My REU partner and I are working on a problem involving iterations of quadratic rational maps over an algebraically closed field $K$ that is complete with respect to a non-trivial non-archimedean absolute value $|\cdot|$. We've reduced it to the problem of determining whether $\phi(x) = k(x + x^{-1})$ is post-critically finite for $0 < |k| < 1$ (that is, determining whether the critical points $\pm 1$ are preperiodic). Our conjecture maintains that $\phi$ is not post-critically finite, but this is where we're stuck.

A simple strategy would be to show that some $n$-periodic point $\gamma$ of $\phi$ strictly attracts a critical point of $\phi$. According to a recent paper (Theorem 1.5, p. 4), if $p$ is the residue characteristic of $K$ and either $p=0$ or $p>2$ (assumptions we're willing to make), then a critical point of $\phi$ will be attracted to the cycle containing an $n$-periodic point $\gamma$ of $\phi$ if the multiplier $m(\phi^n,\gamma)$ is strictly between $0$ and $1$.

Unfortunately, the multipliers of the fixed points $\pm\sqrt{k/(1-k)}$ and $\infty$ are $2k - 1$ and $1/k$, respectively, each of which is at least $1$ in absolute value. The periodic points of minimal period $2$, $\pm\sqrt{-k/(1+k)}$, each have a multiplier of $(2k+1)^2$, which is also at least $1$ in absolute value. We're hesitant to continue this method of checking individual periodic points since the calculations become increasingly cumbersome for minimal period greater than $2$. Does anyone have suggestions for how to either: a) show that a periodic point exists with a multiplier in the desired range, or, more broadly, b) show that $\phi$ is(n't) post-critically finite?

  • $\begingroup$ If $k=1/2$, then $\pm 1$ are fixed points. I suspect a typo here? $\endgroup$ – David E Speyer Jul 26 '15 at 23:19
  • $\begingroup$ How is $2k-1$ greater than $1$ in absolute value? Is $k$ negative or complex, or do you mean a non-Archimedean absolute value $\alpha ^ {v(\cdot)}$? $\endgroup$ – Douglas Zare Jul 27 '15 at 2:52
  • $\begingroup$ @DouglasZare nonarchimedean. I state it in the first sentence, which is quite a wordy sentence so you may have missed it. $\endgroup$ – Justin Case Jul 27 '15 at 3:11
  • $\begingroup$ I also missed it. Of course, there are no non-archimedean norms for which $|1/2|<1$. $\endgroup$ – David E Speyer Jul 27 '15 at 3:26

I have found a counterexample. Take $K = \mathbb{Q}_2$. Let $f(t)$ be the polynomial $2 - t + 6 t^2 + 4 t^3 + 8 t^4$. By looking at the Newton polytope, $f$ has a root in $\mathbb{Q}_2$ of norm $1/2$ (so $2$-adic valuation $1$). Let $k$ be that root.

Then I compute that $f(f(f(f(1))))=f(1)$, where $f(x) = k(x+x^{-1})$. This was found by writing a little Mathematica script to determine the algebraic condition on $k$ which implies that $f^m(k) = f^n(k)$ and looking through the output until I saw one whose Newton polytope (for some $p$) was consistent with $|k|<1$.

I have also found examples with other residue characteristic.

I feel guilty about possibly spoiling the fun of finding this yourself, as this is an REU. But I think there is a lot of fun to be had by (1) learning about Newton polytopes and non-archimedean fields, if you haven't yet and (2) writing a similar script for yourself, and looking for patterns in the output. I suspect there is a lot hiding in that data.

  • $\begingroup$ I wasn't familiar with Newton polytopes, so this gives me some interesting topics to look into! $\endgroup$ – Justin Case Jul 27 '15 at 4:11
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    $\begingroup$ Hensel's lemma is enough in this case: Make the change of variables $t=2u$ and note that the root $u=1$ of $1 - u + 12 u^2 + 16 u^3 + 64 u^4$ lifts to $\mathbb{Q}_2$ by Hensel. But you should learn about Newton polytopes! They make spotting this sort of thing obvious, and they are a lot of fun. $\endgroup$ – David E Speyer Jul 27 '15 at 4:31

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