The n-Category Café

Above I began convincing myself that I should be interested in considering $\mathbb{C}Set$-vector bundles with connection, i.e. pseudo-1-functors $$\mathrm{tra} : P_1(X) \to {}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod} \,.$$

As always, I immediately restricted myself to those $\mathbb{C}\mathrm{Set}$-modules that are of the form $$W = [C^\mathrm{op},\mathbb{C}\mathrm{Set}]$$ for some category $C$.

On these functor categories, $\mathbb{C}\mathrm{Set}$ acts by “pointwise multiplication in $\mathbb{C}\mathrm{Set}$”.

To better see what this means, compare this to ordinary vector spaces with a choice of basis $B$.

These are 0-functors (functions) from the set $B$ to the set $\mathbb{C}$ (I’ll stick to working over the complex numbers, but it’s not relevant.)

$$\mathrm{dim}_\mathbb{C}(V) = |B| \;\; \Rightarrow \;\; V \simeq [B,\mathbb{C}] \,.$$

So what I called “nice” $V$-vector spaces previously, are really $V$-vector spaces with a choice of basis, in a sense.

But notice, crucially, that I am not restricting my basis category $C$ to be a 0-category. If I did so, i.e. would find the “Kapranov-Voevodsky”-version of $\mathbb{C}\mathrm{Set}$-vector spaces, namely those of the form $$W = [\mathrm{Disc}(B),\mathbb{C}\mathrm{Set}] \,.$$

As I don’t tire of emphasizing, we have no need, in general, to restric ourselves to $$\mathrm{KV}-\mathbb{C}\mathrm{Set}-\mathrm{Vect} \hookrightarrow {}_{\mathbb{C}\mathrm{Set}} \mathrm{Mod} \,.$$

You might recall that I offerered the First $n$-Café-Millenium Prize to draw attention to this. .

Anyway. I want to undestand this 2-category $${}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod}$$ of $\mathbb{C}\mathrm{Set}$-vector spaces.

To do so, I should first better understand $\mathbb{C}\mathrm{Set}$ itself, and its role as a category we are enriching over.

First, we need a monoidal structure on $\mathbb{C}\mathrm{Set}$. We want this to be cartesian product of sets, combined with the product in $\mathbb{C}$.

In other words, for $$s : S \to \mathbb{C}$$ and $$t : T \to \mathbb{C}$$ two $\mathbb{C}$-sets, their product should be $$(s : S\to \mathbb{C}) \otimes (t : T\to \mathbb{C}) := ( S \times T \stackrel{s\times t}{\to} \mathbb{C}\times \mathbb{C} \stackrel{\cdot}{\to} \mathbb{C} ) \,.$$

Now comes the first potentially interesting point: what is then the closed structure on $\mathbb{C}\mathrm{Set}$?

Clearly, we expect $$\mathrm{hom}(s : S \to \mathbb{C}, t : T \to \mathbb{C})$$ to be the $\mathbb{C}$-set which as a set is essentially just $\mathrm{Hom}(S,T)$.

But which $\mathbb{C}$-labels $$h : \mathrm{Hom}(S,T) \to \mathbb{C}$$ do we put on these maps?

Let me write $$\{ A_{c}, B_{c_2}, \cdots \}$$ for a $\mathbb{C}$-set, where the $\mathbb{C}$-labels are in the subscript.

Then in particular morphisms

exist if and only if $$c_3 = c_1 \cdot c_2 \,.$$

First assume that $c_2 \neq 0$. Then it follows that $$\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\})$$ must be the 1-element set, with the label of the single element being $$\frac{c_3}{c_2} \,.$$

But what happens if $c_2 = 0$? A short reflection shows that then the internal hom object $$\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\})$$ cannot exist! Because if also $c_3 = 0$ we would find that $$\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\}\otimes \{B_{c_2}\} , \{C_{c_3}\} )$$ is the 1-element set, for any value of $c_1$, but no matter what we would choose for $$\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) \,,$$ the cardinality of the set $$\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\} , \mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) )$$ would depend on the label of $c_{1}$. Hence there would not be a natural isomorphism $$\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\}\otimes \{B_{c_2}\} , \{C_{c_3}\} ) \simeq \mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\} , \mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) ) \,.$$

Therefore, with the above choice of tensor product (which we need if we want to have any chance to recover what we want to recover after applying the $\mathbb{C}$-cardinality map) is not closed!

