Questions about non-associative algebras other than Lie algebras.

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9
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3answers
488 views

Applications of Jordan algebras

Jordan algebras are non-associative algebras satisfying a somewhat strange (to me) list of axioms, see wikipedia. Basic examples are real symmetric and complex hermitian matrices with the product ...
2
votes
0answers
68 views

Is the generated subalgebra of a subset of pairwise operator-commuting element in a JB-algebra associative?

In a Jordan algebra elements $a$ and $b$ are said to operator-commute, whenever $a \circ (b \circ x) = b \circ (a \circ x)$ for every other element $x$. (That is: $T_aT_b = T_bT_a$, writing $T_x(y) = ...
0
votes
0answers
106 views

A question about index of the commutant in a Moufang loop

Let $M$ be a non-commutative Moufang loop and $C(M)$ be its commutant. I can prove that the index of $C(M)$ in $M$, $|M:C(M)|$, is greater than or equal to 4. Also, I can show that $|M:Z(M)|\geq 4$, ...
2
votes
0answers
126 views

Is there a system of quasigroup equations implying non-associativity?

I have read that if 4 quasigroup operations, $\cdot,\circ,\star,\square$, on a set $S$ respect the following equation: $$x\cdot (y\circ z) = (x \star y) \square z$$ for all $x,y,z\in S$, then all 4 ...
7
votes
2answers
261 views

Is there a cohomology for magmas?

Is there a cohomology theory for magmas? Or cohomology theories for any class of non-associative algebras (other than Lie and maybe Jordan)?
7
votes
1answer
213 views

Homotopes of simple Lie algebras

Let $\mathfrak{g}$ be a complex simple Lie algebra with bracket $[x,y]$. For which $z\in \mathfrak{g}$ does the formula $$ \mu(x,y)=ad (z)([x,y])=[z,[x,y]] $$ define another Lie bracket on the same ...
5
votes
0answers
183 views

Reference request: The relationship between norm and trace forms on an Albert algebra

I am interested in either a nice reference, or some clarification. Overview: I am considering $J_3(\mathbb{O})$, the Jordan algebra of $3\times 3$ self adjoint octonionic matrices. This algebra is a ...
11
votes
1answer
240 views

What is flexible about flexible algebras?

A possibly non-associative algebra is flexible if it satisfies the identity $$(xy)x=x(yx).$$ This is clearly a very weak form of associativity —and obviously an associative algebra is flexible— but it ...
2
votes
2answers
128 views

Any results or concise introduction about nonassociative algebra that even does not satisify Power associativity?

Any results or concise introduction about nonassociative algebra that even does not satisify Power associativity?
16
votes
1answer
819 views

The octonions on a bad day

We can define the algebra of quaternions $\mathbb H$ over any field $k$, and depending on the arithmetic of $k$ it is either a division algebra or a matrix algebra. We can also define the algebra of ...