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In <3284D945.101E@math.unifi.it> Biblioteca matematica
 writes: 
>
>Let C be Cantor set.
>can you find a subset D of R that is omeomorph with C, and such that 
>Lebesgue measure of D isn' t zero?
Yes.
The Cantor set is based on taking out the middle third at each stage.
If we remove the middle fourth in the first stage, the middle sixteenth
of each segment in the second, and the middle 4^n in the n-th stage,
we will remove a set with measure less than one.

Suppose $D$ is a diagonal matrix with decreasing non-negative entries
and $A$ is a square matrix with 2-norm bounded by 1 (in other words, 
$I - A^t A$ is positive semi-definite).  Is it possible
to find a bound on the matrix norm of 
$(e^D (I+A) -e^(-D) (I-A)) (e^D (I+A) +e^(-D) (I-A))^(-1)$,
a bound that is uniform in $D$ and $A$ (meaning, that does not depend on
$D$ or $A$) ?
That would be very useful.  Thanks in advance,
Patrice Sawyer                          | sawyer@ramsey.cs.laurentian.ca
Department of Math and Computer Science | phone: (705) 675-1151 x 2167
Laurentian University                   | fax:   (705) 673-6591
Sudbury, Canada P3E 2C6                 |

I and a colleague are looking at bipartite graphs.  We need
to know some information about certain kinds of automorphisms.
Let G be a bipartite graph, and let V(G) = A \cup B be the
partition of the vertices.  Let \pi be an automorphism of G.
We'll call \pi a switching automorphism if \pi takes the 
vertices in A to vertices in B and vice versa.  Clearly any 
switching automorphism is of even order.
Our question:  Can a bipartite graph have switching automorphisms,
but have no switching automorphism of order 2?  If so, can you
give us an example of such a graph?
Thanks in advance for the help.
Robin Sanders
Assistant Prof. of Math
Illinois Wesleyan University
Bloomington, IL 61702-2900
Phone: (309) 556-3146 (office) or (217) 586-3498 (home)
E-mail: sanders@sun.iwu.edu

Consider the following Linear Complementarity Problem:
q + Mz > 0, z'(q + Mz) = 0, z > 0 (where > stands for weak inequality
and the prime denotes transposition), with
    |  0     A' |
M = |           |.
    | -A-B   0  |
I know that for some classes of matrices A, B and vector q, this problem
has an odd number of solutions. I would appreciate any references to
theorems on the number of solutions of this type of LCP.
Guido Erreygers
University of Antwerp
dse.erreygers.g@alpha.ufsia.ac.be

Alex Smith (smithaj@uwec.edu) wrote:
: If f is an isometry between two Banach
: spaces, the must f necessarily be linear?
Not necessarily: take a translation of a Banach space
to itself.
But the following does hold, and is due to Mazur and Ulam:
If F and G are Banach spaces and f is an isometric bijection
such that f(0)=0 , then f is R-linear.
(see Semadeni, Banach spaces of continuous functions, page 130)
Greetings,  
Henno Brandsma

Biblioteca matematica (udini@math.unifi.it) wrote:
: Let C be Cantor set.
: can you find a subset D of R that is omeomorph with C, and such that 
: Lebesgue measure of D isn' t zero?
Yes you can. I'll show two ways to do this in the unit interval:
1. The "non-constructive" way: Let A and B be such that 
A is a dense G_\delta of measure 0, and B is a first category
in [0,1], and A \cup B = [0,1] and A \cap B = \emptyset.
(do this e.g. by taking unions of small intervals around rationals).
Then \mu(B) = 1, and B is the union of countably many nowhere dense
closed subsets. At least one of them has positive measure, take one
and take its Cantor-Bendixson kernel (removing a countable subset).
The result is 0-dimensional (nowhere dense implies this), and has
no isolated points and is compact metric. So it is homeomorphic to C,
and has positive measure.
2. Using a "middle-third" type of construction: to get a Cantor set
of measure 1/2, take a convergent sum of \epsilon_n that converges
to 1/2, all strictly positive of course. 
Remove at step 1 an \epsilon_1 length interval symmetrically 
around 0, at step two remove in each of the two remaining parts
an interval of length \epsilon_2 /2 symmetrically around their
respective middle points, and in the third step remove 4 times
\epsilon_3 /4 etcetera. In the limit we get a Cantor set of measure 1/2.
Hope this helps,
Greetings,    Henno Brandsma

