Timeline for Is $\mathbb{Q}$ the orbit of a rational function under iteration?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2022 at 9:34 | vote | accept | Ivan Meir | ||
| Jan 5, 2022 at 2:03 | answer | added | Joe Silverman | timeline score: 7 | |
| Jan 4, 2022 at 21:42 | comment | added | Emil Jeřábek | @LSpice The $x+1$ map takes $\infty$ to itself, the $-1/x$ map to $0$. Thus, if a path from one rational to another goes through $\infty$, that part of the path must be just $0\mapsto\infty\mapsto0$, and you can eliminate it. | |
| Jan 4, 2022 at 21:35 | comment | added | pregunton | @LSpice If at some point when we apply these generators in sequence we arrive at $\infty$ again, the product $\phi'$ of the remaining generators still sends $\infty$ to $q$, so WLOG we can choose $\phi$ so that on succesively applying the sequence of generators we will always stay inside $\mathbb{Q}$. Then $\phi_0 = S^{-1}\phi = T^aST^bST^c\ldots$ sends $0$ to $q$, as we wanted. | |
| Jan 4, 2022 at 21:35 | comment | added | pregunton | @LSpice I guess it's not completely obvious, but here is how one can see it: by transitivity, there is some $\phi \in SL(2,\mathbb{Z})$ sending $\infty$ to any rational number $q$. We can decompose it as $\phi = ST^aST^bST^c\ldots$, where $S : x\mapsto -1/x$ and $T : x\mapsto x+1$ (the first factor may be taken to to be $S$ WLOG, since $\infty+1=\infty$). | |
| Jan 4, 2022 at 20:14 | comment | added | LSpice | @pregunton, is it obvious that it is possible to generate $\mathbb Q$ by applying those maps only to elements of $\mathbb Q$ (that is, I suppose, that it is not the case that the unique way first to get to $0$ is as $-1/\infty$)? | |
| S Jan 4, 2022 at 17:06 | history | suggested | Dirk Werner | CC BY-SA 4.0 |
some typos
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| Jan 4, 2022 at 17:04 | review | Suggested edits | |||
| S Jan 4, 2022 at 17:06 | |||||
| Jan 4, 2022 at 15:34 | history | edited | Stopple | CC BY-SA 4.0 |
less -> fewer
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| Jan 4, 2022 at 11:50 | comment | added | Algernon | @IvanMeir I see! I thought by rational you meant $f:\mathbb{Q}\to\mathbb{Q}$. My bad. | |
| Jan 4, 2022 at 11:33 | answer | added | Wojowu | timeline score: 19 | |
| Jan 4, 2022 at 11:30 | comment | added | Joe Silverman | As Fedor says, a linear fractional transformation won't work. And if $f(x)\in\mathbb Q(x)$ has degree $d\ge2$ and $a\in\mathbb Q$ has infinite $f$-orbit, then the height $H\bigl(f^{\circ n}(a)\bigr)$ grows like $C^{d^n}$ for some $C>1$. Hence not only is the orbit not dense, it is in fact extremely sparse as a subset of $\mathbb Q$. | |
| Jan 4, 2022 at 11:25 | comment | added | Ivan Meir | @Algernon By rational function I mean a ratio of polynomials so not necessarily continuous but with at worst finitely many discontinuities. | |
| Jan 4, 2022 at 11:25 | comment | added | Wojowu | @Algernon All rational functions are continuous outside of a single set of poles. | |
| Jan 4, 2022 at 11:06 | comment | added | Algernon | Do you mean continuous rational functions? Without continuity, you can obviously construct a single function based on an enumeration of the rationals that does the job. | |
| Jan 4, 2022 at 11:01 | comment | added | Fedor Petrov | $f(x)=(ax+b)/(cx+d)$ does not work, and other functions do not seem to be surjective | |
| Jan 4, 2022 at 10:56 | comment | added | Roland Bacher | A unique such rational function (if it exists) induces cyclic permutations of all elements of the projective line over $\mathbb F_p$ for all (or perhaps almost all) primes 𝑝. I think this is quite an extraordinary property. | |
| Jan 4, 2022 at 10:41 | comment | added | Ivan Meir | Yes, great observation thank you. | |
| Jan 4, 2022 at 10:32 | comment | added | pregunton | It is possible to generate $\mathbb{Q}$ with the two functions $x \mapsto x+1$ and $x \mapsto -1/x$, as the modular group $PSL(2,\mathbb{Z})$ generated by these two functions acts transitively on $\mathbb{QP}^1$. | |
| Jan 4, 2022 at 10:26 | history | edited | Ivan Meir | CC BY-SA 4.0 |
Fixed typo
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| Jan 4, 2022 at 10:19 | history | asked | Ivan Meir | CC BY-SA 4.0 |