# Matrix Representation of Octonions

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?

-
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

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.

-