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.

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 {MSO_*(\ast)}, and the determination of the first couple of stages of a Postnikov decomposition for {MSO}.

In other words, our goal is to show that {MSO_*(X) \rightarrow H_*(X; \mathbb{Z})} is a surjection in degrees {\leq 6}, for any space {X}. We can prove this after localization at each prime {p}, and this is how we will attack the problem. In this post, we’ll handle the case of {p \neq 3}, 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 {X}, the map {MSO_{(2)*}(X) \rightarrow H_*(X; \mathbb{Z}_{(2)})} 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

\displaystyle MSO_{(2)} \rightarrow H \mathbb{Z}_{(2)}

admits a section: indeed, we saw previously that {MSO_{(2)}} splits as a wedge of Eilenberg-MacLane spectra {H \mathbb{Z}/2} and {H\mathbb{Z}_{(2)}}. One of these maps isomorphically onto {H\mathbb{Z}_{(2)}} by the Thom class (as a look at {\pi_0} shows), while the others map by zero.

Anyway, since the map of spectra admits a section, it follows that {\pi_*(MSO_{(2)} \wedge X) \rightarrow \pi_*(H \mathbb{Z}_{(2)} \wedge X)} 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 {MSO_*(X) \otimes \mathbb{Q} \rightarrow H_*(X; \mathbb{Q})} is a surjection. For the splitting of {MSO_{\mathbb{Q}}} is something which holds for every spectrum when rationalized; it does not use any specific properties of {MSO}. So, once we know that {MSO_*(\ast) \otimes \mathbb{Q} \rightarrow H_*(\ast; \mathbb{Q})} is a surjection (which is clear), then we know the analog for any space {X} replacing {\ast}.

2. The case of an odd prime {p > 3}

Let {p} be an odd prime, {p > 3}. Our goal is now to show that

\displaystyle MSO_i(X)_{(p)} \rightarrow H_i(X; \mathbb{Z}_{(p)}), \quad i \leq 7,

is a surjection. This time, we do need the dimensional restrictions, but we can relax them to {7} instead of {6}.

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

\displaystyle \phi: MSO_{(p)} \rightarrow H \mathbb{Z}_{(p)} \oplus H \mathbb{Z}[4]_{(p)}

which is a 7-equivalence (i.e., an equivalence on homotopy groups of degree {\leq 7}). (The fact that such a map might exist is not too surprising given the low-dimensional computations of the cobordism groups.) The Thom class {MSO_{(p)} \rightarrow H \mathbb{Z}_{(p)}} will fit into a commutative diagram:

where the map {\pi} is the evident projection. The map {\pi} has a section and thus induces a surjection of homology theories. If {\phi} is 7-connective, it will follow that for any space {X}, the map {MSO_*(X)_{(p)} \rightarrow (H \mathbb{Z}_{(p)} \oplus H \mathbb{Z}_{(p)}[4])_* X} is an equivalence in degrees {\leq 7}, from which it will follow that the composite

\displaystyle MSO_k(X)_{(p)}\rightarrow H_k(X; \mathbb{Z}_{(p)})

is a surjection for {k \leq 7}.

In order to prove this, let’s define the map {\phi}.

  1. The map {MSO_{(p)} \rightarrow H \mathbb{Z}_{(p)}} classifies the Thom class.
  2. The map {MSO_{(p)} \rightarrow H \mathbb{Z}_{(p)}[4]} classifies an element in {H^4(MSO; \mathbb{Z}_{(p)}) \simeq H^4(BSO; \mathbb{Z}_{(p)})}, which we take to be the Pontryagin class {p_1}.

Our goal is now to show that {\phi} is a 7-equivalence. For this, in view of the known homotopy groups of both spectra up to dimension {7}, it suffices to show that {\phi} is a 6-equivalence. For this in turn, it suffices to show that the map induces an equivalence on mod {p} cohomology in degrees {\leq 7}, since we are localized at {p}. But this is straightforward. The mod {p} cohomology of {MSO_{(p)}} in degrees {\leq 7} is a free vector space on the Thom class and {p_1}. The mod {p} cohomology of {H \mathbb{Z}_{(p)} \oplus H \mathbb{Z}_{(p)}[4]} in these degrees is a free vector space on {\iota_0, \iota_4}: because {p > 3}, terms of the form {\mathcal{P}^1 \iota_0} are out of the dimensional range.

It follows now that {MSO_{(p)} \rightarrow H \mathbb{Z}_{(p)} \oplus H \mathbb{Z}_{(p)}[4]} is an equivalence in these degrees; it now follows that for any space {X} (or connective spectrum),

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

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.