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

[RoNN|3]

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? :smile:

I read about the classical example of L=\omega_1\times[0,1) with lexicographical order and the order topology. It's Hausdorff, not second-countable and locally homeomorphic to \mathbb{R}. (found a nice page about it (http://www.uoregon.edu/~koch/math431/LongLine.pdf)) To make an n-manifold I thought L\times[0,1]^{n-1} could work. But is L a differentiable manifold? Are the gluing maps 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]

Gave it more thought.
To clarify, the definition of an n-differentiable manifold I am using is: 2nd-countable, Hausdorff space X s.t. there's an atlas \{(U_i,h_i,V_i)\} where U_i open in X, \cup{U_i}=X, V_i open in \mathbb{R}^n, h_i: U_i \rightarrow V_i a homeomorphism and gluing maps are C^\infty i.e. \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 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:

\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 (\mathbb{R}^n\times\{x\}, \pi, \mathbb{R}^n ) where \pi is projection (it's a homeomorphism). But I don't know if this is valid since there are no gluing maps: all U_i and U_j 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 :smile: 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 \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. :grumpy:

[hamster143]

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]

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) :)