# Explicit formula for associator of commutative power series

Perhaps this question is too elementary, but if it's written down anywhere, I'd love to know about it. Suppose I have a power series $f\in R[[x,y]]$ for some commutative, unital ring. I've recently been trying to write down an explicit formula for the power series in three variables, $f(f(x,y),z)-f(x,f(y,z))$, written down in terms of the coefficients of $f$, but am wondering if this classical-seeming computation has already been done somewhere else. I'm perfectly happy assuming $f(x,y)=f(y,x)$. Additionally, I'm really only interested in power series such that $f(x,0)=x$ and $f(0,y)=y$. Note that the so-called associator is also a the Gerstenhaber circle product of $f$ with itself, if that helps at all.

Thanks

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I don't know what applications you have in mind, but if you're lucky you can switch your attention to Lazard's universal ring: en.wikipedia.org/wiki/Lazard%27s_universal_ring –  John Wiltshire-Gordon Jul 4 '13 at 19:11
@JohnWiltshire-Gordon Yeah I wish. Haha. That's sort of my starting point. In a sense. –  Jon Beardsley Jul 4 '13 at 19:16
FWIW I have completely determined this power series up to degree 5, and there is definitely a pattern (although the first 3 degrees are 0 if $f$ is commutative). –  Jon Beardsley Jul 4 '13 at 19:30
I would have sent you to Lazard’s 1955 paper, but you evidently are familiar with that. –  Lubin Jul 4 '13 at 21:27
Wow @Lubin I'd love to talk to you more about this if you'd be willing. It specifically concerns the cohomology theory you and John Tate describe the first two degrees of in your 1966 paper, specifically, the third degree of that cohomology theory. I am essentially trying to reprove a theorem of Lazard's from a purely homological standpoint. –  Jon Beardsley Jul 5 '13 at 2:15

## 1 Answer

Okay, so I'm pretty sure I have a sort-of answer for this for $f=x+y+\sum_{i,j>0}a_{ij}x^iy^j$, though it's not a closed form at all.

With a bit of fiddling one can see that $$f\circ f = \sum_{i,j>0}a_{ij}\left(x+y+\sum_{l,k>o}a_{lk}x^ly^k\right)^iz^j-\sum_{i,j>0}a_{ij}x^i\left(y+z+\sum_{l,k>o}a_{lk}y^lz^k\right)^j,$$ and also that all of the monomials of less than 3 variables cancel out.

Now, if one attempts to work this out degree by degree, one quickly sees that the number of coefficients gets enormous quite quickly. As far as I can tell, this is because they are controlled by certain kinds of partitions of integers. Let me make a definition:

Suppose we have a collection of $n$ indistinguishable red stones and $m$ indistinguishable blue stones. Then I'm interested in partitions of this collection which have a certain form. Specifically, the only monochrome blocks have precisely one stone, e.g. $r\vert r\vert b$ and $r\vert rb$ are valid partitions of $\{r,r,b\}$ but $rr\vert b$ is not. I'll call these "restricted two-colored partitions" for the purposes of this answer. I think there are things called colored partitions in combinatorics, and I don't know if they're connected to such ideas. It should be clear that a partition is unchanged by switching two stones of identical color, but changed into a new partition by switching two stones of different color. Let the set of all restricted partitions of of $n$ red stones and $m$ blue stones be denoted by $Q(n,m)$ (since $P(n,m)$ means something else!).

Now, in the above expression for $f\circ f$, let's attempt to determine the coefficient of a given monomial $x^\alpha y^\beta z^\gamma$. On the left hand side of the expression, we can fix $j=\gamma$ and on the right hand side we can fix $i=\alpha$. Let's start with $j=\gamma$. So we're interested in determining all the ways we can get $x^\alpha y^\beta$ out of the product $\left(x+y+\sum_{l,k>o}a_{lk}x^ly^k\right)^i$, and what the associated coefficients will be. I claim that these are determined by the set $Q(\alpha,\beta)$. That is, given a partition $P\in Q(\alpha,\beta)$, there is a map which takes a spot in the partition say, $\vert r^pb^q\vert$ to the monomial $a_{pq}x^py^q$, and multiplies all such spots together, then multiplies that entire monomial by the coefficient $a_{b_P,\gamma}$, where $b_P$ is the number of spots in the partition $P$. In other words, we've defined a map (which I claim is a bijection) $\phi:Q(\alpha,\beta)\to\{\mathrm{monomials~in~}x,y\mathrm{~of~degree~(\alpha,\beta)}\},$ where perhaps the target should be better explained, but I think it's clear. Thus if we sum all the coefficients that exist in the image of $\phi$, we'll get the total coefficient of $x^\alpha y^\beta z^\gamma$ coming from fixing the power of $z$, which I'll denote $A^{\alpha,\beta}_\gamma$. We must do an analogous process with fixed power of $x$, and get another coefficient $A^{\beta,\gamma}_{\alpha}$. So the coefficient of $x^\alpha y^\beta z^\gamma$ is $A^{\alpha,\beta}_\gamma-A^{\beta,\gamma}_{\alpha}$, where I've packed quite a bit of meaning into those little symbols.

Anyway, the reason we're having to use so-called "restricted colored partitions" is because our polynomials are restricted, in the sense that there are no monomials of the form $a_{n0}x^n$ in them, except when $n=1$. However, I suspect that this is really unnecessary, and that if we ignored this restriction, we could actually just get everything in terms of all possible colored partitions of $n$ blue stones and $m$ red stones, or whatever. The cool thing about that is that if you set all your coefficients to be $1$, I suspect you get a sort of generating function for these types of partitions, which happen to subsume non-colored partitions. So I imagine this sort of power series actually has some kind of combinatorial meaning, and I'd be surprised if it wasn't described somewhere in that discipline's literature, which I know nothing about.

Anyway, I'll leave this answer up, though it's pretty unsatisfying. Basically what it tells me is that this power series is really hard to deal with. I tried computing all the coefficients of the monomial $x^3y^3$ and it gets really big really fast. Luckily a huge amount of stuff cancels out, so when you just have a computer calculate these terms, they get (relatively...) simpler.

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