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This is yet more on "algebraic objects in functional analysis".

Since Compact Hausdorff spaces are algebraic over Set, it seems to follow that one can find "Compact Hausdorff objects" in any suitable category representing functors from that category to CompHaus.

An obvious such functor is the spectrum of a unital $C^*$-algebra. This seems to imply that $\mathbb{C}$ is a compact Hausdorff object in the category of unital $C^*$-algebras. So:

Question 1: Is this right?

Followed by the obvious:

Question 2: Are there any other interesting "Compact Hausdorff" objects in other categories?

Similarly, $C^\ast$-algebras is algebraic, and whilst Banach spaces isn't algebraic then it embeds in an algebraic theory (of totally convex spaces). Again, to any compact Hausdorff space one can assign its $C^\ast$-algebra of continuous functions to $\mathbb{C}$. This suggests that $\mathbb{C}$ is a "$C^\ast$-algebra" object in CompHaus - except that $\mathbb{C}$ is not a compact Hausdorff space. However, we have a way out due to the way that $C^\ast$-algebras are algebraic: it's the unit ball that we should be thinking of and this is continuous functions to the closed unit disc in $\mathbb{C}$, which is compact Hausdorff. Thus $\{z \in \mathbb{C} : |z| \le 1\}$ seems to be a $C^\ast$-algebra object in Compact Hausdorff spaces. Again:

Question 3: Is this right?

and

Question 4: Are there any other interesting "$C^\ast$-algebra" objects in other categories?

and

Question 5: Are there any "Banach space" objects (or "totally convex space" objects) floating around anywhere?

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  • $\begingroup$ Andrew, I'm having trouble parsing the sentence that makes up your second paragraph. So, I don't understand what you mean by a "compact Hausdorff object" (though maybe I can guess). Could you clarify? $\endgroup$
    – Tom Leinster
    Commented Feb 10, 2010 at 11:24
  • $\begingroup$ I was extrapolating from "group object" to compact Hausdorff so a "compact Hausdorff object" would represent a functor to CompHaus just as a "group object" represents a functor to Grp. As this is all reasonably new to me, perhaps there's something that is obvious but that I haven't yet "got" which says that whilst "group objects" are fine, "compact Hausdorff objects" aren't. Is there? $\endgroup$
    – Andrew Stacey
    Commented Feb 10, 2010 at 11:42
  • $\begingroup$ Are you allowed to ask five (5!) questions in a single post? ;) $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 10, 2010 at 15:53
  • $\begingroup$ Of course I am! Well, they're all reflections of the same question so I figured it was better to ask them all in one go than clutter up the board with separate ones. $\endgroup$
    – Andrew Stacey
    Commented Feb 10, 2010 at 18:54
  • 1
    $\begingroup$ +1 for linking to a nonexistent nLab page! $\endgroup$
    – Reid Barton
    Commented Feb 10, 2010 at 20:37

3 Answers 3

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This "answer" doesn't even get as far as answering question 1, but I'll go ahead anyway.

All I want to say is how I think "compact Hausdorff space object" should be defined. This should be equivalent to what Sridhar said, though I haven't stopped to think about it.

Let $\mathcal{E}$ be a category with small products. A compact Hausdorff object in $\mathcal{E}$ should be an object $X$ of $\mathcal{E}$ together with, for each set $I$ and ultrafilter $U$ on $I$, a function \[ \xi_U: X^I \to X \] satisfying some axioms that I'm too lazy to write down, but will explain a bit in a moment.

When $\mathcal{E} =$ Set, you can think of $\xi_U$ as specifying the $U$-limit of each $I$-indexed family of points of $X$. (That there's exactly one limit point is the compact Hausdorff property.) One axiom tells you what happens when $U$ is the principal ultrafilter on some $i \in I$: then $\xi_U$ sends a family x to $x_i$. A second says something about limits of limits. A third (and I think there are only three) says something about what happens when you have a map $I \to J$.

This formulation doesn't come out of thin air, you won't be surprised to hear---there's a systematic process for taking a (suitable kind of) monad on Set and producing a definition of its "algebras" in any category with products. But I won't go into that now.

