The n-Category Café

OK, I see. I tried choosing an arbitrary isomorphism from each object to the chosen member of its isomorphism class, but these isomorphisms then have to satisfy coherence equations with respect to the monoidal structure, which won’t happen in general. Perhaps this can be sorted out by introducing associativity and unit natural transformations that ‘cancel out’ these coherence equations — which, of course, means you’re not strict any more.

Thanks, Tom!

Jamie, your proposal is very interesting, though I must admit I am more confused than ever. By the way, you wrote:

the correct choice for the associators for infinite sets just jumps out at you: use (1) to work out $A \times (B \times C)$ and $(A \times B) \times C)$, and the associator is the bijection that maps between them.

Remember, what I’d like to see is not the “correct” associator on Set, as such. Surely that’s the ordinary one $((a,b),c) \mapsto (a,(b,c))$ using your favourite set-theoretic model of ordered pair…gulp! I’d like to see how this associator explicitly pulls back to Card, the category whose objects are cardinalities (i.e. five, seven, aleph zero, aleph one and so on).

Having mentioned things like “cardinality” and “aleph” it’s clear there is some horrible set theory going down here. I guess the point is that for finite sets, there is somehow a canonical choice for a representative in each isomorphism class. It’s the set

For infinite sets, there’s no such canonical choice, right? We can’t write

Well we can… but that’s not canonical. I guess this is just the ordinal numbers vs cardinal numbers thing.

Anyhow, let me throw all that infinite stuff aside, and concentrate on finite sets. Suppose we consider the monoidal category (Numbers, $\times$, 1). Objects are positive natural numbers, morphisms are functions $f: \{1,2,\ldots, n\} \rightarrow \{1,2,\ldots,m\}$. The monoidal product is just product of numbers, and the associator is the identity.

You are saying there is another associator, another gismo which solves the pentagon equation, which is equivalent to the trivial one. It’s the “reindexing” thing you outlned above. It’s certainly not the identity. It’s got a nasty formula, i.e.

is a nontrivial bijection. My thoughts are the following:

1. I don’t this this will work; I can’t see how such a complicated formula could solve the naturality and the pentagon equations… but I may be wrong.

2. I think there is up to monoidal isomorphism only one associator on Numbers… right?

3. In fact I think there is only one associator on Numbers… the trivial one. Nothing else solves naturality and the pentagon equations.

4. I am a bit ashamed I don’t know the answers to this stuff.

Jamie wrote:

I reckon we can also make a category of finite-dimensional vector spaces strict and skeletal… can somebody back me up on this?

Depressingly, I’ve never thought about this either. It’s a very important question.

If the trick you used to make $(\Fin\Set, \times, 1)$ strict and skeletal really works (I haven’t had time to check), the same sort of trick should probably work for $(\Fin\Vect, \otimes, k)$.

I think this is normally called Mat(k), with natural numbers for objects and $k$-matrices for morphisms.

Yeah. But the key issue here, which I’ve never seen discussed, is how we identify the tensor product of an $n \times n$ matrix and an $m \times m$ matrix with an $n m \times n m$. Presumably we want to use your isomorphism of finite ordinals $n \times m \cong n m$, where $n \times m$ is ordered lexicographically.

In fancy jargon, this amounts to taking advantage of the monoidal ‘free vector space’ functor $F: \Fin\Set \to \Fin\Vect$, sending $n$ to $k^n$.

I reckon we can also make a category of finite-dimensional vector spaces strict and skeletal…

I guess there’s an ‘indirect’ proof of this as well. Firstly we observe that there can only be one associator on the ordinary category of vector spaces… the usual one. That’s because an associator is natural, so it is completely determined by its component $a_{\mathbb{C}, \mathbb{C}, \mathbb{C}}$, where it can only be the map

because the pentagon equation has the form

and so it only has one solution… the trivial one (this argument holds water, right? In general recall that there are only finitely many possible associators on any fusion category… this is Ocneanu rigidity.)

