# The exact relationship between Clifford algebras and certain special isomorphic k-algebras

question(s):

Choose any real or complex clifford algebra $\mathcal{Cl}_{p,q}$. It's known that there is some $A \simeq \mathcal{Cl}_{p,q}$, where $A$ is either a matrix ring $M(n,R)$ or a direct sum of matrix rings $M(n,R)\oplus M(n,R)$, for $n \geq 1$ and $R \in \{\mathbb R, \mathbb C, \mathbb H\}$ such that $\dim(\mathcal{Cl}_{p,q}) = \dim(A)$ (as k-algebras).

1. Given an element $X\in \mathcal{Cl}_{p,q}$, defined on a standard basis, how can I find an explicit isomorphism $f:\mathcal{Cl}(p,q)\to A$, that preserves algebraic properties such as grade interaction? What is the "character" (qualitatively or otherwise) of such an isomorphism? Is it part of some group like the general linear or orthogonal groups?

2. Conversely, say $y^{i,j} \in R$ is an entry of some matrix $Y\in A$, and $y^{i,j}_k \in \mathbb R$ is a component of $y^{i,j}$. Let's call it a component of $A$ as well. What is the relationship of $y^{i,j}_k$ to $\mathcal{Cl}_{p,q}$? Does it have a single grade that can be determined by its grade in $R$? Or is there something more nuanced going on? How is it related to other components of $A$?

3. There are cases where a given $A$ is isomorphic to multiple Clifford algebras. Then these Clifford algebras must be isomorphic to one another. For example: $\mathcal{Cl}_{7,0} \simeq \mathcal{Cl}_{5,2} \simeq \mathcal{Cl}_{3,4} \simeq \mathcal{Cl}_{1,6} \simeq M(8,\mathbb C)$. What's going on here? Is there a mathematical term for these kinds of special isomorphisms between Clifford algebras in general?

BONUS QUESTION: how is this problem referred to in the academic math literature? I have scoured, with my amateur math education, the "matrix representations of Clifford algebras" stuff, and mostly found stuff about real matrix representations of real Clifford algebras with real matrix generators, etc, but that's not what I'm looking for. How to distinguish?

context:

I have been using sage and sympy (computer algebra systems) to compute symbolic monomial representations of products for various Clifford algebras. These are then used to generate GPU code for geometric algebra usage.

It works nice for Clifford algebras with up to 7 generating dimensions, and for low grade computations like planar rotation, I can get up to 8 or 9 generating dimensions. I've been successful in pumping out reasonably fast 8 dimensional planar rotation functions in glsl.

Recently, I was reading about Bott periodicity and the classification of Clifford algebras. Using the equations given on the wikipedia page "classification of Clifford algebras", I tried generating some of these isomorphic algebras.

For whatever reason, they are much, much faster to generate, likely due to the ubiquity of matrix multiplication. But I have absolutely no idea how, in general, I would construct a generic multivector in them. For example, generating a symbolic representation of $M(4,\mathbb H)$ is very quick in Sage. Presumably there is some isomorphism between this and the 64 dimensional $\mathcal{Cl}_{2,4}$. But how do I use it?

• that preserves algebraic properties such as grade interaction? to "preserve grade interaction" one would first have to have a grade defined on the matrix rings... what would that be? – rschwieb Apr 28 '17 at 1:24
• @rschwieb, that was my awkward way of distinguishing an isomorphism as algebras from an isomorphism as vector spaces. Really this whole question was about finding how to extract "multivector coefficients" from a matrix of a specific representation. I'm gonna take some time to clarify and summarize what I've learned when I get a chance. – micahscopes Apr 28 '17 at 17:45

I can't say I understand the context you gave behind your question, but I think most of your questions can be answered by a primer on Clifford algebras. One thing I want to point out is that I use the opposite sign convention from you.

When making $\mathbb{C}$, mathematicians adjoined a square root of negative one to $\mathbb{R}$. When making the quaternions $\mathbb{H}$, Hamilton adjoined a new square root of negative one, called $j$, to $\mathbb{C}$, and eventually determined that in order for the obvious choice of norm to be multiplicative we would need $k:=ij$ to be linearly independent of $1,i,j$ (thus jutting out in a fourth dimension) and for $i$ and $j$ to anticommute rather than commute (meaning $ij=-ji$).

