Finally, it’s time to try to understand the computation of the cobordism ring ${\pi_* MU}$. This will be the first step in understanding Quillen’s theorem, that the formal group law associated to ${MU}$ is the universal one. We will compute ${\pi_* MU}$ using the Adams spectral sequence.

In this post, I’ll set up what we need for the Adams spectral sequence, which is a little bit of algebraic computation. In the next post, I’ll describe the actual calculation of the spectral sequence, which will complete the description of $\pi_* MU$.

1. The homology of ${MU}$

The starting point for all this is, of course, the homology ${H_*(MU)}$, which is a ring since ${MU}$ is a ring spectrum. (In the past, I had written reduced homology ${\widetilde{H}_*(MU)}$ for spectra, but I will omit it now; recall that for a space ${X}$, we have ${\widetilde{H}_*(X) = H_*(\Sigma^\infty X)}$.)

Anyway, let’s actually do something more general: let ${E}$ be a complex-oriented spectrum (which gives rise to a homology theory). We will compute ${E_*(MU) = \pi_* E \wedge MU}$.

Proposition 1 ${E_*(MU) = \pi_* E [b_1, b_2,\dots]}$ where each ${b_i}$ has degree ${2i}$.

The proof of this will be analogous to the computation of ${H_* (MO; \mathbb{Z}/2)}$. In fact, the idea is essentially that, by the Thom isomorphism theorem,

$\displaystyle E_*(MU) = E_*(BU) \simeq \pi_* E [b_1, \dots, ]$

where the last equality is because ${E}$ is complex-oriented, and consequently the ${E}$-homology of ${BU}$ looks like the ordinary homology of it.

Let’s just review this again, anyway. We recall that ${MU(1) = \Sigma^{-1} \mathbb{CP}^\infty}$ is the desuspension of the Thom space ${T(\zeta_1)}$ of the line bundle ${\zeta_1 \rightarrow \mathbb{CP}^\infty }$ (the Thom space is homotopy equivalent to ${\mathbb{CP}^\infty}$ again, though). Then we have

$\displaystyle E_*(MU(1)) = \widetilde{E}_{* - 1}(\mathbb{CP}^\infty) = \pi_* E \left\{ \gamma_0 , \gamma_1, \dots\right\}$

where each ${\gamma_i}$ is in degree ${2i}$. In general, we have that

$\displaystyle E_*(MU(n)) = \mathrm{Sym}^n E_* ( MU(1))$

because of the Thom isomorphism and the fact that ${E_*(BU(n)) = \mathrm{Sym}^n E_* (BU(1))}$. The multiplicative structure comes along in the same way as the multiplicative structure for ${E_* ( BU)}$, because the Thom isomorphism is multiplicative (this is part of the condition on ${E}$). In the limit, we have

$\displaystyle E_*(MU) = \varinjlim E_* MU(n) = \pi_* E [b_1, b_2, \dots]$

where ${\gamma_i \in E_*(MU(1))}$ goes to ${b_i}$. Observe that ${\gamma_0}$ becomes the unit element. Anyway, this was more or less the same as before. Observe that this isomorphism is natural in the complex-oriented theory (i.e., for morphisms preserving the complex orientation).

In particular, we have

$\displaystyle H_* (MU) = \mathbb{Z}[b_1, b_2, \dots ] \quad \deg b_i = 2i.$

Since rationally, (stable) homotopy and homology are the same, we find that ${\pi_* MU \otimes \mathbb{Q}}$ is a polynomial ring on a countable set of variables in even degrees. We will see that this is in fact true for ${\pi_* MU}$ itself, although the Hurewicz homomorphism ${\pi_* MU \rightarrow H_* MU}$ is not an isomorphism.

2. The action of the Steenrod algebra

For the purposes of the ASS, we’re now going to specialize to ${E = H\mathbb{Z}/p}$ for a prime ${p}$. We will need not only the structure of ${H_* (MU ; \mathbb{Z}/p)}$ as a ring, but also as a comodule over the dual mod ${p}$-Steenrod algebra ${\mathcal{A}_p^{\vee}}$.

