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In Spivak's Diff Geom (vol.1), p.19, he says a closed manifold is non-bounded and compact (A point in boundary has a neighborhood homeomorphic to half-space). I don't know a non-trivial example of that.

For example, compact subset of $\mathbb{R}^2$ is usually closed set and has boundary, so it's not closed manifold according to Spivak's definition.

An example is the finite set of discrete points of $\mathbb{R}^2$, it's compact and, since no point in it has a neighborhood homeomorphic to half-space, it has no boundary. But the example is trivial.

Does anyone know a non-trivial example of closed manifold (in Spivak's definition)?

[Well I check the definition of closed and compact manifolds here: https://mathworld.wolfram.com/ClosedManifold.html https://mathworld.wolfram.com/CompactManifold.html

It seems what confuses me is that 'compact' here means $\sigma$-compact (locally compact and connected, or say its any open cover has countable sub cover), and I thought it means that its open cover has finite sub cover. Right?]

Charlie Chang
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  • Using bounded/unbounded in this context is bad terminology. The common terms are manifold with boundary and manifold without boundary. Also, the symbol $\times$ should be used for Cartesian product, the symbol $\ast$ is used to denote join. – Michael Albanese Jul 22 '20 at 15:02

3 Answers3

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The standard first examples of closed manifolds are the spheres. For instance, $S^2$, usually imagined as the unit sphere in $\Bbb R^3$. The torus is also a popular example, and you will eventually be very familiar also with the projective plane and the Klein bottle. If you want to make your own examples, the boundary of compact manifolds with boundary will always work.

"Compact" here does not mean $\sigma$-compact. Indeed, $\sigma$-compactness is one of Spivak's requirements for any manifold. If I recall correctly, there is an appendix exploring a few non-$\sigma$-compact examples, like the so-called long line.

BigM
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Arthur
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  • It seems intuitively closed manifolds roughly means (or some of them are) manifolds 'empty inside', and bounded manifolds roughly means ones with something 'inside'. – Charlie Chang Jul 22 '20 at 06:21
  • It seems if we limit the discussion to subsets of $\mathbb{R}$ then bounded open sets are closed manifolds, closed sets bounded manifolds, and unbounded open sets like $\mathbb{R}$ and $(a, +\infty)$ are manifolds that are unbounded and not closed. – Charlie Chang Jul 22 '20 at 06:28
  • @CharlieChang I don't think what you are thinking is right. For example removing a single point from a torus or sphere will not be closed manifolds. Not sure what the etymology is but it could be that it "closes" up or that when embedded in a bounded region of $\mathbb{R}^n$ it is a closed set. Bounded intervals are not closed manifolds (they are not compact). –  Jul 22 '20 at 06:29
  • @CharlieChang Non-empty open subsets of $\Bbb R^n$ are never compact. And closed manifolds must be embedded in higher-dimension Euclidean space, so in $\Bbb R$, the single point set is the only connected example. In $\Bbb R^2$, the only connected example is the circle. In $\Bbb R^3$, we start to get more room to work with, and have the sphere, and the tori with different number of holes, as well as the circle and single points. In $\Bbb R^4$ the world opens up even more, but it starts being very difficult to visualize. – Arthur Jul 22 '20 at 06:34
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Spheres ($S^n$) ,Genus n surfaces with n greater than or equal to 1, are all closed and bounded as subsets of $\Bbb R^n$ and thus compact as topological spaces. We can observe that they are also manifolds without boundary. Thus, they are some examples.

Reader Manifold
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(1) The statement within the above brackets is incorrect. 'Compact' here means what it usually means, according to the feedbacks I get. And what confuses me is actually the def of 'compact', which now I figure out.

We can prove compactness of a manifold by dividing the manifold which has an infinite cover, then dividing the part(s) which has infinite cover, then repeating the process till we get an infinitely small part which tends to a point $p$ of the manifold, and which is included in an open set of the cover (a neighborhood of $p$). From this we see, for example, a sphere (viewed as embedded in higher dimensional space--this view makes it easier to allow us to relate cases of closed manifolds embedded in $\mathbb{R}^n$ or any 'bigger' manifold--or just as itself floating alone in 'vacuum') is compact.

