Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

EDIT: I've tried to alter the question so that its basic nature is clearer, as it's been unclear to a number of people now.

At any prime p, there is a graded polynomial ring $V \cong {\mathbb Z}_{(p)}[v_1, v_2, \ldots]$ carrying two formal group laws. These formal group laws are of the form $$ F(x,y) = \ell^{-1}(\ell(x) + \ell(y)) $$ for a logarithm $\ell(x) = \sum \ell_n x^{p^{n+1}} \in ({\mathbb Q} \otimes V)[\![x]\!]$ (where $\ell_0 = 1$ by convention). Both of these formal group laws have the property that they are universal among so-called $p$-typical formal group laws, and see heavy computational use in stable homotopy theory.

These two are based on choices of recursive definition for the logarithm coefficients in terms of the generators $v_i$ of $V$. The first definition (the Araki generators) satisfies: $$ p \ell_n = \sum_{k=0}^n v_k^{p^{n-k}} \ell_k = v_n + \ell_1 v_{n-1}^p + \cdots + \ell_{n-1} v_1^{p^{n-1}} + \ell_n p^{p^n} $$ The Hazewinkel generators are instead defined by: $$ p \ell'_n = \sum_{k=1}^{n} v_k^{p^{n-k}} \ell'_k = v_n + \ell'_1 v_{n-1}^p + \cdots + \ell'_{n-1} v_1^{p^{n-1}} $$ This gives the ring $V$ with two logarithms $\ell$ and $\ell'$, and two distinct universal formal group laws.

My question is: Are these two formal group laws isomorphic? Strictly isomorphic?

ADDED: Since the ring is torsion free, any isomorphism between them is of the form $f(x) = (\ell')^{-1} (c \ell(x))$ for a unit $c \in \mathbb{Z}_{(p)}^\times = V^\times$. They are therefore isomorphic if and only if they are strictly isomorphic. Therefore, the question is equivalent to the following:

Does the power series $(\ell')^{-1} \circ \ell$ have coefficients in $V \subset V \otimes \mathbb{Q}$?

(The issue was brought up when thinking about truncated Brown-Peterson spectra ${\rm BP}\langle n\rangle$, whose rings of coefficients are $V/(v_{n+1},v_{n+2}, \cdots)$. It then becomes a question as to whether these are equivalent as ring spectra depending on the choice of generators. There are certainly different choices of generators for which they are inequivalent.)

share|cite|improve this question
Dear Tyler: it's been a while since I've thought about these things, so perhaps my memory is foggy, but why doesn't the precise formulation of the universal property solve the problem? And have you looked at the (rather exhaustive) discussion of $p$-typical formal groups in Hazewinkel's book "Formal groups and their applications"? –  BCnrd Sep 4 '10 at 4:18
@BCnrd: The issue is that the universal property doesn't respect the noncanonical structure being imposed (namely, the choice of polynomial generators). You can use the universal property to produce automorphisms of the universal ring that interchange the Hazewinkel and Araki generators, but not much more. I looked in Hazewinkel's book a number of years ago to try and find any results relating interchange of these generators and did not find one - it is mostly further applications in topology that have made use of the non-invariant ideal $(v_{n+1},\ldots)$. –  Tyler Lawson Sep 4 '10 at 5:14

3 Answers 3

up vote 14 down vote accepted

No, at least when $p=2$ the coefficient of $x^8$ in the relevant power series is not 2-locally integral. I have put a Maple worksheet at with a PDF version at

share|cite|improve this answer
And not even a coefficient involving v_3 either, so the Hazewinkel BP<2> isn't the Araki BP<2> as ring spectra. Outstanding. –  Tyler Lawson Nov 10 '10 at 3:21

As a way of additional information - explicit expressions of the $l$s in terms of the generators look like this:

for Hazewinkel generators, $$ l_n=\sum_{\substack{n_1+...+n_k=n,\\1\leqslant k\leqslant n}}\frac1{p^k}v_{n_1}v_{n_2}^{p^{n_1}}v_{n_3}^{p^{n_1+n_2}}\cdots v_{n_k}^{p^{n_1+n_2+...+n_{k-1}}} $$ while for the Araki ones, $$ l_n=\frac1{p^{p^n-1}-1}\sum_{\substack{n_1+...+n_k=n,\\1\leqslant k\leqslant n}}-\frac{(-1)^k}{p^k}\frac{v_{n_1}v_{n_2}^{p^{n_1}}v_{n_3}^{p^{n_1+n_2}}\cdots v_{n_k}^{p^{n_1+n_2+...+n_{k-1}}}}{(p^{p^{n_1}-1}-1)(p^{p^{n_1+n_2}-1}-1)\cdots(p^{p^{n_1+n_2+...+n_{k-1}}-1}-1)}. $$

I would not object if anybody finds this uselessly horrible --- such people may view this as a joke.

Then for the latter, as an additional joke, note that because of these expressions the following becomes tantalizingly close to being true:

Araki's $v_n$ is ``almost'' equal to $-\frac1{\sqrt[p^n]{p^{p^n-1}-1}}$ times the Hazewinkel's $v_n$ :)

share|cite|improve this answer

I happened to see your question. And I don't know whether you've found a good reference. You may want to have a look at Haynes Miller's notes on cobordism. Below is a link for that:

The answer to your question is YES. It's discussed in the 5th section of Chapter 2 in the note.

share|cite|improve this answer
Hi Penguin, thanks for the notes. These notes indeed show the universality of the two sequences of generators, but they fix the logarithm and change the definition of the coefficients $v_i$ - which is the opposite of what I'd like to ask about. I've altered the question to try to make it clearer. –  Tyler Lawson Nov 2 '10 at 12:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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