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.

1. The Postnikov decomposition of ${MSO_{(3)}}$

Instead, consider the cohomology operation ${\Phi: H \mathbb{Z} \rightarrow H \mathbb{Z}[5]}$ which is given by the composite of the Bockstein and ${\mathcal{P}^1}$. In other words, this cohomology operation on a space acts as

$\displaystyle H^n(X; \mathbb{Z}) \rightarrow H^n(X; \mathbb{Z}/3) \stackrel{\mathcal{P}^1}{\rightarrow} H^{n+4}(X; \mathbb{Z}/3) \stackrel{\beta}{\rightarrow} H^{n+5}(X; \mathbb{Z}).$

(${\Phi}$ is a variant of the “Steenrod cube” ${H \mathbb{Z} \rightarrow H \mathbb{Z}[3]}$ obtained by doing the same with ${\mathrm{Sq}^2}$ mod 2 instead of ${\mathcal{P}^1}$.) Let ${F}$ be the homotopy fiber of ${\Phi}$, so that we have an exact triangle

$\displaystyle H \mathbb{Z}[4] \rightarrow F \rightarrow H \mathbb{Z} \rightarrow H \mathbb{Z}[5],$

from which it follows that ${F}$ is a “twisted” product of the Eilenberg-MacLane spectra ${H \mathbb{Z}, H \mathbb{Z}[4]}$. However, we observe that the cohomology of ${F}$ in degrees ${\leq 7}$ has only two generators, one in degree one (the pullback of ${\iota_0}$) and one in degree four (${\mathcal{P}^1 \iota_0}$). So, we might expect ${F}$ to be a better 3-local approximation to ${MSO}$ in low degrees.

Let’s now work entirely 3-locally: ${\Phi}$ will be regarded as a map ${H \mathbb{Z}_{(3)} \rightarrow H \mathbb{Z}_{(3)}[5]}$. The map

$\displaystyle MSO_{(3)} \rightarrow H \mathbb{Z}_{(3)} \stackrel{\Phi}{\rightarrow} H \mathbb{Z}_{(3)}[5]$

is zero (in fact, we’ve seen that ${H^*(MSO_{(3)}; \mathbb{Z}_{(3)})}$ is zero in dimension five), so we can find a lift ${MSO_{(3)} \rightarrow F_{(3)}}$.

Proposition 2 ${MSO_{(3)} \rightarrow F_{(3)}}$ is a 7-equivalence.

Proof: As before, we just have to prove that it is a 6-equivalence, since the homotopy groups of both spaces vanish in dimension ${7}$. So we might as well prove that the map on cohomology

$\displaystyle H^*(F_{(3)}; \mathbb{Z}/3) \rightarrow H^*(MSO_{(3)}; \mathbb{Z}/3)$

is an equivalence in dimensions ${\leq 7}$.

There is a commutative diagram:

Now, as we saw, ${H^*(F_{(3)}; \mathbb{Z}/3)}$ is generated by the pull-back ${\pi^* \iota_0}$ and ${\mathcal{P}^1 \pi^* \iota_0}$. These classes pull back in ${H^*(MSO_{(3)}; \mathbb{Z}/3)}$ to the Thom class ${t}$ and ${\mathcal{P}^1 t}$. Since we know that the cohomology of ${MSO_{(3)}}$ in these dimensions is generated by ${t}$ and ${p_1 t}$, it suffices to show that

$\displaystyle \mathcal{P}^1 t = p_1 t. \ \ \ \ \ (1)$

This is a general fact that holds on the Thom space of an oriented vector bundle. In other words, we have:

Lemma 3 Let ${E \rightarrow X}$ be an oriented vector bundle, and let ${\theta \in H^*(\mathrm{Th}(E); \mathbb{Z}/3)}$ be the Thom class. Then ${\mathcal{P}^1 \theta = p_1(E) \theta}$.

This is analogous to the statement with ${\mathrm{Sq}^2}$ and ${w_2}$, for instance.

By the properties of both ${p_1}$ and ${\mathcal{P}^1}$, we find that if the result is true for vector bundles ${E_1, E_2}$, then it is true for ${E_1 \oplus E_2}$. Using the “splitting principle,” we can reduce to an oriented 2-dimensional bundle, or what is the same thing, a complex line bundle. So it suffices to prove the above lemma in the “universal” complex line bundle case: that is, when one regards ${\mathbb{CP}^\infty}$ as the Thom space of the ${\mathcal{O}(1)}$ over ${\mathbb{CP}^\infty}$.

In this case, if ${x \in H^2(\mathbb{CP}^\infty; \mathbb{Z}/3)}$ is the hyperplane class, then it is also the Thom class, and ${p_1(\mathcal{O}(1)) = x^2}$. So

$\displaystyle \mathcal{P}^1 x = x^3 = x^2 .x = p_1(\mathcal{O}(1)) x,$

which verifies the claim.

