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Stream: learning: questions

Topic: subsets and subgroups

David Egolf (Aug 21 2022 at 16:16):

I learned recently that the subsets of a set $S$ are in 1-1 correspondence with the maps $2^S$, namely those from $S$ to $2$, where $2$ is some set with two elements, say $\{1,0\}$. Then the subset $T \subseteq S$ is specified by sending its elements to 1 and all the other elements of $S$ to 0.

I was wondering if we can do this with groups, so as to get some set that corresponds to all the subgroups of a given group $G$. Let $2$ now be a group with two elements $\{0,1\}$. There are two subgroups of $2$: the one associated with the identity element, and the one associated with all of $2$. A group homomorphism $f: G \to 2$ might be able to, intuitively, send all subgroups except one to the identity element subgroup (and the remaining one of interest to all of $2$). However, a group homomorphism is required to preserve some structure, and so I'm worried that such a group homomorphism won't necessarily exist. As one example, it seems like we can't specify the identity element subgroup by mapping it to all of $2$ - it has to get mapped to the identity element subgroup of $2$. So this seems like it isn't working out.

(In a little more detail, if we have a surjective group homomorphism $f$ from $G$ to 2, then $G/\ker(f) \cong 2$. That's a pretty restrictive condition on the structure of $G$, I think.)

Is there some nice concise way to talk about all the subgroups of a group $G$, in terms of maps from $G$?

Tobias Schmude (Aug 21 2022 at 19:15):

Unfortunately this does not work! But seeing why is quite enlightening. The concept you're after is called a [[subobject classifier]], and it exists in the category $\mathrm{Set}$, but not in $\mathrm{Grp}$. Let's see why that is the case:

In full generality, a subobject classifier is an object $\Omega$ of a category $\mathcal{C}$ (which has to have pullbacks of monomorphisms) such that subobjects of any object correspond to morphisms into $\Omega$ (e.g. $\Omega = \{0, 1\}$ in $\mathrm{Set}$). To phrase it categorically:
We have two presheaves on $\mathcal{C}$. Firstly there is $\mathrm{Sub}$, defined on objects by taking the set of subobjects and on morphisms by pullback (that's why we need them). Secondly there is $\mathcal{C}(-, \Omega)$, defined on objects by taking the set of morphisms into $\Omega$ and on morphisms by composition. We demand these to be naturally isomorphic, i.e. $\mathrm{Sub}$ is represented by $\Omega$.

This gives us a necessary criterion for the existence of a subobject classifier: representable presheaves preserve limits! Due to contravariance this amounts to mapping colimits in $\mathcal{C}$ to limits in $\mathrm{Set}$.
Let's concentrate on the special case of coproducts: this says that for objects $C, D$, subobjects of the coproduct $C \cup D$ have to be in bijection with pairs of subobjects of $C$ and $D$.

In $\mathrm{Set}$ this is clear: a subset $S \subseteq C \cup D$ is equivalently given by the pair of subsets $S \cap C \subseteq C$ and $S \cap D \subseteq D$.

In the category $\mathrm{Grp}$ this is not the case though. Take for example finite groups $G$ and $H$ with nontrivial involutions $g \in G, h \in H$. Then there are only finitely many pairs of subgroups, but the coproduct ("free product" in group theorists' terms) contains infinitely many distinct involutions

$$(g), (h), (g, h, g), (h, g, h), (g, h, g, h, g), (h, g, h, g, h), \dots$$

which accordingly generate infinitely many distinct subgroups.

Tobias Schmude (Aug 21 2022 at 20:32):

Thanks! The nlab mentions another, more conceptual, way to see that $\mathrm{Grp}$ does not have a subobject classifier:
First off, the trivial group is a zero object (both initial and terminal).
In a category with zero object, any subobject classified by a hypothetical subobject classifier turns out to be a kernel of its classifying map (see Prop 4.1 on the nlab).
This is a problem since not every subgroup is a kernel ("normal" subgroup).

David Egolf (Aug 21 2022 at 20:51):

Thanks Tobias! There's a lot there to chew on for me, especially since I'm still shaky on a lot of the category theory you mention. (I remember wrestling with some basic exercises around the subobject classifier some while back, and it all seemed rather hard).

But just from a first skim of the punchline, the concept that you can get this explosion of subobjects when you take a coproduct in $\mathrm{Grp}$, but not in $\mathrm{Set}$, really stands out as beautiful to me! My initial intuition is that this happens because you have interaction between group elements (they can be composed), which allows this explosive interaction between two groups that you are taking a coproduct of. By contrast, my intuition for a coproduct in $\mathrm{Set}$ is that you end up with two sets "sitting side by side". Hopefully this is sort of in the right direction.

