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)


\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:

  1. {\mathbb{Z}}.
  2. {\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:

  1. {\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.
  2. {\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.