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 Euclidean norm on $A$ such that for any $a\in A$ $|a*a|=|a||a|$.

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/octonions.pdf).

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


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 $\left(\sum_i x_i^2\right)^2 = \sum_k \left(x^T A^k x \right)^2$. 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.

[Revised following YCor's comment:] When we account for the $O(n)$ symmetry of the solution space, we get the formula $$ n^3 - \binom{n+3}{4} - \binom{n}{2}$$ which is positive for $2 \leq n \leq 16$ with maximum 299 at $n=13$. When $n$ is large we get more constraints than variables.

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.

  • $\begingroup$ 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? $\endgroup$ – aglearner Jun 10 '12 at 19:55
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    $\begingroup$ 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 $\endgroup$ – S. Carnahan Jun 11 '12 at 0:29
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    $\begingroup$ For the dimension of the "moduli space", this gives a lower bound in the dimension which is $n^3-\binom{n+3}{4}-\binom{n}{2}$, which is positive iff $n\in [2,16]$, and takes its maximum 299 at $n=13$. Notably for $n=2$ this takes the value $2$. Indeed the set of algebra law has 8 dimensions, there are 5 equations, and 1 dimension of symmetry. One should be able to explicitly provide a family in dimension 2. And also provide explicit examples in dimension 3. $\endgroup$ – YCor Dec 6 '18 at 13:24

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.

  • $\begingroup$ 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 ) $\endgroup$ – aglearner Jun 9 '12 at 11:23
  • $\begingroup$ There is a book by Springer $\endgroup$ – Bruce Westbury Jun 9 '12 at 11:37
  • $\begingroup$ The dimensions are 2,5,8,14,26. $\endgroup$ – Bruce Westbury Jun 9 '12 at 11:38
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    $\begingroup$ What is the name of the book? $\endgroup$ – aglearner Jun 9 '12 at 11:59

An absolute valued algebra is a non-zero real algebra $A$ with a multiplicative norm $\|\|:$, i.e. $\|ab\|=\|a\|\|b\|$ for all $a,b\in A$. This condition implies the condition cited in the question. Thus absolute valued algebras are a subclass of the class of algebras in question.

It is known that finite-dimensional absolute valued algebras exist only in dimensions 1, 2, 4 and 8 (Albert, 1947). The same paper shows that their norm is Euclidean. The only unital such algebras are, up to isomorphism, the real numbers, complex numbers, quaternions and octonions. But if the existence of a unity is not required, then there are uncountably many algebras in each of dimension 4 and 8.

In dimension 8, the classification problem is far from being solved, in the sense that there exist no list that contains precisely one algebra from each isomorphism class. Partial classifications and my research experience indicate that the classification problem is hard.

Thus, a fortiori, there is no classification (in this sense) of the algebras about which this question asks.

To add some context on finite-dimensional absolute valued algebras: a finite-dimensional real algebra is absolute valued if and only if it is both a division algebra and a composition algebra.


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