Formal group law over $\mathbb{F}_p$

Let $p$ be a prime. For each $n > 0$ there is a unique 1-dimensional commutative formal group law $F$ over $\mathbf{Z}$, $F(X, Y) = X + Y + \dots \in \mathbf{Z}[[X, Y]]$, whose logarithm function is given by $$l(x) = \sum_{k \ge 0} \frac{x^{p^{nk}}}{p^k}.$$

Let $\bar{F} \in \mathbf{F}_p[[X, Y]]$ be the formal group over $\mathbf{F}_p$ given by reduction of $F$ modulo $p$. (Cf. Prop. 9.25 in http://neil-strickland.staff.shef.ac.uk/courses/formalgroups/fg.pdf.)

Is $\bar{F}$ an element of $\mathbf{F}_p[X] [[Y]]$?

• I think you should make your question self-contained -- it is not reasonable to require people to read other documents in order to understand what you are asking. – Stefan Kohl Jan 31 '15 at 11:30
• I think this is an interesting question and it would be a pity if it got closed, so I've edited it to knock it into shape. – David Loeffler Jan 31 '15 at 14:44
• I don't see any reason why it should be in this smaller ring. A computation supports this: computing $\bar{F}$ to precision 100 for $p = 3$ and $n = 1$, I got that the coefficient of $Y$ was $1 - x^{2} + x^{4} - x^{6} + x^{10} + x^{12} - x^{18} + x^{28} + x^{30} + x^{36} - x^{54} + x^{82} + x^{84} + x^{90} + O(x^{99})$, which certainly doesn't look like it's going to be of finite degree. But I don't see how to prove this. Any takers? – David Loeffler Jan 31 '15 at 15:13
• I think this is a fascinating question, and I thought I had a strategy for attacking it, but Ghassan Sarkis tossed off an elementary proof, which I’m now checking. If it pans out, I’ll put it up as an answer, and if not, I’ll delete this comment. – Lubin Feb 6 '15 at 1:56
• Bakuradze has just put a proof on the arxiv at arxiv.org/pdf/1502.04152v1.pdf. This may just be a coincidence; he does not refer to the discussion here, I don't know if he has seen it. – Neil Strickland Feb 17 '15 at 7:51

Now multinomial-free, I believe that Ghassan Sarkis and I have a proof of the following

Theorem. Let $h\ge2$, and let $L(x)=x + x^{p^h}/p + x^{p^{2h}}/p^2+\cdots$ be the logarithm of the formal group $F(x,y)\in\Bbb Z_p[[x,y]]$. Then $F(x,y)\in\Bbb Z_p\{\{x\}\}[[y]]$, where $\Bbb Z_p\{\{x\}\}$ is the ring of convergent power series: those whose coefficients go to zero.

A word about this ring: it’s the completion of the polynomials with respect to the “Gauss norm”, i.e. the uniform norm on the closed unit disk; or, if you like, the $p$-adic completion of the ring of polynomials.

Since you get $\Bbb F_p[x]$ when you tensor the ring of convergent series with $\Bbb F_p$, Neil Strickland’s guess turns out to be correct, in a very strong way.

Now for an outline of the proof, which depends entirely on $L'(x)$ being a convergent series, but the proof I found depends also on the particular form of the logarithm.

(Perhaps I should say that the cognoscenti may look at all this and say, C’mon, it’s all clear ’cause the invariant differential is a convergent series, and it all drops out automatically from general facts. But I’m no cognoscente in anything, so I have to go through at least some of the motions. I add that Ghassan wonders whether the present result may be in Hazewinkel already, though in some indecipherable formulation.)

Treat $F(x,y)$ as an element of $\Bbb Z_p[[x]][[y]]$, so write it as $$F(x,y)=x +\sum_{m\ge 1}f_m(x)y^m\,.$$ The aim is to show that each $f_m$ is in $\Bbb Z_p\{\{x\}\}$, not just in $\Bbb Z_p[[x]]$. The argument is by induction, starting with $f_1$, which we already know to be $1/L'(x)$, so convergent. We write out the fundamental property of the logarithm: $$L\bigl(F(x,y)\bigr)=L(x)+L(y)\,,$$ and arrange the pieces differently: $$0=\sum_{N\ge0}\Bigl[F(x,y)^{p^{Nh}} - y^{p^{Nh}}\Bigr]\Big/p^N-L(x)\,.$$ In the above, we want to look at the total coefficient-function of $y^s$, knowing inductively that all $f_m(x)$ for $m<s$ are in $\Bbb Z_p\{\{x\}\}$. In this, we’re not interested in the participation of any monomial with $y$-degree greater than $s$, so we may truncate, and again rearrange: $$-(x+\sum_{m=1}^sf_m(x)y^m)\equiv \sum_{N\ge1}\Bigl[(x+\sum_{m=1}^s f_my^m)^{p^{Nh}} - y^{p^{Nh}}\Bigr] - L(x)\pmod{y^{s+1}}\,.$$ Now, when you look at the occurrence of $y^s$ for each piece with $N\ge1$, there’s only one of them, and lo and behold, the coefficient is $p^{N(h-1)}x^{p^{Nh-1}}$, one of the monomials in $L'(x)$. Collect them all on the other side, and get $$-f_s(x)L'(x) = \text{y^s-coefficient in}\sum_{N\ge1}\Bigl[x+\sum_{m=1}^{s-1} f_my^m\Bigr]^{p^{Nh}}\Big/p^N\,,$$ though in case $s=p^{nh}$, one must add on the left $1/p^n$, an inconsequential change. But here, my friends, our tale is almost done.

