## October 18, 2007

### Klein 2-Geometry IX

#### Posted by David Corfield

Seeing Jim Dolan expose that notion of types, predicates and axioms in his lecture, I was reminded of the introduction John gave me to it in Minneapolis. While there, we tried to see what we could make of the idea that a process of categorification moves us up from propositional to predicate to modal logic, (an idea of Jim’s?). What we arrived at was a multi-agent form of S5.

As in the lecture Jim’s example of an axiomatic theory is Euclidean geometry, this led me to a ramshackle series of night thoughts, which is all I’m fit to record at the moment.

So, models of a theory axiomatised in predicate logic assign sets to types, maps from their types to truth values are assigned to typed predicates, and truth values to sentences. The models then form a groupoid. The idea of categorifying predicate logic to modal logic allows metatypes, being assigned groupoids. So,

1) Can we see a degenerate ‘propositional’ 0-geometry?

And,

2) Did we see any sign of something ‘modal’ in our 2-geometry forays?

Maybe in the example I give at the end of this post, doubled-up Euclidean geometry, we might say a single line possibly passes through a point if it passes through it or its twin. While we might say that line-twins necessarily pass through a point, if one of them does.

But doesn’t that doubled-up geometry resemble Connes’s two-sheeted spacetime? And why should this be surprising if many noncommutative algebras arise as the convolution algebra over groupoids? After all, if we’re looking for 2-groups to be symmetries of something, a groupoid seems like a good bet.

If a spectrum is best thought of as a groupoid, what is the spectrum of the convolution algebra on a groupoid?

On a different note, John in a taverna in Delphi suggested thinking about finite 2-groups with a bit of twist to them. So what are the smallest finite groups, $G$ and abelian $A$, such that $H^3(G, A)$ is nontrivial?

Posted at October 18, 2007 1:35 PM UTC

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### Re: Klein 2-Geometry IX

I have some basic questions about modal logic, which I would like to learn more about. Could you (David) or someone else explain some things to me?

I’m looking for appropriate categorical semantics of the modal operator denoted by a box (“it is necessarily true that…”), something I can really sink my teeth into. Looking at the wikipedia article you linked under S5, it looks like box should be interpreted as a product-preserving, maybe even a left exact comonad, maybe satisfying some additional properties. Axiom T would correspond to a counit, axiom S4 would correspond to a comultiplication, and axiom K to product-preservation.

For example, given a topology on a set $X$, there is an operator

$int: 2^X \to 2^X$

which assigns to a subset of $X$ its interior, and this is a meet-preserving comonad. Would that be a reasonable topological semantics of box, or (thinking of elements of $2^X$ as ‘propositions’) an example of a necessity-modality on a propositional theory? (This reminds me of some things Steve Vickers said in the first few pages of his book Topology via Logic.)

Or, given a category $C$ whose set of objects is $X$, there is a left exact comonad $G$ acting on (what Lawvere sometimes calls) the category of attributes of type $X$, $Set/X$. Here $G$ is defined by the fiberwise formula

$[G(p: Y \to X)]_x = \prod_{f: x \to x'} p^{-1}(x')$

where $x, x'$ belong to $X = Ob(C)$ and $f: x \to x'$ belongs to $Mor(C)$. The category of coalgebras is the topos $Set^C$ (assuming I didn’t get the variance screwed up). Is there some appropriate interpretation of this comonad as a modal box operator?

Is any of this connected with what you and John were discussing?

Posted by: Todd Trimble on October 18, 2007 7:14 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

What you say sounds right to me and I would also like to know more about this. I once read that in categorical logic there is an analogy between monad-comonad theory and modal logic. I wonder what this analogy might be?

Does anyone have access to the paper “Monad as Modality” by S. Kobayashi in Theoretical Computer Science 175(1) 1997, pp.27-74 ? Perhaps someone could check this paper in order to answer Todd’s questions.

Posted by: Charlie Stromeyer Jr on October 18, 2007 8:17 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Todd wrote approximately:

I’m looking for appropriate categorical semantics of the modal operator denoted by □ (“it is necessarily true that…”), something I can really sink my teeth into.

For example, given a topology on a set X, there is an operator which assigns to a subset of X its interior, and this is a meet-preserving comonad. Would that be a reasonable topological semantics of □?

That’s the semantics that Awodey and Kishida give for the ‘propositional model logic system S4’. They show this semantics is sound and complete. And then they go from propositional to first-order model logic by ‘replacing the space with a sheaf over it’. You’ll probably like this!

I saw Awodey talk about this stuff at the Mac Lane memorial conference in 2006. It was a nice talk. He may have a full paper out now, but the abstract is fairly detailed.

By the way: the kind of modal logic that Jim and David and I think should correspond to 2-groupoids is not quite this sort. David says it’s a ‘multi-agent form of S4’, but all I know for sure is that it’s the form of logic where any theory has a 2-groupoid of models, and any 2-groupoid is the 2-groupoid of models of a unique theory! One can work backwards from this, keeping the more familiar decategorified analogue firmly in hand, and see what one gets. David and I spent an afternoon doing this, once upon a time.

(By the way, to get that box symbol, I typed

&#9633;

This is one of the UNICODE geometric shapes. A bigger box would be better, but I forget how to get one, either in UNICODE or LaTeX. I must have known once, ‘cause I used to work on the wave equation.)

Posted by: John Baez on October 19, 2007 4:59 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

My memory is worryingly fading of that afternoon. In fact I remember it as an evening. But I thought you noted down what we wrote on the board in one of your notebooks.

One important ingredient in the way philosophers think about modal logic is through possible world semantics. We are to imagine a collection of worlds and an accessibility relation between them.

Then we have various truths holding in the different worlds. So, $P$ is possibly true in this world if $P$ is true in some world accessible to it. $P$ in necessarily true in this world if it is true in every world accessible to it.

Different modal logics correspond to different properties obeyed by the accessibility relation.

If the relation is reflexive, then P implies possibly P. (Necessarily P implies P.) Axiom T.

If the relation is symmetric, then P implies necessarily possibly P. Axiom B.

If it is transitive, then possibly possibly P implies P. (Necessarily P implies necessarily necessarily P.) Axiom 4.

Oh look, it’s here. I was wondering what had happened to Axiom 5. It requires the accessibility to be ‘Euclidean’.

Multi-agent modal logic covers these worlds with different accessibility relations.

I seem to recall our contemplating proper groupoids of worlds, rather than just puny equivalence relations.

Given that possible world semantics is called Kripke semantics, and it’s linked to Kripke-Beth-Joyal semantics in a topos, I guess I ought to be finding out all about that too.

Posted by: David Corfield on October 19, 2007 3:15 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David, these two wikipedia entries on Kripke semantics and possible world semantics say it is widely regarded as a mistake to equate the two. However, I don’t know anything about philosophy so you should read these two entries for yourself.

Posted by: Charlie Stromeyer Jr on October 19, 2007 3:56 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

They’re being a bit nit-picky here:

The term “possible worlds semantics” is often used as a synonym for Kripke semantics, but this is widely regarded as a mistake: Kripke semantics can be used to analyse modes other than alethic modes (that is, it can be used in logics concerned not with truth per se; for example, in deontic logic, which is the logic of obligation and permission); and Kripke semantics does not presuppose modal realism, which the language of possible worlds arguably presupposes.

Right. Modal logic can be used for different kinds of possibility, e.g., as far as I know P is possibly true. Then you might wonder whether if it’s possible that P is possibly true, that then P is possibly true.

In this case, you’re imaging the ‘worlds’ as epistemic states. Something’s possibly true, if there’s a state of knowledge you could reach from here where it is true. In this interpretation, symmetry seems implausible.

There are some people who buy into this possible world business literally when dealing with ‘alethic’ modality. So it’s true that I could have had eggs for breakfast today, when I didn’t, if there’s a possible world close to the actual one in which a counterpart of myself ate eggs.

It’s the sort of thing that gives philosophy a bad name outside the field, but some big names do insist on doing it.

More important for us is to think about sheaves on this category of states/worlds.

Posted by: David Corfield on October 19, 2007 4:43 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David wrote:

Kripke-Beth-Joyal semantics in a topos, I guess I should be finding out about that too.

The books by McClarty and by Mac Lane and Moerdijk talk about this. The latter book is actually a bit more chatty about what’s really going on. Here’s my vague understanding — I hope someone will correct me if I’m wrong.

Any topos has an ‘internal language’, the ‘Mitchell-Benabou language’, where types are objects and terms are morphisms. If you want, you can write down an axiom in this language for each equation in your topos. This lets you merrily scribble down formulas and proofs all day, just like a good logician should, and pretend you’re doing intuitionistic set theory, when you’re actually reasoning within the topos.

As far as I can tell, the ‘Kripke-Joyal-Beth semantics’ for this internal language is basically just the obvious semantics where the symbols stand for what they really mean! The types stand for the objects they secretly are, and terms stand for morphisms, and the axioms assert the equations that actually hold in your topos.

The fun twist is that an ‘element’ of some type $X$ is not necessarily a ‘global’ element $c: 1 \to X$; it can be a generalized element $c: U \to X$. When we have a topos of sheaves, say over a space, these generalized elements can be ‘defined only on some open set, not the whole space’. This has philosophical consequences: it means an ‘individual may not exist in all worlds’.

Hmm — now I seem to be relating modal logic to topos logic, very much like Awodey and Kishida. Surely the use of ‘Kripke semantics’ in both fields is no coincidence! But most topos theorists don’t think they’re doing modal logic, do they? And what does the relation between topos theory and modal logic have to do with our hoped-for analogy:

$propositional:predicate:modal::set:groupoid:2-groupoid$

That’s pretty confusing! Maybe part of the point is that groupoids correspond perfectly to classical predicate logic, and now I’m talking about intuitionistic predicate logic. I’m not quite sure how that helps, but I’m desperate.

