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Is there a well-defined notion of "associativity" when discussing external binary operators? From all the sources I have looked into, it seems that there in fact does not exist such a notion outside the realm of (internal) binary operators; however, this conclusion does not leave me satisfied.

The origin of this question stems from the definition of $R$-modules. In particular, one of the axioms (as listed in Dummit and Foote) states that $(rs)m=r(sm),$ for all $r\in R,m,n\in M.$ This appears to be as close to a generalization to associativity I feel we can get when discussing external binary operators. Additionally, it is this condition which directly corresponds to the definition of associativity when discussing the natural $R$-module structure on $R$ (where scalar operator defines the multiplicative operator).

I appreciate any help I can get. Thanks all.

J.G.131
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Sure. If $\ast$ and $\circ$ are two binary operators we want to make sense of the associativity condition

$$a \ast (b \circ c) = (a \ast b) \circ c.$$

So let's just work through what the domains and codomains of $\ast, \circ$ have to be in order for this to make sense. Let's take $a, b, c$ to have values in three sets $A, B, C$. We see from the LHS that $\circ$ must have domain $B \times C$, and from the RHS that $\ast$ must have domain $A \times B$. Furthermore we have to be able to plug the values of $\circ$ into $\ast$ on the right and vice versa, which gives the codomains

$$\ast : A \times B \to B$$ $$\cdot : B \times C \to B.$$

So associativity can involve three sets $A, B, C$ in which $A$ acts on $B$ from the left and $C$ acts on $B$ from the right, and these actions commute. This actually does happen in mathematics: for example $A, C$ can be two rings and $B$ can be what is called a bimodule, which means it has a left $A$-module structure and a right $C$-module structure, and these actions commute, which explicitly works out to the associativity condition above. There are also many examples in which $A$ and $C$ are groups.

Qiaochu Yuan
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  • We could also introduce a fourth operation, each with suitable target spaces, so that the equation $(ab)c=a(bc)$ makes sense. And indeed as you say, such associativity conditions arise pretty naturally (one example being dealing with vector(bundle)-valued forms as I outline here). – peek-a-boo Aug 13 '24 at 22:16