The Modules over Algebras over Operads are not what they seem.

Question

Operads are a nice framework to model all kinds of different algebras, i.e.

• Monoids are algebras over the operad Assoc in the category of sets
• Associative algebras are algebras over the operad Assoc in the category of k-vector spaces
• Topological monoids are algebras over the operad Assoc in the category of topological spaces

Working with symmetric operads, the $j$-ary operations Assoc(j) in the operad Assoc are given by the "free thing in the base category" generated by the symmetric group in $j$ letters $\Sigma_j$, e.g.

• Assoc(j) in the category of sets is $\Sigma_j$ with the canonical $\Sigma_j$-action
• Assoc(j) in the category of $k$-vector spaces is the free vector space generated by $\Sigma_j$ with the canonical $\Sigma_j$-action
• Assoc(j) in the category of topological spaces is $\Sigma_j$ with the discrete topology and the canonical $\Sigma_j$-action

Mostly, there is also the notion of a module over an algebra, i.e.

• A module over a monoid $M$ is a $M$-set
• A module over an associative algebra $A$ is an $A$-module
• A module over a topological monoid $M$ is a space with an $M$-action

Here, we have the choice of either considering left- or right-modules.

How can we model the notion of a (left or right) module over an algebra in the language of operads?

Searching for "module over an algebra over an operad", I found the following definition here:

Let $O$ be a symmetric operad and $A$ an algebra over $O$. A module over $A$ is an object $M$ in the base category together with maps $$O(j)\otimes A^{j-1}\otimes M\rightarrow M,$$ which satisfies certain associativity, unitality and equivariance conditions.

At first glance, this seems suitable for my needs, but I get confused even plugging in the basic examples mentioned above. For a monoid $A$ considered as an algebra over the operad Assoc, I expected the corresponding modules to be left $A$-sets, so let $M$ be such a gadget. We are in the need of maps $$\Sigma_j\times A^{j-1}\times M\rightarrow M.$$ If I had to define this map just on $\Sigma_{j-1}$, the obvious choice would be to map $(\sigma,a_1,\ldots,a_{j-1},m)$ to $(a_{\sigma(1)}\ldots a_{\sigma(j-1)})\cdot m$, but the fact that the map should be defined on $\sigma_j$ suggests that we also should be able to multiply elements from $A$ to $M$ on the right, so I expect the corresponding notion to be something like a set with an action of $A$ on both sides, i.e. a "bimodule" instead of just a left (or right) module.

Can someone clarify my confusion?

Answer 1

The point here is that you don't have to precise the action of $\Sigma_j$

$f:\Sigma_j\times A^{j-1}\times M\rightarrow M$, it just a map which associates to an element $(\sigma,a_1,..,a_{j-1},m)$ an element $f(\sigma,a_1,...,a_{j-1},m)$ of $M$. You don't have to figure out the action $\sigma.(a_1,..,a_{j-1},m)$ here, the role of $\Sigma_j$ appears in the compatibility diagram:

for example $f(\sigma,a_1,a_2,a_3,m)=a_1(a_2(a_3.m))$ in the definition you don't precise how is acting $\Sigma_j$.

What May precises in the definition here is the action of $\Sigma_j$ in the compatibility diagram. In this case you take $j_1+j_2=3$, so you take elements $s_3\in \Sigma_3$, $s_1\in \Sigma_1, s_2\in \Sigma_2$ then for the first downarrow of the compatibility diagram you you associate to $(s_3,(s_2,a_3,a_2),(s_1,a_1),m)$ an element $(s_3,a_{32}',a_1,',m)$ where $a_{32}'$ is defined by $\Sigma_2\times A_2\rightarrow A_2$.

Answer 2

Yes, a "module over an algebra over $\mathtt{Ass}$" is indeed a bimodule over that associative algebra. There are plenty of ways to see this, you can find this statement in Loday and Vallette ' Algebraic Operads (an excellent book if you're interested in operads for that matter). Another way to see this is to look at the universal enveloping algebra of an algebra $A$ over $\mathtt{Ass}$, which is $U_{\mathtt{Ass}}(A) = A \otimes A^{op}$ (and left modules over this are indeed $(A,A)$-bimodules).

Besides, it's not really surprising – why would the left side be prioritized over the right side? There's nothing asymmetric in the definitions, either in the definitions of the operad $\mathtt{Ass}$ or in the definition of modules over an algebra over an operad.^*

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It is actually possible to recover left modules over $A$, but the theory is a bit ad-hoc (by this I mean that "left modules" isn't canonically associated to the associative operad like "modules" are associated to every operad). I've seen it in Horel's "Operads, modules and topological field theories", Section 3. Given an operad $\mathtt{P}$, consider an associative algebra $\mathtt{M}$ in the category of right $\mathtt{P}$-modules. Then one can define "$\mathtt{M}$-shaped $\mathtt{P}$-modules" from this data (they are actually a "relative" or "Swiss-cheese type" operad).

If you take the enveloping operad $\mathtt{P}[\cdot]$ (see Fresse's Modules over Operads and Functors for example), then you recover the classical notion of a module over an algebra over an operad, and $\mathtt{Ass}[\cdot]$-shaped $\mathtt{Ass}$-modules are bimodules over an associative algebra. But it's also possible to define an algebra $\mathtt{Ass}^+$ in right $\mathtt{Ass}$-modules, such that you get left modules over an associative algebra in this case.

One of the main interests of modules over an algebra over an operad is that you can define a (co)homology theory that takes an algebra $A$ over $\mathtt{P}$, a module over $A$ and spits out $H_*^{\mathtt{P}}(A;M)$, so in this way if you're looking at things from an operadic viewpoint, then bimodules over associative algebras is the natural thing to consider.

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^* I now realize that in the definition of May you quoted, the $M$ is always on the right-most end. It's actually a trick to save space; there is one such map for every position of $m \in M$. So if you're working in an algebraic context, for $p \in \mathtt{P}(n)$, $a_1, \dots, a_{n-1} \in A$, a and $m \in M$, then to define a module structure you need to define $p(a_1, \dots, a_i, m, a_{i+1}, \dots, a_{n-1})$ for all $0 \le i \le n-1$. In other words, there are $n$ different maps of the type: $$\mathtt{P}(n) \otimes A^{\otimes (n-1)} \otimes M \to M,$$ one for each position in which to plug in $M$. It's made more explicit in §12.3.1 of Algebraic Operads.