By the way, a while back you wanted to see some degenerate examples: ‘Klein 0-geometries’. To get one of these, just write down a bunch of formulas in the propositional calculus:

$P \implies (Q \vee R) , \qquad Q \vee (P \iff Q), \qquad R \implies P$

These formulas will have a set of models — the lines in your truth table where they all come out ‘true’. And, if you think of them as statements about ‘geometry’, you’ve got a Klein 0-geometry. In this degenerate game, we lack the ability to do existential or universal quantifiers over figures of a given type, since we don’t even have variables — but I guess we can still make assertions that we think of as assertions about individual figures.

We should keep pushing the idea of ‘modal logic as Klein 2-geometry’…

Posted by: John Baez on October 19, 2007 11:49 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Both modal propositional and intuitionistic proposition logic use topological semantics, so there must be some relationship between them. Perhaps epistemic logic is the bridge.

I remember doing work on semantics for intuitionistic logic for my Master’s thesis. I was interested in the two varieties of semantics: proof theoretic and topological. The proof theoretic interprets propositions as sets of their proofs, where, e.g., a proof of $P$ or $Q$ is a proof of $P$ or a proof of $Q$, and a means to say which. That gives the sense of excluded middle failing, a you may not have a proof of either disjunct.

A proof of $P$ implies $Q$ is a way to convert proofs of $P$ into proofs of $Q$.

The philosopher Michael Dummett believed this to be the ‘right’ semantics, rather than the topological, which has, e.g., negation taking the interior of the complement of an open set.

I came to the conclusion that we didn’t need to say one was right and one wrong, but that category theory was telling us what we needed to know about the semantics and their relation.

Posted by: David Corfield on October 20, 2007 1:15 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

A lot of profound questions, David. I shouldn’t whine about not enough people commenting on Jim’s first lecture, given that you just wrote a whole blog-article in response to it!

You’re right, Jim was the one who first suggested a systematic pattern relating:

propositional logic and sets
predicate logic and groupoids
modal logic and 2-groupoids
.
.
.
$n$-predicate logic and $n$-groupoids
.
.
.

where ‘$n$-predicate logic’ is a stupid name I just invented for logic in which we quantify over worlds of worlds of worlds… of things.

But, Jim has put most of his energy on the relation between predicate logic and groupoids. And we’re seeing the fruits of this labor in his lectures — slightly hidden by his pedagogical focus on the special case of groups.

As you note, any theory in predicate logic gives a groupoid:

So, models of a theory axiomatised in predicate logic assign sets to types, maps from their types to truth values are assigned to typed predicates, and truth values to sentences. The models then form a groupoid.

But now Jim is showing us how to work backwards and get the theory from the groupoid!

He’s actually considering theories written in the predicate calculus that have a unique model up to isomorphism. Then he’s taking this model — on some set $S$, say — and looking at its group $G$ of symmetries. In lecture 2 he forms the orbi-simplex of $G$ acting on $S$. And in lecture 3 he shows how to read off the original theory from the orbi-simplex!

It would be great to do this sort of thing one level up, in modal logic.

But as you note, many easier questions are already fascinating!

A theory in predicate logic can always be seen as a ‘Klein geometry’. The types are ‘types of figures’. The predicates are ‘incidence relations’. And the axioms are… well, axioms for a kind of geometry.

So, if we’re truly going along with the tao of mathematics, a theory in modal logic should be the same as a ‘Klein 2-geometry’. And, we used to know lots of examples of Klein 2-geometries. So, secretly, we should know lots of examples of theories in the language of modal logic!

But, I’m too tired to figure them out right now.

On a different note, John in a taverna in Delphi suggested thinking about finite 2-groups with a bit of twist to them. So what are the smallest finite groups, G and abelian A, such that $H^3(G,A)$ is nontrivial?

Hmm, I used to know. I think it’s just what you’d hope:

$H^3(\mathbb{Z}/2,\mathbb{Z}/2)$

I could be wrong. But think about it! Any guy in here comes from an associator on some skeletal 2-group whose group of objects is $\mathbb{Z}/2$, where every object has $\mathbb{Z}/2$ as its automorphism group. The 3-cocycle condition is just the pentagon identity!

So, given three guys $x,y,z \in \mathbb{Z}/2$, we’re looking for an associator

$a(x,y,z) \in \mathbb{Z}/2$

and we want this to satisfy the pentagon identity. Is there a nontrivial way to do this?

Posted by: John Baez on October 19, 2007 4:38 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

I’ve never quite seen why we’ve made such a meal of finite permutation 2-groups. There ought to be an accessible range acting on nice bite-size groupoids.

We have $S_2$ acting on two points. We can categorify that to the 2-group with $S_2$ of objects and trivial morphisms acting on 2 objects with trivial morphisms.

Then we want to allow the groupoid being acted on to have non-trivial morphisms. So we might try 2 objects each with one non-trivial morphism. That has symmetry 2-group with $G = A = \mathbb{Z}/2$, and trivial action of $G$ on $A$, and presumably trivial third cohomology element.

