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:

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

and

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

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 ${2}$.