The goal of the next few posts is to compute ${\pi_* MU}$:

Theorem 1 (Milnor) The complex cobordism ring ${\pi_* MU}$ is isomorphic to a polynomial ring ${\mathbb{Z}[c_1, c_2, \dots]}$ where each ${c_i}$ is in degree ${2i}$.

We are also going to work out what the image of the Hurewicz map is on indecomposables. The strategy will be to apply the Adams spectral sequence to ${MU}$, at each prime individually.

1. Change-of-rings theorem

In order to apply the ASS, we’re going to need the groups ${\mathrm{Ext}^{s,t}_{\mathcal{A}_p^{\vee}}(\mathbb{Z}/p, H_*(MU; \mathbb{Z}/p))}$ because the spectral sequence runs

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

The ${\mathrm{Ext}}$ groups are computed in the category of (graded) comodules over ${\mathcal{A}_p^{\vee}}$.

In the previous post, we computed

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

as a comodule over ${\mathcal{A}_p^{\vee}}$. In order to compute the ${\mathrm{Ext}}$ groups, we need a general machine. The idea is that ${P \otimes \mathbb{Z}[y_i]_{i + 1 \neq p^k}}$ is almost a coinduced comodule—if it were, the ${\mathrm{Ext}}$ groups would be trivial. It’s not, but the general “change-of-rings” machine will enable us to reduce the calculation of these ${\mathrm{Ext}}$ groups to the calculation of (much simpler) ${\mathrm{Ext}}$ groups over an exterior algebra.

2. Coalgebra

Let ${A}$ be a coalgebra (over a field ${k}$). Let ${B}$ be another coalgebra, and consider a map

$\displaystyle A \rightarrow B$

of ${k}$-coalgebras. Then, there is a functor

$\displaystyle \mathrm{Res}: \mathrm{CoMod}(A) \rightarrow \mathrm{CoMod}(B)$

which sends an ${A}$-comodule ${M}$ to a ${B}$-comodule ${\mathrm{Res}(M)}$; the comodule structure on on ${M}$ comes from

$\displaystyle M \rightarrow A \otimes M \rightarrow B \otimes M.$

In ordinary algebra, given a morphism of algebras ${R \rightarrow S}$, one gets a functor ${\mathrm{Mod}(S) \rightarrow \mathrm{Mod}(R)}$ given by restriction; the left adjoint to restriction is given by ${S \otimes_R: \mathrm{Mod}(R) \rightarrow \mathrm{Mod}(S)}$. In the coalgebra case, everything is dualized.

We will see that that the restriction of scalars has a right adjoint in the dual case. To do this, we will need to define a dual construction to the tensor product in coalgebra.

Definition 2 Let ${M}$ be a right ${A}$-comodule, and ${N}$ a left ${B}$-comodule. Define the cotensor product ${M \square_A N}$ via the equalizer

$\displaystyle M \square_A N = \mathrm{eq}( M \otimes N \rightrightarrows M \otimes A \otimes N ).$

The two maps come from the two comodule structure maps. The unadorned tensor products are over ${k}$.

This is completely dual to the definition of the tensor product of a right ${R}$-module ${F}$ and a left ${R}$-module ${F'}$, as the coequalizer ${F \otimes R \otimes F' \rightrightarrows F \otimes F}$. Then the cotensor product is a ${k}$-vector space, and in general has no additional structure.

However, if ${M}$ is equipped with a dual left ${A}$-comodule structure “commuting” with the right ${A}$-comodule structure, then ${M \square_A N}$ gets a left comodule structure by the left action on ${M}$.

Example 1 If ${A}$ is a coalgebra, then ${A}$ is a left and a right comodule. Given a left ${A}$-comodule ${N}$, the vector space ${A \square_A N}$ acquires the structure of a left ${A}$-comodule and is, in fact, isomorphic to ${N}$.

All this framework is completely dual to the ordinary case of algebra, and in fact, one can formulate the notion of an “algebra object” or a “module” internal to any monoidal category. Then, “coalgebra” just becomes “algebra” when one works in the opposite category to the category of ${k}$-vector spaces.

This duality makes evident the following:

Proposition 3 The restriction functor ${\mathrm{Res}: \mathrm{CoMod}(A) \rightarrow \mathrm{CoMod}(B)}$ has a right adjoint ${N \mapsto B \square_A N}$.

One of the reasons we care about the cotensor product is the following. Suppose we’re working with a Hopf algebra here, not just a coalgebra. Then the one-dimensional vector space ${k}$ becomes a comodule via the unit map ${k \rightarrow A}$. We have the identity

$\displaystyle \hom_A(k, M) \simeq k \square_A M.$

Both represent the “primitive” elements of ${M}$. For the purposes of the ASS, we are interested in the derived functors of ${\hom_A(k, \cdot)}$, and those will be equivalently (by this observation) provided by the derived functors of the cotensor product.

