i+(k-1)n}\pm
e_{a+(j-1)n}\right)$
$\displaystyle=T\left(v(\ell^{(1)})\otimes\cdots\otimes\pi_{0}^{(n)}\left(\phi_{a}\right)v(\ell^{(j)})\otimes\cdots\otimes
v(\ell^{(m)})\right)$
$\displaystyle=T\left(\left(\pi_{0}^{(n)}\right)^{\otimes
m}\left(\phi_{a,j}\right)v(\ell^{(1)})\otimes\cdots\otimes
v(\ell^{(m)})\right).\qed$

Notice that no irreducible module of $\operatorname{Cl}_{q}(n,k)$ is faithful
since

$\dim\operatorname{Cl}_{q}(n,k)=(8k)^{n}=2^{2n}\cdot(2k)^{n}>2^{2n}=\dim\operatorname{End}\left(\bigwedge\nolimits_{q}(\mathbf{k}^{n})\right).$

This can be explained by the nontrivial center of
$\operatorname{Cl}_{q}(n,k)$, which accounts for the difference in dimension
from its classical counterpart
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$.

Despite this, we may combine the inequivalent $\pi_{p}$ to prove
$\operatorname{Cl}_{q}(n,k)$ is isomorphic to a direct sum of matrix algebras.
This means that there are no other irreducible representations other than
those constructed in 4.1. For contrast, recall that the classical
representation
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)\to\operatorname{End}(\bigwedge(V))$
is faithful, so there is an isomorphism
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)\cong\mathrm{Mat}_{2^{n}}(\mathbf{k})$
showing
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$ is
simple.

######  Theorem 4.5.

If $\zeta_{2k}\in\mathbf{k}$, then the quantum Clifford algebra
$\operatorname{Cl}_{q}(n,k)$ is semisimple. In particular,

$\operatorname{Cl}_{q}(n,k)\cong\bigoplus_{p\in\mathbb{Z}_{2k}^{n}}\operatorname{End}\left(\Omega^{p}\right)\cong(2k)^{n}\mathrm{Mat}_{2^{n}}(\mathbf{k}).$

###### Proof.

Each $\pi_{p}$ is irreducible, so it surjects onto
$\operatorname{End}(\Omega^{p})\cong\mathrm{Mat}_{2^{n}}(\mathbf{k})$.
Therefore, the sum $\bigoplus_{p}\pi_{p}$, which maps
$\operatorname{Cl}_{q}(n,k)$ into a matrix algebra of dimension
$(2k)^{n}\cdot(2^{n})^{2}=(8k)^{n}=\dim(\operatorname{Cl}_{q}(n,k))$, must be
an isomorphism. ∎

When $\zeta_{2k}\notin\mathbf{k}$, the semisimplicity of
$\operatorname{Cl}_{q}(n,k)$ depends on the base field $\mathbf{k}$ over which
it is defined. In particular, $\operatorname{Cl}_{q}(n,k)$ is semisimple
exactly when the group ring $\mathbf{k}[\mathbb{Z}_{2k}^{n}]\cong Z_{q}(n,k)$
is semisimple. Since
$\operatorname{Cl}_{q}(n,k)\cong\operatorname{Cl}_{q}(1,k)^{\otimes n}$ by
3.11, we need only consider the semisimplicity of $Z_{q}(1,k)$.

######  Theorem 4.6.

The quantum Clifford algebra $\operatorname{Cl}_{q}(n,k)$ is semisimple if and
only if $Z_{q}(1,k)\cong\mathbf{k}[\mathbb{Z}_{2k}]$ is semisimple.

###### Proof.

We consider a slightly more general problem. Let $R$ be a commutative ring,
and fix some invertible $r\in R^{\times}$. Let $\operatorname{Cl}_{r}(R\oplus
R^{*})$ be the Clifford algebra generated by $\psi,\psi^{*}$ subject to
$\psi\psi^{*}+\psi^{*}\psi=r$. We show that if $R$ is Artinian, then
$\operatorname{Cl}_{r}(R\oplus R^{*})$ is semisimple if and only if $R$ is
semisimple. The claim then follows from 3.12 and 3.11.

We first show there is a one-to-one correspondence between the two-sided
ideals of $\operatorname{Cl}_{r}(R\oplus R^{*})$ and $R$. Clearly any ideal
$J\subseteq R$ can be used to construct an ideal of
$\operatorname{Cl}_{r}(R\oplus R^{*})$. Thus, we just need to show that for
every two-sided ideal $I\subseteq\operatorname{Cl}_{r}(R\oplus R^{*})$, we
have the same form as above for some ideal $J\subseteq R$. It is sufficient to
show this for a principal ideal of $\operatorname{Cl}_{r}(R\oplus R^{*})$.

We start with a generic element

$x=\alpha+\beta\psi+\gamma\psi^{*}+\delta\psi\psi^{*}$

in $\operatorname{Cl}_{r}(R\oplus R^{*})$ and we look at the two-sided ideal
$I$ generated by $x$. We will show by direct computation that this is
equivalent to the ideal $(\alpha,\beta,\gamma,\delta)\subseteq R$. The
following elements belong to $I$:

$\displaystyle\alpha\psi$ $\displaystyle=r^{-1}\psi x\psi^{*}\psi,$
$\displaystyle\alpha\psi^{*}$ $\displaystyle=r^{-1}\psi^{*}\psi x\psi^{*},$
$\displaystyle\alpha\psi\psi^{*}$ $\displaystyle=\psi x\psi^{*},$
$\displaystyle\beta\psi$ $\displaystyle=r^{-2}\psi\psi^{*}x\psi^{*}\psi,$
$\displaystyle\beta\psi^{*}$ $\displaystyle=r^{-1}\psi^{*}x\psi^{*},$
$\displaystyle\beta\psi\psi^{*}$ $\displaystyle=r^{-1}\psi\psi^{*}x\psi^{*},$
$\displaystyle\gamma\psi$ $\displaystyle=r^{-1}\psi x\psi,$
$\displaystyle\gamma\psi^{*}$ $\displaystyle=r^{-2}\psi^{*}\psi
x\psi\psi^{*},$ $\displaystyle\gamma\psi\psi^{*}$ $\displaystyle=r^{-1}\psi
x\psi\psi^{*}.$

Consequently, we have that $\alpha+\delta\psi\psi^{*}\in I$, and since
$(\alpha+\delta\psi\psi^{*})\psi=(\alpha+r\delta)\psi$, we have that
$\delta\psi\in I$. It is easy to see that
$\delta\psi^{*},\delta\psi\psi^{*}\in I$, which in turn implies $\alpha\in I$.
Finally, we have $r^{-1}(\psi^{*}\beta\psi-\beta\psi\psi^{*})=\beta\in I$, and
similarly $\gamma,\delta\in I$.

The Jacobson radical of a (not necessarily commutative) ring $S$ is the
intersection of all maximal left ideals of $S$. As maximal two-sided ideals
are necessarily maximal left ideals, we have that the Jacobson radical of $S$
is contained in the intersection of all two-sided ideals of $S$. Thus if the
Jacobson radical of $R$ is $0$, then the Jacobson radical of
$\operatorname{Cl}_{r}(R\oplus R^{*})$ is also $0$. Since $R$ is Artinian,
$\operatorname{Cl}_{r}(R\oplus R^{*})$ is also Artinian as it is a finite rank
free module over $R$. An Artinian ring is semisimple if and only if its
Jacobson radical is trivial; see e.g., [Lam01, Thm. 4.14]. Therefore, if $R$
is semisimple then $\operatorname{Cl}_{r}(R\oplus R^{*})$ is semisimple as
well.

For the converse we assume $R$ is not a semisimple ring, and so $R$ is not a
semisimple (rank $1$ free) $R$-module. Consider the left
$\operatorname{Cl}_{r}(R\oplus R^{*})$ spinor module $\pi$, with basis
$1,\psi$. Then $\pi\cong R+\psi R$ with $\psi^{*}R=0$ and $\psi^{*}\psi R=rR$.
Since $r\in R^{\times}$, we have $rM=M$ for every $R$-module $M$, and hence,
$\operatorname{Cl}_{r}(R\oplus R^{*})$ is not a semisimple ring. ∎

From the perspective that
$\operatorname{Cl}_{q}(n,k)\cong\operatorname{Cl}_{r}(R\oplus R^{*})^{\otimes
n}$ for $R=Z_{q}(n,k)$, we can construct all irreducible representations when
$R$ is semisimple by inducing an irreducible $Z_{q}(n,k)$-representation. This
allows us to remove the condition that $\zeta_{2k}\in\mathbf{k}$ from 4.1 and
4.5, although the precise descriptions stated there change and involve the
representation theory of $\mathbb{Z}_{2k}$. For instance, if $n=1$, $k=2$, and
$\mathbf{k}=\mathbb{R}$, we could induce from the $2$ dimensional (nonsplit)
irreducible rotation $\mathbb{Z}_{4}$-module (so $z_{1}$ acts by rotating
$\mathbb{R}^{2}$ by $\pi/2$). We thank Gurbir Dhillon for pointing out that
these irreducible representations could also be constructed directly by
inducing from an irreducible representation of the commutative ring
$\langle\omega_{1},\dotsc,\omega_{n}\rangle$.

We conclude this section by showing that the quantum volume element has
important representation-theoretic implications, much like its classical
counterpart. For example, there is a quantum analog of [ML90, Prop. 5.10].