So if you’re right about $H^3(\mathbb{Z}/2, \mathbb{Z}/2)$ that it has a nontrivial element, what would a groupoid look like which the corresponding 2-group could act on in an interesting way?

Murat Alp’s thesis has a list (page I-77) of the smallest crossed-modules.

Posted by: David Corfield on October 19, 2007 10:44 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David wrote:

… and presumably trivial third cohomology element.

Let me just say a well-known fact about the 3rd cohomology class that any 2-group gives. (See HDA5 for definitions and proofs.)

If we have a strict 2-group (associator and unitors are identity morphisms), and it’s skeletal (isomorphic objects are equal), this 3rd cohomology class is trivial.

I think this is the case in your example, namely $AUT(X)$ where $X$ is the groupoid $\mathbb{Z}/2 \sqcup \mathbb{Z}/2$ — the disjoint union of two copies of $\mathbb{Z}/2$. This 2-group has two objects: the identity $1 : X \to X$ and the invertible functor that switches the two copies of $\mathbb{Z}/2$. And, it’s skeletal, since there are no natural isomorphisms between these functors. It’s also strict — 2-groups of the form $AUT(X)$ for $X$ a category always are. So, yeah, the 3rd cohomology class is trivial here.

So if you’re right about $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$ that it has a nontrivial element,…

Let’s see. Taking a peek in Weibel’s Introduction to Homological Algebra, I see that if $A$ is any $\mathbb{Z}/n$-module,

$H^3(\mathbb{Z}/n,A) = \{a \in A: N a = 0\}/(\sigma - 1)A$

Here we write $\sigma$ for the generator of $\mathbb{Z}/n$, and we think of this group multiplicatively, so it consists of $1, \sigma, \sigma^2 , \dots, \sigma^{n-1}$. And:

$N a = (1 + \sigma + \sigma^2 + \cdots + \sigma^{n-1})a$

When I wrote $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$ I was implicitly treating $A = \mathbb{Z}/2$ as a trivial module over $\mathbb{Z}/2$, so that $\sigma$ acts as the identity. So in this case,

$N a = (1 + \sigma)a = 2a = 0$

for all $a \in \mathbb{Z}/2$. On the other hand $\sigma - 1$ acts as zero on $A$. So,

$\{a \in \mathbb{Z}/2: N a = 0\} = \mathbb{Z}/2$

while

$(\sigma - 1)\mathbb{Z}/2 = 0$

so the quotient $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$ is $\mathbb{Z}/2$.

Someone should check this — it’s easy to make mistakes in simple calculations when you’re simultaneously writing them in TeX.

But, if the 3rd cohomology group really is $\mathbb{Z}/2$, there should be some very nice little skeletal 2-group with $\mathbb{Z}/2$ as its group of objects, $\mathbb{Z}/2$ as the automorphisms of any object, and a nontrivial associator — probably something nice like

$a(x,y,z) = x y z$

where again I’m writing $\mathbb{Z}/2$ multiplicatively, and $x,y,z \in \mathbb{Z}/2$ are three objects.

Sorry I’m not answering all your questions — we really need some other people to join in. But, having a few small 2-groups to play with will make Klein 2-geometry more fun.

Posted by: John Baez on October 19, 2007 6:37 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

So it looks like there are two weak 2-groups with $G = \mathbb{Z}/2 = H$. One of them can be made skeletal and strict, the other can’t as the nontrivial third cohomology element shows.

How do you view the latter then? Do you think of all bracketed products of $1$ and, say, $A$ as objects, with morphisms between them, including associators and weak inverses, so they fall into two components, depending on parity of number of $A$s.

But what can be said about what they act on and how? Presumably, one can do the Cayley trick as with groups, and say that a 2-group acts on itself. Every 2-group is a sub-2-group of its own automorphism 2-group.

It looks like our strict 2-group is equivalent to its own automorphism 2-group. What happens with the weak one?

Posted by: David Corfield on October 20, 2007 12:56 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

I don’t understand the idea of an automorphism group for a weak 2-group. Where can we learn more about weak 2-groups?

I think I do understand the concept of an automorphism group (which is an abelian group) for a pseudo 2-group. The definition of a pseudo 2-group is not too hard to grasp. Please see section 3 “Recall on 2-groups and crossed-modules” at the bottom of page 2-3 of this paper by B. Noohi for the definition of a pseudo 2-group.

The underlying category of a pseudo 2-group is a groupoid, and pseudo 2-groups and strict monoidal functors between them form a category which contains the category of strict 2-groups as a full subcategory.

Posted by: Charlie Stromeyer Jr on October 20, 2007 2:58 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Baez and Lauda, Higher-Dimensional Algebra V: 2-Groups.

Try page 52 for automorphism 2-groups.

Posted by: David Corfield on October 20, 2007 3:13 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David almost wrote:

So it looks like there are two weak 2-groups with $G=\mathbb{Z}/2= A$. One of them can be made skeletal and strict, the other can’t as the nontrivial third cohomology element shows.

