Does there exist an algebraic construction of quaternion algebra such that the algebraic properties (such as associativity and distributivity) are immediately obvious? I am looking for something similar to how the complex numbers are constructed in Galois theory, as an extension of reals: $$\mathbb{C} \cong \mathbb{R}[x]/(x^2+1)$$ Apart from the algebraic properties, which are crystal clear in this construction, we also obtain that complex conjugation is an isomorphism without any explicit calculations. Is this possible for quaternions?

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    $\begingroup$ The magic words you're probably looking for are division algebra and Cayley-Dickson construction; see e.g. math.ucr.edu/home/baez/octonions/node5.html $\endgroup$ Aug 9, 2020 at 21:57
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    $\begingroup$ You can also define them as the quotient $\mathbb R[Q_8]/\langle i+(-i),j+(-j),k+(-k)\rangle$, where $Q_8$ is the quaternion group, $\mathbb R[Q_8]$ the respective group ring, and $\pm i/j/k$ are the elements of $\mathbb R[Q_8]$ corresponding to the elements of $Q_8$ (so $-i/j/k$ is not the additive inverse of $i/j/k$). I don't really see what properties are easily deductible this way, though. $\endgroup$ Aug 9, 2020 at 22:10
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    $\begingroup$ @Vercassivelaunos Since $(-i)$ and $-(i)$ may be hard to distinguish, might help to call the central involution of $Q_8$ by $\varepsilon$ instead, in which case $\Bbb H=\Bbb R[Q_8]/(1+\varepsilon)$. $\endgroup$
    – runway44
    Aug 9, 2020 at 22:12
  • $\begingroup$ @runway44: Using $-i/j/k$ is a bit cumbersome to write, but I think it's better at conveying the idea that we take $\mathbb R[Q_8]$, but turn $-i/j/k$ into the additive inverse of $i/j/k$ (since $i+(-i)$ is in the ideal, it is zero in the quotient). But that may be because I have a hard time myself wrapping my head around the fact that $\mathbb R[Q_8]/(1+\varepsilon)$ does the same thing. Probably because of some playing around with the defining relations of $Q_8$? $\endgroup$ Aug 9, 2020 at 22:26
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    $\begingroup$ I don't think that the octonions that were linked are a good analogy because they are nonassociative. The better analogy (or generalization) would be Clifford algebras. $\endgroup$
    – runway44
    Aug 9, 2020 at 22:32

2 Answers 2


First, note if $A$ is any unital associative algebra and $a\in A^{\times}$ a unit, then conjugation ($x\mapsto axa^{-1}$) is an algebra automorphism - we check this the same way we do for groups (but also check linearity).

The Clifford algebra $\mathcal{C}\ell(n)$ is the free associative algebra generated by $n$ anticommuting square roots of negative one. That is, it may be defined as the quotient of a free associative algebra:

$$ \mathbb{R}\langle x_1,\cdots,x_n\rangle/(x_i^2+1,x_ix_j+x_jx_i) $$

so that the relations $x_i^2=-1$ and $x_jx_i=-x_ix_j$ (when $i\ne j$) are imposed. Note the free associative algebra generated by $\{x_1,\cdots,x_n\}$ is basically the "noncommutative" polynomial ring in them, although of course they commute if there's only one variable.

Then $\mathcal{C}\ell(0)=\mathbb{R}$ and $\mathcal{C}\ell(1)=\mathbb{C}$ (generated by $i$) and $\mathcal{C}\ell(2)=\mathbb{H}$ (generated by $i,j$).

Since $Q_8\subset\mathbb{H}^{\times}$ is generated by $i,j$ as well, it spans $\mathbb{H}$, so we can define $\mathbb{H}$ by a quotient of the group algebra $\mathbb{R}[Q_8]$, however this is about as natural as defining $\mathbb{C}$ as a quotient of the group algebra $\mathbb{R}[\{\pm1,\pm i\}]$.

They Cayley-Dickson construction proceeds to build algebras $\mathbb{R}\subset\mathbb{C}\subset\mathbb{H}\subset\mathbb{O}\subset\mathbb{S}$ where the octonions $\mathbb{O}$ lose full associativity (although they retain "alternativity," i.e. every subalgebra generated by $2$ elements is associative), and sedenions $\mathbb{S}$ begin introducing zero divisors. The octonions are $8$-dimensional and the sedenions are $16$-dimensional unital algebras.

Clifford algebras also begin introducing zero divisors; $\mathcal{C}\ell(3)$ is $\cong\mathbb{H}\oplus\mathbb{H}$ (good exercise). There is an interesting "mod $8$" periodicity to them (not up to isomorphism, but up to Morita equivalence) which may be described by what Baez calls the "Clifford clock".

You can also do Clifford algebras over other fields, or describe them in a "coodinate-free" way as a quotient of a tensor algebra over a vector space with a quadratic form, you can describe how they combine as super-algebras with super-tensor products, etc.


The magic words you're probably looking for are division algebra and Cayley-Dickson construction; see e.g. math.ucr.edu/home/baez/octonions/node5.html. - Steven Stadnicki Aug 9 at 21:57


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