######  Proposition 4.7.

Consider the superalgebra grading on $\operatorname{Cl}_{q}(n,k)$, and let
$\operatorname{Cl}_{q}^{+}(n,k)$ denote the subalgebra of
$\operatorname{Cl}_{q}(n,k)$ generated by elements of even degree. Let
$\pi\colon\operatorname{Cl}_{q}(n,k)\to\operatorname{End}(W)$ be an
irreducible representation of $\operatorname{Cl}_{q}(n,k)$. Then there exists
a splitting

$W=W^{+}\oplus W^{-},$

with $W^{\pm}=\bigl{(}1\pm\pi(f_{n})\bigr{)}(W)$, where $W^{\pm}$ is invariant
under $\operatorname{Cl}_{q}^{+}(n,k)$.

###### Proof.

3.9 implies $f_{n}$ commutes with $\operatorname{Cl}_{q}^{+}(n,k)$, so
$\operatorname{Cl}_{q}^{+}(n,k)$ preserves every eigenspace of $f_{n}$. The
claim then follows because $f_{n}^{2}=1$: $W^{\pm}$ is a $\pm 1$ eigenspace of
$f_{n}$ and the only possible eigenvalues of $f_{n}$ are $-1$ and $+1$. ∎

##  5\. Twisting and square roots

We note that nearly all the relations defining $\operatorname{Cl}_{q}(n,k)$
allow us to take $k=1/2$ except for Equation 3.10. Yet, we can adjust this by
considering new generators

$\phi_{a}=\psi_{a}\qquad\text{and}\qquad\phi_{a}^{*}=\psi_{a}^{*}\omega_{a}^{k}.$

Using these generators, our algebra is presented by

$\displaystyle\omega_{a}\omega_{b}$ $\displaystyle=\omega_{b}\omega_{a},\hskip
70.0pt$ $\displaystyle\omega_{a}^{4k}$
$\displaystyle=(1+q^{-2k})\omega_{a}^{2k}-q^{-2k},$ (5.1)
$\displaystyle\omega_{a}\phi_{b}$
$\displaystyle=q^{\delta_{ab}}\phi_{b}\omega_{a},$
$\displaystyle\omega_{a}\phi_{b}^{*}$
$\displaystyle=q^{-\delta_{ab}}\phi_{b}^{*}\omega_{a},$
$\displaystyle\phi_{a}\phi_{b}$ $\displaystyle+\phi_{b}\phi_{a}=0,$
$\displaystyle\phi_{a}^{*}\phi_{b}^{*}$
$\displaystyle+\phi_{b}^{*}\phi_{a}^{*}=0,$
$\displaystyle\phi_{a}\phi_{a}^{*}$ $\displaystyle+\phi_{a}^{*}\phi_{a}=1,$
$\displaystyle\phi_{a}\phi_{a}^{*}$
$\displaystyle+q^{-2k}\phi_{a}^{*}\phi_{a}=\omega_{a}^{2k},$
$\displaystyle\phi_{a}\phi_{b}^{*}$ $\displaystyle+\phi_{b}^{*}\phi_{a}=0$
$\displaystyle\text{if }a\neq b.$

Using this presentation, we can take $k\in\frac{1}{2}\mathbb{Z}_{>0}$. It is
easy to compute analogous reduction rules involving $\phi_{a}$ and
$\phi_{a}^{*}$ instead of $\psi_{a}$ and $\psi_{a}^{*}$. In addition to those
from a simple reformulation of the defining relations, we have the following
reductions analogous to 3.3.

######  Lemma 5.1.

The following relations hold in $\operatorname{Cl}_{q}(n,k)$:

$\omega_{a}^{2k}=(1-q^{-2k})\phi_{a}\phi_{a}^{*}+q^{-2k},\qquad\qquad\phi_{a}^{*}\phi_{a}\phi_{a}^{*}=\phi_{a}^{*},\qquad\qquad\phi_{a}\phi_{a}^{*}\phi_{a}\phi_{a}^{*}=\phi_{a}\phi_{a}^{*}.$
(5.2)

Thus, we have the analog of 3.4 by a similar proof.

######  Theorem 5.2.

Consider the set $\mathcal{B}^{\prime}$ of monomials

$\prod_{a=1}^{n}\phi_{a}^{p_{a}}(\phi_{a}^{*})^{d_{a}}\omega_{a}^{v_{a}},$

where $p_{a},d_{a}\in\\{0,1\\}$ and $v_{a}\in\\{0,1,\dotsc,2k-1\\}$. Then
$\mathcal{B}^{\prime}$ is a basis of $\operatorname{Cl}_{q}(n,k)$ for
$k\in\frac{1}{2}\mathbb{Z}_{>0}$.

Therefore, the dimension formula from 3.5 still holds. In particular, we have
$\dim\operatorname{Cl}_{q}(n,1/2)=4^{n}$.

