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From: Rotwang on 1 Feb 2010 12:23
Rainer Urian wrote:
> The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
> topologically a sphere with two points removed.
> Can anyone tell, how to show this, maybe by an explicit homeomorphism?
>
> Thanks
> Rainer

I see that Jos� has already posted an answer in the time I've been
trying to come up with something. However I have an approach that is
more "visual" (though also more labour-intensive), so you might like to
see it anyway:

First, consider the subset S of C defined by S = C\{x | -1 <= x <= 1}.
It is not hard to see that S is homeomorphic to the open upper
hemisphere with a pole removed, {x ,y, z e S^2 | 0 < z < 1}; for
example, construct a homeomorphism between S and C\B (where B is the
closed unit circle) by shifting points above or below the line segment
[-1,1] up or down, and then compose this with the restriction of the
stereographic projection. Call the resulting homeomorphism h_1, and
define a second homeomorphism h_2 from the open lower hemisphere with
the pole removed to S by h_2(x, y, z) = (h_1(x , y, -z))*.

The point of considering S is that we can define a continuous square
root p(x) of (1 - x^2) on S. To show this, consider what happens when x
is in each of the four quadrants {re x >= 0, im x >= 0}, {re x <= 0, im
x >= 0} etc.; in each case 1 - x^2 lies in either the upper or lower
half plane, and there are two choices of continuous square root on each
half plane. Show that it is possible to make a choice for each quadrant
which is consistent, i.e. if x in S lies in two quadrants at once then
the two expressions for p(x) are equal.

Now for x in the domain of h_1, define g_1(x) = (h_1(x), p(h_1(x))).
Similarly for x in the domain of h_2, define g_2(x) = (h_2(x),
-p(h_2(x))). Then the ranges of g_1 and g_2 are disjoint subsets of the
complex unit circle. Furthermore, each of g_1 and g_2 may be
continuously extended to include the circle {x, y ,z e S^2 | z = 0}.
Unless I've screwed up, the two extensions should coincide on this
circle, and the common extension to the whole of the sphere minus the
poles should be a homeomorphism onto the complex unit circle.
From: Rotwang on 1 Feb 2010 12:45
Rotwang wrote:
> [...]

I forgot to add:

More generally, the way that physicists attempt to picture complex
curves such as the complex unit circle is in terms of "branch cuts"; for
a multivalued function such as sqrt(1 - x^2) = sqrt((1 - x)(1 + x)),
consider two copies of the complex plane, pictured as infinite sheets of
paper, and imagine making a cut along a line between x = -1 and x = 1;
then glue the lower edge of the cut on one sheet to the upper edge of
the cut on the other sheet, and vice-versa. The point of this
construction is that it tells you what happens as you analytically
continue a function element which is locally a square root of 1 - x^2
along a closed path: if the path avoids the cut then you end up with the
same function element you started with, but if it crosses the cut an odd
number of times then you end up with minus the function element you
started with.

The way to make this precise is to consider the complete analytic
function corresponding to sqrt(1 - x^2), which is a Riemann surface
which is, IIRC, homeomorphic to the space of solutions to x^2 + y^2 = 1.
I don't remember much about this stuff, but I do remember liking
Springer's "Introduction to Riemann Surfaces" when I read it some years
ago, and that book describes a more systematic way of understanding the
topology of such curves.
From: José Carlos Santos on 1 Feb 2010 17:52
On 01-02-2010 15:54, Jos� Carlos Santos wrote:

>> The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
>> topologically a sphere with two points removed.
>> Can anyone tell, how to show this, maybe by an explicit homeomorphism?
>
> So you have this set S = {(x,y) in C^2 | x^2 + y^2 = 1}. It has a
> natural group structure:
>
> (x_1,y_1).(x_2,y_2) = (x_1x_2 - y_1y_2,x_1y_2 + x_2y_1).
>
> Now, consider the map _f_ from C into S defined by
>
> f(z) = (cos(z),sin(z)).
>
> It is a surjective group homomorphism (in C, the group operation being
> the sum) and its kernel is 2pi.i.Z. So S is homeomorphic to C/2p.i.Z
> (there are some topological details to fill here).

Correction: the kernel is 2pi.Z.

> Now, it turns out that C/2pi.i.Z is homeomorphic to C\{0}. That can be
> proved by a similar argument, which uses the group homomorphism
>
> exp:(C,+) ---> (C\{0},.),
>
> whose kernel is, again, 2pi.i.Z. I suppose that it is clear that C\{0}
> and a sphere with two points removed are homeomorphic.
>
> In order to turn this into something explicit, consider the map from
> C\{0} onto S define by z |-> (cos(w),sin(w)), where _w_ is such that
> e^w = z. Obviously, the choice of _w_ does not matter.

It should have been z |-> (cos(i.w),sin(i.w)).

Best regards,

Jose Carlos Santos

From: James Dolan on 2 Feb 2010 12:46
in article <hk5v98$4vo$02$1(a)news.t-online.com>,
rainer urian <rainer(a)urian.eu> wrote:

|The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
|topologically a sphere with two points removed. Can anyone tell, how
|to show this, maybe by an explicit homeomorphism?

by the change of variables (a = x+iy, b = x-iy), x^2+y^2=1 is
equivalent to ab=1. a can be any nonzero complex number and then b is
its reciprocal.


--


jdolan(a)math.ucr.edu

From: Pubkeybreaker on 2 Feb 2010 16:05
On Feb 1, 5:52 pm, José Carlos Santos <jcsan...(a)fc.up.pt> wrote:
> On 01-02-2010 15:54, José Carlos Santos wrote:
>
>
>
>
>
> >> The complex unit circle , i.e. complex solutions of x^2+y^2=1 is
> >> topologically a sphere with two points removed.
> >> Can anyone tell, how to show this, maybe by an explicit homeomorphism?
>
> > So you have this set S = {(x,y) in C^2 | x^2 + y^2 = 1}. It has a
> > natural group structure:
>
> > (x_1,y_1).(x_2,y_2) = (x_1x_2 - y_1y_2,x_1y_2 + x_2y_1).
>
> > Now, consider the map _f_ from C into S defined by
>
> > f(z) = (cos(z),sin(z)).
>
> > It is a surjective group homomorphism (in C, the group operation being
> > the sum) and its kernel is 2pi.i.Z. So S is homeomorphic to C/2p.i.Z
> > (there are some topological details to fill here).
>
> Correction: the kernel is 2pi.Z.
>
> > Now, it turns out that C/2pi.i.Z is homeomorphic to C\{0}. That can be
> > proved by a similar argument, which uses the group homomorphism
>
> > exp:(C,+) ---> (C\{0},.),
>
> > whose kernel is, again, 2pi.i.Z. I suppose that it is clear that C\{0}
> > and a sphere with two points removed are homeomorphic.
>
> > In order to turn this into something explicit, consider the map from
> > C\{0} onto S define by z |-> (cos(w),sin(w)), where _w_ is such that
> > e^w = z. Obviously, the choice of _w_ does not matter.
>
> It should have been z |-> (cos(i.w),sin(i.w)).
>
> Best regards,
>
> Jose Carlos Santos- Hide quoted text -


Couldn't we also do it by starting with the Riemann sphere = {C U oo)
i.e.
the Riemann sphere is C plus the point at infinity. Puncturing the
Riemann
sphere removes the point at infinity and just leaves C. Now if we
puncture
the complex plane, show that it is topologically the complex circle.
i.e.
the punctured point just maps to the interior of the circle?
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