### Clifford Algebras as Graded Modules

Now a little post about algebra. Recall that when we have a family of ${A}$-modules ${\{M_i\}_{i\in I}}$, with ${A}$ commutative ring with unit, the direct sum,

is the ${A}$-module with elements of the form ${\alpha=(\alpha_i)_{i\in I}}$, where ${\alpha_i\in M_i}$ and ${\alpha_i=0}$ cofinitely. We also have ${(\alpha + \beta)_i=\alpha_i + \beta_i}$, for ${r\in A}$,  ${(r\alpha)_i=r\alpha_i}$, and sometimes we write by ${\alpha}$, ${\sum \alpha_i}$.

Now the definition of graded module. Take a commutative ring ${A}$ and a ${A}$-module ${M}$. We say that ${M}$ is a graded ${A}$-module when there exists for each ${n\in \mathbb{Z}}$, a ${A}$-submodule ${M_n}$ of ${M}$ such that,

The elements of ${M_n}$ are called homogeneous of degree ${n}$, and we use the notation ${|x|=n}$. Let’s take a look to some examples of this constructions.

Square matrices: Denote ${M(n)}$ the set of ${n\times n}$ matrices over ${\mathbb{R}}$ wich is an ${\mathbb{R}}$-module. For each pair ${(i,j)}$ with ${1\leq i,j \leq n}$, ${M(i,j)}$ is the ${\mathbb{R}}$-submodule of ${M}$ consisting of ${n\times n}$ matrices with all entries equal zero but maybe the entry (i,j). And we have,

And ${M(n)}$ is a graded module

if for example we take ${M_m=M(i,j)}$ for ${0\leq m\leq n^2}$ in some order and ${M_m=0}$ for the others ${m\in \mathbb{Z}}$.

Rings: Every commutative ring ${A}$ with unit is trivially seen as an ${A}$-module graded

when we take ${A_n=0}$ if ${n\neq 0}$ and ${A_0=A}$

Numerable submodules: At this point its clear that when a module ${M}$ is the direct sum of submodules indexed by a countable set, ${M}$ can be seen as a graded module.

The real plane: ${\mathbb{R}^2}$ is a ${R}$-module an any straight line thought the origin is a ${\mathbb{R}}$-submodule of the plane. If we take two different lines ${l_0}$ and ${l_1}$ then

where ${l_n=0\in \mathbb{R}^2}$ for every ${n\neq 1,0}$. In the picture,

the vectors in the line ${l_0}$ are homogeneous of degree 0 and in the line ${l_1}$ the vectors has degree ${1}$. The origin ${0}$ is homogeneous of all degrees. And the vector like ${\vec{v}}$ are not homogeneous.

Special matrices: In the real plane the unit circle can be identified with the set of matrices ${S_2}$, of the form,

Where the parameter ${\varphi}$ correspond to the angle counterclockwise from the ${x}$ axe. ${S_2}$ is an ${\mathbb{R}}$-module when we take ${rB(\varphi)=B(r\varphi)}$ and ${B(\varphi_1)+B(\varphi_2)=B(\varphi_1+\varphi_2)}$.

we can do exactly the same with the circles with radius ${r}$ with values in ${\mathbb{R}_{\geq 0}}$, use the notation ${S_2(r)}$.

Now, if we take the direct sum over ${\mathbb{R}_{\geq 0}}$, the elements of the ${\mathbb{R}}$-module,

could be represented graphically by a horizontal ray from the origin, wich has finite many points dispersed along the circles centered in the origin.

Question: As a ${\mathbb{R}}$-module, ${\underset{r\in \mathbb{R}_{\geq 0}}{\bigoplus} S_2(r)}$ is isomorphic to?

Clifford algebras: With this example we going to explain the concept of Clifford algebra from, as close as I can, basic notions. But, be careful, this is just a sketch.

Take ${A}$ commutative ring and ${E}$ an ${A}$-module. An application ${Q:E\rightarrow A}$ is a quadratic form over ${E}$ if for every ${\alpha \in A}$, ${x\in E }$ ${Q(\alpha x)= \alpha^2 Q(x)}$, and the application ${\theta: E\times E\rightarrow A}$ defined by ${(x,y)\mapsto Q(x+y)-Q(x)-Q(y)}$ is a bilinear form.

