# I Frobenius Theorem - Bresar, Theorem 1.4 ...

1. Nov 23, 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 Theorem 1.4 ... ...

Questions 1(a) and 1(b)

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

" ... ... Suppose $n \gt 4$. Let $i, j, k$ be the elements from Lemma 1.3.

Since the dimension of $V$ is $n - 1$, there exists $v \in V$ not lying in the linear span of $i, j, k$.

Therefore $e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k$

is a nonzero element in $V$ and it satisfies $i \circ e = j \circ e = k \circ e = 0$.... ... "

My questions are as follows:

(1a) Can someone please explain exactly why $e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k$ is a nonzero element in $V$?

(1b) ... ... and further, can someone please show how $e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k$ satisfies $i \circ e = j \circ e = k \circ e = 0$?

Question 2

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

" ... ... However, from the first two identities we conclude $eij = -iej = ije$, which contradicts the third identity since $ij = k$ ... ... "

I must confess Bresar has lost me here ... I'm not even sure what identities he is referring to ... but anyway, can someone please explain why/how we can conclude that $eij = -iej = ije$ and, further, how this contradicts $ij = k$?

Hope someone can help ....

Help will be appreciated ... ...

Peter

The above post refers to Lemma 1.3.

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

In order for readers of the above post to appreciate the context of the post I am providing pages 1-4 of Bresar ... as follows ...

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• ###### Bresar - Page 4 ... ....png
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Last edited: Nov 23, 2016
2. Nov 23, 2016

### Staff: Mentor

If we assume $e := v+\frac{i\circ v}{2}i+\frac{j\circ v}{2}j+\frac{k\circ v}{2}k=0$ then $v=-\frac{i\circ v}{2}i-\frac{j\circ v}{2}j-\frac{k\circ v}{2}k$. Now this would imply (in contrast to our assumption) that $v \in lin_\mathbb{R}\{i,j,k\}$ if the coefficients $\frac{i\circ v}{2},\frac{j\circ v}{2},\frac{k\circ v}{2}$ were all real. But this follows from the last two lines directly ahead of Lemma 1.3. where he shows that they have to be real.
(Hint: As a general rule: It's always helpful to ask "what if not" when one tries to understand a reasoning. I do this quasi permanently when I read a mathematical text. It's the most important question of all.)

What bothers me a lot more, is the theorem itself, because it only holds for associative division algebras $D$, since the octonions build a non-associative division algebra of dimension $8$ over the reals. Thus somewhere in the entire process, there must be a hidden usage of associativity. Where? And why doesn't he mention this? Maybe it's within the line with $eij$. But somewhere it has to be.

The algebra tower goes like this:
$\mathbb{Q}$ - prime field of characteristic $0$
$\mathbb{R}$ - limits gained, field, topological closure of $\mathbb{Q}$, $\dim_\mathbb{R} = 1$
$\mathbb{C}$ - solvability gained, algebraic closure of $\mathbb{R}$, $\dim_\mathbb{R} = 2$
$\mathbb{H}$ - commutativity lost, skew-field, $\dim_\mathbb{R} = 4$
$\mathbb{O}$ - associativity lost, division ring, $\dim_\mathbb{R} = 8$

Last edited: Nov 23, 2016
3. Nov 23, 2016

### Math Amateur

Thanks for the help fresh_42 ...

You write:

" ... ... If we assume $e := v+\frac{i\circ v}{2}i+\frac{j\circ v}{2}j+\frac{k\circ v}{2}k=0$ then $v=-\frac{i\circ v}{2}i-\frac{j\circ v}{2}j-\frac{k\circ v}{2}k$. Now this would imply (in contrast to our assumption) that $v \in lin_\mathbb{R}\{i,j,k\}$ if the coefficients $\frac{i\circ v}{2},\frac{j\circ v}{2},\frac{k\circ v}{2}$ were all real. But this follows from the last two lines directly ahead of Lemma 1.3. where he shows that they have to be real. ... ... "

So ... this shows that v must be nonzero ... but I am still not sure why exactly $e \in v$ ... ... can you help ... ?

Also ,,, I remain unsure on exactly why/how defining e as

$e := v+\frac{i\circ v}{2}i+\frac{j\circ v}{2}j+\frac{k\circ v}{2}$

implies that

$i \circ e = j \circ e = k \circ e = 0$?

