# is there a better classification of a desirable algebra?

Consider a finite-dimensional, associative algebra presented as follows: $$\mathcal{A} = e_1 \mathbb{R}\oplus e_2 \mathbb{R} \oplus \cdots \oplus e_n \mathbb{R}$$ with multiplication $*: \mathcal{A} \times \mathcal{A} \rightarrow \mathcal{A}$ defined by linearly extending $$\color{red}{ e_i *e_j = C_{ij}^ke_k \qquad \text{no summation over k intended}.}$$ For example, the direct product algebra $e_i*e_j = \delta_{ij}e_i$ fits this condition.

The clarity of the multiplication is partly from a good choice of notation. If I took a different basis then the multiplication could get muddled. For example, $\mathbb{R}^2$ with the commutative direct product algebra is described by $e_1 *e_1=e_1$ and $e_2*e_2=e_2$ and $e_1*e_2=0$. If I instead presented the multiplication by $f_1 = 2e_1+3e_2$ and $f_2=e_1+e_2$ then $f_1*f_1 = 5f_1-6f_2$.

Apparently I could be given an algebra where a proper change of notation would induce the $\color{red}{\text{desired property}}$. For the sake of this question, let us define $\mathcal{A}$ as desirable iff there exists some basis $\{f_1,\dots f_n \}$ for $\mathcal{A}$ such that $f_i *f_j = C_{ij}^kf_k$ where no summation is intended over $k$.

Question: is there a more standard term for a desirable algebra?

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To me this seems to be related to the terminology of having a multiplicative basis. – Julian Kuelshammer Sep 13 '12 at 7:09
@JulianKuelshammer that looks reasonable. However, I have yet to locate a good reference. A google search brought out many uses of the term... but, no origin. I have some reading to do. – James S. Cook Sep 14 '12 at 2:59

A multiplicative basis of an algebra $A$ is a basis $B\subseteq A$ such that for all $b_1$, $b_2\in B$ we have either $b_1b_2\in B$ or $b_1b_2=0$ —sometimes, this is extended a bit by allowing in the first possibility that $b_1b_2$ be a scalar multiple of an element of $B$. In any case, this is more general than the semigroup algebras that Bill mentions, because I am allowing here the second option.

This is a pretty standard notion that comes up in the study of finite dimensional algebras. There are lots of algebras with multiplicative bases, the path algebra of a quiver being a basic example. Moreover, multiplicative bases show up in very important results.

One example: it is a theorem that finite dimensional algebras of finite representation type have multiplicative bases; this is a very difficult result due to R. Bautista, P. Gabriel, A. Roiter and L. Salmerón. The converse is not true, and there are group algebras, for example of infinite representation type.

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Thanks! Do you happen to recall a text where this is stated? – James S. Cook Sep 17 '12 at 2:16
The four authors have a paper with this result, but right now I can't give a precise reference because my connection to MathSciNet is not working. This result is probably discussed in the book by Gabriel and Roiter on the representation theory of algebras. – Mariano Suárez-Alvarez Sep 17 '12 at 2:33

If an algebra $\mathcal{A}$ (over $\mathbb{F}$) has a basis $\beta=\{e_1,e_2,\dots\}$ such that for all $i,j$ we have $e_ie_j=e_k$ for some $k$, then $\mathcal{A}=\mathbb{F}[\beta]$ is a semigroup algebra. If $\beta$ contains an identity (so it's a monoid), then $\mathcal{A}=\mathbb{F}[\beta]$ is a monoid algebra. If $\beta$ also has inverses, then (of course) we have a good old group algebra.

As for the $\beta$ itself, like Julian, I would tend to call such a basis multiplicative. But I can't say that this is standard terminology. The term "multiplicative" is quite overused.

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