The n-Category Café

Ugh! I just woke up, at 4:20 a.m., and thought of this embarrassingly simple example… but you beat me to it.

Hi, Thanks for referring to our work, but for quasi-Borel spaces, |hom(1,1+1)|=2, and the Kleisli category of our monad isn’t a ccc.

Nonetheless this does suggest ways to get nice answers to Question 4 and John’s new question, via Kleisli categories of fancy affine monads (even on Set).

That’s a way to make the question more interesting! Or: can you find a category such that $hom(1,1+1)$ has two elements but $hom(1,1+1+1)$ has more than three?

Take the category of totally ordered sets, and pairs of functions (f, g) where forall x, f x <= g x as the morphisms.

Then there are three points in 2: (F,F), (F,T), (T,T).

But is 2 equivalent to 1+1 in this category?

Take the category of meet-semilattices (not necessarily bounded). Then $1+1$ is the free semilattice on two generators, and it has three elements, or equivalently three morphisms $1 \to 1 + 1$: the two generating elements together with their meet.

However, this category is not cartesian closed, since $(1 + 1) \times -$ does not preserve coproducts: $(1 + 1) \times (1 + 1)$ has 9 elements, while $(1 + 1) + (1 + 1)$ has 15 elements.

If your category is cartesian closed (just distributive would do) then 1+1 is an internal Boolean algebra, and so Hom(1,1+1) is an ordinary Boolean algebra. So if it’s finite, it must have 2^n elements for some n.

Hmm, but being cartesian closed means $x\times (-)$ has a right adjoint so preserves colimits.

Now that I think about it, I should have consulted the canonical source before starting to type :-S.

I have a feeling that Richard and Sam’s work starting up here is an alternative route to tackling Conjectures 1 and 2.

Maybe I’m missing something obvious in this discussion, but I thought it was the case that any $[n]$ (a coproduct of $n$ copies of the terminal object $[1]$, with $n \geq 1$) is, in a distributive category with products and coproducts, a retract of some suitable $[2]^k = [2^k]$ (we could take $k = n$ for instance).

In other words, any such $[n]$ belongs to the Cauchy completion of the category consisting of products of $[2]$.

Take for example $[3] = [1] + [1] + [1]$. The first summand $[1]$ is a retract of $[2]$ in a pretty easy way. Applying the functor $- + [1] + [1]$ preserves this retract, so $[3]$ is a retract of $[4]$.

This should mean that $\hom([1], [3])$ is determined by the Boolean algebra $B = \hom([1], [2])$. Indeed, it’s obtained by splitting the idempotent

$$B \times B \stackrel{\langle \pi_1, \wedge\rangle}{\to} B \times B$$

which I found simply by analyzing the previous paragraph.

Here is a theorem which might be folklore but which seems to me to be relevant:

• Let $T$ be a category with finite products, and $\bar{T}$ its Cauchy completion. Then $\bar{T}$ also has finite products, and for any Cauchy complete category with products $D$, the category $Prod(T, D)$ of product-preserving functors $T \to D$ is equivalent to $Prod(\bar{T}, D)$.

We could take $D = Set$ for instance. Now, if we take $T_2$ to be the Lawvere theory given by the full subcategory in $C$ whose objects are finite products of $[2]$, we get the same Cauchy completion $\bar{T}$ as we would if we took $T_n$, the Lawvere theory generated by $[n]$:

$$Mod_{T_2} = Prod(T_2, Set) \simeq Prod(\bar{T}, Set) \simeq Prod(T_n, Set) = Mod_{T_n}.$$

I have a vague memory that we were discussing this sort of thing when Tom was posting about ultrafilters and codensity monads and the like, but I’m having trouble locating it just now.

So in case the message of my last post didn’t transmit well: if we suppose that the objects of the form $[2^n]$ form a category equivalent to the one we get by looking at those objects in the case $C = Fin^k$ for some finite $k$, then the same is true for all the other $[m]$. That is, the category of all the $[m]$ is a full subcategory of the Cauchy completion of the category of the $[2^n]$, and since $Fin^k$ is itself Cauchy complete, it’s the same as full subcategory of $Fin^k$ consisting of the $[m]$, i.e., the finite (nonempty) coproducts of $[1]$ in $Fin^k$.

Part of my point is also that there is nothing special about $2$ here. You could replace $2$ by any $n \geq 2$ and get an analogous statement.

Duh, yes, of course – the equaliser defining $1+1+1$ in terms of $(1+1)^2$ is split, hence absolute. So as Todd rightly says, $C(1,1+1)$ determines $C(1,1+1+1)$ whenever $C$ is distributive. So, to summarise: $C$ is distributive with as objects are the finite copowers of $1$ if and only if it is the category of Boolean-valued matrices for the Boolean algebra $B = C(1,1+1)$.

In fact, in combination with Sam’s remarks above, this turns out to be one of the class of examples in the Johnstone paper on “Collapsed toposes”.

As Sam points out, if $C$ is distributive and its objects are the copowers of $1$, then $E = FP(C^{op},Set)$ is a cartesian closed variety. Moreover, because $C(1,1) = 1$, this variety is affine, meaning that every one-element subset is a subalgebra, or equivalently that the terminal algebra is a projective generator. Conversely, if $E$ is an affine cartesian closed variety, then it is equivalent to $FP(C^op, Set)$ where $C$ is the full subcategory of $E$ on the coproducts of $1$.

So: distributive categories $C$ whose objects are copowers of $1$ are the same as cartesian closed affine varieties. Example 8.8 of “Collapsed toposes” notes that $E$ is a cartesian closed affine (finitary) variety precisely if and only if it is the category of sheaves on a Boolean algebra $B$ which are either empty or have global support. The $B$ in question is, of course, going to $E(1,1+1)$. There is, moreover, a simple syntactic characterisations of such varieties: they are the categories of models of commutative hyperaffine theories (where hyperaffine is defined in the Johnstone paper).

At the end of Example 8.8, reference is made to Bergman’s “Actions of Boolean rings on sets” which basically works out everything that is relevant to this situation. It is good stuff and worth a read.

Another point. If $C$ is a distributive category whose objects are the copowers of $1$, then the models of the theory $C^{op}$ are, as discussed, sheaves of global support on the Boolean algebra $C(1,1+1)$ together with the empty model.

This is related to the theory of skew Boolean algebras. A non-empty model of $C^{op}$ can also be described as a skew Boolean algebra whose Boolean algebra quotient is $C(1,1+1)$. See the paper “Boolean sets, skew Boolean algebras and a non-commutative Stone duality” by Kudryavtseva and Lawson.

Thanks, Todd and Richard!

$C$ is distributive with objects being the finite copowers of $1$ if and only if it is the category of Boolean-valued matrices for the Boolean algebra $B= C(1,1+1)$.

Great!

Boolean-valued matrices sound more like relations than functions to me; I guess you mean matrices of some special sort here? (Like those that have right adjoints in a suitable sense?)