In the previous post, we were trying to show that any homology class of a space X in dimension at most six can be represented by a smooth oriented manifold mapping to X. This statement is a geometric one, but it can be proved via homotopy-theoretic means. In the previous post, we interpreted it in terms of homotopy theory, and we showed that

\displaystyle MSO_*(X)_{(p)} \rightarrow H_*(X; \mathbb{Z}_{(p)})

was a surjection in degrees {\leq 6} (actually, in degrees {\leq 7}) for {p} either {2} or an odd prime {p>3}. In this post, we will handle the case {p = 3}. Namely, we will produce an approximation to {MSO} in the first few homotopy groups (essentially, we’ll work out the first couple of pieces of a Postnikov decomposition). This will give a criterion for when a homology class in low degrees is in the image of {MSO_*}, and we’ll see that it is always satisfied in degrees {\leq 6}. This will complete the proof of:

Theorem 1 For any space {X}, the map {MSO_i(X) \rightarrow H_i(X; \mathbb{Z})} is surjective for { i \leq 6}: that is, any homology class of dimension {\leq 6} is representable by a smooth manifold.

In the case of an odd prime {>3}, we used {H \mathbb{Z}_{(p)} \oplus H \mathbb{Z}_{(p)}[4]} as a 7-approximation to {MSO_{(p)}}. This is not going to work at {3}, because the cohomologies are off. Namely, the cohomology of {MSO} at {3} has two generators in degrees {\leq 8} (namely, the Thom class {t} and {p_1 t} for {p_1} the first Pontryagin class). However, {H \mathbb{Z}_{(3)} \oplus H \mathbb{Z}_{(3)}[4] } has four generators in mod {3} cohomology in these dimensions: {\iota_0, \iota_4, \mathcal{P}^1 \iota_0, \beta \mathcal{P}^1 \iota_0} for {\iota_0, \iota_4} the tautological classes. So the Postnikov decomposition is going to look somewhat different.

Most of this material described in the past few posts comes from a variety of sources: Thom’s original paper (Quelques propriétés globales), Rudyak’s On Thom Spectra, Orientability, and Cobordism, and Stong’s Notes on Cobordism Theory. 


A classical problem in topology was whether, on a (suitably nice) topological space {X}, every homology class can be represented by a manifold. In other words, given a homology class {x \in H_n(X; \mathbb{Z})}, is there an {n}-dimensional oriented manifold {M} together with a continuous map {f: M \rightarrow X} such that

\displaystyle f_*([M]) = x,

for {[M] \in H_n(M; \mathbb{Z})} the fundamental class?

The question can be rephrased in more modern language as follows. There is a spectrum {MSO}, which yields a homology theory (“oriented bordism”) {MSO_*} on topological spaces. There is a morphism of spectra {MSO \rightarrow H \mathbb{Z}} corresponding to the Thom class in {MSO}, which means that for every topological space {X}, there is a map

\displaystyle MSO_*(X) \rightarrow H_*(X; \mathbb{Z}).

Since {MSO_*(X)} can be described (via the Thom-Pontryagin construction) as cobordism classes of manifolds equipped with a map to {X}, we find that that the representability of a homology class {x \in H_n(X; \mathbb{Z})} is equivalent to its being in the image of {MSO_n(X) \rightarrow H_n(X; \mathbb{Z})}.

The case where {\mathbb{Z}} is replaced by {\mathbb{Z}/2} is now straightforward: we have an analogous map of spectra

\displaystyle MO \rightarrow H \mathbb{Z}/2

which corresponds on homology theories to the map {MO_*(X) \rightarrow H_*(X; \mathbb{Z}/2)} sending a manifold {M \rightarrow X} to the image of the {\mathbb{Z}/2}-fundamental class in homology. Here we have:

Theorem 1 (Thom) The map {MO_*(X) \rightarrow H_*(X; \mathbb{Z}/2)} is a surjection for any space {X}: that is, any mod 2 homology class is representable by a manifold.

This now follows because {MO} itself splits as wedge of copies of {H \mathbb{Z}/2}, so the Thom class {MO \rightarrow H \mathbb{Z}/2} actually turns out to have a section in the homotopy category of spectra. It follows that {MO \wedge X \rightarrow H \mathbb{Z}/2 \wedge X} has a section for any space {X}, so taking homotopy groups proves the claim.

The analog for realizing {\mathbb{Z}}-homology classes is false: that is, the map {MSO_*(X) \rightarrow H_*(X; \mathbb{Z})} is generally not surjective for a space {X}. Nonetheless, using the tools we have so far, we will be able to prove:

Theorem 2 Given a space {X} and a homology class {x \in H_n(X; \mathbb{Z})} for {n \leq 6}, {x} is representable by an oriented manifold. In general, any homology class has an odd multiple which is representable by a manifold. (more…)