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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):

  1. 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).

  2. 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.

I will appreciate any relevant reference or any other information.

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.

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.

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Thanks Alexandre. I tried to prove that the map $f$ was not fractional-linear, but could not do it. –  Venkataramana Feb 5 '13 at 5:40

4 Answers 4

up vote 4 down vote accepted

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
$$y^2=x(x-1)(x-\lambda (z)),$$ 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 $\pm 1\quad$).

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.

You can replace $z/3\quad$ by any element of the form $w=az+b \quad$ for $0< a,b < 1/2\quad $. It seems to me that not all these functions $ z\mapsto \lambda^{-1}(x(az+b))\quad $ can be fractional linear.

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$z\mapsto x(z/3):H \rightarrow {\mathbb P}^1\setminus \{0,1,\infty\}$ is a holomorphic map from a simply connected space, and hence, by the lifting criterion lifts to a map $f:H\rightarrow H$ to the universal cover such that $ x(z/3)=\lambda (f(z))$. –  Venkataramana Feb 3 '13 at 4:35
    
The above was in answer toa query by the OP as to why this was true, but he seems to have deleted his query. –  Venkataramana Feb 3 '13 at 4:45
    
Alexandre, Thanks for the edit: i meant $z\mapsto \lambda ^{-1}(x(az+b))$. –  Venkataramana Feb 3 '13 at 5:13
    
I deleted my comment because I understood why it was true:-) It is indeed unlikely that your function $f$ is fractional-linear, but I would like to see a proof. –  Alexandre Eremenko Feb 3 '13 at 5:22
    
I think I can prove that there are no fractional-linear $f$ satisfying the conditions. This implies that Aakumadula's example is not fractional-linear. –  Alexandre Eremenko Feb 3 '13 at 15:31

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.

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Thanks! This is the reference I was looking for. –  Alexandre Eremenko Feb 19 '13 at 2:33

Here is a slightly different version of Aakumadula's answer to my question.

Let's say that $f$ omits $g$ if $f(z)$ is never equal to $g(z)$.

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).

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 $G$.

Proof. $f=\lambda\circ g$.

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.

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.

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.

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 {$\{ |z|<1\}$} to $C\backslash${0,1} omits $0,1,\infty$ and all elements of the Schwarz's group.

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as to naming the group $G$, how about "the little Picard group"? There is a Picard modular group (a discrete subgroup of $SU(2,1)$), and "the Picard group" in algebraic geometry, but no "little Picard group", as far as I know. –  Venkataramana Feb 6 '13 at 14:22
1  
Sounds nice. Did you invent this, or someone used this in the literature? I believe it was Schwarz who introduced it in function theory. And it is Schwarz who constructed it from tiling by triangles with zero angles. –  Alexandre Eremenko Feb 7 '13 at 4:19
    
That bit about Schwartz is very interesting. I thought this would be a good name since it is used in the proof of "little Picard theorem" as well. I don't know if anyone else thought of it, but I would hesitate to claim credit for this! –  Venkataramana Feb 7 '13 at 13:04
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OK, next time we use it, let's call it the Schwarz group, and see whether this can become a standard name. (There is a funny story how Picard proved his Little Theorem. He was related to Hermit (by marriage) and once he dined with his relative. Picard told to Hermit that he was attending a lecture of Weierstrass, who asked whether an entire function can omit two values. Hermit agreed that this is an interesting problem, and explained to Picard another piece of news: how Schwarz constructed his Schwarz modular function using the reflection principle... Picard left the dinner deep in thought :-) –  Alexandre Eremenko Feb 7 '13 at 20:40
    
Very interesting story! –  Venkataramana Feb 8 '13 at 1:37

I think the idea is the Big Picard theorem.

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.

Your problems seem of a similar flavor.

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Picard's theorem is about functions analytic in the whole plane, or in a neighborhood of an isolated singularity. In the context of the problem Picard's theorem gives nothing. –  Alexandre Eremenko Feb 1 '13 at 14:37

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