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 2Given a space and a homology class for , is representable by an oriented manifold. In general, any homology class has anoddmultiple 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 3Given 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.