# N-Dimensional Real Division Algebras

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• Math Amateur
In summary, Bresar's book "Introduction to Noncommutative Algebra" discusses finite dimensional division algebras, which are real vector spaces that contain the field of real numbers. The question of whether it is possible to define multiplication on an n-dimensional real space so that it becomes a real division algebra is explored. For n=1, the only such algebra is up to isomorphism, the field of real numbers itself. For n=2, the only known example is the complex numbers, but the question of whether there are any other examples is relatively easy to answer. For n=3, Bresar rules out this case using Lemma 1.3 and its proof. Questions are posed regarding the rigor and formal understanding of these results
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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 understanding some remarks that Matej Bresar makes in Chapter 1 ...

The relevant text is as follows:

My questions regarding the above text are as follows:Question 1

In the above text from Bresar we read the following:

" ... ... Is it possible to define multiplication on an ##n##-dimensional real space so that it becomes a real division algebra?

For n = 1 the question is trivial; every element is a scalar multiple of unity and therefore up to an isomorphism ##\mathbb{R}## itself is the only such algebra. ... ... "How do we know exactly (rigorously and formally) that up to an isomorphism ##\mathbb{R}## itself is the only such algebra?

Question 2

In the above text from Bresar we read the following:

" ... ... for ##n = 2## we know one example, ##\mathbb{C}##, but are there any other? This question is quite easy and the reader may try to solve it immediately. ... ... "

Question 3

In the above text from Bresar we read the following:

" ... ... what about ##n = 3##? ... ... "Bresar answers this question on page 4 after proving Lemmas 1.1, 1.2, and 1.3 ... (see uploads below)

Bresar writes:

" ... ... Lemma 1.3 rules out the case where ##n = 3##. ... ... "Can someone please help me to understand why/how Lemma 1.3 rules out the case where ##n = 3##?
Lemma 1.3 and its proof read as follows:

Help with the above questions will be much appreciated ... ...

Peter
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So that readers of the above post can reference other parts of Bresar's arguments, Lemmas and proofs ... as well as appreciate the context of my questions I am providing pages 1-4 of Matej Bresar's book ... as follows:

#### Attachments

• Bresar - 1 - After the Complex Numbers - PART 1 ... ....png
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• Bresar - 2 - After the Complex Numbers - PART 2 ... ....png
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• Bresar - Lemma 1.3 ... ... .png
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• Bresar - Page 1.png
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• Bresar - Page 2.png
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• Bresar - Page 3 ... ....png
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• Bresar - Page 4 ... ....png
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As to questions one and two, it's as so often: rigor lies in the precision of wording.

The division algebras ##D## are defined as those "containing" ##\mathbb{R}## (cp. discussion in a previous thread).
To be more precise: division algebras here are real vector spaces, i.e. ##\forall_{r\in \mathbb{R}}\; r\cdot 1_D \in D##. This means a copy of ##\mathbb{R}## is via the mapping ##r \leftrightarrows r\cdot 1_D## a one-dimensional subspace of ##D##. Since ##\dim D = 1## it is already equal.

Of course this mapping is an isomorphism and it is the reason why @andrewkirk spoke of ##\mathbb{R}'## instead of ##\mathbb{R}## in https://www.physicsforums.com/threads/bresar-lemma-1-2-finite-division-algebras.894043/#post-5625075

As to question 3: In Lemma 1.3. we have already four ##\mathbb{R}-##linear independent elements ##1,i,j,k \in D## if ##n>2##. So ##n>2## implies that even ##n\geq 4## holds.

I suggest you work out the case of dimension two as an exercise. You may use Lemmata 1.1.,1.2.,1.3. and the definition of ##V##. ##\mathbb{C}## is already an example. One has to show that any other ##2-##dimensional division algebra over the reals look like it, i.e. has an element ##j## with ##j^2=-1## that can be mapped to the imaginary ##i## in ##\mathbb{C}##.
The main goal here is again to be precise. Students (and not only them) often hide uncertainties by the use of many unnecessary words, in an attempt that truth will be miraculously hidden somewhere if it only were long enough. Guess I've already beaten myself, since this gets definitely too long ...

Math Amateur said:
How do we know exactly (rigorously and formally) that up to an isomorphism ##\mathbb{R}## itself is the only such algebra?
Let another 1D division algebra be V. Consider the map ##f:\mathbb R\to V## that maps 0 to ##0_V##, 1 to ##1_V## and is then extended linearly to be a linear map between vector spaces. It is a result of linear algebra that ##f## is a vector space isomorphism.

