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Stream: theory: category theory

Topic: Classifying spaces of monoids

John Baez (Apr 06 2020 at 07:35):

@Joe Moeller was interested in the idea of an absorbing element in a monoid, i.e. an element $x$ such that $xy = yx = x$ for all $y$.

In this paper:

• Maja D. Rabrenovic, Classifying spaces of monoids: applications in geometric combinatorics.

they say $x$ is a black hole if $xy = x$ for all $y$. They show that if a monoid has a black hole, the geometric realization of its nerve is contractible! Cute.

"Classifying space" here means "geometric realization of the nerve".

John Baez (Apr 06 2020 at 07:40):

The point is that if we think of the monoid as a one-object category, the black hole defines a natural transformation from the identity functor to the constant endofunctor sending all morphisms to $1$.

John Baez (Apr 06 2020 at 07:40):

Thus, on the classifying space of the monoid, the identity map is homotopic to a constant map!

sarahzrf (Apr 06 2020 at 08:11):

cute indeed

Morgan Rogers (he/him) (Apr 06 2020 at 09:51):

John Baez said:

In this paper:

• Maja D. Rabrenovic, Classifying spaces of monoids: applications in geometric combinatorics.

they say $x$ is a black hole if $xy = x$ for all $y$. They show that if a monoid has a black hole, the geometric realization of its nerve is contractible! Cute.

"Black hole" seems a bit extreme! But I suppose it's evocative... I prefer absorbing element (or zero when it's on both sides)
To me this paper says 'classifying spaces are too blunt an instrument to analyse monoids with', but in the other direction I suppose identifying a space with the classifying space of a monoid is a good way to understand its homotopy theory.

Jonathan Beardsley (Apr 06 2020 at 11:22):

I think this sort of thing is a good argument for directed homotopy theory and eilenberg Mac lane spaces of monoids

Joe Moeller (Apr 06 2020 at 16:27):

Right, classifying space kills any non-group-like algebra. I've found monoids like bands (eg left regular bands) useful in applications.

John Baez (Apr 06 2020 at 17:31):

I don't know what "classifying space kills any non-group-like algebra" is supposed to mean, nor "classifying spaces are too blunt an instrument to analyse monoids with".

Any connected homotopy type comes from the classifying space of a monoid, while only connected homotopy types with vanishing homotopy groups $\pi_n$ with $n \gt 1$ come from classifying spaces of groups. These facts show that classifying spaces of monoids that aren't groups can be a lot more interesting than classifying spaces of groups!

This shows any connected CW complex has the same homotopy type as the classifying space of a monoid:

• Dusa McDuff, On the classifying spaces of discrete monoids.

This gives a 5-element monoid whose classifying space is (homotopy equivalent to) $S^2$:

• Zbigniew Fiederowicz, A counterexample to a group completion conjecture of JC Moore.

Moore's conjecture would imply that the classifying space of any monoid is homotopy equivalent to that of its group completion. This would kill any non-group-like behavior - but it's false.

John Baez (Apr 06 2020 at 17:35):

I don't understand Dusa McDuff's proof. She gives an explicit construction.

Morgan Rogers (he/him) (Apr 06 2020 at 17:53):

Well that certainly changed my mind!

Joe Moeller (Apr 06 2020 at 18:08):

Oh, I was working on the assumption that Moore's conjecture was true. I didn't know it was false.

Morgan Rogers (he/him) (Apr 06 2020 at 18:30):

John Baez said:

I don't know what "classifying space kills any non-group-like algebra" is supposed to mean, nor "classifying spaces are too blunt an instrument to analyse monoids with".

I really like this as a non-aggressive way of disagreeing, I'm going to use this in future.
The motivation for my statement is presumably where Moore's conjecture came from in the first place. Based on the group case, one might be led to assume that any interesting information carried into the classifying space should lie in the first homotopy group (also from its construction, since the elements become 1-simplices!)

Morgan Rogers (he/him) (Apr 06 2020 at 18:32):

So the next question is: what do we lose when passing from a monoid to its classifying space? (and what is the relationship between PSh(M) and Sh(X(M)) for X(M) the classifying space of M?)

John Baez (Apr 06 2020 at 19:30):

Morgan Rogers said:

So the next question is: what do we lose when passing from a monoid to its classifying space?

That's a really interesting question to me right now. Also, what don't we lose?

John Baez (Apr 06 2020 at 19:37):

Here's something. If we take a monoid $M$, form its classifying space $BM$, and then take the fundamental group $\pi_1(BM)$, this group is naturally isomorphic to the group completion of $M$.

(The group completion is roughly the group where we throw in formal inverses to the elements of $M$: group completion is the left adjoint to the forgetful functor from monoids to groups.)

This, together with other stuff I said, implies that forming $BM$ is strictly less destructive than group completion: it retains more information about $M$.

Morgan Rogers (he/him) (Apr 07 2020 at 07:54):

I would love to actively work on this, and also to know who else is actively working on this. Does subdividing your simplicial complex before applying McDuff's construction change the resulting monoid? How are special properties of monoids reflected in their classifying spaces (and which are homotopy invariant?) You (@John Baez) implicitly stated earlier that the classifying spaces of groups have vanishing higher homotopy groups; can we strengthen the disproof of Moore's conjecture to a proof that this property distinguishes groups amongst monoids?

