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}.

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