## The Landweber exact-functor theorem

This post assumes familiarity with formal group laws, the definition of exact sequences, the motivation of the Landweber-Ravenel-Stong construction, that the exactness axioms is one of the generalized Eilenberg-Steenrod axioms, and the fact that formal group laws over $R$ are represented by maps from the Lazard ring to $R$.

Recall the the Landweber-Ravenel-Stong Construction: $MU^*(X) \otimes_{L} R \simeq E^*(X)$, where $MU^* \simeq L$ and $R \simeq E^*(pt)$.

We know that in general, tensoring with abelian groups does not preserve exact sequences (e.g., applying $-\otimes_{\mathbb{Z}} \mathbb{Z}/2$ to $0 \to \mathbb{Z} \xrightarrow{\times p} \mathbb{Z} \to \mathbb{Z}/p \to 0$). So, when does the functor $-\otimes_L R: MU^*(X) \to E^*(X)$ preserve exact sequences?

## Die Philosophie der formalen Gruppengesetzen der elliptischen Kurve

Ich lerne Deutsch. Bitte, vergizieh mir und meine Unwissenheit der deutschen Grammatik.

In der Studie von Gruppen mit topogischen Struktur, wir haben die globale Objekt (die Gruppe) mit eine lokalen Objekt (die infinitesimale Gruppe) ersetzen. Wir betrachten dieses Spiel folgt vor:

1. Wir beginnen mit eine Raum.
2. Wir definieren einem binäre Operation an der Menge von Punkten in diesem Raum (einem Operation ist kommutativ, assoziativ, unital, und hat Inversen)
3. Wir leiten eine infinitesimale Gruppe.

z.B. Von Lie-Gruppe (Gruppe internen in die Kategorie der glatten Mannigfaltigkeiten) nach eine Lie-Algebra (Gruppen internen in die Kategorie der unendlich Mannigfaltigkeiten).

1. Wir beginnen mit einer glatten Mannigfaltigkeit (von Geschlect 0 oder 1).
2. Wir ein Produkt Morphismus definieren (Lie-Gruppe).
3. Wir leiten eine infinitesimale Gruppe  (Lie-Algebra).

## Group Law on the (Punctured) Affine Line

There are likely inaccuracies in this post, as I am just beginning to learn the basics of algebraic geometry. Constructive criticism is strongly encouraged.

#### There once was a line…

Let’s look at the affine line over $\mathbb{C}$. This is just the complex line with no distinguished element (i.e., a plane which forgot it’s origin — it’s 2 real dimensions, or, equivalently, 1 complex dimension).

#### $\mathbb{A}^1 \simeq \text{Spec } \mathbb{C}[x]$

As we know, the affine line over the field $\mathbb{K}$ is isomorphic to the spectrum of a ring of single variable polynomials (with coefficients in $\mathbb{K}$). If you aren’t familiar with this isomorphism, I recommend popping over to Spectrum of a Ring. For simplicity, let’s work with the field $\mathbb{C}$, although, I’m pretty sure the rest of this post still works for any $\mathbb{K}$.

#### Is there a reasonable way to take in two points, and ask for a third?

This is generally a fun question to answer when you’re handed a space. So, how do we add two points of $\mathbb{A}^1$ to get a third point in $\mathbb{A}^1$? Continue reading Group Law on the (Punctured) Affine Line

## 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 $F$ has coefficients in $R$. 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 $u(c_{ij}) \in R’$)… …we can get from $F(x,y)$ to any other polynomial $F'(x,y)$. #### What’s a group-y polynomial?

Intuitively, a polynomial is “group-y” if there’s a constraint on our coefficients that forces the polynomial to satisfy the laws of a commutative group.

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

1. $F(x,y) = F(y,x)$                     commutativity
2. $F(x, 0) = x = F(0, x)$                 identity
3. $F(F(x,y), z) – F(x, F(y,z)) = 0$    associativity

We can make sure that our polynomial satisfies these constraints! How? We mod out our coefficient ring $c_{ij}$ by the ideal $I$ — generated by the relations amoung $c_{ij}$ imposed by these constraints.

If you’d like to see the explicit relations, I wrote a cry for help post on stack overflow.

The ring of coefficients that results is called the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$.

It’s important to note here that group-y polynomials are morphisms out of the Lazard ring, not elements of the Lazard ring (i.e., that an assignment of values to each of the $c_{ij}$ describes a group-y polynomial, but the ring of the $c_{ij}$ itself is just a polynomial ring).

In other words, group-y polynomials $f(x,y)$ are morphisms out of the Lazard ring, not elements of the Lazard ring. More formally: for any ring $R$ with group-y polynomial $f(x,y) \in R[[x,y]]$ there is a unique morphism $L \to R$ that sends $\ell \mapsto f$.

$L \to R \simeq F_R$

(where $F_R$ denotes a group-y polynomial with coefficients in $R$)

This makes sense. If it doesn’t, then scroll up a bit! As we saw above, a change of base ring corresponds to a new group-y polynomial.

As we’ve noted, the Lazard ring $L = \mathbb{Z}[c_{ij}]/I$is the quotient of a polynomial ring on the $c_{ij}$ by some relations.

Lazard proved that it is also a polynomial ring (no relations) on a different set of generators. More specifically, $\alpha$ is a graded ring isomorphism:

$\mathbb{Z}[c_{ij}]/I \xrightarrow{\alpha} \mathbb{Z}[t_1, t_2 …]$

(where the degree of $t_i$ is $2i$).

Lurie talks about this a bit (Theorem 4, Lecture 2: The Lazard Ring), but I have yet to understand the proof myself.

Thanks to Alex Mennen for deriving constraints the associativity condition puts on our coefficients; thanks to Qiaochu Yuan and Josh Grochow for kindly explaining some basic details and mechanics of the Lazard ring.

In this post, I have committed two semantic sins in the name of pedagogy. Namely, sins of oversimplification which I’ll attempt to rectify s.t. you aren’t hopelessly confused by the literature:

1. group-y polynomial = “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]]