John Baez (Apr 07 2020 at 17:18):

I'd be happy to keep talking to you here about classifying spaces of monoids, @Morgan Rogers. Maybe we can come up with something interesting.

You (@John Baez) implicitly stated earlier that the classifying spaces of groups have vanishing higher homotopy groups...

Yes: the classifying space of a group is a connected space with vanishing higher homotopy groups, any connected space with vanishing higher homotopy groups is weakly homotopy equivalent to the classifying space of a group, and any connected CW complex with vanishing higher homotopy groups is homotopy equivalent to the classifying space of a group.

can we strengthen the disproof of Moore's conjecture to a proof that this property distinguishes groups amongst monoids?

No, I pointed out that this is false:

John Baez (Apr 07 2020 at 17:19):

John Baez said:

I just got some insight into Puzzle 1 from MathOverflow:

If $G$ is the group completion of a commutative monoid $M$, the
canonical map $BM \to BG$ is a homotopy equivalence; even if $M \to G$
is not injective. (This is easy to prove: think of $M \to G$ as a
functor between one object categories and apply Quillen's Theorem A to
it. There is only one slice category to check and using commutativity
it is easy to see this category is filtered and thus contractible.)

John Baez (Apr 07 2020 at 17:21):

This stuff (which I'd be glad to explain further) says that the classifying space of a commutative monoid is homotopy equivalent to the classifying space of its group completion. For example, the classifying space of $\mathbb{N}$ is homotopy equivalent to the classifying space of $\mathbb{Z}$: they're both homotopy equivalent to a circle.

John Baez (Apr 07 2020 at 17:59):

Here are a few things we might try to do:

1) Understand McDuff's proof that any simplicial complex is homotopy equivalent to the classifying space of some monoid. Can we explain in an intuitive way how she builds this monoid? (She actually builds a category that's equivalent to a monoid, i.e. one where all objects are isomorphic. That's fine.)

2) Understand Fiederowicz's proof that some 5-element monoid has a classifying space homotopy equivalent to $S^2$.

... and some more...

Morgan Rogers (he/him) (Apr 08 2020 at 12:20):

John Baez said:

John Baez said:

I just got some insight into Puzzle 1 from MathOverflow:

If $G$ is the group completion of a commutative monoid $M$, the
canonical map $BM \to BG$ is a homotopy equivalence; even if $M \to G$
is not injective. (This is easy to prove: think of $M \to G$ as a
functor between one object categories and apply Quillen's Theorem A to
it. There is only one slice category to check and using commutativity
it is easy to see this category is filtered and thus contractible.)

Let me break down this proof, since these classifying spaces have simplices of arbitrarily high dimension which seems a little scary a priori, and showing (or eyeballing) that such a space is contractible in general could be difficult.
To apply Quillen's theorem A, for example, we need to show that the classifying space of the comma of the unique object of $G$ over the functor $M \to G$ is contractible. This slice category has:

• An object for each element $g \in G$, call it $\underline{g}$. In commutative world the elements of $G$ can be written as $mn^{-1}$ for $m,n \in M$, but more generally it will be a finite word of elements and their inverses.
• A morphism $\underline{g}_1 \to \underline{g}_2$ for each element $m \in M$ such that $mg_1 = g_2$.

The classifying space of this therefore has a $0$-simplex for each element of $g$ and a (direction-labelled) $1$-simplex for each morphism as described above. In full generality we already get a path-connected $1$-skeleton here, since the identity must be connected to any element along a path that reads the given word expression.
Next we glue a $2$-simplex along the $1$-simplices $m_1,m_2,(m_1m_2)^{-1}$, where the latter is the edge corresponding to the element $m_1m_2$ travelled along backwards. For a general category, that ensures that any path composed of individual $1$-simplices (travelled along in the right direction) is homotopic to the path labelled by the composite morphism. So far so good; but this is where McDuff trails off with "and so on".
After thinking about the definition of the nerve, I understood that the $3$-simplices correspond to composable sequences of $3$ morphisms, so "makes associativity homotopic"; it provides a homotopy between the composite homotopies $m_1,m_2,m_3 \Rightarrow (m_1m_2),m_3 \Rightarrow m_1m_2m_3$ and the one obtained from the alternative bracketing. The higher simplices similarly have faces for all the ways we can bracket expressions of length equal to their dimension.

Morgan Rogers (he/him) (Apr 08 2020 at 12:47):

From a path point of view, I believe we get exactly what we expect/hope for: using some kind of simplicial approximation theorem and then inductively applying homotopies, I can see how any loop in the classifying space of either $M$ or the comma category should be homotopic to one constructed out of the edges (corresponding to morphisms), and hence can be labelled by a composite of elements of $G$, since we have to allow the path to go along some edges backwards. In the case of the comma category, though, a loop which starts and ends at the identity must be labelled by the identity of $G$, and so is homotopic to a point. That's simple connectedness; but how do I get contractibility?
In Rabrenovic's paper, they rely on the fact, which I might need a little more convincing of even if it makes some intuitive sense, that a natural transformation of functors gives a homotopy in the classifying space of the codomain category, so that in particular a natural transformation from the identity to a functor that factors through the one-object category defines a homotopy with a point. In the quote above it's stated that being a filtered category makes the classifying space contractible; why is that?

Amar Hadzihasanovic (Apr 08 2020 at 13:44):

Morgan Rogers said:

In the quote above it's stated that being a filtered category makes the classifying space contractible; why is that?

