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

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.

Here is the simplest way I know of getting matrix representations of even or full Clifford algebras of vector spaces over $\def\|#1{\mathbb#1}\|R$ or $\|C$ with arbitrary nondegenerate signature. There are just two recursive rules and a trivial base case. It may also work over other fields but some details would have to be changed.

Details:

The full algebra in $0$ dimensions and the even algebra in ${\le}1$ dimension are isomorphic to the base field.
If there is a unit pseudoscalar $ω$ that is not a scalar, then any element can be uniquely written as $A+Bω$ where $A$ and $B$ belong to the subalgebra of a subspace of codimension one. If additionally $ω$ commutes with all elements of the subalgebra, then the algebra is isomorphic to $\|C$ (if $ω^2=-1$) or $\|D$ (if $ω^2=1$) tensored with the subalgebra. This works in dimension at least 1, for full algebras in odd dimensions and even algebras in even dimensions. It doesn't work for full algebras in even dimensions because the pseudoscalar anticommutes with odd elements, and it doesn't work for even algebras in odd dimensions because there is no pseudoscalar.
Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for the vector space, and define (if possible) $i=e_1e_2,\;j=e_2\cdots e_n,\;k=ij$. Note that regardless of signature, $i^2j^2k^2=-i^4j^4=-1$. If $i$, $j$, $k$ commute with all elements of the subalgebra of the subspace spanned by $\{e_3, \ldots, e_n\}$, then the algebra is isomorphic to $\|H$ (if $i^2,j^2,k^2$ are all negative) or $\|P$ (if two of them are positive) tensored with the subalgebra. (You may have to permute $i,j,k$ to get a standard basis where $j^2=k^2=ijk$.) This works in dimension at least 2, for full algebras in even dimensions and for even algebras in odd dimensions. It doesn't work for full algebras in odd dimensions because $j$ and $k$ anticommute with odd elements, and it doesn't work for even algebras in even dimensions because $j$ and $k$ don't exist.

Note that precisely one of these rules applies in any given situation. However, you have a choice of subspaces when applying the recursive rules, so there are many possible factorizations.

As for the isomorphisms:

$\|D\cong\|R\oplus\|R{:}\;1\leftrightarrow(1,1),\;i\leftrightarrow(1,-1)$
$\|P\cong M_2(\|R){:}\; 1,i,j,k \leftrightarrow \def\M#1{\bigl(\begin{smallmatrix}#1\end{smallmatrix}\bigr)} \M{1&0\\0&1}, \M{0&1\\-1&0}, \M{0&1\\1&0}, \M{1&0\\0&-1}$
$\|C\otimes\|C\cong\|C\otimes\|D{:}\; i' \leftrightarrow ii'$
$\|C\otimes\|H\cong \|C\otimes\|P{:}\; j', k' \leftrightarrow ij', ik'$
$\|H\otimes\|H\cong \|P\otimes\|P{:}\; j, k, j', k' \leftrightarrow ji', ki', ij', ik'$

Here's a worked example: the even algebra of $\|R^{1,3}$.

The pseudoscalar squares to $-1$ so we factor out $\|C$, leaving us with either $\|R^{0,3}$ or $\|R^{1,2}$. I'll pick the latter.

Let $\{\hat t,\hat x,\hat y\}$ be an orthonormal basis for this subspace with $\hat t^2=1$. Let $i=\hat x\hat y,\;j=\hat y\hat t,\;k=\hat t\hat x$. We factor out $\|P$, leaving us with $\|R^{0,1}$ or $\|R^{1,0}$.

The even Clifford algebra of $\|R^{0,1}$ or $\|R^{1,0}$ is just $\|R$, so our original algebra was isomorphic to $\|C\otimes\|P \cong M_2(\|C)$.

Explicitly, we can take $\hat t\hat x\hat y\hat z = \M{i&0\\0&i},\ \hat x\hat y = \M{0&1\\-1&0},\ \hat y\hat t = \M{0&1\\1&0}$, and this generates the rest of the algebra.

A weakness of this method is that it doesn't give you a chiral basis for the full algebra in even dimensions. You can work around that by using a third recursive rule:

Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for the vector space, and define (if possible) $i=e_1,\;j=e_2\cdots e_n,\;k=ij$. If $i^2j^2k^2=-1$, then the algebra is isomorphic to $\|H$ or $\|P$ tensored with the even subalgebra of the subspace spanned by $\{e_2, \ldots, e_n\}$. This works in nonzero even dimension for full algebras only. It doesn't work in odd dimensions because $i^2j^2k^2=1$.

$\mathcal{Cl}_{7,0} \simeq \mathcal{Cl}_{5,2} \simeq \mathcal{Cl}_{3,4} \simeq \mathcal{Cl}_{1,6} \simeq M(8,\mathbb C)$. What's going on here?

$M(8,\mathbb C)$ only makes sense as a Clifford algebra if you also supply an embedding of the underlying vector space, and these embeddings will necessarily be different for nonisomorphic vector spaces, so I wouldn't say that these algebras are isomorphic in any particularly interesting sense.

For what it's worth, the factorization in this answer can get you explicit isomorphisms fairly easily. You will get for each of these algebras a factor of $\|C$ and three factors of $\|H$ or $\|P$, and after converting the $\|H$s to $\|P$s (or vice versa), the map taking generators to their counterparts in another algebra extends to an isomorphism of the algebras.

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

Tags:

Computational Geometry

Homotopy Theory

Clifford Algebras

Representation Theory

Semi Simple Rings

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