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.