The structure of the algebra ${\mathcal{A}_p^{\vee}}$ is due to Milnor, and was discussed on this blog in the case ${p = 2}$:

1. When ${p = 2}$, ${\mathcal{A}_2^{\vee} = \mathbb{Z}/2[\zeta_1, \zeta_2, \dots]}$ where ${\deg \zeta_i = 2^i - 1}$
2. When ${p > 2}$, ${\mathcal{A}_p^{\vee} = \mathbb{Z}/p[\zeta_1, \zeta_2, \dots ] \otimes E[\tau_0, \tau_1, \dots] }$ where ${E}$ denotes an exterior algebra. Here ${\deg \zeta_i = 2(p^i - 1)}$ and ${\deg \tau_i = 2p^i - 1}$.

The coalgebra structure is somewhat complicated, corresponding to the fact that the Steenrod algebra is messy. One has

1. For ${p = 2}$,

$\displaystyle \Delta \zeta_n = \sum_{0 \leq i \leq n} \zeta_{n-i}^{2^i} \otimes \zeta_i \quad (\zeta_0 = 1).$

2. For ${p}$ odd,

$\displaystyle \Delta \zeta_n = \sum_{0 \leq i \leq n} \zeta_{n-i}^{p^i} \otimes \zeta_i \quad (\zeta_0 = 1)$

and

$\displaystyle \Delta \tau_n = \tau_n \otimes 1 + \sum_{0 \leq i \leq n} \zeta_{n-i}^{p^i } \otimes \tau_i.$

(I just copied these formulas from Ravenel’s “green book.”) There is a “high-concept” way of looking at ${\mathcal{A}_p^{\vee}}$ at least when ${p = 2}$; ${\mathrm{Spec} \mathcal{A}_2^{\vee}}$ is precisely the automorphism group scheme of the formal additive group. I don’t know what the analog is for ${p}$ odd: presumably it relies on “super” algebraic geometry.

The algebra ${H_*(MU; \mathbb{Z}/p)}$ is a comodule over ${\mathcal{A}_p^{\vee}}$. In fact, it is a comodule over something a fair bit smaller. If we dualize to cohomology, we find that the Bockstein has to act trivially, since the cohomology is entirely in even dimensions: in other words, ${H^*(MU; \mathbb{Z}/p)}$ is a module over a quotient of ${\mathcal{A}_p}$. Dually, ${H_*(MU; \mathbb{Z}/p)}$ is a comoduleover a subcoalgebra (cosubalgebra?) of ${\mathcal{A}_p^{\vee}}$.

In fact, define the subcoalgebra ${P \subset \mathcal{A}_p^{\vee}}$ for ${p = 2}$ as that generated by the ${\zeta_i^2}$, and in case ${p \neq 2}$, by the ${\zeta_i}$ (so ignoring the exterior part of it). In either case, it’s immediate from the formulas given that ${P}$ is an honest subcoalgebra. We are going to see that ${H_*(MU; \mathbb{Z}/p)}$ is a coinduced comodule over ${P}$.

First, we need:

Claim: ${H_*(MU; \mathbb{Z}/p)}$ is actually a comodule over ${P}$Proof: The idea is to use the fact that ${H_*(MU; \mathbb{Z}/p)}$ is entirely in even degrees. In fact, the comodule map

$\displaystyle \phi: H_*(MU; \mathbb{Z}/p) \rightarrow H_*(MU; \mathbb{Z}/p) \otimes \mathcal{A}_p^{\vee}$

is a graded map, so any term that occurs must have even degree. Choose a homogeneous basis ${\left\{m_i\right\}}$ of ${H_*(MU; \mathbb{Z}/p)}$ and, for any homogeneous ${m \in H_*(MU;\mathbb{Z}/p)}$ write

$\displaystyle \phi(m) = \sum m_i \otimes a_i, \quad a_i \in \mathcal{A}_p^{\vee}$

(where the ${a_i}$ are uniquely determined). The claim is that the ${a_i}$ live in ${P}$, though what is only clear right now is that the ${a_i}$ are in even degrees. But now apply the comodule structure map again, and use coassociativity. One the one hand, we get

$\displaystyle \sum \phi(m_i) \otimes a_i,$

and on the other, we get

$\displaystyle \sum m_i \otimes \Delta(a_i).$

Using this, we find that the ${\Delta(a_i)}$ have to be of the form ${\sum b'_j \otimes b_j''}$ where the ${b_j', b_{j''}}$ live in even degrees. But this means that the ${a_i}$ have to live in ${P}$, as one checks from the formulas. $\Box$

3. The main result

Using this, we will obtain the desired structure of ${H_*(MU; \mathbb{Z}/p)}$ as an ${\mathcal{A}_p^{\vee}}$-comodule.