“Filtered” gives you that every finite diagram extends to a cocone. So suppose you take some diagram $f$ of shape $J$ whose nerve $NJ$ is a homotopy $n$-sphere; the nerve of the cocone on $J$ is always contractible; thus taking the nerve of the entire “extension” diagram tells you that the class of $Nf$ is trivial.

To conclude you would need to know that every class of $\pi_n(NC)$ is represented by the nerve of some finite diagram. This seems a bit more fiddly but probably not too hard?

Reid Barton (Apr 08 2020 at 13:47):

Morgan Rogers said:

In Rabrenovic's paper, they rely on the fact, which I might need a little more convincing of even if it makes some intuitive sense, that a natural transformation of functors gives a homotopy in the classifying space of the codomain category, so that in particular a natural transformation from the identity to a functor that factors through the one-object category defines a homotopy with a point.

This one's easy: the classifying space construction preserves finite products, a natural transformation between $F : C \to D$ and $G : C \to D$ can be identified with a functor $H : C \times [1] \to D$ (where $[1]$ is the arrow category), and the classifying space of $[1]$ is an interval.

Morgan Rogers (he/him) (Apr 08 2020 at 13:56):

Amar Hadzihasanovic said:

Morgan Rogers said:

In the quote above it's stated that being a filtered category makes the classifying space contractible; why is that?

“Filtered” gives you that every finite diagram extends to a cocone. So suppose you take some diagram $f$ of shape $J$ whose nerve $NJ$ is a homotopy $n$-sphere; the nerve of the cocone on $J$ is always contractible; thus taking the nerve of the entire “extension” diagram tells you that the class of $Nf$ is trivial.

To conclude you would need to know that every class of $\pi_n(NC)$ is represented by the nerve of some finite diagram. This seems a bit more fiddly but probably not too hard?

Yes! The $n$-sphere is homotopic to the "$n$ composable arrows" category (the totally ordered set with $n+1$ elements). So I take it that a simplicial complex is contractible if and only if all of its homotopy groups are trivial..?

Amar Hadzihasanovic (Apr 08 2020 at 14:01):

If by “its homotopy groups” you mean “those of a fibrant replacement”, I guess...

Morgan Rogers (he/him) (Apr 08 2020 at 14:03):

Why do I need a fibrant replacement?

Amar Hadzihasanovic (Apr 08 2020 at 14:04):

I think “the homotopy groups of a simplicial set” are by definition “the combinatorial homotopy groups of a fibrant replacement”.

Amar Hadzihasanovic (Apr 08 2020 at 14:05):

If you avoid passing through geometric realisation, at least.

Morgan Rogers (he/him) (Apr 08 2020 at 14:06):

I said "simplicial complex", intending "geometric realisation of the category", since Quillen's theorem also talks about the classifying space. I don't know homotopy theory well enough to know how a fibrant replacement of a simplicial set is constructed.

Amar Hadzihasanovic (Apr 08 2020 at 14:10):

So what the argument above gives you is that any sphere in a fibrant replacement $NC_f$ of $NC$ which is of the form

• $Nf: NJ \to NC$ for a finite diagram in $C$,
• composed with the replacement $NC \to NC_f$,
  is contractible. If you can prove that any sphere in $NC_f$ is equivalent to one of this form, then it's done.

Amar Hadzihasanovic (Apr 08 2020 at 14:10):

Morgan Rogers said:

I said "simplicial complex", intending "geometric realisation of the category", since Quillen's theorem also talks about the classifying space. I don't know homotopy theory well enough to know how a fibrant replacement of a simplicial set is constructed.

Ah, I guess I just find it easier to never do the geometric realisation and work directly with the nerve of a category...

Morgan Rogers (he/him) (Apr 08 2020 at 14:14):

It's okay, I realised that any simplicial complex is a CW complex, so Whitehead's theorem applies.

Morgan Rogers (he/him) (Apr 08 2020 at 15:45):

Returning from an Ex-citing diversion, I can check when the comma category $(*\downarrow M)$ is filtered.
It is inhabited (true for any monoid whatsoever) :check_mark:
Given $g_1,g_2 \in G$ we need elements $m_1,m_2 \in M$ with $m_1g_1 = m_2g_2$. In a commutative monoid and more generally in a monoid satisfying the left Ore condition, so that for every $p_1,p_2$ there are $q_1,q_2$ with $q_1p_1 = q_2p_2$, we can express any element of $G$ as $m^{-1}n$, because we can express every element of $G$ as a zigzag of elements of $M$ and the left Ore condition allows us to 'push out' all of the zags. Applying that condition a second time we get the cospans we needed.
Finally, taking $m_1,m_2,g$ with $m_1g = m_2g$ we need to have $n \in M$ with $nm_1 = nm_2$. For $M$ commutative, we can just take the element $n$ with $g = m^{-1}n$ from the previous paragraph. A slightly weaker condition is that $M$ satisfies the right Ore condition (dual to the above), so that we can write $g$ as $nm^{-1}$, and the condition that whenever there exists $n$ with $m_1n = m_2n$, we can find $n'$ with $n'm_1 = n'm_2$.

Paolo Capriotti (Apr 08 2020 at 15:53):

are you assuming that $M$ is cancellative?

Morgan Rogers (he/him) (Apr 08 2020 at 15:55):

No, but I have been assuming that one can express the group completion by formally adding inverses satisfying suitable identities.

