reference for list of left-regular representations of real associative algebras

Question

Suppose $\mathcal{A}$ is a unital associative algebra over $\mathbb{R}$. If we identify $\mathcal{A} = \mathbb{R}^n$ then the $\mathcal{A}$ multiplication corresponds to particular linear maps on $\mathbb{R}^n$. Of course any linear map on $\mathbb{R}^n$ corresponds uniquely to its standard matrix hence we obtain a correspondence between vectors in $\mathcal{A}$ and matrices in $\mathbb{R}^{n \times n}$. These square matrices are known as the left regular representation of the algebra. This is not unique unless we add additional data about the correspondence of the $n$-dimensional algebra and its presentation on $\mathbb{R}^n$.

My favorite examples, $\mathbb{C} = \mathbb{R}^2$ is naturally identified with the subalgebra of $2 \times 2$ matrices of the form:

$$ \left[\begin{array}{cc} a &-b \\ b &a \end{array} \right] $$

Or the hyperbolic numbers $\mathbb{R}+j\mathbb{R}$ identified with

$$ \left[ {\begin{array}{cc} a & b \\ b & a \end{array}} \right] $$

Or the direct product of $\mathbb{R}$ with $\mathbb{R}$ the element $(a,b)$ is identified with

$$ \left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right] $$

Up to isomorphism the last two examples are actually the same example. I know of one other two-dimensional algebra up to isomorphism.

Question: where can I find a complete tabulation of the low-dimensional left-regular representations of unital algebras?

I have found many results on google and here, but I can't find one which stands out as a continuation of the list I began at the start of this post. In particular, complete list of complex associative algebra, is great except the base-field is $\mathbb{C}$. If there was a simple theorem that allowed me to extract the list I desire from that list then that would also be a useful answer. But, I'd rather have direct reference for a list of the real associative left regular representations. Ideally this will help me choose a good notation if there already is an agreed notation accepted among those who worked on such classifications.

As always the help of the MO community is greatly appreciated.

Answer 1

Let $A$ be an algebra of dimension $d$ over an alg. closed field $k$, and $r\subseteq A$ its radical.

• If $d=1$, then of course $A\cong k$.

• If $d=2$, then either $\dim r=0$ and then $A\cong k^2$ because of Wedderburn's theorem, or $\dim r=1$. In the latter case, we must have $r^2=0$, so the ordinary quiver $Q$ of $A$ is a loop. The only $2$-dimensional admissible quotient of $kQ$ is $k[X]/(X^2)$.

• Suppose $d=3$. Since $A/r$ is semisimple and $k$ algebraically closed, Wedderburn tells us that $\dim A/r$ is a sum of squares; it is at most $3$, the only possible square is which fits is $1$. If $\dim A=\dim A/r=1+1+1$, then $A\cong k^3$; if $\dim A=1+1$, then $\dim r=1$ and $r^2=0$ (because $r$ is nilpotent) and the ordinary quiver $Q$ of $A$ is then $\bullet\to\bullet$ or that of $k[X]/(X^2)\times k$. In the first case, since $kQ$ is $3$-dimensional, we must have $A\cong kQ$, which is in fact the algebra $T$ of $2\times 2$ upper triangular matrices. In the second case, $A\cong k[X]/(X^2)\times k$.

• Finally, suppose $d=4$. Weedderburn's theorem, as above, tells us that either $A/r\cong M_2(k)$, in which case in fact $A\cong M_2(k)$, or $A/r\cong k^s$ with $s\leq 4$.

  • If $s=4$, of course $A\cong k^4$.

  • If $s=3$, then $\dim s=1$, the ordinary quiver $Q$ has only an arrow, and counting dimensions we see that $A\cong kQ$: if the arrow is a loop, we have $A\cong k[x]/(x^2)\times k^2$, and if it is not a loop, we habe $A\cong T\times k$.

  • If $s=2$, the quver $Q$ has two vertices. If $r^2\neq 0$, we have one arrow which does not square to zero, and this is only possible if $A=k[X]/(X^3)\times k$. If instead $r^2=0$, we have two arrows in $Q$ and $A$ is the quotient of $kQ$ by the square of the arrow-ideal: this gives six possibilities which are inbijection with the possible quivers.

  • if $s=1$, the quiver has one vertex. It has $\ell=\dim r/r^2$ loops, with $1\leq\ell\leq3$ If $\ell=1$, then there is only one loop and the only admissible quotient is $k[X]/(X^4)$. If $\ell=3$, then we have three loops and all products of arrows vanish, so $A\cong k\langle x,y,z\rangle/(x^2,y^2,z^2,xy,yx,xz,zx,yz,zy)$. If $\ell=2$, we have two arrows in $Q$, call them $x$ and $y$, and the products $x^2$, $xy$, $yx$, $y^2$, being in $r^2$, are all multiples of a fixed non-zero element of $A$. Using this one can complete the list —here, for the first time, we get a non-discrete family.

Answer 2

(Too long for a comment). More (modern and not-so-modern) references, some of them may (partially) contain list(s) you are interested in:

• A.A. Albert, Structure of Algebras, AMS, 1939: on page 172 discusses classification of 4-dimensional algebras.

• S.C. Althoen, K.D. Hansen, L.D. Kugler, A survey of four-dimensional C-associative algebras, Algebras Groups Geom. 21 (2004), N1, 9-27

• R. Ballieu, Anneaux finis; systèmes hypercomplexes de rang trois sur un corps commutatif, Ann. Soc. Sci. Bruxelles Sér. I. 61 (1947), 222-227: presumably contains classification of complex 3-dimensional algebras.

• W.A. de Graaf, Classification of nilpotent associative algebras of small dimension, arXiv:1009.5339.

• D. Happel, Klassifikationstheorie endlich-dimensionaler Algebren in der Zeit von 1880 bis 1920, Enseign. Math. 26 (1980), 91-102 DOI:10.5169/seals-51060 : A nice historical survey from the modern viewpoint, with a large bibliography.

• O.C. Hazlett, On the classification and invariantive characterization of nilpotent algebras Amer. J. Math. 38 (1916), N2, 109-138 http://www.jstor.org/stable/2370262

• G. Pickert, Dreidimensionale assoziative nichtkommutative Algebren, J. Algebra 234 (2000), N2, 280-290 DOI:10.1006/jabr.2000.8550

• Scorza, Atti Acad. Sci. Fis. Mat. Napoli 20 (1935), N13 and N14: Classification of 3- and 4-dimensional algebras.

• D.A. Suprunenko and R.I. Tyshkevich, Commutative Matrices, Acad. Press, 1968 (translation from Russian): On p.61 (of the Russian edition) there is a discussion of commutative nilpotent algebras of dimension 5.

Answer 3

I'm not sure how far he got, but it seems that Benjamin Peirce in his 1882 book "Linear Associative Algebra" made the first attempt at classifying associative algebras of low dimensions. The book is available free at http://archive.org/details/linearassociati00peirgoog