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)}.

Sketch proof: If we grant the analogous theorem for {BU} (that {H_*(BU)} is a polynomial ring on the generators of {H_*(\mathbb{CP}^\infty)}, without hypotheses on {R}), then we can deduce the result as follows. There is a “forgetful” map {BU \rightarrow BSO} as well as a “complexification” map {BSO \rightarrow BU}. The composite is homotopy equivalent to multiplication by {2} on the H space {BSO}. Consequently, we have a diagram in homology

\displaystyle H_*(BSO; R) \rightarrow H_*(BU; R) \rightarrow H_*(BSO; R)

where the composite is multiplication by two. It follows that {H_*(BSO; R)} is a retract of {H_*(BU; R)}, because we have assumed {\frac{1}{2} \in R}.

The upshot of all this is that {H_*(BSO)} is a quotient of {H_*(BU)}: it is the image under the realification map {H_*(BU) \rightarrow H_*(BSO)}, or equivalently the image in homology of the operator {f: BU \rightarrow BU} sending a (stable) complex vector bundle {E} to {E \oplus \overline{E}}. In cohomology, the map

\displaystyle H^*(BU; R) \rightarrow H^*(BU; R)

corresponding to the map {f: BU \rightarrow BU} annihilates the odd-dimensional Chern classes: if {E} is any complex vector, then {c_i(E \oplus \overline{E}) =0 } for {i } odd, as one checks by writing {E} as a direct sum of line bundles; by contrast, the even-dimensional Chern classes are pretty much arbitrary. For if {E = \ell_1 \oplus \dots \oplus \ell_n}, then

\displaystyle c(E \oplus \overline{E}) = \prod (1 - c_1(\ell_i)^2),

which shows that the even-dimensional Chern classes (as the symmetric functions of {c_1(\ell_i)^2}) can be algebraically independent, while the odd-dimensional Chern classes are zero.

It follows that {H^*(BSO; R)} is a polynomial ring in the images of the even-dimensional Chern classes {c_{2i}}. The corresponding Hopf algebra {H_*(BSO; R)} is the subalgebra of {H_*(BU; R)} generated by the even-dimensional classes in {H_*(\mathbb{CP}^\infty; R)}. \Box

1. Computing the homology

Our next goal is to show that there is no odd torsion in the oriented cobordism ring {\pi_* MSO}. In order to do this, we will use the Adams spectral sequence (mod an odd prime). Recall that this is a spectral sequence

\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}^{\vee}_p}(\mathbb{Z}/p, H_*(MSO; \mathbb{Z}/p)) \rightarrow \pi_{t-s} MSO \otimes \mathbb{Z}_{p}.

Our claim is that we can completely work out the {E_2} term in the same way as we did for {MU}, by working out the coaction of the dual Steenrod algebra {\mathcal{A}_p^{\vee}.} Recall that, for {p} odd, we have the isomorphism

\displaystyle \mathcal{A}_p^{\vee}\simeq \mathbb{Z}/p[\zeta_1, \zeta_2, \dots ] \otimes E(\tau_0, \tau_1, \dots ), \quad \deg \zeta_i = 2(p^i - 1).

In fact, we have an analogous situation as with complex cobordism:

Proposition 3 Let {P \subset \mathcal{A}_p^{\vee} } be the subHopfalgebra {\mathbb{Z}/p[\zeta_1, \zeta_2, \dots ] } (i.e., not including the {\tau_i}). As a comodule over {\mathcal{A}_p^{\vee}}, we have

\displaystyle H_*(MSO; \mathbb{Z}/p) \simeq P \otimes_{\mathbb{Z}/p} \mathbb{Z}/p [x_{i}]|_{i + 1 \neq 2p^f},

where {x_i} is in degree {4i}.

In fact, let’s start by taking {\mathbb{CP}^\infty}; this has a basis {\gamma_i \in H_{2i}(\mathbb{CP}^\infty)}. Then by the previous result, we have that, under the inclusions {\mathbb{CP}^\infty \hookrightarrow BSO},

\displaystyle H_*(BSO; \mathbb{Z}/p) \simeq \mathbb{Z}/p[\gamma_2, \gamma_4, \dots ].

As we saw earlier, the coaction of the dual Steenrod algebra in {\mathbb{CP}^\infty} is via, for any {f \in \mathbb{Z}_{\geq 1}},

\displaystyle \gamma_{p^f} \mapsto \zeta_f\otimes \gamma_1 + 1 \otimes \gamma_{p^f} + \dots.

This reflects the fact that, when using Steenrod {p}th powers, one can get from the cohomology class in degree {2} to cohomology classes in degrees {2 p^f} but nowhere else. Also,

\displaystyle \gamma_{j} \mapsto 1 \otimes \gamma_j + \dots

if {j \neq p^f}. The “dots” indicate tensor products {a \otimes b} where both {a, b} have smaller degree (“decomposable elements”).

Now, we have an equivalence {\phi: \mathbb{CP}^\infty \simeq \Sigma^{2} MSO(2)} coming from the equivalence {BSO(2) \simeq \mathbb{CP}^\infty} and the fact that the Thom space of {BSO(2)} is homotopy equivalent to it. In particular,

\displaystyle H_*( MSO(2); \mathbb{Z}/p) \simeq \widetilde{H}_{* - 2}(\mathbb{CP}^\infty; \mathbb{Z}/p)

where the element in degree zero is the Thom class, corresponding to {\Gamma_1}.