Ralph Sarkis (Aug 21 2022 at 22:49):

Evident follow-up question: is there a "normal subgroup classifier" in $\mathbf{Grp}$?

Tobias Schmude (Aug 22 2022 at 08:32):

David Egolf said:

My initial intuition is that this happens because you have interaction between group elements (they can be composed), which allows this explosive interaction between two groups that you are taking a coproduct of. By contrast, my intuition for a coproduct in $\mathrm{Set}$ is that you end up with two sets "sitting side by side". Hopefully this is sort of in the right direction.

Exactly! Coproducts of groups can be way larger than each group, so it doesn't feel plausible that they should have so few subgroups.

Tobias Schmude (Aug 22 2022 at 08:32):

If you get stuck on some part of the argument, don't hesitate to ask!

Tobias Schmude (Aug 22 2022 at 08:33):

Ralph Sarkis said:

Evident follow-up question: is there a "normal subgroup classifier" in $\mathbf{Grp}$?

Very good point! I suspect not, but I don't have a counterexample. I'll keep it in the back of my mind.

Oscar Cunningham (Aug 22 2022 at 11:28):

For each element of a group there's a unique homomorphism from $\mathbb{Z}$ to the group sending $1$ to that element. So if there was a normal subgroup classifier it would have one element for each normal subgroup of $\mathbb{Z}$. The normal subgroups of $\mathbb{Z}$ are precisely $n\mathbb{Z}$ for natural $n$, so the normal subgroup classifier must have countably many elements.

John Baez (Aug 22 2022 at 15:48):

David Egolf said:

Thanks Tobias! There's a lot there to chew on for me, especially since I'm still shaky on a lot of the category theory you mention. (I remember wrestling with some basic exercises around the subobject classifier some while back, and it all seemed rather hard).

But just from a first skim of the punchline, the concept that you can get this explosion of subobjects when you take a coproduct in $\mathrm{Grp}$, but not in $\mathrm{Set}$, really stands out as beautiful to me! My initial intuition is that this happens because you have interaction between group elements (they can be composed), which allows this explosive interaction between two groups that you are taking a coproduct of. By contrast, my intuition for a coproduct in $\mathrm{Set}$ is that you end up with two sets "sitting side by side". Hopefully this is sort of in the right direction.

That's right, but there are also categories where the coproduct is not so "explosive" when it comes to elements (in the traditional set-theoretic sense) yet still explosive when it comes to subobjects.

John Baez (Aug 22 2022 at 15:51):

First let me recap Tobias' argument. He said that if a category has a subobject classifier and coproducts, we must have

$\mathrm{Sub}(A + B) \cong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$

where + here means coproduct in our category and $\times$ is product of sets.

John Baez (Aug 22 2022 at 15:56):

The proof is easy so let's do it. Let $\Omega$ be the subobject classifier; then

$\mathrm{Sub}(A+B) \cong \mathrm{hom}(A+B,\Omega) \cong \mathrm{hom}(A,\Omega) \times \mathrm{hom}(B,\Omega) \cong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$

John Baez (Aug 22 2022 at 15:57):

The first and last steps are by the definition of 'subobject classifier', while the middle step is a general fact about coproducts: the definition of coproduct implies

John Baez (Aug 22 2022 at 15:58):

$\mathrm{hom}(A+B,X) \cong \mathrm{hom}(A,X) \times \mathrm{hom}(B,X)$

John Baez (Aug 22 2022 at 15:58):

for any objects $A,B,X$.

John Baez (Aug 22 2022 at 16:00):

Now, look at the category of abelian groups or the category of vector spaces. (Take the one you're most comfortable with.) In this category the product of two objects also serves as its coproduct:

$A + B \cong A \times B$

John Baez (Aug 22 2022 at 16:01):

and we often call this the 'direct sum', or sometimes 'direct product'. An element of $A \times B$ is just a pair of an element of $A$ and an element of $B$, so there's no dramatic 'explosion' as there is in the coproduct of groups.

John Baez (Aug 22 2022 at 16:02):

A subobject of a vector space is called a 'subspace', and a subobject of an abelian group is called a 'subgroup'.

