Split Quaternions

We can repeat this process to obtain the split quaternions, denoted $\HH'$, which satisfies \begin{equation} \HH' = \CC \oplus \CC L \end{equation} In order to keep track of these different algebras, we will use $K$ rather than $i$ for the imaginary unit here. Thus, $\HH'$ consists of linear combinations of $1$, $K$, $L$, and $KL$, and it remains to work out the full multiplication table. We have \begin{align} K^2 &= -1 \qquad L^2 = +1 \\ (K)(L) &= KL = -(L)(K) \\ (KL)^2 &= KLKL = -KKLL = +1 \\ K(KL) &= -L = -(KL)K \\ (KL)L &= K = -L(KL) \end{align} The split quaternions are associative, but not commutative.

A typical element $Q\in\HH'$ thus takes the form \begin{equation} Q = Q_1 + Q_2K + Q_3 KL + Q_4 L \end{equation} and has (squared) norm \begin{equation} |Q|^2 = Q\bar{Q} = Q_1^2 + Q_2^2 - Q_3^2 - Q_4^2 \end{equation} which has signature $(2,2)$, that is, there two of our basis elements have (squared) norm $+1$, namely $1$ and $K$, and two have (squared) norm $-1$, namely $KL$ and $L$.

The split quaternions can also be obtained as \begin{equation} \HH' = \CC' \oplus \CC'K \end{equation} so there are only two 4-dimensional composition algebras over the reals, namely $\HH$ and $\HH'$, not three. That is, it doesn't matter whether the Cayley-Dickson process is done first with $\epsilon=+1$, then with $\epsilon=-1$, or vice versa. Note, however, that the split quaternions $\HH'$ contain both the split complex numbers $\CC'$ and the ordinary complex numbers $\CC$.

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