The n-Category Café

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.

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.

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

Try page 52 for automorphism 2-groups.

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.

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.

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:

ie

That’s clearly true.

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.

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.

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?