Lazard’s theorem II

We are in the middle of proving an important result of Lazard:

Theorem 1 The Lazard ring ${L}$ over which the universal formal group law is defined is a polynomial ring in variables ${x_1, x_2, \dots, }$ of degree ${2i}$ .

The fact that the Lazard ring is polynomial implies a number of results which are not a priori obvious: for instance, it shows that given a surjection of rings ${ A \twoheadrightarrow B}$ , then any formal group law on ${B}$ can be lifted to one over ${A}$ .

We began the proof of Lazard’s theorem last time: we produced a map

$\displaystyle L \rightarrow \mathbb{Z}[b_1, b_2, \dots ], \quad \deg b_i = 2i,$

classifying the formal group law obtained from the additive one ${x+y}$ by the “change of coordinates” ${ \exp(x) = \sum b_i x^{i+1}}$ . We claimed that the map on indecomposables was injective, and that, in fact the image in the indecomposables of ${\mathbb{Z}[b_1, b_2, \dots ]}$ could be determined completely. I won’t get into the details of this (it was all in the previous post), because the purpose of this post is to prove a result to which we reduced last time.

Let ${A}$ be an abelian group. A symmetric 2-cocycle is a “polynomial” ${P(x,y) \in A[x, y] = A \otimes_{\mathbb{Z}} \mathbb{Z}[x, y]}$ with the properties:

$\displaystyle P(x, y) = P(y,x)$

and

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

These symmetric 2-cocycles come up when one tries to classify formal group laws over the ring ${\mathbb{Z} \oplus A}$ , as we saw last time: in fact, we can think of them as “deformations” of the additive formal group law.

The main lemma which we stated last time was the following:

Theorem 2 (Symmetric 2-cocycle lemma) A homogeneous symmetric 2-cocycle of degree ${n}$ is a multiple of ${\frac{1}{d} ( ( x+y)^n - x^n - y^n )}$ where ${d =1}$ if ${n}$ is not a power of a prime, and ${d = p}$ if ${n = p^k}$ .

For a direct combinatorial proof of this theorem, see Lurie’s notes. I want to describe a longer homological proof, which is apparently due to Mike Hopkins and which appears in the COCTALOS notes. The strategy is to interpret these symmetric 2-cocycles as actual cocycles in a cobar complex computing an ${\mathrm{Ext}}$ group. Then, the strategy is to compute this ${\mathrm{Ext}}$ group independently.

This argument is somewhat longer than the combinatorial one, but it has the benefit (for me) of engaging with some homological algebra (which I need to learn more about), as well as potentially generalizing in other directions.

1. Some reductions

To prove the symmetric cocycle lemma, we can reduce to the case of a finitely generated abelian group, since any abelian group a filtered colimit of such. In particular, it suffices to prove it in the two cases:

${\mathbb{Z}}$ .
${\mathbb{Z}/p^m}$ for a prime ${p}$ and an integer ${m \geq 1}$ .

This is because any finitely generated abelian group is a direct sum of such cyclic groups. To prove it for ${\mathbb{Z}}$ , we can prove it for ${\mathbb{Q}}$ and use the fact that the polynomial ${\frac{1}{d} ( (x+y)^n - x^n - y^n)}$ has coefficients which are together relatively prime (this is a straightforward lemma with binomial coefficients).

Finally, we can reduce the case of ${\mathbb{Z}/p^m}$ to the case of ${\mathbb{Z}/p}$ using a filtration argument. Hence, in particular:

Reduction: It suffices to prove the symmetric 2-cocycle lemma when ${A}$ is a prime field. In particular, we will be able to use additional multiplicative structure to interpret these in terms of ${\mathrm{Ext}}$ groups.

In this post, we will only handle the case of a finite prime field ${\mathbb{F}_p}$ . The case of ${\mathbb{Q}}$ will follow from the fact that the morphism

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

is an isomorphism mod torsion: this follows from the (to be proved) fact that a formal group law over a ${\mathbb{Q}}$ -algebra is uniquely isomorphic to the additive one.

2. The relevant coalgebra and cobar complex

Let ${k}$ be a field. We will consider the polynomial ring ${\Lambda = k[x]}$ ; this can be made into a Hopf algebra if ${x}$ is primitive,

$\displaystyle \Delta x = x \otimes 1 + 1\otimes x.$

In other words, we have described the universal enveloping algebra of the one-dimensional abelian Lie algebra, or equivalently the ring of functions on the additive group ${\mathbb{G}_a}$ . The coalgebra map

$\displaystyle k[x] \rightarrow k[y,z]$

sends a polynomial ${f(x)}$ to ${f(y+z)}$ (making, again, the interpretation as ${\Gamma(\mathbb{G}_a)}$ transparent).

This coalgebra will be the protagonist for this story.

Idea: Symmetric 2-cocycles will be literally 2-cocycles for a complex computing ${\mathrm{Ext}^2_\Lambda(k, k)}$ satisfying the additional symmetry condition.

This is not too surprising, because the 2-cocycle condition looks like a cocycle condition. So the problem is to write down the relevant complex. This complex is the cobar complex. Let ${M}$ be an ${\Lambda}$ -comodule. We can write down a concrete and functorial ${\Lambda}$ -free resolution of ${M}$ ,

$\displaystyle 0 \rightarrow M \rightarrow \Lambda \otimes M \rightarrow \Lambda \otimes \Lambda \otimes M \rightarrow \dots .$

In other words, we construct a complex ${C^\bullet(M)}$ of ${\Lambda}$ -free comodules. We have

$\displaystyle C^n(M) = \Lambda^{\otimes (n+1)} \otimes M$

and use the combination of the comodule and coalgebra structure to make ${C^\bullet(M)}$ into a cosimplicial ${\Lambda}$ -comodule. (This is the cobar construction at work: I promised in the comments to cover it more fully sometime, and will try to keep that promise.) Using the Moore complex, we can extract a resolution.

Example: Let’s write down, for instance, what the differential on 2-cochains looks like. A 2-cochain is a sum of terms of the form ${[f_1|f_2|f_3] m}$ where ${f_1, f_2, f_3 \in \Lambda}$ and ${m \in M}$ ; we use the bar notation to denote tensor products.

The coboundary sends this to the 3-cochain

$\displaystyle [f_1'|f_1''|f_2|f_3]m - [f_1 |f_2'|f_2''|f_3]m + [f_1|f_2|f_3'|f_3''] m - [f_1|f_2|f_3|f']m' \ \ \ \ \ (1)$

if, by abuse of notation, we write the coalgebra and comodule structures via

$\displaystyle \Delta(f_i) = f_i' \otimes f_i'', \quad m \mapsto f' \otimes m'.$

In a similar (but easier) way, we can write down the 1-coboundary. This sends

$\displaystyle [f_1|f_2] m \mapsto [f_1'|f_1''|f_2] m - [f_1|f_2'|f_2'']m + [f_1 |f_2|f']m'.$

3. Identification of the 2-cocycles

As promised, let’s now identify the 2-cocycles in the cobar complex. Namely, let’s try to compute ${\mathrm{Ext}^2_{\Lambda}(k, k)}$ : to do so, we form the cobar complex

$\displaystyle C^\bullet(k) = \Lambda \otimes k \rightarrow \Lambda \otimes \Lambda \otimes k \rightarrow \dots$

and take ${\hom_\Lambda(k, \cdot)}$ . Taking ${\hom_\Lambda(k, \cdot)}$ peels off the first factor of ${\Lambda}$ and we get the complex

$\displaystyle \hom_\Lambda(k, C^\bullet(k)) = k \rightarrow k \otimes \Lambda \otimes k \rightarrow k \otimes \Lambda \otimes \Lambda \otimes k \rightarrow \dots.$

What are the 2-cocycles in this complex? The 2-cochains are given by the polynomial ring ${k \otimes \Lambda \otimes \Lambda \otimes k = k[z_1, z_2]}$ . Let ${k[w_1, w_2, w_3]}$ be the ring of 3-cochains. By the earlier formula (1), the coboundary sends

$\displaystyle f(z_1, z_2) \mapsto f(w_2, w_3) - f(w_1 + w_2, w_3) + f(w_1, w_2 + w_3) - f(w_1, w_2) .$

It now follows that a 2-cocycle in this complex which is symmetric is precisely the same thing as a symmetric 2-cocycle in the earlier sense.

We will also identify the 2-coboundaries. A 1-cochain in ${\hom_\Lambda(k, C^\bullet(k))}$ is just an element of ${k[t]}$ , and the coboundary goes

$\displaystyle f(t) \mapsto f(z_2) - f(z_1 + z_2) + f(z_1).$

These are the “uninteresting” 2-cocycles.

Our next goal is to compute the ${\mathrm{Ext}}$ groups of this coalgebra: in particular, we will find out what the “interesting” 2-cocycles are. These will arise when the denominator ${d}$ in the statement of the theorem is not ${1}$ .

4. Divided powers

Since ${A}$ is a coalgebra, its dual (say, its graded dual) is an algebra. This is the divided power algebra on a single variable. Namely, we find that for each ${n}$ , there exists an element ${x^{(n)} \in A^{\vee}}$ (dual to ${x^n}$ in the basis ${\left\{1, x, x^2, \dots\right\}}$ for ${A}$ ). As a vector space, we have

$\displaystyle A^{\vee} = k \left\{1, x^{(1)}, x^{(2)}, \dots\right\} .$

In ${A}$ , we have the coalgebra structure

$\displaystyle \Delta(x^n) = \sum \binom{n}{i} x^i \otimes x^{n-i}.$

Dualizing, we get the algebra structure in ${A}$ ,

$\displaystyle x^{(i)}x^{(j)} = \frac{n!}{i! j!}x^{(n)}.$

This is precisely the defining property of the divided power algebra. The idea is that ${x^{(i)} }$ should be morally ${\frac{x^i}{i!}}$ , but in general one can’t divide by ${i!}$ . The formulas above state that ${x^{(i))}}$ still behave that way.

We will write

$\displaystyle A^{\vee} = \Gamma(x),$

and use the notation ${\Gamma}$ to denote a divided power algebra.

Note that this dualization process is very natural: it comes up in the relation between cohomology and homology. The space ${\mathbb{CP}^\infty}$ is a topological abelian group (or at least, homotopy equivalent to one) as a ${K(\mathbb{Z}, 2)}$ , and its cohomology ring is precisely the Hopf algebra ${A}$ above. Consequently, its homology ring is the divided power algebra on an element in degree ${2}$ .

Example 1 Over a ring containing ${\mathbb{Q}}$ , ${\Gamma(x) = k[x]}$ : in this case, we can legitimately divide by ${n!}$ .

In characteristic ${p}$ , this is completely false. Let’s consider a field ${k}$ of characteristic ${p}$ and the divided power algebra ${\Gamma(x)}$ on one generator over ${k}$ . Then some combinatorics with binomial coefficients shows that the subspaces

$\displaystyle E_n = \left\{1, x^{(p^n)}, x^{(2p^n)}, \dots, x^{( (p-1)p^n)}\right\}$

are actually subalgebras. Moreover,

$\displaystyle E_n \simeq k[y]/y^p, \quad y = x^{(p^n)} ,$

because the first few powers of ${x^{(p^n)}}$ are nonzero multiples of the appropriate divided powers. The basic combinatorial lemma here is that ${\frac{n!}{i!j!}}$ is divisible by ${p}$ (for ${i + j = n}$ ) if and only if the sum ${i+j}$ is computed without “carrying” in base ${p}$ .

In fact,

$\displaystyle \Gamma(x) = E_0 \otimes E_1 \otimes E_2 \otimes \dots,$

because any divided power can be expressed as a product of the divided powers ${x^{(ip^n)}}$ for ${0 \leq i \leq p-1}$ for some ${n}$ .

5. Computing cohomology

In fact, it’s probably easier just to dualize again, and treat the coalgebra ${\Lambda}$ as a tensor product

$\displaystyle \Lambda = \Lambda_0 \otimes \Lambda_1 \otimes \Lambda_2 \otimes \dots,$

where ${\Lambda_n = \left\{1, x^{p^n}, x^{2p^n}, \dots, x^{(p-1) p^n}\right\}}$ . In this way, ${\Lambda_n}$ is dual to the algebra ${E_n}$ (and we can see this just as well using binomial coefficients).

By the Künneth theorem, we are reduced to computing the cohomology of the subcoalgebras ${\Lambda_n}$ , or equivalently that of the algebras ${E_n}$ . ${E_n}$ is a truncated polynomial algebra, and we would like to compute

$\displaystyle \mathrm{Ext}^{\bullet}_{E_n}(k, k) = \mathrm{Ext}^\bullet_{\Lambda_n}(k, k).$

Proposition 3 The algebra ${\mathrm{Ext}^{\bullet}_{E_n}(k, k) }$ is a tensor product of an exterior algebra on one generator of degree one with a polynomial algebra on one generator two.

The algebra structure ${\mathrm{Ext}^{\bullet}_{E_n}(k, k) }$ comes from the Yoneda product. Using a version of the Eckmann-Hilton argument (see this MO post), one sees that it is also the cup-product which comes from the Hopf algebra structure on ${E_n = k[y]/y^p}$ with ${y}$ primitive. However, unwinding, we find that ${E_n}$ is the group algebra of the cyclic group ${\mathbb{Z}/p}$ , and this algebra ${\mathrm{Ext}^{\bullet}_{E_n}(k, k) }$ is then really just ${H^*(\mathbb{Z}/p; k)}$ with the cup product in group cohomology! In particular, this statement is now a classical statement in the cohomology of cyclic groups (or of ${K(\mathbb{Z}/p, 1)}$ ‘s).

