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I don't know much about the fondations of the theory of manifolds, but the way additional structure if defined on manifolds doesn't feel right to me. For example, to define a smooth manifold, we first define what it means for maps in $\mathbb{R}^n$ to be smooth, and then require that the transition maps of the manifold be smooth as $\mathbb{R}^n$ maps.

This approach feels to me like it is singling out $\mathbb{R}^n$ as a uniquely special smooth manifold, like the theory is reliant on first defining $\mathbb{R}^n$, and then defining all other smooth manifolds in relation to $\mathbb{R}^n$, or something along those lines.

To understand why I feel weird about this, immagine the situation with vector spaces rather than manifolds. The usual way of defining a (real, but not nescessarily) vector space is to say that it is a set with a binary opperation and scalar multiplication that satisfies some axioms. No mention of $\mathbb{R}^n$ anywhere, and an alien species that thinks differently than us could potentially think of this definition without having any concept of $\mathbb{R}^n$, other than having $\mathbb{R}$ as the underlying scalar field. We never first define vector adition and scalar multiplication in $\mathbb{R}^n$, and then say that a vector space is a set with a binary operation and scalar multiplication, that also has a structure preserving bijective linear map to $\mathbb{R}^n$.

I'm interested in knowing if there is a similar, intrinsic way of defining smooth manifolds, without first having to invoke the smooth structure on $\mathbb{R}^n$, similarly to how it's done for vector spaces? Any answers and/or resources would be appreciated.

Thank you.

Markus Scheuer
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Carl Chaanin
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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed. – Shaun Dec 22 '24 at 17:26

4 Answers4

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This is a good question. I have some comments:

  1. A manifold can be thought of as a space on which one can do differential and integral calculus. Since calculus is most easily developed on $\mathbb{R}^n$, it is natural to define a manifold as a space that is locally equivalent to an open set in $\mathbb{R}^n$.

  2. It is in fact possible to define a manifold without explicitly using $\mathbb{R}^n$ and defining a manifold as being locally equivalent to an open subset of affine space. This approach is in fact necessary for defining an infinite-dimensional manifold. The main challenge, as many have already commented, is defining a sufficiently strong topology on the affine space so that you can do calculus. One straightforward way is to use a Banach norm on the vector space associated with an affine space to define open sets and differentiation (ADDED: See @peek-a-boo's link to Michor's slides for other infinite-dimensional models). In finite dimensions, any two Banach norms are equivalent, so you can prove that the topology of the manifold does not depend on the norm used to define it. I do like this approach and have even taught manifolds using it. But However, I believe that for most students, using open subsets in $\mathbb{R}^n$ is much easier to understand when first learning about manifolds.

  3. To me, the ugliest part of the definition of a manifold is the use of an atlas. A definition of a manifold would be purer if it does not use atlases. But I do not know how to do this, and I suspect that if there is such a definition it would be harder to understand and work with. ADDED (thanks to @MoisheKahan for his comment): One way is to define a manifold as a locally ringed space modeled on smooth functions on $\mathbb{R}^n$. But this is a hard definition to understand at first, even though its perspective is sometimes useful, after manifolds have already been defined.

  4. In the end, we want to use definitions that make our work easier, not harder. At this point, using local coordinates to define a manifold and then building the abstraction on top of that appears to be the best way to achieve this goal.