Clifford algebras extend this idea. I call the Clifford algebra $C\ell(n)$ the free associative algebra generated by $n$ anticommuting square roots of negative one. If we call them $e_1,\cdots,e_n$ and let $v$ be any vector in their span, we easily compute $v^2=-\|v\|^2$ (where $\|\cdot\|^2$ comes from the standard basis here). This inspires a generalization: if $(V,q)$ is any quadratic space (so $q$ is a quadratic form on $V$, meaning $q(x)=b(x,x)$ for some symmetric bilinear form $b$), then $C\ell(V,q)$ is the tensor algebra on $V$ modulo (the two-sided ideal generated by) the relations $v\otimes v=-q(v)1$.

We usually only consider nondegenerate bilinear forms on a vector space. At the other extreme is the completely degenerate form which always equals $0$; this gives the algebra isomorphism $C\ell(V,0)\cong\Lambda V$, known better as the exterior algebra on $V$. Even if $q$ is not identically $0$, there is still always a canonical vector space isomorphism $\Lambda V\to C\ell(V,q)$, given by

$$v_1\wedge \cdots\wedge v_k\mapsto v_1\cdots v_k$$

whenever $v_1,\cdots,v_k$ are orthogonal (meaning $b(v,w)=0$ or $q(v+w)=q(v)+q(w)$).

While the tensor algebra $TV$ is $\mathbb{N}$-graded, $C\ell(V)$ is not. Indeed, $v^2=-q(v)$ seems to lie in both the ostensible $2$ and $0$ graded components; in general, $C\ell(V)$ is only $\mathbb{Z}/2\mathbb{Z}$-graded: the even component are those elements expressible as a sum of products of evenly many vectors, and similarly for the odd component. This makes it something called a superalgebra. While $\Lambda V$ is supercommutative, $C\ell(V,q)$ isn't in general since odd elements commute with themselves.

Quadratic spaces together with quadratic-form-preserving linear maps as the morphisms form a category $\mathsf{QVect}$. It has a monoidal operation on it: $(V_1,q)\oplus(V_2,q_2)=(V_1\oplus V_2,q_1\oplus q_2)$, with quadratic form defined by $(q_1\oplus q_2)(v_1,v_2)=q_1(v_1)+q_2(v_2)$ (so it contains $V_1$ and $V_2$ as orthogonal subspaces). The assignment $(V,q)\mapsto C\ell(V,q)$ is functorial from $\mathsf{QVect}$ to $\mathsf{SAlg}$, the category of superalgebras. The latter category has its own monoidal operation, the super tensor product with $(a_1\otimes b_1)(a_2\otimes b_2)=\pm (a_1a_2\otimes b_1b_2)$ with $(-)$ sign if both $b_1,a_2$ are odd and $(+)$ sign otherwise inside $A\widehat{\otimes}B$. Then the $C\ell$ functor is actually monoidal;

$$C\ell(V_1\oplus V_2,q_1\oplus q_2) \cong C\ell(V_1,q_1) ~\widehat{\otimes}~ C\ell(V_2,q_2).$$

Clifford algebras, like tensor products, satisfy a universal property. If $(V,q)$ is a quadratic vector space and $A$ is any algebra and we have a linear map $\phi:V\to A$ satisfying $\phi(v)^2=-q(v)1_A$, then it extends to an algebra homomorphism $C\ell(V,q)\to A$ (via $v_1\cdots v_k\mapsto \phi(v_1)\cdots\phi(v_k)$).

Sylvester's law of inertia classifies nondegenerate quadratic forms on real vector spaces according to an invariant called its signature $(p,q)$: there is a basis in which

$$q(x)=x_1^2+\cdots+x_p^2-x_{p+1}^2-\cdots-x_{p+q}^2 .$$

We call the corresponding clifford algebra $C\ell(p,q)$ (there are other naming conventions too).

$\bullet$ Vacuously, $C\ell(0,0)=\mathbb{R}$.

