## A quick comparison of Lie algebras and formal group laws

This post assumes that you are familiar with the definition of Lie group/algebra, and that you are comfortable with the Lazard ring. Note: This is less of an expository post and more of an unfinished question.

Why care about formal group laws? Well, we want to study smooth algebraic groups, but Lie algebras fail us in characteristic p (for example, $\frac{d}{dx}(x^p) = 0$), so, rather than a tangent bundle, we take something closer to a jet bundle.

Lie algebras are to smooth groups over $\mathbb{R}$ or $\mathbb{C}$ as formal group laws are to smooth algebraic groups over any ring $R$.

I want to apply this analogy! I want this deeply. I’m trying to puzzle out how to see if this analogy is deep or superficial. How deep does the rabbit hole go? Let’s look at an example.

Given the universal enveloping algebra of a Lie algebra $\mathfrak{g}$, we might think of this as a deformation of the Symmetric algebra:

$$U_{\epsilon}(\mathfrak{g}) := T(\mathfrak{g})/(x \otimes y – y \otimes x – \epsilon[x, y])$$

$$\mathbb{C}[\mathfrak{g}^*] \simeq Symm[\mathfrak{g}] := T(\mathfrak{g})(x \otimes y – y \otimes x)$$

Action on all of this is the adjoint action, that is, the action which takes an element $g$ of a Lie group $G$ sends $X \mapsto gXg^{-1}$. Orbits of this action stratify the dual Lie algebra, and there is a symplectic form that lives on each orbit.

I want to think of the adjoint action as directly analogous to the compositional conjugation action on the spectrum of the Lazard ring (over a ring $R$). Continue reading A quick comparison of Lie algebras and formal group laws

## 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]]