Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Since quaternions $\mathbb{H}$ have a matrix representation as elements of $\text{SU}(2,\mathbb{C})$ as the following

$$ 1 \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \mathrm i \mapsto \begin{pmatrix} \mathrm i_{\mathbb C} & 0 \\ 0 & -\mathrm i_{\mathbb C} \end{pmatrix},\quad \mathrm j \mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad \mathrm k \mapsto \begin{pmatrix} 0 & \mathrm i_{\mathbb C} \\ \mathrm i_{\mathbb C} & 0 \end{pmatrix}, $$ I always wondered if there is also matrix representation of the octonions?

How is the non-associativity realised with matrices?

share|improve this question
You shpuld probably be explicit about the sense in which quaternions have a matrix representation as SU(2,C)... H is an algebra and SU(2,C) is a group! –  Mariano Suárez-Alvarez Jan 4 '12 at 18:47
Matrix multiplication is associative, so what do you hope for? –  Grigory M Jan 4 '12 at 18:48
@GrigoryM: I don't hope, I wonder. So the answer is NO? –  draks ... Jan 4 '12 at 19:01
What you mean, probably, is that there is a representation of $\mathbb H$ as a real subalgebra of the algebra $M_2(\mathbb C)$ of $2\times 2$ matrices. $\mathrm{SU}(2,\mathbb C)$ has little to do with it. –  Mariano Suárez-Alvarez Jan 4 '12 at 19:08
@MarianoSuárez-Alvarez: Would $\text{su}(2,\mathbb{C})$ suit better? –  draks ... Jan 4 '12 at 22:09
show 2 more comments

2 Answers

up vote 4 down vote accepted

A $\mathbb R$-linear function $\phi:\mathbb O\to A$ to any real associative algebra which is multiplicative and unitary (so that $\phi(xy)=\phi(x)\phi(y)$ for all $x$, $y\in \mathbb O$, and $\phi(1)=1$) has to vanish on the bilateral ideal $I$ generated by the elements of the form $$(x\cdot y)\cdot z-x\cdot(y\cdot z)$$ with $x$, $y$, $z\in\mathbb O$. Now, this ideal is not zero because $\mathbb O$ is not associative, and therefore $I=\mathbb O$, because $\mathbb O$ has no non-trivial bilateral ideals (it is a division thing).

It follows that the map $\phi$ is in fact zero.

share|improve this answer
add comment

M. Zorn (Abh.Mat.Sem.Hamburg,9,395, 1933) modified matrix multiplication for 2x2 "vector-matrices" with two real-numbers as diagonal entries and two 3d-vectors off the diagonal.

See also Matrix Representation of Octonions and Generalizations by Jamil Daboul and Robert Delbourgo, (9 June 1999)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.