Morgan Rogers (he/him) (Apr 08 2020 at 15:56):

Some observations:

• Passing to the slice category introduces an asymmetry which is not present in the spaces. That is, the classifying space of $M^{op}$ is by inspection isomorphic to the classifying space of $M$, so the duals of the conditions I describe above would also be sufficient.
• We're losing an unquantified amount of information here: Quillen's theorem A and the filteredness condition are both sufficient conditions. A complete, but perhaps less computationally useful result is mentioned in the comments of the MO post: "In Fiedorowicz, Z. Classifying spaces of topological monoids and categories. Amer. J. Math. 106 (1984), no. 2, 301–350 it is shown that the natural map $BM \to BG$ is a homotopy equivalence if and only if $H_n(M,\mathbb{Z}G)=0$ for $n\geq 1$. – Benjamin Steinberg"

Paolo Capriotti (Apr 08 2020 at 15:56):

when $M$ is not cancellative, I think you need to quotient as well

Morgan Rogers (he/him) (Apr 08 2020 at 15:56):

Yes, that's what I meant by "suitable identities", sorry.

Morgan Rogers (he/him) (Apr 08 2020 at 16:07):

The latter result is very interesting, though, since it suggests that the separation between $BM$ and $BG$ might be quantified by homology, albeit with potentially scary coefficients, yet we can use spaces of this form to talk about homotopy! I mustn't get too excited about this, though, especially when I haven't yet looked at the paper which is the source: it could just be that there is a Hurewicz-type coincidence of trivial homotopy and trivial homology when $BM \to BG$ is a homotopy-equivalence.

Amar Hadzihasanovic (Apr 08 2020 at 16:10):

It sounds like it could be that way, considering that the map always induces an isomorphism of the fundamental groups.

Morgan Rogers (he/him) (Apr 08 2020 at 16:15):

Hmm, this would in particular imply that the homotopy theory of a simply connected space can be quantified by the $\mathbb{Z}$ homology of a suitable monoid, which would seem a little too good to be true? (this is me taking homology to be "hard, but much easier than homotopy")

Amar Hadzihasanovic (Apr 08 2020 at 16:17):

Sorry, I meant: it sounds like it could be a case of Hurewicz/Whitehead...

Amar Hadzihasanovic (Apr 08 2020 at 16:18):

Isn't there a relative version which works with maps that induce an isomorphism of fundamental groups, and the isomorphism in homology takes coefficients in the group ring of the fundamental group of one of the spaces?

Reid Barton (Apr 08 2020 at 16:20):

Since it was (apparently?) used in the second paper @John Baez linked to earlier, can someone tell me why a simply-connected space with the homology of $S^2$ is homotopy equivalent to $S^2$? (To apply Whitehead/Hurewicz you need a map...)

Reid Barton (Apr 08 2020 at 16:20):

I assume this is a special fact for spheres or in low dimension.

John Baez (Apr 08 2020 at 18:18):

Hmm, I'll have to think about that. Whose paper is the "second" one?

Morgan Rogers (he/him) (Apr 08 2020 at 18:27):

The Fiedorowicz one; top of the second page.

Reid Barton (Apr 08 2020 at 19:44):

I mean, I didn't try to follow the rest of the argument very closely either, but that jumped out at me.

John Baez (Apr 09 2020 at 00:00):

I'm not very good at this but here's my attempt to get started. Suppose $X$ is a connected and simply connected CW complex with the same homology as $S^2$. We want to show $X$ is homotopy equivalent to $S^2$.

By Hurewicz's theorem $X$ has the same $\pi_2$ as $S^2$:

$\pi_2(X) \cong H_2(X) \cong \mathbb{Z}$

So, pick a map

$f : S^2 \to X$

that's a representative of the generator of $\pi_2(X)$. Note that $f$ induces an isomorphism from $\pi_2(S^2)$ to $\pi_2(X)$.

If we can show this map induces an isomorphism on homology we'll be done, by a corollary of Whitehead's theorem:

a continuous map $f\colon X\to Y$ between simply connected CW complexes that induces an isomorphism on all integral homology groups is a homotopy equivalence.

But it's easy to show $f$ induces an isomorphism on homology since if $n \ne 2$ we have $H_n(X) \cong H_n(S^2) = \{0\}$, while for $n = 2$ we've chosen $f$ to induce an isomorphism on $\pi_2$, so by the Hurewicz isomorphism, which is natural, it induces an isomorphism on $H_2$ as well.

Hey, I did it!

John Baez (Apr 09 2020 at 00:02):

Of course all this stuff is so elementary for Fiederowicz that he didn't bother to mention it. :smirk:

Reid Barton (Apr 09 2020 at 01:03):

Oh right, thanks. For some reason I tried building a map the other way.

Morgan Rogers (he/him) (Apr 09 2020 at 15:50):

John Baez said:

1) Understand McDuff's proof that any simplicial complex is homotopy equivalent to the classifying space of some monoid. Can we explain in an intuitive way how she builds this monoid? (She actually builds a category that's equivalent to a monoid, i.e. one where all objects are isomorphic. That's fine.)

To get to grips with this proof, I'm going to present the construction for the case of the circle $S^1$, or more precisely the boundary of a triangle.

