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I am totally new to differential geometry and am having trouble understanding a very basic idea. In what follows, I apologize for being gratuitously pedantic, but I want to be sure I clearly understand what's going on.

If $M$ is a set and $T$ is a topology on $M$ such that $(M,T)$ is Hausdorff and second countable, then $M$ is a topological manifold if for all $p\in M$ there exists an ordered pair $(U,x)$ such that $U \subset M$ is $T$-open and $x:U\rightarrow \mathbb{R}^d$ is a homeomorphism whose image is an open subset of $\mathbb{R}^d$ in the standard topology.

Ordered pairs $(U,x)$ that satisfy the conditions in the above paragraph are called charts on the manifold. An atlas for $M$ is a collection of charts on $M$, $A = \{(U_a,x_a)\colon a \in I\}$, such that $\cup_{\alpha\in I}U_a = M$.

Question 1: Does every manifold have at least one atlas?

My answer: I believe so, since by the definition of a manifold there exists at least one chart for each point, and the collection of either all or at least one of the charts at each point can be taken as an atlas. Perhaps however there is some technical problem in set theory with this construction.

Question 2: Does an atlas uniquely define a manifold? That is, if $A$ and $A'$ are atlases and $A \neq A'$, is it necessary true that the manifolds with $(X,T)$ as their underlying space but with atlases $A$ and $A'$ respectively are different? (In the naive sense--not considering the possibility that they are diffeomorphic)

I believe the core concept I'm struggling with here is what the naive notion of equivalence is for manifolds. (For example, for topological spaces "naive equivalence" means that the two underlying sets are equal and the two topologies have exactly the same open sets, rather than the existence of a homeomorphism, which is a more sophisticated notion of equivalence.)

If instead we define a topological manifold as an ordered triple $(M,T,A)$, where $A$ is an atlas, my confusion vanishes. But then naive equivalence requires exactly the same charts in the atlas, which might be too much to reasonably say that two manifolds are the same. I've also not seen this definition in any of the references I'm using. This brings up the following question.

Question 3: Is it possible to define a manifold as an ordered triple, as in the paragraph above?

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  • $\begingroup$ "Diffeomorphic" is the wrong word if you're talking about topological manifolds--the non-naive notion of equivalence is just homeomorphism. $\endgroup$ – Eric Wofsey Mar 29 at 2:47
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For Question 1, you are right. For instance, you can just take the set of all charts on $(M,T)$ and they will be an atlas.

For Questions 2 and 3, as you have defined a topological manifold, a topological manifold is just a topological space which satisfies certain properties. So, an atlas doesn't actually have anything to do with what a topological manifold is (an atlas just happens to exist on any topological manifold). Two manifolds are equal iff they are equal as topological spaces.

That said, no one actually cares about equality of manifolds. What people actually care about is whether two manifolds are homeomorphic (or more specifically, whether specific maps between them are homeomorphisms). In other words, the "naive equivalence" you are asking about is not important for any applications. As a result, it's perfectly fine to use a definition as you propose in Question 3, where an atlas is part of what a manifold is. This will change what equality of manifolds means (i.e., "naive equivalence") but will not change the notion of equivalence that actually matters, which is homeomorphism.

In the language of category theory, you can define a category $Man$ whose objects are topological manifolds (according to your original definition) and whose maps are continuous maps. You can also define a category $Man'$ whose objects are topological manifolds together with an atlas and whose maps are continuous maps. There is a forgetful functor $F:Man'\to Man$ which forgets the atlas. This functor is not an isomorphism of categories, but it is an equivalence of categories, which is good enough for everything people ever want to do with manifolds.


As a final remark, atlases are pretty irrelevant to the study of topological manifolds. The reason atlases are important is to define smooth manifolds, which impose some additional conditions on what kind of atlases are allowed. A smoooth manifold cannot be defined as just a topological space, but instead must be defined as a topological space together with an atlas satisfying certain assumptions (or a topological space together with some other additional structure equivalent to an atlas).

For smooth manifolds, although an atlas must be included in the definition, there is still an issue similar to your Questions 2 and 3. Namely, multiple different atlases can give "the same" smooth manifold, in the sense that the identity map is a diffeomorphism. This means that if you define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a smooth atlas on $(M,T)$, then the "naive equivalence" is not the equivalence you actually care about, similar to if you used the definition for topological manifolds you proposed in Question 3.

To avoid this, many authors instead define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a maximal smooth atlas on $(M,T)$ (or alternatively, $A$ is an equivalence class of smooth atlases on $(M,T)$). This makes the choice of $A$ unique, in the sense that if $(M,T,A)$ and $(M,T,A')$ are smooth manifolds such that the identity map $M\to M$ is a diffeomorphism between them, then $A=A'$. As with topological manifolds, though, it doesn't really matter whether you use this definition or the previous one, since all that changes is what it means for two smooth manifolds to literally be equal and that's not what we actually care about.

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  • $\begingroup$ Thank you, Eric! I ultimately want a definition I can use for smooth manifolds, but I thought this was a more basic issue with the definition of 'manifold' itself. May I make an addendum to the question, basically asking (2) and (3) for smooth manifolds? Thank you again. $\endgroup$ – Mortified Through Math Mar 29 at 12:13
  • $\begingroup$ I have added some more comments on the smooth case. $\endgroup$ – Eric Wofsey Mar 29 at 15:14
  • $\begingroup$ Awesome. Thanks! $\endgroup$ – Mortified Through Math Mar 29 at 15:54

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