I Bresar, Lemma 1.2 - Finite Division Algebras ...

1. Nov 19, 2016

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 1.2 ... ...

My questions related to the above proof by Bresar are as follows:

Question 1

In the above text from Bresar we read the following:

" ... ... Since $u, v, 1$ are linearly independent, this yields $\lambda + \mu = \lambda - \mu = 0$, hence $\lambda = \mu = 0$, and $u + v \in V$ follows from the first paragraph. ... ... "

My question is ... ... how exactly does it follow that $u + v \in V$?

Question 2

In the above text from Bresar we read the following:

" ... ... Again, using the observation from the first paragraph we see that $x + \frac{v}{2} \in V$. Accordingly, $x = - \frac{v}{2} + ( x + \frac{v}{2} ) \in \mathbb{R} \oplus V$. ... ... "

My question is ... ... how exactly does it follow that $x + \frac{v}{2} \in V$ and, further, how exactly does it then follow that $x = - \frac{v}{2} + ( x + \frac{v}{2} ) \in \mathbb{R} \oplus V$ ... ... ?

Hope someone can help ...

Peter

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In order for readers of the above post to appreciate the context of the post I am providing pages 1-2 of Bresar ... as follows ...

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2. Nov 20, 2016

Staff: Mentor

The whole thing is quite tricky. One can easily get lost (my opinion).
Q1:
I'm not sure where to start here. If you came to $\lambda = \mu =0$, then the situation is as follows:
$(u+v)^2 \in \mathbb{R}$ because $\lambda =0$ (1st condition for elements of $V$). Now let's assume $(u+v)$ is not in $V$, which means $(u+v)^2>0$. By the first paragraph (with $(u+v)$ in the role of $x$) this means that $(u+v) \in \mathbb{R}$. But then $(u+v)=:r \in \mathbb{R} \Rightarrow 0=1\cdot r + (-1)\cdot u + (-1)\cdot v$ with non-zero coefficients, contradicting the linear independence of $\{1,u,v\}$. Thus our assumption was wrong and $(u+v) \in V$.

Q2:
This is basically the same trick again this time with $(x+\frac{v}{2})$ in the role of $x$ in the first paragraph and again by contradicting the assumption $(x+\frac{v}{2}) \notin V$

Last edited: Nov 20, 2016
3. Nov 20, 2016

andrewkirk

I don't understand how the author is using notation. Under Lemma 1 he writes $x^2+\lambda x\in\mathbb R$, but the left-hand side is an element of $D$, a $n$-dimensional vector space, which has no intersection with $\mathbb R$, unless some specific construction work has been done, that is not evident in the supplied material. I suspect what he means is that the LHS is in the 1D subspace of vector space D generated (as vector space) by $1_D$, the multiplicative identity of the division algebra. Do you think that's what he means?

Also (and this may be related) in the statement of Lemma 1.2 what does he mean by $v^2\leq 0$? Since $v^2\in D$, which is a vector space, and no order has been specified for that space, there is no apparent meaning for the $\leq$ symbol. Perhaps he means $v^2=\lambda\cdot 1_D$ and $\lambda\leq 0$?

Also, do you know what the justification is for the first sentence in the proof of Lemma 1:

'Since the dimension of $D$ is $n$, the elements $1,x,....,x^n$ are linearly independent'

It is not obvious to me why this should be the case.

EDIT: d'oh! He didn't write that they're linearly independent, as I incorrectly read it. He said they are linearly dependent. Since there are $n+1$ items in the list and the vector space has only $n$ dimensions it is immediate that they are linearly dependent. As you were.

Last edited: Nov 20, 2016
4. Nov 20, 2016

Staff: Mentor

@andrewkirk Does your post #3 intend to help the OP or may we discuss these questions?

5. Nov 20, 2016

andrewkirk

I don't think the post, considered in isolation, will help him, but if I can find answers to the questions, it will help me to help him.

6. Nov 20, 2016

Staff: Mentor

I also noticed that the author didn't define what he means by a division algebra, except he doesn't require it to be associative. But I think he only considers $\mathbb{R}-$division algebras with $1_D=1_\mathbb{R}$ and thus containing $\mathbb{R}$. With $\mathbb{C}\, , \, \mathbb{H}$ and $\mathbb{O}$ in mind, the Lemmata make sense. I suppose he wants to show that these are the only ones (I think).
What's unusual, too, is that in Lemma 1.1 he uses $\omega$ as indeterminate and $x$ as an element. Linear dependency is meant over $\mathbb{R}$.
In addition the first two lines of the proof of Lemma 1.2. contain an $x$ which is subsequently used multiple times within the proof for different elements and in one case even by an element $x+\frac{v}{2}$. A little bit confusing. But again with the examples above in mind, the definition of $V$ makes sense, e.g. $i\, , \,j\, , \,k \in \mathbb{H}$ are elements of $V$.

