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The monoid algebra $R[M]$, for a commutative ring $R$ and a monoid $M$, can be described as the free $R$-algebra on $M$. We think of $R[M]$ as the set of finite formal sums of elements of $M$ with coefficients in $R$, where we use $M$ for multiplication. In the most familiar case, $M$ is the free commutative monoid on a set $X$ (written multiplicatively), and $R[M]$ is the algebra of polynomials over $R$ in the indeterminates $X$. If $U$ is the forgetful functor from $R$-algebras to monoids which takes an algebra to its multiplicative monoid, $M \mapsto R[M]$ appears to be a left adjoint to $U$.

More generally, there is a concept of the category algebra or categorical algebra $R[C]$ for a category $C$. For example, choose $n \ge 1$ and consider the category $C$ whose objects are $\{0, 1, ..., n - 1\}$, with a single morphism from $i$ to $j$ for each pair $(i, j)$. An order-$n$ matrix over $R$ is a formal sum of these morphisms with coefficients in $R$. Regarding this matrix as a map from Mor$(C)$ to $R$, we can define the product of two matrices $\alpha$, $\beta$ by the convolution formula

$$(\alpha \beta)(m) = \sum_{m_0 m_1 = m} \alpha(m_0) \beta(m_1)$$

So $R[C]$ is the familiar matrix algebra. If instead we choose $C$ to be the totally ordered set with $n$ elements, $R[C]$ is the algebra of upper triangular matrices.

Question 1: What's the most general definition of a category algebra?

For example, suppose $U$ is the forgetful multiplicative-monoid functor from $R$-algebras to the category of all small categories. Is there a left adjoint to $U$ in this case?

We need some finiteness condition in order to define $R[C]$. If we require the formal sum to be finite, we will not have a multiplicative unit when $C$ is infinite. But if the formal sums can be infinite, the product of two elements may not be defined.

The matrix example above works because $C$ is finite. Fei Xu seems to be taking this approach in the paper A Mini Course on Category Algebras. This suggests that a suitable left adjoint can be defined in that case.

When $C$ is a partially ordered set, the category algebra is called an incidence algebra, and the usual requirement is that $C$ is locally finite: each morphism can be written as a product of two other morphisms in only a finite number of ways.

However, these approaches fail to include simple monoid algebras such as group algebras over an infinite group. My best guess is to define $R[C]$ as the set of formal sums $\alpha = \sum r_m m$ such that the morphisms $m$ involve only a finite number of objects. That is, for each $\alpha$, the set $\{m: r_m \neq 0\}$ of morphisms has a finite number of domain objects and a finite number of codomain objects. This does give us an algebra, but it's not clear to me that it has the required universal property.

Question 2: What other interesting algebras arise from this construction?

Presumably there are other matrix subalgebras. How can we describe those? And what about other small categories?

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This is not what you are looking for, but a suggestion which is probably a more natural generalization of monoid algebras.

There is a forgetful functor from $R$-algebras to monoids, which forgets the $R$-module structure. The left adjoint is the monoid algebra $R[-]$.

We want to categorify this.

There is a forgetful $2$-functor from $R$-linear categories to categories, which forgets the $R$-module structure on the hom-sets (the case of one-object categories is precisely the functor above). It has a left adjoint, mapping $C$ to $R[C]$, which is an $R$-linear category with the same objects as $C$, but hom-modules $\hom_{R[C]}(x,y) = $ free $R$-module on $\hom_C(x,y)$. The composition is the $R$-linear extension of the one in $C$.

It is not really natural to make an $R$-algebra out of an $R$-linear category.

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Maybe this article will be of interest for you:

Zbl 1224.16036 Khripchenko, N.S. Finitary incidence algebras of quasiorders. (English) Mat. Stud. 34, No. 1, 30-37 (2010).

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  • $\begingroup$ Which can be found here. Thanks! $\endgroup$
    – Hew Wolff
    Commented Dec 1, 2013 at 16:28
  • $\begingroup$ No, this is other paper. $\endgroup$ Commented Dec 1, 2013 at 16:31
  • $\begingroup$ I can send it, if it is of interest to you. $\endgroup$ Commented Dec 1, 2013 at 16:39
  • $\begingroup$ Oh, I think I see it here. $\endgroup$
    – Hew Wolff
    Commented Dec 1, 2013 at 16:39
  • $\begingroup$ Yes, this is it. $\endgroup$ Commented Dec 1, 2013 at 17:04

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