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I would like to know if there exists any kind of classification of finite dimensional (possibly non-associative) algebras $A$ over $\mathbb R$ that satisfy the following condition:

There exists a norm on $A$ such that for any $a\in A$ $|a*a|=|a||a|$ (the norm on $A$ is Euclidean).

Of course, complex numbers, quaternions and octonians provide examples of such algebras. But there are further examples of dimension $2^n$ that are given by Cayley-Dickson construction (pages 8-10 in http://math.ucr.edu/home/baez/octonions/oct.pdf).

Is there a classification of such algebras, at least in small dimensions (over $\mathbb R$)? What are possible dimensions of such algebras?

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2 Answers

up vote 3 down vote accepted

I don't think these objects can be classified in a manner similar to the normed unital division algebras, if you take "algebra" to mean "vector space $V$ equipped with a bilinear map $V \otimes V \to V$". In particular, I suspect you end up with high-dimensional moduli spaces of such structures in all large dimensions.

Here is a naive calculation of degrees of freedom:

Let $a_{i,j}^k$ be the structure constants of our algebra, assembled into matrices $A^k$, and consider a point $x = (x_1, \ldots, x_n)$. We may write $x \ast x = (x^T A^1 x,\ldots, x^T A^n x)$. The length condition becomes $\left(\sum_i x_i^2\right)^2 = \left(\sum_k x^T A^k x \right)^2$. Writing this out in terms of the coordinates of $x$, we obtain an identity of homogeneous polynomials in $x_i$ of total degree 4, with coefficients that are quadratic in the structure constants. In other words, the space of solutions is an intersection of $\binom{n+3}{4}$ quadric hypersurfaces in $n^3$-dimensional space. Subtracting, we get positive dimensions for $2 \leq n \leq 17$ with maximum 377 at $n=13$, but after that, we get more constraints than variables. As it happens, the positivity holds for $n \leq 16$ when we account for the $O(n)$ symmetry of the solution space.

Since the Cayley-Dickson construction exists and provides solutions in arbitrarily large dimension, it is clear that the constraints are highly non-generic. This does not completely eliminate the possibility that in some dimensions there are no solutions, but I think it is at least discouraging as far as classification is concerned.

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Dear Scott, thanks a lot for a thoughtful answer! The dimension calculation is nice :). Unfortunately after the answer of Bruce I realised that I am not 100% sure that Cayley-Dickson construction works... Do you think these algebras are indeed examples satisfying $|a*a|=|a|^2$? I understand that these algebras (sedonians, ect) are power associative, and each element has an inverse. But it seems to me that in order to prove that $|a*a|=|a|^2$ holds one has to show that $a$ and $a^{-1}$ together generate a (multiplicative) subgroup of the ring isomorphic to $\mathbb Z$. Is this indeed true? –  aglearner Jun 10 '12 at 19:55
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If you have a power-associative algebra, then any invertible element generates a cyclic group. Norm one elements may be torsion, but that's not a big deal. The Cayley-Dickson algebras satisfy the norm condition, by the properties of the norm-preserving anti-involution $x \mapsto x^\ast$. See the last section of: en.wikipedia.org/wiki/Cayley-Dickson_construction –  S. Carnahan Jun 11 '12 at 0:29
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The commutative algebras with this property are the imaginary parts of the cubic Jordan algebras. Let $\mathbb{A}$ be a composition algebra and take $A$ to be $3\times 3$ trace-free Hermitian matrices with entries in $\mathbb{A}$. Define multiplication to be $(a,b)\mapsto ab+ba-\frac23 tr(ab)$. Define the inner product by $\langle a,b\rangle=tr(ab)$. Then the condition $|a^2|=|a|^2$ is satisfied.

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Dear Bruce, thank you very much for the answer. I would like to ask you to give a bit more details. Would you advise some reference? After some googling I seem to deduce from your answer that the dimension over $\mathbb R$ of imaginary parts of the cubic Jordan algebras is bounded. At least as far as I got, over every field there are only 4 composition algebras (this is stated here : en.wikipedia.org/wiki/Composition_algebra ) –  aglearner Jun 9 '12 at 11:23
    
There is a book by Springer –  Bruce Westbury Jun 9 '12 at 11:37
    
The dimensions are 2,5,8,14,26. –  Bruce Westbury Jun 9 '12 at 11:38
    
What is the name of the book? –  aglearner Jun 9 '12 at 11:59
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