We also have the analog of 3.6 when $q^{2k}\neq 1$. Furthermore, when
$q^{2k}\neq 1$, we also note that (3.10) and Equation 3.8 imply

$\displaystyle\phi_{a}\phi_{a}^{*}=\frac{\omega_{a}^{2k}(q^{2k}\omega_{a}^{2k}-1)}{q^{2k}-1}=\frac{q^{2k}\omega_{a}^{2k}-1}{q^{2k}-1},\qquad\phi_{a}^{*}\phi_{a}=\frac{q^{4k}\omega_{a}^{2k}(1-\omega_{a}^{2k})}{q^{2k}-1}=\frac{q^{2k}(1-\omega_{a}^{2k})}{q^{2k}-1},$
(5.3)

yielding another alternative relation

$\phi_{a}\phi_{a}^{*}+q^{-4k}\phi_{a}^{*}\phi_{a}=\omega_{a}^{4k}=(1+q^{-2k})\omega_{a}^{2k}-q^{-2k}.$

It is also easy to see that the center computation (3.12) extends to the case
$k\in\mathbb{Z}_{\geq 0}+\frac{1}{2}$ as Equation 3.15 becomes

$z_{a}=q\omega_{a}-(q-1)\phi_{a}\phi_{a}^{*}\omega_{a}.$

In particular, when $k=\frac{1}{2}$, we have

$\omega_{a}=(1-q^{-1})\phi_{a}\phi_{a}^{*}+q^{-1}$ (5.4)

and $z_{a}=1$. Thus, the center of $\operatorname{Cl}_{q}(n,\frac{1}{2})$ is
trivial as claimed by 3.12.

Let us further explore the case when $k=\frac{1}{2}$. As we saw, these quantum
Clifford algebras do not have a nontrivial center and start looking like the
classical Clifford algebra as Equation 5.4 says $\omega_{a}$ belongs to the
subalgebra generated by $\psi_{a}$, $\psi_{a}^{*}$. In fact, by examining the
relations in Lemma 5.1, we see that $\operatorname{Cl}_{q}(n,\frac{1}{2})$ can
be given the presentation of
$\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$.

######  Theorem 5.3.

We have

$\operatorname{Cl}_{q}(n,1/2)\cong\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right).$

5.3 is also another reflection of the principle that the quantum Clifford
algebra is a classical Clifford algebra over its (group ring) center.

With a little bit of work to express everything in terms of the new
generators, it can be shown that 3.10, 4.5, and 4.6 still hold for all
$k\in\frac{1}{2}\mathbb{Z}_{>0}$. In particular, we build the modified quantum
volume element
$\check{f}_{r}=[\phi_{1},\phi_{1}^{*}]\cdots[\phi_{r},\phi_{r}^{*}]$, which
behaves analogously to the classical volume element (or 3.9) by the defining
relations (5.1). We leave the details for the interested reader.

The quantum Clifford algebra for $k=\frac{1}{2}$ has also appeared in [DF94]
in connection with an FRT construction, where we note that [DF94, Eq.
(5.3.1)–(5.3.4)] are the defining relations (using Lemma 5.1 for rewriting
$\omega_{a}^{2k}$). In particular, Ding and Frenkel [DF94] start with Theorem
5.3 and construct $\operatorname{Cl}_{q}(n,\frac{1}{2})$.

We could consider the limit $k\to\infty$ (in contrast to the case $n\to\infty$
considered in [Hay90]) when, say, $\lvert q\rvert>1$. In this case our center
tends towards a product of circle groups, $\omega_{a}^{2k}\to 1$, any spinor
representation is no longer irreducible (although it remains indecomposible),
and its “dual” (in the sense of 4.3) is $1$ dimensional. Indeed, we would have
$\psi_{a}^{*}\psi_{a}v=0$ (or $\phi_{a}^{*}\phi_{a}v=0$) for any $v$ in any
spinor representation.

##  6\. Clifford algebra morphisms

We examine the remark by Mitsuhiro Takeuchi included at the end of [Hay90]. We
assume that $q^{2k}\neq 1$ and $\zeta_{2k}\in\mathbf{k}$. For brevity, we set
$\operatorname{Cl}_{(n)}=\operatorname{Cl}\left(\mathbf{k}^{n}\oplus(\mathbf{k}^{n})^{*}\right)$.
We denote the generators of $\operatorname{Cl}_{(n)}$ by $v_{i},v_{i}^{*}$,
for $i=1,\dotsc,n$.

######  Proposition 6.1 (M. Takeuchi [Hay90]).

Suppose $\zeta_{2k}\in\mathbf{k}$. Consider the map
$\vartheta\colon\operatorname{Cl}_{q}(n,k)\to\operatorname{Cl}_{(n)}^{(2k)^{n}}$
defined on any $y\in\operatorname{Cl}_{q}(n,k)$ by

$\vartheta(m)=\bigl{(}\vartheta_{\mathbf{z}}(y)\bigr{)}_{\mathbf{z}\in\\{\zeta_{2k}^{j}\mid
j=1,\dotsc,2k\\}^{n}}$

where for each $\mathbf{z}\in\\{\zeta_{2k}^{j}\mid j=1,\dotsc,2k\\}$ the
$\mathbb{Z}^{n}$-graded $\mathbf{k}$-algebra morphism
$\vartheta_{\mathbf{z}}\colon\operatorname{Cl}_{q}(n,k)\to\operatorname{Cl}_{(n)}$
is defined on generators by