Let ${A}$ a commutative ring with unit. A set ${E}$ is called an algebra over ${A}$(or ${A}$-algebra) if ${E}$ is equipped with a structure of ${A}$-module and a bilinear application ${\theta_E:E\times E \rightarrow E}$.

In this case if ${E}$ is a graded ${A}$-module of type ${\mathbb{N}}$ (the decomposition of ${E}$ in direct sum of submodules of ${E}$ is indexed by ${\mathbb{N}}$), we say that ${E}$ is a graded algebra of type ${\mathbb{N}}$ if the algebra’s product satisfies for every pair of homogeneous elements x, y of ${E}$, of degree ${p}$ and ${q}$ resp. the product ${xy}$ is homogeneous of degree ${p+q}$.

bilateral ideal ${I}$ of the algebra over ${A}$, ${E}$, is a ${A}$-submodule of ${E}$ such that ${a\in I}$ and ${x\in E}$ implies ${ax\in I}$ and ${xa\in I}$. With ${I}$ we have an equivalence relation over ${E}$ defined by ${x\equiv x'}$ if only if ${x-x'\in I}$.

By the bilateral property of ${I}$ as ideal, if ${x\equiv x'}$ and ${y\equiv y'}$ then ${xy \equiv x'y'}$. And this defines a bilinear application ${(E/I)\times (E/I)\rightarrow (E/I)}$.

So, we can see the quotient ${A}$-module ${E/I}$ as an ${A}$-algebra, this algebra is called the quotient algebra of ${E}$ by the bilateral ideal ${I}$.

The tensor algebra of an ${A}$-module is the ${A}$-module

where ${T^n(M)}$ is the tensor product of ${M}$ by himself ${n}$ times, with ${T^0(M)=A}$, ${T^1(M)=M}$, ${T^2(M)=M\otimes M}$, etc.

The adjective algebra of ${Tens(M)}$ is because it is a graded ${A}$-module of type ${\mathbb{N}}$ and it has a binary product that make him a graded algebra of type ${\mathbb{N}}$. For the detail of this operations i refer the reader to the chapter III, page 55, of the book Algèbre by the imaginary mathematician Bourbaki, this reference is to the french version of the book, but there is an english version. I don’t give more details because it is the notion that I want to explain for understand this little example.

Now, the definition of Clifford algebra.

Let ${Q}$ a quadratic form over the ${A}$-module ${E}$, recall ${Q:E\rightarrow A}$. This define a bilinear ideal in ${E}$, noted ${I(Q)}$, wich elements has the form ${x\otimes x- Q(x)\cdot 1}$, with ${x\in E}$. Take the tensor algebra ${E}$, ${Tens(E)}$ and the bilinear ideal ${I(Q)}$ to form the quotient algebra of ${Tens(E)}$ by ${I(Q)}$. This quotient algebra is noted ${C(Q)}$ and is called the Clifford algebra of ${Q}$.

Finally the example: in the book Algebra of Bourbaki, chapter IX, page 136-137, french version, you can see this definition of Clifford algebra and the definition of two submodules of ${C(Q)}$, noted ${C^+(Q)}$ and ${C^-(Q)}$. This two submodules form over ${C(Q)}$ a graduation of type ${\mathbb{Z}/2\mathbb{Z}}$.

If you don’t have any Bourbaki’s books, maybe google can help you, if you give him the right words.

References

• [1] Alain Prouté, Introduction à la logique catégorique, Université Paris VII, CNRS, 2010,
http://people.math.jussieu.fr/~alp/cours_2010.pdf
• [2] H. Blaine Lawson, JR. and Marie-Louise Michelsohn,  Spin Geometry, Princeton University Press, 1989.
• [3] Joseph J. Rotman, An introduction to the Theory of Groups, Springer, 1994.
• [4] N. Bourbaki, Algèbre, Chapitres I-III, Springer, 2007.
• [5] N. Bourbaki, Algèbre, Chapitre IX, Springer, 2007.
• [6] Saunders Mac Lane, Categories for the Working Mathematician, Springer, 1971.

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