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Stream: learning: questions

Topic: Non-symmetric monoidal tensors on Set

Dan Doel (Oct 05 2020 at 02:03):

Is it possible to have a non-symmetric monoidal structure on $\mathsf{Set}$? I asked someone and they mentioned that it seemed like it would have to do with formal power series, and that the answer was probably 'no'. But I thought I'd ask here in case someone knew of a more definitive answer.

Dan Doel (Oct 05 2020 at 02:06):

I found some stuff about how monoidal structures could be generated by $A \otimes B = \displaystyle \sum_{i,j} S_{i,j} × A^i × B^j$, but it seemed like the people talking about it didn't know definitively whether all monoidal structures were generated that way.

Antonin Delpeuch (Oct 05 2020 at 06:57):

Interesting question! What are the conditions on the $S_{i,j}$ to get a monoidal structure? It does not look like any choice of these would give you associativity at least. Do you have a link to share about these monoidal structures?

Matteo Capucci (he/him) (Oct 05 2020 at 08:52):

Dan Doel said:

I found some stuff about how monoidal structures could be generated by $A \otimes B = \displaystyle \sum_{i,j} S_{i,j} × A^i × B^j$

This looks like the evaluation of a bivariate polynomial on $(A, B)$

Dan Doel (Oct 05 2020 at 11:42):

@Antonin Delpeuch I'm not exactly sure. The discussion is here. I guess they're at least supposed to be 'natural numbers'. But I'm not sure if something like $S_{i,j} = S_{j,i}$ needs to hold. Possibly they aren't assuming that, but if it doesn't, the tensor won't actually be associative.

Antonin Delpeuch (Oct 05 2020 at 12:37):

Wow that's a really nice discussion (which uses all sorts of notions I don't know at all), thanks! The proposed connection with analytic functions is exciting but also a bit daunting…

Nathanael Arkor (Oct 05 2020 at 12:41):

I haven't checked this in detail, so I may have made a simple oversight, but the following looks like a (nonsymmetric) semicartesian monoidal structure on Set. Let $A \otimes B = A$ if $|A| \neq 1$ and $B$ otherwise, and let $f \otimes g$ be $f$ if neither its domain nor codomain has cardinality 1, and $g$ otherwise. The operation picks the "leftmost non-singleton set", and is associative and unital, I think naturally.

Dan Doel (Oct 05 2020 at 12:41):

Isn't that not functorial?

Nathanael Arkor (Oct 05 2020 at 12:42):

It may not be. I just realised I forgot to check.

Dan Doel (Oct 05 2020 at 12:43):

I think stuff like that likely wouldn't be, even if it can be defined.

Dan Doel (Oct 05 2020 at 12:45):

I'm actually asking due to this, and someone there suggested a similar thing with $0$ instead of $1$, which certainly fails to be functorial.

Nathanael Arkor (Oct 05 2020 at 12:46):

I would have thought if it doesn't work with $0$ as the unit, it won't work with $1$ as the unit either.

Nathanael Arkor (Oct 05 2020 at 12:53):

I wonder if there's an argument to be made that, being the free cocompletion of a point, the only monoidal operations must be expressible in terms of colimits, which must therefore be symmetric by the distributivity of colimits over colimits.

Dan Doel (Oct 05 2020 at 12:53):

Well, for $0$, I think the problem is that $0 \otimes B = B$, but there is a unique map $0 → A$, which should give you a choice of $B → A$ for every $A$. However, composing with any $A → C$ gives the unique map $0 → C$, and must give the choice for $C$. However, first getting the choice for $A$ and then composing with an $A → C$ will not give the same result in general.

Dan Doel (Oct 05 2020 at 12:54):

So I don't think that argument will work with $1$, but my suspicion is that something else is wrong with it.

Nathanael Arkor (Oct 05 2020 at 12:57):

I don't actually think what I wrote is well-defined :sweat_smile: If we have $1 \xrightarrow{f} A$ and $B \xrightarrow{g} C$ then we want a map $1 \otimes B \to A \otimes C$ = $B \to A$, but what I wrote gives a map from $B \to C$.

