Hiya,
I found one, which is pretty good evidence they do exist.
As soon as I’ve crossed the eyes and dotted the tease I’ll write
it up. Moreover, it is possible to tile the surface of the sphere
with a family of such maps, hence possible to construct a just-
touching IFS (Iterated Function System) for a sphere, using as few
as two invertible contraction mappings (or as many more as you like),
and secure thereby the blessings of "Liberte, Egalite, Homologie".
But wait, there’s more: analogous results extend not only to
n-dimensional hyperspheres, but, for each dimension, to any
"surface" topologicaly equivalent to a (hyper)sphere.
The trick is to use a suitable coordinate system, e.g.
something like, but not quite the same as, lattitude and
longitude; and a suitable metric, something like, but not quite
the same as, the Manhattan Metric…
Hey, I may be missing something, but I don’t think they can exist…
Consider the diameter of the set (i.e. the max distance between any two
points on the sphere). Now, if this is a contraction mapping, then after
the sphere is mapped onto the sphere this distance will be smaller than it
was originally. Furthermore, no other points after the mapping could be
further apart than the original diameter. Thus, if the mapping is one to
one and onto, (which is what I assume you had in mind), then the diameter
of the set will shrink at every mapping. Now, consider the image of the
points at either end of the diameter. They must be closer together than they
were originally. Thus, the metric would have to change (which is cheating, of
course), or the mapping couldn’t be a contraction mapping.
Problems with this: if the ‘sphere’ is an open set (given whatever metric
you have defined), then there may not be any points separated by the
diameter of the set, there will be simply points arbitrarily close to being
separated by the diameter. I would suspect you might be able to get by
with some sleight of hand here, but I would suspect that a sphere will be a
closed set under any metric. (Since it’s closed under arc length).
Granted, there are people on the net who may pick holes in this arguement,
but I’d appreciate the education. You may be able to tell from the form of
my arguement that I’m not a mathematician by profession… (but I do enjoy
it!)
Mike Olsen
NASA Ames Research Center
Disclaimer: I speak for myself! (Also, this is outside my professional
expertise, in this area I am amatuer!).
In article <1993Dec15.201347.21…@news.arc.nasa.gov>,
mol…@kyzyl.arc.nasa.gov (Michael E. Olsen) wrote:
> Hey, I may be missing something, but I don’t think they can exist…
> Consider the diameter of the set (i.e. the max distance between any two
> points on the sphere). Now, if this is a contraction mapping, then after
> the sphere is mapped onto the sphere this distance will be smaller than it
> was originally. etc…
I may be trying to bat when the big boys are pitching, but… I think you
can extend this argument a bit beyond the idea that the points must be on
one side of a diameter or equator, namely:
Since you can map one to one and onto betwixt a plane and all but the
"north pole" of a sphere, and since contractive transforms definitely do
exist on the plane, you may be able to define a metric (don’t know if it
breaks any of the rules for metrics but I don’t think so) where the
distance between points on the sphere is defined to be the distance between
the points when mapped onto the plane. In this way, with the exclusion of
1 point, there absolutely *would* be contractive mappings on the sphere. I
don’t see how you could get around the restriction of excluding one point
though. Another thought was to define a metric on the sphere by requiring
that the distance between two points be measured along the circle defined
by the two points in question and an arbitrary third point (let’s call it
the origin) but that turns out to break the triangle inequality rule for
metrics.
doc@miracle
the opinions i express are no doubt utterly contrary to those of my
employer
/*
Hey, I may be missing something, but I don’t think they can exist…
Consider the diameter of the set (i.e. the max distance between any two
points on the sphere). Now, if this is a contraction mapping, then after
the sphere is mapped onto the sphere this distance will be smaller than it
was originally. Furthermore, no other points after the mapping could be
further apart than the original diameter. Thus, if the mapping is one to
one and onto, (which is what I assume you had in mind), then the diameter
of the set will shrink at every mapping. Now, consider the image of the
points at either end of the diameter. They must be closer together than they
were originally. Thus, the metric would have to change (which is cheating, of
course), or the mapping couldn’t be a contraction mapping.