The following must be known.  Can anyone give me a reference?
An open convex subset of R^n is C^\infty diffeomorphic to R^n.
Please answer me by email.
Michael Barr
barr@math.mcgill.ca

Let f : X -> Y be a surjective morphism between complete algebraic
surfaces, where X is nonsingular and Y is normal. I allow f to have
degree greater than 1. Let p be a singular point of Y (necessarily
isolated) and let C_1,...,C_n be the irreducible components of the
preimage of p in X. Let M be the intersection matrix of the curves
C_1,..,C_n, i.e. the i,j entry of M is the intersection number of
C_i and C_j.
True or false: M is nonsingular.
Please provide a proof or a reference. Thanks.
Allan Adler
adler@pulsar.cs.wku.edu

This Week's Finds in Mathematical Physics - Week 94
John Baez
Today I want to talk a bit about asymptotic freedom.
First of all, remember that in quantum field theory, studying
very small things is the same as studying things at very high
energies.  The reason is that in quantum mechanics you need to
collide two particles at a large relative momentum p to make sure 
the distance x between them gets small, thanks to the uncertainty 
principle.  But in special relativity the energy E and momentum p 
of a particle of mass m are related by
                    E^2 = p^2 + m^2,
in God's units, where the speed of light is 1.  So small x also 
corresponds to large E.   
"Asymptotic freedom" refers to the fact that the fact that some forces
become very weak at high energy scales, or equivalently, at very short
distances.   The most interesting example of this is the so-called
"strong force", which holds the quarks together in a hadron, like a
proton or neutron.  True to its name, it is very strong at distances
comparable to the radius of proton, or at energies comparable to the
mass of the proton (where if we don't use God's units, we have to use 
E = mc^2 to convert units of mass to units of energy).  But if we smash 
protons at each other at much higher energies, the constituent quarks act 
almost as free particles, indicating that the strong force gets weak when 
the quarks get really close to each other.   
Now in "week76" I talked about another phenomenon, called
"confinement".  This simply means that at lower energies, or larger
distance scales, the strong force becomes so strong that it is
*impossible* to pull a quark out of a hadron.  Asymptotic freedom and
confinement are two aspects of the same thing: the dependence of 
the strength of the strong force on the energy scale.  Asymptotic 
freedom is better understood, though, because the weaker a force is, 
the better we can apply the methods of perturbation theory --- a widely 
used approach where we try to calculate everything as a Taylor series in 
the "coupling constant" measuring the strength of the force in question.
This is often successful when the coupling constant is small, but not
when it's big.
The interesting thing is that in quantum field theory the coupling 
constants "run".  This is particle physics slang for the fact that 
they depend on the energy scale at which we measure them.  "Asymptotic 
freedom" happens when the coupling constant runs down to zero as we 
move up to higher and higher energy scales.  If you want to impress 
someone about your knowledge of this, just mutter something about 
the "beta function" being negative --- this is a fancy way of saying 
the coupling constant decreases as you go to higher energies.  You'll 
sound like a real expert.
Now, Frank Wilczek is one of the original discoverers of asymptotic
freedom.  He *is* a real expert.  He recently won a prize for this work,
and he gave a nice talk which he made into a paper:
1) Frank Wilczek, Asymptotic freedom, preprint available as
hep-th/9609099.  
Among other things, he gives a nice summary of the work of Nielsen
and Hughes, which gave the first really easy to understand explanation
of asymptotic confinement.  For the original work, try:
2) N. K. Nielsen, Am. J. Phys. 49, 1171 (1981).
3) R. J. Hughes, Nucl. Phys. B186, 376 (1981). 
Why would a force get weak at short distance scales?  Actually it's
easier to imagine why it would get *strong* --- and sometimes that is
what happens.  Of course there are lots of forces that decrease with
distance like 1/r^2, but I'm talking about something more drastic: I'm
talking about "screening".
For example, say you have an electron in some water.  It'll make an
electric field, but this will push all the other negatively charged
particles little bit *away* from your electron and pull all the
positively charged ones a little bit *towards* your electron:
                                   -
                                     +
                         your electron: -        +-
                                            +
                                              -
In other words, it will "polarize" all the neighboring water molecules.
But this will create a counteracting sort of electric field, since it
means that if you draw any sphere around your electron, there will be a
bit more *positively* charged other stuff in that sphere than negatively
charged other stuff.  The bigger the sphere is, the more this effect
occurs --- though there is a limit to how much it occurs.  We say that