Morgan Rogers (he/him) (Apr 09 2020 at 15:54):

I'll write $P$ for this complex, to match McDuff.
"The category $F(P)$ has one object $u$ for every vertex $u$ of $P$."
So we start with objects $a < b < c$, which will shortly be connected by isomorphisms; I forgot to mention that the complex is required to be ordered, hence the ordering on these objects (which is not yet expressed in the structure of the category).

Morgan Rogers (he/him) (Apr 09 2020 at 15:57):

"Its morphisms are generated by:
(a) an invertible morphism $m(u_0, u_1): u_0 \to u_1$ for each $1$-simplex $(u_0, u_1)$ in P with $u_0< u_1$."
These make our category into the free groupoid with 3 objects.

Morgan Rogers (he/him) (Apr 09 2020 at 15:59):

"(b) a morphism $h(u_0, \dotsc , u_k): u_k \to u_k$ for each $k$-simplex $(u_0, \dotsc , u_k)$ in $P$ with $k > 1$ and $u_0< \cdots<u_k$;"
There are none of these in our current example; maybe I'll look at $S^2$ later!

Morgan Rogers (he/him) (Apr 09 2020 at 16:03):

"and are subject to the following relations:
(c) $h(u_1, \dotsc , u_k)^2 = h(u_1, \dotsc , u_k)$;
(d) $h(u_1, \dotsc , \hat{u}_i, \dotsc, u_k)h(u_1, \dotsc , u_k) = h(u_1, \dotsc , u_k)$ if $k \geq 3$ and $i \neq k$;
(e) $m(u_{k-1}, u_k)^{-1}h(u_0, \dotsc , u_{k-1})m(u_{k-1}, u_k)h(u_0, . . . , u_k) = h(u_0,\dotsc, u_k)$ if $k \geq 3$;"
It sounds like it will be worth looking into $S^3$ too, but still nothing for our present example here.

Morgan Rogers (he/him) (Apr 09 2020 at 16:11):

"(f) $m(u_0, u_2)^{-1}m(u_0, u_1)m(u_1, u_2)h(u_0, u_1, u_2) = h(u_0, u_1, u_2)$."
This relation encodes the gluing of 2-cells, again none of these in our first example. :octopus:

Morgan Rogers (he/him) (Apr 09 2020 at 16:22):

To recover the monoid from this category, we just look at the endomorphisms of any object. Looking at the object $c$ in this case, it takes only a little eyeballing to see that we recover the abelian group $\mathbb{Z}$, with generator $m(a, c)^{-1}m(a, b)m(b, c)$; we also see that if we glue in a 2-simplex to our triangle, relation (f) above makes the resulting idempotent of $c$ absorb this generator, so that the induced monoid of that complex is the monoid $\mathbb{Z} \cup \infty$ (where anything added to infinity is infinity), which we know from other results in this discussion has contractible classifying space.

Morgan Rogers (he/him) (Apr 09 2020 at 16:39):

Let's move up to $S^2$, realised as the boundary of a 2-simplex, so that we have some of these idempotents showing up;
This time we have four objects, $a<b<c<d$.
(a) again gives us an invertible morphism for each 1-simplex.
(b) gives us one generator $h(a,b,c)$ at $c$ and three, $h(a,b,d), h(a,c,d), h(b,c,d)$, at $d$.
(c) tells us that these generators are idempotent.
(d) and (e) don't apply.
(f) tells us that these generators absorb the cycles that are obtained by going around the 2-cells they correspond to.
There are no other relations, so we have a lot of free stuff happening. It's easiest to unpack the resulting monoid at $d$, but I'll save that for tomorrow :innocent: then we'll see how it compares to Fiedorowicz' construction!

John Baez (Apr 09 2020 at 18:08):

Thanks for starting this! I think this is the way to understand what McDuff is doing.

John Baez (Apr 09 2020 at 18:08):

We do need to go up at least to $S^2$ or $S^3$ to see how her marvelous construction really works.

Morgan Rogers (he/him) (Apr 10 2020 at 10:35):

A correction for my analysis yesterday: the idempotent element introduced when gluing in the interior of a triangle only absorbs things on its left, but we still know that this is enough to make the classifying space contractible thanks to the paper that John kicked off this topic with:
John Baez said:

• Maja D. Rabrenovic, Classifying spaces of monoids: applications in geometric combinatorics.

they say $x$ is a black hole if $xy = x$ for all $y$. They show that if a monoid has a black hole, the geometric realization of its nerve is contractible!

Morgan Rogers (he/him) (Apr 10 2020 at 10:57):

A further thing that was bugging me, and the reason that my generator supposedly based at $c$ in the construction from yesterday looks wrong:
McDuff writes composition in the "read along the arrows" direction rather than the "function composition" direction I'm used to :grimacing:
In order to avoid further confusion as we analyse her paper, I'll just stick with her convention, but bear it in mind!

Morgan Rogers (he/him) (Apr 10 2020 at 10:58):

So what monoid do we get for $S^2$ in the end? I'm looking at the monoid of endomorphisms at the object $d$ from the category constructed yesterday, which is generated by morphisms corresponding to the simplices of the boundary of a 3-simplex.
The loops formed by the invertible morphisms going around the faces adjacent to $d$ give us invertible generators,
$x = m(c,d)^{-1}m(a,c)^{-1}m(a,d)$, $y = m(a,d)^{-1}m(a,b)m(b,d)$ and $z = m(b,d)^{-1}m(b,c)m(c,d)$.
Any loop based at $d$ following the 1-simplices are all expressible in terms of these generators; for example, the loop around the final face, based at $c$ and conjugated by $m(c,d)$, is precisely the composites $xyz$.

