Hurwitz Theorem Composition Algebras Proof I've been reading the proof to Hurwitz's Theorem that says that the only finite-dimensional  real  division  algebras have orders 1,2 4 or 8.


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*I read that it is considering non-associative algebras, in which case how do you multiply elements together?


It gets to a point where it considers a finite group G generated by elements $a_1, a_2... a_{m-1}, \epsilon$ with m > 2 that follow the equations $a_i^2=\epsilon\neq1, \epsilon^2=1, a_ia_j=\epsilon a_j a_i$ for $i \neq j$. They say that G is clearly a group of order $2^m$.


*Is this because all the $a_i$ have order 4 and $\epsilon$ has order 2? 


Then, as $a_i\epsilon=\epsilon a_i, \epsilon$ is in the centre. 


*Does the equation come from finding the two ways of writing $a_i$ inverse and from that the inverse is unique?


At this point, they work out that the commutator subgroup is $\{1,\epsilon\}$. 


*How do you work that out for non-associative algebras?


Then it says that the centre consists of 1 and $\epsilon$ when m is odd, and 1, $\epsilon , a_1 a_2... a_{m-1}$ and $\epsilon a_1 a_2... a_{m-1}$ when m is even. Again, I don't know how to work that out for non-associative groups, and am a bit dodgy even for associative groups.


*How is that worked out for both cases? 


If I let the group be associative, then I get that each element can be written in the form $\epsilon^{b_0} , a_1^{b_1} a_2^{b_2}... a_{m-1}^{b_m-1}$ where $b_i$ is either 0 or 1, and that if an element g has a an even number of $a_i$ with $b_i = 1$, then $g^{-1}=g$ and if it has an odd number $g^{-1} = \epsilon g$. This is as far as I got for the associative case.
It also says that if an element g is not in the centre, then its conjugacy class consists of g and $\epsilon$ g, which I get if I let it be associative, but have no idea where to start for non associative.
I'd written all this out before I realised that the wiki page has the proof there.
The wiki page has the following paragraph;

When N is even, there are two and their dimension must divide the order of the group, so is a power of two, so they must both have dimension $2^{(N − 2)/2}$. The space on which the $a_i's$ act can be complexified. It will have complex dimension N. It breaks up into some of complex irreducible representations of G, all having dimension $2^{(N − 2)/2}$. In particular this dimension is ≤ N, so N is less than or equal to 8. If N = 6, the dimension is 4, which does not divide 6. So N can only be 1, 2, 4 or 8.

From what I've gleamed from wikipedia, complexification basically means creating a space from our space, say V, into $V^\mathbb{C} \cong V \otimes Vi$. 


*What is a complex dimension? How does it into break the representations? Where do they get that N is less than or equal to 8? When they get 1,2,4,8, are they saying they are the only possible ones, but not guaranteeing at this point that they exist?

 A: 
I've been reading the proof to Hurwitz's Theorem that says that the only finite-dimensional real division algebras have orders 1,2 4 or 8.

Some terminology:


*

*A division algebra is one in which there are no zero divisors. It is not necessarily associative. (Note that without the associative property, existence of multiplicative inverses and existence of zero divisors are not necessarily mutually exclusive.)

*A composition algebra, or Hurwitz algebra, is one  which admits a nondegenerate symmetric bilinear form which induces a multiplicative quadratic form.

*A Euclidean Hurwitz algebra is one in which the quadratic form is positive-definite.


Hurwitz's theorem specifically concerns Euclidean Hurwitz algebras. Kervaire and Milnor generalized this to all finite-dimensional real division algebras. Hopf had earlier proven that every finite-dimensional real commutative division algebra has dimension 1 or 2. (Indeed, if I recall correctly, there are uncountably many distinct isomorphism classes.) Frobenius' theorem classifies finite-dimensional associative real division algebras. As for composition algebras, in addition to Hurwitz ones there are also split composition algebras, where the quadratic form has indefinite signature (so can be negative), and there are zero divisors.

I read that it is considering non-associative algebras, in which case how do you multiply elements together?

The lack of associativity does not change the fact that there is assumed to be a binary operation $A\times A\to A$ called multiplication. To multiply two elements, you apply this binary operation to it.
Now let's suppose we have 
$$G=\langle a_1,\cdots,a_{m-1},\epsilon\mid a_i=\epsilon, \epsilon^2=1,a_ia_j=\epsilon a_ja_i\rangle$$

Does the equation [$a_i\epsilon=\epsilon a_i$] come from finding the two ways of writing $a_i$ inverse and from that the inverse is unique?

