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This is perhaps an ill-proposed question. Any way, thank you guys.

We have a lot of bi-stuffs, such as bi-module, bi-bundle, etc. They are basically two commuting actions from two sides, left and right. This is somehow related to we are writing horizontally. If we are writing vertically, people may talk top module, under module.

I am wondering if we have three (or 4, any other) actions on an object by three different things. Imaging We have a triangle, acted on tree side. Is it possible to keep them compatible in a reasonable nice way? The uninteresting way is they are mutually commutative. If one can define an "tri-module", is it just un-useful toy?

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  • $\begingroup$ The actions should be thought of as being covariant and contravariant acting on the same side and commuting. A higher number of "directions" than two would probably appear in higher category theory using the higher opposites. $\endgroup$
    – Harry Gindi
    Commented Sep 16, 2011 at 10:00
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    $\begingroup$ For instance, let G be a 2-group. Then a top-left G-action on a category C is just a 2-functor $G->Cat$ whose underlying category is C. A top-right G-action is a 2-functor $G^op -> Cat$ whose underlying category is C. The bottom-left and bottom-right actions correspond respectively to $G^{co}$ and $G^{co,op}$. $\endgroup$
    – Harry Gindi
    Commented Sep 16, 2011 at 10:05
  • $\begingroup$ (By the way, I am identifying G with its classifying 2-category BG). $\endgroup$
    – Harry Gindi
    Commented Sep 16, 2011 at 10:17
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    $\begingroup$ The answer to your first question is "yes". The theory of operads gives you a set of tools for defining actions with direction. For example, an $E_n$ module of an $E_n$ algebra has a space of actions parametrized by the sphere $S^{n-1}$. When $n=1$, you get bimodules. $\endgroup$
    – S. Carnahan ♦
    Commented Sep 16, 2011 at 11:21
  • $\begingroup$ Bi-the-way, not everything is bi... Dilogarithm, for instance. And trilogarithm is much more than a toy. $\endgroup$
    – Wadim Zudilin
    Commented Sep 16, 2011 at 11:44

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If you work with a single associative algebra $A$, then there is not so much sense to try to define a notion of $A$-tri-module. Namely, $A$-bimodules appear naturally as operadic $A$-modules.

To give you a picture, an associative algebra $A$ is in particular an algebra over the colored operad given by intervals of the real line and their disjoint union inclusions (a-k-a prefactorization algebras on the real line).

It has the property to be locally constant: the space associated to any interval is the same ($A$ itself).

Let me now relax a bit this locally constant property. Assume that you have such a (pre)factorization algebra over the real line such that it is locally constant outside the origin. Then you will get two associative algebras (one from $(0,+\infty)$ and the other from $(-\infty,0)$) together with a bunch of bimodules (One for each interval containing the origin).

There is nothing against the idea of having a factorization algebra on the graph consisting of three half-edges attached to a single vertex. Things like this might have appeared in $2d$ lattice models.

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Here's a "low-level" way answer your question. As you've very correctly pointed out, a bimodule is simply a module with two commuting actions. More precisely, if $R$ and $S$ are rings and $M$ is an $(R,S)$-bimodule (by which I mean it has a left $R$-action and a right $S$-action), then one can equivalently view $M$ as a left $R \otimes _{\mathbb{Z}} S^{op}$-module. Here $S^{op}$ is the opposite ring of $S$. The action is defined by

$(r \otimes s^{op}) \cdot m := rms \in M$

From this perspective, one could think of a "trimodule" as a left $R \otimes S \otimes T$-module. This just means that we have commuting actions of the three rings $R$, $S$, and $T$ on a module, as you indicated. Granted, by deleting the "opposite" construction, we lose the left-to-right distinction. I wouldn't know of an analogue of this construction to have rings acting on three different "sides" of a module. But perhaps the operadic perspective also mentioned here accomplishes this? If so, I'd be very interested if someone could illustrate exactly how that works.

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  • $\begingroup$ If we forget the directions, i.e. turn all to left action, then the question may become how to get a nice 'tensor' from more than two rings? $\endgroup$
    – Ma Ming
    Commented Sep 16, 2011 at 14:08

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