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I Unknown hypercomplex numbers

  1. Apr 19, 2017 #1
    Some time ago, I stumbled upon an interesting set of hypercomplex numbers. I thought that somebody else might have discovered them ( it was too facile a construction ) and forgot about them for many years.
    Lately, I searched on the web and did not find any mention of their existence. I must admit I did not know where to look and if they had been discovered, I did not know their name.
    So, being here in this great forum for a while, I am asking your permission to disclose them. Shall I go ahead?
    I am giving myself a 99.999% chance that they've been already known and I wouldn't bother, but am too old now to keep them for myself.
     
  2. jcsd
  3. Apr 19, 2017 #2

    mathman

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    You need to describe them. How do you define "hypercomplex numbers"? As is, it is just words.
     
  4. Apr 19, 2017 #3

    jedishrfu

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    Also can you provide some context like what problem you were trying to solve that helped you to discover these puppies.
     
  5. Apr 19, 2017 #4
    Ok, I take it that permission has been granted now for me to post :


    They are constructed somewhat similar to the Quaternions, the only difference being : ## k = ij = 1- ji ##, hence they are neither commutative nor anti-commutative, (associativity still holds good.)
    By this rule only , if ## i^2 = -1 ## and ## j^2 = -1 ## then ## k^3 = -1 ##. Notice this is a cube, not a square.
    In the process, the rest of the combinations of the basis elements are recovered, namely :
    ## ki = jk = i + j ##,
    ## ik = -j ##,
    ## kj = -i ##,
    ## k^2 = k -1 ##.
    Conjugation: The conjugate of ## a + bi + cj + dk ## is taken to be : ## a + d - bi - cj - dk ## ,
    and the quadratic form is : ## a^2 + b^2 + c^2 + d^2 + ad - bc ## ,
    thus giving rise to an unique left or right inverse ( like the Quaternions, they are both the same ) :
    ## \frac { a + d - bi - cj - dk } { a^2 + b^2 + c^2 + d^2 + ad - bc } ##

    I call them the Quaternionoids ( a variant of the Quaternions. )
     
  6. Apr 19, 2017 #5

    jedishrfu

    Staff: Mentor

    What would you use them for?

    Quaternions have been used in physics. At one time, they were the dominant framework for Classical Mechanics championed by Hamilton until vectors came into the picture and superseded them primarily because the notation and concepts were simpler. More recently, quaternions have made somewhat of a comeback because they incorporate rotational orientation into the picture and are useful in computer graphics systems.

    https://en.m.wikipedia.org/wiki/Quaternion
     
  7. Apr 20, 2017 #6

    fresh_42

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    They have another nice property: ##SU(2,\mathbb{C}) \cong U(1,\mathbb{H})##. :smile:
     
  8. Apr 20, 2017 #7

    Stephen Tashi

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  9. Apr 20, 2017 #8

    fresh_42

    Staff: Mentor

  10. Apr 20, 2017 #9
    You may take my word for it, that by a laborious checking I have proven that my Quaternionoids are in their most general case, associative.
     
  11. Apr 21, 2017 #10
    From what I could gather from Wikipedia, I have the feeling that Frobenius theorem hinges on definite quadratic forms of this kind : ## a_1x_1^2 + a_2x_2^2 +..+ a_nx_n^2 ##.
    Now ## x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_1x_4 - x_2x_3 ##, the squared norm I described above, is certainly positive definite but not of this kind: ## a_1x_1^2 + a_2x_2^2 +..+ a_nx_n^2 ##.
    I cannot see how my algebra is isomorphic to ## \mathbb{H}##. Enlightment please.
     
    Last edited: Apr 21, 2017
  12. Apr 21, 2017 #11

    lavinia

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    Yes. If the quaterniods are a real division algebra then they must be isomorphic to the quaternions. There must be a third element whose square is -1
    Not sure either.

