Computing the complex cobordism ring

We finally have all the computational tools in place to understand Milnor’s computation of ${\pi_* MU}$ , and the goal of this post is to complete it. Let’s recall what we have done so far.

We described the Adams spectral sequence, which ran
$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee}}(\mathbb{Z}/p, H_*(MU; \mathbb{Z}/p)) \implies \widehat{\pi_{t-s}(MU)},$

where the hat denotes ${p}$ -adic completion.
We worked out the homology of ${MU}$ . ${H_*(MU; \mathbb{Z}/p)}$ , as a comodule over ${\mathcal{A} _p^{\vee}}$ , is ${P \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}}$ where ${P}$ is a suitable subHopf-algebra of ${\mathcal{A}_p^{\vee}}$ .When ${p = 2}$ , we had that ${P = \mathbb{Z}/p[\zeta_1^2, \zeta_2^2, \dots ] \subset \mathcal{A}_2^{\vee}}$ . When ${p}$ is odd, ${P = \mathbb{Z}/p[\zeta_1, \zeta_2, \dots]}$ .
We worked out a general “change-of-rings isomorphism” for ${\mathrm{Ext}}$ groups.

Now it’s time to put these all together. The ${E_2}$ page of the Adams spectral sequence for ${MU}$ is, as a bigraded algebra,

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee}}(\mathbb{Z}/p, P ) \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}.$

The polynomial ring can just be pulled out, since it’s not relevant to the comodule structure. Consequently, the ${y_i}$ are in bidegree ${(0, 2i)}$ : the first bidegree comes from the ${\mathrm{Ext}}$ , and the second is because we are in a graded category. The ${y_i}$ only contribute in the second way.

Now, by the change-of-rings result from last time, we have the ${E_2}$ page:

$\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee} // P}(\mathbb{Z}/p, \mathbb{Z}/p ) \otimes \mathbb{Z}/p[y_i]_{i + 1 \neq p^k}.\ \ \ \ \ (1)$

This is actually an isomorphism of bigraded algebras. It’s not totally obvious, but the Adams spectral sequence is a spectral sequence of algebras for a ring spectrum like ${MU}$ .

Now observe that ${\mathcal{A}_p^{\vee} // P}$ is an exterior algebra ${E}$ on the generators ${\tau_0, \tau_1, \dots}$ for ${p}$ odd, and on the ${\zeta_i}$ for ${p = 2}$ . Using the formulas for the codiagonal in ${\mathcal{A}_p^{\vee}}$ in this post, we find that the generators for ${E}$ are primitive.

1. The cohomology of an exterior algebra

So what we need now is compute ${\mathrm{Ext}^{s,t}_E(\mathbb{Z}/p, \mathbb{Z}/p)}$ : that is, compute the cohomology of an exterior algebra. That turns out to be fairly straightforward. We will work it out in the case of an exterior algebra on one generator.

Proposition 1 Let ${E}$ be an exterior algebra over ${\mathbb{Z}/p}$ on one primitive generator ${x}$ in an odd degree ${n}$ . Then

$\displaystyle \mathrm{Ext}^{s,t}(\mathbb{Z}/p, \mathbb{Z}/p)$

is a polynomial algebra on a generator in bidegree ${(1, n)}$ .

In fact, this result suffices to prove an analogous result for a tensor product of exterior algebras on primitive generators (by the Künneth theorem).

To prove it, we can write down an explicit resolution of ${\mathbb{Z}/p}$ by ${E}$ -free comodules. It looks like

$\displaystyle 0 \rightarrow \mathbb{Z}/p \rightarrow E \rightarrow E[n] \rightarrow E[2n] \rightarrow E[3n] \rightarrow \dots.$

Here ${E[n]}$ denotes ${E}$ shifted by ${n}$ , and the maps are the obvious ones. Taking maps of ${\mathbb{Z}/p}$ into each piece, we get the additive structure of the ${\mathrm{Ext}}$ groups: in fact, if ${C^\bullet}$ is the complex ${E \rightarrow E[n] \rightarrow \dots}$ , then ${\hom(\mathbb{Z}/p, C^\bullet)}$ is the complex

$\displaystyle \mathbb{Z}/p \rightarrow \mathbb{Z}/p[n] \rightarrow \mathbb{Z}/p[2n] \rightarrow \dots.$

We can also put a multiplicative structure on this resolution to get the multiplicative structure on ${\mathrm{Ext}}$ . In fact, we can think of the resolution as the algebra

$\displaystyle E(x) \otimes \mathbb{Z}/p[y], \quad \deg x = (0, n), \deg y = (1, n).$

We make this algebra into a differential graded algebra via ${dx = y, dy = 0}$ . Then this is a dga resolution of ${\mathbb{Z}/p}$ , and applying ${\hom_E(\mathbb{Z}/p, \cdot)}$ and taking cohomology gives ${\mathbb{Z}/p[y]}$ , as desired.

2. The ${E_2}$ page, completely

We saw that the ${E_2}$ page of the Adams spectral sequence was, by (1),

$\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee} // P}(\mathbb{Z}/p, \mathbb{Z}/p ) \otimes \mathbb{Z}[y_i]_{i + 1 \neq p^k},$

where ${y_i}$ is in bidegree ${(0, 2i)}$ . As we’ve seen, the quotient Hopf algebra

$\displaystyle \mathcal{A}_p^{\vee}//P$

is an exterior algebra on primitive generators of degrees:

When ${p = 2}$ , degrees ${2^i -1}$ for ${i = 1, 2, \dots}$ .
When ${p }$ is odd, degrees ${2p^i - 1}$ for ${i = 0, 1, \dots}$ .