Right. But, I changed your ‘$H$’ to an ‘$A$’ here.

I like to use $A$ for the (abelian!) group of endomorphisms of the unit object in a 2-group, especially when describing 2-groups in terms of cohomology classes, as we are here. Every 2-group can be made skeletal and then described as a quadruple $(G,A,\alpha,a)$ — group, abelian group, action and 3-cocycle (coming from the associator).

And, I like to use $H$ for the group of morphisms whose source is the identity object of our 2-group, especially when describing 2-groups in terms of crossed modules. Every 2-group can be made strict and then described as a crossed module $(G,H,\alpha, t)$ — group, group, action and homomorphism $t: H \to G$ (coming from the target map).

We’re playing the first of these two games here. So, I’m making your notation fit my conventions.

How do you view the latter then? Do you think of all bracketed products of 1 and, say, $-1$ as objects, with morphisms between them, including associators and weak inverses, so they fall into two components, depending on parity of number of $-1$s.

(Again I changed your notation to not collide with my conventions.)

What you say is okay, but note: the bracketing is unnecessary! When we have a skeletal 2-group, as we do here, isomorphic objects are equal, so $(x \otimes y) \otimes z) = x \otimes (y \otimes z)$ for all objects $x,y,z$, even if the associator is nontrivial!

So, you can remove the brackets in your expressions without causing any ambiguity. Our 2-group coming from the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$ has just two objects: $1$ and $-1$. It has

$1 \otimes -1 = -1 \otimes 1 = -1$

and

$1 \otimes 1 = -1 \otimes -1 = 1$

In other words, the group of objects is just $\mathbb{Z}/2$. Each object has $\mathbb{Z}/2$’s worth of automorphisms. The only really exciting thing about this 2-group is the associator, which I still haven’t figured out explicitly — though I made a guess.

But what can be said about what they act on and how? Presumably, one can do the Cayley trick as with groups, and say that a 2-group acts on itself. Every 2-group is a sub-2-group of its own automorphism 2-group.

(Here you mean its automorphisms as a groupoid, not as a 2-group.)

Hmm. This is a bit subtle for weak 2-groups. After all, the 2-category of groupoids, functors and natural transformations is strict. So, even if your 2-group $X$ is weak, the 2-group of automorphisms of its underlying groupoid, which I’ll call $AUT(X)$, is strict.

I think we nonetheless get a homomorphism $X \to AUT(X)$ by letting $X$ act on itself by left translations. $X$ is weak, $AUT(X)$ is strict, but there’s no paradox: the trick is that the homomorphism itself is weak!

However, I forget if this homomorphism makes $X$ into a ‘sub-2-group’ of $AUT(X)$ in some sense.

You may recall that we had a big discussion in an earlier Klein 2-geometry thread, where I worked out different notions like ‘monic’ for homomorphisms of strict 2-groups, and tried to guess the correct (non-evil) definition of ‘sub-2-group’ in the strict case. All this needs to be revisited in the weak case, to see if the Cayley embedding really works.

(This was the summer before last, shortly before we started the $n$-Category Café. I did the calculations in a Starbucks in Shanghai.)

It looks like our strict 2-group is equivalent to its own automorphism 2-group. What happens with the weak one?

I’m not sure. If worst comes to worst, we can replace it with an equivalent strict one… which can’t be skeletal. But, that’ll make it bigger.

Posted by: John Baez on October 21, 2007 11:45 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

I think we nonetheless get a homomorphism $X\to\mathrm{AUT}(X)$ by letting $X$ act on itself by left translations. $X$ is weak, $\mathrm{AUT}(X)$ is strict, but there’s no paradox: the trick is that the homomorphism itself is weak!

However, I forget if this homomorphism makes $X$ into a ‘sub-2-group’ of $\mathrm{AUT}(X)$ in some sense.

Every monoidal category $X$ is monoidally equivalent to the monoidal category whose objects are endomorphisms $E:X\to X$ with a natural transformation $\delta:EA\otimes B\to E(A\otimes B)$ satisfying some coherence conditions, and whose arrows are natural transformations $E\Rightarrow F:X\to X$ compatible with $\delta$. The tensor is the composition of endomorphisms , with a corresponding composition of the $\delta$s. You send an object $A$ of $X$ to the endofunctor $A\otimes -:X\to X$, with $\delta$ given by the associativity. This is an application of the 2-categorical Yoneda lemma and is used to prove that each monoidal category is equivalent to a strict one. See Tom Leinster’s book, pages 15 and 32.

So there is a faithful (but not full, because of the compatibility condition with $\delta$) forgetful functor to the monoidal category of endomorphisms of $X$.

If you start with a 2-group, the functor $A\otimes -$ is an equivalence, so the faithful forgetful functor goes into the automorphism 2-group of $X$, and in this sense $X$ is a sub-2-group of $\mathrm{AUT}(X)$.

