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

Theorem 1The Lazard ring over which the universal formal group law is defined is a polynomial ring in variables of degree .

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 , then any formal group law on can be lifted to one over .

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

classifying the formal group law obtained from the additive one by the “change of coordinates” . We claimed that the map on indecomposables was injective, and that, in fact the image in the indecomposables of 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 be an abelian group. A **symmetric 2-cocycle** is a “polynomial” with the properties:

and

These symmetric 2-cocycles come up when one tries to classify formal group laws over the ring , 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 is a multiple of where if is not a power of a prime, and if .

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 group. Then, the strategy is to compute this 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:

- .
- for a prime and an integer .

This is because any finitely generated abelian group is a direct sum of such cyclic groups. To prove it for , we can prove it for and use the fact that the polynomial has coefficients which are together relatively prime (this is a straightforward lemma with binomial coefficients).

Finally, we can reduce the case of to the case of using a filtration argument. Hence, in particular:

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

In this post, we will only handle the case of a finite prime field . The case of will follow from the fact that the morphism

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

**2. The relevant coalgebra and cobar complex**

Let be a field. We will consider the polynomial ring ; this can be made into a **Hopf algebra** if is primitive,

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 . The coalgebra map

sends a polynomial to (making, again, the interpretation as transparent).

This coalgebra will be the protagonist for this story.

**Idea:** Symmetric 2-cocycles will be literally 2-cocycles for a complex computing 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 be an -comodule. We can write down a concrete and functorial -free resolution of ,

In other words, we construct a complex of -free comodules. We have

and use the combination of the comodule and coalgebra structure to make into a *cosimplicial* -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 where and ; we use the **bar notation** to denote tensor products.

The coboundary sends this to the 3-cochain

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

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

**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 : to do so, we form the cobar complex

and take . Taking peels off the first factor of and we get the complex

What are the 2-cocycles in this complex? The 2-cochains are given by the polynomial ring . Let be the ring of 3-cochains. By the earlier formula (1), the coboundary sends

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 is just an element of , and the coboundary goes

These are the “uninteresting” 2-cocycles.

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

**4. Divided powers**

Since 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 , there exists an element (dual to in the basis for ). As a vector space, we have

In , we have the coalgebra structure

Dualizing, we get the algebra structure in ,

This is precisely the defining property of the divided power algebra. The idea is that should be morally , but in general one can’t divide by . The formulas above state that still behave that way.

We will write

and use the notation 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 is a topological abelian group (or at least, homotopy equivalent to one) as a , and its cohomology ring is precisely the Hopf algebra above. Consequently, its *homology* ring is the divided power algebra on an element in degree .

Example 1Over a ring containing , : in this case, we can legitimately divide by .

In characteristic , this is completely false. Let’s consider a field of characteristic and the divided power algebra on one generator over . Then some combinatorics with binomial coefficients shows that the subspaces

are actually subalgebras. Moreover,

because the first few powers of are nonzero multiples of the appropriate divided powers. The basic combinatorial lemma here is that is divisible by (for ) if and only if the sum is computed without “carrying” in base .

In fact,

because any divided power can be expressed as a product of the divided powers for for some .

**5. Computing cohomology**

In fact, it’s probably easier just to dualize again, and treat the *coalgebra* as a tensor product

where . In this way, is dual to the algebra (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 , or equivalently that of the algebras . is a truncated polynomial algebra, and we would like to compute

Proposition 3The algebra 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 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 with primitive. However, unwinding, we find that is the group algebra of the cyclic group , and this algebra is then really just with the **cup product** in group cohomology! In particular, this statement is now a classical statement in the cohomology of cyclic groups (or of ‘s).

Let’s translate all this back into the cobar complex, for for . 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:

- consists of polynomials satisfying the cocycle condition but which involve only monomials for . Mod the coboundaries, that is: the “uninteresting” solutions to the equation.
- consists of the “additive” polynomials in involving the same monomials.

The claim is that the 2-cocycle

spans : 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 ). That is, we need to show that cannot be written as for a polynomial of one variable. If so, we’ll have produced a basis for .

*Proof:* We have to show that there does not exist a polynomial of degree involving the terms such that . But the point is that this polynomial involves a nontrivial coefficient of terms of degree , and these can’t arise from differences when involves the above monomials only.

The only additive polynomials are the constant multiples , so these span the 1-cocycles (i.e. ).

**6. Putting it all together**

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

by the Künneth theorem.

For each , let be a basis for (representing the polynomial ) and let be a basis for (representing . By the Künneth theorem, the vector space is free on the products and the various .

Now, the ‘s give the cocycles which we claimed earlier: the , in the cases . (When , the cocycle is a coboundary.) The products correspond to the polynomials , which are also 2-cocycles. However, they are not symmetric: the symmetrized versions 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.

May 28, 2012 at 7:04 pm

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