For the rest of the computation, we will assume that ${p}$ is odd; the case ${p = 2}$ is similar. We find that, by the computation of ${\mathrm{Ext}}$ of an exterior algebra,

$\displaystyle E_2 = \mathbb{Z}/p[y_i, z_j]_{i + 1 \neq p^k} \quad \deg y_i = (0, 2i), \deg z_j = (1, 2p^j - 1). \ \ \ \ \ (2)$

In particular, we find that the entire spectral sequence is concentrated in even degrees: thus, the ASS degenerates.

3. Determination of the structure at a prime

We are now ready to use the spectral sequence to compute the ${p}$ -adic completion ${\widehat{\pi_* MU}}$ . In fact, first let’s note that there is an element ${a_0 \in E_2^{1,1}}$ given by the element ${z_0}$ in (2). This element corresponds to an element of Adams filtration one in the associated graded of ${\widehat{\pi_0 MU} = \mathbb{Z}_p}$ : in particular, it must represent ${p}$ or a unit times ${p}$ . This we know just from ${\pi_0 MU}$ .

With this in mind, choose elements of ${\pi_* MU}$ representing each of the generators ${y_i}$ and ${z_j, j > 0}$ ; let’s call them ${\widetilde{y_i}}$ and ${\widetilde{z_j}}$ . Then ${\widetilde{y_i}}$ must live in degree ${2i}$ (where ${i + 1 \neq p^k}$ ) and ${\widetilde{z_j}}$ lives in degree ${2(p^j - 1)}$ .

Proposition 2 ${\widehat{\pi_* MU}}$ is a polynomial ring over ${\mathbb{Z}_p}$ on the ${\widetilde{y_i}}$ and ${\widetilde{z_j}, j > 0}$ .

Proof: In fact, fix ${k}$ , and choose an element ${x \in \widehat{\pi_k MU}}$ . Then the image of ${x}$ in the associated graded of ${\widehat{\pi_k MU}}$ that comes from the Adams filtration is a polynomial in the ${y_i, z_j}$ (where the ${i}$ , ${j}$ admissible are controlled by ${k}$ ). So there exists a polynomial ${Q}$ such that

$\displaystyle x - a_0^r Q(y_i, z_j)$

is in one step lower in the Adams filtration. We can keep doing this (subtract ${p^r Q(\widetilde{y}_i, \widetilde{z}_j)}$ from ${x}$ ) to approximate ${x}$ arbitrarily closely by a polynomial, and eventually make ${x }$ equal to a polynomial with ${\mathbb{Z}_p}$ -coefficients in the ${y_i, z_j}$ .

It follows that the map

$\displaystyle \mathbb{Z}_{p}[\widetilde{y_i}, \widetilde{z}_j] \rightarrow \widehat{\pi_* MU}$

is a surjection, and it is an injection because, again, we can use the associated graded to see algebraic independence of the ${\widetilde{y}_i, \widetilde{z}_j}$ .

$\Box$

It follows now that for each prime ${p}$ , we have a sequence ${\left\{x_i\right\} \subset \pi_* MU}$ such that after ${p}$ -adic completion, the ${\left\{x_i\right\}}$ become polynomial generators for ${\pi_* MU}$ .

Corollary 3 For any prime ${p}$ , ${\pi_* MU}$ has no ${p}$ -torsion.

In particular, if we consider the Hurewicz map

$\displaystyle \pi_* MU \rightarrow H_* MU,$

it must be injective: in fact, it is an isomorphism after tensoring with ${\mathbb{Q}}$ , and we have just seen that the map ${\pi_* MU \rightarrow \pi_* MU \otimes \mathbb{Q}}$ is an injection.

4. Determining the global structure

We’re still not completely there. We know that ${\pi_* MU}$ right now is very close to looking like a polynomial ring on even-degree generators. We also have seen that ${\pi_* MU \rightarrow H_* MU}$ is injective.

Let ${I}$ be the augmentation ideal in ${\pi_* MU}$ , and consider the indecomposable quotient ${I/I^2}$ . The degree ${k}$ part ${(I/I^2)_k}$ maps to the corresponding part ${(J/J^2)_k}$ of the indecomposable quotient ${J/J^2}$ of ${H_* MU}$ .

After ${p}$ -adic completion, ${I/I^2}$ becomes a free module of rank one (we’ve seen that ${\widehat{\pi_* MU}}$ is a polynomial ring), and as it is finitely generated we conclude that ${I/I^2}$ is itself isomorphic to ${\mathbb{Z}}$ . Moreover, the map

$\displaystyle \mathbb{Z} \simeq (I/I^2)_k \rightarrow (J/J^2)_k \simeq \mathbb{Z}$

is an injection, since it is an isomorphism mod torsion.

For each ${k}$ , choose an element ${x_k \in \pi_{2k} MU}$ which generates the indecomposable quotient ${(I/I^2)_k}$ . Then, we have:

Theorem 4 (Milnor) ${\pi_* MU}$ is a polynomial ring on the ${x_k}$ , that is

$\displaystyle \pi_* MU = \mathbb{Z}[x_1, x_2 , \dots ] \quad \deg x_i = 2i.$

Proof: This is now going to be straightforward from the analysis already done. We have a map

$\displaystyle \mathbb{Z}[x_1, x_2 , \dots ] \rightarrow \pi_* MU,$

which is a surjection since all the indecomposables are hit. Moreover, it is an injection because it is an injection after tensoring with ${\mathbb{Q}}$ . After tensoring with ${\mathbb{Q}}$ , it is a surjection of graded algebras of the same dimension in each degree. $\Box$

Climbing Mount Bourbaki