Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quaternion ring

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that the quaternion division ring H has infinitely many u satisfying u[tex]^{2}[/tex]=-1

    2. Relevant equations
    Elements of H is of the form a.1 +bi+cj+dk where a, b, c, d in [tex]\textsl{R}[/tex] ( reals) and i[tex]^{2}[/tex]= j[tex]^{2}[/tex]= k[tex]^{2}[/tex]=ijk = -1.

    3. The attempt at a solution
    Let u = a.1 +bi+cj+dk then u[tex]^{2}[/tex]=a[tex]^{2}[/tex]-b[tex]^{2}[/tex]-c[tex]^{2}[/tex]-d[tex]^{2}[/tex]+2a(bi+cj+dk)+2bicj+2bidk+2cjdk and this = -1 provided a=0 and -b[tex]^{2}[/tex]-c[tex]^{2}[/tex]-d[tex]^{2}[/tex]=1 and 2bicj+2bidk+2cjdk= 0 but I do not see how 2bicj+2bidk+2cjdk= 0.
  2. jcsd
  3. Mar 11, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Two comments:

    (1) You do know what the product [itex]i \cdot j[/itex] is, right?

    (2) -b²-c²-d²=1 can never be satisfied.
  4. Mar 12, 2008 #3
    Thanks for the comments
    (1) The product of i.j =k but this ring is not a commutative ring and so I do not have 2bicj = 2bcij=2bck.
    ij=k, jk=i, ki=j, ji=-k, kj=-i, ik=-j
    (2) My bad. I meant -b²-c²-d²=-1

    But still, I do not see why 2bicj+2bidk+2cjdk= 0. Is there anything that I am missing here?
  5. Mar 12, 2008 #4
    if its not a commutative ring then how do you have 2bicj?
    Shouldnt that term then be :
    i j (b*c) + j i (c*b)
    If ij=-ji, and c*b=b*c then these cancel, right?
  6. Mar 12, 2008 #5
    I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?
  7. Mar 12, 2008 #6
    Yes!!! Thank you very much. I see it now. Poor me.
  8. Mar 12, 2008 #7
    I do not know what [bi, cj] or {bi, cj} stand for.
  9. Mar 12, 2008 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    But fortunately, the real numbers are in the center of the ring -- a fact that is usually explicitly given by describing the quaternions as an algebra over the reals. However (unless I made an error), you can actually derive this fact from the identities given.

    (I note that you already assumed this fact when you simplified bibi to -b²)
  10. Mar 12, 2008 #9
    commutator and anticommutator:
    [A,B] = AB-BA
    {A,B} = AB+BA
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook