2.2 The Cayley-Dickson Construction

It would be nice to have a construction of the normed division algebras $\R,\C,\H,\O$ that explained why each one fits neatly inside the next. It would be nice if this construction made it clear why $\H$ is noncommutative and $\O$ is nonassociative. It would be even better if this construction gave an infinite sequence of algebras, doubling in dimension each time, with the normed division algebras as the first four. In fact, there is such a construction: it's called the Cayley-Dickson construction.

As Hamilton noted, the complex number $a+bi$ can be thought of as a pair $(a,b)$ of real numbers. Addition is done component-wise, and multiplication goes like this:

$$(a,b)(c,d) = (ac - db,ad + cb)$$

We can also define the conjugate of a complex number by

$$(a,b)^* = (a,-b).$$

Now that we have the complex numbers, we can define the quaternions in a similar way. A quaternion can be thought of as a pair of complex numbers. Addition is done component-wise, and multiplication goes like this:

$$(a,b)(c,d) = (ac - db^*, a^* d + cb)$$

This is just like our formula for multiplication of complex numbers, but with a couple of conjugates thrown in. If we included them in the previous formula nothing would change, since the conjugate of a real number is just itself. We can also define the conjugate of a quaternion by

$$(a,b)^* = (a^*,-b).$$

The game continues! Now we can define an octonion to be a pair of quaternions. We add and multiply them using the same formulas that worked for the quaternions. This trick for getting new algebras from old is called the Cayley-Dickson construction.

Why do the real numbers, complex numbers, quaternions and octonions have multiplicative inverses? I take it as obvious for the real numbers. For the complex numbers, one can check that

$$(a,b) (a,b)^* = (a,b)^* (a,b) = k (1,0)$$

where $k$ is a real number, the square of the norm of $(a,b)$. This means that whenever $(a,b)$ is nonzero, its multiplicative inverse is $(a,b)^*/k$. One can check that the same holds for the quaternions and octonions.

But this, of course, raises the question: why isn't there an infinite sequence of division algebras, each one obtained from the preceding one by the Cayley-Dickson construction? The answer is that each time we apply the construction, our algebra gets a bit worse. First we lose the fact that every element is its own conjugate, then we lose commutativity, then we lose associativity, and finally we lose the division algebra property.

To see this clearly, it helps to be a bit more formal. Define a $\ast$-algebra to be an algebra $A$ equipped with a conjugation, that is, a real-linear map $\ast \maps A \to A$ with

$$a^{**} = a, \quad \quad (ab)^* = b^* a^*$$

for all $a,b \in A$. We say a $\ast$-algebra is real if $a = a^*$ for every element $a$ of the algebra. We say the $\ast$-algebra $A$ is nicely normed if $a + a^* \in \R$ and $aa^* = a^* a > 0$ for all nonzero $a \in A$. If $A$ is nicely normed we set

$$\Re (a) = (a + a^\ast)/2 \in \R, \qquad \Im (a) = (a - a^\ast)/2 ,$$

and define a norm on $A$ by

$$% latex2html id marker 1540 \Vert a\Vert^2 = aa^\ast .$$

If $A$ is nicely normed, it has multiplicative inverses given by

$$% latex2html id marker 1541 a^{-1} = a^\ast / \Vert a\Vert^2 .$$

If $A$ is nicely normed and alternative, $A$ is a normed division algebra. To see this, note that for any $a,b \in A$, all 4 elements $a,b,a^\ast,b^\ast$ lie in the associative algebra generated by $\Im (a)$ and $\Im (b)$, so that

$$% latex2html id marker 1542 \Vert ab\Vert^2 = (ab)(ab)^\ast... ... a^\ast) = a(bb^\ast)a^\ast = \Vert a\Vert^2 \Vert b\Vert^2 .$$

Starting from any $\ast$-algebra $A$, the Cayley-Dickson construction gives a new $\ast$-algebra $A'$. Elements of $A'$ are pairs $(a,b) \in A^2$, multiplcation is defined by

$$(a,b)(c,d) = (ac - db^*, a^* d + cb)$$

and conjugation is defined by

$$(a,b)^* = (a^*,-b).$$

The following propositions show the effect of repeatedly applying the Cayley-Dickson construction:

Proposition 1.   $A'$ is never real.

