3
$\begingroup$

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 $u(F(x,y))f(F(x,y))=\phi^*F(u(x)f(x),u(y)f(y)).$

$\endgroup$
4
  • $\begingroup$ What's a finite subgroup of a formal group? $\endgroup$ Sep 22, 2020 at 5:35
  • 1
    $\begingroup$ @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 $(f(x),f(y))$. $\endgroup$
    – kiran
    Sep 22, 2020 at 7:26
  • $\begingroup$ 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? $\endgroup$
    – Lubin
    Sep 22, 2020 at 17:47
  • $\begingroup$ @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 $G/\text{ker} T\hookrightarrow \phi^*G$. $\endgroup$
    – kiran
    Sep 22, 2020 at 22:54

2 Answers 2

2
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – kiran
    Sep 23, 2020 at 19:05
  • $\begingroup$ 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. $\endgroup$
    – Lubin
    Sep 23, 2020 at 19:34
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$
    – kiran
    Mar 24, 2021 at 15:41
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 24, 2021 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.