Timeline for A classification of non-associative algebras with a norm?
Current License: CC BY-SA 4.0
6 events
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| Dec 6, 2018 at 17:39 | history | edited | S. Carnahan♦ | CC BY-SA 4.0 |
Corrected length condition, revised O(n) discussion following comment.
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| Dec 6, 2018 at 13:24 | comment | added | YCor | 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. | |
| Jun 12, 2012 at 11:20 | vote | accept | aglearner | ||
| Jun 11, 2012 at 0:29 | comment | added | S. Carnahan♦ | 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 | |
| Jun 10, 2012 at 19:55 | comment | added | aglearner | 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? | |
| Jun 10, 2012 at 11:36 | history | answered | S. Carnahan♦ | CC BY-SA 3.0 |