3. Quotienting by a normal subHopf algebra

We’ll be interested in a special case of this operation of “coextending” scalars. Let ${A}$ be a connected, graded Hopf algebra over the field ${k}$. “Connected” means that in degree zero, ${A}$ is just ${k}$.

Let ${B \subset A}$ be a subHopf algebra (graded). We say that ${B}$ is normal if ${AB^+ = B^+ A }$, where ${B^+ \subset B}$ consists of the augmentation ideal. If this case, we define:

Definition 4 We define the Hopf algebra ${ A // B = A/A B^+}$ as a vector space. The algebra and coalgebra structures are induced from that of ${A}$.

In the case we are interested in, we will be taking the dual Steenrod algebra ${\mathcal{A}_p^{\vee}}$ as ${A}$, and ${B}$ will be the subalgebra ${P}$ defined earlier; normality is automatic since the Hopf algebras are commutative. Observe that ${A // B}$ is a quotient Hopf algebra of ${A}$, and we have a (non-exact) sequence of Hopf algebra maps

$\displaystyle B \rightarrow A \twoheadrightarrow A// B.$

The main result we will need is the following, which states that we can recover ${B}$ from ${A//B}$:

Lemma 5 (Change of rings) We have ${B = A \square_{A//B} k}$.

Here we treat ${k}$ as a comodule via the unit map ${k \rightarrow A//B }$: that is, the comodule structure is somewhat uninteresting.

Proof: We need to compute the equalizer of the two maps

$\displaystyle A \rightrightarrows A \otimes A // B$

where the first map is ${a \mapsto a \otimes 1 }$ (this comes from the ${k}$-comodule structure) and the second map is ${a \mapsto \Delta(a)}$ where

$\displaystyle \Delta: A \rightarrow A \otimes A // B$

is the comultiplication, reduced mod ${AB^+}$.

So, in other words, the claim is that if ${a \in A}$ is an element such that ${\Delta(a) - a \otimes 1 \in A \otimes A B^+}$, then ${a \in B}$. Let’s prove this, by induction on the degree. When the degree is one, then we have

$\displaystyle \Delta(a) = a \otimes 1 + 1 \otimes a,$

and if this is equal to ${a \otimes 1}$ in ${A \otimes A// B}$, then we find that ${a \in A B^+}$. Since ${a}$ has degree one, we must have ${a \in B^+}$ itself.

Suppose ${a \in A}$ is a homogeneous element and ${\Delta(a) = \sum a_i \otimes a_i' \in A \otimes A}$ actually lies in ${a \otimes 1 + A \otimes AB^+}$. We can assume the ${a_i}$ are linearly independent in ${A}$, and that one of the ${a_i}$ is ${1}$ and one is ${a}$: in fact, we have

$\displaystyle \Delta(a) = a \otimes 1 + 1 \otimes a + \dots$

where the ${\dots}$ indicate terms where both degrees are positive.

Then by coassociativity,

$\displaystyle \sum a_i \otimes \Delta(a_i') \in \Delta(a) \otimes 1 + A \otimes A \otimes AB^+ .$

This means, that if we consider the sum ${\sum a_i \otimes a_i'}$, then all the ${a_i'}$ except for the one ${a \otimes 1}$ have the property that ${\Delta(a_i') \in a_i' \otimes 1 + A \otimes AB^+}$. Using induction on the dimension, we find that each of the ${a_i' }$ must belong to ${B^+}$ itself, so ${\Delta(a) \in a \otimes 1 + A \otimes B^+}$. Since ${1 \otimes a}$ occurs as a term as well, we find that actually ${a \in B^+}$. $\Box$

4. The change of rings theorem

The main computational tool for handling the ASS for ${MU}$ is going to be a derived version of the cotensor adjunction together with with lemma of the previous section.

Proposition 6 (Change-of-rings theorem) Let ${A}$ be a graded, connected Hopf algebra, ${B \subset A}$ a connected subHopf algebra. Then we have, for an ${A}$-comodule ${M}$ and a vector space ${V}$,

$\displaystyle \mathrm{Ext}_{A, i}(M, B \otimes V) \simeq \mathrm{Ext}_{A//B, i}(M, V),$

where ${V}$ is considered as an ${A//B}$-comodule in the trivial way.

This is now a consequence of the cotensor-restriction adjunction: for ${i = 0}$, we have

$\displaystyle \hom_{A, i}(M, B \otimes V) = \hom_{A, i}(M, A \square_{A//B} (V)) = \hom_{A//B, i}(M, V),$

by the analysis of the previous section. We can now derive this equality to get the desired statement.