# 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.