Morgan Rogers (he/him) (Apr 10 2020 at 11:18):

Now for the idempotents: I'll write $h_x,h_y,h_z$ and $h_{xyz}$ for $h(a,c,d),h(a,b,d),h(b,c,d)$ and $m(c,d)^{-1}h(a,b,c)m(c,d)$ respectively. These satisfy:
$x^kh_x = h_x$,
$y^kh_y = h_k$,
$z^kh_z = h_z$ and
$(xyz)^kh_{xyz} = h_{xyz}$
respectively for every integer $k$. Unless I've missed something (which I could have, feedback would be appreciated!), there are no further relations constraining these, so we've ended up with a monoid presented with 7 generators, of which 3 are invertible and 4 are idempotent. It's also worth observing that while this monoid looks quite big, it has a decidable word problem: all of the rules are reduction rules, so that I can always reduce a given composite of generators to a canonical form.

Morgan Rogers (he/him) (Apr 10 2020 at 11:28):

Now that we have a few examples to hand, let's have a look McDuff's 'easy observations'.
"(i) For any (ordered) $k$-simplex $\sigma$ of dimension $k \geq 2$, $F(\sigma) = F(\partial \sigma)^*$,"
where the latter is the monoid (resp. 'semigroupoid') obtained by freely adding a left absorbing element $h$, so that $xh = h$ for all elements $x$ in the original monoid (resp. adding such an idempotent to every object such that they conjugate to one another). I have already described this in the case of the 2-simplex.

Morgan Rogers (he/him) (Apr 10 2020 at 11:34):

"(ii) if $P = Q \cup_{\partial \sigma} \sigma$, where $dim \sigma \geq 2$, then $F(P) = F(Q) *_{F(\partial \sigma)} F(\sigma)$,"
this is a formal way of saying that adding a face to a simplicial complex freely adds an idempotent subject only to the relations imposed by its boundary. Of course, we must again conjugate this idempotent to understand the effect this has on the endomorphism monoids of all of the other objects in the semigroupoid, which is how we got $h_{xyz}$ in the construction of the monoid from the boundary of the 3-simplex above.

Morgan Rogers (he/him) (Apr 10 2020 at 11:43):

"(iii) For any directed set $Q^{\alpha}$ of (connected, ordered) complexes, $F(\mathrm{colim}_{\alpha} Q^{\alpha}) = \mathrm{colim}_{\alpha} F(Q^{\alpha})$,"
This is the claim that we can build up $F(P)$ from the semigroupoids of its connected subcomplexes, and that the order in which we do this does not affect the result. Stated on its own this is by no means clear to me, but by inductively applying (ii) and carefully examining the interaction of amalgamated products of monoids/semigroupoids, I see how one might prove it. Perhaps there is a neater argument out there for why this should be the case?

Morgan Rogers (he/him) (Apr 10 2020 at 11:56):

"(iv) corresponding to every inclusion $Q \hookrightarrow P$ there is a right-free inclusion $F(Q) \hookrightarrow F(P)$".
The concept of "right-freeness" is important for the preliminary results introduced by McDuff before this point: it means that $F(P)$ is the disjoint union of the left cosets of $F(Q)$ in $F(P)$, which is to say that "there is a subset $\bar{M} \subseteq F(P)$ containing the identity such that $F(P)$ is the union over $m \in \bar{M}$ of the disjoint left cosets $mF(Q)$". To formally prove this, one would again need to work inductively, but we can see from the form of the construction conditions (a)-(f) both where the name "right-free" comes from and why this claim 'should' hold.

Morgan Rogers (he/him) (Apr 10 2020 at 11:57):

That will do for today. All comments welcome. Tomorrow (or "next time", I don't want this to dominate my time too much or overwhelm people with walls of text!) I plan to think about Fiedorowicz' construction for $S^2$, and whether their 5-generator construction can be obtained from the 7-generator one I described today.

Morgan Rogers (he/him) (Apr 13 2020 at 14:12):

Having a full read-through of McDuff's proof, I reached enlightenment :stuck_out_tongue_wink:
The aim of her construction is precisely to ensure, with minimal extra effort, that point (i)-(iv) I quote above hold. From there, she inductively defines a mapping from the original complex $P$ into the classifying space of the constructed monoid, $BF(P)$, by first observing that this construction makes it easy to define the map (which is a homotopy equivalence) for 1-dimensional complexes, and building $P$ and $BF(P)$ up inductively maintain that homotopy equivalence in a natural way.

Morgan Rogers (he/him) (Apr 13 2020 at 14:16):

One conclusion, therefore, is that this is by no means the most efficient construction of a monoid whose classifying space is homotopic to a given fixed simplicial complex (or other space). Indeed, Fiedorowicz' construction gives a simpler, non-isomorphic monoid. But as far as making the general proof straightforward, McDuff gets the job done. Her justification of points (i)-(iv) is very limited; I'm not so sure that "It is not hard to check that..." cuts it here, but at least we have arrived at the intended outline of the proof.

Morgan Rogers (he/him) (Apr 17 2020 at 10:04):

@John Baez are you still actively interested in this? I tailed off because I didn't want to be talking to myself in an empty room, but if you had a particular direction you wanted to go with this investigation beyond understanding the proof, perhaps we can explore it together?

