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There is a well known result that every one dimensional topological manifold without boundary is homeomorphic either to the circle or to the whole real line. However there is one detail hidden: manifold is understood to be second countable (or paracompact). If we drop this assumption it becomes possible to construct different example, so called open long line or Aleksandroff line. It is defined as $\omega_1 \times [0,1) \setminus \{(0,0)\}$ with suitable order topology. What might be surprising, is that replacing $\omega_1$ by bigger ordinal does not produce manifold anymore (this would produce points with uncountable neighbourhood system). There is also a variant of long line "in both directions". So the natural question is: if we drop the assumption for (one dimensional) manifolds to be second countable, is it possible to characterise all of them?$ Edit: what about two dimensional case?

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Duplicate: mathoverflow.net/questions/21673/… –  John Pardon Mar 27 at 21:09

2 Answers 2

The one-dimensional case is well known: you have the circle, the line $\mathbb{R}$, the long line $L$ and the long ray $R$. The proof is not that easy to find in the literature since non-metrizable manifolds are (in my opinion) underestimated. It is described in a previous answer, so let me give a few pointers for the two-dimensional case. Before that, let me stress that there is another hypothesis in the usual definition of manifolds that is often overlooked: being Hausdorff. Non-Hausdorff manifolds are interesting too because they appear naturally, e.g. space of leaves of a foliation. I do not know if anyone has studied non-Hausdorff non-metrizable manifolds. That would really be looking for trouble.

Back to the point. There is very, very little hope to classify non-metrizable surfaces: there are many of them, and of various kind.

First one think about easy examples: product of two one-dimensional non-metrizable manifolds; that makes 4 examples.

But there are more: take the first octant of the product of two long lines (i.e. $\{(x,y)\in R\times R | y \le x \}$). Or glue a bunch of octants together along their edges. This already makes you a (small) bunch of examples. At that point, I should stress that in the product of two long lines, the diagonal is very different from the coordinate axes. Each axis $A$ is homotopic to a copy of $L$ that is disjoint from $A$, while any embedding of $L$ which is homotopic to the diagonal must meet it on an unbounded set. This comes from the homotopy theory of $L$ and $R$: e.g. there are two homotopy classes of maps $R\to R$: the ones homotopic to a constant, which are exactly the bounded maps, and the ones homotopic to identity, which have an unbounded set of fixed points.

But there are more: one can produce many different "long pipes", which are obtained as increasing unions of annuli indexed by $\omega_1$, where at each non-limit ordinal the inclusion is as trivial as one might want, but at each limit ordinal the lower annuli can accumulate to a circle with a segment attached, or to something worse. By choosing the shape of such singularities, and at which limit ordinal they do appear gives you a very large range of long pipes.

The good news is that classifying long pipes is sufficient to get a classification of $\omega$-bounded surfaces (i.e. surfaces in which any countable sequence has an accumulation point; e.g. the long line is $\omega$-bounded but the long ray is not). This is thank to the beautiful "bagpipe theorem" of Nyikos (The theory of non-metrizable manifolds, in K. Kunen and J. Vaughan, eds, “Handbook of Set-Theoretic Topology” (Elsevier, 1984), 633–684) which says that any $\omega$-bounded surface is obtained by gluing finitely many long pipes (the pipes, obviously) to a compact surface with some disks removed (the bag). The bad news is that even a classification of long pipes seems out of reach. If I remember well, it is an open question whether every long pipe contains an embedded long line.

The worse news is: but there are more. $\omega$-bounded surfaces are a very particular kind of surfaces. A non-metrizable surface which is very different from everything above is the Prüfer manifold. Basically, you glue a bunch (i.e., one for each real number) of planes to a half plane in a way that maps half infinite strips to cones, so that the different planes do not interfere two much one with the other. This is a huge, weird space.

But I guess that there are more (if I remember well, it has been proved that there are $2^{\aleph_1}$ pairwise non-homeomorphic non-metrizable surfaces, but I do not know in which axiom system it holds).

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Actually, there are long pipes that do not contain a copy of the long line, for instance take a tangent bundle of the longline, remove the $0$ section and take one of the two components. This does not contain a copy of the longline but is not $\omega$-bounded. You can then take the quotient under the $\mathbb{Z}$-action $n\mapsto 2^n\cdot x$ on each fiber, this gives a longpipe without embedded longline. What is ZFC-independant is whether there is a copy of $\omega_1$ in each such longpipe. The paper by Nyikos in the Handbook contains a construction under $\diamondsuit$ of a longpipe (cont...) –  Mathieu Baillif Mar 28 at 15:26
without copy of $\omega_1$, and PFA implies that a longpipe possesses a copy of $\omega_1$ (due to Balogh, I think). And yes, there are $2^{\aleph_1}$ non pairwise homeomorphic surfaces, as shown by Nyikos in the same article. –  Mathieu Baillif Mar 28 at 15:30
And let me add that apparently the classification of $1$-manifolds was first worked out by Hellmuth Kneser in "Sur les variétés connexes de dimension 1, Bull. Soc. Math. Belg. 10 (1958)" (though it was certainly known before, but it seems nobody had written it down). –  Mathieu Baillif Mar 28 at 15:34

The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions) here (p.2, just after Main Theorem). No proof is given in the above paper (just saying it is easy). Here is a sketch, say the manifold $M$ has an endpoint, and call it "the leftmost point", $a_0$, and then pick a sequence $a_n$ of points "going to the right". If $M=\cup_{n}[a_0,a_n)$ then we are done. Else keep adding points $a_\omega$, $a_{\omega+1}$, etc. Then either we are done at some countable ordinal $\gamma$ (and then we are done, use that for every countable ordinal $\gamma$ there is a subset $T$ of the reals that is order-isomorphic to $\gamma$), or else we define $a_\beta$ for all $\beta<\omega_1$, so we get the long ray. It mist be the case that $M=\cup_{\beta<\omega_1}[a_0,a_\beta)$ since otherwise we may define $t=\sup_{\beta<\omega_1} a_\beta \in M$ and our manifold would not be first-countable at $t$, a contradiction. To see that each $\omega<\gamma<\omega_1$ could be thought a subset of $\mathbb R$ fix a bijection $f:\gamma\to \omega$ and any sequence $c_n>0$ with $\sum_n c_n<\infty$, and for each $\beta<\gamma$ define $r_\beta=\sum_{\delta\le\beta}c_{f(\delta)}\in\mathbb R$, then the set $\{r_\beta:\beta<\gamma\}\subset\mathbb R$ is order-isomorphic to $\gamma$.

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The fact is mentioned without proof in Nyikos´ paper. He says it is "easy to show"; is it? –  Ramiro de la Vega Mar 27 at 18:32
@Ramiro I added a sketch of the proof. –  user48481 Mirko Swirko Mar 27 at 19:50

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