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  • $\begingroup$ That's very much what I thought should happen. It sounds from this as though there is no problem with "compact Hausdorff objects" in any category with small products - do I have that right? In which case, there should be an equivalence between such and representable functors to CompHaus (my interpretation of Sridhar's answer is that he says this is only guaranteed in one direction, I expected both). So then (part of) my question is as to the existence of naturally occurring such objects - for some definition of "naturally", naturally. $\endgroup$
    – Andrew Stacey
    Commented Feb 10, 2010 at 12:26
  • $\begingroup$ I agree that what you (Tom) said is presumably equivalent to what I said. However, I have a question about the part where you say "(suitable kind of) monad on Set". What do you mean by "suitable kind of"? It seems to me this general idea should work for every monad on Set. [That is, pulling it through the correspondence between monads on Set and categories with set-sized products generated by a single object [the (Set-)algebras of the former also corresponding to the Set-models of the latter], and then using the latter to give an account of such algebras in other categories with products] $\endgroup$
    – Sridhar Ramesh
    Commented Feb 10, 2010 at 18:38
  • $\begingroup$ Sridhar, to be honest I was just hedging. I wasn't certain that it would work for all monads on Set, and I didn't want to overstate my case. But maybe it does work for all of them, in the way you describe. $\endgroup$
    – Tom Leinster
    Commented Feb 11, 2010 at 0:59
  • $\begingroup$ Actually, it occurs to me that there's a much easier way to describe this than the detour through infinitary Lawvere theories I've been using. Given a monad $M$ on $Set$, an algebra for this in the category $C$ should be an algebra (in the standard sense) for the monad $M^{C^{op}}$ whose carrier is in the range of the Yoneda embedding, where $M^{C^{op}}$ is the monad on presheaves on $C$ induced by postcomposition with $M$. This should be equivalent to what we've both been saying, but seems to me now much clearer (though perhaps others will differ). $\endgroup$
    – Sridhar Ramesh
    Commented Feb 11, 2010 at 1:38
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The "Bohrification" paper arXiv:0905.2275 may be relevant to Question 4. As I understand, they discuss the notion of $C^\ast$-algebra objects in a given topos.

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Question 1: If I understand you correctly, you're proposing that $\mathbb{C}$ should be a compact Hausdorff object in some category because it represents a functor from that category to the category CH of compact Hausdorff spaces (in something like the sense that the functor $Hom(-, \mathbb{C})$ into Set factors through the forgetful functor from CH to Set). But I don't see why this should be sufficient to make $\mathbb{C}$ a compact Hausdorff object.

That is, presumably, from the approach of functorial semantics, a compact Hausdorff object in a category C should be a product-preserving functor from L to C, where L is the dual of the Kleisli category for the ultrafilter monad on Set (that is, L is the Lawvere theory whose category of (Set-)models is the category of compact Hausdorff spaces). I can see how, more generally, for any Lawvere theory L and category C, every C-model of L (i.e., a product-preserving functor F from L to C) induces a representable functor Hom(-, F(1)) from C to Set which factors through the forgetful functor from Set-models of L to Set. But it's not obvious to me that the converse of this holds as well (that every representable functor from C to Set with this factorization property arises from some C-model of L).

Perhaps I'm missing something and your reasoning for $\mathbb{C}$ being a compact Hausdorff object is something more than this. Perhaps I'm hopelessly confused. But, tentatively, I think the answer to question 1 is "No" or at least "Not necessarily".

(Edit: As seen below, the correspondence does go both ways, so the last line is retracted, leaving the second-to-last line...)

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  • $\begingroup$ But it all works if I replace "compact Hausdorff" by "group", doesn't it? Maybe I'm missing something there as well. If so, perhaps I should ask the more basic question about the difference between finitary and infinitary theories first. $\endgroup$
    – Andrew Stacey
    Commented Feb 10, 2010 at 11:43
  • $\begingroup$ Egads, no, you're right. The correspondence does go both ways. I failed to see it before, being so used to viewing things one way, but if M is a monad on Set, then product-preserving functors from the dual of M's Kleisli category to C (what I was thinking of as an M object) are in correspondence with tuples of the form <contravariant functor F from C to the category of Set-algebras of M, object c in C such that the product of any set of copies of c exists, and natural isomorphism between Hom_C(-, c) and UnderlyingSet(F(-))> (what you were thinking of as an M object). So, I retract my "No". $\endgroup$
    – Sridhar Ramesh
    Commented Feb 10, 2010 at 19:31
  • $\begingroup$ (Why? The latter amounts to just putting the structure of a Set-algebra for M on Hom(x, c) for each x, such that precomposition is a homomorphism of such algebras. For every element in M(k), thought of as a k-ary operation, we obtain a morphism from c^k to c by applying that operation to the k many projections in Hom(c^k, c), from which the result of that operation on arbitrary Hom(x, c) is determined... [continued in next comment] $\endgroup$
    – Sridhar Ramesh
    Commented Feb 10, 2010 at 19:50
  • $\begingroup$ Such morphisms will automatically satisfy the appropriate commutative diagrams (by virtue of the appropriate equations holding in each algebra Hom(c^k, c)). Thus, they can be combined into what I thought of as an M object. I haven't sat down and checked all the details, but I am quite confident now that it works and I was wrong.) $\endgroup$
    – Sridhar Ramesh
    Commented Feb 10, 2010 at 19:51
  • $\begingroup$ Thanks for the clarifications. I wondered if there was some issue with finite/infinite stuff that I was completely unaware of. I feel reassured that at least I didn't miss something completely obvious - should I be feeling that? Or is there still something I've missed in my setup? $\endgroup$
    – Andrew Stacey
    Commented Feb 10, 2010 at 20:48

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