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Triplets and quaternions

  1. Jan 27, 2007 #1
    Only recently I discovered that there is a class of complex numbers named hypercomplex numbers.

    Hamilton invented (or discovered) the quaternions in 1843.
    i² = j² = k² = ijk = -1

    I understand how complex numbers work, but I don't understand how hypercomplex numbers work.

    Can someone explain the following:
    First, Hamilton tried to expand the two dimensional complex plane (Argand plane) to three dimensions. But he failed to work with tripleds.
    Why?
    for example, why does i² = j² = ij = -1 not work?

    Second, how did he discover that in 4 dimensions he could work with entities called quaternions?
    how did he invent i² = j² = k² = ijk = -1
    ij = k
    jk = i
    ki = j
    ji = -k
    kj = -i
    ik = -j


    Third, can someone give me an example to work with quaternions? what can you do with them?

    Dimsun
     
  2. jcsd
  3. Jan 27, 2007 #2

    mathman

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    Start with j2=ij, multiply both on the right by j, then using associative law, you get i=j.
     
  4. Jan 27, 2007 #3

    Hurkyl

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    Let v = (a, b, c) and w = (d, e, f) be two Euclidean 3-vectors.

    We can represent them as quaternions:

    s = ai + bj + ck
    t = di + ej + fk

    What is st?
     
  5. Jan 27, 2007 #4

    D H

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    Quaternions are very useful for representing rotations.

    I will use the notation
    [tex]\mathcal Q = \bmatrix q_s \\ \vec q_v \endbmatrix[/tex]
    to denote a quaternion whose real part is [itex]q_s[/itex] and "imaginary" part is [itex]\vec q_v[/itex].

    Suppose [itex]\vec x[/itex] is a three-vector and [itex]\mathcal Q[/itex] is a quaternion. The quaternion product
    [tex]Q\bmatrix 0\\ \vec x \endbmatrix Q^{-1} = \bmatrix 0\\ \vec x^R \endbmatrix[/tex]
    where [itex]\vec x^R[/itex] is the original vector [itex]\vec x[/itex] rotated in three-space.

    In particular, the unit quaternion
    [tex]\mathcal Q =
    \bmatrix \cos \frac \theta 2 \\ \hat u \sin \frac \theta 2 \endbmatrix[/tex]
    will rotate a vector about some axis [itex]\hat u[/itex] by an angle [itex]\theta[/itex].

    Edited to change \vect in LaTeX to \vec
     
    Last edited: Jan 28, 2007
  6. Jan 28, 2007 #5
    Thanks mathman, i understand. maybe it is also the answer on my second question, Hamilton must have had a real flash of genius in which he questioned the commutative law.

    But what if associative law doesn't work in the case of 3 dimensions?


    hurkyl, so it is no different then calculating with ordinary vectors.
    -(ad + be + cf) + (ae - bd)k + (cd - af)j + (bf - ce)i



    DH, It is still a bit difficult, I gues i have to work on it.

    Thanks, Dimsun
     
  7. Jan 28, 2007 #6

    D H

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    One way to look at complex numbers is to decompose them into real and imaginary parts. Quaternions can similarly be decomposed into real and imaginary parts, but with quaternions the imaginary part is a three-vector.

    Let
    [tex]\mathcal Q_1 = \bmatrix q_{s_1} \\ \vec q_{v_1} \endbmatrix[/tex]

    [tex]\mathcal Q_2 = \bmatrix q_{s_2} \\ \vec q_{v_2} \endbmatrix[/tex]
    be two quaternions with real parts [itex]q_{s_1}, q_{s_2}[/itex] and imaginary vector parts [itex]\vec q_{v_1}, \vec q_{v_2}[/itex]. With this notation, the quaternion product is
    [tex]\mathcal Q_1 \, \mathcal Q_2 =
    \bmatrix q_{s_1} q_{s_2} - \vec q_{v_1} \!\cdot \vec q_{v_2}\\ \vec q_{s_1} \vec q_{v_2} + q_{s_2} \vec q_{v_1} + q_{v_1} \times \vec q_{v_2}\endbmatrix[/tex]

    The vector cross product can thusly be expressed in terms of the quaternion product:
    [tex]\bmatrix 0 \\ \vec a \times \vec b \endbmatrix =
    \frac 1 2 \left(
    \bmatrix 0 \\ \vec a \endbmatrix \, \bmatrix 0 \\ \vec b \endbmatrix -
    \bmatrix 0 \\ \vec b \endbmatrix \, \bmatrix 0 \\ \vec a \endbmatrix\right)[/tex]

    The development of the quaternions by Hamilton preceded and motivated the three-vector cross product. Maxwell's equations were originally expressed in quaternion notation. Oliver Heaviside re-expressed these quaternion forms in vector notation, using the now-familiar vector dot product and cross product.
     
    Last edited: Jan 28, 2007
  8. Jan 28, 2007 #7

    Hurkyl

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    If I rememer my history right, quaternion arithmetic is how the dot and cross products were discovered!
     
  9. Jan 28, 2007 #8

    D H

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  10. Jan 31, 2007 #9
    Heaviside must have more credit for his representation of the Maxwell equations.


    I have some more questions about the hypercomplex:

    What kind of hypercomplex numbers (or what kind of algebra) are used in supersymmetry?

    complex numbers are also used in the quantum description of probability waves, why not hypercomplex numbers?

    In special relativity X = iCt.
    Can special relativity also be writen in hypercomplex numbers?

    and are there ultimate hypercomplex numbers?
    thus are they limited to n-dimensional and not 'bigger' than n-dimensions?
     
  11. Feb 7, 2007 #10
    I try to be more specific.

    In the book "Hyperspace" by Micho Kaku, he writes in chapter 6 about supergravity. he tells that it is based on a new kind of number system.

    if a and b are supernumbers then a X b = - b X a
    and a X a = 0 even when a ≠ 0.

    On wikipedia I have read about different kinds of (hyper-)complex numbers.
    (and please correct me if I am wrong):

    After Hamilton invented the quaternions in 1843, James Cockle came with 4-dimensional tessarines. this concept includes complex numbers and split-complex numbers.

    2 dimensional
    Complex numbers have an imaginary base with i² = -1
    Split-complex numbers have an non-real base with i² = 1.
    Dual numbers have a nilpotent i² = 0.
    Dual numbers complements complex numbers and split-complex numbers.

    more dimensional
    the extend of Complex nummers are the Cayley-Dickson construction.
    4 dimensional => quarternions
    8 dimensional => octonions
    16 dimensional => sedenions

    multiplication of quaternions is not commutative. multiplication of octonions is not commutative and not associative. multiplication of sedenions is not commutative and not associative, and the form not a normed space with multiplicative norm (whatever that means) but they still have the property of power associativity.
    Because of this falling apart of operations I wonder if hypercomplex numbers are limited to a fixed number of dimensions.
    Maybe this question is related to the hypercomplex number definiton by Kantor and Solodovnikov.

    The extend of split-complex numbers is the split-complex algebra

    For me it's not clear what clifford algebra is.

    My goal is to understand which and how these hypercomplex algebras are used in supersymmetry.

    Dimsun
     
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