This is essentially a request for counterexamples, since I know so few $n$-buds (or as some might say, formal group law $n$-chunks). One notices that the only $1$-bud of maximal degree 1 is the additive one, $x+y$, which also happens to be a formal group law. Also, the only 2-buds that I know of over a ring $R$ are of the form $x+y+cxy$ for some $x\in R$, which also defines a formal group law (here I'm assuming $R$ is commutative and unital). Presumably this pattern does not continue infinitely? That is to say, there is a finite polynomial $f=x+y+h.o.t.$ of total degree $n$ such that $f(f(x,y),z)-f(x,f(y,z))$ is equivalent to zero modulo degree $n+1$ terms (i.e. it defines an $n$-bud) but is not equal to zero?
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asked May 29, 2013 at 1:38
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$\begingroup$ I should mention, I have been informed that there is a theorem somewhere (and I'm pretty sure I read it at some point) that the only finite length formal group laws are the additive and multiplicative ones. $\endgroup$– Jonathan BeardsleyCommented Jun 26, 2013 at 19:41
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1 Answer
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If $f(x, y)$ is a formal group law then so is $g(f(g^{-1}(x), g^{-1}(y))$ where $g$ is an invertible (under composition) formal power series. This suggests a strategy for writing down $n$-buds, namely pick a polynomial $g$, a group law $f$, and a polynomial approximation $h$ to $g^{-1}$ and then compute $g(f(h(x), h(y))$. If $h$ agrees with $g^{-1}$ modulo degree $n+1$ terms then this gives an $n$-bud.
For $g$ let's pick $g(x) = x + x^2 + x^3$ and for $f$ let's pick $f(x, y) = x + y$. For $h$ let's pick $h(x) = x - x^2 + x^3$, which agrees with the compositional inverse modulo degree $4$ terms, so $g(h(x) + h(y))$ is a $3$-bud. Expanding, this is (modulo degree $4$ terms)
$$x + y + 2xy + x^2 y + y^2 x.$$
But this shouldn't give a formal group law. (I tried to check this in Sage but it's not happy with expanding the associator here.)
answered May 29, 2013 at 2:33
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$\begingroup$ It's interesting that what you come up with is basically the sum of symmetric functions. I guess I'll trust you that this isn't an FGL. :-) $\endgroup$ Commented May 29, 2013 at 20:21
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$\begingroup$ Also, do you happen to know of good code (I guess in Sage, which I have little experience with) for computing the homogeneous degree n terms of the associator? I'd like to show that the associator is a cocycle in a certain cohomology, but am having a difficult time writing down these rather large polynomials for anything higher than homogeneous degree 2. $\endgroup$ Commented May 29, 2013 at 20:26
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1$\begingroup$ Sage can symbolically manipulate multivariate polynomials (sagemath.org/doc/constructions/…) although SageMathCloud wasn't happy with the above example for some reason. $\endgroup$ Commented May 29, 2013 at 20:50
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1$\begingroup$ A detail: your expansion modulo degree 4 terms is incorrect. The terms x^3 and y^3 cannot appear in a formal group law (since it satisfies $f(x,0)=x$ and $f(0,y)=y$). $\endgroup$ Commented Oct 20, 2014 at 13:33
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2$\begingroup$ The associator seems to be $-4x^2yz + 4xyz^2 - 2x^3yz - 7x^2y^2z + 7xy^2z^2 + 2xyz^3$ modulo degree six. $\endgroup$– LubinCommented Oct 20, 2014 at 21:19