Finishing the proof of Lazard’s theorem

This is the third in the series of posts intended to work through the proof of Lazard’s theorem, that the Lazard ring classifying the universal formal group law is actually a polynomial ring on a countable set of generators. In the first post, we reduced the result to an elementary but tricky “symmetric 2-cocycle lemma.” In the previous post, we proved most of the symmetric 2-cocycle lemma, except in characteristic zero. The case of characteristic zero is not harder than the cases we handled (it’s easier), but in this post we’ll complete the proof of that case by exhibiting a very direct construction of logarithms in characteristic zero. Next, I’ll describe an application in Lazard’s original paper, on “approximate” formal group laws.

After this, I’m going to try to move back to topology, and describe the proof of Quillen’s theorem on the formal group law of complex cobordism. The purely algebraic calculations of the past couple of posts will be necessary, though.

1. Formal group laws in characteristic zero

The last step missing in the proof of Lazard’s theorem was the claim that the map

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

classifying the formal group law obtained from the additive one by “change-of-coordinates” by the exponential series ${\exp(x) = \sum b_i x^{i+1}}$ is an isomorphism mod torsion. In other words, we have an isomorphism

$\displaystyle L \otimes \mathbb{Q} \simeq \mathbb{Z}[b_1, b_2, \dots ] \otimes \mathbb{Q}.$

In fact, we didn’t really need this: we could have proved the homological 2-cocycle lemma in all cases, instead of just the finite field case, and it would have been easier. But I’d like to emphasize that the result is really something elementary here. In fact, what it is saying is that to give a formal group law over a ${\mathbb{Q}}$ -algebra is equivalent to giving a choice of series ${\sum b_i x^{i+1}}$ .

Definition 1 An exponential for a formal group law ${f(x,y) }$ is a power series ${\exp(x) = x + b_1 x^2 + \dots}$ such that

$\displaystyle f(x,y) = \exp( \exp^{-1}(x) + \exp^{-1}(y)).$

The inverse power series ${\exp^{-1}(x)}$ is called the logarithm.

That is, a logarithm is an isomorphism of ${f}$ with the additive formal group law.

So another way of phrasing this result is that:

Proposition 2 A formal group law over a ${\mathbb{Q}}$ -algebra has a unique logarithm (i.e., is uniquely isomorphic to the additive one).

Note that this is very false in characteristic ${p}$ . Formal group laws can be wildly non-isomorphic (we’ll see that the height is an invariant in characteristic ${p}$ which can be used to distinguish FGL’s) and, moreover, the additive group law itself has nontrivial automorphisms.

Proof: In fact, given a formal group law ${f(x,y)}$ , we want to find a formal power series ${\log(x) = x + m_1 x^2 + \dots}$ such that

$\displaystyle \log f(x,y) = \log x + \log y.$

The exponential series will be the inverse to this. The strategy is to “do calculus” with this equation, just as one defines the usual logarithm as an integral.

In fact, if we differentiate with respect to ${x}$ , we find the differential equation:

$\displaystyle \log' ( f(x,y)) f_1(x,y) = \log' x.$

(Here the subscript ${1}$ denotes partial differentiation with respect to the first variable.)

Setting ${x = 0}$ , we get

$\displaystyle \log' ( f(0, y)) f_1(0, y) = 1,$

which becomes (since ${f(0, y) = y}$ ),

$\displaystyle \log' y f_1(0, y) = 1.$

So we should define

$\displaystyle \log y = \int_0^y \frac{1}{f_1(0,t) }dt,$

where the integral is done formally. Note that in order to define the integral, we need to introduce denominators and thus the fact that we are over a ${\mathbb{Q}}$ -algebra.

If we do this, we still have to check that the basic equation ${\log f(x,y) = \log x + \log y}$ is actually satisfied. In fact, since both sides have no constant term and are symmetric in both variables, we can check this by differentiation with respect to ${y}$ : we find the derivatives are

$\displaystyle \log' f(x,y) f_1(x,y) = \frac{1}{f_1(0, f(x,y))} f_1(x,y) \quad \text{and } \frac{1}{f_1(0, x)}.$

So to check our basic equation, we have to prove

$\displaystyle f_1(0, f(x,y)) = f_1(x,y) f_1(0, x).$

This, however, is an infinitesimal version of associativity: differentiate ${f(z, f(x,y)) = f(f(z, x), y)}$ with respect to ${z}$ at ${z =0}$ .

