Proof of ij = -ji = k in Bresar's Lemma 1.3

[Math Amateur]

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:

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 ...

 

[andrewkirk]

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.

 

[fresh_42]

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 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.

 

[Math Amateur]

Andrew, fresh_42 ... thanks for the posts ...

Just reflecting on what you have written ...

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

Peter