$\bullet$ We already know $C\ell(1,0)=\mathbb{C}$,

$\bullet$ We already know $C\ell(2,0)=\mathbb{H}$,

$$1\leftrightarrow 1, \quad i\leftrightarrow e_1, \quad j\leftrightarrow e_2, \quad ij\leftrightarrow e_1e_2.$$

$\bullet$ What about $C\ell(0,1)$? This is $\mathbb{R}[x]/(x^2-1)\cong\mathbb{R}[x]/(x-1)\oplus\mathbb{R}[x]/(x+1)$ by the Chinese Remainder Theorem. If $C\ell(0,1)$ is generated by $f$ satisfying $f^2=1$, then $(1+f)(1-f)=0$ and one checks $(1\pm f)^2=2(1\pm f)$, so $(1\pm f)/2$ are orthogonal idempotents in which case

$$a+bf = (a+b)\frac{1+f}{2}+(a-b)\frac{1-f}{2} \leftrightarrow (a+b,a-b)$$

establishing $C\ell(0,1)\cong\mathbb{R}^2$ (our notation for direct product of rings).

$\bullet$ How about $C\ell(1,1)$ generated by $e,f$ with $e^2=-1$, $f^2=+1$, $ef=-fe$? In this case,

$$1\leftrightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \phantom{f}e\leftrightarrow \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},$$

$$f\leftrightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad ef\leftrightarrow \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix},$$

establishing $C\ell(1,1)\cong\mathbb{R}(2)$ (common notation for $M_2(\mathbb{R})$ in the literature).

$\bullet$ And then $C\ell(0,2)$ generated by $f_1,f_2$ with $f_1^2=f_2^2=1$, $f_1f_2=-f_2f_1$. In this case,

$$1\leftrightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad f_1\leftrightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},$$

$$f_2 \leftrightarrow \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \quad f_1f_2\leftrightarrow \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},$$

establishing $C\ell(0,2)\cong\mathbb{R}(2)$ as well.

So far, this is our table of Clifford algebras:

$$\begin{array}{|c|c|c|c|c|} \hline (p,q) & ~~~0~~~ & 1 & 2 & ~~~3~~~ \\ \hline 0 & \mathbb{R} & \mathbb{R}^2 & \mathbb{R}(2) & \\ \hline 1 & \mathbb{C} & \mathbb{R}(2) & & \\ \hline 2 & \mathbb{H} & & & \\ \hline 3 \\ \hline \end{array}$$

At this point, we should look at (normal, nonsuper) tensor products of $\mathbb{R},\mathbb{C},\mathbb{H}$. The obvious cases are $\mathbb{R}\otimes\mathbb{K}\cong\mathbb{K}$ for each $\mathbb{K}$. The first nontrivial one is

$$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}=\mathbb{C}\otimes_{\mathbb{R}}\frac{\mathbb{R}[x]}{(x^2+1)} \cong \frac{\mathbb{C}[x]}{(x^2+1)} \cong \frac{\mathbb{C}[x]}{(x-i)}\oplus \frac{\mathbb{C}[x]}{(x+i)} \cong \mathbb{C}\oplus\mathbb{C}$$

by the Chinese Remainder Theorem. Indeed,

$$1\otimes 1\leftrightarrow (1,1), \quad i\otimes i\leftrightarrow (-1,1) \\ i\otimes 1\leftrightarrow (i,i), \quad 1\otimes i\leftrightarrow (i,-i)$$

establishes $\mathbb{C}\otimes\mathbb{C}\cong\mathbb{C}^2$ as algebras.

Now, $\mathbb{H}$ is a module over itself from both the left and the right, and these actions commute (this is the associative property, $a(xb)=(ax)b$), which induces a map $\mathbb{H}\otimes\mathbb{H}\to \mathbb{R}(4)$; I'll let you compute what this does to a basis (and thus by dimensions estasblihes an algebra isomorphism); I do a similar trick, regarding $\mathbb{H}$ as a right vector space over $\mathbb{H}$, here in order to establish the algebra isomorphism $\mathbb{H}\otimes\mathbb{C}\cong\mathbb{C}(2)$. Thus, we have this full table:

$$\begin{array}{c||c|c|c} \otimes & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \hline \mathbb{R} & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \hline \mathbb{C} & \mathbb{C} & \mathbb{C}^2 & \mathbb{C}(2) \\ \hline \mathbb{H} & \mathbb{H} & \mathbb{C}(2) & \mathbb{R}(4) \end{array}$$

Moreover, we have $\mathbb{R}(n)\otimes\mathbb{K}\cong\mathbb{K}(n)$ for any algebra $\mathbb{K}$ (including $\mathbb{K}=\mathbb{R}^2$ and $\mathbb{H}^2$, for example, in which case e.g. $\mathbb{C}^2(n)\cong\mathbb{C}(n)^2$). I will assume you can figure out what these isomorphisms are.