We can also prove compactness by proving that limit points of any infinite sequences on the manifold belong to the manifold. From this we can intuitively tell if a manifold is compact, e.g. a sphere w a point removed is not compact.

(2) I try to summarize examples, which is mentioned and that I can think of, of compact manifolds, closed manifolds and non-compact manifolds without boundary (all connected) as follows:

-closed (compact and without boundary) manifolds:

  1. in $\mathbb{R}$: $S^0$ (point)
  2. in $\mathbb{R}^2$: $S^1$ (cicle)
  3. in $\mathbb{R}^3$: $S^1\times S^1$ (spheres, toruses), (by removing from several toruses disks without boundary and connect them we can get extra closed 2-manifolds, i.e. genus n surfaces ).
  4. in $\mathbb{R}^4$: I guess should be something like $S^1\times S^1\times S^1$, and 'connected' disk-removed manifolds.

-non-compact manifolds without boundary: The 'interiors' of manifolds above.

-compact (not closed) manifolds with boundary: Put the above two manifolds together. (Note cases like a $\mathbb{R}^3$-embedded circle, together with an $\mathbb{R}^3$-embedded surface of which the boundary is the circle, also count.)

Moreover, non-compact manifolds with boundary are easily found, such as $[a,+\infty)$, but such manifolds except those homeomorphic to closed half space remain to be enumerated.

(3) From the above examples I see there seem to be other ways (or general principles) to judge if a manifold belongs to the above three kinds or enumerate examples of manifolds of the three kinds. I will limit my discussions to connected manifolds.

-we can try to cut the manifold (as embedded in $\mathbb{R}^n$ with a straight line, if we get on the line a closed set, then it's compact; if we get discrete points only, then it's closed; if we get an open interval, then it's non-compact and without boundary.

-we can also generate cases in $\mathbb{R}^n$ from cases in $\mathbb{R}^{n-1}$ and ultimately from cases in $\mathbb{R}^1$ by the following process:

  1. we know closed manifolds in $\mathbb{R}^1$ is a point;

  2. then by moving around (I intended to say 'rotating (around an axis)' but it seems not to exhaust cases in higher dimensions)--without letting the locus crosses itself--the point to form a ‘closed' (i.e. the point ends where it starts) locus, we get closed manifolds in $\mathbb{R}^2$, a cycle;

  3. similarly by moving around the cycle, without letting its locus self-crossed, in $\mathbb{R}^3$ (or even $\mathbb{R}^4$) we get closed manifolds in $\mathbb{R}^3$ or ($\mathbb{R}^4$, Klein bottle as an example);

  4. and similarly by moving around these closed manifolds in $\mathbb{R}^4$ we get closed manifolds in $\mathbb{R}^4$; such and such...

(compact manifolds with boundary / non-compact manifolds without boundary can be generated by replacing 'point' in 1. with a closed interval / open interval.)

I myself is interested in the 3rd case, since it's easier to imagine; it's like twisting a tube (of finite or infinite length) to form a surface. Thinking this way I guess perhaps even a knot in $\mathbb{R}^n$ ($n>=3$) may be closed manifolds.

(And I'm thinking of cases that the two ends of a tube is not connected, (e.g. a spiral with one end capped and with another end extending to infinitely far regions) and that a tube crosses itself, are they closed manifolds? I intend to explore more before I form more specific questions to ask here.)

-we can, by removing from some of closed n-manifolds (as just mentioned) disks without boundary and connect them, get extra closed n-manifolds.

I am not sure the two principles above will enumerate all closed (and corresponding compact manifolds with boundary and non-compact manifolds without boundary) embedded in Euclidean manifolds of any dimension. Probably not I guess, probably there are special cases and in higher dimension the situation may be very complicated and common cases thereof may be left out. In particular, I haven't consider any projective spaces since I know not much about them.

Charlie Chang
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