$\vartheta_{\mathbf{z}}(\psi_{i})=z_{i}v_{i},\qquad\vartheta_{\mathbf{z}}(\psi_{i}^{*})=z_{i}v_{i}^{*},\qquad\vartheta_{\mathbf{z}}(\omega_{1})=z_{i}(v_{i}v_{i}^{*}+q^{-1}v_{i}^{*}v_{i})=z_{i}(1-q^{-1})v_{i}v_{i}^{*}+z_{i}q^{-1}.$

Then $\vartheta$ is a $\mathbf{k}$-algebra isomorphism.

###### Proof.

By either directly noting the compatible anticommutator relations or using our
structure results (2.2 and 3.10), it is sufficient prove the $n=1$ case. Since
$(v_{i}v_{i}^{*})^{2}=v_{i}v_{i}^{*}$, by a straightforward induction
argument, we have

$\vartheta_{\mathbf{z}}(\omega_{i})^{m}=z_{i}^{m}(1-q^{-m})v_{i}v_{i}^{*}+z_{i}^{m}q^{-m}.$

By verifying the relations of $\operatorname{Cl}_{q}(1,k)$, we see that
$\vartheta_{\mathbf{z}}$ is a surjective $\mathbb{Z}^{n}$-graded
$\mathbf{k}$-algebra morphism for any fixed $\mathbf{z}$ .

Now we show $\vartheta$ is surjective. To this end, let
$\varphi:=\vartheta(\psi_{1})$, $\varphi^{*}:=\vartheta(\psi_{1}^{*})$, and
$\varpi:=\vartheta(\omega_{1})$. We fix the order of the sequence defining
$\vartheta$ as $(1,\zeta_{2k},\dotsc,\zeta_{2k}^{2k-1})$. Then, we have

$q^{m}\varphi\varpi^{m}=(\zeta_{2k}^{j(m+1)}v_{1})_{j=0}^{2k-1},\qquad\qquad
q^{m}\varpi^{m}\varphi^{*}=(\zeta_{2k}^{j(m+1)}v_{1}^{*})_{j=0}^{2k-1}.$ (6.1)

From basic linear algebra (or the representation theory of $\mathbb{Z}_{2k}$)
as this is essentially giving a discrete Fourier transform, we can obtain
$v_{(j)}:=(\delta_{jm}v_{1})_{m=0}^{2k-1}$ and
$v_{(j)}^{*}:=(\delta_{jm}v_{1}^{*})_{m=0}^{2k-1}$, for any fixed
$j=0,\dotsc,2k-1$, through linear combinations of the elements (6.1). Thus, we
have obtained all of the generators of $\operatorname{Cl}_{(1)}^{2k}$, and
hence $\vartheta$ is surjective.

We can define the inverse map $\vartheta^{-1}$ by the inverse discrete Fourier
transform

$\vartheta(v_{(j)})=\frac{1}{2k}\sum_{m=0}^{2k-1}\zeta_{2k}^{-j(m+1)}q^{m}\psi_{1}\omega_{1}^{m},\qquad\qquad\vartheta(v_{(j)}^{*})=\frac{1}{2k}\sum_{m=0}^{2k-1}\zeta_{2k}^{-j(m+1)}\psi_{1}^{*}\omega_{1}^{m}.$

A direct computation shows this is a $\mathbf{k}$-algebra morphism and the
inverse of $\vartheta$. ∎

In particular, we note that
$\dim\operatorname{Cl}_{(n)}^{(2k)^{n}}=(2k)^{n}4^{n}=(8k)^{n}$ as there are
$(2k)^{n}$ such tuples $\mathbf{z}$. 6.1 gives an alternative proof that
$\mathcal{B}$ is linearly independent (without relying on Theorem 3.4). The
appearance of the discrete Fourier transform comes from the fact that we can
consider the quantum Clifford algebra $\operatorname{Cl}_{q}(n,k)$ as a
Clifford algebra over the center $Z_{q}(n,k)$ (see 4.6). Indeed, we are
effectively splitting the natural adjoint representation of the center into
its irreducible ($1$ dimensional) modules and taking the corresponding
Clifford algebra $\operatorname{Cl}(\mathbf{k}\oplus\mathbf{k}^{*})$ on each
such module. Compare this with 4.5.

