Confusion about division in Clifford Algebra

Question

On page 202 of The Road to Reality, Penrose claims that if we want to generalize Quaternions to n dimensions using Clifford Algebra, we must abandon the division property. I have a hard time believing this claim. Would someone who knows about this mind following my reasoning and telling me if I'm wrong?

I recently spent quite a bit of time studying Clifford Algebra, and at it's core is the 'geometric product'. The fact that this product is associative in the algebra is an axiom. The product looks like

\begin{equation} uv=u \cdot v +u \wedge v \end{equation}

Hence, \begin{equation} u^2=u \cdot u +u \wedge u = u \cdot u \end{equation}

Now consider

\begin{equation} w=uv \end{equation}

I can divide w by v to get back u as follows (if |v|=/=0):

\begin{equation} \frac{w}{v}=\frac{wv}{v^2 }=\frac{(uv)v}{v^2 } \end{equation}

Now, since our geometric product is associative,

\begin{equation} \frac{w}{v}=\frac{(uv)v}{v^2 }= \frac{u(vv)}{v^2}=u \end{equation} Voila!

Nowhere in that argument did I need to restrict the dimensionality of my clifford algebra, I needed only the most basic properties of the geometric product, it's associativity and decomposition in the scalar and wedge products.

Can anyone weigh in on why people like Penrose seem to think that one can't divide in n dimensional Clifford algebra?

Answer 1

Not every element of a Clifford algebra has the form $v$ for $v$ a vector. In general an element of a Clifford algebra is a sum of products of such things, and most of these aren't invertible.

Explicitly, let $\text{Cliff}(p, q)$ be the Clifford algebra over $\mathbb{R}$ generated by $p$ generators $e_1, \dots e_p$ satisfying $e_i^2 = -1$ and $q$ generators $f_1, \dots f_q$ satisfying $f_i^2 = 1$, all of which anticommute. If $q \neq 0$ then $f_i^2 = 1$ implies $(f_i + 1)(f_i - 1) = 0$, so $f_i \pm 1$ fail to be invertible.

So now let $q = 0$; the corresponding Clifford algebra is often denoted $\text{Cliff}(p)$. When $p = 1$ we get $\mathbb{C}$ and when $p = 2$ we get $\mathbb{H}$; these are division algebras. It turns out that for $p = 3$ we have

$$\text{Cliff}(3) \cong \mathbb{H} \times \mathbb{H}$$

which is not a division algebra. Explicitly, once you have three anticommuting square roots of $-1$, namely $e_1, e_2, e_3$, you can now write down

$$(e_1 e_2 e_3)^2 = 1$$

and so $e_1 e_2 e_3 \pm 1$ fails to be invertible.

Answer 2

You can check for yourself that in $G^3$ that is the Pauli algebra with basis $ e_1, e_2, e_3$ $ (1+ e_1) (1 - e_1) =0 $ . The claim is that there are zero divisors, not it is impossible in principle to divide two elements.