The last display exhibits $f_s$ as a $\Bbb Q_p$-series in the series $f_1,\dots, f_{s-1}$. But look at the tail-end of the outer sum: because the degrees in $y$ are bounded, the binomial coefficients for the $p^{Nh}$-powers far overwhelm the denominators, and the total coefficients go to zero. So we know that the tail-end is convergent, just as a series of elements of $\Bbb Z_p\{\{x\}\}$. And the part before the tail-end? That is a polynomial with $\Bbb Q_p$-doefficients in the $s-1$ series in $\Bbb Z_p\{\{x\}\}$. Let’s call it $g(x)$ for the moment. We now have $-f_sL'=g$, and thus $f_s=-f_1g$, an element of $\Bbb Q_p\otimes_{\Bbb Z_p}\Bbb Z_p\{\{x\}\}$ that’s also in $\Bbb Z_p[[x]]$, since of course we know that $F$ has its coefficients in $\Bbb Z_p$. Thus $f_s(x)\in\Bbb Z_p\{\{x\}\}$, as desired.

What this result says is that there is an action of the formal group $F$ on the closed disk. Again, maybe the cognoscenti have known this all along, but I certainly didn’t. You certainly don’t expect such a thing for a random formal group, even (as here) of height greater than $1$.

This is really a comment, but a bit too long.

The coefficient of $y$ in $F(x,y)$ is $1/l'(x)$. If $n=1$ then $l'(x)=\sum_ix^{p^i-1}$ and $1/l'(x)$ is not polynomial mod $p$, so the answer is negative.

However, to my surprise, it seems like the answer might be positive for $n>1$. (Note that $l'(x)=1\pmod{p}$ in that case.) For example, when $n=p=2$ one can calculate that the coefficient of $y^8$ in $\overline{F}(x,y)$ is $x^{14}+x^{20}+x^{26}+x^{56}\pmod{x^{1024}}$. As there are no terms $x^i$ with $56<i<1024$, one might reasonably guess that there are no terms $x^i$ with $i>56$, in which case the coefficient of $y^8$ would be polynomial. However, I do not know how to prove this.

• I have a program that I think computes these Honda formal group laws (see mathoverflow.net/questions/124048/… for some pictures). I'm getting $x^{14}+x^{20}+x^{26}+x^{56}+x^{98}+x^{164}+x^{176}+x^{188}+x^{200}+x^{212}+x^{218}+\cdots$. Also, the pictures don't support the conjecture; but I'm not 100% sure my program's correct... – Christian Nassau Jan 31 '15 at 15:46
• @ChristianNassau: Are you sure that this is the same FGL? The integral version of mine has $_F(x)=\exp_F(2x)+_Fx^4$ but people also consider variants with $_F(x)=2x+_Fx^4$ or $_F(x)=2x+x^4$. I did my calculation quickly in Maple today but then I dug out some files that I generated more systematically a few years ago and the result was the same. – Neil Strickland Jan 31 '15 at 16:38
• It's probably not the same: on inspection it seems my program computes an arbitrary formal group $F$ with $_F(x)=x^4$. Sorry for the confusion. – Christian Nassau Jan 31 '15 at 16:56
• I have slightly changed my program and computed two new formal groups for $p=2$ with heights 2 (nullhomotopie.de/fg2_1024.pdf) and 3 (nullhomotopie.de/fg3_9920.pdf). These pictures now do support the idea that there might be a formal group $\bar F$ of height $n\ge 2$ in $\mathbb F_2[X][[Y]]$. – Christian Nassau Jan 31 '15 at 19:43
• Update to my comment to the original question: Ghassan and I think we have a proof that for $n\ge2$ the formal group in characteristic $p$ is indeed in $\Bbb F_p[x][[y]]$. It’s a fussy messy computation involving multinomial coefficients, though, and has to be checked carefully. – Lubin Feb 9 '15 at 17:04