Deane
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    Related to your point 2, Peter W. Michor has some slides about the various model spaces for manifolds. The first few pages are a pretty nice overview. – peek-a-boo Dec 21 '24 at 16:14
  • One can work without atlases by using the notion of a locally ringed space. This is how algebraic geometers do it. But as a differential geometer, I dislike this approach. – Moishe Kohan Dec 21 '24 at 16:19
  • @MoisheKohan, indeed defining a manifold as a locally ringed space does not use an atlas explicitly, But still depends on using $\mathbb{R}^n$ as a local model. – Deane Dec 21 '24 at 16:25
  • @peek-a-boo, thanks. Michor's slides contain an impressive survey of infinite-dimensional manifolds. – Deane Dec 21 '24 at 16:42
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    For the definition you don't need atlasses for sure. And there is no need to involve the general theory of locally ringed spaces, since here the sheaf will consist just of special functions. Basically you define manifolds via the notion of a smooth function. I won't explain the details, it's well-known. – Martin Brandenburg Dec 21 '24 at 17:50
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    I don't understand why the detour via Banach spaces should answer the question. You just expand your class of local models. For them you still need to define smooth maps. It's the same. – Martin Brandenburg Dec 21 '24 at 17:54
  • (see https://mathoverflow.net/questions/88056/is-there-a-sheaf-theoretical-characterization-of-a-differentiable-manifold for the sheaf-theoretic definition) – Martin Brandenburg Dec 21 '24 at 18:07
  • I agree with @MartinBrandenburg. While much of this stuff is interesting and new to me, it still doesn't quite answer my question on how to abandon local models for the definition. I don't know any sheaf theory or locally ringed spaces, but I'm finding the Michor slides ineresting so I will consider the question answered. Thank you for your inputs! – Carl Chaanin Dec 22 '24 at 04:35
  • @MartinBrandenburg, regarding your comment "You just expand your class of local models". As far as I can know, the only local model used for a smooth manifold is the sheaf of smooth functions on an open subset of $\mathbb{R}^n$. I mentioned Banach spaces for two reasons: 1) It's how you define one particular class of infinite dimensional manifolds. 2) It's an easy way to define the local model of a finite-dimensiional manifold without ever mentioning $\mathbb{R}^n$ explicitly. – Deane Dec 22 '24 at 16:31
  • @MartinBrandenburg, the answers and comments on the MathOverflow question are quite amusing. The accepted answer is quite unsatisfying, because it's essentially a translation of each part of the standard definition into sheaf-theoretic language. As far as I can tell, the definition adds no novel insights or tools for studying manifolds. – Deane Dec 22 '24 at 16:48
  • Locally ringed spaces were introduced originally to study singularities of algebraic varieties. At some point, it was recognized that, unlike manifolds, a singular variety could not be studied purely as a parameterized set of points and that more local data needed to be included. It was recognized that the right way to do this was to define the space of "good functions" in open subsets of the variety. This led to an abstract definition of varieties that does not refer to a set of points at all. The concept of a point is derived from the definition and is not part of it. – Deane Dec 22 '24 at 16:54
  • A singular algebraic variety has many possible local models, corresponding to fundamentally different geometric structures. Local algebraic geometry is largely an attempt to understand and classify all possible local models. In contrast, a manifold has only one local model. As we all know, the local structure of a manifold is always the same, an open subset of $\mathbb{R}^n$, and therefore one never studies the local structure of a manifold. Without any additional geometric structure (e.g., Riemannian metric), the study of manifolds is all global and called differential topology. – Deane Dec 22 '24 at 16:59
  • The MathOverflow answer reminded me that I also dislike the standard definition of a manifold that starts by assuming $M$ is a topological space with certain topological properties and then defining atlases compatible with the topological structure. I prefer to start with a set $M$ that an atlas that satisfies natural and reasonable properties and deriving the topological properties (Hausdorff, second countable) as a consequence of the definition of a manifold. – Deane Dec 22 '24 at 17:04
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From my perspective, this is the "wrong question" for someone just learning the theory. Let me explain.

Consider the development of projective space. You want to say "any two distinct lines in the plane meet at a point," but that's false in the Euclidean plane. In the projective plane, parallel lines "meet at infinity," fixing the problem. However, the projective plane is only locally Euclidean--you've glued some affine pieces together. You had to enlarge your universe to say what you wanted to say.

Suppose you now become interested in curves in projective space, so you naturally want to do calculus there. You'll reinvent the notion of smooth manifolds to do so. Why stop there? Why not do calculus on a Klein bottle? Or calculus on spaces of matrices? Or on hyperbolic spaces? Or...

The point is, the abstract definition was invented to capture a zoo of examples people with many different perspectives were already interested in studying. The basic idea is almost embarassingly simple: we can glue together locally Euclidean spaces to make a larger space. Tons of work in the field is a variation on, "what genuinely different spaces does this get us?"

You're welcome to glue together other spaces, by the way. This is at the heart of algebraic geometry, where commutative rings take the place of spaces (really, their spaces of smooth functions), and scheme theory is precisely gluing together such algebro-geometric spaces.

With any of these highly refined theories, it's unfortunately easy to get lost in the formalism. There's so much to cover that modern expositions often ignore or downplay the historical motivation for a subject. It gives the impression that everything somehow comes from heaven, or at least some Platonic realm, when in fact almost everything was motivated by a desire to unify and extend some key families of examples people were already interested in.

Joshua P. Swanson
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    This is not a "wrong" question per se, just poorly stated and written by a user unprepared for complexities of the manifold theory. There are fruitful questions along these lines about characterizing, say, Lie groups among topological groups (Hilbert-V), topological manifolds among locally compact locally contractible spaces ("resolution conjecture"), Riemannian manifolds among metric spaces (Alexandrov). – Moishe Kohan Dec 22 '24 at 14:35
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    Well, as I say, it's my perspective. Experts are welcome to probe the boundaries of characterization and axiomatization. It can be a lot of fun. But when somebody's just learning the theory, they should keep the canonical examples front and center. I've clarified my first line to emphasize this. – Joshua P. Swanson Dec 22 '24 at 14:42
  • That I fully agree with. – Moishe Kohan Dec 22 '24 at 14:57
  • In fact, the manifold structure of projective space and more generally Grassmannians are easily defined without any explicit use of coordinates. Given an $n$-dimensional vector space, let $G(V,k)$ be the set of all $k$-dimensional subspaces of $V$. Given $K_0 \in G(V,k)$ and a transversal $(n-k)$-dimensional subspace $L_0$, let $\mathcal{M}(K_0,L_0)$ be the vector space of all linear maps from $K_0$ to $L_0$. The graph of each such map is a $k$-plane. This defines a cover of $G(V,k)$ by coordinate maps, where the transition maps are affine. – Deane Dec 24 '24 at 21:44
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If there was a simple definition, we'd know it. It took 100 years between Riemann and Whitney to arrive at the current definition, and still it works in some cases and not in others. In algebraic geometry, for example, we are quickly limited by the usual definition when we are dealing with singular points, and then we have to define schemes when we have to take arithmetic aspects into account.

