After describing the computation of {\pi_* MU}, I’d now like to handle the remaining half of the machinery that goes into Quillen’s theorem: the structure of the universal formal group law.

Let {R} be a (commutative) ring. Recall that a formal group law (commutative and one-dimensional) is a power series {f(x,y) \in R[[x,y]]} such that

  1. {f(x,y) = f(y,x)}.
  2. {f(x, f(y,z)) = f(f(x,y), z)}.
  3. {f(x,0) = f(0,x) = x}.

It is automatic from this by a successive approximation argument that there exists an inverse power series {i(x) \in R[[x]]} such that {f(x, i(x)) = 0}.

In particular, {f} has the property that for any {R}-algebra {S}, the nilpotent elements of {S} become an abelian group with addition given by {f}.

A key observation is that, given {R}, to specify a formal group law amounts to specifying a countable collection of elements {c_{i,j}} to define the power series {f(x,y) = \sum c_{i,j} x^i y^j}. These {c_{i,j}} are required to satisfy various polynomial constraints to ensure that the formal group identities hold. Consequently:

Theorem 1 There exists a universal ring {L} together with a formal group law {f_{univ}(x,y)} on {L}, such that any FGL {f} on another ring {R} determines a unique map {L \rightarrow R} carrying {f_{univ} \mapsto f}.

Another way of saying this is that the functor sending a ring {R} to the set {\mathrm{FGL}(R)} of formal group laws on {R} is corepresentable.

Proof: In fact, {L} is just {\mathbb{Z}[c_{i,j}]/I}, where {I} is the ideal generated by the polynomial constraints on the {\left\{c_{i,j}\right\}} necessary to make a formal group law out of {f(x,y) = \sum c_{i,j} x^i y^j}. \Box

This ring {L} is called the Lazard ring. {L} has the property that mapping out of {L} is the same as specifying a formal group law. In particular, this means that if {E} is a complex-oriented cohomology theory, then there is a unique map

\displaystyle L \rightarrow \pi_* E

classifying the formal group law on {\pi_* E}.

It’s worth noting, and will be necessary for the sequel, that {L} is canonically a graded ring where the {c_{i,j}} is graded to be degree {2(i+j) - 1}. This grading passes to the quotient, and one sees that the map {L \rightarrow \pi_* E} that one then obtains is actually a map of graded rings, because of the way the formal group law on {\pi_* E} comes about. The whole point of this grading is that if the variables {x,y} have degree {-2}, then {f(x,y)} should also have degree {-2}. This, again, is reasonable given the topological motivations: there the Chern classes live in degree {2}, and we just flip everything.

Our goal is to determine the structure of the Lazard ring.

Theorem 2 (Lazard) The Lazard ring {L} is a polynomial ring {\mathbb{Z}[x_1, x_2, \dots, ]} on infinitely many variables {x_i} of degree {2i}.

In particular, it is the same structure as {\pi_* MU}. It will turn out in fact that the map {L \rightarrow \pi_* MU} is an isomorphism.

This is the goal of the next couple of posts.

1. Strategy of proof

The strategy of proving Lazard’s theorem is to produce a map

\displaystyle \phi: L \rightarrow \mathbb{Z}[b_1, b_2, \dots ]

which will be an imbedding of {L} in a sufficiently large polynomial ring. This will turn out to be the algebraic analog of the Hurewicz homomorphism {\pi_* MU \rightarrow H_*(MU)}.

To produce such a map {\phi}, we need a formal group law on {\mathbb{Z}[b_1, b_2, \dots ]}. There is an easy way to get one: we can use the {\left\{b_i\right\}} as a change of coordinates of any other formal group, for instance the additive one. That is, the strategy is to take the formal power series

\displaystyle \exp(x) = \sum_{i=0}^\infty b_i x^{i+1} \quad (b_0 = 1)

and use that to define a new formal group law

\displaystyle \exp( \exp^{-1}(x) + \exp^{-1}(y)).

Here {\exp^{-1}(x)} is the inverse power series. (The analogy is with the ordinary exponential and logarithm functions, of course, which translate multiplication into addition: here {\exp} is to be thought of as translating a more exotic formal group law to the additive one.)

Note that if {b_i} has degree {2i}, then the map produced is a map of graded rings.

The strategy is to prove that the map {L \rightarrow \mathbb{Z}[b_1, \dots, ]} is injective, and moreover to determine the image on indecomposables. How might we do this? Let {I} be the augmentation ideal of {L} and {J} the augmentation ideal of {\mathbb{Z}[b_1, \dots]}, so that {I/I^2, J/J^2} are the “indecomposable” quotients.

Our goal is to get a handle on {I/I^2}, or at least on the graded pieces {(I/I^2)_{2k}}. Here we are helped by noting that, if {M} is any abelian group, we have an isomorphism

\displaystyle \hom_{\mathrm{graded}}(L, \mathbb{Z} \oplus M[2k]) \simeq \hom_{\mathrm{Ab}}( (I/I^2)_{2k}, M).

