The Cayley-Dickson construction is a way to systematically construct, starting from the real numbers, a sequence of ever higher-dimensional real algebras $A_k$ which starts with complex numbers and quaternions.

The rules are as follows:

• Each algebra comes with an operation called conjugation, where the conjugate of $a$ is denoted as $a^*$.

• The algebra $A_0$ is simply the real numbers themselves, with conjugation defined as the identity.

• The algebra $A_{n+1}$ is given by pairs of elements of $A_n$, with addition, multiplication and conjugation defined as

  \begin{align} (a, b) + (c, d) &= (a+c, b+d)\\ (a, b) \cdot (c, d) &= (a c - d^* b, d a + b c^*)\\ (a, b)^* &= (a^*, -b) \end{align}

Now one property of the construction is that each algebra $A_n$ can be identified with the subalgebra of $A_{n+1}$ where the second element is $0$; it is easily verified that $(a,0) + (b,0) = (a+b, 0)$, $(a, 0) \cdot (b,0) = (ab, 0)$ and $(a,0)^* = (a^*, 0)$.

This means you can generate the direct limit to arrive at an infinite-dimensional algebra $A$. The elements of $A$ are given by sequences of real numbers with finite support (that is, they eventually turn constant zero). Addition is per element, conjugation changes all signs except the first element.

Where it gets complicated is multiplication. Let's define $i_n$ as the sequence which has $a_n=1$ and $a_k=0$ for $k\ne n$. So when e.g. looking at the embedding of the quaternions (that is, the sequences with $a_n=0$ for all $n\ge 4$, we have $i_0=1$, $i_1=i$, $i_2=j$, $i_3=k$).

Now it is obvious that if we determine an explicit formula for the products $i_m i_n$, that can be used to directly calculate every product in $A$.

What is easy to determine is that if we denote with $m\oplus n$ the bitwise exclusive or of the natural numbers $m$ and $n$, then we must have $i_m i_n = \pm i_{m\oplus n}$. What I couldn't figure out is a general formula for the sign. Let's call that $\sigma(m, n)$, such that in general, we can say $$i_m i_n = \sigma(m,n) i_{m\oplus n},\quad |\sigma(m,n)| = 1$$

What is obvious is that, since $i_0=1$ is the multiplicative identity of $A$, for all $n$ we have $\sigma(0,n) = \sigma(n,0) = 1$. Also, all imaginary units square to $-1$, therefore for any $n\ne 0$, $\sigma(n,n)=-1$.

Also, from the product definition, one can derive a recursion rule for $\sigma$: If both $m$ and $n$ are less than $2^k$, then you have (note the varying order of the arguments to $\sigma$):

\begin{align} \sigma(m, 2^k + n) &= \sigma(n, m)\\ \sigma(2^k + m, n) &= \begin{cases} 1 &\text{if $n=0$}\\ -\sigma(m,n) &\text{otherwise} \end{cases}\\ \sigma(2^k + m, 2^k + n) &= \begin{cases} -1 & \text{if $n=0$}\\ \sigma(n, m) &\text{otherwise} \end{cases} \end{align}

What I didn't find, however, is a closed formula. Can anyone help me?