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Can quaternion group be represented by 3x3 matricies?

  1. Aug 30, 2015 #1
    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. jcsd
  3. Aug 30, 2015 #2

    HallsofIvy

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    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 [tex]\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}[/tex] to [tex]\begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex] or to [tex]\begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}[/tex]

    But to what purpose?
     
    Last edited: Aug 31, 2015
  4. Aug 30, 2015 #3

    Geofleur

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    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.
     
  5. Aug 30, 2015 #4
    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.
     
  6. Aug 30, 2015 #5

    micromass

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  7. Aug 30, 2015 #6


    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.
     
  8. Aug 30, 2015 #7

    WWGD

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    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.
     
  9. Aug 30, 2015 #8
    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
  10. Aug 30, 2015 #9

    WWGD

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    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.
     
  11. Aug 30, 2015 #10
    Yes, I meant a non-trivial representation into ##GL_{n+1}(\mathbb{R})## (without just padding a row an column with zeros).
     
  12. Aug 30, 2015 #11

    WWGD

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

    Ben Niehoff

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