Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


I stumbled upon the following commutative, non-associative, three-dimensional algebra (with basis $\{A, B, C\}$):

$A\times A = 0$
$A\times B = A$
$A\times C = 2B$
$B\times A = A$
$B\times B = B$
$B\times C = C$
$C\times A = 2B$
$C\times B = C$
$C\times C = 0$

Is anything known about its irreducible representations? In particular, how many nonequivalent irreps do exist? What are their dimensions? Can we construct explicit representation matrices for a given irrep?

share|improve this question
Algebra over what ring? How exactly is "representation" defined if the algebra is non-associative? First observation: The algebra is commutative and $B$ is its neutral element. –  Johannes Hahn Jul 22 '11 at 20:15
Second observation: If you make this associative, then $0 = A^2 C^2 = A(2B)C=2AC=4B$. Since $B$ is the multiplicative identity, this means that $4x=0$ for any x in your ring. If you don't make it associative, then I don't know what you mean by a representation. –  David Speyer Jul 22 '11 at 20:52
More over, in an associative version $0=AAC=A2B=2A$ holds. Since $\lbrace A,B,C\rbrace$ is assumed to be a basis, the characteristic must be two. Therefore the algebra is just $R[A,C]/(A^2,AC,C^2)$ with whatever base ring $R$ is choosen to be. The simple modules of this are just the simple $R$-modules with $A$ and $C$ acting as zero. –  Johannes Hahn Jul 22 '11 at 22:03
To summarize the above comments: there is no (sensible) notion of representation for non-associative algebras, and therefore your questions doesn't make sense. –  André Henriques Jul 22 '11 at 22:42
I believe there is a notion of module over an algebra over an operad, or something along those lines, and I believe that specializing that definition to this case gives the following: if $A$ is a non-associative algebra (a magma internal to the category of $k$-vector spaces for some field $k$), then an $A$-module is a $k$-vector space $M$ equipped with a bilinear map $A \times M \to M$ satisfying... no axioms! Not a very interesting thing. –  Qiaochu Yuan Jul 23 '11 at 4:12

1 Answer 1

Your algebra is a Jordan algebra.

Armed with Jacobson's Structure and representations of Jordan algebras, for example, you will be able to find the sensible notion of representation of your algebra (there are, in fact, a couple of sensible notions...) Using that, and since the algebra is of dimension $3$, which is hopefully small, one can possibly describe the irreducible ones.

Of course, this is more or less arbitrary: you have to choose in what sense you want to represent your algebra—this is what the comments above are hinting at—but, well, looking at it as a Jordan algebra makes this choice for you.

share|improve this answer
A representation of a Jordan algebra $(\mathbb J,\cdot)$ is a map $\rho:\mathbb J\to \mathrm{End}(V)$ such that $\rho(A\cdot B)=\rho(A)\rho(B)+\rho(B)\rho(A)$ for any $A,B\in \mathbb J$. Another option is to require the map $\rho$ to land in the subset of self-adjoint operators. –  André Henriques Jul 23 '11 at 20:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.