the further you go from your electron, the more its electric charge is
"screened", or hidden, behind the effect of the polarization.
This effect is very common in materials that don't conduct electricity,
like water or plastics or glass.  They're called "dielectrics", and the
dielectric constant, epsilon, measures the strength of this screening
effect.  Unlike in math, this epsilon is typically bigger than 1.  If
you apply an electric field to a dielectric material, the electric field
inside the material is only 1/epsilon as big as you'd expect if this
polarization wasn't happening.  
What's cool is that according to quantum field theory, screening occurs
even in the vacuum, thanks to "vacuum polarization".  One can visualize
it rather vaguely as due to a constant buzz of virtual particle-antiparticle
pairs getting created and then annihilating --- called "vacuum bubbles"
in the charming language of Feynman diagrams, because you can draw them
like this:
                  /\
               e+/  \e-
                /    \
                \    /
                 \  /
                  \/
Here I've drawn a positron-electron pair getting created and then
annihilating as time passes --- unfortunately, this bubble is square,
thanks to the wonders of ASCII art.  
There is a lot I should say about virtual particles, and how despite
the fact that they aren't "real" they can produce very real effects
like vacuum polarization.  A strong enough electric field will even
"spark the vacuum" and make the virtual particles *become* real!  But 
discussing this would be too big of a digression.  Suffice it to say 
that you have to learn quantum field theory to see how something that 
starts out as a kind of mathematical book-keeping device --- a line in a 
Feynman diagram --- winds up acting a bit like a real honest particle.  
It's a case of a metaphor gone berserk, but in an exceedingly useful way.
Anyway, so much for screening.   Asymptotic freedom requires something 
opposite, called "anti-screening"!   That's why it's harder to understand.
Nielsen and Hughes realized that anti-screening is easier to understand
using magnetism than electricity.   In analogy to dielectrics, there
are some materials that screen magnetic fields, and these are called
"diamagnetic" --- for example, one of the strongest diamagnets is bismuth.
But in addition, there are materials that "anti-screen" magnetic fields ---
the magnetic field inside them is stronger than the externally applied
magnetic field --- and these are called "paramagnetic".  For example,
aluminum is paramagnetic.  People keep track of paramagnetism using
a constant called the magnetic permeability, mu.  Just to confuse you,
this works the opposite way from the dielectric constant.  If you
apply a magnetic field to some material, the magnetic field inside it is 
mu times as big as you'd expect if there were no magnetic effects going
on.   
The nice thing is that there are lots of examples of paramagnetism
and we can sort of understand it if we think about it.   It turns
out that paramagnetism in ordinary matter is due to the spin of the 
electrons in it.  The electrons are like little magnets --- they
have a little "magnetic moment" pointing along the axis of their spin.
Actually, purely by convention it points in the direction opposite
their spin, since for some stupid reason Benjamin Franklin decided
to decree that electrons were *negative*.  But don't worry about this ---
it doesn't really matter.  The point is that when you put electrons in a
magnetic field, their spins like to line up in such a way that
their magnetic field points the same way as the externally applied
magnetic field, just like a compass needle does in the Earth's magnetic
field.  So they *add* to the magnetic field.  Ergo, paramagnetism.
Now, spin is a form of angular momentum intrinsic to the electron,
but there is another kind of angular momentum, namely orbital angular
momentum, caused by how the electron (or whatever particle) is moving 
around in space.  It turns out that orbital angular momentum also
has magnetic effects, but only causes diamagnetism.  The idea 
that when you apply a magnetic field to some material, it can also make
the electrons in it tend to move in orbits perpendicular to the
magnetic field, and the resulting current creates a magnetic field.
But this magnetic field must *oppose* the external magnetic field.
Ergo, diamagnetism.  
Why does orbital angular momentum work one way, while spin works
the other way?  I'll say a bit more about that later. Now let
me get back to confinement.
I've talked about screening and antiscreening for both electric
and magnetic fields now.  But say the "substance" we're studying
is the *vacuum*.  Unlike most substances, the vacuum doesn't look
different when we look at it from a moving frame of reference.  We
say it's "Lorentz-invariant".  But if we look at an electric field 
in a moving frame of reference, we see a bit of magnetic field
added on, and vice versa.   We say that the electric and magnetic
fields transform into each other... they are two aspects of single
thing, the electromagnetic field.  So the amount of *electric* screening
