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Background: I've been working through Guillemin and Pollack's "Differential Topology." They take the approach of defining smooth manifolds as "concrete" submanifolds of some ambient $\mathbb R^N$, as opposed to "abstract" topological spaces with charts and atlases. In my opinion, this approach leads to better intuition about the subject. Moreover, even in the standard "abstract" approach we can establish the Whitney embedding theorem, meaning that we aren't losing generality by working "concretely."

With this viewpoint of "abstract" vs "concrete" definitions, I realized that the Whitney embedding theorem with its ambient $\mathbb R^N$, and Cayley's theorem with its ambient symmetric group $S_N$, are of the same flavor. I know that there are many other representation theorems in math like these two, and that many can be viewed as generalization of the Yoneda embedding.

However, I can't find any sources that include the geometric representation theorems like Whitney and Nash in this Yoneda or categorical framework. So, I'm asking:

Question: Can the Yoneda lemma or other category theory be used to unify the Whitney embedding theorem with Cayley's theorem and/or other representation theorems in math?

My background consists of the category theory that showed up in the first year grad school, e.g. in an algebraic topology sequence and commutative algebra. However, any comments on this circle of ideas is greatly appreciated. Thank you in advance!

Anne Bauval
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Sam Freedman
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  • It's really a different kind of embedding. The Whitney embedding theorem embeds the manifold within $\mathbb{R}^N$, whereas the Yoneda lemma 'embeds' the functor of points into the functor category. – Jeroen van der Meer Feb 01 '21 at 16:43
  • That is not to say that you can't apply the Yoneda lemma to the category of (smooth) manifolds. You certainly can, and people have surely tried to consider manifolds from this functorial point of view. But it's a different thing. – Jeroen van der Meer Feb 01 '21 at 16:44
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    Hi Sam, in response to your question concerning the relation between $\mathbf{R}^n$ and $S_n$, one thing you can say is that they are somehow co-final in their respective categories, i.e. the category of finite groups and the category of smooth manifolds. – shubhankar Feb 01 '21 at 21:28
  • Here's what co-final means https://en.wikipedia.org/wiki/Cofinal_(mathematics) – shubhankar Feb 01 '21 at 21:29
  • Hi Shubhankar, good to hear from you again — just emailed you! – Sam Freedman Feb 01 '21 at 21:35
  • Yes, I agree that a good way to say it is that $\mathbb R^N$ and $S_N$ are interesting or unexpected examples of cofinal objects in their respective categories. – Sam Freedman Feb 01 '21 at 21:36
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    I do think the philosophy of the results is somewhat similar: your idea of concrete models vs abstract axiomatization is very reasonable. The same thing could be said about vector spaces and free vector spaces (although here the situation is even better - or worse, depending on your point of view :) ), or about general rings and endomorphism rings, or yet again, categories and presheaf categories – Maxime Ramzi Feb 03 '21 at 18:56
  • Yes, all great additions of the abstract/concrete paradigm! I think your point that (locally small) categories can be viewed as a presheaf subcategory via the Yoneda embedding is close to what I was looking for--thanks for sharing – Sam Freedman Feb 03 '21 at 19:04
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    I have a vague sense that in algebraic geometry, when embedding varieties into "big, standard spaces", things are slightly more formal and closer to "just" being an application of the Yoneda lemma than in diferential topology. – tcamps Feb 15 '21 at 19:00

2 Answers2

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It's really a different kind of embedding. The Whitney embedding theorem embeds the manifold within $\mathbb{R}^N$, whereas the Yoneda lemma 'embeds' the functor of points into the functor category. The reason why Cayley's lemma fits within the Yoneda lemma is because groups happen to be reinterpretable as one-object categories, and there's no such thing for smooth manifolds (well, as far as I know).

That is not to say that you can't apply the Yoneda lemma to the category of (smooth) manifolds. You certainly can, and in this way a smooth manifold $M$ can be captured completely by the functor $h_M \colon \operatorname{Man}^{\mathrm{op}} \to \operatorname{Set}$ sending a manifold $N$ to $\operatorname{Hom}(N,M)$. I'm sure people have tried to consider manifolds from this functorial point of view ('diffeological spaces', 'smooth sets', are some buzzwords that come up for me). But it's a different thing.

Side remark: A vaguely related question, to which I don't know the answer, is whether you can detect smooth embeddings in the category of smooth manifolds entirely in categorical terms.

Shaun
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Jeroen van der Meer
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  • Thanks for the response. It's a good point that "groups are one-object categories" lends them to the Yoneda embedding. I've shifted the focus off the phrase "Yoneda embedding" as a result. – Sam Freedman Feb 01 '21 at 16:56
  • I'm still struck by the similarity between the ambient $\mathbb R^N$ and $S_N$. Do you know if these objects play the same formal role in their respective categories? – Sam Freedman Feb 01 '21 at 16:57
  • Hmm... I can't think of any relation but I like the question! – Jeroen van der Meer Feb 01 '21 at 17:18
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I do not see any relationship between these ideas at all.

Cayley's theorem exploits a structural idea about groups. Groups acting on sets give homomorphisms to the symmetric group. And a group's action on itself is a particularly natural choice of action. $S_n$ is the correct place to think about group actions.

Whitney's theorem is proved by hacking together a map based on charts and partitions of unity. It does not exploit any clever structural ideas about manifolds, nor teach you new ways of thinking about them. There is nothing special about $\Bbb R^N$ except that it is the domain of a chart and it is a vector space.

so... the proofs are different. The reason these objects are special are different. the way these teach you to think about the objects are different. The only relationship seems to be the word "embedding". What could possibly be the relationship? (Of course, "is there a connection" is not something that can be answered negatively because it is not a precise mathematical question which I can disprove.)

user882544
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