The category $Mod^{E\_n}\_A(\mathcal{C})$ of $E\_n$-modules for an $E\_n$-algebra in a symmetric monoidal $\infty$-category $\mathcal{C}$ is defined in Lurie's Higher Algebra as a special case of a more general definition (just replace the little $n$-cubes operad with your favorite $\infty$-operad). The definition is rather obscure, however, and requires a lot of terminology I'm not entirely comfortable with. I was wondering if there is a more intuitive way to define/think about this category. For example, what's the best way to think of $E\_n$-modules for an $E\_n$-algebra in the category of spaces?


There is also a proof in Lurie that $Mod^{E\_1}\_A(\mathcal{C})$ for an $E\_1$-algebra $A$ is equivalent to the category of bimodules for $A$ in $\mathcal{C}$, but again the proof is quite long and difficult. Is there an intuitive way to think about this result? I'd also be happy if there was a short proof for $A$ an ordinary associative algebra. 



Definition of $E_n$-modules for an $E_n$-algebra
$E\_n$ algebras have compatible multiplications for every way of placing a bunch of elements into a collection of balls in $\mathbb{R}^n$. A module for an $E\_n$ algebra has an action for every way of placing a bunch of elements of the algebra into balls not at the origin, and an element of the module into a ball at the origin.


Here's what this looks like in a few examples to help you gain some intuition.


* $E\_1$ algebras in vector spaces are ordinary (non-commutative) algebras. An $E\_1$-module for $A$ is just an $A$-$A$ bimodule. I.e. there's actions for every way of placing another ball not at the origin (so either to the left or right) and you get an associativity axiom for every way of placing two balls not at the origin (both left is the left module axiom, both right is the right module axiom, and one on each side is the bimodule axiom).
* $E\_2$ algebras in vector spaces are commutative algebras (because you can continuously deform two discs in the plane into the opposite order, so $xy = yx$). An $E\_2$-module for $A$ is just an $A$-module (which is also automatically a right $A$-module by using commutativity to turn the left action into a right action).
* $E\_1$ algebras in categories are monoidal categories. $E\_1$-modules for monoidal categories are bimodule categories in the sense of [Chapter 7 of tensor categories](http://www-math.mit.edu/~etingof/egnobookfinal.pdf). I.e. you have bifunctors $\vartriangleright: A \times M \rightarrow M$ and $\vartriangleleft: M \times A \rightarrow M$ together with natural associators for each triple product (e.g. $\alpha\_{a,b,m}: a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m)$ satsifying some hexagon axioms for quadruple products.
* $E\_2$ algebras in categories are braided monoidal categories. Both horizontal and vertical composition are the monoidal structure, but the braiding tells you how 180-degree rotation gives a braiding iso $\beta\_{x,y}: x \otimes y \rightarrow y \otimes x$ satisfying a hexagon axiom for triple tensor products. Note that you can either rotate clockwise or counterclockwise, yielding the braiding and its inverse, but these need not agree. Now let's think about $E\_2$ modules. The action comes from a pair of discs, one of which is a marked disc at the origin. I.e. we want a left module category $M$. The important isotopy here is no longer the swap, but instead a full rotation around the marked disk (i.e. the generator of the affine braid group on one strand). So we need a full-twist isomorphism $\eta\_{a,m}: a \vartriangleright m \rightarrow a \vartriangleright m$. But looking at three discs, this needs to satisfy a compatibility axiom related to the defining relation of the 2-strand affine braid group which states that $\eta\_{a\otimes b,m}: (a \otimes b) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$ is equal to the composite:
$$(a \otimes b) \vartriangleright m \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m \rightarrow (b \otimes a) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$$
where the first and fourth maps are associators and the second and third maps are the full-twists $\mathrm{id}\_a \vartriangleright \eta\_{b,m}$ and $\eta\_{a, b\vartriangleright m}$ (or possibly vice-versa), and the fifth and sixth maps are both the braiding (or possibly both the inverse braiding). This structure is called a "braided module category" by [Enriquez](https://arxiv.org/abs/math/0408035) (see also [Brochier](https://arxiv.org/abs/1209.0417)) and the translation between $E\_2$-modules and braided module categories is given in Theorem 3.11 of [Ben-Zvi-Brochier-Jordan](https://arxiv.org/abs/1209.0417).
* $E\_3$ algebras in categories are symmetric monoidal categories. I.e. braided tensor categories where the braiding and the inverse braiding agree. Similarly, for $E\_3$-modules the full-twist maps are automatically trivial by untying them in the extra dimension. So an $E\_3$-module for a symmetric tensor category is just a module category.


(Note that in vector spaces $E\_2$-algebras are $E\_\infty$-algebras, while in categories $E\_3$-algebras are $E\_\infty$-algebras. So in both cases we see that $E\_\infty$-modules for $E\_\infty$-algebras agree with the usual notion of left-module = right-module, but below the stable range the notion is subtler and is a bimodule with the appropriate amount of compatibility between the left and right module structures.)