or antiscreening in the vacuum has to equal the amount of the 
*magnetic* screening or antiscreening.  In other words, thanks to
the silly way we defined epsilon differently from mu, we must have
                       epsilon = 1/mu
in the vacuum.  
Now the cool thing is that the Yang-Mills equations, which describe
the strong force, are very similar to Maxwell's equations.  In 
particular, the strong force, also known as the "color" force, 
consists of two aspects, the "chromoelectric" field and "chromomagnetic"
field.  Moreover, the same argument above applies here: the vacuum
must give the same antiscreening for the chromoelectric field as
it does for the chromomagnetic field, so epsilon = 1/mu here too.
So to understand asymptotic freedom it is sufficient to see why the
vacuum acts like a paramagnet for the strong force!   This depends
on a big difference between the strong force and electromagnetism.
Just as the electromagnetic field is carried by photons, which are
spin-1 particles, the strong force is carried by "gluons", which
are also spin-1 particles.  But while the photon is electrically 
uncharged, the gluon is charged as far as the strong force goes: we 
say it has "color".  
The vacuum is bustling with virtual gluons.  When we apply a chromomagnetic
field to the vacuum, we get two competing effects: paramagnetism thanks
to the *spin* of the gluons, and diamagnetism due to their *orbital
angular momentum*.  But --- the spin effect is stronger.  The vacuum
acts like a paramagnet for the strong force.  So we get asymptotic
freedom!
That's the basic idea.  Of course, there are some loose ends.
To see why the spin effect is stronger, you have to calculate a bit.  
At least I don't know how to see it without calculating --- but Wilczek 
sketches the calculation, and it doesn't look too bad.   It's also true 
in most metals that the spin effect wins, so they are paramagnetic. 
You might also wonder why spin and orbital angular momentum work
oppositely as far as magnetism goes.  Unfortunately I don't have any
really simple slick answer.   One thing is that it seems any answer
must involve quantum mechanics.  In volume II of his magnificent 
series:
4) Richard Feynman, Robert Leighton, and Matthew Sands, "The
Feynman Lectures on Physics", Addison-Wesley, Reading, Mass., 1964. 
Feynman notes: "It is a consequence of classical mechanics that
if you have any kind of system - a gas with electrons, protons, and
whatever - kept in a box so that the whole thing can't turn, there
will be no magnetic effect.  [....]  The theorem then says that if
you turn on a magnetic field and wait for the system to get into
thermal equilibrium, there will be no paramagnetism or diamagnetism -
there will be no induced magnetic moment.  Proof: According to statistical
mechanics, the probability that a system will have any given state
of motion is proportional to exp(-U/kT), where U is the energy of
that motion.  Now what is the energy of motion.  For a particle moving
in a constant magnetic field, the energy is the ordinary potential energy
plus mv^2/2, with nothing additional for the magnetic field.  (You
know that the forces from electromagnetic fields are q(E + v x B),
and that the rate of work F.v is just qE.v, which is not affected by
the magnetic field.)  So the energy of a system, whether it is in
a magnetic field or not, is always given by the kinetic energy plus
the potential energy.  Since the probability of any motion depends only
on the energy - that is, on the velocity and position - it is the same
whether or not there is a magnetic field.  For *thermal* equilibrium,
therefore, the magnetic field has no effect."   
So to understand magnetism we really need to work quantum-mechanically.
Laurence Yaffe has brought to my attention a nice path-integral argument
as to why orbital angular momentum can only yield diamagnetism; this
can be found in his charming book:
5) Barry Simon, "Functional Integration and Quantum Physics", Academic
Press, 1979.
This argument is very simple if you know about path integrals, but
I think there should be some more lowbrow way to see it, too.  I think
it's good to make all this stuff as simple as possible, because
the phenomena of asympotic freedom and confinement are very important
and shouldn't only be accessible to experts.  
I'd like to thank Douglas Singleton, Matt McIrvin, Mike Kelsey, and
Laurence Yaffe for some remarks on sci.physics.research that helped
me understand this stuff.
-----------------------------------------------------------------------
Previous issues of "This Week's Finds" and other expository articles on
mathematics and physics, as well as some of my research papers, can be
obtained by anonymous ftp from math.ucr.edu; they are in the
subdirectory pub/baez.  The README file lists the contents of all the
papers.  On the World-Wide Web, you can get these files by going to
http://math.ucr.edu/home/baez/
A complete index of the old issues of "This Week's Finds" is available
at
http://math.ucr.edu/home/baez/twf.html
but if you are cursed with a slow connection and just want a jumping-off
place to the olds issues, go to
http://math.ucr.edu/home/baez/twfshort.html
For the latest issue, go to
http://math.ucr.edu/home/baez/this.week.html