Since $\epsilon$ is a power of $a_i$ for each $i$, we know $\epsilon$ commutes with each $a_i$. It also commutes with itself, which means it commutes with all of the elements of the generating set $\{a_1,\cdots,a_{m-1},\epsilon\}$, hence it commutes with all of $G$, i.e. it is central.

They say that $G$ is clearly a group of order $2^m$. Is this because all the $a_i$ have order $4$ and $\epsilon$ has order $2$?

Every element of $G$ may be written as a product of powers of the generators. Since all the generators have finite order, we may assume the powers are positive. Moreoever, since all the $a_i$ have order $4$, we may assume their exponents are among $\{0,1,2,3\}$. Since $a_i^2=\epsilon$ and $a_i^3=\epsilon a_i$, we may without loss of generality assume all the exponents of the $a_i$ are $0$ or $1$. Moreover, using the relation $a_ia_j=\epsilon a_ja_i$, we can exchange $a_i$s until the word simplifies to $\epsilon^{e_{\large 0}}a_1^{e_{\large 1}}\cdots a_{m-1}^{e_{\large m-1}}$. These words are all distinct, since an equality of two of them would give a relation that does not follow from the given ones. Since the $e_i$s are each chosen in $\{0,1\}$, that gives $2^m$ elements.

How do you work that out for non-associative algebras?

Note that $G$ is not a group of elements of the algebra, it is a group of linear operators on the algebra (defined by left-multiplication by elements of the algebra). It doesn't matter if the algebra is associative or not, as composition of linear operators is associative regardless. There is no such thing as a "non-associative group." (You might be thinking of loops.)

How is that worked out for both cases?

Commutator subgroup: Since $[a_i,a_j]=\epsilon$, anything in the commutator subgroup $[G,G]$ (also called the derived subgroup $G'$) will have both a containing-$\epsilon$-version and a not-containing-$\epsilon$ version. So really, we should focus on computing $[G,G]$ mod $\epsilon$. Given any $g,h\in G$, we know $[g,h]\langle \epsilon\rangle=[g\langle \epsilon\rangle,h\langle\epsilon\rangle]$. Since $G/\langle\epsilon\rangle$ is abelian (from $a_ia_j=\epsilon a_ja_i$), its commutator subgroup is trivial, so $[g\langle\epsilon\rangle,h\langle\epsilon\rangle]=\langle\epsilon\rangle$, and the relation $[g,h]\langle\epsilon\rangle=\langle\epsilon\rangle$ implies $[g,h]\in\langle\epsilon\rangle$.
Center: The element $\epsilon^{e_{\large 0}}a_1^{e_{\large 1}}\cdots a_{m-1}^{e_{\large m-1}}$ is central if and only if it commutes with every generator. It automatically commutes with $\epsilon$, so we want it to commute with each $a_i$. Use the rule $a_ia_j=\epsilon a_ja_i$ to slide $a_j$ past all the letters of our word. For each $j$, we need $\sum_{i\ne j}e_i$ to be even. Since the difference between two such sums is $(\sum_{i\ne j}e_i)-(\sum_{i\ne k}e_i)=e_k-e_j$ and must be even, all exponents must be equal. If the exponents are all $1$, the element commutes with each $a_i$ if and only if $m-1$ is odd, so $m$ is even. Thus we have


*

*If $m$ is odd, $Z(G)=\{1,\epsilon\}$,

*If $m$ is even, $Z(G)=\{1,\epsilon,a_1\cdots a_{m-1},\epsilon a_1\cdots a_{m-1}\}$.



What is a complex dimension?

The dimension of a complex vector space, treated as a complex vector space. For example, $\mathbb{C}^n$ has complex dimension $n$ and real dimension $2n$. Every complex vector space could also be treated as a real vector space if you forget about multiplication-by-$i$, and so we need to distinguish between real and complex dimension. If we take a real vector space of dimension $n$ and complexify, we get a complex vector space of complex dimension $n$. For instance $\mathbb{C}\otimes\mathbb{R}^n\cong\mathbb{C}^n$.

How does it into break the representations?

For the purposes of the proof, it doesn't matter.

Where do they get that N is less than or equal to 8? 

The relation $2^{(N-2)/2}\le N$ holds for $N\le 8$ and fails for $N\ge10$.

When they get 1,2,4,8, are they saying they are the only possible ones, but not guaranteeing at this point that they exist?

That's correct.
