MHB 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 $$n \gt 4$$. 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 ... ...

PeterThe above post refers to Lemma 1.3.

Lemma 1.3 reads as follows:View attachment 6224
=====================================================

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 $$n \gt 4$$. 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 $$V$$?

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 $$e := v + \frac{i \circ v}{2} i + \frac{j \circ v}{2} j + \frac{k \circ v}{2} k$$ satisfies $$i \circ e = j \circ e = k \circ e = 0$$?

v doesn't "satisfy" that, there is no v in it. $$i \circ e = j \circ e = k \circ e = 0$$ 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 $$eij = -iej = ije$$, 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?

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

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

is a nonzero element in $$V$$ ...

Can you help further ... ?

You also write:

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

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

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

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

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

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

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