# Some special subgroups of formal groups

Let $$G$$ be a 1-dimensional, commutative formal group over a ring $$R$$. Give $$G$$ a coordinate $$x$$ and let $$A\subset R$$ be the subring generated by the coefficients of the corresponding formal group law $$F(x,y)= \sum_{ij}a_{ij}x^iy^j$$. So $$G$$ is really defined over $$A$$.

Call a finite subgroup $$K\subset G$$ special if it is the kernel of a homomorphism $$T:G\rightarrow \phi^*G$$ for some ring map $$\phi:A\rightarrow R$$. ($$\phi^*G$$ is the formal group over $$R$$ with formal group law $$\phi^*F(x,y)=\sum_{ij}\phi(a_{ij})x^iy^j$$.)

What's an example of a finite subgroup $$K\subset G$$ that is not (isomorphic to) a special subgroup? (I guess that would be the same as asking that $$G/K$$ does not inject into any $$\phi^*G$$.)

In all the cases I've tried (including the additive, multiplicative, and universal formal group laws) I seem to have convinced myself that all subgroups are special. Which leads me to suspect that maybe special subgroups are not so special.

$$\textbf{Edit}$$: To be explicit, a subgroup $$K$$ corresponds to a monic polynomial $$f_K(x)\in R[x]$$ with nilpotent lower order coefficients and such that $$f(F(x,y))\equiv 0\ \text{mod}\ (f(x),f(y))$$. That subgroup $$K$$ is special if furthermore there is some invertible power series $$u(x)\in R[[x]]$$ and a ring map $$\phi:A\rightarrow R$$ as above such that $$u(F(x,y))f(F(x,y))=\phi^*F(u(x)f(x),u(y)f(y)).$$

• What's a finite subgroup of a formal group? Sep 22, 2020 at 5:35
• @QiaochuYuan It's a closed formal subscheme that's finite and free and is also a group. That's equivalent to the data of a monic polynomial $f(x)$ with nilpotent lower order coefficients such that $f(F(x,y))=0$ mod $(f(x),f(y))$. Sep 22, 2020 at 7:26
• Your description of specialness in the Edit seems to be demanding that there would be an endomorphism $u\cdot f$ of $F$, vanishing on the kernel of $T$, so that in some sense $G/(\ker T)\cong G$ as formal groups. Have I misconstrued? Sep 22, 2020 at 17:47
• @Lubin So $u\cdot f$ "is" $T$ - it is not quite an endomorphism of $F$ but rather sends $F$ to $\phi^*F$. And instead of $G/\text{ker}T\cong G$ all I have is $G/\text{ker} T\hookrightarrow \phi^*G$. Sep 22, 2020 at 22:54

Let me specialize heavily to the case of formal groups (group laws) of dimension one over a $$p$$-adic ring $$\mathfrak o$$, i.e. the ring of integers of a finite extension $$k$$ of $$\Bbb Q_p$$.

I still am uncertain about what category you’re thinking of. If we restrict further to formal groups of finite height (the endomorphism $$[p]$$ being of finite degree $$p^h$$), then these things become $$p$$-divisible groups, or, if you like, ind-finite objects. For instance the kernel of $$[p^n]$$ will be a finite $$\mathfrak o$$-group-scheme, $$K_n=\ker([p^n])=\mathrm{Spec}(\mathfrak o[[x]]/([p^n](x))\,)$$, and you have natural maps $$K_n\hookrightarrow K_{n+1}$$, and you see that $$\projlim\mathfrak o[[x]]/([p^n](x))\cong\mathfrak o[[x]]$$. In this sense, your $$G$$, if indeed a formal group of finite height over $$\mathfrak o$$, is the union of its finite subgroups. This is the viewpoint that I tend to work with.

Now, let’s consider just one fairly simple case, where the formal group law has all its coefficients in an unramified extension $$A$$ of $$\Bbb Z_p$$, even in $$\Bbb Z_p$$ itself, and suppose the height is $$h=2$$ for simplicity. This means that $$[p](x)\equiv px+ux^{p^2}\pmod{x^{p^2+1}}$$, where $$u$$ is a unit of $$A$$, and the congruence ignores all terms in the power series of degree $$>p^2$$. Look at the Newton polygon and see that all the $$z\in\overline k$$ with $$v_p(z)>0$$ and $$[p](z)=0$$ have $$v_p(z)=\frac1{p^2-1}$$, plus of course $$0$$. So $$p^2$$ in all, and thus they form an elementary $$p$$-group of order $$p^2$$.

Now take any of the cyclic subgroups of $$\ker[p]$$, call it $$\Gamma$$. One proves that $$\pi_\Gamma(x)=\prod_{\gamma\in\Gamma}F(x,\gamma)\,,$$ which is defined over a totally ramified extension $$A'$$ of $$A$$ (actually of degree $$p+1$$), is a morphism into another formal group, which I will abuse language in calling $$G/\Gamma$$.

