Apologies if the question is too elementary/something well-known.

I believe it is a well-known fact that the rational formal power series $F(z)=\frac{P(z)}{Q(z)}$ which have finite order under composition are (certain) Möbius transformations.

(By finite order I mean $F^{\langle n \rangle}(z) = z$ for some $n$, where the exponent denotes composition. Of course for this to be defined we should assume the power series has zero constant term.)

Is there some kind of classification of algebraic formal power series with finite order under composition?

EDIT: By "algebraic power series" I mean $P_n(z)F(z)^n + P_{n-1}(z)F(z)^{n-1} + \cdots + P_{0}(z) = 0$ for some $n\geq 1$ and polynomials $P_{n}(z), \ldots, P_0(z)$ (not all zero), where now the multiplication is the usual multiplication of power series, not composition: see e.g. Stanley's EC2, Definition 6.1.1. In certain contexts algebraic power series are the next natural step after rational power series.

EDIT 2: I guess a more precise way to formulate the question would be, say someone hands you an algebraic power series. Can you easily decide if it has finite order? One issue with this formulation is how do we “present” algebraic power series: with rational power series it is clear how we can compactly encode them, but with algebraic you need a little more than the minimal polynomial because there will be multiple roots in general.

  • $\begingroup$ I guess I'm being vague about the coefficients here, but feel free to assume whatever (e.g. $\mathbb{C}$) if it's convenient... $\endgroup$ Jul 1 at 19:53
  • $\begingroup$ I'm confused by your notion of algebraic. One interpretation, surely the wrong one, is that you mean that $P(x) \in \mathbb{C}[x]$ (or $\mathbb{C}$ replaced by your ground field), and that $P(F(z))$ means the composition $(P \circ F)(z)$ of power series, and that $=0$ means "is the zero power series". But then the only "values" of $F$ are roots of $P$, and the roots are discrete. So $F$ is a constant valued at some root of $P$. (I put "values" in quotes because I am well aware that $F$ is merely a power series. But this argument can be implemented by first looking at $F(0)$ and then looking... $\endgroup$ Jul 1 at 22:15
  • $\begingroup$ ... at derivatives.) $\endgroup$ Jul 1 at 22:15
  • $\begingroup$ Since I am confident that you mean a nontrivial notion of "algebraic", you must mean something other than that $P(x) \in \mathbb{C}[x]$. Since you mention rational functions, I suspect that you mean that $P(x) \in R[x]$ where $R = \mathbb{C}[z]$? So for example any rational function is the root of a "linear" equation. $\endgroup$ Jul 1 at 22:17
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    $\begingroup$ @TheoJohnson-Freyd: Sorry, the notion of "algebraic formal power series" is very standard in combinatorics but I wrote something slightly wrong. I will fix it. For example, the ordinary generating function $C(x) = \sum_{n \geq 0} C_n x^n$ of the Catalan numbers is algebraic since $C(x) = 1+xC(x)^2$. $\endgroup$ Jul 1 at 22:19

2 Answers 2


The well-known fact you cite is easy to prove, since for a rational function $\deg(F^n)=(\deg F)^n$, so if $F$ has finite order, then $\deg F=1$. You don't mention the characteristic of your field $k$ of coefficients. In the case that $k=\mathbb F_p$, the group in question is called the Nottingham group, where Wikipedia notes that "for every finite group of order a power of p there is a closed subgroup of the Nottingham group isomorphic to that finite group." In particular, for every $n\ge0$ there is an element of order $p^n$. However, finding a natural construction for such elements is non-trivial, see for example "Automata and finite order elements in the Nottingham group" (J. Algebra 602 (2022), 484–554). Another good reference is "Torsion in the Nottingham group" (Bull. Lond. Math. Soc. 43 (2011), no. 3, 547–560).

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    $\begingroup$ @SamHopkins Sorry, I did not understand what you meant by "algebraic" power series. So I thought you were asking about elements of finite order in the general Nottingham group $z+z^2k[\![z]\!]$, and I was attempting to answer that at least for $k$ a finite field, the answer is complicated, and to provide some references. But now I see from your edit what you meant by asking for algebraic power series. Of course, your algebraic power series of finite order will live in the NG, so the references may still be helpful. Anyway, if you feel my answer isn't useful, let me know and I'll delete it. $\endgroup$ Jul 2 at 0:37
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    $\begingroup$ No, please don't delete it, it is still useful! $\endgroup$ Jul 2 at 0:38

As I mentioned in a comment, the paper Elements of Finite Order in the Group of Formal Power Series Under Composition by Marshall M. Cohen explains that over $\mathbb{C}$ every formal power series of finite compositional order is conjugate to $\omega z$ for some root of unity $\omega$ (and probably this is a "classical" result).

This is perhaps a kind of answer to my question.

But it also suggests that there is probably no hope of classifying (or recognizing) algebraic power series of finite order, because if we have any algebraic power series $G(z)$ for which $G^{\langle -1 \rangle}(z)$ is defined then $G(z) \circ \omega z \circ G^{\langle -1 \rangle}(z)$ will have finite order and be algebraic.

(Here we use that the composition, and compositional inverse, of algebraic power series remains algebraic, which I think should be true. Of course the compositional inverse of a rational power series is not rational and that is the key difference with algebraic ones.)

E.g., with $G(z) = 1-\sqrt{1+z}$ so $G^{\langle -1 \rangle}(z) = z^2-z$, we can set $$F(z) = (1-\sqrt{1+z}) \circ (-z) \circ (z^2-z) = 1-\sqrt{-z^2+2z+1}$$ and then $F^{\langle 2 \rangle}(z)=z$.


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