Covering spaces
Jeff Murugan
doubly punctured plane, say R^2 / (+-1,0) is?
John Baez
Jeff Murugan <mur...@maths.ox.ac.uk> wrote:
>Does anybody out there know what the universal covering space of the
>doubly punctured plane, say R^2 / (+-1,0) is?
Yeah. It's usually called "the universal covering space of the
doubly punctured plane." What do you want to know about it?
John Baez
>Jeff Murugan <mur...@maths.ox.ac.uk> wrote:
>>Does anybody out there know what the universal covering space of the
>>doubly punctured plane, say R^2 / (+-1,0) is?
>Yeah. It's usually called "the universal covering space of the
>doubly punctured plane."
That was a pretty dumb answer - I should have said it's usually called
the plane! It's a connected, simply connected noncompact 2-dimensional
manifold, so it's diffeomorphic to the plane.
Marc Nardmann
> Does anybody out there know what the universal covering space of the
> doubly punctured plane, say R^2 / (+-1,0) is?
I'm a bit irritated: You mean "\" instead of "/", right? (Both versions
make sense, but "doubly punctured plane" sounds like the "\" version.)
And you want to remove two points, don't you? (+-1,0 looks like three
points to me.) Anyway:
Let M be the manifold (R^2 minus a discrete set). The universal cover of
M is a noncompact connected 2-manifold with trivial fundamental group.
There's a classification of 2-manifolds, and a comparatively easy
special case of it says that such a manifold is homeomorphic (when you
consider smooth structures, even diffeomorphic) to R^2.
-- Marc Nardmann
Marc Nardmann
I wrote:
> I'm a bit irritated: You mean "\" instead of "/", right?
My poor mastery of the English language again: The English word
"irritated" has a slightly different meaning than the German word
"irritiert". I guess I wanted to say "I'm a bit confused".
-- Marc Nardmann
Michael Weiss
: [the universal covering space of the doubly punctured plane is]
: usually called
: the plane! It's a connected, simply connected noncompact 2-dimensional
: manifold, so it's diffeomorphic to the plane.
Hey, if you're going to go diffeomorphic on us, why not go all the way and
be analytic. The universal covering space of the doubly punctured complex
plane is the upper half plane, or equivalently, the interior of a disk, and
the covering map is the elliptic modular function lambda. See Ahlfors or
Lang or almost any other text on complex analysis.
This fact is the basis of the very beautiful proof of Picard's little
theorem: an entire function whose range omits two values is constant.
Proof: lift the function to the covering space (use the disk), whereupon we
have an entire function whose range lies in a disk. That is, it's bounded,
hence constant. QED.
Jeff Murugan
Marc Nardmann wrote:
> I'm a bit irritated: You mean "\" instead of "/", right? (Both versions
> make sense, but "doubly punctured plane" sounds like the "\" version.)
> And you want to remove two points, don't you? (+-1,0 looks like three
> points to me.) Anyway:
Yes, you're right. Forgive me but it was late when I was typing it. I did,
of course, mean C \ (+-1).
> Let M be the manifold (R^2 minus a discrete set). The universal cover of
> M is a noncompact connected 2-manifold with trivial fundamental group.
> There's a classification of 2-manifolds, and a comparatively easy
> special case of it says that such a manifold is homeomorphic (when you
> consider smooth structures, even diffeomorphic) to R^2.
Thanks. This helps a lot. Do you have any references?
Jeff
Marc Nardmann
> Let M be the manifold (R^2 minus a discrete set). The universal cover of
> M is a noncompact connected 2-manifold with trivial fundamental group.
> There's a classification of 2-manifolds, and a comparatively easy
> special case of it says that such a manifold is homeomorphic (when you
> consider smooth structures, even diffeomorphic) to R^2.
Jeff Murugan replied:
> Thanks. This helps a lot. Do you have any references?
No perfect one, I fear. The "easy special case" I mentioned is exercise
9.3.3 on p. 207 in [Hirsch: Differential topology]; Hirsch gives some
hints. If you need more details, just ask. (The idea is that the
manifold can be exhausted by compact manifolds-with-boundary, which are
classified in Hirsch's chapter 9; it is thus a nested union of disks.)
If you want a classification of all (paracompact) 2-manifolds (maybe you
don't, because it's not so easy), try
Richards, I.: On the classification of noncompact surfaces.
Trans. Am. Math. Soc. 106, 259-269 (1963).
or
Goldman, M.E.: An algebraic classification of noncompact 2-manifolds.
Trans. Am. Math. Soc. 156, 241-258 (1971).
(If these papers talk about triangulated manifolds or PL (piecewise
linear) manifolds instead of smooth or topological manifolds, that
doesn't matter. All these objects have the same classification in
dimension 2.)
Michael Weiss told us that we could regard your problem as a problem in
complex analysis (but he didn't go all the way and say what the
universal cover of the doubly punctured open unit disk is :-)). The book
Ahlfors/Sario: Riemann surfaces ,
for example, should contain the stuff you need, too; I don't have it
handy right now. If you're just interested in C without two points, look
up Michael's reading suggestions. The point here is that the *Riemann
mapping theorem* tells us that there are exactly three simply connected
Riemann surfaces up to biholomorphy, namely the Riemann sphere, C, and
the open unit disk in C. The latter two are diffeomorphic to R^2. (By
the way, every Riemann surface is paracompact; we don't have to include
this property into the definition.) The universal cover of any Riemann
surface is thus diffeomorphic to either S^2 or R^2. And, by the way,
every (paracompact) connected orientable 2-manifold admits the structure
of a Riemann surface. But these complex-analytic arguments are a big
hammer for a little nail.
-- Marc Nardmann