Can you help?

Peter

4. Nov 24, 2016

### Staff: Mentor

It implies $e \neq 0$. $v\neq 0$ is already guaranteed by the choice of $v \notin lin_\mathbb{R}\{i,j,k\}$.
Lemma 1.2.: $V$ is a linear subspace. So with $v,i,j,k \in V$ ($v$ by assumption, $i,j,k$ by their definition), hence also $e = v + c_1\cdot i +c_2 \cdot j +c_3 \cdot k \in V$ holds.
We have to use the following rules here:
• $u \circ (r \cdot w) = r \cdot (u \circ w)$ for all $r \in \mathbb{R}$ and $u,w \in V$
• The coefficients $c_1,c_2,c_3$ in the definition of $e$ are all real. (See the last two lines before Lemma 1.3.)
• Finally $i \circ j = j \circ k = k \circ i = 0$ by their definition and $u \circ u = -2$ for all $u \in \{i,j,k\}$
Now
\begin{align*} i \circ e &= i \circ v + i \circ (c_1\cdot i) + i \circ (c_2\cdot j) + i \circ (c_3\cdot k) \\ &= i \circ v + c_1 \cdot (i \circ i) + c_2 \cdot (i \circ j) + c_3 \cdot (i \circ k) \\ &= i \circ v - 2 \cdot c_1 \\ &= iv+vi-2 \cdot (\frac{iv}{2}+\frac{vi}{2}) \\ &= 0\end{align*}
It's been now the second time Bresar has hidden the essential hint within these two lines ahead of Lemma 1.3. Would have been a lot easier to see, if he had numbered this as some Corollary. Anyway. I admit it's been a bit tricky to see, but I think it would have helped you a lot to "play" with those formulas on your own: for practice and for the feeling of success if you found out. This way, I thank you for giving me this feeling, but it would have been better placed on your side. I know it's frustrating to star at things with the "knowledge" in mind: 'This should be obvious, why can't I see?' - and be ensured I know this from an uncounted number of own experiences, however, I'm afraid it's part of the game.

5. Nov 24, 2016

### Math Amateur

Thanks so much for the assistance fresh_42 ... your post was extremely helpful ...

But I need some further help ...

You write:

" ... ... $u \circ (r \cdot w) = r \cdot (u \circ w)$ for all $r \in \mathbb{R}$ and $u,w \in V$ ... ..."

I am having trouble understanding why this is true ...

My thoughts are as follows:

$u \circ (r \cdot w) = u (r \cdot w) + ( r \cdot w) u$

$= u( r \cdot w ) + r \cdot ( wu)$ ... ... ... ... ... (1)

and

$r \cdot ( u \circ w ) = r \cdot ( uw + wu )$

$= r \cdot (uw ) + r \cdot (wu)$ ... ... ... ... ... (2)

Now, we require expression (1) to be equal to expression (2) ...

For this to be true we require ...

$u( r \cdot w ) = r \cdot (uw )$ ... ... but ... ... why is this true ... the scalar $r$ cannot commute with vectors $u$ or $w$ ... ???

Can you help?

Peter

***EDIT***

SOLVED?

I have been doning some checking of the axioms for an associative algebra with identity such as a real division algebra D nd have found that a fundamental axiom involving multiplication of elements of D and multiplication by a scalar (i.e. a real ) is the following:

$r \cdot (uw) = ( r \cdot u) w = u ( r \cdot w )$

and this, I think, answers my problem in needing the equality

$u( r \cdot w ) = r \cdot (uw )$

Is my reasoning correct?

Peter

Last edited: Nov 24, 2016
6. Nov 25, 2016

### Staff: Mentor

Yes, this is correct. To be honest, I hadn't thought about it very much. Because we don't distinguish between left and right division algebras (rings), scalars (elements of $\mathbb{R}$) can be applied from both sides with identical result (and $\circ$ is defined with ordinary multiplications).

(You're on a good track questioning everything! )

Last edited: Nov 25, 2016
7. Nov 25, 2016

### Math Amateur

Thanks for all your help fresh_42 ... it has assisted me a great deal in trying to understand division algebras ...

Peter