Introduction
This project provides functionality for constructing and working with Clifford algebras associated with quadratic spaces. Currently, it supports base rings such as the rationals, algebraic number fields, and their respective rings of integers.
Definitions and notations
Let $R$ be a ring and $(M,q)$ a quadratic $R$-module, i.e. $M$ is an $R$-module that carries an $R$-valued quadratic form $q.$ For a positive integer $m$, we write $\underline{m} = \lbrace 1, \ldots, m \rbrace$.
A Clifford algebra for $(M,q)$ is a pair $(C,g)$, that consists of an associative unital $R$-algebra $C$ and an $R$-module homomorphism $g : M \rightarrow C$ such that $g(x)^2 = q(x) \cdot 1_C$, for all $x \in M$, which has the following universal property: For any other pair $(B,f)$ of an associative unital $R$-algebra $B$ and an $R$-module homomorphism $f : M \rightarrow B$ with $f(x)^2 = q(x) \cdot 1_B$ for all $x \in M$, there exists a unique $R$-algebra homomorphism $h : C \rightarrow B$ such that $f = h \circ g$.
Since the Clifford algebra of a quadratic $R$-module $M = (M,q)$ is defined via a universal property, the Clifford algebra $C = C(M)$ is unique up to isomorphism of $R$-algebras. The Clifford algebra can be constructed as a quotient of the free associative $R$-algebra on a generating set of $M$, which shows existence.
In this project, we make the following assumptions:
- $R$ is a Dedekind domain of characteristic zero with field of fractions $K$, where we allow $R = K$ as a trivial example of a Dedekind domain
- $M$ is a projective $R$-module of finite rank $n$
Note that if $M$ is projective, then the homomorphism $g$ is injective, so we always identify $M$ as a subset of $C(M)$ and simply write $x \in C(M)$ instead of $g(x) \in C(M)$.
Clifford algebras over fields
Over the field $R = K$, the quadratic $K$-space $V = (V,q)$ is obviously free and $C = C(V)$ can be constructed as the quotient of the tensor algebra of $V$ by the two-sided ideal generated by elements of the form $x^2 - q(x) \cdot 1_C$, $x \in V$. It is well known that $C$ is free of rank $2^n$. More specifically, if
\[(e_i \mid i \in \underline{n})\]
is a $K$-basis of $V$, then a $K$-basis of $\mathcal{C}$ is given by
\[(e_I \mid I \subseteq \underline{n}),\]
with $e_I = e_{i_1} \ldots e_{i_k}$, where $I = \lbrace i_1, \ldots, i_k \rbrace$, $i_1 < \ldots < i_k$ for some $0 \leq k \leq n$. For example, we have
\[e_\emptyset = 1_\mathcal{C}, \quad e_{\lbrace 1 \rbrace} = e_1, \quad e_{\lbrace 1,3,4 \rbrace } = e_1 e_3 e_4, \quad etc.\]
From now on, we will drop the brackets in the indices, so we simply write $e_{1,3,4}$ instead of $e_{\lbrace 1,3,4 \rbrace}$ and so on.
The Clifford algebra is commutative if and only if $n \in \lbrace 0, 1 \rbrace$, so in general, the order in the above product matters; see is_commutative.
Relations in the Clifford algebra
Note that the (free) quadratic $K$-space $V$ of rank $n$ is determined up to isometry by a symmetric $n \times n$-matrix with entries in $K$ and main diagonal in $2K$ (this is trivially satisfied as $K$ has characteristic zero). This matrix $G$ coincides with the Gram matrix (with respect to some fixed basis) of the so-called polarisation, the symmetric bilinear form on $V$ defined by
\[b_q(x,y) = q(x + y) - q(x) - q(y), \quad x,y \in V.\]
As a consequence, the structure of $C$ is uniquely determined by $G$ in the sense that the multiplication table of $C$ is recursively obtained from it. Specifically, if $G$ is the Gram matrix of $b_q$ with respect to the basis $(e_i \mid i \in \underline{n} )$ of $V$ then the following relations hold in $C$:
- $e_i e_j + e_j e_i = b_q(e_i, e_j)\cdot 1_C = G_{i,j} \cdot 1_C, \quad i,j \in \underline{n}, \; i \neq j$
- $e_i^2 = q(e_i) \cdot 1_C = \dfrac{1}{2}G_{i,i} \cdot 1_C, \quad i \in \underline{n}$
In all of the above, we only used the fact that $V$ is a free $R$-module, but not that $R = K$ is a field. As a consequence, all of the above holds as well for Clifford algebras over the integers or, more generally, for Clifford algebras over principal ideal domains. Just note that over these rings, the condition that the main diagonal of the Gram matrix $G$ be in $2R$ is not trivially satisfied.
Lattices and orders over Dedekind domains
Let $R$ be a Dedekind domain in characteristic zero that is not a field with field of fractions $K$. All fractional ideals of $R$ are implicitly assumed to be non-zero. Given a $K$-space $V$ an $R$-lattice is a finitely generated $R$-submodule $L$ of $V$. The lattice $L$ is called full in $V$ if it contains a $K$-basis of $V$, i.e. if $KL = V$. Equivalently, an $R$-lattice is a finitely generated projective $R$-module $L$. Note that in the latter version, $L$ can be regarded as a full $R$-lattice in the ambient space $V \coloneqq K \otimes_R L$. On this page, we will use the first point of view and assume that $L$ is full in $V$.
It is well known that every $R$-lattice $L$ has a pseudo-basis, which is a finite sequence $(e_i, \mathfrak{a}_i)_{i \in \underline{n}}$, where the $\mathfrak{a}_i$ are fractional ideals of $R$ and $(e_i \mid i \in \underline{n})$ is a $K$-basis of $V$, such that
\[L = \bigoplus\sum\limits_{i = 1}^n \mathfrak{a}_i e_i.\]
The rank $n$ of $L$ is well-defined and $n = \mathrm{rank}(L) = \mathrm{dim}(V)$.
If $V = (V,q)$ is a quadratic $K$-space then by means of restriction, we obtain the quadratic $R$-lattice $L = (L,q)$. We always assume that $L$ is an even lattice, which is to say that the form $q$ is $R$-valued.
In general, the quadratic form $q$ need not be $R$-valued. In this case, $L$ is not a quadratic $R$-module as defined above. In particular, the Clifford algebra of $L$ does not exist.
Let $A$ be a $K$-algebra. An $R$-order in $A$ is a full $R$-lattice $\Lambda$ in $A$ that is also a subring with the same identity element. Let $L$ be an even $R$-lattice with ambient space $V$. The following examples of $R$-orders are relevant to this project:
- The Clifford algebra $C(L)$ is an $R$-order in $C(V)$, called the Clifford order of $L$.
- The even Clifford algebra $C_{0}(L) \coloneqq C_{0}(V) \cap C(L)$ of $L$ is an $R$-order in $C_{0}(V)$.
- The centroid $\mathcal{Z}(L) \coloneqq \mathcal{Z}(V) \cap C(L)$ of $L$ is an $R$-order in $\mathcal{Z}(V)$.
Clifford orders over rings of integers
Let $L$ be a full even $R$-lattice in $V$ of rank $n$ with pseudo-basis $(e_i, \mathfrak{a}_i)_{i \in \underline{n}}$. Then the Clifford order $C(L)$ is a full $R$-lattice of rank $2^n$ in the Clifford algebra $C(V)$. More specifically, a pseudo-basis of $C(L)$ is given by $(e_I, \mathfrak{a}_I)_{I \subseteq \underline{n}}$, with $e_I$ defined as in the algebra case and $\mathfrak{a}_I = \mathfrak{a}_{i_1}\ldots \mathfrak{a}_{i_k}$, where $I = \lbrace i_1, \ldots, i_k \rbrace$, for some $0 \leq k \leq n$.
Relations in the Clifford order
The Clifford order $C(L)$ inherits all relations that hold within $C(V)$ because it is a subset. Thus, we still have:
- $e_i e_j + e_j e_i = b_q(e_i, e_j)\cdot 1_C = G_{i,j} \cdot 1_C, \quad i,j \in \underline{n}, \; i \neq j$
- $e_i^2 = q(e_i) \cdot 1_C = \dfrac{1}{2}G_{i,i} \cdot 1_C, \quad i \in \underline{n}$
Moreover, as in the field case, the even $R$-lattice $L$ is determined by the Gram matrix $G$ of $V$ with respect to the $K$-basis $(e_i \mid i \in \underline{n})$. However, we need the additional conditions $G_{i,j} \in (\mathfrak{a}_i\mathfrak{a}_j)^{-1}$ and $G_{i,i} \in 2\mathfrak{a}_i^{-2}$ to ensure that $G$ actually defines an even lattice.
Content
This project is structured as follows:
- The section Clifford algebras over fields contains functionality for Clifford algebras over fields (e.g.,
QQor other algebraic number fields) - The section Clifford orders over rings of integers contains functionality for Clifford orders (i.e. orders within Clifford algebras over the rings of integers, such as
ZZor $\mathcal{O}_K$)
Status
This part of OSCAR is in an experimental state; please see Adding new projects to experimental for what this means.
Contact
Please direct questions about this part of OSCAR to the following people:
You can ask questions in the OSCAR Slack.
Alternatively, you can raise an issue on github.