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I am interested in any adjunctions between any of the familiar categories of Groupoids and the category of finite dimensional Hilbert spaces. Do any exist? Are there any well know monads on the category of groupoids of this type, ie, generated by such an adjunction? For instance, are there any interesting adjunctions between the category of finite groupoids and FDHilb?

This question is a bit vague, as has been pointed out. We can restrict the problem to exactly the group of finite groupoids, and the category of finite dimensional Hilbert spaces with all linear maps.

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  • $\begingroup$ To clarify: are you changing your question to ask whether there exist any interesting adjunctions between the category of finite groupoids and the category of finite-dimensional Hilbert spaces? And what are the morphisms in the category of finite-dimensional Hilbert spaces? I can think of several natural choices -- all linear maps, contractive linear maps, linear maps which preserve the inner product, ... $\endgroup$
    – Tim Campion
    Aug 15, 2018 at 21:32
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    $\begingroup$ Tim, first let me thank you for answering in the first place. I guess the question was not posed very precisely. I didnt even know there were different categories of groupoids. The quick answer to your question is, yes. I want to open this question up where we allow for any category of groupoids. $\endgroup$
    – Ben Sprott
    Aug 15, 2018 at 21:50
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    $\begingroup$ It seems the answer to my question was in fact no -- you want to be vague about what category of groupoids you are considering. Since you still haven't indicated what the morphisms are in the category of finite-dimensional Hilbert spaces, am I to suppose that you want to be vague on that point as well? In any event I think the question in this form is far too vague to be definitively answered, but my impression is that the answer is no, there are no such interesting adjunctions. $\endgroup$
    – Tim Campion
    Aug 15, 2018 at 22:02
  • $\begingroup$ Tim, apologies for the vagueness. I have restricted the question and if you still feel like answering, I am sure your answer will suite me just fine. $\endgroup$
    – Ben Sprott
    Aug 15, 2018 at 22:24

2 Answers 2

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One suggested variant was to ask whether there are any interesting adjunctions between the category $FinGpd$ of finite groupoids and the category $FinHilb$ of finite-dimensional Hilbert spaces, which is still undefined. The answer is still no.

First let me suppose that the morphisms of $FinHilb$ are all linear maps. Then the automorphism group of $\mathbb C \in FinHilb$ is $\mathbb C^\ast$, a divisible abelian group. As such, it admits no nontrivial homomorphisms to any finite group. So if $F: FinHilb \to FinGpd$ is a functor, then the induced map $Aut(\mathbb C) \to Aut(F\mathbb C)$ is trivial; in particular, $F(1) = F(2)$ where $1,2: \mathbb C \to \mathbb C$ are the scalar maps. Moreover, if $F$ is a right adjoint, then it preserves abelian group objects, so we may subtract $1$ from $2$ to see that $F(0) = F(1)$. From this it follows that $F$ is constant at the terminal object. A similar argument using cogroup objects shows there are no interesting left adjoints $FinHilb \to FinGpd$.

On the other hand, if the morphisms of $FinHilb$ are chosen differently, then it's unlikely that $FinHilb$ has finite products / finite coproducts, so there can't be a nontrivial right / left adjoint functor $FinGpd \to FinHilb$.

The only exception I can think of is if one takes the morphisms of $FinHilb$ to be contractive linear maps, i.e. those maps $f$ such that $\|f(x)\| \leq \|x\|$. With this definition, $FinHilb$ still doesn't have finite products, so there can't be a right adjoint functor $FinGpd \to FinHilb$, but it has finite coproducts. But any right adjoint functor $FinHilb \to FinGpd$ will still identify all nonzero scalars for the same reasons as above, which I think already makes any such functor uninteresting.

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  • $\begingroup$ Please do not hate me, ahah. $\endgroup$ Aug 15, 2018 at 22:43
  • $\begingroup$ @IvanDiLiberti No worries, it was kind of interesting thinking this one through. $\endgroup$
    – Tim Campion
    Aug 15, 2018 at 22:52
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I don't know what morphisms you intend for the category of finite-dimensional Hilbert spaces, but it doesn't actually matter. The answer is no, there are no interesting adjunctions between the category of groupoids and the category of finite-dimensional Hilbert spaces.

More generally,

If $C$ is a complete and cocomplete category and $D$ is a small category then there are typically no interesting adjunctions between $C$ and $D$.

More precisely,

Observation 1: Let $C$ be a category with all small coproducts and $D$ a small category. Then for any $c \in C, d \in D$ there is at most one morphism $L(c) \to d$.

Proof: For any set $S$, the coproduct $\amalg_{s \in S} c$ exists in $C$. So the coproduct $\amalg_{s \in S} L(c)$ exists in $D$. The result follows from the next observation.

Observation 2: Let $D$ be a small category and $d \in D$. Suppose that the coproduct $\amalg_{s \in S} d$ exists for any set $S$. Then there is at most one morphism $d \to d'$ for any $d' \in D$.

Proof: This is the same argument as the proof that no small category is cocomplete unless it is a poset: if there are two distinct morphisms $d \to d'$, then the cardinality of $Hom(\amalg_{s \in S} d, d')$ is at least $2^{|S|}$, which eventually exceeds the cardinality of $D$.

In your case, the only object of $FinHilb$ which has at most one morphism to any other object is $0$. So the only left adjoint functor $Gpd \to FinHilb$ is the constant functor at $0$.

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    $\begingroup$ I upvoted, but I am not in love with this answer. It might very well be the case that there is an interesting adjunction with finite groupoids, and this answer is not really eliminating this possibility. Technically you proved that the answer to the question in no, but this argument does not provide a "no" for $\epsilon$-natural variations of the question. $\endgroup$ Aug 15, 2018 at 21:05
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    $\begingroup$ @IvanDiLiberti If a calculus student asks whether $x^n + y^n = (x+y)^n$ in $\mathbb R$, I explain to them why the answer is definitively no. I wait for their follow-up questions to judge whether I should discuss the situation over $\mathbb F_p$. $\endgroup$
    – Tim Campion
    Aug 15, 2018 at 21:30
  • $\begingroup$ Of course you are right. $\endgroup$ Aug 15, 2018 at 21:53

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