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. 

(more…)

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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…)

This is the fourth in a series of posts on the oriented cobordism ring. In the previous posts, we showed that the oriented cobordism ring had no odd torsion, and determined the structure of the mod 2 cohomology. The purpose of this post is to tie this results together by describing MSO_{(2)} and then applying that to compute a few low-dimensional cobordism groups.

I stated earlier that we would prove the following result using the Adams spectral sequence:

Theorem 12 (Wall) All the torsion in the oriented cobordism ring {\Omega_{SO} \simeq \pi_* MSO} has order two: that is, each cobordism group {\Omega_{SO}^k} is a direct sum of copies of {\mathbb{Z}} and {\mathbb{Z}/2}.

In fact, once we know that there is no odd torsion (which we proved using the ASS), we don’t need the Adams spectral sequence to prove this. We can prove directly the following result, which will imply Wall’s theorem: the spectrum {MSO_{(2)}} is equivalent to a wedge of shifts of {H \mathbb{Z}_{(2)}} and {H \mathbb{Z}/2}. To see this, we use the main result of the previous post: the mod 2 cohomology of {MSO_{(2)}} (or {MSO}, same thing) is a direct sum of copies of {\mathcal{A}_2} and {\mathcal{A}_2/\mathcal{A}_2 \mathrm{Sq}^1}.

Now, to show that {MSO_{(2)}} is as claimed, it suffices to show that there is a map

\displaystyle MSO_{(2)} \rightarrow \bigoplus_{i \in F} H \mathbb{Z}_{(2)}[d_i] \oplus \bigoplus_{j \in G} H \mathbb{Z}/2[d_j],

inducing an isomorphism in mod 2 homology: in fact, one such is necessarily an equivalence in the since everything is localized at 2 (at least for connective spectra). (more…)

This is the third in a series of posts on oriented cobordism. In the first post, we analyzed the spectrum MSO at odd primes; in this post, we will analyze the prime 2. After this, we’ll be able to deduce various classical geometric facts about manifolds.

The next goal is to  determine the structure of the homology {H_*(MSO; \mathbb{Z}/2)} as a comodule over {\mathcal{A}_2^{\vee}}. Alternatively, we can determine the structure of the cohomology {H^*(MSO; \mathbb{Z}/2)} over the Steenrod algebra {\mathcal{A}_2}: this is a coalgebra and a module.

Theorem 8 (Wall) As a graded {\mathcal{A}_2}-module, {H^*(MSO; \mathbb{Z}/2)} is a direct sum of shifts of copies of {\mathcal{A}_2} and {\mathcal{A}_2/\mathcal{A}_2\mathrm{Sq}^1}.

This corresponds, in fact, to a splitting at the prime 2 of MSO into a wedge of Eilenberg-MacLane spectra.

In fact, this will follow from the comodule structure theorem of the previous post once we can show that if {t \in H^0(MSO; \mathbb{Z}/2)} is the Thom class, then the action of {\mathcal{A}_2} on {t} has kernel {J = \mathcal{A}_2 \mathrm{Sq}^1}: that is, the only way a cohomology operation can annihilate {t} if it is a product of something with {\mathrm{Sq}^1}. Alternatively, we have to show that the complementary Serre-Cartan monomials in {\mathcal{A}_2} applied to {t},

\displaystyle \mathrm{Sq}^{i_1} \mathrm{Sq}^{i_2} \dots \mathrm{Sq}^{i_n} t, \quad i_k \geq 2i_{k-1}, \quad i_n \neq 1,

are linearly independent in {H^*(MSO; \mathbb{Z}/2)}. (more…)

The next goal of this series of posts (started here) is to analyze the oriented cobordism spectrum {MSO} at the prime 2; the main result is that there is a splitting of {MSO_{(2)}} into a direct sum of copies of {H\mathbb{Z}_{(2)}} (the torsion-free part) and {H \mathbb{Z}/2} (the torsion-part). In particular, it will follow that there is only torsion of order two in the cobordism ring — since we showed last time that there was no odd torsion. We will see this using the Adams spectral sequence at the prime {2}, once we’ve figured out what {H_*(MSO; \mathbb{Z}/2)} looks like as a comodule over the dual Steenrod algebra. This, however, is apparently somewhat tricky to do directly.

In order to get there, we’ll need a bit of algebraic machinery (which we state in a dual context). Recall that a graded vector space {V} is called connected if {V_0} is one-dimensional and {V_i = 0} for {i < 0}. The next result provides a sufficient criterion for a module over a graded, connected Hopf algebra to be free.

Theorem 5 (Milnor-Moore) Let {A} be a connected, graded Hopf algebra over a field {k}, and let {M} be a graded, connected {{A}}-module which is simultaneously a coalgebra (in such a way that {M \rightarrow M \otimes_k M} is an {A}-homomorphism). Let {1 \in M_0} be a generator, and suppose the map of {A}-modules

\displaystyle A \rightarrow M, \quad a \mapsto a . 1

is a monomorphism. Then {M} is a free graded {A}-module.

This is a pretty surprising result, as a relatively minor hypothesis (coalgebra, and the action on {1} is free) leads to freeness of the whole thing. The idea of the proof is going to be to produce generators of {M} by lifting a vector space basis of {\overline{M} = M \otimes_A k}. The fact that these generators are forced to be linearly independent is an unexpected consequence of the coalgebra structure; the graded connectedness will be used to make certain inductive arguments. (more…)

Let {MSO } be the Thom spectrum for oriented cobordism, so {MSO} can be obtained as a homotopy colimit

\displaystyle MSO = \varinjlim MSO(n) = \varinjlim \Sigma^{-n}\mathrm{Th}(BSO(n))

where the {n}th space is the Thom space of the (universal) vector bundle over {BSO(n)}. By the Thom-Pontryagin theorem, the homotopy groups {\pi_* MSO} are isomorphic (as a graded ring) to the cobordism ring {\Omega_{SO}} of oriented manifolds. The goal of the next few posts is to discuss some of the classical results on the oriented cobordism ring. This post will handle the easiest case; since it is somewhat analogous to the situation for complex cobordism, it is a bit brief.

In the past, we described Milnor’s computation of {\pi_* MU} (the complex cobordism ring), and showed that {\pi_* MU \simeq \mathbb{Z}[x_2, x_4, \dots ]}. The situation for {MSO} is somewhat more complicated, as there will be torsion; as we will see, however, it is only 2-torsion that occurs, and not too wild of a sort. Let’s start by noting what {\Omega_{SO} \otimes \mathbb{Q}} is.

Theorem 1 (Thom) We have

\displaystyle \pi_* MSO \otimes \mathbb{Q} \simeq \mathbb{Q}[x_4, x_8, x_{12}, \dots ],

where the {x_i} can be taken to be the even-dimensional {\mathbb{CP}^{2i}}.

This result is a corollary of Serre’s work, which shows that rational stable homotopy is equivalent to rational homology. In other words, we have an isomorphism

\displaystyle \pi_* MSO \otimes \mathbb{Q} \simeq H_*(MSO; \mathbb{Q}) \simeq H_*(BSO; \mathbb{Q})

(where the last isomorphism is the Thom isomorphism). Now {H_*(BSO; \mathbb{Q})} is a polynomial ring on variables as above; this follows from the following computation:

Proposition 2 Let {R} be a ring containing {1/2}. Then {H_*(BSO; R)} is a polynomial ring on generators {\alpha_{i} \in H_{4i}(BSO; \mathbb{R})}. These generators are obtained as the image of generating elements in {H_{4i}(BSO(2); R) = H_{4i}(\mathbb{CP}^\infty; R)}. (more…)