Morgan Rogers (he/him) (Apr 17 2020 at 10:18):

For example, I wonder (but haven't yet attempted to calculate) if barycentric subdivision is a nice operation relative to this construction, to the end that one could translate the simplicial approximation theorem into the language of McDuff monoids and give a meaningful (if still rather convoluted) way to compute the homotopy groups. Even if it's not, modifying her construction to make this the case would seem like a natural step to take.

John Baez (Apr 18 2020 at 02:31):

Morgan Rogers said:

John Baez are you still actively interested in this? I tailed off because I didn't want to be talking to myself in an empty room, but if you had a particular direction you wanted to go with this investigation beyond understanding the proof, perhaps we can explore it together?

I haven't understood the proof yet; I watched you starting to figure out the proof and then you said you understood it... but this sort of thing is not really a spectator sport.

John Baez (Apr 18 2020 at 02:43):

So, I'd still like to understand her proof - or maybe some "better", for example more conceptual, proof.

I'm not really sure what the good questions are, but I bet there must be some, so it would be good to figure out what they are. (Good questions are ones that have interesting answers.)

When you mention barycentric subdivision, I'm instantly reminded of the proof that every simplicial complex is the nerve of a poset. Do you know that proof? There could be some interesting relation to what we're thinking about.

I also wonder how this "classifying space of a monoid" stuff is connected to the work of Eilenberg and many others on monoids and classes of languages. A good (but big) introduction is

• Jean-Éric Pin, Mathematical Foundations of Automata Theory, 2019.

Don't feel you need to read this! This just seems to be a big region of work on monoids.

Morgan Rogers (he/him) (Apr 18 2020 at 08:52):

John Baez said:

Morgan Rogers said:

John Baez are you still actively interested in this? I tailed off because I didn't want to be talking to myself in an empty room, but if you had a particular direction you wanted to go with this investigation beyond understanding the proof, perhaps we can explore it together?

I haven't understood the proof yet; I watched you starting to figure out the proof and then you said you understood it... but this sort of thing is not really a spectator sport.

There are degrees of understanding; I feel like I understand the structure of the proof enough to explain it to someone else, but I'm not at the stage of being able to present the finer details of how a formal proof would run. That's mostly because McDuff claims (i)-(iv) are "not hard" without further details.

Morgan Rogers (he/him) (Apr 18 2020 at 08:59):

John Baez said:

When you mention barycentric subdivision, I'm instantly reminded of the proof that every simplicial complex is the nerve of a poset. Do you know that proof? There could be some interesting relation to what we're thinking about.

I don't! Do you know where I can find it? I've been trained to look at constructions like the classifying space or simplicial complexes as bridges (not only connecting different structures from which they are derived, but with those structures providing different perspectives from which to understand their properties) so having another way to present any simplicial complex as the nerve of a poset seems like very useful extra info!

Gershom (Apr 18 2020 at 15:47):

John Baez said:

I don't! Do you know where I can find it? I've been trained to look at constructions like the classifying space or simplicial complexes as bridges (not only connecting different structures from which they are derived, but with those structures providing different perspectives from which to understand their properties) so having another way to present any simplicial complex as the nerve of a poset seems like very useful extra info!

My understanding is that it goes something like this: By McCord, geometric simplicial complexes are weakly equivalent to Alexandroff spaces (where the intersection of any opens is also open). Alexadroff spaces are equivalent to posets, done! (Or do you have another construction in mind John?)

This is covered in Raptis' "Homotopy theory of posets" https://projecteuclid.org/euclid.hha/1296223882 which is a lovely piece of work, not least because it shows that the Thomason model structure on Cat is "really" just a model structure on Pos (i.e. it only sees the structure on Cat that comes via Pos anyway).

Reid Barton (Apr 18 2020 at 16:04):

In the case of simplicial complexes this construction is elementary and very visual. The barycentric subdivision of a simplex is, by definition, the nerve of a poset: the poset of all nonempty subsets of the vertices, or equivalently of all the faces of the simplex, ordered by inclusion. Geometrically, given a nonempty subset of the vertices of $\Delta^n$ (which we think of as geometrically embedded in a Euclidean space), we map it to the average (or barycenter) of those vertices. The chains in the poset make up the faces of the barycentric subdivision and the geometric realization of the subdivision is the same as the original simplex.

Reid Barton (Apr 18 2020 at 16:06):

If we start with a simplicial complex we can do the same thing, namely take the all those nonempty subsets of vertices which span faces, or equivalently just the faces of the complex itself, and order them by inclusion. It's the same as applying the barycentric subdivision to every face simultaneously.

Reid Barton (Apr 18 2020 at 16:06):

In conclusion, if you start with a simplicial complex, then its subdivision is the nerve of a poset, and it has the same geometric realization as the original complex.

Reid Barton (Apr 18 2020 at 16:14):

Basically the point is that when you glue the face posets of two simplices along that of a common face, nothing bad happens. (The nerve is a right adjoint so it usually doesn't preserve colimits, but it does preserve this kind of colimit.)

Reid Barton (Apr 18 2020 at 16:20):

If you start with a simplicial set, then you need to do more general kinds of gluing and bad things can happen. For example take $\Delta^1 / \partial \Delta^1$, a 1-simplex with its endpoints identified. The face poset of $\Delta^1$ looks like $\{0\} \to \{0, 1\} \leftarrow \{1\}$. When you identify the endpoints $\{0\}$ and $\{1\}$, you "should" get two distinct maps from the new thing into $\{0, 1\}$, but since you're in posets you can only have one, and so the geometric realization ends up looking like an interval, when it should be a circle.

