Jon, there are also the theory of dendroidal sets and its variants, among which dendroidal complete Segal spaces

@SeanTilson so they have that book on wheeled infinity properads, which is probably a good thing for me to know about.

i think in general that uses the dendroidal set-up, which is also something i should learn more about

@DenisNardin do you have a reference for that? there's that thomason-may paper on operator categories right? is this in there?

I mean Clark's operator categories, that you can read about in his paper here: arxiv.org/abs/1302.5756

Basically the notion of operad over an operator category is a generalization of the notion of complete Segal spaces

(I think that May-Thomason's are called categories of operations)

Ah sure.

Hrmmmm... quasi-multicategories...? Anyone?

I.e. algebras for the "free monoid" monad on simplicial sets...?

Ah... I think I need to understand globular operads...

is there a name for a monoid in the category of commutative algebras over a ring?

lentic if you mean a monoid with respect to the product I think any commutative algebra has this structure

the product in commutative algebra is the product in R-modules which is also the coproduct

I mean with respect to the tensor product

that does not work either

cause the tensor product is the coproduct

and any object in a category with coproducts is uniquely a monoid with respect to the symmetric monoidal structure given by the coproduct

so I guess the name you are looking for is commutative algebra

I'm confused. Rigs can be characterized as monoids in commutative monoids...

@Geoffroy: I mean: Set is a symmetric monoidal category with the product. I can take a commutative monoid here, and then a monoid therein, and that's a rig. Why can't I do the same with the symmetric monoidal category of k-modules?

the categorical fact I have said is correct. This is a very easy exercise

I guess that characterization of a rig is wrong...

maybe

there is also the problem of units

do you mean unital monoid ?

yes

then what I have said is fine

so maybe the problem is with this characterization of rigs

I guess another way of seeing it is Eckmann-Hilton. A monoid in monoids is a commutative monoid. A commutative monoid in monoids is in particular a monoid in monoids...

A relevant difference between the category of commutative monoids and commutative algebras is that commutative monoids have a closed symmetric monoidal structure (the tensor product) and commutative algebras aren't closed symmetric monoidal - the tensor product is already the coproduct

and of course it's difficult to imagine an internal hom of commutative rings

oh the above characterization of a rig is right, if interpreted correctly. It's a monoid in commutative monoids all right, but where commutative monoids have the monoidal structure the tensor product of commutative monoids, not the cartesian one.

yes, any commutative monoid has a unique structure of commutative monoid for the cartesian product

since as Geoffroy said, the product is the coproduct

but we don't have to take that one when we consider rigs, we take the tensor product (a similar one to the one in abelian groups)

yes

so, suppose I take the "group algebra" of a ring. i.e., fix a commutative ring k and take a ring S. I can consider k[S]. It is an augmented commutative k-algebra, but this only uses the structure of commutative monoid of S. If I see k[S] as a bialgebra where all elements are grouplike, the invertibility of + in S gives me the antipode. I still haven't used the product of S. Is it reflected in k[S] somehow?

Anyone know, are strict $\omega$-categories basically incompatible with quasicategories?

I know that they are equivalent to complicial sets, but I don't know how this interacts with quasicategories either.

@lenticcatachresis k[S] surely has extra structure of some sort, but I don't know how to axiomatize it

yeah, that's my issue (and the problem that motivated the above discussion)

Ah... I guess I want weak $\omega$-categories...

This stuff is heavvvvy.

if you take the free commutative k-algebra on the elements of S, which has k[S] as a quotient, you get what's known as a "plethory"

sounds obscure

@JonBeardsley aren't omega-categories a model for (oo, oo)-categories?

@lenticcatachresis it's as obscure as you make it

I like plethories a lot, personally

do you have any preferred reference? I had never heard of them

I like this paper a lot arxiv.org/pdf/math/0407227v1.pdf

I think there's something in Hazewinkel's Witt vectors book, as well

@SaulGlasman i don't really know!

Looking at this, trying to understand: arxiv.org/pdf/math/0604414v3.pdf

reading between the lines here, I think this is (oo, oo)-category theory

ie non-invertible n-morphisms are allowed for all n

i see

i guess it can give you quasicategories tho

that's kind of what i'm interested

i'll tell you my ideas this weekend.

yeah, just like there are adjoint pairs between oo-categories and oo-groupoids

there are adjoint pairs between (oo, n)-categories and (oo, m)-categories for any m and n

right

i sort of think of these as looping/delooping types of things, but maybe this is not the right way to think about it

to be more specific, for m < n, you can regard an (oo, m)-category as an (oo, n)-category, and this functor has both adjoints

is this a situation like... one of the adjoints just freely inverts the higher things, and the other just ignores stuff that isn't inverted?

exactly

I don't know whether there's a connection with looping/delooping

@lenticcatachresis I second Saul's suggestion. I also like the beginning of the talk here: msri.org/workshops/689/schedules/18234 if you'd rather see someone explain a bit about what's going on.

