Some facts about formal group laws

 Conference topic

Here we solve some problems from the 2016 West Coast Algebraic Topology Summer School (WCATSS) at The University of Oregon. Thanks to Piotr Pstragowski and Carolyn Yarnall for the solutions. First we recall some definitions.

Definition: Let $R$ be a commutative ring with unit. A formal group law $F$ over $R$ is an element $F\in R[[x,y]]$ satisfying

1. $F(x,y) = F(y,x)$ (symmetry), 
2. $F(x,0) =x$ and $F(0,y)=y$ (uniticity),
3. $F(F(x,y),z) = F(x,F(y,z))$ (associativity).

It follows from these three properties that $F(x,y)=x+y+($higher order terms$)$ for all $F$.

Proposition: For any formal group law $F(x,y)$ over $R$, $x$ has a formal inverse. That is, there exists an element $i(x)\in R[[x]]$ such that $F(x, i(x)) = 0$.

Proof: Consider $F(x,y+z)$, with $|z| = n$. Note that
$$\begin{align*} F(x,y+z) & = x+y +z +\sum_{i,j\geqslant 1} a_{ij} x^i(y+z)^j \\ & = x+y +z+\sum_{i,j\geqslant 1} a_{ij} x^i \sum_{k=0}^j \binom jk y^k z^{j-k} \\ & = x+y+z+\sum_{i,j\geqslant 1} a_{ij} x^i \left(y^j + \sum_{k=0}^{j-1} \binom jk y^k z^{j-k} \right)\\ & = x+y+z+\sum_{i,j\>1} a_{ij} x^i y^j  + \underbrace{\sum_{i,j\geqslant 1} a_{ij} x^i}_{\text{deg }\geqslant \ 1}\underbrace{\sum_{k=0}^{j-1} \binom jk y^k z^{j-k}}_{\text{deg }=\ k+n(j-k)\geqslant n}\\ & = F(x,y) + z + (\text{terms of deg }\geqslant\ n+1). \end{align*}$$
First choose $z_1$ to be the negative of all the degree-1 terms of $F(x,0)$, so that $F(x,z_1)$ has terms of degree 2 and higher. Now choose $z_2$ to be the negative of all the degree-2 terms of $F(x,z_1)$, so $F(x,z_1+z_2)$ has terms of degree 3 and higher. Continue in this manner ad infinitum to get a formal inverse $\sum_i z_i$ (this will be a power series) of $x$. $\square$

Recall that we call $f_a(x,y) = x+y$ the additive formal group law and $F_m(x,y) = x+y+xy$ the multiplicative formal group law. Via the universal Lazard ring of formal group laws, these turn out to be the formal group laws of ordinary singular cohomology theory (additive) and complex $K$-theory $KU$ (multiplicative). Recall also nested notation: for $F$ a formal group law, we write
$$\begin{align*} [1]_F(x) & = x, \\ [2]_F(x) & = F(x,x), \\ [3]_F(x) & = F(F(x,x),x), \\ [4]_F(x) & = F(F(F(x,x),x),x), \end{align*}$$
and so on.

Definition: Let $F$ be a formal group law over $R$. A morphism of formal group laws is an element $\varphi\in R[[u]]$, giving a formal group law $\varphi F\in R[[x,y]]$ by $\varphi F(x,y):= F(\varphi(x),\varphi(y))$.

An isomorphism of formal group laws is a morphism where the formal power series $\varphi$ is an isomorphism.

Proposition: The additive formal group law and the multiplicative formal group law are not isomorphic over $F_p$.

Proof: We compare $[p]_{F_m}(x)$ and $[p]_{F_a}(x)$ and show they are not the same. If there were an isomorphism $\varphi$ between $F_a$ and $F_m$, we should have that
$$F_m(x,x) = F_a(\varphi(x),\varphi(x)) = \varphi(F_a(x,x)) \ \ \implies\ \ [p]_{F_m}(x) = \varphi([p]_{F_a}(x)),$$
since $\varphi$ is a homomorphism. However, we first see that
$$[1]_{F_a}(x) = x ,\hspace{1cm} [2]_{F_a}(x) = F_a(x,x) = 2x ,\hspace{1cm} [3]_{F_a}(x) = F_a(F_a(x,x),x) = 3x,$$
and so continuing this pattern we get that $[p]_{F_a}(x) = px = 0$ in $F_p$. Next, for the multiplicative formal group law we find that
$$[1]_{F_m}(x) = x, ,\hspace{1cm} [2]_{F_m}(x) = F_m(x,x) = 2x + x^2 ,\hspace{1cm} [3]_{F_m}(x) = F_m(2x+x^2,x) = 3x + 3x^2 + x^3.$$
Here the pattern  is not immediate, but continuing these small examples we find that $[p]_{F_m}(x) = (x+1)^p-1 = 1+x^p-1 = x^p$ in $F_p$. An isomorphism sends only 0 to 0, but in this case $\varphi$ should send $x^p\neq 0$ to $0$, a contradiction. Hence no such isomorphism exists over $F_p$. $\square$