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I Bresar, Lemma 1.3 - Real Quaternions ... Division Algebras

  1. Nov 20, 2016 #1
    I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

    I need help with some aspects of the proof of Lemma 1.3 ... ...

    Lemma 1.3 reads as follows:


    ?temp_hash=34c5196f407ac73e36866c16ee629634.png




    In the above text by Matej Bresar we read the following:

    " ... ... Set ##i := \frac{1}{ \sqrt{ -u^2 } } u , \ \ j := \frac{1}{ \sqrt{ -v^2 } } v## , and ##k := ij##.

    It is straightforward to check that (1.1) holds ... ... "



    I need some help in proving that ##ij = -ji = k## ... sadly I cannot get past the point of substituting the relevant formulas ...

    Hope someone can help ...


    Peter

    EDIT ... I must admit that as I reflect further on Lemma 1.3 I am more confused than I first thought ... why is Bresar defining another 'multiplication' in V ... that is why define ##\circ## ... we already have a multiplication from D ... and how does the new definition ##\circ## play out when validating 1.1 ...


    ============================================================================

    In order for readers of the above post to appreciate the context of the post I am
    providing pages 1-3 of Bresar ... as follows ...



    ?temp_hash=34c5196f407ac73e36866c16ee629634.png
    ?temp_hash=34c5196f407ac73e36866c16ee629634.png
    ?temp_hash=34c5196f407ac73e36866c16ee629634.png
     

    Attached Files:

    Last edited: Nov 21, 2016
  2. jcsd
  3. Nov 21, 2016 #2

    andrewkirk

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    Science Advisor
    Homework Helper
    Gold Member

    That ##ij=k## is true by definition of ##k##. So all we have to prove is that ##ij=-ji##, ie that ##ij+ji=0_D##. Here goes:

    \begin{align*}
    ij+ji&=\left(\frac1{\sqrt{-u^2}}u\right)\left(\frac1{\sqrt{-v^2}}v\right)
    +\left(\frac1{\sqrt{-v^2}}v\right)\left(\frac1{\sqrt{-u^2}}u\right)\\
    &\textrm{[I've now edited the above line to remove the typo pointed out by fresh_42 below. Thank you]}\\
    &=\frac{uv+vu}{\sqrt{u^2v^2}}
    \end{align*}
    which is zero iff the numerator is zero.

    But the numerator is ##uv+vu=u\circ v=0##, as claimed in the second line of the proof.

    To see that that claim is true, we calculate as in the alignment below. But first, note that ##v^2## commutes with everything, since ##v^2\in\mathbb R'## so that ##v^2=a1_D## for some ##a\in\mathbb R,\ a\leq 0##, and ##1_D## commutes with everything, as does scalar multiplication:

    \begin{align*}
    u\circ v&=uv+vw\\
    &=\left(w-\frac{w\circ v}{v\circ v}v\right)v+v\left(w-\frac{w\circ v}{v\circ v}v\right)\\
    &=wv+vw-\frac{(w\circ v)v^2+v(w\circ v)v}{v\circ v}\\
    &=wv+vw-\frac{(wv+vw)v^2+v(wv+vw)v}{2v^2}\\
    &=\frac{2wvv^2+2vwv^2-(wv+vw)v^2-v(wv+vw)v}{2v^2}\\
    &=\frac{2wvv^2+2vwv^2-wvv^2-vwv^2-vwv^2-v^2wv}{2v^2}\\
    &=\frac{2wvv^2+2vwv^2-wvv^2-vwv^2-vwv^2-wvv^2}{2v^2}\\
    &=\frac{0}{2v^2}=0
    \end{align*}
    where we have repeatedly used the ability of ##v^2## to commute with anything else, to move it about.
     
    Last edited: Nov 21, 2016
  4. Nov 21, 2016 #3

    fresh_42

    Staff: Mentor

    Just a few remarks. Firstly, ignore the last bracket in @andrewkirk 's second line, it's a cut+paste typo. Secondly, I'll try to answer
    I think we have to consider the following: Goal of all this is, to show there are division algebras over ##\mathbb{R}##, namely the complex numbers ##\mathbb{C}=\mathbb{R}+V=\mathbb{R}+i\cdot\mathbb{R}##, the quaternions ##\mathbb{H}=\mathbb{R}+V=\mathbb{R}+i\cdot\mathbb{R}+j\cdot\mathbb{R}+k\cdot\mathbb{R}## and possibly the octonions as well.

    Therefore Lemma 1.2 which proves that ##V## (the imaginary part of our ##D##) is a linear subspace.

    Many of the techniques here, e.g. Lemma 1.1., are simply a reduction to formulas in the real numbers, where we know how to do calculations, whereas in a potential division algebra ##D## we don't - yet. Furthermore, we have to use the specialties of the real numbers somehow, namely that it is an Archimedean ordered field, which Lemma 1.1. is closely related to, or that squares in ##\mathbb{R}## are always non-negative.

    So Bresar uses the structure in ##\mathbb{R}## to deduce insights on ##D##. As he also knows, where this ends, namely at ##\mathbb{C},\mathbb{H}## and ##\mathbb{O}## and that ##i\,\cdot\,j=k=-j\,\cdot\,i## holds, it makes sense to examine the "zeroness" of ##i \circ j = i\cdot j + j\cdot i## which explains the additional (now made commutative again(!)) multiplication. In addition ##i,j,k## have to be defined somehow, only by the use of a construction with real numbers. The entire difficulty here is not to show that ##\mathbb{C},\mathbb{H},\mathbb{O}## are division algebras, but to show they are the only ones. Therefore we need necessary conditions for ##D## being a division algebra over ##\mathbb{R}##, the more the better.

    At least this is what I hope for, because to show that the examples are actually division algebras, we could simply perform some calculations instead.
     
  5. Nov 21, 2016 #4
    Andrew, fresh_42 ... thanks for the posts ...

    Just reflecting on what you have written ...

    Thanks again ... appreciate your help and your insights ...

    Peter
     
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