# Hopf-algebras in associative ring spectra

I'm interested in a definition of cocommutative Hopf-algebra objects in the $\infty$-category of associative (read: $A_\infty$) ring spectra. One thought I had was to think of cocommutative Hopf-algebras in this setting as functors $H:Alg(\mathbb{S})\to E_\infty Spc$, in other words, functors from associative algebras that land in the $\infty$-category of infinite loop spaces. This, in my mind, generalizes the notion of what one does to produce a cocommutative cogroup object in algebra. One of the things I'd like to get from this definition is that if we have a one-old loop space $\Omega X$, then its suspension spectrum should be such a functor (with the cogroup structure coming from the diagonal on $\Omega X$ and the algebra structure coming from the fact that it's a loop space). However, from a bit of discussion in the homotopy theory chat room, it seems that this is not attainable by the above given definition.

Another idea suggested in the homotopy theory chat room was the notion of some kind of cosimplicial object in associative algebras. Embarassingly, this also sort of trips me up, since cosimplicial objects look more like monoids than comonoids to me (e.g. we take a space which is naturally a comonoid and apply the cobar construction to get a monoid, a.k.a. loops on that space).

Do people have workable notions for this idea, or definitions that they know are written down anywhere? One can do relatively straightforward things in $E_\infty$-rings because the tensor product is the coproduct, but this is not the case for associative ring spectra.

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So, okay there are clearly some things wrong with what I've said above, and some bad intuition (e.g. my confusion about the cosimplicial structure, and about the functorial point of view for associative algebras) but I would still very much like to know if anyone has a good working definition for Hopf-algebras in associative ring spectra and moreover whether or not this includes the sort of example I give above.

• A cocommutative Hopf algebra is not a cocommutative cogroup object in algebras even in the ordinary setting, precisely because the tensor product of associative algebras is not the coproduct. What is true is that a commutative Hopf algebra is a cogroup object in commutative algebras, and that a cocommutative Hopf algebra is a group object in cocommutative coalgebras. – Qiaochu Yuan Nov 14 '14 at 2:31
• Point taken. It's true that I was just sort of generalizing willy nilly from commutative algebra which, as you say, doesn't make any sense. – Jonathan Beardsley Nov 14 '14 at 2:39
• You can talk about algebras or coalgebras with respect to symmetric monoidal products other than the categorical product or coproduct, so I don't think this vitiates his comments. Case in point: the free R-module functor from sets to R-modules sends a group in sets to a group w.r.t. tensor in R-Mod. – Robert Bruner Nov 14 '14 at 13:25
• In fact, products/coproducts w.r.t. categorical product/coproduct trivialize. This is one reason that people often don't notice that groups are actually Hopf algebras in (Sets, x): there is only one possible coproduct w.r.t. the categorical product, so the monoid structure and inverse are all you have to check. – Robert Bruner Nov 14 '14 at 13:28
• @Robert: yes, of course you can talk about monoid or comonoid objects with respect to an arbitrary monoidal structure (not necessarily symmetric, even). What you cannot do is talk about group objects or cogroup objects via a functor of points taking values in groups; that requires that your monoidal structure be the categorical product or coproduct respectively. (And the reason I don't like that definition of groups is that it isn't compatible with the functor of points perspective on what a group object is.) – Qiaochu Yuan Nov 14 '14 at 16:53

Okay, I'll take a crack at it, although this is basically just a translation of Tyler's comments from the chat room. Recall that given a topological group (or group like $A_\infty$-space) $A$, we can form the bar complex associated to $A$, $BA$ such that $\Omega BA\simeq A$. $BA$ is what is commonly known as the delooping of $A$. It is a defining characteristic of algebra objects that we can form this delooping. Moreover, the simplicial object of which this delooping is the geometric realization satisfies a special property often called a Segal condition. Intuitively, this condition says that at level $n$, this simplicial object looks like the $n$-fold tensor product (in whatever way tensor product makes sense to us) of the first level of the simplicial object. Note also that we're taking the tensor product relative to the zeroth level of the simplicial object. As a result, if we were interested considering group-objects in some category it would suffice to look at simplicial objects satisfying the Segal condition whose zeroth level is a terminal object (if we don't have this last condition, we would be looking at group-oid objects).

Now, dualizing, we should say that an object is a coalgebra if it admits a looping which we model with the cobar complex (which we can always construct given a comonoidal structure). And the cobar complex satisfies a suitable coSegal condition. At first glance this seems kind of silly, since we can ALWAYS loop a space. But I claim that this is precisely because spaces are ALWAYS coalgebras! This of course is just by using the diagonal map.

So, a coassociative comalgebra in $A_\infty$-ring spectra should be precisely the data of a cosimplicial object in $A_\infty$-rings satisfying a coSegal condition. I think that just composing the suspension spectrum functor with the cosimplicial space defining the comonoidal structure on a given space $X$, we can obtain that, using this definition, every space yields a coalgebra in spectra. What's more, if this space is an $n$-fold loop space, then this functor can be lifted to $E_n$-ring spectra.

Now, we haven't touched the issue of cocommutativity at all. For the extent of this answer, we'll just have to take on faith that the right place for cocommutative coalgebras of symmetric monoidal $\infty$-category $C$ to live is $Alg_{E_\infty}(C^{op})$. The motivation for one to believe this is essentially section 5.2 of Lurie's "Higher Algebra" where it is shown that this is the place that the iterated cobar construction lands (producing a Koszul dual monoid to a commutative algebra) and that the iterated bar construction gets us back here.

So, the point is that, all things considered, I believe that the right definition of a "cocommutative Hopf-algebra in associative ring spectra" is precisely an $E_\infty$-algebra in $(Alg(Spectra)^{op})$.

Now, how do we know that the suspension spectrum of a loop space satisfies this? Well, first we need to notice that the suspension spectrum functor lands in $Alg(Spectra)$, as expected. Next, we need to notice that it is a monoidal functor with respect to the smash product in both categories, so there is a suitable monoidal op-ification of it (living inside the category of infinity categories) which takes commutative monoids in the opposite category of loop spaces (which, by the way, is ALL OF THEM), to commutative monoids in the opposite category of associative ring spectra.

• Actually this answer is rife with mistakes and misunderstandings. I'll try to update it soon... – Jonathan Beardsley Nov 17 '14 at 16:10