Why do quaternions not only use two imaginary numbers. Can we not simplify quaternions

$$q = a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} \tag{1}$$

to the form

$$ \begin{align} q & = a + b\mathbf{i} + (c + d\mathbf{i})\mathbf{j} \tag{2}\\ & = e + f\mathbf{j} \end{align} $$ Where $e,f \in \mathbb{C}$.

Thus we only introduce one new imaginary number. Where we note that $\bf{ij} = k$, which holds true.

This would make quaternions seem more like complex numbers and we only need to define the interactions between $\bf i$ and $\bf j$.

The same would also hold true for octonions being represented as $s = g + h\mathbf{k'}$ where $g, h \in \mathbb{H}$. Where we see that we now have only 3 imaginary numbers instead of 7.

This 'factorized' representation seems to simplify things and make these numbers look more familiar.

  • $\begingroup$ Then we don’t even need one imaginary number, just write $e+f\mathbf j = g$. $\endgroup$
    – Vishu
    Apr 21, 2021 at 14:21
  • $\begingroup$ In the context of using quaternions to model 3D and 4D rotations, there is no reason to privilege the $\mathbf{i}$ direction in space over any other; the complex number format is arbitrary and obfuscates the geometry. $\endgroup$
    – anon
    Apr 23, 2021 at 5:19
  • $\begingroup$ As explained in KCd's answer the role of $\mathbf{j}$ here is that conjugating a complex number with it amounts to the usual complex conjugation. This is an instance of the so called Skolem-Noether theorem, implying (among other things) that any two embeddings of a field into a division algebra differ by conjugation. I find those results fascinating even though I haven't fully internalized the reason why this happens. $\endgroup$ May 5, 2021 at 15:40

1 Answer 1


This is already done. It uses constructions going back over 100 years.

For quaternions built from complex numbers, look up cyclic algebras. This leads to a central simple algebra using an arbitrary cyclic Galois extension, which for quaternions comes from the Galois extension $\mathbf C/\mathbf R$.

Note that $j$ does not commute with all of $\mathbf C$: $jz = \overline{z}j$. You have to decide if you want to treat the quaternions as a left $\mathbf C$-vector space or a right $\mathbf C$-vector space. As a vector space over $\mathbf R$, that subtlety does not arise. So a more compact representation brings its own new features.

The compact construction you describe for octonions as pairs of quaternions (as well as for sedenions as pairs of octonions) can be found in books or online under the name "Cayley-Dickson doubling process" or "Cayley-Dickson construction".

  • $\begingroup$ "So a more compact representation brings its own new features" But quaternions are already non-commutative. $\endgroup$ Apr 21, 2021 at 14:41
  • $\begingroup$ Yes I have corrected the sedenions to octonions $\endgroup$ Apr 21, 2021 at 14:43
  • $\begingroup$ This is starting to make a lot of sense, but why do you never see or use the compact representation? $\endgroup$ Apr 21, 2021 at 14:45
  • $\begingroup$ The quaternions overall are noncommutative, but real numbers commute with all of them. That was my point. You do not have to make a distinction about left or right vector space when treating the quaternions as a real vector space. No amount of alternate notation is going to change the actual overall structure in the quaternions; it can only emphasize certain features. $\endgroup$
    – KCd
    Apr 21, 2021 at 14:47
  • $\begingroup$ Ah now I understand what you meant there. Thanks a lot! $\endgroup$ Apr 21, 2021 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .