Morphisms of formal group laws $\,F_a \rightarrow F_m\,$ and $\,F_m\to F_m$

Question

While studying cohomology theories on the stable homotopy setting, I have come up with the following basic question:

Consider the additive formal group law, $F_a$, and the multiplicative formal group law, $F_m$, both defined over a ring $R$.

If $R$ is a $\mathbb{Q}$-algebra, the exponential series defines a morphism of formal group laws $$F_a \to F_m \quad , \quad x \mapsto e^x := \sum_{k=0}^\infty x^k / k! $$ and it is not difficult to check that any other morphism $\,f\colon F_a \to F_m\,$ is of the form $f (x) = e^{ax}$, for some $a\in R$.

I can extend the previous reasoning in case $R$ is an integral domain–of any characteristic–, and I am wondering how many morphisms of formal group laws $F_a \to F_m$ are there over a general ring. That is to say:

The question is to describe all the series $\,f \in R[[x]]\,$ such that $$ f (s+t) = f(s)\cdot f(t) \qquad \mbox{ and } \qquad f(0) = 1 \ . $$

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On the other hand, I am also interested in endomorphisms of the multiplicative group law $F_m$. The obvious ones are rising to the $k^{th}$ power, for $k \in \mathbb{Z}$:

$$ F_m \to F_m \quad , \quad x \mapsto (1+x)^k \ , $$ and the analogous question would be:

Describe all the series $\,f \in R[[x]]\,$ such that $$ f( x + y + xy) = f(x) \cdot f(y) \quad \mbox{ and } \quad f(0) = 1 \ . $$

I think that there only exist these "rising to the $k^{th}$-power" maps, no matter the ring of coefficients.

Answer 1

Let $D$ be the divided power ring $$ D = \mathbb{Z}[a_0,a_1,a_2,\dotsc]/(a_0-1,a_na_m-(n,m)a_{n+m}) $$ (where $(n,m)$ denotes the binomial coefficient $(n+m)!/(n!\,m!)$). Then ring maps from $D$ to $R$ biject with power series $f(t)=\sum_ka_kx^k\in R[[x]]$ satisfying $f(s+t)=f(s)f(t)$ and $f(0)=1$, and thus with homomorphisms $F_a\to F_m$ over $R$. Now $\mathbb{Q}\otimes D$ is just the polynomial ring $\mathbb{Q}[a_1]$ and you recover the description that you mentioned for homomorphisms $F_a\to F_m$ over $\mathbb{Q}$-algebras. On the other hand, for any prime $p$ one can check that $$ D/p = \mathbb{Z}/p[a_1,a_p,a_{p^2},\dots]/(a_{p^i}^p=0) $$ This gives rise to a description of the exponential series over any $\mathbb{Z}/p$-algebra $R$. More explicitly, given $b\in R$ with $b^p=0$ we get an exponential series $$ E_b(x)=\sum_{k=0}^{p-1}(bx)^k/k! $$ and we find that every exponential series hs the form $$ f(x) = \prod_i E_{b_i}(t^{p^i}) $$ for some sequence of coefficients $b_i$ such that $b_i^p=0$.

Answer 2

To complement Neil Strickland's answer, one can lift the description of $\mathbb{F}_{(p)} \otimes D$ to a description of $\mathbb{Z}_{(p)} \otimes D$ where $\mathbb{Z}_{(p)}$ is the localisation of $\mathbb{Z}$ at $p$. One has $$ \mathbb{Z}_{(p)} \otimes D = \mathbb{Z}_{(p)}[b_0,b_1,b_2,\dots]/(b_i^p - p b_{i+1}). $$ If we set $\sum_{i} c_i X^i = \prod_{r} \left(\sum_{0 \leq \ell < p} (b_rX^{p^r})^{\ell} \right)$ then the corresponding morphism $F_a \rightarrow F_m$ is given by $\sum_{i} \frac{p^{v_p(i!)}}{i!} c_i X^i$.

Answer 3

For endomorphisms of $F_m$ one should consider the ring of numerical polynomials: $$ P = \{f(a)\in \mathbb{Q}[a]: f(\mathbb{Z})\subseteq\mathbb{Z}\} $$ The functions $b_i(a)=\left(\begin{array}{c}a\\ i\end{array}\right)$ give a basis for $P$ over $\mathbb{Z}$, with $$ b_ib_j = \sum_{m=\max(i,j)}^{i+j} \frac{m!}{(m-i)!(m-j)!(i+j-m)!} b_m. $$ We have a series $f(x)=\sum_{i>0}b_ix^i\in P[[x]]$ (which is morally "$(1+x)^a-1$") satisfying $$ f((1+x)(1+y)-1)=(1+f(x))(1+f(y))-1. $$ In other words, $f$ is an endomorphism of $F_m$. Any endomorphism of $F_m$ over any ring $R$ arises by applying some homomorphism $P\to R$ to the coefficients of $f(x)$.

Now let $R$ be an algebra over $\mathbb{Z}/p$. Consider the map $\mathbb{Z}\to R[[x]]$ given by $a\mapsto (1+x)^a-1$. This is continuous for the $p$-adic topology on $\mathbb{Z}$ and the $x$-adic topology on $R[[x]]$. This allows us to define $(1+x)^a-1$ for all $a\in\mathbb{Z}_p$, and this is still an endomorphism of $F_m$.

For a more complete story, we should consider $P/p$. One can identify this with the ring of continuous maps $\mathbb{Z}_p\to\mathbb{Z}/p$ (where $\mathbb{Z}_p$ has the $p$-adic topology, and $\mathbb{Z}/p$ has the discrete topology). To understand this, put $T=\{u\in\mathbb{Z}_p:u^p=u\}$. It is known that the reduction map $T\to\mathbb{Z}/p$ is bijective; the inverse map $\tau\colon\mathbb{Z}/p\to T$ is the Teichmuller lift map, given by $\tau(u)=\lim_ku_0^{p^k}$ for any lift $u_0$ of $u$. Given any element $a\in\mathbb{Z}_p$ there are unique elements $c_k(a)\in\mathbb{Z}/p$ such that $a=\sum_k\tau(c_k(a))p^k$. The functions $c_k$ are continuous, so they give elements of $P/p$. In fact, we find that $$ P/p = \mathbb{Z}/p[c_0,c_1,c_2,\dotsb]/(c_k^p-c_k). $$ Any homomorphism from this ring to $R$ will give an endomorphism of $F_m$ defined over $R$.

All of the above can be extracted from the literature on operations and cooperations in complex $K$-theory and its $p$-adic completion.