Let {\Gamma_{j} \in H_{4j}(MSO(2); \mathbb{Z}/p)} be the image of {\gamma_{2j+1}} under this equivalence {\phi}; somewhat confusingly, under the Thom isomorphism with {H_*(\mathbb{CP}^\infty)}, it actually corresponds to {\gamma_{2j}}. When computing the coaction of {\mathcal{A}_p^{\vee}}, the Thom isomorphism should be avoided, though, at least here. Then, we find that {H_*(MSO; \mathbb{Z}/p) \simeq \mathbb{Z}/p[\{\Gamma_{2j}\}]}, where {\Gamma_0 = 1 } in the homology ring (as it corresponds to the Thom class). This follows by the Thom isomorphism and the analogous computation for {H_*(BSO; \mathbb{Z}/p)}. The coaction is thus given by a shift of the earlier formulas:

\displaystyle \Gamma_{(p^{f}-1)/2} \mapsto \zeta_f \otimes 1 + 1 \otimes \Gamma_{(p^{f}-1)/2} + \dots \ \ \ \ \ (1)


\displaystyle \Gamma_j \mapsto 1 \otimes \Gamma_j + \dots , \quad j \neq ( p^f - 1)/2. \ \ \ \ \ (2)

Motivated by this, we can define a map

\displaystyle H_*(MSO; \mathbb{Z}/p) \rightarrow P \otimes \mathbb{Z}/p[x_{i}]_{i + 1 \neq 2 p^f}

which is defined via

\displaystyle H_*(MSO; \mathbb{Z}/p) \rightarrow P \otimes H_*(MSO; \mathbb{Z}/p) \rightarrow P_* \otimes \mathbb{Z}/p[x_{i}]|_{i + 1 \neq p^f};

here the first map is the coaction (observe that {H_*(MSO; \mathbb{Z}/p)} is a comodule over {P}, as it has even degrees) and the second map sends {\Gamma_j \rightarrow x_j} for {j \neq (p^f - 1)/2} and {\Gamma_j \mapsto 0} for {j = (p^f - 1)/2}. This is a morphism of comodule algebras, and it is an isomorphism on indecomposables by the above formulas (1) and (2).

In general, it’s worth noting that comodule algebras tend to have a simple structure; we’ll see more of this in the next post.

2. The Adams spectral sequence for MSO

We can now continue more or less repeating the story for {MU} for {MSO} (but only at an odd prime). Namely, we saw in the previous section that {H_*(MSO; \mathbb{Z}/p) \simeq P_* \otimes \mathbb{Z}/p[x_i]|_{i + 1 \neq 2 p^f}}. The change-of-rings formula now gives, as before:

\displaystyle \mathrm{Ext}^{s, t}(\mathbb{Z}/p, H_*(MSO; \mathbb{Z}/p)) \simeq \mathrm{Ext}^{s,t}_{E(\tau_0, \tau_1, \dots)}(\mathbb{Z}/p, \mathbb{Z}/p) \otimes \mathbb{Z}/p[x_i]|_{i + 1 \neq 2p^f} .\ \ \ \ \ (3)

Here, in the exterior algebra {E(\tau_0, \tau_1, \dots, )}, we have

\displaystyle \deg \tau_i = 2p^i - 1.

As before, {\mathrm{Ext}} of an exterior algebra is a bigraded polynomial algebra: {\mathrm{Ext}_{E(\tau_i)}(\mathbb{Z}/p, \mathbb{Z}/p) \simeq \mathbb{Z}/p[y_i]} for {y_i } bigraded via { (1, 2p^i - 1)}. Here {y_i} can be represented in the cobar complex of {E(\tau_i)} via {[\tau_i]}. In particular, we get from (3) and the Künneth formula that

\displaystyle \mathrm{Ext}^{\bullet, \bullet}(\mathbb{Z}/p, H_*(MSO; \mathbb{Z}/p)) \simeq \mathbb{Z}/p[x_i]|_{i + 1 \neq p^f} \otimes \mathbb{Z}/p[y_f]|_{f =0, 1, 2, \dots},

where {\deg x_i = (0, 4i)} and {\deg y_f = (1, 2p^f - 1)}.

Dimensional considerations now force it to degenerate. Observe that the vertical lines are chains of multiplication by {y_0 = [\tau_0]}, which corresponds to multiplication by {p} in homotopy: it follows that the extension problems which arise are all of the same sort, and that the homotopy groups are devoid of {p}-torsion. We conclude in fact:

Theorem 4 {\pi_*MSO \otimes \mathbb{Z}_{p} \simeq \mathbb{Z}_p[t_4, t_8, \dots ]}.

In fact, we can argue as before: the indecomposable elements in {\pi_* MSO \otimes \mathbb{Z}_p} in each degree form an abelian group which is either free of rank one over {\mathbb{Z}_p} or zero. This itself forces the isomorphism, as taking the {t_i} to be indecomposable generators produces a map

\displaystyle \mathbb{Z}_p[t_4, t_8, \dots ] \rightarrow \pi_* MSO \otimes \mathbb{Z}_p,

which is surjective as it hits indecomposables, and is injective because when tensored with {\mathbb{Q}} both rings have the same graded dimension.

In the next post, we’ll begin analyzing the situation at the prime {2}.