Mathematicians are always insisting that manifolds need not be embedded, and by Occams Razor, are best thought without the surrounding ambient space.

For example, the surface of a table is a 2d manifold that is embedded in space. A line on a page is a 1d manifold embedded in a 2d space.

To visualise them without embedding one has to remove the table, or the page. Rather like the Cheshire cats smile without the Cheshire cat.

It appears that all manifolds that we can lay are hands on naturally occur in an embedding.

Are there any that don't?

Well, spacetime seems to be the only one. In what would spacetime be embedded in?

Are there in fact any others?

3 Answers 3


A manifold is a space, not an object in space. The line you draw on paper is not a manifold, although it may be an attempt to visualise one.

The reason you should not think of manifolds as embedded is that it leads to wrong intuition about shortest paths, distances, curvature: if you visualise a 2D sphere embedded in 3D space, you might think the »real« shortest path between its poles is along the axis. This is wrong because the axis is not part of the manifold.

As for manifolds really existing, this depends on whether you are a platonist. If by »naturally« you mean occurring in our space-time continuum, then by definition everything is embedded in space-time.

Edit: Mathematically, just as you can have »subspaces« in Euclidian space, you can have »sub-manifolds«. In fact, there are infinitely many 2D spherical manifolds »embedded« in Euclidean space. But that's maths. And in even in mathematics you abstract from the fact that a manifold is or might be embedded, because otherwise you would not examining the manifold but the thing it is embedded in. Can a manifold be a useful to describe real objects? Obviously. But only as long as we abstract from the fact that they aren't. The surface of the earth can be seen as a manifold; but still, the shortest route from the north pole to the south pole is through the centre of the earth.

  • You get a very interesting alternative metric by approximating how long it would take you to travel a particular way and return. Given that we know of no way to tunnel through the centre of the earth, the path metric simply doesn't go there.
    – AndrewC
    Nov 4, 2014 at 2:07
  • 1
    That is were things get interesting, all right. I'm not so sure my edit (in response to now delete comments) clarifies anything. The point I was trying to make is this: A manifold doesn't have to be embedded. While you are examining a manifold, you disregard its potential embeddedness. If you consider the embedding, you no longer consider the manifold itself.
    – olaf b
    Nov 4, 2014 at 2:46
  • Very good points, yes, sorry if I clouded them.
    – AndrewC
    Nov 4, 2014 at 3:34
  • @olafb: I think those were my comments that I deleted. I didn't want to look unduly critical as they were pretty long; I'll add them to the main question. Nov 12, 2014 at 16:15

Mathematician here. It is very common for manifolds to arise that do not embed in Euclidean space in any obvious (or particularly natural) way. Indeed, the reason that mathematicians think of manifolds without a surrounding ambient space isn't necessarily Occam's razor -- it's just that it wouldn't be practical to develop the theory of manifolds in a context where an ambient space is always required.

The simplest examples of manifolds that don't embed naturally into an ambient space are quotients of existing manifolds, i.e. manifolds obtained from other manifolds by identifying (or gluing) together certain sets of points. For example, the projective plane is a surface that can be obtained from a sphere by gluing together every pair of antipodal points. The result is a perfectly good surface (it's just like a sphere, except you only need to go halfway around to get back to where you started), but it's not immediately clear how to embed this surface into Euclidean space. Similarly, the projective special linear group PSL(2,R) is obtained from the special linear group SL(2,R) by identifying pairs of matrices that are negatives of one another. Again, SL(2,R) embeds into R^4 using the entries of the matrices, but there is no immediately obvious embedding of PSL(2,R) into a Euclidean space. Both of these examples are quite important in mathematics --- the projective plane is the simplest non-orientable closed surface and the main setting of plane projective geometry, and PSL(2,R) is a simple Lie group and is the group of all isometries (i.e. symmetries) of the hyperbolic plane.

Part of what's important to understand here is that mathematicians don't use manifolds primarily as a way of describing shapes of physical objects. For a simple example, parabolas are quite important in mathematics, but not because you tend to see that many actual parabolas in everyday life. Instead, parabolas are important for a theoretical reason -- they are graphs of quadratic equations, and studying the graphs helps you understand quadratic equations better. As a general rule, the primary motivation in mathematics for studying geometry isn't to model physical reality directly -- it's because understanding geometry helps us to understand equations and other mathematical objects, which can then in turn be used to model reality. Abstract geometric manifolds like the projective plane and PSL(2,R) are very important in this regard, even if they don't necessarily describe a shape that you're likely to run into in real life.