######  Example 6.2.

Consider the case $n=1$ and $k=1$. The isomorphism
$\vartheta\colon\operatorname{Cl}_{q}(1,1)\to\operatorname{Cl}_{(1)}\times\operatorname{Cl}_{(1)}$
from 6.1 is given explicitly on basis elements by

$\displaystyle 1$ $\displaystyle\mapsto(1,1),$ $\displaystyle\omega_{1}^{m}$
$\displaystyle\mapsto\bigl{(}(1-q^{-m})v_{1}v_{1}^{*}+q^{-m},(-1)^{m}(1-q^{-m})v_{1}v_{1}^{*}+(-1)^{m}q^{-m}\bigr{)},$
$\displaystyle\psi_{1}$ $\displaystyle\mapsto(v_{1},-v_{1}),$
$\displaystyle\psi_{1}\omega_{1}$ $\displaystyle\mapsto(q^{-1}v,q^{-1}v_{1}),$
$\displaystyle\psi_{1}^{*}$ $\displaystyle\mapsto(v_{1}^{*},-v_{1}^{*}),$
$\displaystyle\psi_{1}^{*}\omega_{1}$
$\displaystyle\mapsto(v_{1}^{*},v_{1}^{*}),$

for $m=1,2,3$. For fixed $x=\pm 1$, the morphism $\vartheta_{x}$ is surjective
but not injective since
$\vartheta_{1}(\psi_{1}^{*}-x\psi_{1}^{*}\omega_{1})=0$. However, looking at,
e.g., $(\psi_{1}^{*},\psi_{1}^{*}\omega_{1})$, we get the discrete Fourier
transform matrix $\begin{bmatrix}1&1\\\ -1&1\end{bmatrix}$. We can see that

$\displaystyle\vartheta^{-1}(v_{1},0)$
$\displaystyle=\psi_{1}-q\psi_{1}\omega_{1},$
$\displaystyle\vartheta^{-1}(v_{1}^{*},0)$
$\displaystyle=\psi_{1}^{*}-\psi_{1}^{*}\omega_{1},$
$\displaystyle\vartheta^{-1}(0,v_{1})$
$\displaystyle=\psi_{1}+q\psi_{1}\omega_{1},$
$\displaystyle\vartheta^{-1}(0,v_{1}^{*})$
$\displaystyle=\psi_{1}^{*}+\psi_{1}^{*}\omega_{1},$

is coming from the inverse discrete Fourier transform matrix.

We continue to examine the case $n=1$ and $k=1$ in a bit more detail. Let
$\operatorname{Cl}_{a,b}$ denote the Clifford algebra given by the quadratic
form with signature $(a,b)$; that is, the form whose Hessian matrix is
diagonal with $a$ entries being $1$ followed by $b$ entries being $-1$. We can
change the basis in $\operatorname{Cl}_{(1)}:=\langle v,v^{*}\rangle$ by
writing

$\xi=v+v^{*},\qquad\qquad\xi^{*}=v-v^{*},$ (6.2)

which satisfy

$\xi^{2}=1,\qquad(\xi^{*})^{2}=-1,\qquad\xi\xi^{*}=1-2vv^{*},\qquad\xi\xi^{*}+\xi^{*}\xi=0.$

(Note $\xi=\epsilon_{2}$ and $\xi^{*}=\epsilon_{1}$ from Equation 2.6.) Thus,
$\operatorname{Cl}_{(1)}\cong\operatorname{Cl}_{1,1}$, where the inverse
isomorphism is given by

$v=\frac{\xi+\xi^{*}}{2},\qquad\qquad v^{*}=\frac{\xi-\xi^{*}}{2},\qquad\qquad
vv^{*}=\frac{1-\xi\xi^{*}}{2}.$ (6.3)

Next, we have
$\operatorname{Cl}_{2,1}\cong\operatorname{Cl}_{1,1}\times\operatorname{Cl}_{1,1}$
as $\mathbf{k}$-algebras by the isomorphism

$e_{1}\mapsto(\xi,\xi),\qquad\qquad
e_{2}\mapsto(\xi\xi^{*},-\xi\xi^{*}),\qquad\qquad
e_{3}\mapsto(\xi^{*},\xi^{*}),$

where $e_{1}^{2}=e_{2}^{2}=1$ and $e_{3}^{2}=-1$. Note that this is not a
filtered $\mathbf{k}$-algebra isomorphism under the natural filtrations.
Hence, we have an isomorphism
$\operatorname{Cl}_{q}(1,1)\cong\operatorname{Cl}_{2,1}$ as
$\mathbf{k}$-algebras. We can write out this isomorphism explicitly:

$\displaystyle\begin{aligned}
\frac{e_{1}+e_{3}}{2}&=(v,v)=q\psi_{1}\omega_{1},&\frac{e_{1}-e_{3}}{2}&=(v^{*},v^{*})=\psi_{1}^{*}\omega_{1},\\\
\frac{e_{1}e_{2}+e_{2}e_{3}}{2}&=(-v^{*},v^{*})=-\psi_{1},&\frac{e_{1}e_{2}-e_{2}e_{3}}{2}&=(v,-v)=\psi_{1},\end{aligned}$
$\displaystyle\begin{aligned}
\omega_{1}^{m}&=\left(\frac{(q^{-m}-1)\xi\xi^{*}+q^{-m}+1}{2},(-1)^{m}\frac{(q^{-m}-1)\xi\xi^{*}+q^{-m}+1}{2}\right)\\\
&=\begin{cases}\frac{q^{-m}-1}{2}e_{1}e_{3}+\frac{q^{-m}+1}{2}&\text{if
}m=2,\\\
\frac{q^{-m}-1}{2}e_{2}-\frac{q^{-m}+1}{2}e_{1}e_{2}e_{3}&\text{otherwise}.\end{cases}\end{aligned}$

For general $n$, with $k=1$, we note that all of the other relations are
supercommuting relations with considering $\omega_{a}$ as an even element and
$\psi_{a}$ and $\psi_{a}^{*}$ as odd elements. Therefore, we can form an
isomorphism by taking the $n$-fold supersymmetric product of
$\operatorname{Cl}_{2,1}$ with $e_{2}$ being an even element and $e_{1},e_{3}$
being odd elements. However, this is not an isomorphism of supercommutative
superalgebras.

## References

  * [Abo22a] Willie Aboumrad.  Skew Howe duality for types $\mathbf{BD}$ via quantum Clifford algebras, 2022.  URL: https://arxiv.org/abs/2208.09773, doi:10.48550/ARXIV.2208.09773.
  * [Abo22b] Willie Aboumrad.  Skew Howe duality for $U_{q}(\mathfrak{gl}_{n})$ via quantized Clifford algebras, 2022.  URL: https://arxiv.org/abs/2208.08979, doi:10.48550/ARXIV.2208.08979.
  * [Ari96] Susumu Ariki.  On the decomposition numbers of the Hecke algebra of $G(m,1,n)$.  J. Math. Kyoto Univ., 36(4):789–808, 1996.  doi:10.1215/kjm/1250518452.
  * [Bum04] Daniel Bump.  Lie Groups.  Springer, second edition, 2004.  doi:10.1007/978-1-4757-4094-3.
  * [BZ05] Arkady Berenstein and Sebastian Zwicknagl.  Braided symmetric and exterior algebras.  2005\.  URL: https://arxiv.org/abs/math/0504155, doi:10.48550/ARXIV.MATH/0504155.
  * [CKM14] Sabin Cautis, Joel Kamnitzer, and Scott Morrison.  Webs and quantum skew Howe duality.  Math. Ann., 360(1-2):351–390, 2014.  doi:10.1007/s00208-013-0984-4.
  * [CP94] Vyjayanthi Chari and Andrew Pressley.  A guide to quantum groups.  Cambridge University Press, 1994.  doi:10.1007/BF00750760.
  * [DF94] Jin Tai Ding and Igor B. Frenkel.  Spinor and oscillator representations of quantum groups.  In Lie theory and geometry, volume 123 of Progr. Math., pages 127–165. Birkhäuser Boston, Boston, MA, 1994.  URL: https://doi.org/10.1007/978-1-4612-0261-5_5, doi:10.1007/978-1-4612-0261-5\\_5.
  * [Dri85] V. G. Drinfel′d.  Hopf algebras and the quantum Yang-Baxter equation.  Dokl. Akad. Nauk SSSR, 283(5):1060–1064, 1985.
  * [FH91] William Fulton and Joe Harris.  Representation theory, volume 129 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 1991.  A first course, Readings in Mathematics.  doi:10.1007/978-1-4612-0979-9.
  * [Ger19] Thomas Gerber.  Heisenberg algebra, wedges and crystals.  J. Algebraic Combin., 49(1):99–124, 2019.  doi:10.1007/s10801-018-0820-8.
  * [GW09] Roe Goodman and Nolan Wallach.  Symmetry, representations, and invariants.  Springer, first edition, 2009.  doi:10.1007/978-0-387-79852-3.
  * [Hay90] Takahiro Hayashi.  Q-analogues of Clifford and Weyl algebras-spinor and oscillator representations of quantum enveloping algebras.  Communications in Mathematical Physics, 127(1):129–144, 1990.  doi:10.1007/bf02096497.
  * [Jan96] Jens Carsten Jantzen.  Lectures on quantum groups, volume 6 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 1996.  doi:10.1090/gsm/006.
  * [Jim85] Michio Jimbo.  A $q$-difference analogue of $U({\mathfrak{g}})$ and the Yang-Baxter equation.  Lett. Math. Phys., 10(1):63–69, 1985.  URL: http://dx.doi.org/10.1007/BF00704588, doi:10.1007/BF00704588.
  * [JMMO91] Michio Jimbo, Kailash C. Misra, Tetsuji Miwa, and Masato Okado.  Combinatorics of representations of $U_{q}(\widehat{\mathfrak{gl}}(n))$ at $q=0$.  Comm. Math. Phys., 136(3):543–566, 1991.  URL: http://projecteuclid.org/euclid.cmp/1104202436.
  * [JMO00] Naihuan Jing, Kailash C. Misra, and Masato Okado.  $q$-wedge modules for quantized enveloping algebras of classical type.  J. Algebra, 230(2):518–539, 2000.  doi:10.1006/jabr.2000.8325.
