We are in the middle of proving an important result of Lazard:
Theorem 1 The 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 1 Over 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 3 The 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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