From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: donuts etc
Date: 19 Dec 1994 07:05:23 GMT
In article ,
Laura Helen wrote:
[an enormous collection of thoughtful questions from geometric and algebraic
topology.]
Good heavens, the topics you're discussing would fill a semester or two of
graduate courses. I'll let others pick and choose their favorite points,
but I will respond to a couple of things here that seem to be of interest
to students being exposed to this material.
>Some questions about homology & cohomology groups. The number of free
>abelian coefficients in the n'th dimensional homology group measures the
>number of n'th dimensional holes in the space. So what does a cohomology
>group measure? Co-holes? Is a co-hole the same as a hole?
This thread (what does homology measure?) came up a couple of months ago;
I won't repeat the discussion except to remark that you have to define
"hole" pretty carefully to make it measured by homology. In particular,
it can't be the same kind of thing counted by homotopy groups since
H_2(torus) = Z but \pi_2(torus) = 0. To make sure it's what you are counting
with homology you pretty much have to define "hole" as a cycle which is
not a boundary, in which case, sure, cohomology counts co-holes, which
are co-cycles which are not coboundaries. Informative, huh?
It's also true that "homology" is insufficiently precise; you really ought
to say what the coefficient groups are. The canonincal choice is to use
integer coefficients, but you'd be better off taking coefficients in an
field because then homology and cohomology are the same (well, dual).
This corroborates your observation,
>So this makes me wonder: are the number of free abelian coefficients --
>the "betti number" -- in the homology and cohomology groups of a given
>dimension always the same? But torsion coefficients such as Z_2 above
>might switch dimensions?
I also enjoy this question:
>To what extent do the homology and cohomology groups determine the
>topology of a space?
Well, not at all in some ways. After all, all the Euclidean spaces
R^n have the same homology groups. At best you can hope to classify your
spaces up to homotopy equivalence, not homeomorphism.
The second thing to concede is that homology, while easier to compute than
homotopy, is less informative. For example, the first homology group of
a space (using integer coefficients) is the abelianization (G/[G,G]) of
the fundamental group. Poincar\'e 's original "poincare conjecture" asked
if a 3-dimensional manifold with trivial homology had to be a sphere;
he himself gave a counterexample by finding a quotient of S^3 by a
perfect group (G=its own commutator subgroup). (Incidentally, if you can
find a 3-d manifold which is simply connected and compact but not the
sphere, I'll buy you lunch. Even dinner.)
The next thing to realize is that even the homotopy groups of a space
can't do everything for you. The reason in a nutshell is that homotopy
groups (and homology groups, for that matter) attempt to compare the
space to Euclidean-like spaces: intervals, spheres, simplices, etc. How
like such spaces are p-adic metric spaces? totally disconnected spaces?
function spaces? So you can't expect homotopy groups to determine the
topology of any larger portion of a space than is more or less Euclidean.
Not surprisingly, many topologists consider only spaces which are, say,
CW-complexes, or manifolds; in that case you have a chance at a complete
topological characterization by algebraic invariants.
One more thing stands in your way: roughly speaking, you might have two
spaces whose homology groups (say) are the same, one dimension at a time,
but you can't find a homeomorphism (say) between them because every time
you look for a map between them which respects the homology in one
dimension, it'll mess up the isomorphism you had somewhere else. A partial
example of what I'm thinking about may be found in the real projective
spaces. You know you have inclusions RP^n --> RP^(n+1); are there any
reasonable maps going backwards? After all, the cohomology is the same
(Z/2Z) in all dimensions except the last. Well, what stands in your way
is that the cohomology groups form a ring, in this case
H_2(RP^n)=(Z/2Z)[X]/(X^{n+1} = 0) [I'm fudging in dimension zero]
Unfortunately there are no non-constant homomorphisms between a
truncated polynomial ring and another with more non-zero groups.
And there's a lot more topology yet not even reflected in the homology
groups as groups or as rings (look up the Steenrod algebra, for example).
OK, well if you accept all these limitations, you'll arrive at what
really is a theorem: if X and Y are CW complexes, and if their homotopy
groups are not only the same but in fact there is
a continuous map f: X--> Y such that f induces isomorphisms
f_i: \pi_i(X) --> \pi_i(Y), then X and Y are homotopy equivalent
(in fact, f is a h.equivalence.)
I'll add just one more example from the cohomology of groups: there are
two spaces, not homotopy equivalent, whose homology rings are identical,
and whose homotopy groups are also identical (in fact, all zero!) except
the first one. The existence of such an example was much studied and the
discovery of this example brought its owner many a lunch. Even dinner.
>If you have a finite sequence of arbitrary finitely generated abelian groups,
>is it possible to construct a space with the n'th homology group isomorphic to
>the n-1'th group in the sequence? ruling out the 0'th homology group?
Sure, this is no problem, in fact you can take the direct product of spaces
which have all homology zero except in one dimension. You can do it with
homotopy groups too (in which case you ought even to ask about non-abelian
groups for \pi_1) and the answer is still "yes"; the groups are certain
CW complexes called Eilenberg-MacLane spaces. These are the building blocks
for making spaces in homotopy theory.
dave
[Who wishes you luck and much endurance; you've bitten off quite a bit
to chew here.]