Clifford Algebras next up previous
Next: Spinors and Trialities Up: Constructing the Octonions Previous: The Cayley-Dickson Construction


2.3 Clifford Algebras

William Clifford invented his algebras in 1876 as an attempt to generalize the quaternions to higher dimensions, and he published a paper about them two years later [20]. Given a real inner product space $V$, the Clifford algebra $\Cliff (V)$ is the associative algebra freely generated by $V$ modulo the relations

\begin{displaymath}
% latex2html id marker 1545
v^2 = -\Vert v\Vert^2 \end{displaymath}

for all $v \in V$. Equivalently, it is the associative algebra freely generated by $V$ modulo the relations

\begin{displaymath}vw + wv = -2\langle v,w\rangle \end{displaymath}

for all $v, w \in V$. If $V = \R^n$ with its usual inner product, we call this Clifford algebra $\Cliff (n)$. Concretely, this is the associative algebra freely generated by $n$ anticommuting square roots of $-1$. From this we easily see that

\begin{displaymath}\Cliff (0) = \R, \qquad\qquad \Cliff (1) = \C, \qquad\qquad
\Cliff (2) = \H .\end{displaymath}

So far this sequence resembles the iterated Cayley-Dickson construction -- but the octonions are not a Clifford algebra, since they are nonassociative. Nonetheless, there is a profound relation between Clifford algebras and normed division algebras. This relationship gives a nice way to prove that $\R,\C,\H$ and $\O$ are the only normed dvivision algebras. It is also crucial for understanding the geometrical meaning of the octonions.

To see this relation, first suppose $\K$ is a normed division algebra. Left multiplication by any element $a \in \K$ gives an operator

\begin{displaymath}
% latex2html id marker 1548
\begin{array}{ccccc}
L_a &\maps & \K & \to & \K \\  & & x &\mapsto& ax .
\end{array} \end{displaymath}

If % latex2html id marker 2643
$\Vert a\Vert = 1$, the operator $L_a$ is norm-preserving, so it maps the unit sphere of $\K$ to itself. Since $\K$ is a division algebra, we can find an operator of this form mapping any point on the unit sphere to any other point. The only way the unit sphere in $\K$ can have this much symmetry is if the norm on $\K$ comes from an inner product. Even better, this inner product is unique, since we can use the polarization identity

\begin{displaymath}
% latex2html id marker 1549
\langle x, y\rangle = {1\over 2}(\Vert x+y\Vert^2 - \Vert x\Vert^2 - \Vert y\Vert^2) \end{displaymath}

to recover it from the norm.

Using this inner product, we say an element $a \in \K$ is imaginary if it is orthogonal to the element $1$, and we let $\Im (\K)$ be the space of imaginary elements of $\K$. We can also think of $\Im (\K)$ as the tangent space of the unit sphere in $\K$ at the point $1$. This has a nice consequence: since $a \mapsto L_a$ maps the unit sphere in $\K$ to the Lie group of orthogonal transformations of $\K$, it must send $\Im (\K)$ to the Lie algebra of skew-adjoint transformations of $\K$. In short, $L_a$ is skew-adjoint whenever $a$ is imaginary.

The relation to Clifford algebras shows up when we compute the square of $L_a$ for $a \in \Im (\K)$. We can do this most easily when $a$ has norm $1$. Then $L_a$ is both orthogonal and skew-adjoint. For any orthogonal transformation, we can find some orthonormal basis in which its matrix is block diagonal, built from $2 \times 2$ blocks that look like this:

\begin{displaymath}
% latex2html id marker 1550
\left( \begin{array}{cc} \cos \...
...in \theta \\  -\sin \theta & \cos \theta
\end{array} \right) \end{displaymath}

and possibly a $1 \times 1$ block like this: $\left( 1 \right)$. Such a transformation can only be skew-adjoint if it consists solely of $2 \times 2$ blocks of this form:

\begin{displaymath}
% latex2html id marker 1551
\pm \left( \begin{array}{cc} 0 & 1 \\  -1 & 0
\end{array} \right). \end{displaymath}

In this case, its square is $-1$. We thus have $L_a^2 = -1$ when $a \in \Im (\K)$ has norm 1. It follows that

\begin{displaymath}
% latex2html id marker 1552
L_a^2 = -\Vert a \Vert^2 \end{displaymath}

for all $a \in \Im (\K)$. We thus obtain a representation of the Clifford algebra $\Cliff (\Im (\K))$ on $\K$. Any $n$-dimensional normed division algebra thus gives an $n$-dimensional representation of $\Cliff (n-1)$. As we shall see, this is very constraining.

We have already described the Clifford algebras up to $\Cliff (2)$. Further calculations [50,73] give the following table, where we use $A[n]$ to stand for $n\times n$ matrices with entries in the algebra $A$:


$n$ $\Cliff (n)$
$0$ $\R$
$1$ $\C$
$2$ $\H$
$3$ $\H \oplus \H$
$4$ $\H[2]$
$5$ $\C[4]$
$6$ $\R[8]$
$7$ $\R[8] \oplus \R[8]$

Table 2 -- Clifford Algebras


Starting at dimension 8, something marvelous happens: the table continues in the following fashion:

\begin{displaymath}\Cliff (n+8) \iso \Cliff (n) \tensor \R[16] .\end{displaymath}

In other words, $\Cliff (n+8)$ consists of $16 \times 16$ matrices with entries in $\Cliff (n)$. This `period-8' behavior was discovered by Cartan in 1908 [13], but we will take the liberty of calling it Bott periodicity, since it has a far-ranging set of applications to topology, some of which were discovered by Bott.

