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Centre of the real Hamilton Quaternions H

  1. Jan 20, 2012 #1
    Can anyone help me with the following exercise from Dummit and Foote?


    Describe the centre of the real Hamilton Quaternions H.

    Prove that {a + bi | a,b R} is a subring of H which is a field but is not contained in the centre of H.


    Regarding the problem of describing the centre - thoughts so far are as follows:

    Let h = a + bi + cj + dk

    Then investigate conditions for i and h to commute!

    i [itex] \star [/itex] h = i [itex] \star [/itex] ( a + bi + cj + dk)
    = ai + b[itex]i^2[/itex] + cij + dik
    = ai - b + ck - dj

    h [itex] \star [/itex] i = ( a + bi + cj + dk) [itex] \star [/itex] i
    = ai + b[itex]i^2[/itex] + cji + dki
    = ai - b -ck + dj

    Thus i and h commute only if c = d = 0

    Proceeding similarly we find that

    j and h commute only if b = d = 0


    k and h commute only if b = c = 0

    Thus it seems as if I am being driven to the conclusion that the only Hamilton Quaternions that commute with every element of the ring of Hamilton Quaternions are elements of the form

    a + 0i + 0j + 0k

    But I am unsure of how to formally and validly argue from the facts established above to conclude this!

    Can anyone help or at least confirm that I am on the right track!
  2. jcsd
  3. Jan 20, 2012 #2
    That is correct. What exactly is your problem?
  4. Jan 20, 2012 #3


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    Science Advisor

    you've proven that only quaternions of the form:

    a + bi commute with i (this makes sense if you think about it, as neither j nor k commute with i).

    well if q = a + bi + cj + dk is in Z(H), it has to commute with i, since i is a quaternion. so that alone narrows down the possibilities right there.

    what you've done with j and k is fine, although you could also show that if a + bi commutes with j, then b = 0, and likewise for k.

    this means that only real quaternions (b = c = d = 0) can possibly be in the center. it's not hard to show that all real quaternions are indeed central, which settles the matter.

    EDIT: a little reflection should convince you that the center of the quaternions has to be a field. the only possibilities for the dimension (as a vector space over R) of this field is either 1 or 2 (if it was 2, it would have to be an isomorph of the complex numbers). the fact that i doesn't commute with j or k, effectively kills this possibility. it actually makes more sense to think of i,j and k being "3 identical copies of √-1", because there's no real way to tell them apart from one another. this is why H is sometimes viewed as "scalars+vectors", the "pure quaternion (non-real)" part, acts very much like a vector in R3, which was actually Hamilton's original goal-to find an "algebra" for 3-vectors.

    the fact that Z(Q8), the center of the group of quaternion units, is equal to {-1,1} should reassure you that your conclusion is correct.
    Last edited: Jan 20, 2012
  5. Jan 20, 2012 #4

    Thanks so much for the help!
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