From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: homotopy and homology
Date: 19 Jun 1999 05:36:23 GMT
Newsgroups: sci.math,alt.math.moderated
Keywords: homotopy classes of maps between spheres
On Fri, 18 Jun 1999, "Nick Halloway" wrote:
> can anyone give a simple example of a map f: S^n, n > 1 --> a space X
> such that f isn't homotopic to the constant map, yet f is a boundary?
and in a latter message continued,
>m ... S^2 --> P^2, the projective plane, by the map identifying
>a point in S^2 with its antipodal point. pi_2(P^2) = Z.
[... and indeed this particular map generates pi_2(P^2).]
I don't know what you mean by "is a boundary" then. One interpretation
is to say that the map f : S^n -> X extends to a map F : D^(n+1) -> X
on the disk of which S^n is the boundary. As it turns out, this
interpretation is exactly equivalent to "homotopic to the constant map",
so the answer to your original question is "no". Note by the way that
these maps into X are the ones in the equivalence class of pi_n(X)
which we use as the identity element, that is, these are the maps f
for which the induced homomorphism f* : pi_n(S^n) -> pi_n(X) is zero.
Perhaps the interpretation you intended (given the Subject line) is that
f is a boundary if when viewed as a singular n-chain on X, f lies
in the image of the boundary operator from the group of singular (n+1)-chains.
Rather than think about these huge groups it's perhaps easier to
describe this condition as "yields the zero element of H_n(X)
under the natural (Hurewicz) homomorphism", mapping pi_n(X) -> H_n(X).
Hurewicz proved that this map is injective if n>1 and pi_k(X)=0 for
k=1, 2, ..., n-1. In dimension 1 it's the map which demonstrates that
H_1(X) = pi_1(X)/(pi_1(X))' (the abelianization of the fundamental group),
so you can easily get nontrivial maps from S^1 to X which are trivial
in H^1(X), as soon as you can concoct a space with a nonabelian
fundamental group. Perhaps you knew S^1 was too easy, which is why you
insisted n > 1.
The example you suggested isn't quite what you need. H_2(P^2) is cyclic
of order 2, and the generator of pi_2(P^2) maps to the nonzero
homology class. But, obviously, _twice_ that generator will be a
nonzero element of pi_2(P^2) = Z and map to zero in H_2(P^2) = Z/2Z.
Just for the record, for n > 1, pi_n(X) is naturally isomorphic to pi_n(X~),
where X^~ is its universal covering space. In particular, there are no
"interesting" homotopy classes for any 1- or 2-dimensional manifold other
than S^2 and P^2 (all others have X~ = R^n). So that leads to...
>How about a map: S^3 --> S^2 that isn't homotopic to the constant map,
>yet is a boundary?
I am told that early in the century it was mentioned as "obvious" in
some textbook or other that there could be no such map, that in fact
all maps S^n -> S^m were null-homotopic if n > m. Well, it may be
obvious, but it's wrong! Such a map is an element of pi_n(S^m), and
thus maps to an element in H_n(S^m); if n>m _this_ group is zero. So
yes indeed, you get an example of what you seek as soon as you can show
some pi_n(S^m) <> 0 for some n>m. This cannot happen when m=1,
but indeed pi_3(S^2) is NOT the zero group (it happens to be the
group of integers). THere are plenty of other nonzero homotopy groups
pi_n(S^m) with n>m, although we'll probably never really know them all.
One construction of the map you're looking for is the Hopf map. There
are many good descriptions, although it's probably easiest to write it as
S^3 = SU(2) -> SO(3) -> S^2, the last found simply by sending a rotation
R to the point R(North pole). For other nice descriptions, see
http://www.math-atlas.org/98/hopf_map
Readers who are still stuck wondering whether mathematicians are "nerds" and
whether or not {{}}={} might appreciate knowing that this is a discussion
of Algebraic Topology; see http://www.math-atlas.org/index/55-XX.html
dave