All that being said, it does turn out that every topological manifold can be embedded into a Euclidean space of sufficiently high dimension --- this is the content of the Whitney embedding theorem. According to this theorem, it's possible to embed the projective plane into four-dimensional Euclidean space, and it's possible to embed PSL(2,R) into six-dimensional Euclidean space. (Actually, PSL(2,R) embeds into three-dimensional Euclidean space -- it turns out to be homeomorphic to an open solid torus.) However, this theorem takes some work to prove, and you shouldn't have to prove such a hard theorem just to get theory of manifolds off the ground.

In addition, there are some other contexts than topology where embedding theorems are either very difficult or impossible to obtain. For example, if you consider Riemannian manifolds (the main objects of study in differential geometry), it can be very difficult to show that there's an isometric embedding of a given Riemannian manifold into a Euclidean space (see the Nash embedding theorems). For example, the hyperbolic plane is very easy to define, but it's very hard to find a smooth isometric embedding of it into a Euclidean space, and I think it's still an open question whether there exists an analytic embedding (i.e. an embedding defined by a single power series formula).


If there are no instances of non-embedded manifolds, then every manifold is embedded in a more complex manifold. If you took the union of all of those enclosing manifolds, you would get a single manifold. In what would it be embedded? So if infinite-nonsense math counts, we are done.

If not, I guess the question comes down to "What do you mean by naturally-occurring?"

Most folks would presume that the whole of navigable space, as we imagine it naively, is a manifold. If that is not naturally occurring, then there are no naturally occurring manifolds at all, and you have your answer.

But it is very difficult for us to presume that all of space is supported inside another real and not imaginary manifold. We can imagine our accessible space is embedded. We theorize the 11-D wrapper of String Theory. I myself favor a model of space with a secondary time dimension (a la Hawking's imaginary time). But even when we do imagine these explanatory embeddings, we still get to a largest one that can be considered 'natural' and still seem to legitimately 'occur'.

One of the points behind a manifold model is that it can be embedded in a larger imaginary space for modelling, so we can do the mathematical equivalent of approaching it from the outside. But the model does not imply faith in the reality of the larger manifold. The model of reality where color is three-dimensional, and time is reversable, so a painted mobile is seven-dimensional, is not something we consider naturally occurring, even as a side-effect of the construction of the eye. But it does model nature, if your main goal is drawing graphics that evolve in time, with colored lighting.

  • The argument in your first paragraph cannot be right because in fact every manifold is imbeddable in a higher-dimensional manifold. If your argument proves otherwise, then your argument proves too much. For a concrete example, consider a line, imbedded in a plane, imbedded in euclidean 3-space, imbedded in euclidean 4-space, etc. The union is a) not a manifold (which is all by itself enough to kill your argument) and b) easily imbeddable in larger spaces (which is also and separately all by itself enough to kill your argument).
    – WillO
    Nov 6, 2014 at 2:41
  • @WillO OK, first of all, he is asking about naturally-occurring somethings. Naturally-occuring infinite-dimensional anything seems inconsistent to me. (That may be presumptuous.) And I have been out of Topology for a long time -- How is the union in that case not a manifold? Isn't it the union of Cartesian products of a pile of lines? Do the combination rules break down that easily? I would think you could take the ball around the last embedding where the coordinates end for that given point and extend with zeroes. (This is a union of finite-dimensional spaces, right? So there is one.)
    – user9166
    Nov 6, 2014 at 14:55
  • If I can't find the Euclidean ball around the last manifold a point was in in its embedding, then we have already broken the rule for manifolds, right? Don't we have, at worse, a manifold with its closure as a boundary?
    – user9166
    Nov 6, 2014 at 15:05
  • Did you try applying your arguments to the counterexample I gave you?
    – WillO
    Nov 6, 2014 at 15:27
  • Look at the note right above you. Is that an attempt? Your tone is offensive, and I am done responding to you.
    – user9166
    Nov 6, 2014 at 15:32

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