Can quaternion group be represented by 3x3 matricies?

1. Aug 30, 2015

jackmell

Hi,

The Quaternion group, $Q=\{1,-1,i,-i,j,-j,k,-k\}$, can be realized by $2x2$ matricies:

\begin{align*} 1=\begin{bmatrix} 1,0 \\ 0,1\end{bmatrix} &\hspace{10pt} i=\begin{bmatrix} \omega,0 \\ 0,-\omega\end{bmatrix} & \hspace{10pt}j=\begin{bmatrix} 0,1 \\ -1,0\end{bmatrix} & \hspace{10pt}k=\begin{bmatrix} 0,\omega \\ \omega,0\end{bmatrix} \end{align*}

with $\omega^2=-1$.

I was told $Q$ can also be represented (non-trivially)by $3x3$ or $4x4$ matricies but could not find any source explaining this and was hoping someone here could either provide a reference or explain this a bit.

Thanks,
Jack

2. Aug 30, 2015

HallsofIvy

I can't see any reason to want to represent them in a more complicated way! Of course, you can always convert a 2 by 2 matrix to 3 by 3 by appending a new row and column consisting entirely of 0s (and to 4 by 4 by appending two new rows and columns consisting entirely of 0s).

For example, change $$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$$ to $$\begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ or to $$\begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$$

But to what purpose?

Last edited by a moderator: Aug 31, 2015
3. Aug 30, 2015

Geofleur

I've never tried to do this, but here is an idea:

The important thing is that the 3x3 (or 4x4) matrices that represent the unit quaternions satisfy the quaternion multiplication table. If you write a 3x3 matrix for each unit quaternion, filled with undetermined constants, and write down all of the entries in the multiplication table that need to be satisfied, you'll get a system of equations that you could solve for the undetermined constants. It may turn out that there is not a unique 3x3 or 4x4 matrix that will work, but a whole bunch of them.

4. Aug 30, 2015

jackmell

Ok thanks. That's an idea worth trying. However maybe I should code it first for 2x2 matrix and see if I come up with the right answer then I should be able to scale it up to 3x3.

5. Aug 30, 2015

micromass

6. Aug 30, 2015

jackmell

Afraid I don't see that micromass although the reference you cited is interesting for further study. Might you explain a little further? Perhaps I'm just not understanding the concept well enough. I thought there may be a set of four 3x3 matricies:
\begin{align*} 1=\begin{bmatrix} a_1,b_1,c_1 \\ d_1,e_1,f_1\\g_1,h_1,j_1\end{bmatrix}&\hspace{20pt}i=\begin{bmatrix} a_2,b_2,c_2 \\ d_2,e_2,f_2\\g_2,h_2,j_2\end{bmatrix}&j=\begin{bmatrix} a_3,b_3,c_3 \\ d_3,e_3,f_3\\g_3,h_3,j_3\end{bmatrix}&k=\begin{bmatrix} a_4,b_4,c_4 \\ d_4,e_4,f_4\\g_4,h_4,j_4\end{bmatrix} \end{align*}
such that they obey quaternion arithmetic. For example $i^2=-I$ that is without embedding the associated 2x2 matrices in 3x3 matricies and padding with zeros.

7. Aug 30, 2015

WWGD

But aren't there, in the quoted source? If you want one for each "basis quaternion" , use the matrix R in the link and for each of a,b,c,d, let the others be 0, e.g., for a rotation by a, use a =a+0i+0j+ 0k , etc.

8. Aug 30, 2015

jackmell

Here's a $GL_4(\mathbb{R})$ representation of the quaternion group:

$1=\begin{bmatrix} 1,0,0,0 \\ 0,1,0,0 \\ 0,0,1,0 \\ 0,0,0,1\end{bmatrix}\quad i=\begin{bmatrix}0,1,0,0 \\-1,0,0,0\\0,0,0,-1\\0,0,1,0\end{bmatrix}\quad j=\begin{bmatrix}0,0,1,0\\0,0,0,1\\-1,0,0,0\\0,-1,0,0\end{bmatrix}\quad k=\begin{bmatrix}0,0,0,1\\0,0,-1,0\\0,1,0,0\\-1,0,0,0\end{bmatrix}$

and if you check the algebra, these satisfy the quaternion group relations. They come from the Wikipedia article on quaternions.

However I do now know how to construct an equivalent (without padding) one for a 3x3. Can someone figure this one out? Actually, how is the 4x4 derived anyway? Can I just continue with larger matricies? Can I construct a set of say 10x10 matricies that satisfy the Quaternion group relations? What about a set of nxn matricies?

Last edited: Aug 30, 2015
9. Aug 30, 2015

WWGD

Maybe you should include additional conditions for your embedding or for your representation into $Gl( n, \mathbb R)$, otherwise, the fact that $Gl (n, \mathbb R)$ embeds in $Gl(n+k, \mathbb R)$ will give you a trivial yes answer.

10. Aug 30, 2015

jackmell

Yes, I meant a non-trivial representation into $GL_{n+1}(\mathbb{R})$ (without just padding a row an column with zeros).

11. Aug 30, 2015

WWGD

Last edited: Aug 30, 2015
12. Aug 31, 2015

Ben Niehoff

What you should be looking up is the representation theory of SU(2), of which the quaternions are a subgroup.