5
$\begingroup$

Let $p$ be a prime, let $\mathbb K$ a field with $\operatorname{char}(\mathbb K)=p$ and $\mathbb{K}(x)$ be the field of fractions of the polynomial ring $\mathbb{K}[x]$, i.e. $$\mathbb{K}(x)=\left\{\frac{g(x)}{h(x)}:g,h \in\mathbb{K}[x], h\neq 0\right\}.$$

My central question is this:

How to characterize elements of order 3 in $\mathbb{K}(x)$, that is, rational functions $f\in \mathbb{K}(x)$ such that $f^3(x):=f\circ f\circ f(x)=x$? Are there explicit formulas for such $f$ and/or conditions that (the coefficients of) $f$ must satisfy?

Also, any input on elements of finite order $n> 3$ would be greatly appreciated!

What I found out so far: In the case of linear fractional transformations, we can write $$f(x) = \frac{ax+b}{x+d}, \qquad a,b,d\in \mathbb{K},$$ and from the condition that $f^3(x)=x$ we can explicitly derive $$b=-a^2-ad-d^2,$$ thus giving us a very clear picture of how these elements look like. Unfortunately, my approach (essentially consisting of symbolic calculations in Sage) becomes impractical for $\mathrm{deg}(g), \mathrm{deg}(h)>1$.

Also there is this paper by E. F. Allen that characterizes linear fractional transformations of any order, but it only applies to polynomials with real coefficients.

Improve this question
$\endgroup$
11
  • 4
    $\begingroup$ I thought that by definition an involution has order two. $\endgroup$
    – Dave Benson
    Commented Mar 11 at 13:08
  • 4
    $\begingroup$ It is known that for every field $K$, the only $f\in K(t)$ that are invertible under composition are homographies $(at+b)/(ct+d)$ (which form the group $\mathrm{PGL}_2(K)$. So the whole picture reduces to classifying torsion elements (here, of order $3$) in this group. The latter follows from a discussion: char 3 vs char $\neq 3$, and whether $t^2+t+1$ is irreducible in $K[t]$. $\endgroup$
    – YCor
    Commented Mar 11 at 13:08
  • 3
    $\begingroup$ Since $\deg(f^n)=\deg(f)^n$, the assumption that $f^n(x)=x$ and $n\ge2$ implies that $\deg(f)=1$. For linear maps, you're asking for the elements if finite order in $\operatorname{PGL}_2(\mathbb K)$, which for order prime to $p$, can be analyzed by looking at the eigenvalues. And of course there are maps such as $f(x)=x+a$ that has order $p$. $\endgroup$
    – Joe Silverman
    Commented Mar 11 at 13:08
  • 2
    $\begingroup$ @Mersn Calling elements of order $3$ involutions is 100% inaccurate :-) By the way, your elements in $R$ are rational functions, you shouldn't call them polynomials. $\endgroup$
    – Peter Mueller
    Commented Mar 11 at 15:06
  • 3
    $\begingroup$ @NickS: This is the wrong way of thinking about this problem. Rational functions are not honest functions $f\colon \mathbb{K}\to \mathbb{K}$ (for example their denominators can have roots, so they are not defined for certain values of $x$); they are just elements of the ring $\mathbb{K}(x)$. $\endgroup$
    – Sam Hopkins
    Commented Mar 11 at 16:25

1 Answer 1

Reset to default
4
$\begingroup$

Since you ask, there has been much research on the analogous question of elements of finite order in the Nottingham group, which is the group of power series under composition: $$ N(\mathbb K) = \left\{ x + \sum_{n\ge2} c_nx : c_n\in \mathbb K\right\}. $$ especially for $\mathbb K$ a finite field. Your example $-\frac{2x}{\sqrt{x^2-4}}$ is almost of this form, it's expansion starts $ix+\text{h.o.t.}$ See this question, which is related to what you're asking about, and my answer to that question gives some references that discuss elements of finite order in the Nottingham group.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I understand this question is asking about positive characteristic, but is it not true that if $\mathbb{K}=\mathbb{C}$ then the only rational functions of finite order are Möbius transformations? $\endgroup$
    – Sam Hopkins
    Commented Mar 11 at 15:49
  • 1
    $\begingroup$ @SamHopkins This was pointed out by Joe Silverman in his comment for the OP's question, no matter what the field $\mathbb K$ ist. $\endgroup$
    – Peter Mueller
    Commented Mar 11 at 15:59
  • $\begingroup$ Okay, I see. Then I'm confused what the OP's question actually is. Doesn't that answer it? $\endgroup$
    – Sam Hopkins
    Commented Mar 11 at 16:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .