I hope I am forgiven for my noob question. But, does it make sense to think of Julia sets using other fields? More precisely I would like to think of fields in which closed and bounded isn't necessarily a compact set. I am not sure what this will give us, but some results that we know in complex numbers wouldn't hold (e.g. will the Julia sets and the filled Julia sets still remain compact and nonempty?).
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$\begingroup$ Silverman's book "The Arithmetic of Dynamical Systems" deals with fields of characteristic p. In that situation there are maps with empty Julia sets and various other properties that differ from the situation over the complex numbers. $\endgroup$– FabianNov 16, 2011 at 10:19
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2$\begingroup$ @Fabian I don't think Silverman's book does any local field of characteristic p. He certainly does the p-adics, but these have characteristic zero. They are also locally compact, so closed and bounded is compact there too. But, yes, sometimes Julia sets are empty and other things change. $\endgroup$– Felipe VolochNov 16, 2011 at 10:22
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$\begingroup$ Indeed. I guess I got carried away by the reductions mod p. $\endgroup$– FabianNov 16, 2011 at 10:35
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$\begingroup$ I mean, I think, its easy to restrict the julia set to the real algebraic numbers and arrive to some non-compact set. But why would it be interesting to look at Julia sets this way? $\endgroup$– Jose CapcoNov 16, 2011 at 11:57
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3$\begingroup$ More promising, perhaps, would be fields $\mathbb C_p$. Complete metric, algebraically closed, but NOT locally compact. Why not start by studying the maps $z^2+c$ which were so interesting in $\mathbb C$... $\endgroup$– Gerald EdgarNov 16, 2011 at 13:41
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3 Answers
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As noted, $\mathbb{Q}_p$ and its finite extensions are locally compact, but the Julia set is often empty. Indeed, in this case that the Fatou set is always non-empty, another difference from the compact case. However, since $\mathbb{Q}_p$ isn't algebraically closed, one might best compare dynamics over $\mathbb{Q}_p$ as being analogous in some ways to dynamics over $\mathbb{R}$. So the analogue of $\mathbb{C}$ is $\mathbb{C}_p$, the completion of the algebraic closure of $\mathbb{Q}_p$. Unfortunately, $\mathbb{C}_p$ is not locally compact (and of course, totally disconnected), so one can't use measure-theoretic arguments. For example, it's not easy to make sense of equidistribution. The modern solution is to instead look at Berkovich space. This is a locally compact and connected space that includes $\mathbb{P}^1(\mathbb{C}_p)$ as a sort of boundary. Two good introductions to Berkovich spaces are listed below. And in Berkovich space over $\mathbb{C}_p$, we're back to the situation where the Julia set is always non-empty.
I'll mention one other interesting difference between the complex and $p$-adic cases. A famous theorem of Sullivan says that a rational map has no wandering domains in $\mathbb{P}^1(\mathbb{C})$. In opposition to this, Benedetto has constructed rational maps that do have wandering domains in $\mathbb{P}^1(\mathbb{C}_p)$. However, it is not known if wandering domains can exist in $\mathbb{P}^1(\mathbb{Q}_p)$.
answered Nov 16, 2011 at 14:29
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When you say "other fields", I assume that you're thinking that $\mathbb{C}$ is where "ordinary" Julia sets live. If you're interested only in Julia sets of polynomials (over $\mathbb{C}$) then that's not a bad point of view. On the other hand, if you're interested in Julia sets of rational functions (over $\mathbb{C}$) then the ambient space that you're working with is really $\mathbb{C} \cup \{\infty\}$, the Riemann sphere. Of course, this isn't a field at all.
If you have a look at Milnor's book Dynamics in One Complex Variable, you'll see Julia sets developed not for arbitrary fields, but for arbitrary Riemann surfaces. So, the general situation is that you have a Riemann surface $X$ and a holomorphic map $f: X \to X$; any such $f$ has an associated Julia set $J(f) \subseteq X$. Taking $X$ to be the Riemann sphere, this means that $f$ is a rational function and $J(f)$ is the "ordinary" Julia set.
answered Nov 16, 2011 at 15:03
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You may also construct Julia sets in the Quanternions $\mathbb{H}$. I guess that not all results about the Julia sets in $\mathbb{C}$ hold since this algebra is not kommutativ. I am sorry not to know details, but a think there are paper..
answered Sep 10, 2012 at 15:21
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$\begingroup$ The obvious 'Mandelbrot set' over the quaternions is somewhat trivial: It is obtained by rotating the ordinary Mandelbrot set along imaginary directions. I guess Julia sets might be somewhat more interesting. $\endgroup$ Mar 29, 2022 at 15:27