John Baez (Aug 22 2022 at 16:03):

It's easy to see that we don't have

$\mathrm{Sub}(A+B) \cong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$

John Baez (Aug 22 2022 at 16:03):

in either of these examples.

John Baez (Aug 22 2022 at 16:05):

In both cases, you can see there's a map

$\mathrm{Sub}(A) \times \mathrm{Sub}(B) \to \mathrm{Sub}(A+B)$

That is, you can get a subobject of $A + B$ from a subobject of $A$ and a subobject of $B.$

But, you don't get all of them this way.

John Baez (Aug 22 2022 at 16:07):

For example, say we're doing abelian groups, and $A = B = \mathbb{Z}/2$. Then $A$ and $B$ each have two subgroups, and these give us 2 $\times$ 2 = 4 subgroups of $\mathbb{Z}/2 \times \mathbb{Z}/2$.

John Baez (Aug 22 2022 at 16:08):

But there's also a fifth subgroup, namely $\{ (0,0), (1,1) \}$.

John Baez (Aug 22 2022 at 16:09):

The moral: this subgroup sits 'diagonally' in the product $\mathbb{Z}/2 \times \mathbb{Z}/2$, so we don't get it just by taking the product of a subgroup in each separate copy of $\mathbb{Z}/2$.

David Egolf (Aug 22 2022 at 16:51):

John Baez said:

The proof is easy so let's do it. Let $\Omega$ be the subobject classifier; then

$\mathrm{Sub}(A+B) \cong \mathrm{hom}(A+B,\Omega) \cong \mathrm{hom}(A,\Omega) \times \mathrm{hom}(B,\Omega) \cong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$

Oh, that makes sense! Thanks for explaining it like that.

John Baez said:

The moral: this subgroup sits 'diagonally' in the product $\mathbb{Z}/2 \times \mathbb{Z}/2$, so we don't get it just by taking the product of a subgroup in each separate copy of $\mathbb{Z}/2$.

I see... Let me see how this works for vector spaces. Let's take $A = B = \mathbb{R}$, the vector space of the real line with the real numbers as the scalars. Then $A + B \cong \mathbb{R}^2$, the vector space of pairs of real numbers where the scalars are real numbers. This has a lot of subobjects: the origin, the whole plane, and one subobject for each line through the origin. However, $A$ only has two subobjects: the real line, and the origin. So $\mathrm{Sub}(A) \times \mathrm{Sub}(B)$ only has four elements, while $\mathrm{Sub}(A+B)$ has infinitely many. To make an analogy with the "diagonal-ness" you described above, there are a lot of lines through the origin in a plane besides the x and y axis - and these correspond (intuitively) to extra subobjects that appear in the coproduct $A + B$.

John Baez (Aug 22 2022 at 17:37):

Right!

John Baez (Aug 22 2022 at 17:40):

So we've learned that if $\mathrm{Sub}(A + B) \ncong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$ then our category can't have a subobject classifier. It's fun to go through categories (that have coproducts) and think about when this criterion rules out the possibility of a subobject classifier. I think it happens for a lot of 'algebraic' gadgets like groups, abelian groups, rings, commutative rings, vector spaces, etc.

John Baez (Aug 22 2022 at 17:43):

If one wanted to struggle a bit, it might be a good challenge to find a category with coproducts and natural isomorphism $\mathrm{Sub}(A + B) \ncong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$ which still doesn't have a subobject classifier.

John Baez (Aug 22 2022 at 17:43):

I don't know one offhand, and I'm not in the mood for trying to find one, but I feel confident one should exist.

Reid Barton (Aug 22 2022 at 17:55):

I would venture a guess that the category of topological spaces has those properties, if "subobject" is taken with respect to the class of all monomorphisms (as opposed to the regular monomorphisms).

Reid Barton (Aug 22 2022 at 18:00):

Then a subobject of $X$ is a subspace of $X$, but with possibly a finer topology than the induced topology.

Sam Speight (Aug 22 2022 at 18:05):

Cat fits the bill, right?

David Egolf (Aug 22 2022 at 18:17):

John Baez said:

So we've learned that if $\mathrm{Sub}(A + B) \ncong \mathrm{Sub}(A) \times \mathrm{Sub}(B)$ then our category can't have a subobject classifier. It's fun to go through categories (that have coproducts) and think about when this criterion rules out the possibility of a subobject classifier. I think it happens for a lot of 'algebraic' gadgets like groups, abelian groups, rings, commutative rings, vector spaces, etc.

Interesting... I wonder - if so many categories we care about don't have a subobject classifier, could it make sense to think about "powering up" how we're examining objects in terms of maps out of them? For example, instead of just considering maps to a single fixed classifying object $\Omega$, we could consider pairs of maps out of an object $A$, to a (fixed) pair of objects $\Omega_1$ and $\Omega_2$. Maybe one could describe more subobjects of $A$ in terms of these pairs of maps from $A$?

John Baez (Aug 22 2022 at 18:46):

The Yoneda Lemma says we know everything about an object $x$ if we know all the morphisms out of it, and how a morphism $f: x \to y$ can compose with a morphism $g : y \to z$ to give a morphism $gf : x \to z$.

John Baez (Aug 22 2022 at 18:47):

So, we can determine the subobjects of $x$ if we know "everything about the morphisms out of it" in this strong sense.

John Baez (Aug 22 2022 at 18:49):

Having a subobject classifier is a wonderful thing which tends to happen for categories of "things like spaces", not "things like algebras".

John Baez (Aug 22 2022 at 18:49):

That's why it's part of the definition of [[topos]].

David Egolf (Aug 22 2022 at 19:31):

John Baez said:

So, we can determine the subobjects of $x$ if we know "everything about the morphisms out of it" in this strong sense.

That makes sense. In categories with a subobject classifier, to my understanding, we can associate the subobjects of an object $A$ with the maps from it to $\Omega$. So - in this special case, we don't need all the morphisms from $A$ (and how they can be converted to others morphisms from $A$ by morphisms between other objects) - we just need those to a single object.
So, on one extreme the Yoneda Lemma tells us that maps to all other objects from $A$ (and how they can be converted to one another via morphisms between other objects) is enough to tell us about the subobjects of $A$. And on the other extreme in some cases (when we have a subobject classifier) the maps to just one of the objects from $A$ is enough to talk about subobjects of $A$.

Maybe there is a whole ascending ladder of complexity of categories in between these two cases, though! (Where knowing everything about the morphisms out of $A$ to two, or three, or n special objects is enough to determine the subobjects of $A$).

Mike Shulman (Aug 22 2022 at 19:39):

An [[extensive category]] is a category with coproducts and equivalences of slice categories $\mathcal{C}/(A+B) \simeq \mathcal{C}/A \times \mathcal{C}/B$. Since ${\rm Sub}(A)$ is the poset of subterminal objects in $\mathcal{C}/A$, any extensive category satisfies ${\rm Sub}(A+B) \cong {\rm Sub}(A)\times {\rm Sub}(B)$. But there are plenty of extensive categories without a subobject classifier, including (as mentioned) $\rm Cat$ and $\rm Top$.

Mike Shulman (Aug 22 2022 at 19:41):

Maybe also worth noting that a subobject classifier also implies similar facts about other colimits. For instance, any coequalizer diagram $A \rightrightarrows B \to C$ is a category with a subobject classifier implies an equalizer diagram ${\rm Sub}(C) \to {\rm Sub}(B) \rightrightarrows {\rm Sub}(A)$. This tends to fail in categories like $\rm Cat$ and $\rm Top$ that are extensive but not toposes.

John Baez (Aug 23 2022 at 00:43):

I wasn't thinking very hard but I figured there should be tons of categories where $\mathrm{Sub}$ gets along with binary coproducts but not coequalizers.

John Baez (Aug 23 2022 at 00:44):

There should also be some where $\mathrm{Sub}$ gets along with finite colimits but not arbitrary ones. I don't think you get much for free in this particular game!

Tobias Schmude (Aug 23 2022 at 17:42):

Ralph Sarkis said:

Evident follow-up question: is there a "normal subgroup classifier" in $\mathbf{Grp}$?

There isn't, here's the smallest possible counterexample in all glory:
Let's take $G = H = \mathbb{Z}_2$, and rename the 1 from each cofactor to $a$ and $b$ respectively. Then the coproduct consists of the words of alternating a's and b's, where pairs of a's and b's cancel. What does conjugation do? If the length of the word is even, conjugation with either $a$ or $b$ reverses (= inverts) the word, so there's nothing new to gain. This means that for each natural number $n$ we have a normal subgroup consisting of the words of length divisible by $2n$.
Just for "fun": on a word with length $2n+1$, conjugation produces either the word with length $2(n+1)+1$ starting with the other letter, or the word with length $2(n-1)+1$ starting with the other letter in case $n \geq 1$. So immediately any normal subgroup containing a word of odd length contains the words of length $4n+1$ starting with either $a$ or with $b$, as well as the words of length $4n+3$ starting with the other. By conjugation and multiplication it also contains all words of length divisible by 4. Multiplication doesn't lead out of those, so we get two further normal subgroups consisting of the words of length divisible by 4 as well as either of the two families of words of odd length.
If we further assume that it contains a word of length $4n+2$ or a word from the other family of words of odd length, we get all words, so that's it.

