NetNews Usenet Archive 1992 #23/NN_1992_23.iso/spool/sci/math/13272

Text File | 1992-10-15 | 2.9 KB | 62 lines

Newsgroups: sci.math
Path: sparky!uunet!cs.utexas.edu!wupost!ukma!darwin.sura.net!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!mp.cs.niu.edu!rickert
From: [e-mail] (Neil Rickert)
Subject: Re: What a weird group!!!
Message-ID: <[e-mail]>
Organization: Northern Illinois University
References: <[e-mail]>
Date: Thu, 15 Oct 1992 20:27:38 GMT
Lines: 51

In article <[e-mail]> [e-mail] (John C. Baez) writes:
>
>From an exercise in a book:  The Bohr compactification of the real
>numbers is a compact topological group which contains the real numbers
>as a dense set.  The embedding of the reals into the Bohr group is not a
>homeomorphism.
>
>Well, actually that in itself is not so weird now that I think of it.
>One can embed the real numbers in the torus as a subgroup, namely a line
>of irrational slope, and then it's dense but the image of the reals in
>the torus with the induced topology is not homeomorphic to the reals
>with their usual topology.  (I think this is what was meant by the
>exercise above, by the way.)
>
>But I bet the Bohr group is a lot weirder than the torus.
>
>Here's how you get it. ...

I believe this amounts to the same as the following:

  Start with the real numbers as a discrete group (i.e. every point is
  both open and closed.  Now take the dual group.  That group is the
  compact, since the dual of a discrete group is compact.  The dual
  group is just the group of continuous characters (i.e. representations
  into the circle group).

  The continuous reals are self dual, with the characters w(t) = exp(iwt).
  Every such character is also a character on the discrete reals, which
  gives the embedding.

Yes, it is a little weirder than a torus.

Consider, that the discrete reals is an infinite direct sum of copies of
the discrete rationals.  When we go to the dual group, the Bohr group is
an infinite product of copies of the dual of the rationals.  Now if it
were an infinite product of copies of the dual of the integers it would
be an infinite dimensional torus.  So, in some ways, the weirdness comes
in going from the dual of the integers (the circle group, or reals mod 1)
to the dual of the discrete rationals.

The rationals have, as a subgroup, the integers.  The quotient group is
one in which every element is of finite order (some multiple of a fraction
is an integer).  This means that in the dual group there is a subgroup
which is a totally disconnected compact group (topologically much like
the cantor set), such that the quotient group is the circle group.  Look
at it this way - take the line, and wrap it around (mod 1) to get the
circle group.  Now instead wrap it (mod 2) to get another circle group,
which contains the first circle as a quotient.  Do the same (mod 2*3).
Keep going using factorial n in the nth step.  In the limit, you have
the dual of the discrete rationals.