From: foltinek@math.utexas.edu (Kevin Foltinek)
Subject: Re: n-manifolds using hypercubes
Date: 05 Dec 2000 11:54:31 -0600
Newsgroups: sci.math
Summary: [missing]

In article <90cove$f6u$1@nnrp1.deja.com> deepthought042@my-deja.com
writes:

> I was just wondering if there's a semi-nice formula for the number of n-
> manifolds that can be constructed by orienting the hypersides of an n-
> hypercube.  To explain what I mean let n=2.  Up to rotation we have the
> torus, the klein bottle, and the projective plane.  This is out of
> 2^4=16 possible combinations of orientations.  3-manifolds gets very
> confusing and that's what I'm currently stuck with.  Since every face
> can be turned in four directions there's a total of 4^6 orientations,
> but up to rotation I'm sure these narrow down.  Anyone have any idea
> how to do this?

More or less an idea, but I don't know if it'll give you the answer in
the form you're looking for.

Go back to n=2 and study it some more.  Draw a square.  Each edge is
currently unlabelled.  Repeat the following algorithm until each edge
is labelled:
1. Count labelled edges; divide by 2, add 1, and let this be k.
2. Choose an unlabelled edge; label this edge k.  (Or, label it with
   the k-th letter of the alphabet.)
3. Choose one of the two possible orientations for the edge, and
   orient this edge.
4. Choose another unlabelled edge; label this edge k.
5. Choose one of the two possible orientations for the edge, and
   orient this edge.
6. Repeat.

This procedure can be used (with appropriate choices) to generate
every possible method of "identifying edges" of the square (or any
polygon with an even number of edges).  (It looks like you're doing
something slightly less than this, namely, identifying opposite edges;
whether this was your intent or not, I don't know.  It might make a
difference in higher dimensions, but I'm not sure.  If you really
wanted just opposite sides, then the above algorithm simplifies in the
obvious way.)

Now you have to figure out which of these edge identifications form
equivalent topological objects.  The edge identifications can be
expressed by a "word"; for example, aba'b' represents a torus.  There
is a set of simplification rules on these words (which are equivalent
to various topological manipulations) which can be used to bring the
word into a "normal form".  (I think Stillwell's "Classical Topology
and Combinatorial Group Theory" describes these rules.)  I don't
remember if these rules will let you say that, for example, aba'b' is
the same as a'b'ab, or ab'a'b, or bab'a', but that's not too hard
(just relabel things so that things are alphabetically sorted, and
that a primed letter cannot occur before an unprimed).

(These rules, by the way, are applicable to polygons with more sides
than the square, and that's where they're really useful.)

It's likely that, by understanding where these rules came from, you
can extend them to the n=3 or higher cases, and then making the
obvious modifications to my algorithm above (or the modification for
the simplified problem of only opposite identifications), reduce the
question to an essentially combinatorial question.  (Of course you
won't be dealing with "words" any more.)

Kevin.