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.

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