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Question. Is there a category $\mathcal{C}$ that is equivalent (or anti-equivalent) to the category of smooth manifolds $\mathbf{Man}$ such that the definition of the objects of $\mathcal{C}$ does not involve a set of local models?

This is an attempt to frame the question Is there an intrinsic approach to defining manifolds? in the language of category theory, whose accepted answer I find rather unsatisfying. I am asking this question just out of curiosity. I am not saying that there is anything wrong with the usual definition (this is another difference to the linked question).

If it helps to answer the question, feel free to drop properties such as smooth, paracompact, Hausdorff from the manifolds, but of course not the property of being locally Euclidean. Feel also free to answer the question to manifold-like objects, but whose definition still a priori involves local models. But I will require that we should not limit the dimension, since otherwise potential classification results are available, which goes in a different direction. I am not looking for generalizations such as diffeological spaces which are defined precisely by using local models. I am looking for something more global (as the OP did in the linked question, I assume).

I am aware that such a definition might not exist.

Something rather silly but which goes in that direction is the observation that $\{\mathbb{R}\}$ is dense in $\mathbf{Man}$. It follows formally that $\mathbf{Man}$ is equivalent to a full subcategory of $M{-}\mathsf{Set}$, where $M$ is the monoid of all continuous maps $\mathbb{R} \to \mathbb{R}$. But (a) $M$ is vast, (b) it is not clear which $M$-sets lie in the image (without essentially repeating the definition of a manifold), (c) it still involves the local model $\mathbb{R}$ in some sense.

I understand that this is a rather soft question since I didn't specify exactly when a definition does not involve local models. But I claim that every mathematician will immediately recognize a suitable definition as such. Let me therefore give you an example how a global definition might look like. Assume we have constructed a functor $T : \mathbf{Top} \to \mathbf{Top}$ just by working abstractly with topological spaces. Then the category $T{-}\mathbf{Alg}$ of $T$-algebras (objects are morphisms $T(X) \to X$) has a global definition. Of course, $\mathbf{Man}$ will not be of this form since it lacks fiber products. The name $T$ here intentionally looks like the tangent bundle, which does satisfy a universal property for manifolds.

The category $\mathsf{LRS}$ of locally ringed spaces also counts as global (even though it involves some gluing of local data, we don't have a set of local models), and $\mathsf{Man}$ is a full subcategory of $\mathsf{LRS}$, so it will be sufficient to describe the image without referring to local models, but probably that is not possible, and we should rather go in a different direction.

By Gelfand duality, the category of compact topological manifolds is anti-equivalent to a full subcategory of the category of $C^*$-algebras. But I am not sure if the image can be described without local models, i.e., how to detect the $C^*$-algebra $A := C(M)$ as coming from a manifold without just saying that $\mathrm{Spec}(A)$ is a manifold.

Martin Brandenburg
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  • Are we allowed to use the topology of $\Bbb R$? – Lukas Heger Dec 22 '24 at 15:29
  • @LukasHeger Hard to answer this question. On the one hand, I am inclined to say Yes, because otherwise there is no hope. But If I say yes, then you basically have $\mathbf{R}^n$ already and can "cheat" your way out. I can only say for sure that an answer without the topology on the real numbers (or the real numbers at all) would be amazing. Maybe too much to hope for. – Martin Brandenburg Dec 22 '24 at 15:40
  • I would give the solution of Hilbert 5th as a successful example (a locally compact separable topological group without small subgroups is a Lie group). – Moishe Kohan Dec 22 '24 at 16:23
  • The simplest way I know how to do this is by using the definition of a submanifold: A subset $S \subset \mathbb{R}^n$ is a connected $k$-manifold manifold if there exists a smooth map $f: \mathbb{R}^n \rightarrow \mathbb{R}^{n-k}$ such that $S$ is a connected component of $f^{-1}(0)$ and for each $p \in S$, $df(p)$ has maximal rank. – Deane Dec 22 '24 at 17:25
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    You might be interested in Connes’s result that oriented compact smooth manifolds can be characterized by commutative spectral triples with certain conditions. See Section 11 and Theorem 12.1 of https://arxiv.org/abs/0810.2088. The result in question is not framed as a categorical equivalence, but I would assume getting one is possible. For example, it would seem that Theorem 11.4 implies the category of oriented compact smooth manifolds with smooth maps is equivalent to the opposite of the category of involutive algebras satisfying the conditions listed therein with algebra homomorphisms. – David Gao Dec 22 '24 at 18:24
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    (I’m by no means an expert on spectral triples though, so take what I said with a grain of salt - I might have misinterpreted the result.) – David Gao Dec 22 '24 at 18:25
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    @DavidGao Current state of affairs is such that there's still no reasonable definition of a morphism between spectral triples. There are different notions of submersions proposed, but even conjectural descriptions of things that should correspond to general smooth maps are absent. (...And if you want to look at smooth maps of noncompact manifolds, which do not have a biLipschitz representative, you're just out of luck.) At best (if you assume that Connes' results are true) you can recover maximal subgroupoid of the category of manifolds. – xsnl Dec 24 '24 at 09:16
  • @xsnl Thanks for the info! But is it not the case that an algebra homomorphism $\pi: C^\infty(M) \to C^\infty(N)$ always comes from a smooth map $N \to M$ for compact $M, N$? I thought if you consider local coordinate functions on $M$ and then pull them to $N$ using $\pi$, then you can retrieve a smooth map $N \to M$. – David Gao Dec 25 '24 at 01:15
  • @DavidGao Manifolds are indeed a full subcategory of rings. The issue is that in Connes approach the commutative algebra is not the same algebra of smooth functions! It is a certain subalgebra of operators on it. So you get kind of a Galois reconstruction, similar to Neukirch theorem for number fields. – xsnl Dec 25 '24 at 05:46

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Here is an answer for Lie groups (As you mention: Feel also free to answer the question to manifold-like objects, but whose definition still a priori involves local models.)

In the direction of Hilbert's fifth problem, Yamabe has shown the following characterization:

Theorem (Yamabe) A Hausdorff, locally compact, separable, connected topological group is a Lie group iff it has no small subgroups.

Here we say that a topological group has no small subgroups if there is an open neighborhood of the identity element that contains no non-trivial subgroups.

From there, we can characterize all Lie groups: they are locally connected, Hausdorff, locally compact topological groups with at most countably many connected components such that the identity component admits no small subgroups.

Lukas Heger
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  • Thanks! That's a fascinating result I forgot about. – Martin Brandenburg Dec 22 '24 at 16:45
  • Is it perhaps possible to leverage this into a definition of manifolds as spaces whose local model consists of neighborhoods of the identity of a Hausdorff, locally compact, separable, connected topological group? Then Yamabe's theorem says that these are all equivalent to the standard class of smooth or even real analytic manifolds? – Deane Dec 22 '24 at 17:14
  • @Deane how do get general smooth (or analytic) maps between such neighborhoods of the identity from just Lie group homomorphisms? – Lukas Heger Dec 22 '24 at 18:08
  • @LukasHeger Awesome! Thanks for the great answer! I asked the original question to which this post links to, and this answer is definetly in the direction I was looking for. – Carl Chaanin Dec 22 '24 at 18:41
  • @LukasHeger, good question. My question was speculative. I haven't tried to make it precise at all. – Deane Dec 22 '24 at 18:42