Posted by: Mathieu Dupont on October 22, 2007 11:29 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Thanks a lot, Mathieu! I was thinking this issue seemed a bit Yoneda-esque.

So, David, 2-groups act on themselves quite nicely by left translation, even weak ones. We can then start pondering ‘homogeneous spaces’ for these 2-groups, and their Klein 2-geometries, and so on. It should be fun to do with that little weak 2-group coming from $H^3(\mathbb{Z}/2,\mathbb{Z}/2) \cong \mathbb{Z}/2$.

Posted by: John Baez on October 22, 2007 8:58 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

What you say is okay, but note: the bracketing is unnecessary!

I think I was trying to go strict but not skeletal. Given a weak 2-group is it the case that one can opt to make it skeletal OR strict, but only both if that third cohomology element is trivial?

Let’s say we went skeletal as you suggest. So we have $\mathbb{Z}/2$’s worth of objects. So then all associators are morphisms from an object to itself. And your guess was that the associators on 1 are all $Id_{1}$, and those on $-1$ are all $-1_{-1}$, i.e., the nontrivial morphism at $-1$.

Posted by: David Corfield on October 23, 2007 10:45 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

I know, e.g., that one can always strictify a weak 2-groupoid without losing any homotopy theoretic information and without altering derived mapping spaces, but I don’t know what is meant by the term “skeletal”.

Is this term also discussed, e.g., in the Baez and Lauda paper? Thanks.

Posted by: Charlie Stromeyer Jr on October 23, 2007 1:27 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

On p. 5 they talk about a skeletal 2-group as

…containing just one representative from each isomorphism class of objects.

That corresponds in our example to just having two objects.

Posted by: David Corfield on October 24, 2007 2:05 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

A category is ‘skeletal’ if isomorphic objects are equal.

Every category is equivalent to some skeletal subcategory (also known as a ‘skeleton’). To show this, you pick a representative for each isomorphism class of objects, pick an isomorphism from each object to the representative of its isomorphism class, and follow your nose from there.

Working with skeletal categories is never necessary, precisely since every category is equivalent to a skeletal one. But occasionally it’s convenient. Choosing a skeletal subcategory of a category is a little bit like choosing a basis for a vector space. (Just a little bit.)

Posted by: John Baez on October 25, 2007 1:30 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

$a(x, y, z) = x y z$

doesn’t satisfy the pentagon identity.

We would need

$(w x y z) \cdot (w x y z) = (w x y) \cdot (w x y z) \cdot (x y z)$

for all $w, x, y, z \in \mathbb{Z}/2$. But if $x = -1$, and the others are $+1$, then this condition fails.

$a(x, y, z) = x z?$

Then the pentagon identity requires that

$(w x z) \cdot (w y z) = (w y) \cdot (w z) \cdot (x z),$

which is right.

Posted by: David Corfield on October 24, 2007 2:25 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

I’m joining this discussion at David Corfield’s invitation. Mikael Vejdemo Johansson and I have looked at this and reckon we have an answer. The nontrivial 3-cocycle is $a(x,y,z)=xyz$. I too got the pentagon identity wrong the first time. Actually, the pentagon identity says $x y z - (w+x)y z + w(x+y)z - w x(y+z) + w x y = 0$ and fortunately this is true. The reason that the pentagon identity is like this is that addition is the group structure on $\mathbb{Z}/2$, not multiplication.

We still need to prove that this is nontrivial. One way is to say that the cohomology ring is known and this class is the cube of the polynomial generator in degree one. The diagonal approximation for the standard resolution is $\Delta[g_1|...|g_n] = \sum_{r=0}^{n} [g_1|...|g_r]\otimes g_1...g_r[g_{r+1}|...|g_n]$ as you may read e.g. on p. 108 of Brown’s book (Springer GTM 87).

Or one can check by hand. Suppose $a = \delta b$. Then $b(y,z) - b(x+y,z) + b(x,y+z)- b(x,y)=x y z.$ Putting $y=0$ and $z=0$ we have $b(x,0)=b(0,0)$. For similar reasons we have $b(0,x)=b(0,0)$. Now putting $x=y=z=1$ we get $0=1$, a contradiction. So $a$ isn’t a coboundary.

Posted by: David Green on October 25, 2007 3:05 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Ah. I think you’ve confused additive and multiplicative notation here.

You want to calculate $H^3(\mathbb{Z}/2, \mathbb{Z}/2)$, but the two $\mathbb{Z}/2$s are different, so let’s actually call the latter $F_2$, the field with two elements (we will want to use the extra structure in the field). I will think of both $\mathbb{Z}/2$ and $F_2$ as consisting of the elements $\{0,1\}$, so the group structure is written additively.

I would tend to consider the group cohomology $H^*(\mathbb{Z}/2,F_2)$ as the cohomology of real projective space (which is a classifying space for $\mathbb{Z}/2$) and I know that that cohomology is just the polynomial ring on one generator in degree one.

