Re: Dense subgroup of R^n

jane wrote:
What are the dense abelian subgroups of R^n ?

These should be of the form Z^m, but i'm not
sure
how to prove
that.

Thanks.


Ooops, sorry. Instead of "dense" i meant
"discrete". Here is the
question: What are the discrete abelian
subgroups
of R^n ?

Thanks.

Well, it's not difficult to show that such a
subgroup
must be torsion-free, since R^n is. The only thing
left
is to show it must be finitely generated:
finitely-generated
torsion free abelian groups are free abelian.

Suppose you have a discrete subgroup A of R^n. If
no
finite
subset of A generates A, there are two
possibilities:
the norms
(as elements of R^n) are unbounded, or there is a
sequence of
generators that has an accumulation point. The
second
case can't
happen, since you could translate this
accumulation
point to
occur arbitrarily close to each point of A by
adding
a suitable
element of A. This would contradict the assumption
that A is
discrete.

What about the possibility of every generating set
taking
elements of unbounded norm? I think that also
cannot
happen;
I don't have a ready proof off the top of my head,
but imagine
that you could show that any set of generators
could
be replaced
by a set having bounded norms, by noting that any
n
linearly
independent (over R) generators {a_1 ... a_n} of A
produce a
lattice in R^n with fundamental domain (i.e., the
n-dimensional
parallelepiped with vertices given by

{0, a_1 ... a_n, a_1+a_2, a_1+a_3, ...
and so forth, on to a_1 + ... + a_n} )

having finite volume in R^n, and then using such a
lattice to
replace every generator not in {a_1 ... a_n} by an
element
of the fundamental domain, and so with norm less
than
the diameter
of that domain. Of course, if there is no set of n
R-linearly
independent generators, there is some smaller set,
and one would
proceed with whatever rank one had (with the same
end).

I think that would do it. I'll leave the large
number
of details
to you, and will surely appreciate the correction
of
my betters
in this area.

Thanks a lot for you answer. Now i want to find ALL
closed subgroups of R^n : i have the following idea:

Take U - a neighborhood of 0 in our group. Let k be
the maximal number of linearly independent vectors in
U. The since our group if closed and contains, then i
should be R^k (here is the point i'm not sure
about).
Projecting our group onto the complement of U we will
get a discrete subgroup of R^{n-k} which is Z^m for
some
m <= n-k according to you proof above.

This is quite vaguous, do you think it is more or
less close to the truth ?

Any comments ?

Thanks in advance.

Thanks.



Dale
.

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