Ralph Sarkis (Aug 24 2022 at 01:21):

Cool, thank you!

Todd Trimble (Mar 27 2023 at 00:54):

(I'm still catching up on messages after months of being away.)

An interesting fact is that if a finitely complete category has a subobject classifier, then all subobject lattices (I mean posets) are cartesian closed (in particular, are distributive lattices if they have joins). This quickly rules out many categories as having subobject classifiers. For example, for categories of algebras, it is rarely the case that subobject lattices are distributive. And this has nothing to do with whether the category itself is cartesian closed (i.e., there's no microcosm principle being invoked here!).

Another consequence of having a subobject classifier is that every monomorphism is a regular monomorphism (this is not true for $Top$!), being the equalizer of the pair consisting of the classifying map and the constant map at $t: 1 \to \Omega$. This in turn implies that the ambient category must be balanced (i.e., mono + epi implies iso).

All this is mentioned here.

John Baez (Mar 27 2023 at 23:27):

Nice facts, Todd!

One thing some people used to say is that in quantum logic "and" doesn't distribute over "or". What they meant, basically, is that the lattice of subspaces of a vector space meet doesn't distribute over join. This is easy to see with a low-dimensional example. Combined with your comment, @Todd Trimble, this implies that Vect doesn't have a subobject classifier. This is also easy to see directly, but still it's nice to see that it follows.

People working on quantum logic occasionally try to define a "quantum topos", but so far all the attempts I've seen are unconvincing. For starters, it's supposed to be some sort of symmetric mooidal category, and Vect with its usual tensor product, or maybe Hilb with its usual tensor product, is supposed to be an example. But as we've just seen, these categories are so un-topos-like that taking inspiration from topos theory is almost hopeless!

John Baez (Mar 27 2023 at 23:30):

As far as I can tell, the people who try to define a quantum topos tend to not know category theory very well, much less topos theory - and they're hoping you can stick "quantum" in front of a concept and get a new concept.

Todd Trimble (Mar 28 2023 at 00:07):

Thanks, John. Of course you know, but maybe not everyone here has considered, simple examples of how distributivity fails in the lattice of subspaces of a vector space: take a plane $P$ and three distinct lines $L, M, N$ in $P$ through the origin, giving the join $M \vee N = P$ and the meets $L \wedge M = 0, L \wedge N = 0$, hence a strict inequality

$(L \wedge M) \vee (L \wedge N) = 0 \subset L = L \wedge (M \vee N)$

that shows failure of distributivity. Similar examples can be produced for other linear categories like abelian groups.

Anyone who has thought even a little about so-called "quantum logic" has to know this fact.

John Baez (Mar 28 2023 at 01:01):

Yes, they know that, probably even the unnamed people I was casting aspersions against.

Morgan Rogers (he/him) (Mar 28 2023 at 08:15):

Thanks for the example, @Todd Trimble !

John Baez (Mar 28 2023 at 14:38):

There's another canonical example, from physics. Take the Hilbert space $L^2(\mathbb{R})$$ and take

• $L$ = functions supported on $[0,\infty)$.

• $M$ = functions whose Fourier transform is supported on $[0,\infty)$

• $N$ = functions whose Fourier transform is supported on $(-\infty,0]$

In other words

• $L$ is the space of states of a particle to the right of the origin.

• $M$ is the space of states of a particle moving right.

• $N$ is the space of states of a particle moving left.

Then $M \vee N$ is the whole Hilbert space so

$L \wedge (M \vee N) = L$

but $L \wedge M = 0$ and $L \wedge N = 0$ by the Paley-Wiener theorem so

$(L \wedge M) \vee (L \wedge N) = 0$

In other words, it's impossible to have a particle to the right of the origin and moving right. It's impossible to have a particle to the right of the origin and moving left. But it's possible to have a particle to the right of the origin and moving right or left!

John Baez (Mar 28 2023 at 14:44):

Todd's example also has a physical interpretation if we replace the plane $\mathbb{R}^2$ by $\mathbb{C}^2$, the space of states of a spin-1/2 particle. We can take our 3 propositions to be

L = the particle's spin is pointing in the x direction
M = the particle's spin is pointing in the y direction
N = the particle's spin is pointing in the z direction.