With the lemma proved, the proposition is proved as well. $\Box$

2. Completion of the proof

Now suppose ${X}$ is a space. We have a map

$\displaystyle \pi_i(X \wedge MSO_{(3)}) \rightarrow \pi_i(X \wedge F_{(3)}),$

which is an isomorphism for ${i \leq 6}$ and a surjection for ${i = 7}$, because the cofiber of ${MSO_{(3)} \rightarrow F_{(3)}}$ has homotopy groups concentrated in dimension ${8}$ and above (and thus so does its smash product with ${X}$). This is good, but we do not know that ${\pi_i(X \wedge F_{(3)} )}$ surjects onto ${H_i(X; \mathbb{Z}_{(3)})}$. Rather, there is an exact sequence for each ${i}$

$\displaystyle \pi_i(X \wedge F_{(3)}) \rightarrow \pi_i(X \wedge H\mathbb{Z}_{(3)}) \rightarrow \pi_i(X \wedge H \mathbb{Z}_{(3)}[5]).$

So, the upshot of all this is that an element in ${H_i(X; \mathbb{Z}_{(3)})}$ is hit by an element of ${\pi_i(X \wedge F_{(3)})}$ if and only if the dual homology operation

$\displaystyle \Phi^t: H_i(X; \mathbb{Z}_{(3)}) \rightarrow H_{i-5}(X; \mathbb{Z}_{(3)})$

annihilates it. We find as a result:

Proposition 4 For ${i \leq 7}$, a homology class in ${H_i(X; \mathbb{Z}_{(3)})}$ is in the image of ${MSO_i(X)_{(3)}}$ if and only if it is annihilated by the dual homology operation ${\Phi^t}$.

Our goal is now to show that ${\Phi^t}$ is always zero in degrees ${\leq 6}$. In dimensions ${\leq 4}$, this is evident, since ${\Phi^t}$ decreases degree by ${5}$. Let us now handle dimension ${5}$.

Proposition 5 For any space ${X}$, ${\Phi^t}$ acts as the zero operation on ${H_5(X; \mathbb{Z}_{(3)})}$.

Proof: In fact, ${\Phi^t}$ is the composite of the Bockstein ${\beta}$ and the dual to ${\mathcal{P}^1}$, which will be written as ${\mathcal{P}^{1,t}}$ (this decreases homology degree by four). Namely, it is the composite:

$\displaystyle H_i(X; \mathbb{Z}_{(3)}) \rightarrow H_i(X; \mathbb{Z}/3) \stackrel{\mathcal{P}^{1,t}}{\rightarrow} H_{i-4}(X; \mathbb{Z}/3) \stackrel{\beta}{\rightarrow} H_{i-5}(X; \mathbb{Z}_{(3)}),$

where ${\beta}$ is the homology version of the Bockstein.

When ${i = 5}$, the relevant Bockstein is going from ${H_1(X; \mathbb{Z}/3)}$ to ${H_0(X; \mathbb{Z}_{(3)}}$. But this Bockstein is always zero: that is, consider the exact sequence

$\displaystyle H_1(X; \mathbb{Z}/3) \stackrel{\beta}{\rightarrow} H_0(X; \mathbb{Z}_{(3)}) \stackrel{3}{\rightarrow} H_0(X; \mathbb{Z}_{(3)}) \rightarrow H_0(X; \mathbb{Z}/3) \rightarrow 0.$

Since multiplication by ${3}$ is injective on ${H_0(X; \mathbb{Z}_{(3)})}$, the Bockstein is forced to be zero. $\Box$

Slightly more subtle is the next fact, which will complete the proof of the theorem:

Proposition 6 For any finite CW complex ${X}$, ${\Phi^t}$ acts as the zero operation on ${H_6(X; \mathbb{Z}_{(3)})}$.

Proof: This is a little trickier. The strategy is to pick a class ${x \in H_6(X; \mathbb{Z}_{(3)})}$. If ${\Phi^t x \in H_1(X; \mathbb{Z}_{(3)})}$ is not zero, we can evaluate ${\Phi^t x}$ on a class ${y}$ in ${H^1(X; \mathbb{Z}/3^n) = \hom(H_1(X; \mathbb{Z}_{(3)}) , \mathbb{Z}/3^n)}$ and get a nonzero element of ${\mathbb{Z}/3^n}$ if ${n}$ is appropriate; this is because ${H_1(X; \mathbb{Z}_{(3)})}$ is a finitely generated ${\mathbb{Z}_{(3)}}$-module. Consequently, for every nonzero element of ${H_1(X; \mathbb{Z}_{(3)})}$, we can choose a map

$\displaystyle H_1(X; \mathbb{Z}_{(3)}) \rightarrow \mathbb{Z}/3^n,$

for some ${n}$, which does not annihilate it.

Choose such a class ${y}$. This amounts to pairing ${\Phi y}$ with ${x }$, so we can say that ${(\Phi y, x) \neq 0}$. Now, ${y}$ is pulled back from the tautological class ${\iota_1}$ in ${K(\mathbb{Z}/3^n, 1)}$ via a map ${f: X \rightarrow K(\mathbb{Z}/3^n, 1)}$. In other words, we find that

$\displaystyle (\Phi y, x) = (\Phi \iota_1, f_* x) = (\iota_1, \Phi^t f_* x) \neq 0.$

In other words, we are reduced by naturality to the “universal” case of a ${K(\mathbb{Z}/3^n, 1)}$. But ${H_6( K(\mathbb{Z}/3^n, 1); \mathbb{Z}) = 0}$, so the proposition is trivial in this case. This can be proved as follows: ${H_6(K(\mathbb{Z}/3^n, 1); \mathbb{Z})}$ is the sixth group homology of the cyclic group ${\mathbb{Z}/3^n}$, and one can compute this by an explicit resolution to see that it is zero in positive even dimensions (see for instance this page). $\Box$

This, finally, completes the proof that homology classes in degrees ${\leq 6}$ are representable by smooth manifolds. Note that we also get a necessary and sufficient condition in dimension seven:

Corollary 7 A homology class ${x \in H_7(X; \mathbb{Z})}$ is representable by a smooth manifold if and only if ${\Phi^t x = 0}$.