The generator for $H^1(\mathbb{Z}/2,F_2)$ is the “identity map” $\mathbb{Z}/2\mapsto F_2$. Now the ring structure on $H^*(\Z/2,\mathbf{F}_2)$, I think, is just the naive thing, so for example if $\theta$ and $\phi$ are $1$-cocycles then $\theta.\phi(x,y)=\theta(x)\phi(y)$ (here we’re using the product in $\F_2$). This means that the generator of $H^3(\mathbb{Z}/2,F_2)$ can be defined by $b(x,y,z)=xyz$ as was claimed. Now we have to check the cocycle condition (remembering that the group product is written additively), this is the following:

(1)$b(x,y,z)+b(w,x+y,z)+b(w,x,y)=b(w+x,y,z)+b(w,x,y+z)$

ie

(2)$x y z+w(x+y)z+w x y=(w+x)yz +w x(y+z).$

That’s clearly true.

Posted by: Simon Willerton on October 25, 2007 3:10 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

My “Ah. I think you’ve confused additive and multiplicative notation here.” did, of course, refer to David C’s comment and not David Green’s response immediately above: the latter was posted while I was writing my comment!

Posted by: Simon Willerton on October 25, 2007 3:26 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David wrote:

$a(x,y,z)= x z?$

Great! Annoyingly asymmetrical compared to my guess… but at least it’s not clearly wrong!

You showed $a$ is a 3-cocycle. To finish proving your answer is ‘right’, we need to show it’s not a 3-coboundary. This will mean our non-strict skeletal 2-category isn’t equivalent to a strict skeletal one.

(Since the cohomology group in question is just $\mathbb{Z}/2$, there’s just one 3-cocycle mod coboundaries — so while there’s more than one right answer, they’re all equivalent.)

How to show $a$ isn’t a cocycle? If I remember correctly, this means we need to check that $a$ is not of the form

$a(x,y,z) = b(y,z) \; b(x y,z)^{-1} \; b(x,y z) \; b(x,y)^{-1}$

(To remember this formula, you just need to know the answer to the riddle “what do you do if you’ve got a schoolbus that comfortably holds one kid per seat, and one day an extra kid shows up when all the seats are full?” If you give up, read page 28 here.)

I’m too lazy to check all $2^{2^2} = 16$ possible choices of $b$ and check that none of them works. Given that $a(x,y,z)$ is independent of $y$, maybe it’s good to separately consider two cases, $y = +1$ and $y = -1$. But I’m still too lazy. There should be some slick trick…

Still, it would be nice to understand this example. Here’s one thing it would be good for: it would let us put a funny monoidal structure on the category of $\mathbb{Z}/2$-graded vector spaces, with the usual tensor product, but an associator that included a minus sign sometimes, governed by this 3-cocycle $a$.

That would be amusing, since supersymmetry involves a funny symmetric monoidal structure on the category of $\mathbb{Z}/2$-graded vector spaces.

Posted by: John Baez on October 25, 2007 2:06 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

The obvious $b$ to check for is $b(x, y) = x y$. Then your formula gives us

$(y z) \cdot (x y z)^{-1} \cdot (x y z) \cdot (x y)^{-1}.$

But in our case this is equal to $(x z)$, so my suggestion is a coboundary.

I can’t believe there isn’t a way to derive the right cocycle.

Posted by: David Corfield on October 25, 2007 9:54 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

A quick online search for 3-cocyles yielded nothing. But it did throw up some potential physics interest in the shape of three Bernard Grossman papers.

1) The meaning of the third cocycle in the group cohomology of nonabelian gauge theories.

2) A 3-cocycle in quantum mechanics

3) Three-cocycle in quantum mechanics II. Abstract:

We offer a new topological interpretation of the three-cocycle in a consistent quantum mechanics. Our basis is homotopy theory. We show how higher cocycles may appear in quantum mechanics. Moreover, a three-cocycle gives rise to a nonassociative algebra describing defects in a quantum-mechanical system. We also show the relation of the three-cocycle to axions in string theory.

Posted by: David Corfield on October 25, 2007 11:36 AM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

As cohomology forms a ring, two elements of degree 3 can be added. Does this correspond to any operation on corresponding 2-groups?

And what would the product element of degree 6 mean in terms of anything?

Posted by: David Corfield on October 25, 2007 12:42 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Thanks for straightening things out, David and Mikael and Simon!

There’s a bit of residual confusion, which I should try to undo. David and I were using multiplicative notation for the group operation in $\mathbb{Z}/2 = \{+1, -1\}$, since multiplicative notation is what folks use for composing morphisms.

So, David and I wrote the 3-cocycle condition multiplicatively. and my guess at a solution:

$a(x,y,z) = x y z$

was also supposed to be interepreted in a way where multiplication stands for the group operation in $\mathbb{Z}/2$.

So, David didn’t get the cocycle condition wrong… but it turns out that for solving this problem, it’s nicer to write the group operation in $\mathbb{Z}/2$ as addition. Then the cocycle condition gets written additively, and a guess that looks just like mine:

$a(x,y,z) = x y z$

turns out to be correct — but only if we now intepret multiplication as multiplication in the ring $\mathbb{Z}/2$.

