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

In summary, the conversation between Peter and Andrewkirk revolves around proving Lemma 1.3 in Matej Bresar's book "Introduction to Noncommutative Algebra". Peter is struggling with the proof and is seeking help from Andrewkirk, who provides a step-by-step explanation and calculation to prove that ##ij=-ji=k##. The discussion also touches upon the purpose of defining the new multiplication ##\circ## and its connection to the structure of the real numbers in order to deduce insights on potential division algebras.
  • #1
Math Amateur
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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
 

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  • #2
Math Amateur said:
I need some help in proving that ##ij=−ji=k## ... sadly I cannot get past the point of substituting the relevant formulas
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.
 
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  • #3
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
Math Amateur said:
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 ...
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.
 
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  • #4
Andrew, fresh_42 ... thanks for the posts ...

Just reflecting on what you have written ...

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

Peter
 

FAQ: Bresar, Lemma 1.3 - Real Quaternions .... Division Algebras

What is Bresar, Lemma 1.3?

Bresar, Lemma 1.3 is a mathematical theorem that states that in a real quaternion division algebra, the norm of a nonzero element is always positive.

What are real quaternions?

Real quaternions are a type of number in mathematics that extends the concept of complex numbers. They are written in the form a + bi + cj + dk, where a, b, c, and d are real numbers and i, j, and k are imaginary units.

What is a division algebra?

A division algebra is a type of algebra in which every nonzero element has a unique multiplicative inverse. In other words, division is always possible in a division algebra.

How is Bresar, Lemma 1.3 significant?

Bresar, Lemma 1.3 is significant because it provides a necessary and sufficient condition for a real quaternion algebra to be a division algebra. This has important implications in various areas of mathematics, such as algebraic geometry and topology.

Can Bresar, Lemma 1.3 be applied to other types of division algebras?

Yes, Bresar, Lemma 1.3 can be extended to division algebras over fields other than the real numbers, such as the complex numbers or quaternions. However, the specific conditions may differ depending on the type of field being used.

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