Frobenius Theorem - Bresar, Theorem 1.4 ....

[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 ... ...

Theorem 1.4 reads as follows:
View attachment 6223Questions 1(a) and 1(b)


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

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

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

Therefore \(\displaystyle e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k\)

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


My questions are as follows:

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

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

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

" ... ... However, from the first two identities we conclude \(\displaystyle eij = -iej = ije\), which contradicts the third identity since \(\displaystyle ij = k\) ... ... "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 \(\displaystyle eij = -iej = ije\) and, further, how this contradicts \(\displaystyle ij = \)k?
Hope someone can help ...Help will be appreciated ... ...

PeterThe above post refers to Lemma 1.3.

Lemma 1.3 reads as follows:View attachment 6224
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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 ...View attachment 6225
https://www.physicsforums.com/attachments/6226
View attachment 6227
View attachment 6228

 

[HallsofIvy]

Questions 1(a) and 1(b)


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

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

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

Therefore \(\displaystyle e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k\)

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


My questions are as follows:

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

This is under the assumption that the space has dimension greater than 4. i, j, and k are three independent vectors so span a three dimensional subspace. Since the entire space has dimension greater than 4, there must exist non-zero vectors that are not in that subspace.
Let v be any of those vectors.

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

v doesn't "satisfy" that, there is no v in it. \(\displaystyle i \circ e = j \circ e = k \circ e = 0\) is true because i, j, and k are defined to be orthogonal to each other.

Question 2

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

" ... ... However, from the first two identities we conclude \(\displaystyle eij = -iej = ije\), which contradicts the third identity since \(\displaystyle ij = k\) ... ... "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 \(\displaystyle eij = -iej = ije\) and, further, how this contradicts \(\displaystyle ij = \)k?

How is "e" defined? You don't seem to have included that. However since, in this space, multiplication is defined to be "anti-symmetric", ab= ba, for any vector, e, ei= -ie so eij= (ei)j= (-ie)j= -iej= -i(ej)= -i(-je)= ije.

Hope someone can help ...Help will be appreciated ... ...

PeterThe above post refers to Lemma 1.3.

Lemma 1.3 reads as follows:
=====================================================

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 ...

 

[Math Amateur]

This is under the assumption that the space has dimension greater than 4. i, j, and k are three independent vectors so span a three dimensional subspace. Since the entire space has dimension greater than 4, there must exist non-zero vectors that are not in that subspace.
Let v be any of those vectors. v doesn't "satisfy" that, there is no v in it. \(\displaystyle i \circ e = j \circ e = k \circ e = 0\) is true because i, j, and k are defined to be orthogonal to each other.

How is "e" defined? You don't seem to have included that. However since, in this space, multiplication is defined to be "anti-symmetric", ab= ba, for any vector, e, ei= -ie so eij= (ei)j= (-ie)j= -iej= -i(ej)= -i(-je)= ije.

Hi HallsofIvy ... Thanks for the help ...

... BUT ... just some clarifications ...
You write:

" ... ... This is under the assumption that the space has dimension greater than 4. i, j, and k are three independent vectors so span a three dimensional subspace. Since the entire space has dimension greater than 4, there must exist non-zero vectors that are not in that subspace.
Let v be any of those vectors."I understand that there must exist at least one element v \in V that does not belong to the linear span of i, j, k ... but I cannot see why e defined by

\(\displaystyle e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k\)

is a nonzero element in \(\displaystyle V\) ...

Can you help further ... ?

You also write:

" ... ... v doesn't "satisfy" that, there is no v in it. \(\displaystyle i \circ e = j \circ e = k \circ e = 0\) is true because i, j, and k are defined to be orthogonal to each other. ..."My question 1b involved e defined as

\(\displaystyle e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k\)

and Bresar's assertion was that e (not v) satisfied

\(\displaystyle i \circ e = j \circ e = k \circ e = 0\)

I am still unsure why e satisfies the above ... can you help further ...

Further, you write:

" ... ... How is "e" defined? You don't seem to have included that. However since, in this space, multiplication is defined to be "anti-symmetric", ab= ba, for any vector, ... ... ... "As I mentioned above e is defined by

\(\displaystyle e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k\)

You mention that "in this space, multiplication is defined to be "anti-symmetric", ab= ba, for any vector" ... ...

BUT ... where does Bresar state that multiplication is "anti-symmetric", ab= ba, for any vector ... can you help?Peter