Secondly, it is easy to make $Vect$ skeletal… we just choose the full subcategory $Vect_0$ having objects $\mathbb{C}^n$. We pull-back the tensor-product and associator from $\Vect$ to make $\Vect_0$ a monoidal category (this is the ‘indirect’ step because sneakily we’re making an ‘uncardinal’ number of choices in order to construct an explicit equivalence of categories $\Vect \cong \Vect_0$, which we need in order to pull-back this data).

So here is our candidate strict monoidal category, $(\Vect_0, \otimes, a)$. It must indeed be strict, because if it wasn’t we’d have a contradiction: it would mean there would be two solutions to the pentagon equations (because we always have the trivial associator on a strict category which is equipped with a tensor product $\otimes$).

Bruce wrote:

(this argument holds water, right?)

The first stage of your argument argument seems to hold water. You’re using some special features of $Vect$ to show that $Vect$ with its usual tensor product can have only one associator satisfying naturality and the pentagon identity: namely, the usual associator.

The second stage of your argument doesn’t seem to use anything special about $Vect$. You note that, just like any monoidal category $Vect$ is monoidally equivalent to a skeletal monoidal category $Vect_0$. It’s easy to prove this result if we use the axiom of choice, just by choosing any skeleton

$$Vect_0 \hookrightarrow Vect$$

and showing there’s a functor going the other way

$$Vect \to Vect_0$$

which gives an equivalence of categories

$$Vect \simeq \Vect_0$$

Then, using this, we can transport the tensor product on $Vect$ over to $Vect_0$. Then, we can show that different choices of associator for $Vect_0$ correspond in a 1-1 way with different choices of the associator for $Vect$.

Combining these two stages of your argument, we can conclude that the skeletal category $Vect_0$, equipped with the tensor product coming from $Vect$ as above, admits only one associator satisfying the pentagon identity.

But then, in the third stage of your argument, you seem to magically conclude that this associator must be the identity!

I don’t see that. It seems to me that we still need Jamie’s explicit construction to get the associator on $Vect_0$ to be the identity.

Here’s how you magically jump to your final conclusion:

we always have the trivial associator on a strict category which is equipped with a tensor product $\otimes$

I don’t know what this means. First, I don’t know what a ‘strict category’ is. Second, if you mean to say ‘strict monoidal category’, that’s a category that has a trivial associator — and yes, such a category always admits a trivial associator. But…

Oh, whoops — I bet you meant to say ‘skeletal category’. Hmm, is that true? You’re saying that given a monoidal category, and picking a skeletal subcategory and equipping it with a tensor product functor as above, we can always choose the identity

$$1 : (x_1 \otimes x_2) \otimes x_3 \to x_1 \otimes (x_2 \otimes x_3)$$

Is that right? It’s certainly well-defined because both sides are isomorphic (since we started with a monoidal category) hence equal (since now we’re in a skeleton). It’s natural. And, it satisfies the pentagon.

Hmm, so maybe this argument is right. But I still feel there should be a flaw, since there’s only one point where you used any special feature about $Vect$, right near the beginning… and I don’t see why $Set$ lacks this special feature! So, shouldn’t your argument prove that $(Set, \times, 1)$ admits a strict skeletal version, even though our betters have told us otherwise?

If it’s true that $FinSet$ is monoidally equivalent to a skeletal strict monoidal category, but $Set$ is not, there should be something about your argument that breaks down in the latter case. But right now it seems to work equally well in both cases!

Every monoidal functor $F:\mathbf{A}\to \mathbf{B}$ can be factorized in the following way:

$$\mathbf{A}\overset{E}\longrightarrow\mathrm{Im}_{\mathrm{pl}}F\overset{M}\longrightarrow \mathbf{B},$$

where $E$ is a monoidal functor which is an identity on objects and $M$ is a full and faithful monoidal functor.

