After describing the computation of , 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 be a (commutative) ring. Recall that a formal group law (commutative and onedimensional) is a power series such that
 .
 .
 .
It is automatic from this by a successive approximation argument that there exists an inverse power series such that .
In particular, has the property that for any algebra , the nilpotent elements of become an abelian group with addition given by .
A key observation is that, given , to specify a formal group law amounts to specifying a countable collection of elements to define the power series . These are required to satisfy various polynomial constraints to ensure that the formal group identities hold. Consequently:
Theorem 1 There exists a universal ring together with a formal group law on , such that any FGL on another ring determines a unique map carrying .
Another way of saying this is that the functor sending a ring to the set of formal group laws on is corepresentable.
Proof: In fact, is just , where is the ideal generated by the polynomial constraints on the necessary to make a formal group law out of .
This ring is called the Lazard ring. has the property that mapping out of is the same as specifying a formal group law. In particular, this means that if is a complexoriented cohomology theory, then there is a unique map
classifying the formal group law on .
It’s worth noting, and will be necessary for the sequel, that is canonically a graded ring where the is graded to be degree . This grading passes to the quotient, and one sees that the map that one then obtains is actually a map of graded rings, because of the way the formal group law on comes about. The whole point of this grading is that if the variables have degree , then should also have degree . This, again, is reasonable given the topological motivations: there the Chern classes live in degree , and we just flip everything.
Our goal is to determine the structure of the Lazard ring.
Theorem 2 (Lazard) The Lazard ring is a polynomial ring on infinitely many variables of degree .
In particular, it is the same structure as . It will turn out in fact that the map 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
which will be an imbedding of in a sufficiently large polynomial ring. This will turn out to be the algebraic analog of the Hurewicz homomorphism .
To produce such a map , we need a formal group law on . There is an easy way to get one: we can use the 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
and use that to define a new formal group law
Here is the inverse power series. (The analogy is with the ordinary exponential and logarithm functions, of course, which translate multiplication into addition: here is to be thought of as translating a more exotic formal group law to the additive one.)
Note that if has degree , then the map produced is a map of graded rings.
The strategy is to prove that the map is injective, and moreover to determine the image on indecomposables. How might we do this? Let be the augmentation ideal of and the augmentation ideal of , so that are the “indecomposable” quotients.
Our goal is to get a handle on , or at least on the graded pieces . Here we are helped by noting that, if is any abelian group, we have an isomorphism
In other words, graded maps (which means the graded ring with in degree ), are the same as maps from the indecomposables into .
However, a graded map is just a formal group law over of the form
In fact, a map 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 amounts to understanding formal group laws over of the above form (1).
We are interested not just in the indecomposables in , but relating them to those of . But to give a graded homomorphism amounts to giving an element of , which then by “changeofcoordinates” determines a formal group law over . So we need to compare these formal group laws with all possible ones over of the form (1).
2. Symmetric 2cocycles
Let be an abelian group. Consider a “polynomial” as in (1). We would like to ask the question: when does (1) define a formal group law on ?
In other words, when does
satisfy the associativity and commutativity of a FGL? We see:
 For commutativity, is required.
 by unitality.
 For associativity, we have and similarly

So the condition is
These are called symmetric 2cocycles with coefficients in . We can get particular examples of symmetric 2cocycles by using maps : that is, by choosing an image of , and then considering the formal group law obtained by conjugation:
These formal group laws are explicitly computed to be:
In particular, they correspond to the cocycles
Recall that our goal was to compute the image of the map on indecomposables from . Equivalently, we might compute the homomorphism
We have seen the first group is , and the second group is the group of symmetric 2cocycles homogeneous of degree .
The map sends each to the symmetric 2cocycle . So, equivalently, our goal is now to understand what symmetric 2cocycles with coefficients in look like.
Here is the main result:
Proposition 3 Suppose is not a power of any prime . Then symmetric 2cocycles on are uniquely of the form . If is a power of , then symmetric 2cocycles are uniquely of the form .
In other words, we find that
so that . Moreover, we have determined the image of
it is an isomorphism for not a power of any prime, and has index for a power of .
This will be the goal of the next post. With this in mind, we will be able to deduce Lazard’s theorem straightforwardly.
May 27, 2012 at 1:05 am
Hi Akhil – nice post. I’ve ‘done’ this calculation a few times, and I always wonder why is $P(x,y+z+P(y,z)) = P(x,y+z)$?
May 27, 2012 at 7:49 am
has coefficients which are square zero, so changing the input to by an element of doesn’t change the output.
May 27, 2012 at 6:30 pm
[…] began the proof of Lazard’s theorem last time: we produced a […]
May 28, 2012 at 7:04 pm
[…] and so we will determine what looks like, in analogy with the analysis of the Lazard ring. […]
May 28, 2012 at 10:11 pm
[…] purely algebraic terms: was the Lazard ring, and the Hurewicz map was the map classifying the formal group law for the “change of […]
April 9, 2013 at 11:00 pm
Hi Akhil,
What property of formal group law dictates that the $i+j = k+1$. In other words, how does “the condition on the grading” work exactly.
April 10, 2013 at 3:00 pm
Ans: look at grading
woops
February 7, 2015 at 5:48 pm
You’ve listed that we are looking at the Lazard ring under the grading 2(i+j)1, but I think the grading we want is 2(i+j1), cf. http://goo.gl/O9V263.