Non 2nd-countable, Hausdorff, differentiable n-manifold?

**RoNN|3** · Nov3-09, 02:38 PM

I am trying to find a Hausdorff topological space that is not second-countable but otherwise a DIFFERENTIABLE n-manifold. I can't figure it out. Does it exist?

I read about the classical example of $LaTeX Code: L=\\omega_1\\times[0,1)$ with lexicographical order and the order topology. It's Hausdorff, not second-countable and locally homeomorphic to $LaTeX Code: \\mathbb{R}$ . (found a nice page about it) To make an n-manifold I thought $LaTeX Code: L\\times[0,1]^{n-1}$ could work. But is

a differentiable manifold? Are the gluing maps $LaTeX Code: C^\\infty$ ? Can the maps be constructed so that they are?

If this doesn't work I have no clue what could it be. Does anyone know an example?

**RoNN|3** · Nov3-09, 10:27 PM

Gave it more thought.
To clarify, the definition of an n-differentiable manifold I am using is: 2nd-countable, Hausdorff space

s.t. there's an atlas $LaTeX Code: \\{(U_i,h_i,V_i)\\}$ where

open in

, $LaTeX Code: \\cup{U_i}=X$ ,

open in $LaTeX Code: \\mathbb{R}^n$ , $LaTeX Code: h_i: U_i \\rightarrow V_i$ a homeomorphism and gluing maps are $LaTeX Code: C^\\infty$ i.e. $LaTeX Code: \\forall{i,j}\\ { }(h_j\\circ{h_i^{-1}}): h_i(U_i\\cap{U_j})\\rightarrow{h_j(U_i\\cap{U_j})}$ is $LaTeX Code: C^\\infty$ .

My first idea was to try something with discrete topology but I discarded it when I saw that every book/website sites the Long Line as an example. I thought "It has to be the long line. Why would it exist in the first place if you can do it much easier?!". Anyway, I worked with the first idea a bit and got this:

$LaTeX Code: \\mathbb{R}^n\\times\\mathbb{R}$ with the product topology of standard and discrete. It is Hausdorff and not second-countable. I give the atlas of $LaTeX Code: (\\mathbb{R}^n\\times\\{x\\}, \\pi, \\mathbb{R}^n )$ where $LaTeX Code: \\pi$ is projection (it's a homeomorphism). But I don't know if this is valid since there are no gluing maps: all

and

are disjoint so this is.. trivial. Does this count??

Now I realize why the long line is always cited: it's path connected. And I don't require any connectedness in my definition. But then this makes one lousy manifold

I also remember reading that some definitions exclude spaces with uncountably many connected components. What is the property that guarantees that? If I make my space "countably many copies of $LaTeX Code: \\mathbb{R}^n$ " I believe it becomes second-countable but I don't want that.

I am confused and frustrated that there are many different definitions for this sort of thing. Wasted a lot of time on this.

**hamster143** · Nov4-09, 05:05 AM

One big problem with dropping the 2nd-countable requirement from the definition of a differentiable manifold is that your atlas is no longer required to be countable, that could be inconvenient.

As far as I can tell, it's possible to create an atlas and differentiable gluing maps for the long line in the same manner as the proof of theorem 7 from your link.

**janrozendaal** · Nov5-09, 03:34 AM

I think it's a wonderful example of such a manifold. If I guess correctly I know who you are, Ron, since it is a little too convenient that you are trying to find such a space at exactly the same time as I am. So you're probably trying to make the same homework exercises as I am.
I didn't have much time to make them, and was just surfing the net trying to find such a manifold, and the first link I found was your post. It's exactly the answer which I suppose we are to give, it's really nice!
(applause) :)