Since Clifford algebras are built from matrix algebras over $\R,\C$ and $\H$, it is easy to determine their representations. Every representation is a direct sum of irreducible ones, or irreps. The only irrep of $\R[n]$ is its obvious one via matrix multiplication on $\R^n$. Similarly, the only irrep of $\C[n]$ is the obvious representation on $\C^n$, and the only irrep of $\H[n]$ is the obvious one on $\H^n$.

Glancing at the above table, we see that unless $n$ equals $3$ or $7$ modulo $8$, $\Cliff (n)$ is a real, complex or quaternionic matrix algebra, so it has a unique irrep. For reasons to be explained later, this irrep is known as the space of pinors and denoted $P_n$. When $n$ is $3$ or $7$ modulo $8$, the algebra $\Cliff (n)$ is a direct sum of two real or quaternionic matrix algebras, so it has two irreps, which we call the positive pinors $P_n^+$ and negative pinors $P_n^-$. We summarize these results in the following table:


$n$ $\Cliff (n)$ irreps of $\Cliff (n)$
$0$ $\R$ $P_0 = \R$
$1$ $\C$ $P_1 = \C$
$2$ $\H$ $P_2 = \H$
$3$ $\H \oplus \H$ % latex2html id marker 2839
$P^+_3 = \H,\, P^-_3 =\H$
$4$ $\H[2]$ $P_4 = \H^2$
$5$ $\C[4]$ $P_5 = \C^4$
$6$ $\R[8]$ $P_6 =\R^8$
$7$ $\R[8] \oplus \R[8]$ % latex2html id marker 2863
$P_7^+ = \R^8,\, P_7^- =\R^8$

Table 3 -- Pinor Representations


Examining this table, we see that in the range of dimensions listed there is an $n$-dimensional representation of $\Cliff (n-1)$ only for $n
= 1,2,4,$ and $8$. What about higher dimensions? By Bott periodicity, the irreducible representations of $\Cliff (n+8)$ are obtained by tensoring those of $\Cliff (n)$ by $\R^{16}$. This multiplies the dimension by 16, so one can easily check that for $n > 8$, the irreducible representations of $\Cliff (n-1)$ always have dimension greater than $n$.

It follows that normed division algebras are only possible in dimensions $1, 2, 4,$ and $8$. Having constructed $\R,\C,\H$ and $\O$, we also know that normed division algebras exist in these dimensions. The only remaining question is whether they are unique. For this it helps to investigate more deeply the relation between normed division algebras and the Cayley-Dickson construction. In what follows, we outline an approach based on ideas in the book by Springer and Veldkamp [83].

First, suppose $\K$ is a normed division algebra. Then there is a unique linear operator $\ast \maps \K \to \K$ such that $1^\ast = 1$ and $a^\ast
= -a$ for $a \in \Im (\K)$. With some calculation one can prove this makes $\K$ into a nicely normed $\ast$-algebra.

Next, suppose that $\K_0$ is any subalgebra of the normed division algebra $\K$. It is easy to check that $\K_0$ is a nicely normed $\ast$-algebra in its own right. If $\K_0$ is not all of $\K$, we can find an element $i \in
\K$ that is orthogonal to every element of $\K_0$. Without loss of generality we shall assume this element has norm 1. Since this element $i$ is orthogonal to $1 \in \K_0$, it is imaginary. From the definition of the $\ast$ operator it follows that $i^\ast = -i$, and from results earlier in this section we have $i^2 = -1$. With further calculation one can show that for all $a,a' \in \K_0$ we have

\begin{displaymath}a(ia') = i(a^* a') , \qquad
(ai)a' = (aa'^*)i, \qquad
(ia)(a'i^{-1}) = (aa')^* \end{displaymath}

A glance at equation (1) reveals that these are exactly the relations defining the Cayley-Dickson construction! With a little thought, it follows that the subalgebra of $\K$ generated by $\K_0$ and $i$ is isomorphic as a $\ast$-algebra to $\K'_0$, the $\ast$-algebra obtained from $\K_0$ by the Cayley-Dickson construction.

Thus, whenever we have a normed division algebra $\K$ we can find a chain of subalgebras $\R = \K_0 \subset \K_1 \subset \cdots \subset
\K_n = \K$ such that $\K_{i+1} \iso \K_i'$. To construct $\K_{i+1}$, we simply need to choose a norm-one element of $\K$ that is orthogonal to every element of $\K_i$. It follows that the only normed division algebras of dimension 1, 2, 4 and 8 are $\R,\C,\H$ and $\O$. This also gives an alternate proof that there are no normed division algebras of other dimensions: if there were any, there would have to be a 16-dimensional one, namely $\O'$ -- the sedenions. But as mentioned in Section 2.2, one can check explicitly that the sedenions are not a division algebra.


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Next: Spinors and Trialities Up: Constructing the Octonions Previous: The Cayley-Dickson Construction

© 2001 John Baez

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