Contents

Idea

The Lazard ring_ is a commutative ring which is

and by Quillen's theorem also

Definition

The Lazard ring can be presented as by generators $a_{i j}$ with $i,j \in \mathbb{N}$

$L = \mathbb{Z}[a_{i j}] / (relations\;1,2,3\;below)$

and relations as follows

1. $a_{i j} = a_{j i}$

2. $a_{10} = a_{01} = 1$; $\forall i \neq 0: a_{i 0} = 0$

3. the obvious associativity relation

the universal formal group law is the formal power series

$\ell(x,y) = \sum_{i,j} a_{i j} x^j y^j \in L[[x,y]]$

in two variables with coefficients in the Lazard ring.

Properties

As classifying ring for formal group laws

For any ring $S$ with formal group law $g(x,y) \in S[ [x,y] ]$ there is a unique morphism $L \to S$ that sends $\ell$ to $g$.

Lazard’s theorem

Lazard's theorem states:

Theorem

The Lazard ring is isomorphic to a graded polynomial ring

$L \simeq \mathbb{Z}[t_1, t_2, \cdots]$

with the variable $t_i$ in degree $2 i$.

(e.g. Lurie lect 2, theorem 4)

As the complex cobordism cohomology ring

By Quillen's theorem on MU the Lazard ring is the cohomology ring of complex cobordism cohomology theory.

Theorem

Let $M P$ denote the peridodic complex cobordism cohomology theory. Its cohomology ring $M P(*)$ over the point together with its formal group law is naturally isomorphic to the universal Lazard ring with its formal group law $(L,\ell)$.

Remark

This can be used to make a cohomology theory out of a formal group law $(R,f(x,y))$. Namely, one can use the classifying map $M P({*}) \to R$ to build the tensor product

$E^n(X) := M P^n(X) \otimes_{M P({*})} R,$

for any $n\in\mathbb{Z}$. This construction could however break the left exactness condition. However, $E$ built this way will be left exact if the ring morphism $M P({*}) \to R$ is a flat morphism. This is the Landweber exactness condition (or maybe slightly stronger). See at Landweber exact functor theorem.

References

Revised on November 25, 2013 06:22:56 by Zoran Škoda (161.53.130.104)