From: israel@math.ubc.ca (Robert Israel)
Subject: Re: "simultaneous diophantine approximatioms"
Date: 26 Oct 1999 19:20:43 GMT
Newsgroups: sci.math

In article <7urrg2$at6$1@nnrp1.deja.com>,
 David Bernier <bernier@my-deja.com> writes:

> Let p_n in [0,1[^2 be defined by:
>     p_n:= (frac(n*pi),frac(n*pi*pi))  for any integer n.
> Let A={p_n}_{n in Z}.  So A \subsetof [0,1[^2.


> What does A look like?  Might A be dense in [0,1[^2 ?

Yes.  More generally, consider a translation of the torus T^n = R^n/Z^n
defined by F(x)_j = frac(x_j + w_j), for any fixed w in R^n.  Then
every orbit of F is everywhere dense iff {1, w_1, ..., w_n} are linearly
independent over the rationals.  In your case {1, pi, pi^2} are linearly
independent over the rationals because pi is transcendental.

Proof: It suffices to consider the orbit A = { F^j(0): j in Z} of 0 
(all others are translates of it). A is a subgroup of T^n, and its
closure C is a closed subgroup.  If C is not all of T^n, the quotient
T^n/C is a compact abelian group with more than one element, and so it
has a nontrivial character, i.e. a homomorphism T^n/C -> T which
is not identically 1.  This corresponds to a nontrivial character of T^n 
which is 1 on C, i.e. p in Z^n such that exp(2 pi i p.a) = 1 for all
a in A. In particular, p.w is in Z, which says that 1,w_1,...,w_n
are not linearly independent over the rationals.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2