We only need to prove that ##f(r\times s)=f(r)\times_Vf(s)##, since the rest of the necessary properties are obtained from the fact that it is a vector space isomorphism.

See if you can do that, using the fact that ##r=r\cdot 1,s=s\cdot 1## and that the vector space homomorphism property gives us that ##f(r\cdot 1)=r\cdot f(1)##. You will need to use the 'compatibility with scalars' axiom for V, as set out here.

In the above I am using ##\times## or ##\times_V## to indicate multiplication of vectors and ##\cdot## to indicate left-multiplication of a vector by a scalar.

Math Amateur
Just out of curiosity - the results of lemma 1.1 depend on factoring a polynomial in ##\mathbb{R}[\omega]##, so what happens when we study finite dimensional division algebras over ##\mathbb{Q}## instead of ##\mathbb{R}## ?

Stephen Tashi said:
Just out of curiosity - the results of lemma 1.1 depend on factoring a polynomial in ##\mathbb{R}[\omega]##, so what happens when we study finite dimensional division algebras over ##\mathbb{Q}## instead of ##\mathbb{R}## ?
It will get a little bit messy, because every field extension of ##\mathbb{Q}## is a division algebra, ##\mathbb{Q}(\sqrt[17]{83},i)## an algebraic extension, ##\mathbb{Q}(\pi, e, 2^{\sqrt{2}})## a transcendental. Plus the same division algebras as over ##\mathbb{R}##, only with ##\mathbb{Q}##.

Edit: Cancel the transcendental.

fresh_42 said:
every field extension of ##\mathbb{Q}## is a division algebra

It hard to understand that phenomena intuitively. Why does less capability in factoring ( by ##\mathbb{Q}[\omega_1, \omega_2,...] ## vis-a-vis ##\mathbb{R}[\omega]## lead to more versatility in the possible finite dimensional division algebras ?

Here, I'm thinking of ##\mathbb{Q}[\omega]## and ##\mathbb{R}[\omega] ##as being polynomials involving the "indeterminate" ##\omega##, not as polynomials in some specific mathematical object like ##\omega = \sqrt{3}##.

I suppose it relates to the fact that there is more diversity in the types of factors that irreducible polynomials over ##\mathbb Q## can have. Over ##\mathbb R## the factor polynomials can have order only 1 or 2, whereas over ##\mathbb Q## they can have any order.

Less capability in factoring goes along with a richer variety of types of factors.

At the extreme, in ##\mathbb C## all factors are the same type: ##(x-z_k)##.

Stephen Tashi

## 1. What is an N-Dimensional Real Division Algebra?

An N-Dimensional Real Division Algebra is a mathematical structure that extends the properties of real numbers to higher dimensions. It is a set of elements that can be added, subtracted, multiplied, and divided, where the division operation is well-defined and every non-zero element has a unique multiplicative inverse. It is also known as an N-Dimensional Real Division Ring.

## 2. What are the properties of N-Dimensional Real Division Algebras?

N-Dimensional Real Division Algebras have several key properties, including closure, associativity, commutativity of addition, distributivity, and the existence of a multiplicative identity. They also have the property of division, which means that every non-zero element has a unique inverse under multiplication. Additionally, they follow the properties of the real numbers, such as the existence of additive and multiplicative inverses and the distributive property of multiplication over addition.

## 3. How are N-Dimensional Real Division Algebras different from other types of algebras?

N-Dimensional Real Division Algebras are different from other types of algebras in that they have the property of division, which is not present in all algebras. This means that every non-zero element has a unique inverse under multiplication, allowing for the possibility of division operations. They also extend the properties of real numbers to higher dimensions, making them useful in fields such as physics and engineering.

## 4. What are some examples of N-Dimensional Real Division Algebras?

The most well-known example of an N-Dimensional Real Division Algebra is the set of n-by-n matrices with real entries, denoted as Matn(R). Other examples include the set of quaternions (n=4), the set of octonions (n=8), and the set of sedenions (n=16). These algebras have applications in various areas, such as physics, computer graphics, and robotics.

## 5. What is the significance of N-Dimensional Real Division Algebras in mathematics?

N-Dimensional Real Division Algebras play an important role in mathematics as they provide a natural extension of the properties of real numbers to higher dimensions. They also have applications in various fields, such as geometry, topology, and physics. Additionally, they serve as a foundation for more advanced mathematical structures, such as Lie algebras and Jordan algebras. The study of N-Dimensional Real Division Algebras also leads to a deeper understanding of abstract algebra and its applications.

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