A classical problem in topology was whether, on a (suitably nice) topological space , every homology class can be represented by a manifold. In other words, given a homology class
, is there an
-dimensional oriented manifold
together with a continuous map
such that
for the fundamental class?
The question can be rephrased in more modern language as follows. There is a spectrum , which yields a homology theory (“oriented bordism”)
on topological spaces. There is a morphism of spectra
corresponding to the Thom class in
, which means that for every topological space
, there is a map
Since can be described (via the Thom-Pontryagin construction) as cobordism classes of manifolds equipped with a map to
, we find that that the representability of a homology class
is equivalent to its being in the image of
.
The case where is replaced by
is now straightforward: we have an analogous map of spectra
which corresponds on homology theories to the map sending a manifold
to the image of the
-fundamental class in homology. Here we have:
Theorem 1 (Thom) The map
is a surjection for any space
: that is, any mod 2 homology class is representable by a manifold.
This now follows because itself splits as wedge of copies of
, so the Thom class
actually turns out to have a section in the homotopy category of spectra. It follows that
has a section for any space
, so taking homotopy groups proves the claim.
The analog for realizing -homology classes is false: that is, the map
is generally not surjective for a space
. Nonetheless, using the tools we have so far, we will be able to prove:
Theorem 2 Given a space
and a homology class
for
,
is representable by an oriented manifold. In general, any homology class has an odd multiple which is representable by a manifold.
This theorem (which is due to Thom and Wall, I think), will be an application of the structure theorems we have discussed so far on , and the determination of the first couple of stages of a Postnikov decomposition for
.
In other words, our goal is to show that is a surjection in degrees
, for any space
. We can prove this after localization at each prime
, and this is how we will attack the problem. In this post, we’ll handle the case of
, which is the trickiest.
1. The 2-local problem
We have a stronger result here; the restriction on the dimension is irrelevant.
Theorem 3 Given a space
, the map
is surjective (in all degrees).
In other words, given a homology class, an odd multiple is always representable by a manifold. The proof of this is, after all the work done so far, fairly straightforward: the point is that the map of spectra
admits a section: indeed, we saw previously that splits as a wedge of Eilenberg-MacLane spectra
and
. One of these maps isomorphically onto
by the Thom class (as a look at
shows), while the others map by zero.
Anyway, since the map of spectra admits a section, it follows that always admits a section, and so is surjective.
The result is quite a bit more elementary if we rationalized: that is, we can show with much less machinery that is a surjection. For the splitting of
is something which holds for every spectrum when rationalized; it does not use any specific properties of
. So, once we know that
is a surjection (which is clear), then we know the analog for any space
replacing
.
2. The case of an odd prime
Let be an odd prime,
. Our goal is now to show that
is a surjection. This time, we do need the dimensional restrictions, but we can relax them to instead of
.
We do not have a splitting of spectra as before, but it turns out that we do have an approximate one. The proof will use a map
which is a 7-equivalence (i.e., an equivalence on homotopy groups of degree ). (The fact that such a map might exist is not too surprising given the low-dimensional computations of the cobordism groups.) The Thom class
will fit into a commutative diagram:
where the map is the evident projection. The map
has a section and thus induces a surjection of homology theories. If
is 7-connective, it will follow that for any space
, the map
is an equivalence in degrees
, from which it will follow that the composite
is a surjection for .
In order to prove this, let’s define the map .
- The map
classifies the Thom class.
- The map
classifies an element in
, which we take to be the Pontryagin class
.
Our goal is now to show that is a 7-equivalence. For this, in view of the known homotopy groups of both spectra up to dimension
, it suffices to show that
is a 6-equivalence. For this in turn, it suffices to show that the map induces an equivalence on mod
cohomology in degrees
, since we are localized at
. But this is straightforward. The mod
cohomology of
in degrees
is a free vector space on the Thom class and
. The mod
cohomology of
in these degrees is a free vector space on
: because
, terms of the form
are out of the dimensional range.
It follows now that is an equivalence in these degrees; it now follows that for any space
(or connective spectrum),
is a surjection in the desired range.
The case of the prime 3 is a little trickier. This will be done in the next post.
June 27, 2012 at 5:16 pm
Excuse me, what is
?
June 27, 2012 at 5:17 pm
I meant the integers localized at the prime ideal
.
June 27, 2012 at 5:20 pm
Thank you. I’m not acquainted with some notations in algebra.