'Tis the season to be surveying: arxiv.org/abs/1902.05046

I was confused for about a year that there is a Toby Bartels and a Tobias Barthel

both of whom are at least peripherally related to higher category stuff

If both of them were in the same room, would you address them collectively as "Tobies Barthels" like in French...

@HarryGindi The way I remember it is that there is a conservation of the letter "s". Tobias goes by Tobias for this reason, to avoid the confusion, which happens more than one would like.

I figured, heh

I am confused by some computations of cyclic bar constructions of monoids. In KN, proposition 8.1, they claim that ther is a map from the cyclic bar construction associated with a monoid $M$ to the cyclic object $k\mapsto\operatorname{Fun}(S_k,BM)^\sim$ which induces an equivalence on geometric realizations. (I guess that the tilde denotes the maixmal groupoid)

$S_k$ is the category generated by the directed cycle of $k$ elements.

They claim that $\operatorname{Fun}(S_k,BM)^\sim$ is equivalent to the homotopy quotient of $M^k$ by componentwise action of the grouplike monoid $(M^\times)^k$ of invertible elements in $M$, without explaining the action. I don't know whether this is a very well known proposition?

Hi Andrea!

@FrankScience I don't know if it is well known, but it follows quickly from the definitions. Since categorical equivalences don't change the equivalence type of the functor category, I'm going to assume that $S_k$ is the simplicial set with $k+1$ 0-simplices labelled $0,1,\dots,k$ and $k+1$ 1-simplices $e_{i,i+1}$ going from $i$ to $i+1$, and nothing else

That is we can realize $S_k$ as the coproduct $(\Delta^1)^{\amalg k+1} \amalg_{(\partial \Delta^1)^{\amalg k+1}} (\Delta^0)^{k+1}$

where the map $(\partial \Delta^1)^{\amalg k+1}→(\Delta^0)^{k+1}$ is the one sending $(0,i)$ to $i$ and $(1,i)$ to $i+1$

That is $\mathrm{Fun}(S_k,BM)^\sim\cong \mathrm{Map}(S_k,BM)\cong (\mathrm{Ar}^\sim\,BM)^{k+1}\times_{(BM^\times \times\, BM^\times)^{k+1}} (BM^\times)^{k+1}$

This is because $\mathrm{Map}(\Delta^0,BM)=(BM)^\sim\cong BM^\times$ and similarly for the arrows

Now note that $\mathrm{Ar}^\sim BM\to B(M^\times \times M^\times)$ is just the Kan fibration corresponding to the functor $B(M^\times\times M^\times)→Cat_∞$ describing the action of $M^\times\times M^\times$ on the arrows of $BM$, that is on $M$ (by left and right multiplication). Hence the pullback is the fibration describing the action of $(M^\times)^k$ on $M^k$ via a certain amounts of left and right multiplications

Now the colimit of the action described by a Kan fibration is just the total space

This looks a lot more complicated than it actually is

And just now I realized that all $k+1$'s should be $k$'s because I misread the notation...

Anyway if you work with $M$ a discrete monoid you'll see that $\mathrm{Fun}(S_k,BM)^\sim$ is as a simplicial set the nerve of the action groupoid of the action they describe. A lot of the complication in my argument comes from dealing with the general case of $M$ a simplicial monoid

Essentially the point is that there is a map $(\Delta^0)^{\amalg k}→S_k$ that induces a fibration $\mathrm{Fun}(S_k,BM)^\sim→ B(M^\times)^k$ and the fiber is $M^k$

Actually that's probably a simpler way of doing what I'm doing

I'm halfway tempted to erase the complicated argument above, but I'll leave as a sign that sometimes I should work out the arguments before I start writing in the chat

Thanks. I beginned with understand elements of $\operatorname{Fun}(S_k,BM)$. It seems to me that you view $S_k$ glued from several directed segments. It was very weird to me because if $M$ is discrete, the classical model of $BM$ has only one point.

I suggest you to work out the case where $BM$ is discrete

$S_k$ is defined as the ∞-category generated by various glued segments

We gotta use the definition of $S_k$ somewhere, right? :)

So a functor $S_k→BM$ (when $M$ is discrete) is determined by where the segments go (since, as you point out, the points have only one place to go), hence it's the datum of $k$ elements of $M$

Morphisms are natural transforms, and given by commutative diagrams, The action in question should be something like $(x_0,\dots,x_{k-1})\mapsto(g_0x_0g_1^{-1},g_1x_1g_2^{-1},\dots,g_{k-1}x_{k-1}g_‌​0^{-1})$. Sorry, I was completely stupid.

Nah don't worry. I was worse :).

Although in your defense I don't think it's fair to call this action "componentwise". Doesn't matter though, the final monoid that acts is contractible so it can acts as it well pleases, it doesn't change the quotient

I don't understand the final monoid.

Is that the geometric realization?