The objects of the full image of $F$, $\mathrm{Im}_{\mathrm{pl}}F$, are the objects of $\mathbf{A}$ and its arrows are the arrows of $\mathbf{B}$ :

$$(\mathrm{Im}_{\mathrm{pl}}F)(A,A^'):=\mathbb{B}(F A,F A^').$$

The composition and identities are those of $\mathbf{B}$. The tensor product is defined on object as in $\mathbf{A}$; the tensor product of $b:A_1\to A_2$ and $b^':A^'_1\to A^'_2$ (i.e. $b:F A_1\to F A_2$ and $b^':F A^'_1\to F A^'_2$ in $\mathbf{B}$) is defined as the following composite:

$$F(A_1\otimes A^'_1)\overset{\varphi}\longrightarrow F A_1\otimes F A^'_1\overset{b\otimes b'}\longrightarrow F A_2\otimes F A^'_2\overset{\varphi}\longrightarrow F(A_2\otimes A^'_2),$$

where $\varphi$ is the natural transformation of the monoidal structure of $F$. The unit of the tensor product is the unit of $\mathbf{A}$ and the associativity, right unit and left unit natural transformations are defined as the image by $F$ of the corresponding transformations in $\mathbf{A}$.

The monoidal functor $E$ sends an object $A$ to $A$ and an arrow $a:A\to A^'$ to $F a:A\to A^'$. The natural transformation of monoidal functor is the identity. The monoidal functor $M$ sends an object $A$ to $F A$ and an arrow $b:A\to A^'$ to $b$ itself. The natural transformations of monoidal functor are those of $F$.

Now, if $F$ is a surjective (up to isomorphism) monoidal functor, then $M$ is a monoidal equivalence since it is full, faithful and surjective. If moreover the domain $\mathbf{A}$ is a monoid seen as a discrete monoidal category, the full image of $F$ is a strict monoidal category : the associativity and left and right unit natural transformations are the image by $F$ of those of $\mathbf{A}$, which are identities since $\mathbf{A}$ is a monoid. So, in this situation, $\mathbf{B}$ is monoidally equivalent to a strict monoidal category whose objects are the elements of the monoid $\mathbf{A}$.

We can now apply this result to the monoidal functor $[-]:\mathbb{N}\to \mathrm{FinSet}$ (both equipped with the product), which sends a natural number $n$ to the set $[n]:=\{k \lt n\}$. The structure of monoidal functor on $[-]$ is given by $$\varphi_{m,n}:[m]\times[n]\to[m n]:(j,k)\mapsto j n+k.$$ It is easy to check that $\varphi$ satisfies the hexagon condition of monoidal functor. The inverse of $\varphi_{m,n}$ is given by Euclidean division.

So the full image of $[-]$ is a strict monoidal category monoidally equivalent to $(\mathrm{FinSet},\times,1)$ whose objects are the natural numbers. In fact, the tensor product on this full image is given on objects by the product of natural numbers, and the tensor product of two functions $f:[m]\to[m']$ and $f^':[n]\to[n^']$ is, thanks to the second equation above, the function $[m n]\to[m^'n^']$ which sends a number $l$ to $f(j)n'+g(k)$, where $j$ and $k$ are the unique numbers such that $l=j n+k$.

For finite dimensional vector spaces, we can take the full image of the functor $$\mathbb{N}\overset{[-]}\longrightarrow\mathrm{FinSet}\overset{L}\longrightarrow\mathrm{FinVect},$$ where $L$ is the free vector space functor, which is surjective (and monoidal, I suppose).

By the way, we have talked about directed homotopy theory and Grandis’ work on it a (tiny) bit here.

The reason that I ask is that there’s an exciting conjecture that could be made, linking ‘commutative quantum $n$-algebras’ (i.e. commutative C*-algebras, symmetric 2-H*-algebras and so on) and $n$-groupoids for all $n$. It’s then natural to wonder whether the extension to noncommutative algebras on the one side might mirror an extension to non-groupoidal $n$-categories on the other.