*/
Note: > if the mapping is one to one and onto…
One to one, yes; onto, no – it cannot be both onto and a contraction
mapping, under any metric. I was looking for one to one functions that were
contraction mappings under some, any, metric. They do exist (though may not
be as much use as I had hoped). Here is an example:
Specify points on the surface of, say, a unit sphere, by "lattitude"
and "longitude", where lattitude ranges over [0, PI]; 0 at the north pole,
PI at the south pole; and longitude ranges over [0, 2PI] – 0 at some arbitrary
meridian. Now, consider the map (lat, long) -> (1/2 lat, 1/2 long). This
map is not onto, but it is clearly one to one; it has an inverse; and it IS
a contraction mapping under the Manhattan metric over lat and long: i.e.
D((a,b), (c,d)) = |c-a| + |d-b|
So, there you have an invertible mapping of the points on the surface of a
sphere that is a contraction mapping under some metric – hence such mappings
exist. In fact, under the coordinate-system conventions given above, here
is a "just touching" IFS whose attractor is the surface of a sphere:
(lat, long) -> 1/2(lat, long)
(lat, long) -> 1/2(lat, long) + (PI/2, 0)
(lat, long) -> 1/2(lat, long) + (0, PI)
(lat, long) -> 1/2(lat, long) + (PI/2, PI)
These maps tile the surface of the sphere with images of itself, they are
all contraction mappings under the Manhattan metric over lattitude and
longitude; hence this IFS is guaranteed (by the collage theorem) to converge
to an attractor – the surface of a sphere. Trouble is, the points generated
by this IFS do not rain down evenly on the surface of the sphere – the
distribution of generated points is not uniform over the surface of the
sphere. In fact, the given IFS is just the IFS of a rectangle on a plane,
masquerading as a sphere by a coordinate transformation fiat – hence no
surprise that the distribution is not uniform. But, as far as the question
whether invertible mappings from the surface of a sphere into itself exist,
where the mappings are contraction mappings under some metric, these notions
of uniform distribution are beside the point: I just gave you four mappings
that meet the criteria. So they do exist.
A more interesting question arises if the metric is constrained:
Are there any invertible (i.e. one to one) mappings from points
on the surface of a sphere into points on the surface of the same sphere,
such that the mapping is a contraction mapping under the "great circle"
metric? Where the great circle distance is defined for non antipodal points
on a sphere as the length of the shorter of the two great circle paths
connecting such points, and defined as PI (assuming a radius of one) for
antipodal points. I think the answer is NO – no such maps exit; but to
prove as much I probably need to assimilate the machinery of algebraic
topology (as suggested in mail I got from Rochester)…
So, in brief – invertible does not imply onto.
In article <1993Dec15.201347.21…@news.arc.nasa.gov>,
Michael E. Olsen <mol…@kyzyl.arc.nasa.gov> wrote:
>Problems with this: if the ‘sphere’ is an open set (given whatever metric
>you have defined), then there may not be any points separated by the
>diameter of the set, there will be simply points arbitrarily close to being
>separated by the diameter. I would suspect you might be able to get by
>with some sleight of hand here, but I would suspect that a sphere will be a
>closed set under any metric. (Since it’s closed under arc length).
I think the sphere, when considered as its own metric space, is both open and
closed. Remember "sets are not like doors"; they can be both open and
closed. If I remember correctly, since a sphere is connected, it has this
open-and-closed property.
In my not-too-well-thought-out opinion, no such function exists. We want
something that is 1-1 and invertible, so it must be onto. I don’t think it
has to be continuous, though. My "proof" goes something like this:
(this is probably a rephrasing of the first proof given)
If such a function exists:
consider two points y1 and y2, such that dist(y1,y2)=diam(sphere)
(this is possible because the sphere is bounded; this assumption is
a weak point of the proof.)
since f is invertible, define x1 and x2 such that f(x1)=y1, f(x2)=y2
now certainly dist(x1,x2) <= diam(sphere) by definition of diam.
so dist(x1,x2) <= dist(f(x1),f(x2))
so f is not a contraction.
My assumption that the sphere is bounded is based on a quick scan of the
sphere metrics I know. There might be one out there that produces an
unbounded sphere.
Andrew M. Ross
(who might have just failed Real Analysis, so take this with a grain of salt)
- Hide quoted text — Show quoted text -
>Granted, there are people on the net who may pick holes in this arguement,
>but I’d appreciate the education. You may be able to tell from the form of
>my arguement that I’m not a mathematician by profession… (but I do enjoy
>it!)
>Mike Olsen
>NASA Ames Research Center
>Disclaimer: I speak for myself! (Also, this is outside my professional
>expertise, in this area I am amatuer!).