Reid Barton (Apr 18 2020 at 16:22):

However, I think when you subdivide a simplicial set once you get a simplicial complex, and that's why there are two subdivisions in the construction of the Thomason model structure.

Morgan Rogers (he/him) (Apr 18 2020 at 16:56):

Great explanations, thanks!

Morgan Rogers (he/him) (Apr 18 2020 at 17:17):

Reid Barton said:

In conclusion, if you start with a simplicial complex, then its subdivision is the nerve of a poset, and it has the same geometric realization as the original complex.

However, I think when you subdivide a simplicial set once you get a simplicial complex, and that's why there are two subdivisions in the construction of the Thomason model structure.

I imagined the same tricky case of the circle as "simplicial set with one 1-simplex and one 0-simplex" when I was trying to work out what barycentric subdivision might do to monoids.

One issue is that the poset we get from the nerve of a generic monoid is monstrously big, because there are composites of arbitrary length. For a finitely presented monoid I think we can argue a homotopy equivalence with a finite simplicial complex (informally, given a finite presentation of a monoid it's enough to have a 1-cell for each generator and to glue in just enough cells to realise the relations), and that's probably the kind of example I'll use to understand these things.

Morgan Rogers (he/him) (Apr 18 2020 at 17:18):

(That said, I am curious about the classifying space of $(\mathbb{N},{\times})$, which is concertedly a big'un)

John Baez (Apr 18 2020 at 17:31):

Reid Barton gave the argument I had in mind, for why every simplicial complex is homotopy equivalent - in fact homeomorphic - to the nerve of a poset.

John Baez (Apr 18 2020 at 17:33):

If you include $0$ in your definition of $\mathbb{N}$, then the classifying space of $(\mathbb{N},\times)$ is contractible, since $0$ is what we called a "black hole" earlier in this discussion ($0x = x0 = 0$) for all $x$, and the classifying space of a monoid with a black hole is contractible.

John Baez (Apr 18 2020 at 17:37):

If you don't include $0$ in your definition of $\mathbb{N}$, you deserve to die. But apart from that, then $(\mathbb{N}, \times)$ becomes the free commutative monoid on countably many generators: it's the coproduct of countably many copies of $(\mathbb{N}, +)$, one for each prime number. Since the classifying space of $(\mathbb{N}, +)$ is homotopy equivalent to a circle, the classifying space of $(\mathbb{N}, \times)$ is probably built from a countable collection of circles in some manner. Anyone want to work this out?

Morgan Rogers (he/him) (Apr 18 2020 at 17:38):

John Baez said:

If you don't include $0$ in your definition of $\mathbb{N}$, you deserve to die.

A strong stance, but indeed, I should probably have put a ${>}0$ subscript on my $\mathbb{N}$

Morgan Rogers (he/him) (Apr 18 2020 at 17:40):

John Baez said:

$(\mathbb{N}, \times)$ is the free commutative monoid on countably many generators: it's the coproduct of countably many copies of $(\mathbb{N}, +)$, one for each prime number. Since the classifying space of $(\mathbb{N}, +)$ is homotopy equivalent to a circle, the classifying space of $(\mathbb{N}, \times)$ is probably built from a countable collection of circles in some manner. Anyone want to work this out?

Does it end up still being contractible by the same argument as for the unit ball in an infinite-dimensional vector space?

John Baez (Apr 18 2020 at 17:48):

You seem to be using some argument to guess that we're getting the smash product of infinitely many circles, i.e. $S^\infty$. Is that what you're guessing?

John Baez (Apr 18 2020 at 17:48):

Are you saying coproducts in $\mathsf{CommMon}$ get sent to smash products in the category of pointed simplicial sets?

John Baez (Apr 18 2020 at 17:49):

(I haven't yet figured out how all this should work, so these are honest questions, not "quiz questions".)

vikraman (Apr 18 2020 at 18:37):

The group completion of $(\mathbb{N}^+,\times,1)$ should be $(\mathbb{Q}^+,\times,1)$, so I suppose you want a description of the $K(-,1)$ of this group.

John Baez (Apr 18 2020 at 22:04):

Yes. But that sounds harder to me than figuring out what the classifying space construction does to coproducts!

John Baez (Apr 18 2020 at 22:05):

$(\mathbb{N}^+, \times, 1)$ is the coproduct of countably many copies of $(\mathbb{N}, +, 0)$ - this is the fundamental theorem of arithmetic.

John Baez (Apr 18 2020 at 22:06):

However, this is the coproduct in the category of commutative monoids.

John Baez (Apr 18 2020 at 22:06):

So we may have to look at what the classifying space construction does to coproducts of monoids and then what it does to abelianization. Both of these sound like good things to think about!

Morgan Rogers (he/him) (Apr 19 2020 at 09:49):

John Baez said:

Yes. But that sounds harder to me than figuring out what the classifying space construction does to coproducts!

:joy: of course, it's a coproduct, not a product. And even if we take the coproduct in the full category of monoids or in the category of categories, there's no reason to assume that the nerve preserves infinite coproducts (although it seems to preserve finite ones), since in the last case the nerve has a left adjoint but not necessarily a right adjoint, for example.