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Let $R$ be a commutative ring. Let $\mathcal C$ be a category that has finitely many objects. The category algebra $R[\mathcal C]$ of $\mathcal C$ consists of finite sums $\sum a_i f_i$, where $f_i$ are morphisms of $\mathcal C$ and $a_i$ are elements of the ring $R$, with a multiplication operation $$ \sum a_i f_i \cdot \sum b_j g_j =\sum a_ib_j f_i\circ g_j $$ by using the composition $\circ$ in the category $\mathcal C$.

On the other hand, a multicategory $\mathcal D$ consists of objects, arrows, a composition operation, and identities, just like an ordinary category, the difference being that the domain of an arrow is not just a single object but a finite sequence of them.

Question: Could we imitate the construction of category algebras to define certain "multicategory algebra" $R[\mathcal D]$? If this sounds like a natural generalization, it seems strange to me since I can't find any reference. So, I'm afraid I'm overlooking something.

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The category algebra is best understood in terms of the category algebroid: the free $R$-linear category generated by $C.$ The algebra just adds a bunch of zero products for non-composable morphisms because algebroids make some people nervous. You can generate an $R$-linear multicategory from a multicategory just as from a category, but it’s not very obvious how to make it into an $R$-algebra. Consider multimorphisms $f:(a,a)\to b$ and $g:(c)\to a.$ Then there’s no natural choice of $g\cdot f$ in your putative multicategory algebra. The problem is that multicategorical composition is just not a binary operation, not even a partial one.

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  • $\begingroup$ Thank you! Probably, we should assume the domains of multimorphisms are all different (in particular, we may exclude $(a, a)$ as you mentioned). Then, we may likely make it into an $R$-algebra, I guess. $\endgroup$
    – Hang
    Commented Nov 18, 2023 at 18:57
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    $\begingroup$ @Hang I guess. It’s not a very natural condition on a multicategory. Better to let the object you’re studying guide the construction rather than the other way around, I think. $\endgroup$
    – Kevin Carlson
    Commented Nov 19, 2023 at 0:35
  • $\begingroup$ Thanks for your answer again. I was wondering if the following idea might work, and would greatly appreciate your comment: First, we take the free module generated by the set of all multi-morphisms, denoted by F. But, F may be too large to have an unambiguous product. So, a naive approach is simply taking the quotient of F with respect to the following equivalent relation: we say two multi-morphisms are equivalent if they are compositions of the same set of multimorphisms, differing only in the positions where the compositions occur. But, I am not sure if this is well-defined. $\endgroup$
    – Hang
    Commented Sep 26 at 2:22

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