  * [Kas90] Masaki Kashiwara.  Crystalizing the $q$-analogue of universal enveloping algebras.  Comm. Math. Phys., 133(2):249–260, 1990.  URL: https://projecteuclid.org/getRecord?id=euclid.cmp/1104201397.
  * [Kas91] Masaki Kashiwara.  On crystal bases of the $q$-analogue of universal enveloping algebras.  Duke Math. J., 63(2):465–516, 1991.  doi:10.1215/S0012-7094-91-06321-0.
  * [Kas93] Masaki Kashiwara.  The crystal base and Littelmann’s refined Demazure character formula.  Duke Math. J., 71(3):839–858, 1993.  doi:10.1215/S0012-7094-93-07131-1.
  * [Kas95] Christian Kassel.  Quantum groups, volume 155 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 1995.  doi:10.1007/978-1-4612-0783-2.
  * [KMS95] M. Kashiwara, T. Miwa, and E. Stern.  Decomposition of $q$-deformed Fock spaces.  Selecta Math. (N.S.), 1(4):787–805, 1995.  doi:10.1007/BF01587910.
  * [Knu91] Max-Albert Knus.  Quadratic and Hermitian forms over rings, volume 294 of Grundlehren der mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences].  Springer-Verlag, Berlin, 1991.  With a foreword by I. Bertuccioni.  doi:10.1007/978-3-642-75401-2.
  * [Kwo14] Jae-Hoon Kwon.  Q-deformed Clifford algebra and level zero fundamental representations of quantum affine algebras.  J. Algebra, 399:927–947, 2014.  doi:10.1016/j.jalgebra.2013.10.026.
  * [Lam01] T. Y. Lam.  A first course in noncommutative rings, volume 131 of Graduate Texts in Mathematics.  Springer-Verlag, New York, second edition, 2001.  doi:10.1007/978-1-4419-8616-0.
  * [Lam05] T. Y. Lam.  Introduction to quadratic forms over fields, volume 67 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 2005.  doi:10.1090/gsm/067.
  * [LLT96] Alain Lascoux, Bernard Leclerc, and Jean-Yves Thibon.  Hecke algebras at roots of unity and crystal bases of quantum affine algebras.  Comm. Math. Phys., 181(1):205–263, 1996.  URL: http://projecteuclid.org/euclid.cmp/1104287629.
  * [LT96] Bernard Leclerc and Jean-Yves Thibon.  Canonical bases of $q$-deformed Fock spaces.  Internat. Math. Res. Notices, (9):447–456, 1996.  doi:10.1155/S1073792896000293.
  * [Lus90] G. Lusztig.  Canonical bases arising from quantized enveloping algebras.  J. Amer. Math. Soc., 3(2):447–498, 1990.  doi:10.2307/1990961.
  * [LZZ11] G. I. Lehrer, Hechun Zhang, and R. B. Zhang.  A quantum analogue of the first fundamental theorem of classical invariant theory.  Comm. Math. Phys., 301(1):131–174, 2011.  doi:10.1007/s00220-010-1143-3.
  * [ML90] Marie-Louise Michelsohn and H. Blaine. Lawson.  Spin Geometry.  Princeton University Press., 1990.
  * [O’M00] O. Timothy O’Meara.  Introduction to quadratic forms.  Classics in Mathematics. Springer-Verlag, Berlin, 2000.  Reprint of the 1973 edition.
  * [QS19] Hoel Queffelec and Antonio Sartori.  Mixed quantum skew Howe duality and link invariants of type $A$.  J. Pure Appl. Algebra, 223(7):2733–2779, 2019.  doi:10.1016/j.jpaa.2018.09.014.
  * [RTF89] N. Yu. Reshetikhin, L. A. Takhtadzhyan, and L. D. Faddeev.  Quantization of Lie groups and Lie algebras.  Algebra i Analiz, 1(1):178–206, 1989.
  * [Sag22] The Sage Development Team.  Sage Mathematics Software (Version 9.7), 2022.  https://www.sagemath.org.
  * [ST19] Antonio Sartori and Daniel Tubbenhauer.  Webs and $q$-Howe dualities in types BCD.  Trans. Amer. Math. Soc., 371(10):7387–7431, 2019.  doi:10.1090/tran/7583.
  * [TFV20] Marco A. Trindade, Sergio Floquet, and J. David Vianna.  Clifford algebras, algebraic spinors, quantum information and applications.  Modern Physics Letters A, 35(29):205–239, 2020.  doi:10.1142/s0217732320502399.
  * [Wen12] Hans Wenzl.  On centralizer algebras for spin representations.  Communications in Mathematical Physics, 314(1):243–263, 2012.  doi:10.1007/s00220-012-1494-z.