Proposition 2 As an ${\mathcal{A}_p^{\vee}}$-comodule algebra, one has

$\displaystyle H_*(MU; \mathbb{Z}/p) \simeq P \otimes \mathbb{Z}/p[y_i]|_{i+1 \neq p^k}.$

This is the analog of the computation of ${H_*(MO; \mathbb{Z}/2)}$ as a comodule over ${\mathcal{A}_2^{\vee}}$.

Maybe it’s worth trying to motivate this result. As with ${H_*(MO; \mathbb{Z}/2)}$, the point is that the coaction of ${\mathcal{A}_p^{\vee}}$ is essentially determined by the coaction on ${H_*(MU(1); \mathbb{Z}/p)}$, which is easy to work out. There are general Hopf-algebra structure theorems that I don’t really understand which can be used to show that the homology is free in this sense. As more motivation, note that if we had an isomorphism as claimed, then because of the structure of ${P}$ as a polynomial algebra, we would get precisely the right number of variables for ${H_*(MU; \mathbb{Z}/p)}$.

Anyway, note that ${H_*(MU; \mathbb{Z}/p)}$ is not a coinduced comodule anymore, though it’s somewhat close. This is going to make the Adams spectral sequence nontrivial, but it is at least going to degenerate at ${E_2}$.

Proof: We deal with the case of ${p}$ odd.

As before, we need to work out the coaction on ${\mathbb{CP}^\infty}$. This space has a basis ${\left\{\gamma_i\right\}_{i \in \mathbb{Z}_{\geq 0}}}$ for homology where ${\deg \gamma_i = i}$. The coaction ${H_*(\mathbb{CP}^\infty; \mathbb{Z}/p) \rightarrow H_*(\mathbb{CP}^\infty; \mathbb{Z}/p) \otimes \mathcal{A}_p^{\vee}}$ is given by

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

where the ${\dots}$ indicate terms involving ${\gamma_i, 1 < i < p^f}$. This reflects the fact that if ${x \in H^2(\mathbb{CP}^\infty; \mathbb{Z}/p)}$ is the dual to ${\gamma_1}$, then one can get from ${x}$ to ${x^{p^f}}$ using an appropriate product of Steenrod powers in precisely one way—and that way is dual to ${\zeta_f}$.

Consequently, if ${H_*(MU; \mathbb{Z}/p) = \mathbb{Z}/p[b_i]_{i >0}}$ as before, we get

$\displaystyle b_{p^f - 1} \mapsto 1 \otimes \zeta_f + b_{p^f - 1} \otimes 1 + \dots.$

This follows from using the formulas in ${\mathbb{CP}^\infty}$ and shifting.

OK, now let’s define the map

$\displaystyle \Psi: H_*(MU; \mathbb{Z}/p) \rightarrow P \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}.$

The map goes ${H_*(MU; \mathbb{Z}/p) \rightarrow P \otimes H_*(MU; \mathbb{Z}/p) \rightarrow P \otimes \mathbb{Z}[y_i]_{i + 1 \neq p^k}}$, where the last map sends ${b_i \mapsto y_i}$ and annihilates some of the ${b_i}$. The first map is the coaction of ${P}$ on ${H_*(MU; \mathbb{Z}/p)}$.

Now we observe that this map ${\Psi}$, which is both a map of comodules and of algebras, behaves the right way on indecomposables. We have that

$\displaystyle \Psi(b_i) = \begin{cases} 1 \otimes y_i + \dots & \text{if } i + 1\neq p^f \\ \zeta_f \otimes 1 + \dots & \text{if } i + 1 = p^f \end{cases}.$

Here the ${\dots}$ indicate decomposables. Consequently, if we look at indecomposables, we find that we hit all the polynomial generators in ${P \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}}$, both the ones coming from ${P}$ and the ${y_i}$. It follows that the map ${\Psi}$ is an isomorphism, by dimension-counting. $\Box$