7. Nov 20, 2016

Stephen Tashi

It's worth commenting on the fact that Bresar's writes as if for each finite dimensional division ring $D$, we have $R \subset D$.

People who think of the real numbers as a specific thing can object to that because they think of $\mathbb{R}$ as something having exactly the properties satisfied by the axioms of "the real numbers" and no other properties in addition to those. (e.g. A real number does not have a "dimension" or a "row" or "column". )

People who think of any two mathematical structures that are isomorphic as being the "same" structure can agree $R \subset D$. I assume that somewhere in his text Bresar states that when we speak of $R \subset D$, we mean the set of elements in $D$ that can be expressed in the form $(r)(I)$ where $I$ is the identity element of $D$ and $r$ is an element of "the real numbers" as we normally think of them. For example if $D$ consists of a ring of 2x2 matrices then an element of $D$ such as $(4.6)\begin{pmatrix} 1&0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} 4.6 & 0 \\ 0 & 4.6 \end{pmatrix}$ is considered to be a member of the "$R$" that is a subset of $D$.

Similarly when we write somthing like $u^2 = w < 0 \in D$ we mean $w$ can be expressed as $r I$ with $r < 0$ (without any implication that $u$ itself can be written as some real number times the identity matrix).

8. Nov 20, 2016

Math Amateur

Hi Andrew,

Thanks for the post ...

You write:

" ... ... in the statement of Lemma 1.2 what does he mean by $v^2\leq 0$? Since $v^2\in D$, which is a vector space, and no order has been specified for that space, there is no apparent meaning for the $\leq$ symbol. Perhaps he means $v^2=\lambda\cdot 1_D$ and $\lambda\leq 0$? ... ... "

I think Bresar answers this when he writes (see notes on page 2 above Lemma 1.1)

" ... we assume that $D$ is an $n$-dimensional division algebra. ... ...

... ... ... First we recall the notational convention. For $\lambda \in \mathbb{R}$ , we write $\lambda 1 \in D$ simply as $\lambda$. In fact we identify $\mathbb{R} with \mathbb{R} 1$, and in this way consider $\mathbb{R}$ as a subalgebra of $D$. ... ... "

So, $\mathbb{R}$ is essentially embedded in $D$ ... and in Lemma 1.2, although $v \notin \mathbb{R}$ (indeed $\mathbb{R} \cap V = 0$ ) $v^2 \in \mathbb{R}$ and hence the usual order relation in $\mathbb{R}$ applies and we can write $v^2 \le 0$ ...

Do you agree ...?

By the way ... for justification for $x^2 + \lambda x \in \mathbb{R}$ ... see Lemma 1.1 ...

Peter

9. Nov 20, 2016

Math Amateur

Hi all ...

Some of the posts above question or refer to Matej Bresar's definition of a division algebra ...

In his section on prerequisites, gives the following information on Algebras ... ...

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10. Nov 20, 2016

andrewkirk

Having now understood Bresar's peculiar mode of expression I can now understand @fresh_42 's post #2, which all makes perfect sense. The only thing I think I can add is to expand a little on the contradiction that is used in (2), as I think that contradiction is slightly different from the one used in (1) - it doesn't appear to use linear independence. In (2) we have that, for $y\in D\smallsetminus \mathbb R'$, Lemma 1.1 is used to choose $\lambda\in\mathbb R$ such that $y^2+\lambda y\in\mathbb R'$.

I am eschewing Bresar's (in my view, needlessly confusing) notation and using $y$ where he uses $x$ (as $x$ means something different in the first para), $\mathbb R'$ for the 1D subspace $\{k1_D\ :\ k\in\mathbb R\}$, and $\lambda$ instead of $v$ for the real scalar (since he uses $v$ in the previous sentence to refer to a non-scalar element of $D$).

Then we have $y^2+\lambda y=(y+\frac{\lambda}21_D)^2-\frac{\lambda^2}4 1_D=:r\in\mathbb R'$.

Hence $(y+\frac{\lambda}21_D)^2=\frac{\lambda^2}4 1_D+r\in\mathbb R'$.

Therefore either $(y+\frac{\lambda}21_D)\in V$ or $(y+\frac{\lambda}21_D)^2>0$. We assume the latter.

Then from the first para we infer that $(y+\frac{\lambda}21_D)=:s\in\mathbb R'$. But then we have $y=s-\frac{\lambda}21_D\in\mathbb R'$, which contradicts our assumption that $y\in D\smallsetminus \mathbb R'$.

Hence we conclude that $(y+\frac{\lambda}21_D)=:v\in V$. So $y=v-\frac{\lambda}21_D$ which is in $V\oplus \mathbb R'$ since $v\in V$ and $-\frac{\lambda}21_D\in \mathbb R'$.