In short: a very happy ending. A nice symmetrical-looking associator.

Posted by: John Baez on October 25, 2007 7:51 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Ah, this is good. If you guys can figure out what 2-Klein geometry is then what would you be able to know about 2-Cartan geometry?

I ask because Cartan geometries are deformed analogues of Klein geometries, and because this weekend I will try to figure out if or how such Cartan geometries are related to the bundle representation of von Neumann’s continuous geometry (which is a generalization of projective geometry which is an example of parabolic geometry which is a type of Cartan geometry).

Posted by: Charlie Stromeyer Jr on October 25, 2007 8:52 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

David wrote:

I can’t believe there isn’t a way to derive the right cocycle.

There must be. To figure it out, we’d need to study how people compute the cohomology of cyclic groups. It’s very explicit and it’s all presented quite nicely in — for example — Weibel’s Introduction to Homological Algebra. (It’s probably in every book on homological algebra going back to Mac Lane’s Homology.) There’s a nice efficient projective resolution of any $\mathbb{Z}/n$-module $A$ and from this one computes the cohomology groups

$H^k(\mathbb{Z}/n, A)$

They turn out to depend only on $k mod \; 2$, except for $k = 0$, which acts a bit differently.

Staring closely at this computation, we could find explicit formulas for all the nontrivial $k$-cocycles.

However, we don’t need to! The big boys have joined our game and beat us to the finish line — I’m talking about David and Mikael and Simon.

So, we can do what we do best, which is to ponder the results of the computation.

A quick online search for 3-cocyles yielded nothing.

It helps to spell it right.

But seriously, the idea I mentioned — using a 3-cocycle to ‘twist the associator’ of a monoidal category — can be found in papers by Yetter and (secretly) Majid.

(Instead of talking about monoidal categories, Majid tends to talk about the bialgebras whose categories of representations are those monoidal categories. But, Tannaka–Krein reconstruction means there’s no big difference — once you’ve inserted suitable technical fine print.)

Yetter and Majid also consider twisting the braiding in a braided monoidal category. That requires a different sort of cocycle.

All this is part of a huge general pattern that spans the Periodic Table: you can twist various ‘-ators’ in your $k$-tuply monoidal $n$-category, and the condition for the new twisted ‘-ator’ to still satisfy all the right coherence laws will always be some sort of ‘cocycle condition’. In nice cases you can do computations in group cohomology (or more generally, the cohomology of topological spaces!) to work out the complete set of possibilities.

To understand this fully, one needs to ponder the connection between $n$-categories and cohomology. I’ve spend a lot of time doing that — it provides a delicious sort of mixture of abstraction and concrete computations, grand traditional themse and novel $n$-categorical ideas, which could easily keep a mathematician happy for a lifetime.

Except there’s so much other stuff to do, and I’m only expecting to have one lifetime.

As cohomology forms a ring, two elements of degree 3 can be added. Does this correspond to any operation on corresponding 2-groups?

That’s a really interesting idea! I guess the idea is that you can ‘add’ the associators of two 2-groups that are the same in every other respect, and get the associator for a new 2-group. But it would be nice to have a more ‘conceptual’ interpretation. I think there’s one that involves ‘tensoring principal bundles’. But how about one in terms of Klein geometry?

And what would the product element of degree 6 mean in terms of anything?

As explained in those notes on $n$-categories and cohomology, you can think of cohomology classes as classifying bundles. Then I think the product in cohomology corresponds to the fiberwise product of bundles.

(Bundles of spaces, bundles of $n$-groupoids — there’s no real difference.)

It would again be nice to have an interpretation in terms of Klein geometry or (equivalently) logic. I’m not really good at Klein 5-geometry, though.

Posted by: John Baez on October 25, 2007 8:05 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

There are a lot of fun calculations to do here… but not this morning!

Let me just say, David, that you could really show those analytic philosophers a thing or two — beat them at their own game, as it were — by writing a little paper that established a nontrivial connection between modal logic and 2-groupoids, and used the cohomology of finite groups to get some interesting examples. I don’t know if you’d want to do that; maybe you’d consider it ‘going over to the dark side’. But, it would show them the advantages of knowing some real math. And, it could be a very interesting paper!

Posted by: John Baez on October 19, 2007 5:55 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Yes, it would be good to write such a paper, whatever it had to with metaphysics.

If there’s anyone to relate such ideas to it’s the philosopher of physics Jeremy Butterfield. He’s done work on the modality involved in classical mechanics, Lagrangian and Hamiltonian.

Posted by: David Corfield on October 20, 2007 12:29 PM | Permalink | Reply to this

### Re: Klein 2-Geometry IX

Possible paths, like possible worlds?

I bet he didn’t point out that assigning a “possibility” (0 or 1) to each path is a lot like assigning a probability or an amplitude or an action.

Posted by: John Baez on October 22, 2007 8:14 AM | Permalink | Reply to this

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