@SaulGlasman have you heard of inverters and coinverters?

No, what's that?

thank you both

That looks interesting

I don't really understand 2-limits, let alone weighted 2-limits

Nor I. This is, unsurprisingly, something Emily told me about.

I kind of wonder if these two things from oo,n-cats to oo,m-cats are of this form.

I guess it's nice to know that these ideas can be expressed as limits of some flavor

I kind of wonder if fixed points and orbits can be expressed in this way.

and if there's a general coinverter/inverter norm map.

fixed points and orbits are already ordinary limits and colimits, so I'm not sure

how useful is a functor being representable, if the category doesn't have anything useful to preserve? e.g. if it's the free category on a graph

A random question: what limits does group completion (from E_oo-spaces to group-like E_oo-spaces) preserve? It clearly preserve products, but it seems like it should preserve also limits indexed over finite spaces (e.g. $\Map(X,BO\times Z)$ is the group completion of the groupoid of vector bundles over $X$)

I finally have all of my thoughts, related to all the questions I always ask here, together into one question: mathoverflow.net/questions/233338/…

(and happily it is making its way as a valid question on the MO site)

@barron Personally the reason why I care about representable functor is that you can study the functor by studying the object in your category (e.g. you see that real line bundles are the same things as classes in $H^1(-;\mathbb{Z}/2)$ because they are both classified by $\mathbb{RP}^\infty$ for unrelated reasons). This is independent from whether your category has limits/colimits (see for example the category of schemes that

has very few colimits, but where representable functors are pretty interesting)

The specific example I have in mind is that (re mathoverflow.net/questions/227452/… ), in monoids factorization sets wrt a presentation P are a representable functor on FCP the free category on the cayley graph of P; you can do factorization theory just as well in semigroups (no identity), but factorization sets aren't a representable functor there, because there's no identity element to start all the factorizations from

but in this setting i'm not seeing that this matters at all

I don't know enough about this specific case to say intelligent things. The only thing barely related is that natural transformations between representable functors are fairly easy to study thanks to the Yoneda lemma, but that's all I can say.

In general you don't study representable functors because they preserve limits, so your first question sounded a bit strange to me

@DenisNardin I don't have explicit examples, but I would guess that taking group completion of commutative monoids does not preserve kernels. (When a homotopy theory question sounds too hard to me I always start with the case of discrete spaces.) EDIT: Ignore me: I just noticed you said finite spaces, the shape of the equalizer is not a groupoid.

OK, at least it really doesn't preserve equalizers: consider the monoid $A_n$ generated by $x$ subject to $x^{n+1} = x$. Its group completion is $C_n$, the cyclic group of order $n$. The morphism $A_4 \to A_2$ sending $x \in A_4$ to $x \in A_2$ has trivial kernel, but its group completion is the canonical projection $C_4 \to C_2$ which does have a kernel.

@Omar If it preserved equalizers it would preserve all limits (since it clearly preserves products), so I wasn't really hoping for that :)

@DenisNardin Ha! Of course. I obviously didn't think this through.

Anyone know where I can find some discussion of the monads of quasicategories associated to oo-operads? I.e. the "free ___-monoid" monad?

Is this in Lurie's HA?

I'm pretty sure that in Lurie's HA there's a discussion of monads associated to adjunctions

(that's when he proves the monadicity theorem)

Right.

Then in some other chapter he studies the free/forgetful adjunction for O-algebras

Ok, right, and there's an adjunction between $C$ and $Alg_O(C)$

I don't know if that's enough

right.

In the past I've found his treatment of monads to be a little bit unsatisfying, but maybe there's enough in there.

ah right, the free algebras are basically just given as left operadic kan extensions yes?

Yes, that's it

ugh. now i'm trying to construct infinity operads from scratch but have to assume the existence of infinity operads, lol.

I don't remember seeing anything explicit about the monad structure for the free O-algebra functor in HA (which is not to say I'm at all sure it isn't there), but I think that you will probably find the lemmas you need to prove that the forgetful functor from O-algebras to C is monadic using the monadicity theorem.