Finding a homeomorphism from circle to square with rotation

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Finding a homeomorphism from circle to square with rotation

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SkanderH

Posted: Tue Feb 22, 2005 1:34 am

Joined: 25 Nov 2004 Posts: 40 Location: Tokyo

Could somebody help me with this math problem.

A circle and a square are homeomorphic (isomorphic?). No suppose you have a set of vectors on the unit circle ( a.i + b.j with a^2+b^2=1), and the rotations of these vectors around the circle, described by 2*2 unitary matrices.
what would be the equivalent of these rotations on the square? e.g. what transformation would happen to this matrices?
Is there some proceedure for finding the equivalent transformations or operations on the image of a set by a homeomorphism.
Please forgive my sloppy language, I am but a lowly computer engineer with little training in math.
Thanks

Willem H. de Boer

Posted: Tue Feb 22, 2005 3:15 am

Guest

Hiya,

Quote:

what would be the equivalent of these rotations on the square?

That depends on the particular homeomorphism you're using.

There's a paper by two computer scientists called "A Low Distortion Map
between Disk and Square", that discusses such a homeomorphism (they
call it a bicontinuous map). It even comes with source-code!

http://www.cs.indiana.edu/~chiuk/pubs/JGT-97.ps.Z

You could use the inverse of their map to go from the circle to the square.
A rotation of n radians in the circle domain would account for a rotation of
n
radians in the square domain. This is because their map can be seen as
an isometry "upto a constant"; which is, in their case, pi/4.

Hope this helps,
Willem

"SkanderH" <skander@hrt.dis.titech.ac-dot-jp.no-spam.invalid> wrote in
message news:421ae08e$1_1@127.0.0.1...

Quote:

Could somebody help me with this math problem.

A circle and a square are homeomorphic (isomorphic?). No suppose you
have a set of vectors on the unit circle ( a.i + b.j with a^2+b^2=1),
and the rotations of these vectors around the circle, described by 2*2
unitary matrices.
what would be the equivalent of these rotations on the square? e.g.
what transformation would happen to this matrices?
Is there some proceedure for finding the equivalent transformations or
operations on the image of a set by a homeomorphism.
Please forgive my sloppy language, I am but a lowly computer engineer
with little training in math.
Thanks

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SkanderH

Posted: Thu Feb 24, 2005 11:31 pm

Joined: 25 Nov 2004 Posts: 40 Location: Tokyo

Thanks,

However, there's something wrong with the gzp file I downloaded, and I can't get a copy from the journal where the paper was originaly published. You know of anything else.
In the meantime of thought some more about the problem. Basically it can be broken down to two subproblems.
1- finding a bijection from the circle to the square.
2- Combining this bijection with the desired rotation. More formally, if there is a bijection between two sets, is there some procedure for finding equivalents in the second set of unary and binary operations under which the first set is closed?

Tobias Fritz

Posted: Fri Feb 25, 2005 9:46 am

Guest

Quote:

A circle and a square are homeomorphic (isomorphic?).

Two mathematical objects of the same kind (here: topological spaces)

are called isomorphic, when there are structure-preserving bijections
between them that are the inverses of each other. Then the two objects
are equal with respect to the structure we are interested in - at
least they cannot be distinguished.
This is the case here when we take the circle and the square as the
objects and their topology as the structure. In the case of
topological spaces, the term "homeomorphic" is more common than
"isomorphic".

Quote:

No suppose you
have a set of vectors on the unit circle ( a.i + b.j with a^2+b^2=1),
and the rotations of these vectors around the circle, described by 2*2
unitary matrices.

It is not really clear what you mean by this. If you just mean that

the circle is rotated by some given angle, then this is not described
by 2*2 unitary matrices.
There are two possibilities to describe the rotation:

1) over the complex numbers:
The complex plane is *1-dimensional* over the complex numbers - you
need only one (complex) number to specify a point. So its linear
mappings are given by 1*1-matrices, which are just numbers themselves.
So any rotation of the complex unit circle is just ordinary
multiplication by a fixed complex number characteristic for the
rotation. As you can figure out easily, the number itself lies on the
unit circle because a rotation preserves absolute values and you can
read off the angle of rotation in the obvious way. For example,
multiplication by i is rotation around 90 degrees.

2) over the real numbers:
Now we regard each point of the plane as a pair of real numbers.
Then rotations are described by *orthogonal* 2*2-matrices - having
real entries. When the angle of rotation is p, the corresponding
matrix is given by

( cos p -sin p )
( )
( sin p cos p )

Quote:

Is there some proceedure for finding the equivalent transformations or
operations on the image of a set by a homeomorphism.

Suppose you have a fixed rotation of the circle. Then there is exactly

one transformation of the square obtained in the following way:
when given a point on the square, we have to determine its image under
the transformation. Therefore, we look at the corresponding point on
the circle and rotate it. We get a new point on the circle, which we
transport back on the square - this is where our original points gets
mapped to.
But it will be pretty improbable that you can find a linear
transformation of the plane that yields this map on the square. So, in
general, this cannot be described by a matrix.

HTH,
Tobias

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