Analytic function avoiding elements of the modular group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:00:04Z http://mathoverflow.net/feeds/question/120369 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120369/analytic-function-avoiding-elements-of-the-modular-group Analytic function avoiding elements of the modular group Alexandre Eremenko 2013-01-30T23:18:24Z 2013-02-19T20:18:21Z <p>A friend recently told me the following two facts, for which he cannot recall a proof or a reference (but he remembers seeing them in the literature):</p> <ol> <li><p>Let $f$ be a holomorphic function mapping the upper half-plane $H$ into itself. Let $G$ be the group of fractional linear transformations $(az+b)/(cz+d)$ where $ad-bc=1$, $a,d$ are odd integers and $b,c$ are even integers. Suppose that for every $g\in G$ and for every $z\in H$, $f(z)$ is not equal to $g(z)$. Then $f$ is fractional-linear. (Or maybe such $f$ just does not exist).</p></li> <li><p>Let $f$ be the same as before. Suppose that for all integers $m,n$ and all $z\in H$ we have $f(z)\neq mz+n$. Then $f$ is fractional-linear.</p></li> </ol> <p>I will appreciate any relevant reference or any other information. </p> <p>EDIT: There is no $f\in Aut(H)$ that satisfies the condition of Problem 1. This implies that $f$ constructed by Aakumadula is NOT fractional-linear.</p> <p>To prove this, we write $(az+b)/(cz+d)=(xz+y)/(uz+t)$, where $a,b,c,d$ are given real numbers, and we want to find integers $x,y,u,t$, where $x,t$ are odd, and $y,u$ are even, so that this has non-real roots $z$. This is to show that certain quadratic form in $a,b,c,d$ is indefinite. And this is performed by an elementary calculation.</p> http://mathoverflow.net/questions/120369/analytic-function-avoiding-elements-of-the-modular-group/120519#120519 Answer by Junkie for Analytic function avoiding elements of the modular group Junkie 2013-02-01T13:29:42Z 2013-02-01T13:29:42Z <p>I think the idea is the Big Picard theorem.</p> <p>There is a similar problem I think in Halmos (Problems for Mathematicians Young and Old), concerning $z$, $f(z)$, and $f(f(z))$ being always distinct (it is phrased as saying that $f\circ f$ has no fixed points maybe). Then one shows $f$ is a translation. You go by forming the function $g(z)={f(f(z))-z\over f(z)-z}$ or the like, which now omits 0, 1, and $\infty$ from its image. After messing around (consider $g'(z)$ somehow?) this should give the desired result.</p> <p>Your problems seem of a similar flavor.</p> http://mathoverflow.net/questions/120369/analytic-function-avoiding-elements-of-the-modular-group/120644#120644 Answer by Aakumadula for Analytic function avoiding elements of the modular group Aakumadula 2013-02-03T03:06:29Z 2013-02-03T08:42:30Z <p>This is an answer to part of question 1. Such a function $f:H \rightarrow H$ does exist. To see this, let $z\in H$ and $P(w)$ be the meromorphic function on the plane which is the Weierstrass $P$ function corresponding to the lattice $L_z= {\mathbb Z}\oplus {\mathbb Z}z\quad$. I wish to add that $P(w)=P(w,z)\quad$ is a function of two variables. Let $x(w)= \frac{P(w)-P(1/2)}{P(z/2)-P(1/2)}=x(w,z)\quad$, and $y(w)$ a suitable multiple of $P'(w)$. Then we have the Legendre form of the equation of the elliptic curve<br> $$y^2=x(x-1)(x-\lambda (z)),$$ where $\lambda :H \rightarrow {\mathbb P}^1\setminus {0,1,\infty } \quad$ is the Picard covering map. The deck transformation group is precisely $G$ (modulo $\pm 1\quad$). </p> <p>Then, by the properties of the elliptic function $P(w)$, the function $P(w)-P(1/2)\quad$ has a double zero at the $2$ division point $1/2$ and hence does not vanish anywhere else. Similarly for $z/2$ and $(1+z)/2\quad$. Consequently, if we specialise $w=z/3$, then the function $x(z/3)$ does not take the value $0,1,\lambda (z) \quad$. By the lifting criterion, $$x(z/3)=x(z/3,z)= \lambda (f(z))$$ for some $f:H \rightarrow H \quad$. Clearly, $f(z)\neq g(z)\quad$ for any $z\in H\quad$ and for any $g\in G\quad$ where $G$ is the congruence subgroup of level $2$, since their lambda values are distinct. </p> <p>You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0&lt; a,b &lt; 1/2\quad$. It seems to me that not all these functions $z\mapsto \lambda^{-1}(x(az+b))\quad$ can be fractional linear. </p> http://mathoverflow.net/questions/120369/analytic-function-avoiding-elements-of-the-modular-group/120728#120728 Answer by Alexandre Eremenko for Analytic function avoiding elements of the modular group Alexandre Eremenko 2013-02-04T04:16:53Z 2013-02-19T20:18:21Z <p>Here is a slightly different version of Aakumadula's answer to my question.</p> <p>Let's say that $f$ omits $g$ if $f(z)$ is never equal to $g(z)$.</p> <p>Let $G$ be the group of fractional linear transformations such that the unit disc $U$ modulo $G$ is C\ { 0,1 } . (Can anyone suggest a short and recognizable name for this group?? This is a truly fundamental object of complex analysis, and the shortest name for it that I know is the "principal congruence subgroup of level 2 of the modular group". Sounds scary for many people). </p> <p>Proposition. TFAE: There exists $f$ holomorphic in $U$ that omits $0,1,\infty$ and $\lambda(z)$, and: there exists $g: U\to U$ that omits all elements of $G$.</p> <p>Proof. $f=\lambda\circ g$.</p> <p>Now it is well-known that there exists $f$ holomorphic in $U$ which omits $0,1,\infty$ and $\lambda$. This is by "extension of holomorphic families of injections" of Slodkowski.</p> <p>The theorem of Slodkowski says that whenever you have any number of holomorphic functions in $U$ with disjoint graphs, you can add one whose graph is disjoint from those given functions. And even prescribe the value of this one added function at one point.</p> <p>Of course, Aakumadula's answer is better because it gives an explicit construction. This new answer shows that this is a special case of a well-known and important general principle. Somehow I did not figure this out before the Aakumadula's answer.</p> <p>EDIT on Feb 19 2013. Multi-valued function $f(z)=\sqrt{z}$ omits $0,1,\infty$ and $f(z)=z$ has no solutions in $C\backslash${0,1}. Thus the composition of $f$ with the universal cover {${ |z|&lt;1}$} to $C\backslash${0,1} omits $0,1,\infty$ and all elements of the Schwarz's group.</p> http://mathoverflow.net/questions/120369/analytic-function-avoiding-elements-of-the-modular-group/122167#122167 Answer by ahmed sulejmani for Analytic function avoiding elements of the modular group ahmed sulejmani 2013-02-18T13:34:52Z 2013-02-18T13:34:52Z <p>Concerning question 2 see Earle, Clifford J. On holomorphic families of pointed Riemann surfaces. Bull. Amer. Math. Soc. 79 (1973), 163–166 and Theorem 3 there. </p>