Understanding the Lazard Ring

When I define a polynomial, I am simply handing you an indexed collection of coefficients.

A polynomial with two variables, $x, y$ and coefficients $c$, is of the form:

$F(x, y) = \sum\limits_{ij} c_{ij} x^i y^j$

The coefficients of a polynomial form a ring. In other words, the coefficients $c_{ij}$ are members of a coefficient ring $R$. When we say $F$ is over $R$, we mean that $F$ is a formal power series with coefficients in $R$.

Screenshot from 2015-01-03 16:13:31

Example: The polynomial
$F(x,y) = 7 + 5xy^2 + 2x^3$ can be written out as
$F(x,y) =7x^0y^0 + 5x^1y^2 + 2x^3y^0$ such that
$c_{00} = 7$, $c_{12} = 5$, $c_{30} = 2$, and the rest of $c_{ij} = 0$.

Alright, now let’s change the coefficients; reassign $c_{00} = 4$, $c_{78} = 3$, and all other $c_{ij} = 0$.

Out pops a very different polynomial $P(x,y) = 4 + 3x^7y^8$.

In other words, By altering the coefficients $c_{ij}$ of $F(x,y)$ via a ring homomorphism $u: R \to R’$ (from the coefficent ring $c_{ij} \in R$ to a coefficient ring $u(c_{ij}) \in R’$)…

Screenshot from 2015-01-03 16:15:03

…we can get from $F(x,y)$ to any other polynomial $F’(x,y)$.

Screenshot from 2015-01-03 16:17:59

What’s a Group-y Tuple?

Intuitively, a group-y tuple is a polynomial + a constraint on our coefficients that forces the polynomial to satisfy group laws.

Concretely, a group-y tuple is an operation of the form $F(x,y) = \sum\limits_{ij}c_{ij}x^iy^j$ such that

  1. $F(x, 0) = F(0, x) = x$
  2. $F(F(x,y), z) – F(x, F(y,z)) = 0$

We can make sure that our polynomial satisfies the constraints of a group-y tuple by modding out our coefficient ring $c_{ij}$ by the ideal $I$ generated by the relations amoung $c_{ij}$ defined above. The ring of coefficients that results is called the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$.

What does this have to do with cobordism?

Let $F^L(x,y)$ be a group-y tuple over $L$.

The premise of $F^L(x,y)$ is that any group-y tuple $F(x, y)$ can be generated from a ring homomorphism $u: L \to R$ such that $F(x,y) = u \circ F^L(x,y)$.

Quillen’s theorem: For any group-y tuple $F(x,y)$ over any commutative ring $R$ there is a unique ring homomorphism $u: MU^*(pt) \to R$ such that $F(x, y) = u \circ F^Q(x, y)$.

Quillen’s proof technique was to show that: $F^L \simeq F^Q$.

For your ventures ahead…

In this post, I have committed two semantic sins in the name of pedagogy.

  1. group-y tuple = “1-dimensional abelian formal group law”
  2. polynomial = “formal power series”

Conventionally, a “polynomial” is a special case of a formal power series (in which we expect that our variables evaluate to a number – useful if we care about convergence).

polynomials $\subset$ formal power series

The polynomial ring $R[x]$ is the ring of all polynomials (in two variables) over a given coefficient ring $R$.

The ring of formal power series $R[[x]]$ is the ring of all formal power series (in two variables) over a given coefficient ring $R$.

polynomial ring $\subset$ ring of formal power series
R[x] $\subset$ R[[x]]


Groupes de Lie formels à un paramètre
Groupes analytiques en caractéristique 0
Groupes de Lie algébriques (travaux de Chavelley)
Formal Group Laws – Ravenel

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