My thoughts / background information:

It is easy to find an inclusion homomorphism for complex matrices into real matrices: considering the one dimensional case note that multiplying a complex number $z$ by a complex number of unit length is the same as rotation the complex number $z$ by a certain angle. Writing down the rotation matrix in $\mathbb R^2$ makes it obvious that mapping $a + ib$ to $\displaystyle \left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$ is the desired inclusion homomorphism in one dimension and generalising it to $n$ dimensions is not difficult.

My question:

What is not clear to me is generalising this to quaternionic matrices: given a quaternion $p = a + jb$ (caution! $a$ and $b$ are now in $\mathbb C$) and a unit quaternion $q = \cos {\theta \over 2} + (u_x i + u_y j + u_z k)\sin {\theta \over 2}$ I am aware that $qpq^{-1}$ is a rotated quaternion by angle $\theta$.

But how, from that, can I derive that the homomorphism in $1$ dimension should be

$$ a + bj \mapsto \left(\begin{array}{cc} a & b \\ -\overline{b} & \overline{a} \end{array}\right)$$

How to derive this map?


I think you're mislead by focussing on the "rotation" aspect, instead of just the operation of "multiplication by a quaternion (or complex number)".

For a fixed quaternion $a+bj$, the map $$ (x+yj) \mapsto (x+yj)(a+bj) $$ is a $\mathbf{C}$-linear map from $\mathbf{H}$ to $\mathbf{H}$. Here we are viewing $\mathbf{H}$ as a two-dimensional complex vector space where scalar multiplication is from the left; this is an important detail since if $\lambda \in \mathbf{C}$ and $q \in \mathbf{H}$, then $\lambda q \neq q \lambda$ in general.

Multiplying out the right-hand side and simplifying (using that $jz=\bar{z}j$ for complex numbers $z$), we get $(xa-y\bar{b})+(xb+y\bar{a})j$, which means that the matrix representation of this linear map is $$ (x,y) \mapsto (x,y) \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} . $$ (This looks a little unusual, since normally we let matrices act on column vectors rather than row vectors, but it has to do with the choice of putting scalars on the left.)

Now, composing two such maps shows that $$ (x+yj) \mapsto (x+yj)(a+bj)(c+dj) $$ corresponds to $$ (x,y) \mapsto (x,y) \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} \begin{pmatrix} c & d \\ -\bar{d} & \bar{c} \end{pmatrix} , $$ so the product of the matrices indeed corresponds to the product of the corresponding quaternions.

  • $\begingroup$ So there is no way to generalise what I did in the complex case (the rotation thing) to the case of quaternions? $\endgroup$ – learner Nov 19 '14 at 11:33
  • 1
    $\begingroup$ Doesn't the rotation formula $p \mapsto qpq^{-1}$ presuppose that $p$ is a pure quaternion (no real part), so that $p$ can be interpreted as a vector in $\mathbf{R}^3$? Then you're using quaternions to represent rotations of real three-dimensional vectors, and it seems difficult to start from real $3 \times 3$ rotation matrices and from that somehow discover the $2 \times 2$ complex representation. $\endgroup$ – Hans Lundmark Nov 19 '14 at 12:46
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    $\begingroup$ However, there is this whole "double cover of $SO(3)$ by $SU(2)$" thing, so perhaps there is some way to do what you want, but it seems a bit far-fetched to go down that path when the explanation above is so simple. (By the way, it also explains the matrix representation for complex numbers, if you replace complexes and quaternions by reals and complexes). $\endgroup$ – Hans Lundmark Nov 19 '14 at 12:47
  • $\begingroup$ Thank you, you have helped me greatly! $\endgroup$ – learner Nov 19 '14 at 23:36
  • $\begingroup$ You're welcome! $\endgroup$ – Hans Lundmark Nov 20 '14 at 8:01

It is better here to view $\mathbb{H}$ as a real vector space with basis $\{1,i,j,k\}$. Then we have an algebra isomorphism with the complex matrices

$$ \mathbb{H}\cong \left\{ \begin{pmatrix} w & z \\ -\overline{z} & \overline{w}\end{pmatrix}\mid z,w\in \mathbb{C}\right\}. $$

via $$a+bi+cj+dk \mapsto \begin{pmatrix} a+bi& c+di\\ - c+di & a-bi \end{pmatrix}. $$

For more details see here.

  • $\begingroup$ Ok but then my question is: how did martini in the linked thread derive the matrices for $I,J$ and $K$? $\endgroup$ – learner Nov 19 '14 at 10:55
  • $\begingroup$ By choosing $a,b,c,d$ so that the matrices, except $\pm Id$ have order $4$ and satisfy the quaternion relations. $\endgroup$ – Dietrich Burde Nov 19 '14 at 10:59
  • $\begingroup$ So the quaternion case is not analogous to the complex case, that is, in one dimension this is not a rotation? $\endgroup$ – learner Nov 19 '14 at 11:01
  • $\begingroup$ I mean, (sorry for not expressing myself well), in the complex case I could start by observing one dimension and then easily find the morphism. Do I really have to do trial and error to find these matrices? Is there no way of "seeing" it? $\endgroup$ – learner Nov 19 '14 at 11:03
  • $\begingroup$ I understand that you're saying "determine the homomorphism by determining what it maps the basis elements to". But this is not what I did in the complex case. $\endgroup$ – learner Nov 19 '14 at 11:05

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