Dan Doel (Oct 05 2020 at 12:57):

Oh yeah, that'd be a problem. :)

fosco (Oct 06 2020 at 08:47):

Dan Doel said:

I found some stuff about how monoidal structures could be generated by $A \otimes B = \displaystyle \sum_{i,j} S_{i,j} × A^i × B^j$, but it seemed like the people talking about it didn't know definitively whether all monoidal structures were generated that way.

Where do the indices $i,j$ run over? Depending on the answer, this might be interpreted as some sort of Day convolution. It really reminds me the convolution induced on $[{\cal J}, {\sf Set}]$ by a promonoidal structure $\mathfrak{P} : {\cal J}^\text{op}\times {\cal J}^\text{op} \times {\cal J} \to \sf Set$: for all $x \in \cal J$,

$\displaystyle A *_{\mathfrak P} B(x) = \int^{ij} A^i \times B^j \times P^x_{ij}$

Morgan Rogers (he/him) (Oct 06 2020 at 08:56):

So what goes wrong if we take $A \otimes B \cong A$ whenever $A$ is non-empty and $A \otimes B \cong B$ when $A$ is empty? At first glance it seems functorial, associative and unitial with unit $\emptyset$

Reid Barton (Oct 06 2020 at 13:24):

At first glance it doesn't seem functorial to me... what are the maps supposed to be?

Morgan Rogers (he/him) (Oct 06 2020 at 13:32):

Given $f:A \to A'$ and $g: B \to B'$, $f \otimes g$ is $f$ if $A$ is inhabited and $g$ if $A$ is empty.

Morgan Rogers (he/him) (Oct 06 2020 at 13:33):

Ah, damn, if $A'$ is inhabited but not $A$, we need a morphism $B \to A'$, hmm

Morgan Rogers (he/him) (Oct 06 2020 at 14:08):

I suspect this could be settled by some case-analysis. Here is some lazy speculation... Consider $1 \otimes 1$, for example: if it's empty, then since any pair of objects has a morphism $! \otimes !: A \otimes B \to 1 \otimes 1$, the monoidal product would have to be empty everywhere, which is not allowed. If $1 \otimes 1 \cong 1$, (if I'm calculating this correctly), we get a comparison transformation from the cartesian product that sends $(a,b) \in A \times B$ to $a \otimes b: 1 \otimes 1 \to A \otimes B$. This doesn't immediately preclude asymmetry, but perhaps there is some argument that can extend this line of reasoning.

If $1 \otimes 1 \cong 1 + 1$, we should get a comparison morphism to the cocartesian monoidal structure, although actually constructing this requires some powerset shenanigans. If all of this serves to prevent any of these cases from being asymmetric, then we arrive at the final case which is that $1 \otimes 1$ is "big" (has more than two elements). On the one hand, there might be room for asymmetry here, but I suspect that the unit will again cause problems, and possibly enough to prevent this case too.

Dan Doel (Oct 06 2020 at 14:23):

@fosco I think it's supposed to be finite naturals, because it's like a formal power series.

Dan Doel (Oct 06 2020 at 14:39):

@[Mod] Morgan Rogers One thing that discussion I linked mentions is that almost every one of those 'analytic' functors must have unit $0$, because only $A×B$ can have unit $1$. I'm not sure if there's an argument for more exotic cases, though.

Dan Doel (Oct 06 2020 at 14:43):

($0$ and $1$ are the only possibilities for reasons that escape me.)

Reid Barton (Oct 06 2020 at 14:45):

Dan Doel said:

($0$ and $1$ are the only possibilities for reasons that escape me.)

I wonder if it's something like this. Suppose the unit is $S$ where $S$ has at least two elements. Then $S$ has nontrivial self-maps and those induce natural transformations $A \cong A \otimes S \to A \otimes S \cong A$. But the only natural transformation from the identity functor to itself is the identity. So, now if we plug in $A = S$ do we get a contradiction somehow?

Reid Barton (Oct 06 2020 at 15:11):

I think this does work, although it requires paying more attention to the axioms of a monoidal category than I'm used to.

Reid Barton (Oct 06 2020 at 15:12):

Something about the left and right unitors agreeing on the unit object, so when you plug in $A = S$ to this map, on one hand, it has to produce the identity map as I argued above, while on the other hand it also has to produce the original (non-identity) map $S \to S$.

Dan Doel (Oct 06 2020 at 15:25):

Ah, interesting.