What to do now? I need closedness because I want to enrich over $\mathbb{C}\mathrm{Set}$, which in turn I need in order to talk about $\mathbb{C}\mathrm{Set}$-modules in a sensible way!

At the moment, I can see only one conclusion that this is forcing upon as:

We need to quit using $\mathbb{C}$ and start using $\mathbb{C}^\times$.

While I’d have to see where this leads us, it might be potentially interesting: this will imply that our quantum phases are non-vanishing complex numbers. Which indeed they are supposed to be.

Well, they are even supposed to be unimodular complex numbers, i.e. elements of $U(1) \subset \mathbb{C}^\times \subset \mathbb{C}$.

And indeed, that’s what Jeffrey Morton considers: $U(1)\mathrm{Set}$ instead of $\mathbb{C}\mathrm{Set}$.

Maybe I am just doing nothing but finding (another?) argument for why we indeed are forced to use $U(1)\mathrm{Set}$.

But for the moment I should probably continue with $\mathbb{C}^\times\mathrm{Set}$.

Okay. In that case, we now know that $\mathbb{C}^\times\mathrm{Set}$ is indeed closed. The internal Hom-object is $$\mathrm{hom}_{\mathbb{C}^\times\mathrm{Set}}( \mathbf{S}, \mathbf{T} ) = ( h : \mathrm{Hom}(S,T) \supset H \to \mathbb{C}^\times ) \,,$$ where the set $H$ contains all those maps $S \to T$ that induce a fixed quotient on the labels of their image and preimage elements. By means of $h$ these maps are labeled by exactly that quotient in $\mathbb{C}^\times$.

There is maybe a more elegant way to say this. But anyway.

Unless I made a mistake, I now have a closed monoidal category $$V := \mathbb{C}^\times\mathrm{Set} \,.$$ I guess I should also check completeness and cocompleteness. But right now I won’t.

So next I want to understand $${}_{\mathbb{C}^\times\mathrm{Set}}\mathrm{Mod} \,.$$

The category of $\mathbb{C}^\times\mathrm{Set}$-vector space has the following bimonoidal structure.

It has a coproduct $$[C,\mathbb{C}^\times\mathrm{Set}] \oplus [D,\mathbb{C}^\times\mathrm{Set}] \simeq [C \cup D,\mathbb{C}^\times\mathrm{Set}] \,,$$ inherited from $\mathrm{Cat}$, which corresponds to taking the disjoint union of bases of vector spaces as a new generating basis.

Similarly for the product $$[C,\mathbb{C}^\times\mathrm{Set}] \times [D,\mathbb{C}^\times\mathrm{Set}] \simeq [C \times D,\mathbb{C}^\times\mathrm{Set}] \,,$$ which corresponds to the tensor product of vector spaces.

(Really, this is to be read as the product of $\mathbb{C}^\times\mathrm{Set}$-enriched categories. But I guess that does not make a difference.)

Good. So given a 1-functor $$\mathrm{tra} : P_1(X) \to \mathrm{Vect}_{\mathbb{C}^\times\mathrm{Set}} \,,$$ and assuming it couples to a 1-particle $$\mathrm{par} = \{\bullet\}$$ whose configurations are given by $$\mathrm{conf} = \mathrm{Disc}(\mathrm{Obj}(P_1(X))) \simeq \mathrm{Disc}(X) \,.$$ Then, a state of this 1-particle is a morphism $$\psi : 1_* \to \mathrm{tra}_* \,,$$ where $\mathrm{tra}_*$ is the restriction of $\mathrm{tra}$ to $\mathrm{Disc}(X)$ and $1_*$ is the tensor unit in the category of such functors, which means it is the functor that sends every point of $X$ to the $\mathbb{C}^\times\mathrm{Set}$-vector space $$[\{\bullet\},\mathbb{C}^\times\mathrm{Set}]$$ (where the category $\{\bullet\}$ really denotes the 1-object $\mathbb{C}^\times\mathrm{Set}$-enriched category whose objects of morphisms is $I_{\mathbb{C}^\times\mathrm{Set}} = \{\bullet_1\}$.)

