# Irreps of this algebra?

Hello!

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?

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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

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.
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