The last thing we have to see is that the logarithm is unique. But if the logarithm weren’t unique, then we’d have a nontrivial automorphism of the formal additive group in characteristic ${0}$ : that is, a power series ${g(x)}$ satisfying

$\displaystyle g(x+y) = g(x) +g(y).$

In particular, this gives ${g(2x) = 2g(x)}$ , which forces ${g}$ to be the trivial automorphism. $\Box$

We have, in particular, completed the proof of Lazard’s theorem.

2. Some other corollaries

There are a few other isolated results which we can pick up by virtue of the analysis of Lazard’s theorem. One of them concerns “partial” formal group laws, which Lazard chooses to call “buds.”

Definition 3 Let ${q \geq 1}$ . A ${q}$ -bud over a ring ${R}$ is a polynomial ${f(x,y) \in R[x,y]}$ of degree ${\leq q}$ which satisfies the equations of a formal group law modulo ${(x,y)^{q+1}}$ .

Just as with the Lazard ring, there is a “universal” ${q}$ -bud for each ${q}$ . We thus get a series of rings ${L_q}$ for each ${q}$ , together with maps

$\displaystyle \dots \rightarrow L_q \rightarrow L_{q+1} \rightarrow \dots$

which are obtained by restricting terms.

Theorem 4 (Restricted Lazard theorem) Each ${L_q}$ is a polynomial ring on variables ${\mathbb{Z}[x_1, x_2, \dots, x_{q-1}]}$ and the maps ${L_{q} \rightarrow L_{q+1}}$ are the inclusions

$\displaystyle \mathbb{Z}[x_1, \dots, x_{q-1}] \rightarrow \mathbb{Z}[x_1, \dots, x_{q}].$

In particular, every ${q}$ -bud can be extended to a ${q+1}$ -bud.

We can prove this in the same way as Lazard’s theorem. Namely, we find that, as before, ${L_q}$ is a graded ring, and there is an augmentation ideal ${I_q \subset L_q}$ as before. To understand ${L_q}$ in relation to ${L_{q+1}}$ , we may as well understand the map on indecomposables

$\displaystyle (I_q/I_q^2)_{2k} \rightarrow (I_{q+1}/I_{q+1}^2)_{2k}$

for each ${k}$ , as well as what these indecomposables are. To do so, we might as well understand for each ${k}$ ,

$\displaystyle \hom( (I_q/I_q^2)_{2k}, M) ,$

which is identified with the set of ${q}$ -buds in ${\mathbb{Z} \oplus M}$ of appropriate grading, so of the form

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

But when ${k \leq q-1}$ , this is just the set of all symmetric 2-cocycles on ${M}$ . When ${k \geq q}$ , the group ${(I_q/I_q^2)_{2k}}$ is zero. In particular, we find (by the symmetric 2-cocycle lemma) that graded pieces of the indecomposables ${(I_q/I_q^2)_{2k}}$ are ${\mathbb{Z}}$ for ${k \leq q-1}$ and ${0}$ elsewhere, and the map ${(I_q/I_q^2)_{2k} \rightarrow (I_{q+1}/I_{q+1}^2)_{2k}}$ is either an isomorphism or zero.

From this, we want to claim that ${L_q}$ is actually a polynomial ring, and the map ${L_{q} \rightarrow L_{q+1}}$ is as claimed. But we can prove this by choosing elements ${x_1, \dots, x_{q-1}}$ which generate the indecomposables in the appropriate degrees of ${L_q}$ . Then we get a map

$\displaystyle \phi: \mathbb{Z}[x_1, \dots, x_{q-1}]\rightarrow L_q,$

which is an isomorphism on indecomposables. There is a map

$\displaystyle L_q \rightarrow L,$

and, by the above analysis in terms of cocycles, we find that ${L_q \rightarrow L}$ induces either an isomorphism or zero on indecomposables in the relevant degrees. It follows that ${\phi}$ must be an isomorphism (i.e., first that ${\phi}$ is surjective, and second that the ${\left\{x_i\right\}}$ are algebraically independent in ${L_q}$ ). This proves that ${L_q}$ is a polynomial ring, and even that the map ${L_q \rightarrow L_{q+1}}$ is as claimed.