Let’s translate all this back into the cobar complex, for ${\mathrm{Ext}^i_{\Lambda_n}(k, k)}$ for ${i = 1, 2}$ . The point is that both these spaces are one-dimensional.

We now want to translate this back into the earlier language of the cobar complex.

These spaces, as one easily checks, can be described as follows:

${\mathrm{Ext}^2_{\Lambda_n}(k, k)}$ consists of polynomials ${f(x,y)}$ satisfying the cocycle condition but which involve only monomials ${x^{ a p^n}y^{b p^n}}$ for ${a, b \in [0, p-1]}$ . Mod the coboundaries, that is: the “uninteresting” solutions to the equation.
${\mathrm{Ext}^1_{\Lambda_n}(k, k)}$ consists of the “additive” polynomials in ${x,y}$ involving the same monomials.

The claim is that the 2-cocycle

$\displaystyle \frac{1}{p} ( (x+y)^{p^{n+1}} - x^{p^{n+1}} - y^{p^{n+1}})$

spans ${\mathrm{Ext}^2}$ : to see this, we just need to show that it is not a coboundary (since it is evidently a cocycle, as it is even in ${\mathbb{Z}}$ ). That is, we need to show that ${\frac{1}{p} ( (x+y)^{p^n} - x^{p^n} - y^{p^n}) }$ cannot be written as ${f(x+y) - f(x) - f(y)}$ for a polynomial ${f}$ of one variable. If so, we’ll have produced a basis for ${\mathrm{Ext}^2}$ .

Proof: We have to show that there does not exist a polynomial ${f(x)}$ of degree involving the terms ${1, x^{p^n}, \dots, x^{(p-1)p^{n}}}$ such that ${f(x+y) - f(x) - f(y) = \frac{1}{p} ( (x+y)^{p^{n+1}} - x^{p^{n+1}} - y^{p^{n+1}}) }$ . But the point is that this polynomial involves a nontrivial coefficient of terms of degree ${(p-1)p^{n}}$ , and these can’t arise from differences ${f(x+y) - f(x) - f(y)}$ when ${f}$ involves the above monomials only. $\Box$

The only additive polynomials are the constant multiples ${x^{p^n}}$ , so these span the 1-cocycles (i.e. ${\mathrm{Ext}^1}$ ).

6. Putting it all together

With this in mind, we can compute the symmetric 2-cocycles over a field ${k}$ of characteristic ${p}$ . In fact, we see that

$\displaystyle \mathrm{Ext}^\bullet_\Lambda(k, k) = \bigotimes \mathrm{Ext}^\bullet_{\Lambda_n}(k, k)$

by the Künneth theorem.

For each ${n}$ , let ${\alpha_n}$ be a basis for ${\mathrm{Ext}^1_{\Lambda_n}(k, k)}$ (representing the polynomial ${x^{p^n}}$ ) and let ${\beta_n}$ be a basis for ${\mathrm{Ext}^2_{\Lambda_n}(k, k)}$ (representing ${\frac{1}{p}( (x+y)^{p^{n+1}} - x^{p^{n+1}} - y^{p^{n+1}})}$ . By the Künneth theorem, the vector space ${\mathrm{Ext}^2_{\Lambda}(k, k)}$ is free on the products ${\alpha_i \alpha_j}$ and the various ${\beta_j}$ .

Now, the ${\beta_j}$ ‘s give the cocycles which we claimed earlier: the ${\frac{1}{d} ( (x+y)^n - x^n - y^n)}$ , in the cases ${d = p}$ . (When ${d =p }$ , the cocycle is a coboundary.) The products ${\alpha_i \alpha_j}$ correspond to the polynomials ${x^{p^i} y^{p^j}}$ , which are also 2-cocycles. However, they are not symmetric: the symmetrized versions ${x^{p^i} y^{p^j} + x^{p^j}y^{p^i}}$ are the other symmetric 2-cocycles to consider. But these are zero in cohomology because the cup product is skew-commutative.

Anyway, this proves the symmetric 2-cocycle lemma in the torsion case. In the non-torsion case, one can just run the same computation and see that everything is zero instead. I’d also give a direct proof next time, by showing that Lazard’s theorem is true mod torsion, by a comparatively straightforward argument.

Climbing Mount Bourbaki