Of course $$[\{\bullet\},\mathbb{C}^\times\mathrm{Set}] \simeq \mathbb{C}^\times \mathrm{Set} \,.$$

Is the space of all these states $$[1_*,\mathrm{tra}_*]$$ equivalent to $$\oplus_{x \in X} \mathrm{tra}(x)$$ or is it not?

The answer to that should go in a separate comment.

I wrote:

What I don’t understand yet is what you say afterwards

Toby replied:

OK, I’ll talk about that now.

and did so.

Thanks, that was very illuminating!

It’s really good that somebody (you, that is) takes the time to think about issues like that. It’s the kind of issue which may appear too irrelevant to bother with on first sight, but it’s not.

But in fact, a category enriched over (−1)-categories is a poset, not simply a set!

Okay. Maybe for the record and for the benefit of those following this I’ll recall that

a category enriched over

$$-1\mathrm{Cat} = (\{\mathrm{false}, \mathrm{true}\},\otimes = \mathrm{and}, I = \mathrm{true})$$

is a category whose Hom-objects “are” either the empty set or the 1-element set (where saying “are” really means applying the canonical monoidal isomorphism $-1\mathrm{Cat} \simeq \{\{\}, \{1\}\} \subset \mathrm{Set}$), therefore these are categories with at most one morphism for any ordered pair of objects, hence these are posets.

And how do you define a 0-poset?

They’re the Hom-objects of 1posets, of course. That is, they’re truth values. But now if I recreate a 1poset as a category enriched over 0posets, I get the correct result!

Okay.

What is the difference between 0-categories and 0-groupoids?

There’s no difference of course; both of these are just sets. So (−1) groupoids are defined to be the Hom-objects of 0groupoids, that is (once again) truth values.

Okay.

But now if I recreate 0groupoids as groupoids enriched over (−1) groupoids, I get the correct answer back.

This is your central point. Let me see if I understand that.

A groupoid enriched over (-1)groupoids is a poset in which all morphisms are invertible.

But a poset with all morphisms invertible is a disjoint union of codiscrete categories.

Right?

Such a category is equivalent to the the discrete category over its set of connected components.

It is this equivalent discrete category (= 0-category = set) which you are thinking of when saying that a groupoid enriched in (-1)groupoids is a set.

Right?

While I’m pontificating, I might add that I not only find the concept of ncategory lacking (in general —obviously specific cases are fine if it happens to be what you really want); but I also think that the numbering makes more sense for nposet than for ngroupoid (and ncategory).

So the basic concepts should be nposet and nset. In more familar terms, an nposet is an ncategory in which parallel nmorphisms are always equivalent (usually called ‘equal’ in this context); and an nset is an (n − 1)­groupoid. (I already referred to the fact that a 2set is a groupoid here.) Of course, for n := ∞, the level shift (and the notion of top-level morphism) disappears, so an ∞poset is an ∞category, and an ∞set is an ∞groupoid. And for n := −1 (the lowest that I understand), both (−1)­posets and (−1)­sets are simply truth values.

For that matter, I think that ngroupoids are the really interesting concept philosophically (you can blame Jim Dolan for this, whether or not he actually agrees), particularly for refining the notion of identity. And the basic idea behind ‘categorification’ should be something like enrichment over or internalisation in groupoids, rather than over/in categories, unless you’re categorifying something that already has enough noninvertibility to suggest that the categorified version needs an extra level of that. Even the notion of categorified category is best identified with (2,1)­category (that is a 2category whose 2morphisms are all invertible) rather than 2category (although that answer isn’t too bad; fortunately, categorification isn’t a precise concept, so mutliple answers are OK).

Is that enough idiosyncratically radical opinion? ^_^