    The Froebenius argument shows that the subspace of elements whose square is negative must have codimension 1 - in this case dimension 3. The argument is that the trace of the R-linear map ##x→ax## is zero if and only if ##a^2## is negative - that is: the subspace of elements whose square is negative is the kernel of a linear map into the reals.

    This suffices to demonstrate that the division algebra is isomorphic to the quaternions. So either the quaternioids are not a real division algebra or there is another real basis that satisfies the usual relations for the quaternions.
     
    Last edited: Apr 21, 2017
  13. Apr 21, 2017 #12

    Stephen Tashi

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    I don't claim to understand the proof, but I see nothing in the proof that relies on a particular quadratic form. In particular the "##Q##" in the proof is something that can be defined without any dependence on a particular norm.


    We haven't established that you have a divison algebra yet. (Pehaps a digression, but I'm curious how "fast and loose" one can be in defining algebras by defining a set of products of "generators" and then declaring that multiplication of linear combinations of such expressions shall be done by the distributive law. Does that prove the distributive law works by definition? And what properties of the products do we need in order to guarantee we are defining a finite dimensional algebra? )

    Assuming, you do have a finite dimensional divison algebra, can you represent an element of it as a 4x4 matrix of real numbers? That would move the problem of determining an isomorphism into the familiar territory. We would ask if your matrix is similar (in the technical sense of "similar" matrices) to the standard 4x4 matrix representation of a quaternion.

    Your norm gives some hint about this. If we want to transform ##a^2 + b^2 - ab## to ##x^2 + y^2## we can look for ways to express ##a## and ##b## as a linear combinations of ##x## and ##y##.
     
  14. Apr 21, 2017 #13
    No need for all this!

    The substitution : ## I = \frac { i + k } { \sqrt {2} } ## , ## J = - \frac { j + k } { \sqrt {2} } ##, where i,j,k are the usual quaternion units gives the isomorphic quaternionic algebra with the above rules:

    ## I^2 = -1 ## , ## J^2 = -1 ## , ## K^3 = -1 ##,
    ## K = IJ = 1- JI ## etc., that I described.

    R.I.P. Quaternionoids and many thanks to all of you!
     
    Last edited: Apr 21, 2017
  15. Apr 22, 2017 #14

    lavinia

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    [QUOTE="Stephen Tashi, post: 5745452, member: 186655" (Pehaps a digression, but I'm curious how "fast and loose" one can be in defining algebras by defining a set of products of "generators" and then declaring that multiplication of linear combinations of such expressions shall be done by the distributive law. Does that prove the distributive law works by definition? And what properties of the products do we need in order to guarantee we are defining a finite dimensional algebra? )[/QUOTE]

    Here is one way to generate finite dimensional associative algebras. Given a finite group ##G## and a field ##F## the elements of ##G## can formally be taken as a basis for a finite dimensional vector space over ##F##. Vectors are all formal linear combinations ##Σf_{i}[g_{i}]## where the sum is taken over the elements of ##G## - here denoted as ##[g_{i}]##. This vector space becomes an associative algebra under the multiplication rule ##Σf_{i}[g_{i}]Σe_{j}[g_{j}]= Σf_{i}e_{j}[g_{i}g_{j}]##. Modding out by an ideal produces another associative algebra.

    In the case of the quaternions the field is the reals and the group is generated by two elements ##i## and ##j## with the relations ##i^2 = j^2 = (ij)^2 = z, z^2 = 1## with ##z## not equal to the identity. Since this a group of order 8 the vector space over ##R## is 8 dimensional. Modding out by the relations [##j##] + [##jz##] = [##1##]+[##z##] = [##i##] + [##iz## ]= 0 gives the quaternions (I think).
     
    Last edited: Apr 22, 2017
  16. Apr 22, 2017 #15

    Stephen Tashi

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    Yes, a group has an associated "group algebra" over the reals (or another field). It's disappointing that the group algebra of the quaternion group isn't the quaternion algebra ##\mathbb{H}##. (I'm thinking about starting a thread on the topic of "Verbal Disappointments in Mathematics". That failure would be one of them!
     
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