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](http://arxiv.org/pdf/0910.0932v2.pdf), 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. 



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


