# Why abstract manifolds?

If we can use Whitney embedding to smoothly embed every manifold into Euclidean space, then why do we bother studying abstract manifolds, instead of their embeddings in $\mathbb{R}^n$? A vague explanation I have heard is that from this abstract viewpoint, we gain understanding into the intrinsic behavior of the manifold, without knowing anything about the ambient space in which it can be embedded. Can anyone give some examples of this, or any other reason why abstraction is necessary?

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If you insist that a manifold has to be embedded in $\mathbb R^n$, then you need to show that whatever property you are interested in is independent of that embedding. – Grumpy Parsnip Mar 12 '11 at 15:33
By the way, PL manifolds DO come embedded in $\mathbb{R}^n$. You need an embedding on $\mathbb{R}^n$ to define a polyhedron. So this is a feature of the smooth and topological categories. – Daniel Moskovich Nov 27 '11 at 14:25

The same reason we study abstract groups instead of their embeddings into symmetry groups. If we know some properties of a manifold $M$ which are intrinsic and don't depend on embeddings, and we later encounter that manifold $M$ somewhere else in a different embedding than the one we're used to, then we automatically know a list of things we can say about this new copy of $M$. The point is that the study of things like groups and manifolds naturally separates into the study of their abstract structure and the study of their concrete representations, and failing to make this separation is unnecessarily confusing.

For example, topological and smooth manifolds don't have a notion of length of paths or volume: these things are dependent on an embedding into $\mathbb{R}^n$, so if I meet a manifold in a different embedding I can't assume that things like length and volume are the same, and I will only get confused if I do assume this. I can assume, however, that things like the number of connected components, the homotopy groups, the cohomology, etc. are the same, so whenever I meet a manifold whose cohomology groups I know I can always apply that information.

I wouldn't say that the abstraction is "necessary." Nothing in mathematics is necessary. We make the definitions we make so we can more easily understand certain mathematical phenomena, and the phenomena we are interested in with regard to manifolds are easier to understand if we define abstract manifolds first and then worry about their embeddings later.

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Or imagine graphs as being defined as 1-dimensional polyhedra in $\mathbb R^3$. – Ryan Budney Mar 13 '11 at 3:40

The canonical example of the value of the intrinsic viewpoint is probably General Relativity. How would one make physical sense of a 4-dimensional curved space-time if one had to embed it in a higher-dimensional Euclidean space? What would you embed the universe in? It's (by definition) everything.

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A practical point to complement the answers already given: A lot of things can relatively easily be described as manifolds, e.g. as quotient spaces of other manifolds, but their embedding in a higher-dimensional Euclidean space is more difficult to think about or imagine or even to come up with. For instance, the Klein bottle is easily described as a square with certain edge identifications, and projective space is easily described as the set of all directions, and it's rather immediate that these are manifolds, and yet I wouldn't know off the top of my head how to embed either of them in a higher-dimensional Euclidean space. If the theory of manifolds were built on embeddings, you'd first have to figure out how to embed these things before you could start investigating them.

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If you restricted the definition of manifolds to be "submanifolds of Euclidean space", you'd run into the problem that sometimes you want to do calculus on an object -- natural objects like homogeneous spaces, Grassmann manifolds, cut-and-paste objects like the "pac man" torus and such. But the formal restrictions on usage of manifolds would mean you would have to embed the manifold in Euclidean space before you could apply the machinery of calculus to it. In that sense the submanifold restiction is like a type of formal red-tape that gets in the way of understanding. Abstract manifolds dispense with that.

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Many manifolds don't have a nice, "natural" embedding into a Euclidean space. For example, the usual definition of projective spaces is not as a submanifold of $\mathbb R^n$. Using the abstract definition, it is easy to see that a projective space is a manifold. However, it is very non-trivial to find an embedding of it into a Euclidean space and I have never heard of such an embedding being useful.

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Semi-serious: You don't spend time with proving the Whitney embedding theorem although it is a largely useless for any of your purposes.

Serious: You don't know whether such an embedding exists if you impose further structure onto the manifold (such as a metric tensor) and demand a structure-preserving embedding into $\mathbb R^n$. Vice versa, in case you generalize the concept of manifold (say, singular manifolds), it is again unclear whether such an embedding exists or has any meaning, whereas the instrinsic point of view appears much more natural and allows for convenient mathematics.

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nash embedding theorem – yoyo Mar 12 '11 at 16:06

As noted by Yuan, I think, the main reason is intrinsic properties of manifolds. A very nice result related to this is Theorema Egregium of Gauss. There are quantities which can be measured by the people of the manifold without passing to another realms. When you embed the manifold into another space, you also add extra structure to your manifold.

A second factor, I think, is the definition of manifold itself. A manifold is an equivalence classes of atlases. We then think it as a set of points in our head. More precisely, what makes $S^1$ $S^1$ is not the set of points of $S^1$ but the atlas you abstractly consider. Then we say that the point set $S^1$ is a manifold with that atlas since $S^1$ admits that atlas. Since we have an abstract theory that work for our atlas, results are independent of the uniformization.

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