1
$\begingroup$

Let $f:\mathbb{C}\to\mathbb{C}$ be meromorphic or even entire. Let $z:\mathbb{C}\to\mathbb{C}$ be such that it holds $$z(t+1) = f(z(t)).$$ If $f$ is entire and we choose carefully $z$ can be constructed to be also entire. This is done usually by choosing an unstable direction of a fixed point and then propagate analyticity using $f$.

So my question is the following: Given $f$ can we find $g$ such that $$z(t+i) = g(z(t))?$$ Is this something known? Does this $g$ depend on $z$? Is it unique? What can we say about $g$ without knowing $z$ explicitly?

The obvious property of $g$ is that it "commutes" with $f$, i.e. $f\circ g = g\circ f$. What else can we say?

One way to look at the function $z$ is as a way to conjugate $f$ to the map $t\mapsto t+1$. So $g$ is conjugated to $t\mapsto t+i$ which is orthogonal to $t\mapsto t+1$, hence the name.

I am interested in this because it changes fixed points that are centres to nodes.

For a trivial example let's consider the map $f(z)=\frac{z}{1-z}$. The function $\Phi(z)=-z^{-1}$ conjugates it to a translation, i.e. $$ \Phi\circ f\circ\Phi(z) = z+1. $$ Let $T(z)=z+i$, then $$ \Phi\circ T\circ\Phi(z) = \frac{i z}{i+z}. $$

Improve this question
$\endgroup$
3
  • $\begingroup$ Is there any reason why $z(t+i)$ would only depend on $z(t)$? If not, then it would seem that typically there won't be such a $g$. $\endgroup$
    – Christian Remling
    Commented Jul 24, 2017 at 0:45
  • $\begingroup$ The function $z(t)$ is the solution in the sense that it conjugates $f$ to the translation $t \to t+1$. Given any point in $\mathbb{C}$ with some mild conditions, there is always a solution $z(t)$ locally. The example that I wrote above is special not only because $z$ can be written explicitly, but also because it is unique. In general this doesn't have to be the case. That's why I'm asking about uniqueness. $\endgroup$
    – tst
    Commented Jul 24, 2017 at 2:27
  • $\begingroup$ Why the downvote? Care to explain? $\endgroup$
    – tst
    Commented Jul 24, 2017 at 14:07

1 Answer 1

Reset to default
3
$\begingroup$

If $f$ is rational, then $g$ is also rational. It is a very rare, exceptional situation when such two functional equations are satisfied. All such cases have been explicitly described by J. Ritt in Permutable rational functions, Trans. Amer. Math. Soc. 25 (1923), no. 3, 399–448.

If $f$ is entire transcendental, the situation is unclear. No single example (of such pair $f$, $g$) is known to me. Under some strong a priori conditions on $f$, I. N. Baker proved that this is impossible.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Wow, thanks! That is more than what I hoped for. One question on that. What if we look at it locally? Is it known if there is always a $g$ such that the above is true in a neighbourhood of a fixed point? $\endgroup$
    – tst
    Commented Jul 24, 2017 at 14:06
  • 1
    $\begingroup$ @tst: No. This was proved by Ecalle MR0340569 and Liverpool MR0364611. $\endgroup$
    – Alexandre Eremenko
    Commented Jul 24, 2017 at 22:20

You must log in to answer this question.

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