John Baez (Mar 28 2023 at 14:45):

Then the "and" of any two of these proposition is, but the "or" of any two of these propositions is true, so distributivity fails.

Emily (Mar 28 2023 at 17:32):

Although $\mathsf{Grp}$ doesn't have a subobject classifier, there's an analogous result for the correspondence between subsets of $S$ and functions $S\to 2$. Namely, $2$ admits a poset structure where $0\to 1$, $0\to 0$, and $1\to 1$ (note that this is precisely logical implication if we write $0=\mathrm{false}$ and $1=\mathrm{true}$). Then, this poset has meets, and admits a Cartesian monoidal structure. Writing $\{\mathrm{t},\mathrm{f}\}$ for the Cartesian monoidal category associated to this poset, we have the following results:

• 0-Categorical Day Convolution. Let $(A,\mu_A,\eta_A)$ be a monoid. We have a bijection between the following data:
  • Submonoids of $A$;
  • Lax monoidal functors $(A,\mu_A,\eta_A)\to\{\mathrm{t},\mathrm{f}\}$
  • Monoids in the category $(\mathcal{P}(A),\circledast,1_{\mathcal{P}(A)})$, where $\circledast$, the 0-categorical version of Day convolution, is given by

$$U\circledast V\overset{\mathrm{def}}{=}\{uv\in A\ |\ u\in U,v\in V\}$$

(in fact if you view $U$ and $V$ as functions from $A$ to $2$ (write $\chi_U$ and $\chi_V$ for them) you can write $\circledast$ as a 0-categorical coend, namely $\int^{x,y\in A}\mathrm{Hom}_{A}(-,xy)\times\chi_U(x)\times\chi_V(y)$)

• The same but for groups. Similarly, given a group $(G,\mu_G,\eta_G,\chi_G)$, there's a bijection between the following:
  • Subgroups of $G$
  • Hopf monoidal functors $(G,\mu_G,\eta_G)\to\{\mathrm{t},\mathrm{f}\}$
  • Group objects in $(\mathcal{P}(G),\circledast,1_{\mathcal{P}(G)})$

Emily (Mar 28 2023 at 17:35):

Another cool thing about this point of view is that free submonoids on a subset are computed via the usual formula, see this image (copied from my notes):

Emily (Mar 28 2023 at 17:35):

image.png

Emily (Mar 28 2023 at 17:37):

Also the monoids in $\mathcal{P}(A\times A)\overset{\mathrm{def}}{=}\mathbf{Rel}(A,A)$ are precisely the congruence relations but without the equivalence part, i.e. just the $x\sim x'$ and $y\sim y'$ implies $xx'\sim yy'$ condition

Morgan Rogers (he/him) (Apr 25 2023 at 08:21):

John Baez said:

People working on quantum logic occasionally try to define a "quantum topos", but so far all the attempts I've seen are unconvincing. For starters, it's supposed to be some sort of symmetric mooidal category, and Vect with its usual tensor product, or maybe Hilb with its usual tensor product, is supposed to be an example. But as we've just seen, these categories are so un-topos-like that taking inspiration from topos theory is almost hopeless!

This has been lying in my drafts since March but a conversation I had at SYCO last week reignited the thought in my mind - I think there is something here. Rather than taking the category Hilb as the prototypical quantum topos, we could take a category of sheaves on Hilb (which retains and extends the monoidal structure on the representable sheaves via Day convolution). This is a 'monoidal topos', where we can apply all of the tools from topos theory directly, while the 'linear' structure is preserved in a subcategory. I haven't seriously studied such things yet, but I will in the not-too-distant future in the context of extending my work on monoid actions; Ross Duncan gave me some references for the quantum topos ideas that you alluded to, John.

Please do not interpret this message as me planting my flag in this topic, since I know I will not have time to work on it myself, let alone organise a collective research effort, for many months. Rather, if you are reading this and decide to look into this topic (eg if you are a student looking for research questions), I am happy to provide ideas and direction, since I have many that are shelved at the moment.

Cole Comfort (Apr 25 2023 at 08:38):

deleted

Morgan Rogers (he/him) (Apr 25 2023 at 09:38):

@Cole Comfort no, Abramsky and co show how quantum effects can be characterized by the failure of certain sheaves to have global sections. It's a positive application of sheaf theory, not a negative result!