I ask you to believe that I have shown you a formal group $$G/\Gamma$$ that, as far as I can see, will prove to you that $$\Gamma$$ is not a special subgroup of $$G$$, once you see that the formal group law of $$G/\Gamma$$ is not isomorphic to that of $$G$$, not even with a morphism $$\varphi^*$$ of the type you allow. (I think, because I’m not sure what properties you allow $$\varphi^*$$ to have.)

How do I know that $$G/\Gamma$$ is nothing like $$G$$? By Newtonian magic, you see that the Newton polygon of $$[p]_{G/\Gamma}$$ has vertices at $$(1,1)$$, $$(p,\frac1{p+1})$$, and $$(p^2,0)$$. The important fact is that this polygon is not the same as that of $$[p]_G$$; and since the shape of the Newton polygon of $$[p]$$ is an invariant, it follows that there is no way for $$G/\Gamma$$ to be isomorphic to $$G$$.

In the appropriate category, the map from $$G$$ to $$G/\Gamma$$ is onto. You can show, for instance, that if $$v_p(\eta)>0$$, there is $$\xi$$ in a finite extension of $$k(\eta)$$ such that $$v_p(\xi)>0$$ and $$\pi_\Gamma(\xi)=\eta$$.

(All of this is in an old and poorly-written paper of mine, Finite subgroups and isogenies…. EDIT:The “Newtonian magic” involves the “Newton copolygon”, also called the valuation function. I’ll bet a nickel that somebody else has explained it better than I can, but it’s in a later paper of mine, Canonical subgroups of formal groups, and I fear that it’s at most a little better-written than the other. Copolygon talk begins on p. 109.)

• Thanks alot for this! Some questions: what is the smallest ring over which $G/\Gamma$ is defined (I think that's $A'$ in the above?) and how does it relate to $A$ and $\mathfrak{o}$? How can I show that $G/\Gamma$ does not even inject into any $\phi^*G$? ($\phi^*$ is any map $Spf\mathfrak{o}\rightarrow SpfA$? Also, where can I learn Newtonian magic (preferably without selling my soul)? I looked through your paper and couldn't find "Newton" anywhere. Sep 23, 2020 at 19:05
• Yes, if $A$ was unramified over $\Bbb Z_p$, $A'$ is the smallest. Without changing the Newton polygon, the only maps are isomorphisms, if you’re talking the category of one-dimensional formal groups. I thought that the paper I quoted has the Newtonian magic explained, but really it’s in a later paper. I’ll edit the answer itself to include the reference. Sep 23, 2020 at 19:34

Consider the case of a formal group $$G$$ of finite height over a complete local Noetherian ring $$R$$ of residue characteristic $$p>0$$. For each $$m$$ there is a finite $$R$$-algebra $$S$$ that classifies finite subgroups of $$G$$ of order $$p^m$$ in the sense that $$R$$-algebra homomorphisms $$S\to T$$ biject with subgroup-schemes $$A<\text{spec}(T)\times_{\text{spec}(R)}G$$ such that $$\mathcal{O}_A$$ is free of rank $$p^m$$ over $$T$$. The structure of this classifying ring is described in my paper Finite subgroups of formal groups; there is another version on my home page with additional exposition. In another paper I showed how this ring arises from a calculation in algebraic topology, in the case where $$G$$ is the universal deformation of a formal group $$G_0$$ over a finite field $$F$$. In the case where $$F$$ is of prime order, every finite subgroup of $$G_0$$ is the kernel of a power of Frobenius, and we can use the universal deformation property to deduce a kind of specialness property for $$G$$. Specifically, given $$\alpha\colon R\to T$$ and a finite subgroup $$A<\alpha^*G$$ there is another map $$\beta\colon R\to T$$ with $$(\alpha^*G)/A\simeq\beta^*G$$. There is a similar but slightly more involved statement in the case where $$|F|$$ is not prime. In algebraic topology this is all closely bound up with the theory of power operations in $$H_\infty$$ ring spectra. This is explained in a paper by Charles Rezk. The similar specialness property of Lazard's universal FGL is similarly bound up with the $$H_\infty$$ structure of the complex cobordism spectrum $$MU$$, via Quillen's fundamental theorem that the homotopy ring $$\pi_*(MU)$$ is canonically isomorphic to the Lazard ring.

• Thanks, is the claim you're making that any (or many?) formal group coming from a ring spectrum that admits an H-infinity complex orientation has only "special" subgroups? Because the reason I actually posted this question is that I had convinced myself that something like that is true! Mar 24, 2021 at 15:41
• That's certainly the right idea; I'm not sure exactly what technical assumptions you need to support it. It should work for $K(n)$-local even periodic $H_\infty$ ring spectra. If $E$ is even periodic and $p$-complete then in many cases it is a retract of $\prod_nL_{K(n)}E$. I don't remember what is the maximum generality in which that is known. Using that fact, one should be able to generalise the statement about special subgroups. Mar 24, 2021 at 15:52