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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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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. –  Jon Beardsley Jun 26 '13 at 19:41

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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.)

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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. :-) –  Jon Beardsley May 29 '13 at 20:21
    
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. –  Jon Beardsley May 29 '13 at 20:26
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Sage can symbolically manipulate multivariate polynomials (sagemath.org/doc/constructions/…) although SageMathCloud wasn't happy with the above example for some reason. –  Qiaochu Yuan May 29 '13 at 20:50
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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 $f(0,y)=y$). –  Baptiste Calmès Oct 20 at 13:33
    
@Baptiste: you're right, I just checked the calculation again and the correct calculation just has the $x^3$ and $y^3$ terms removed. Not sure how those got there. Thanks! –  Qiaochu Yuan Oct 20 at 16:20

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