I am working on numerical modelling of elastic-wave propagation in
disordered 2D fiber networks (a thin paper sheet is a good example
of such a network structure). I have also made some simulations
with simpler lattice models (bond diluted square and triangular
lattices).
I have now run into the problem of (Anderson) localization of sound
waves. My questions are:
        1) Are elastic waves ALWAYS localized in a 2D random medium?
        2) If not always, in what cases are they localized?
I'd be grateful if you can help me.
* Markku Kellomaki, mmk@jyu.fi, http://www.phys.jyu.fi/~kellomaki   
* Department of Physics, University of Jyvaskyla 
* P.O. Box 35, FIN-40351 Jyvaskyla, FINLAND      
* Phone: +358-14-602378, Fax: +358-14-602351

In article <3284D945.101E@math.unifi.it> Biblioteca matematica  writes:
>From: Biblioteca matematica 
>Subject: measure
>Date: Sat, 09 Nov 1996 11:19:33 -0800
>Let C be Cantor set.
>can you find a subset D of R that is omeomorph with C, and such that 
>Lebesgue measure of D isn' t zero?
Of course you can, and thatīs well known. The Cantor set is constructed by 
removing "middle thirds" infinitely often, and the resulting set has measure 
zero because (1-1/3)*(1-1/3)*... = 0. All you have to change is this infinite 
product; replace it by one that converges (i.e. to some positive number).

Alex Smith wrote:
> 
> If f is an isometry between two Banach
> spaces, the must f necessarily be linear?
   No. f(x)=x+c
   But usually when one talks of isometries of Banach spaces, it is
meant linear isometries.
-- 
Jan Rosenzweig
e-mail: rosen@math.mcgill.ca
office:                                        home:
Department of Mathematics and Statistics       539 Rue Prince Arthur O. 
Burnside Hall, room 1132, mbox F-10            Montreal
805 Rue Sherbrooke O.                          Quebec H2X 1T6
Montreal, Quebec H3A 2K6 
    "It is unworthy of excellent men to lose hours, like slaves, in
     the labors of calculation"    ..... Leibnitz

In article <562s4a$827@senator-bedfellow.MIT.EDU>,
Lones A Smith  wrote:
>Let U(x,y) be a bounded function on [0,1]X[0,1], with U(x,0)=0 for all x,
>and U(x,y) monotone nondecreasing in all y>=0. Suppose that the sum of 
>all discontinuities of U in x, the sum taken over all (x,y) in [0,1]X[0,1],
>is at most 1. I presume this means that there are countably many such 
>discontinuities (x_1,y_1),(x_2,y_2),... with the left or right jump in x 
>at (x_i,y_i) equal to a_i, with |a_1|+|a_2|+... <= 1. 
>
>CLAIM: For all epsilon>0 there exists delta>0 s.t. for almost all x, 
>U(x,y) 0, U(0,0) = 0 and U(0,y) = 1
for y > 0.  
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4

In article <3284D945.101E@math.unifi.it>,
Biblioteca matematica   wrote:
>Let C be Cantor set.
>can you find a subset D of R that is omeomorph with C, and such that 
>Lebesgue measure of D isn' t zero?
Yes.  This is a very standard example.  Do the standard construction
of C, except that instead of taking out the middle thirds at each
stage take out the middle (2^n)ths at stage n.
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4

	In article <3284D945.101E@math.unifi.it>,
		Biblioteca matematica   wrote:
> Let C be Cantor set.
> can you find a subset D of R that is omeomorph with C, and such that 
> Lebesgue measure of D isn' t zero?
	Instead of deleting middle thirds at each step, delete middle open
intervals of rapidly decreasing lengths at each step, so that the sum of
the lengths of the deleted intervals is less than 1.
	Mike Hardy
Michael Hardy
hardy@stat.umn.edu