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.