In other words, graded maps {L \rightarrow \mathbb{Z} \oplus M[k]} (which means the graded ring { \mathbb{Z} \oplus M} with {M} in degree {k}), are the same as maps from the indecomposables into {M}.

However, a graded map {L \rightarrow \mathbb{Z} \oplus M[k]} is just a formal group law over {\mathbb{Z} \oplus M[k]} of the form

\displaystyle x + y + \sum_{i + j = k+1} m_{i,j} x^i y^j. \ \ \ \ \ (1)

In fact, a map {L \rightarrow \mathbb{Z} \oplus M[k]} is the same thing as a formal group law, and the condition on the grading ensures precisely that it is of the above form.

In other words, we have: Understanding the indecomposables in {L} amounts to understanding formal group laws over {\mathbb{Z} \oplus M[k]} of the above form (1).

We are interested not just in the indecomposables in {L}, but relating them to those of {\mathbb{Z}[b_1, \dots]}. But to give a graded homomorphism {\mathbb{Z}[b_1, \dots ] \rightarrow \mathbb{Z} \oplus M[k]} amounts to giving an element of {M}, which then by “change-of-coordinates” determines a formal group law over {\mathbb{Z} \oplus M[k]}. So we need to compare these formal group laws with all possible ones over {\mathbb{Z} \oplus M[k]} of the form (1).

2. Symmetric 2-cocycles

Let {M} be an abelian group. Consider a “polynomial” {P(x,y) =\sum_{i+j = k+1} m_{i,j} x^i y^j} as in (1). We would like to ask the question: when does (1) define a formal group law on {\mathbb{Z} \oplus M[k]}?

In other words, when does

\displaystyle f(x,y) = x + y + P(x,y)

satisfy the associativity and commutativity of a FGL? We see:

  1. For commutativity, {P(x,y) = P(y,x)} is required.
  2. {P(x,0) = x} by unitality.
  3. For associativity, we have   and similarly
  4. \displaystyle f(f(x,y), z) = x + y + z + P(x,y) + P(x+y, z).

    So the condition is

    \displaystyle P(x, y+z) + P(y,z) = P(x,y) + P(x+y, z).

These are called symmetric 2-cocycles with coefficients in {M}. We can get particular examples of symmetric 2-cocycles by using maps {\mathbb{Z}[b_1, b_2, \dots ] \rightarrow \mathbb{Z} \oplus M[k]}: that is, by choosing an image {m} of {b_k}, and then considering the formal group law obtained by conjugation:

\displaystyle \exp( \exp^{-1}(x) + \exp^{-1}(y)), \quad \exp(x) = x + m x^{k+1}, \exp^{-1}(x) = x - m x^{k+1}.

These formal group laws are explicitly computed to be:

\displaystyle \exp( x - m x^{k+1} + y - m y^{k+1}) = x + y + m(x+y)^{k+1} - m x^{k+1} - m y^{k+1}.

In particular, they correspond to the cocycles

\displaystyle m \left( (x+y)^{k+1} - x^{k+1} - y^{k+1} \right).

Recall that our goal was to compute the image of the map {(I/I^2)_k \rightarrow (J/J^2)_k} on indecomposables from {L \rightarrow \mathbb{Z}[b_1, b_2, \dots ]}. Equivalently, we might compute the homomorphism

\displaystyle \hom((J/J^2)_k, M) \rightarrow \hom( (I/I^2)_k, M).

We have seen the first group is {M}, and the second group is the group of symmetric 2-cocycles homogeneous of degree {k+1}.

The map {M \rightarrow \hom( (I/I^2)_k, M)} sends each {m} to the symmetric 2-cocycle {m \left( (x+y)^{k+1} - x^{k+1} - y^{k+1} \right)}. So, equivalently, our goal is now to understand what symmetric 2-cocycles with coefficients in {M} look like.

Here is the main result:

Proposition 3 Suppose {k+1} is not a power of any prime {p}. Then symmetric 2-cocycles on {M} are uniquely of the form {m ( (x+y)^{k+1} - x^{k+1} - y^{k+1})}. If {k + 1} is a power of {p}, then symmetric 2-cocycles are uniquely of the form {m \frac{1}{p} ( (x+y)^{k+1} - x^{k+1} - y^{k+1} )}.

In other words, we find that

\displaystyle \hom ( (I/I^2)_k, M) \simeq M,

so that {(I/I^2)_k \simeq \mathbb{Z}}. Moreover, we have determined the image of

\displaystyle (I/I^2)_k \rightarrow (J/J^2)_k;

it is an isomorphism for {k+1} not a power of any prime, and has index {p} for {k+1} a power of {p}.

This will be the goal of the next post. With this in mind, we will be able to deduce Lazard’s theorem straightforwardly.

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