Thomas
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$\newcommand\R{\mathbb{R}}$I'm posting this as a separate answer, since my first one is already too long and rambling. Below I give a definition that makes no explicit mention of $\mathbb{R}^n$ and, I believe, is valid for infinite dimensional manifolds. Corrections are, however, welcome. I've also used a bit of category theory.

Let $V$ be a topological vector space Let $\mathcal{V}$ be a category whose objects are the open subsets of $V$ and morphisms comprise a specified subcollection of continuous bijections such that the inverse of each morphism is a morphism.

Example: If $V$ is an $n$-dimensional vector space, given $b_0 \in V$ and a basis $B=\{b_1, \dots, b_n\}$, let $$ P_{b_0,B} = \{ b_0+a^1b_1+\cdots+a^nb_n:\ 0 < a_1, \dots, a_n < 1 \}. $$ The collection of all such parallelotopes is a basis of the standard topology of $V$. The two most common choices for morphisms of $\mathcal{V}$ are homeomorphisms or smooth diffeormorphisms.

An atlas $\mathcal{A}$ of a set $M$ is defined to be a cover of $M$ and, for each $O$ in the cover, a map $F_O: O \rightarrow V$, called a coordinate map, where

  1. $F_O$ is injective
  2. $F_O(O)$ is open
  3. If $O_1, O_2$ are in the cover, $F_1, F_2$ are the respective coordinate maps, and $O_1\cap O_2 \ne \emptyset$, then $F_1(O_1\cap O_2)$, $F_2(O_1\cap O_2)$ are open
  4. The map $$ F_2\circ\left(\left.F_1\right|_{O_1\cap O_2}\right)^{-1}: F_1(O_1\cap O_2) \rightarrow F_2(O_1\cap O_2) $$ is a morphism in $\mathcal{V}$.

A manifold is defined to be a set $M$ with an atlas $\mathcal{A}$ that satisfies all of the properties listed above.

If we also assume

  1. $V$ is a finite-dimensional vector space with the standard topology
  2. The cover has countably many subsets of $M$
  3. For any $x_1, x_2 \in M$, there exist $O_1, O_2 \in \mathcal{A}$, and open subsets $O'_1 \subset O_1$ and $O_2' \subset O_2$ such that $x_1 \in O_1'$, $x_2 \in O_2'$, and $O'_1\cap O'_2 = \emptyset$,

then if the morphisms of $\mathcal{V}$ are homeomorphisms, $M$ is a topological manifold and if the morphisms of $\mathcal{V}$ are smooth diffeomorphisms, $M$ is a smooth manifold.

It is now straightforward to define the category of $n$-dimensional topological or smooth manifolds, where the morphisms are continuous or smooth maps.

One reason I like this definition is because the topology of $M$ is not an assumed property but arises naturally from the definition of a manifold.

The tangent and cotantant bundles, as well as the pushforward and pullback maps, can be expressed using functors.

Deane
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  • Can you please explain this here? "and morphisms comprise a specified subcollection of continuous bijections such that the inverse of each morphism is a morphism." – Martin Brandenburg Dec 24 '24 at 21:30
  • @MartinBrandenburg, could you specify which aspect of this needs to be clarified? The idea is the following: The collection of all homeomorphisms is obviously a valid choice for the morphisms of the category. That leads to the definition of a topological manifold. However, it is also possible to restrict morphisms to a subcollection of homeomorphisms. This leads to a more restrictive class of manifolds. – Deane Dec 24 '24 at 21:35
  • @Deane I appreciate the the different perspectives that you are presenting, but for me the probleme still persists that the manifold is defined by maps to a differrent space. For vector spaces, there are plenty of things we can study while never mentioning coordinates or even any map to $\mathbb{R}^n$, and importantly this includes the definition itself. I understand that historically the whole point of manifolds is that they have local coordinate maps to $\mathbb{R}^n$, but there are plenty of fields that were redifined throughout their history, and I want to see if the same can be done here. – Carl Chaanin Dec 25 '24 at 23:25
  • @CarlChaanin, your observation shows that a vector space is a more fundamental concept than that of a manifold. The concept of a manifold is derived from that of a vector space, and I do not believe is manifold can be defined without some reference to a vector or affine space. I think of a manifold as a nonlinear space that can be approximated by a linear space well enough to do calculus. It’s worth noting that it is impossible to do calculus without the concept of linear functions. – Deane Dec 26 '24 at 00:01