Think about it this way: the homotopy quotient of X by M can be calculated as the geometric realization of a certain simplicial object X⇐X×M ⇐ X×M×M... We are exploding the simplicial object $Fun(S_\bullet,BM)^\sim$ as a bisimplicial object and then realizing in the other order

We have an action of $(M^\times)^k$ on $M^k$ for each $k$ in $\Delta^{\operatorname{op}}$. What we got is that $\operatorname{colim}_{B(M^\times)^k}$ of a certain functor is $\operatorname{Fun}(S_k,BM)^\sim$.

Well, sure. But homotopy quotients can also be computed as a certain geometric realization (this is a special case of what's called the Bousfield-Kan formula)

So the idea is to consider the following functor $Δ^{op}×Δ^{op}→Space$

$([k],[n])$ is sent to $(M^\times)^{nk}×M^k$

Does that work for general $M$?

I mean, under their assumption that $M$ is just a monoid in the $\infty$-category of spaces.

So that the arrows $[n]→[n']$ are given by multiplication (either $M^\times\times M^\times→M^\times$ or $M^\times×M→M$ depending on the map) and arrows $[k]→[k']$ do a complicated dance I'm not trusting myself to write correctly

Yes, it always works

The simplicial diagram might be a coherent diagram though

This diagram is rigged so that if you take the colimit in the second variable you get the simplicial diagram $\mathrm{Fun}(S_k,BM)^\sim$, while if you take the colimit in the first variable you get the cyclic bar construction on $M$ (up to homotopy)

It is a standard technique: you want to show that the colimits of two simplicial diagrams are equivalent? Find a bisimplicial diagram that realizes to both of them

Thanks. When $M$ is group-like, they claim that $\operatorname{Fun}(S_k,BM)^\sim$ is equivalent to the free loop space of $BM$.

Indeed. When $M$ is group-like, $BM$ is an ∞-groupoid. The left adjoint of the inclusion of ∞-groupoids into ∞-cats sends an ∞-category to its classifying space. In the case of $S_k$, this is $S^1$

I can believe this, but I don't know how the general theorem looks like. Well, we first replace $BM$ by a Kan complex.

If $M$ is group-like, $BM$ is a Kan complex already, I believe

It's a quasi-category where every morphism is an equivalence

Now, the quickest way to prove that the classifying space of $S_k$ is $S^1$ is using the colimit I wrote previously: $S_k = (Δ^1)^{⊔k}⊔_{(\partial Δ^1)^{⊔k}} (Δ^0)^{⊔k}$

Since taking the classifying space commutes with homotopy colimits (it's a left adjoint!), we have that the classifying space of $S_k$ is just $(Δ^0)^{⊔k}⊔_{(\partial Δ^1)^{⊔k}}(Δ^0)^{⊔k}$ and if you write down the mapping cylinder you'll see it's an $S^1$

Similarly, you can see that the maps induced by the $S_k→S_{k'}$ are homotopic to the identity

Well, ok you have to do the $k=0$ as a special case, since then you have 1 arrow, not 0

But still

Is it easy to see that it is $S^1$-equivariant?

How are you defining the $S^1$-action on the realization of the cyclic bar construction?

Anyway, I think that the answer is: it's elementary, you just have to work your way carefully along the constructions. But unfortunately this doesn't mean "easy"

If you define the S^1-action using that the classifying space of the cyclic category is BS^1 you can proceed as follows:

extend the functor $k\mapsto S_k$ to a functor $\Lambda^\op\to \mathrm{Cat}_\infty$. Show that the proof of proposition 8.1 respects this additional functoriality (essentially that the bisimplicial set I wrote above can be interpreted as a functor $\Lambda^{op}\times Δ^{op}→\mathrm{Space}$) and then show that the resulting action on the geometric realization of the classifying spaces of $S_k$ is the canonical one

None of these steps are hard, although some are annoying to write (especially the second)

I am most confused by the second step.

Hrmm, wait something's not right

So: disclaimer. I haven't thought this through (I never read those notes before and I probably never thought about this beyond the case of a discrete monoid). I might say something confusing or outright false

There seem no problem.

I'm not confident I can do all the details now. But the way to go is certainly extend the bisimplicial set $Δ^{op}×Δ^{op}→\mathrm{Space}$ to some functor on a bigger indexing category $I$ that projects to $\Lambda^{op}$ and recover the $S^1$ action by exploiting that $|I|=BS^1$. I think you can choose $I=\Lambda^{op}×Δ^{op}$ but I'm not sure I'd bet a lot of money on it

I would like to visualize the action of $(M^\times)^k$ on $M^k$ as follows: we draw the picture for $S_k$ as an oriented circle with some end points, and the gadgets from $M^\times$ lives on endpoints, and gadgets from $M^k$ on segments. In this way it seems apparent that the action is compatible with rotations and "killing endpoints", this corresponds to your choice of $I$.

Note that this has nothing to do with ∞-categories. In fact, if I wanted to look for a reference I'd take a look at how Madsen does it in Algebraic K-theory and traces, page 202 (for a topological group, but still the proof is essentially the same)

Thanks