… ultimately I want all concepts to combine in some sort of cosmic implosion — the mathematical ‘Big Crunch’

Well, to me that’s what’s so exciting about a lot of the things that we talk about on this blog — for example, your recent paper with Mike Stay is certainly very Crunchy.

Jamie almost wrote:

The reason that I ask is that there’s an exciting conjecture that could be made, linking ‘commutative quantum $n$-algebras’ (i.e. commutative C*-algebras, symmetric 2-C*-algebras and so on) and $n$-groupoids for all $n$.

Right: the categorified Gelfan’d–Naimark hypothesis, setting up a duality between compact Hausdorff $n$-groupoids $X$ and symmetric $(n+1)$-C*-algebras $A$: $$A \simeq hom(X,n Hilb)$$ $$X \simeq Spec(A)$$ or something like that.

(The $n = 0$ case is the usual Gelfan’d–Naimark theorem; already for $n = 1$ I only know how to handle a limited special case of this hypothesis!)

It’s then natural to wonder whether the extension to noncommutative algebras on the one side might mirror an extension to non-groupoidal $n$-categories on the other.

Is it? I don’t see why. It’s probably good to ponder some $n = 1$ examples.

… ultimately I want all concepts to combine in some sort of cosmic implosion — the mathematical ‘Big Crunch’

Well, to me that’s what’s so exciting about a lot of the things that we talk about on this blog — for example, your recent paper with Mike Stay is certainly very Crunchy.

Thanks! That’s a nice-sounding adjective.

It’s interesting that mathematicians seem generally less grandiose than physicists when it comes to Dreams of a Final Theory. Perhaps it’s just more obvious that math is an infinite endeavor. But anyone who believes in god, or even in the depth of nature, should be skeptical of claims that we’ll figure out all the fundamental laws of physics anytime soon. I wouldn’t be surprised if they go on forever too.

John corrected me to:

The reason that I ask is that there’s an exciting conjecture that could be made, linking ‘commutative quantum $n$-algebras’ (i.e. commutative C*-algebras, symmetric 2-C*-algebras and so on) and $n$-groupoids for all $n$.

Hmm… I’m pretty sure I haven’t seen a definition of a 2-C*-algebra before. Do you not think 2-H*-algebras will be good enough for the hypothesis? Or is 2-C*-algebra the same thing as a 2-H*-algebra? I don’t see how the difference between H*-algebras and C*-algebras categorifies!

It’s then natural to wonder whether the extension to noncommutative algebras on the one side might mirror an extension to non-groupoidal $n$-categories on the other.

Is it? I don’t see why. It’s probably good to ponder some $n=1$ examples.

Well, if there is any sort of extension along these lines, I’m not saying it’s easy to find. We fall at the first hurdle if we naively try and look at the unitary representations of general $n$-categories in $n$Hilb, because ‘unitary’ doesn’t mean anything. So it’s difficult to say outright that the generalisation doesn’t work, because it can’t even be posed.

All I know is that people study sophisticated generalised spaces that have non-commutative ‘algebras of functions’ on them in an attempt to formulate a non-commutative spectral theorem, and it doesn’t seem impossible to me a priori that these could have something to do with ‘1-way topological spaces’.

Will the final thing to disappear in the mathematical Big Crunch be the gulf between the Two Cultures?

David wrote:

Will the final thing to disappear in the mathematical Big Crunch be the gulf between the Two Cultures?

That might even persist after the Crunch! There will always be lots of tricky problems for the problem-solvers to chew on, so I can imagine this science fiction scenario: the theory-builders unify all the beautiful abstract concepts they know and decide mathematics is done, but the problem solvers disagree and keep on working. The theory-builders would say the problem-solvers had been reduced to playing with shallow puzzles, while the problem-solvers would say the theory-builders had become trapped in a deadly edifice of incomprehensible abstraction.

In terms of astrophysics analogies, this would resemble the Big Crunch less than a black hole that swallowed up some but not all mathematicians.

It would make a fun story, but I hope it doesn’t actually happen — though some might argue it’s already well underway.