For a general Clifford algebra $C\ell(p,q)$ generated by $e_1,\cdots,e_p$ (square roots of $-1$) and $f_1,\cdots,f_q$ (square roots of $+1$), all pairwise anticommuting, we may call $\omega=e_1\cdots e_p f_1\cdots f_q$ the orientation element (which element of the algebra this represents is actually independent of choice of basis for the vector space; it only depends on the orientation the basis induces on the space). Check

$$\omega^2=\begin{cases} +1 & p-q\equiv 0,3 \mod4 \\ -1 & p-q\equiv 1,2 \mod 4 \end{cases}$$

Suppose we have quadratic spaces $(V_1,q_1),(V_2,q_2)$ and the orientation element $\omega$ of $C\ell(V_1,q_1)$ anticommutes with all $v_1\in V_1$ (so $\dim V_1$ is even). Consider the map

$$\phi:V_1\oplus V_2\to C\ell(V_1,q_1)\otimes C\ell(V_2,q_2)$$

(usual tensor product) given by extending

$$v_1 \mapsto v_1\otimes 1 \\ v_2\mapsto \omega\otimes v_2$$

Check this satisfies

$$\begin{array}{ll} \phi(v_1,v_2)^2 & =(v_1\otimes 1+\omega\otimes v_2)^2 \\ & = v_1^2\otimes 1+v_1\omega\otimes v_2+\omega v_1\otimes v_2+\omega^2\otimes v_2^2 \\ & = (q_1+\omega^2q_2)(v_1,v_2).\end{array}$$

The universal property gives an algebra isomorphism $$C\ell(V_1\oplus V_2,q_1\oplus\omega^2q_2)\to C\ell(V_1,q_1)\otimes C\ell(V_2,q_2).$$

If we pick $(V_1,q_1)$ to have one of the signatures $(2,0),(1,1),(0,2)$ we get

$$\begin{array}{l} C\ell(2,0) \otimes C\ell(r,s) \cong C\ell(s+2,r) \\ C\ell(1,1)\otimes C\ell(r,s)\cong C\ell(r+1,s+1) \\ C\ell(0,2)\otimes C\ell(r,s)\cong C\ell(s,r+2). \end{array}$$

(This is why it was important we worked out $C\ell(p,q)$ for $p+q\le 2$.) You should be able to use this to fill out the table of Clifford algebras for $0\le p,q\le 8$. If you do, you will notice some patterns.

For example, if you ignore the matrices part and just focus on scalars, they are $8$-periodic, proceeding $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{H}^2,\mathbb{H},\mathbb{C},\mathbb{R},\mathbb{R}^2$. (The "ignoring the matrices part" can be formalized using Morita equivalence.) John Baez calls this the "Clifford clock." Notice the axis of symmetry of this clock is offset by $1$; this is related to the fact the even subalgebra of $C\ell(p,q)$ is isomorphic to $C\ell(p-1,q)$, and spin representations thus use Clifford algebra representations from one less space dimension.

Anyway, that's a long enough answer for the classification I think.

• question 1: your answer goes a long way in helping me understand . Strangely enough, the most helpful part was the high level part about super algebras. Really, if I can define the product on two bases of lower dimensional dimensional algebras, I should be able to keep track pretty easily of how those bases relate to the implied basis of their super algebra, regardless of the representation. Conceptually, that's what's going on. – micahscopes Apr 27 '17 at 3:46
• question 2: now I see that this one's more about basic representation theory, which I've avoided wholeheartedly (e.g. the bonus question). But now I have no choice but to look into it because I've gotten intrigued. Intrigued, why?? Because I've realized that $1 \in \mathcal{Cl}$ must map to the identity matrix. There's no other matrix like it. You alluded, I believe, to something similar w.r.t the pseudoscalar. It might seem obvious, but it's an epiphany to me: there's no way of representing these elements as single entry matrices w/o abandoning the algebraic part of the isomorphism. – micahscopes Apr 27 '17 at 4:14
• If one "forgets" also the complex numbers and quternions there is an algorithm to represent an arbitrary element of $Cl(p,q)$ as a matrix over R. – user48672 May 22 '17 at 13:56