Proposition 2.   $A$ is real (and thus commutative) $\iff$ $A'$ is commutative.

Proposition 3.   $A$ is commutative and associative $\iff$ $A'$ is associative.

Proposition 4.   $A$ is associative and nicely normed $\iff$ $A'$ is alternative and nicely normed.

Proposition 5.   $A$ is nicely normed $\iff$ $A'$ is nicely normed.

All of these follow from straightforward calculations; to prove them here would merely deprive the reader of the pleasure of doing so. It follows from these propositions that:

$\R$ is a real commutative associative nicely normed $\ast$-algebra $\implies$

$\C$ is a commutative associative nicely normed $\ast$-algebra $\implies$

$\H$ is an associative nicely normed $\ast$-algebra $\implies$

$\O$ is an alternative nicely normed $\ast$-algebra

and therefore that $\R,\C,\H,$ and $\O$ are normed division algebras. It also follows that the octonions are neither real, nor commutative, nor associative.

If we keep applying the Cayley-Dickson process to the octonions we get a sequence of $\ast$-algebras of dimension 16, 32, 64, and so on. The first of these is called the sedenions, presumably alluding to the fact that it is 16-dimensional [62]. It follows from the above results that all the $\ast$-algebras in this sequence are nicely normed but neither real, nor commutative, nor alternative. They all have multiplicative inverses, since they are nicely normed. But they are not division algebras, since an explicit calculation demonstrates that the sedenions, and thus all the rest, have zero divisors. In fact [21,68], the zero divisors of norm one in the sedenions form a subspace that is homeomorphic to the exceptional Lie group $\G _2$.

The Cayley-Dickson construction provides a nice way to obtain the sequence $\R,\H,\C,\O$ and the basic properties of these algebras. But what is the meaning of this construction? To answer this, it is better to define $A'$ as the algebra formed by adjoining to $A$ an element $i$ satisfying $i^2 = -1$ together with the following relations:

$$a(ib) = i(a^* b) , \qquad (ai)b = (ab^*)i, \qquad (ia)(bi^{-1}) = (ab)^*$$ (1)

for all $a,b \in A$. We make $A'$ into a $\ast$-algebra using the original conjugation on elements of $A$ and setting $i^* = -i$. It is easy to check that every element of $A'$ can be uniquely written as $a + ib$ for some $a,b \in A$, and that this description of the Cayley-Dickson construction becomes equivalent to our previous one if we set $(a,b) = a + ib$.

What is the significance of the relations in (1)? Simply this: they express conjugation in terms of conjugation! This is a pun on the double meaning of the word 'conjugation'. What I really mean is that they express the $\ast$ operation in $A$ as conjugation by $i$. In particular, we have

$$a^\ast = (ia)i^{-1} = i(ai^{-1})$$

for all $a \in A$. Note that when $A'$ is associative, any one of the relations in (1) implies the other two. It is when $A'$ is nonassociative that we really need all three relations.

This interpretation of the Cayley-Dickson construction makes it easier to see what happens as we repeatedly apply the construction starting with $\R$. In $\R$ the $\ast$ operation does nothing, so when we do the Cayley-Dickson construction, conjugation by $i$ must have no effect on elements of $\R$. Since $\R$ is commutative, this means that $\C = \R'$ is commutative. But $\C$ is no longer real, since $i^* = -i$.

Next let us apply the Cayley-Dickson construction to $\C$. Since $\C$ is commutative, the $\ast$ operation in $\C$ is an automorphism. Whenever we have an associative algebra $A$ equipped with an automorphism $\alpha$, we can always extend $A$ to a larger associative algebra by adjoining an invertible element $x$ with

$$\alpha(a) = xax^{-1}$$

for all $a \in A$. Since $\C$ is associative, this means that $\C' = \H$ is associative. But since $\C$ is not real, $\H$ cannot be commutative, since conjugation by the newly adjoined element $i$ must have a nontrivial effect.

Finally, let us apply the Cayley-Dickson construction to $\H$. Since $\H$ is noncommutative, the $\ast$ operation in $\H$ is not an automorphism; it is merely an antiautomorphism. This means we cannot express it as conjugation by some element of a larger associative algebra. Thus $\H' = \O$ must be nonassociative.

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