Given a Hopf-algebra $H$ (over a commutative ring), it is a classical fact that its category of (left) modules is monoidal, even if $H$ is not commutative. Given two left modules $M$ and $N$, we can form a new left module structure on $M\otimes N$ via the structure map $$H\otimes M\otimes N\overset{\Delta\otimes 1\otimes 1}\to H\otimes H\otimes M\otimes N\overset{1\otimes\tau\otimes 1}\to H\otimes M\otimes H \otimes N\to M\otimes N.$$


When working in a derived setting (let's assume $H$ is an object in a symmetric monoidal quasicategory $\mathscr{C}$), things can be slightly more complicated, and we should probably have $H$ with monoidal structure and comonoidal structure given by some operads like the little $n$-disk operads $\mathbb{E}\_n.$ It's basically formal when working in quasicategories to say that $H$ is an $\mathbb{E}\_n$-algebra with a compatible $\mathbb{E}\_m$-coalgebra structure, making it into an $\mathbb{E}\_n/\mathbb{E}\_m$-bialgebra in $\mathscr{C}$. We just say that $H$ is an $\mathbb{E}\_n$-algebra object in the quasicategory of $\mathbb{E}\_m$-coalgebra objects in $\mathscr{C}$.


It's known that, in general, given an $\mathbb{E}\_n$-algebra, the category of left modules over it is $\mathbb{E}\_{n-1}$-monoidal. This is why, for instance, left modules over a noncommutative ring (i.e. an $\mathbb{E}\_1$-algebra) are not monoidal at all. So my question is, to what extent can we perform the above trick in a "derived" way? Obviously it does not suffice to simply write down the structure map, since we need a whole lot of coherent data to write down a module structure now, but is there some other way to do it?


A good example would be, I think, the example of an $n$-fold loop space $X$. Any space, via the diagonal map, is an $\mathbb{E}\_\infty$-coalgebra. In fact there's an equivalence of quasicategories $CoAlg\_{\mathbb{E}\_\infty}(Top)\simeq Top$. So an $n$-fold loop space is definitely an $\mathbb{E}\_n$-algebra in $\mathbb{E}\_\infty$-coalgebras in $Top$. So, is the category of modules in $Top$ over $X$ somehow "more monoidal than it should be?" In general, how well does this type of thing work?



Monoidal structures on modules over derived coalgebras
I couldn't make the above answer work, so here's an approach explained to me by Rune Haugseng (of course any errors are entirely my own). Let $C$ be symmetric monoidal and $p\colon C^\otimes\to Fin\_\ast$ be the cocartesian fibration witnessing this. First notice that $CoAlg(C)^{op}\simeq Alg(C^{op})$ has a "pointwise" symmetric monoidal structure which is given by [HA.3.2.4.3](http://people.math.harvard.edu/%7Elurie/papers/HA.pdf#page=370), and therefore $CoAlg(C)$ also has a symmetric monoidal structure i.e. there is a cocartesian fibration $q\colon CoAlg(C)^\otimes\to Fin\_\ast$. Here I'm using the fact that $C^{op}$ has a symmetric monoidal structure induced by taking the fiberwise dual of $p$ in the sense of [this paper](https://arxiv.org/pdf/1409.2165.pdf). Similar considerations give a symmetric monoidal structure to $LCoMod(C)\simeq LMod(C^{op})^{op}$. Moreover, there is a cartesian fibration $LMod(C^{op})\to Alg(C^{op})$ by [HA.4.2.3.2](http://people.math.harvard.edu/%7Elurie/papers/HA.pdf#page=527) which gives a fiberwise cartesian fibration $LMod(C^{op})^\otimes\to Alg(C^{op})^\otimes$. By taking fiberwise duals again we get a fiberwise cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$. It's a little tricky, but one can check that this fiberwise cocartesian fibration satisfies the condition A.1.8 of [this paper](https://arxiv.org/pdf/1904.11312.pdf#page=40) and is therefore a cocartesian fibration. On a fixed fiber this says that if I've got a map of coalgebras $A\to B$ (or a finite list of maps of coalgebras) then I get a left adjoint functor $LCoMod\_A(C)\to LCoMod\_B(C)$ which takes the coaction $M\to A\otimes M$ to $M\to A\otimes M\to B\otimes M$ (this is the "opposite" of restriction of scalars). But now we've lifted it up to $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ so that it plays well with the monoidal structure, which we'll need.


Now suppose that $H$ is a bialgebra in $C$, i.e. $H$ is an algebra object in $CoAlg(C)\simeq Alg(C^{op})^{op}$ . In other words, $H$ is determined by a functor of $\infty$-operads $H^\otimes\colon Assoc^\otimes\to CoAlg(C)^\otimes$. Then we can pull back the cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ along $H^\otimes$ to obtain a cocartesian fibration $LCoMod\_H(C)^\otimes\to Assoc^\otimes$. There's a little bit of checking to do if you want to make sure that this pullback really is equivalent to $LCoMod\_H(C)^n$ over each $\langle n\rangle\in Assoc^\otimes$, but it's not too bad.


I should add that you can fully analyze the cocartesian morphisms in $LCoMod\_H(C)^\otimes$ and check that the $H$-coaction on the tensor product of two $H$-comodules $M$ and $N$ is indeed $M\otimes N\to M\otimes H\otimes N\otimes H\to M\otimes N\otimes H\otimes H\to M\otimes N\otimes H$.


