6
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ @user127776 E_n-modules are not the same as algebras under A. Rather, E_n-algebras in E_n-modules over A are equivalent to E_n algebras with an E_n-map from A. $\endgroup$ – Dylan Wilson Jul 19 '19 at 17:12
  • $\begingroup$ I see, you are right. I automatically assumed we are dealing with $E_n$ algebras in the category defined above. $\endgroup$ – user127776 Jul 19 '19 at 17:32
9
$\begingroup$

$E_n$ 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. 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 (see also Brochier) and the translation between $E_2$-modules and braided module categories is given in Theorem 3.11 of Ben-Zvi-Brochier-Jordan.
  • $E_3$ 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.)

$\endgroup$
3
  • $\begingroup$ My thanks to David Ben-Zvi for pointing out that my earlier answer was totally wrong, since I confused what $E_n$ modules with a different notion of modules over $E_n$ algebras (which generalize left modules, rather than generalizing bimodules). Hopefully this answer is truthier. $\endgroup$ – Noah Snyder Jul 19 '19 at 4:50
  • 1
    $\begingroup$ Nice answer! For examples and discussion of E2 (aka braided) module categories over braided tensor categories see arxiv.org/abs/1606.04769 $\endgroup$ – David Ben-Zvi Jul 19 '19 at 13:33
  • 1
    $\begingroup$ @DavidBen-Zvi: Thanks for the reminder, I've incorporated that reference into my answer. I really have no excuse for not remembering it was there, I was thinking last night I must have seen this definition somewhere, but got confused because I couldn't remember where it would be in the ENO literature or the subfactor literature. $\endgroup$ – Noah Snyder Jul 19 '19 at 15:31
6
$\begingroup$

$\newcommand{\E}{\mathbf{E}} \newcommand{\Mod}{\mathrm{Mod}} \newcommand{\cc}{\mathcal{C}}$Here's one way to think about $\E_n$-modules. Let $R$ be an $\E_n$-ring (in a presentable symmetric monoidal category $\cc$), and let $\Mod^{\E_n}_R(\cc)$ denote the category of $\E_n$-$R$-modules. There is a forgetful functor $\Mod^{\E_n}_R(\cc) \to \cc$, whose left adjoint is the free $\E_n$-$R$-module; then, $\Mod^{\E_n}_R(\cc)$ is equivalent to $\mathrm{L}\Mod_{\mathrm{Free}^{\E_n}_R(1_\cc)}(\cc)$. In the case $\cc = \mathrm{Sp}$, one way to see this is as follows: the image of the unit in $\cc$ (i.e., the sphere spectrum $\mathbb{S}$) is a compact generator of $\Mod^{\E_n}_R(\mathrm{Sp})$, so that category is equivalent to the category of left modules over the endomorphism ring of $\mathrm{Free}^{\E_n}_R(\mathbb{S})$ in $\Mod^{\E_n}_R(\mathrm{Sp})$. But this endomorphism ring is just $\mathrm{Free}^{\E_n}_R(\mathbb{S})$.

Returning to the general case, we're reduced to understanding $\mathrm{Free}^{\E_n}_R(1_\cc)$. This object is what's known as the "enveloping algebra"; it's denoted by $U_R$ in Section 3 of this paper. There, it's shown that $\mathrm{Free}^{\E_n}_R(1_\cc) \simeq \int_{S^{n-1}} R$ (factorization homology). The intuitive interpretation of this identification is that an $\E_n$-$R$-module (i.e., a left $\mathrm{Free}^{\E_n}_R(1_\cc)$-module) is an object of $\cc$ with a collection of commuting left actions of $R$, where these actions are parametrized by $S^{n-1}$.

When $n=1$, this factorization homology is easy to describe. Indeed, $S^{n-1} = S^0$ is just a disjoint union of two points, and then $\mathrm{Free}^{\E_1}_R(1_\cc) \simeq \int_{S^0} R = R\otimes R^{op}$. In other words, $\E_1$-$R$-modules (for an $\E_1$-algebra $R$ in $\cc$) are the same thing as left $R\otimes R^{op}$-modules, i.e., $R$-$R$-bimodules in $\cc$.

$\endgroup$
5
$\begingroup$

If you accept an intuition that comes from plain (non-$\infty$) operads, then this is relatively easy. For an $E_n$-algebra $A$, an $E_n$-module over $A$ is an object $M$ equipped with structure maps $E_n(k+1+l) \otimes A^{\otimes k} \otimes M \otimes A^{\otimes l} \to M$ that satisfy relatively obvious associativity and equivariance conditions if you draw pictures with trees.

For example with $n=1$, you have a configuration of intervals in $[0,1]$. In one of the intervals, you can plug an element of $M$, and in the others you plug elements of $A$. Up to homotopy, all that matters is the order of the intervals, so you get maps $A^{\otimes k} \otimes M \otimes A^{\otimes l} \to M$ for $k,l \ge 0$. The $A$'s on the left gets you a (homotopy) left $A$-module structure, and the $A$'s on the right get you a (homotopy) right $A$-module structure. The axioms for an $E_1$-module tell you that these actions are compatible with the $E_1$-algebra structure of $A$. If $A$ is an ordinary algebra, then these axioms are telling you that $(a \cdot a') \cdot m \simeq a \cdot (a' \cdot m)$, the same thing on the other side, and something like $(a \cdot m) \cdot a' \simeq a \cdot (m \cdot a')$, for example. If all these higher operations are trivial, then you really get an $(A,A)$-bimodule structure.

$\endgroup$
1
  • $\begingroup$ The previous version of this answer was a bit hasty: even if the algebra is strict, the module can be up to homotopy (just like an algebra with $d = 0$ can have a nontrivial $A_\infty$-structure I guess). $\endgroup$ – Najib Idrissi Jul 19 '19 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.