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Let $\hat{\mathbb{R}}$ be $\mathbb{R} \cup \{ \infty \}$, $\hat{\mathbb{I}} = i \hat{\mathbb{R}}$ and $\hat{\mathbb{C}}$ be the Riemann sphere.

Let $P$ be the set of all meromorphic functions from $\hat{\mathbb{C}} \to \hat{\mathbb{C}}$ such that for all $p \in P$, $p: \hat{\mathbb{R}} \to \hat{\mathbb{R}}$ is bijective and $p: \hat{\mathbb{I}} \to \hat{\mathbb{I}}$ is bijective.

It is clear that multiplication by a real non-zero scalar is in $P$ and so is reciprocation. But are all functions in $P$ of the form $x z^{\pm 1}$ (where $x$ is real)? How can I prove this?

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I assume you mean $p$ is a holomorphic function. What do you mean by $p : \hat{\mathbb{R}} \to \hat{\mathbb{R}}$? And what do you mean by scalar multiplication or inversion for a function that takes values in an arbitrary compact Riemann surface? Do you actually want $\mathcal{M}$ to be the Riemann sphere? –  Qiaochu Yuan Jan 3 '12 at 5:24
    
Also, it's mildly confusing to write $x$ for a real scalar since $x$ typically denotes the real part of $z$. –  Qiaochu Yuan Jan 3 '12 at 6:01
    
Doh! How stupid of me. $\mathcal{M}$ should indeed be $\hat{\mathbb{C}}$. Thankyou! It is now fixed. –  Alexander Jan 5 '12 at 15:54
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You still don't need to say "meromorphic." A holomorphic function from the Riemann sphere to itself is precisely a meromorphic function on $\mathbb{C}$ which has a pole of finite order at infinity. –  Qiaochu Yuan Jan 5 '12 at 18:06
    
@QiaochuYuan: but what if he allows poles somewhere else? –  savick01 Jan 5 '12 at 19:14

1 Answer 1

up vote 2 down vote accepted

(1) Note that $z^{2k+1} \ (k\in\mathbb{Z})$ is ok. It is obviuosly ok for for $z\in \hat{\mathbb{R}}$ and for $z\in \hat{\mathbb{I}}$ we get $+/- \ iy^{2k+1}$, which is also bijective.

(2) Moreover if $f$ is ok, then $\frac{1}{f}$ is ok, too, so from now I will consider functions that are finite at zero.

(3) Let's try to add some functions defined in (1). $f(z)=\sum r_kz^{2k+1}$. For $z=yi$ we get $$f(z)=ig(y)=i\sum (-1)^k r_ky^{2k+1}$$ and for $z=x$: $$f(z)=h(x)=\sum r_kx^{2k+1}.$$ So for the function $f$ to be bijective for $iy$ and $x$, we need $h$ and $g$ to be bijective (note that by $f,g,h$ I mean some series or their extensions).

Note that $g,h$ do not have poles in finite points, because those functions are odd so a pole in finite $a$ implies a pole in $-a$ and we don't get a bijection.

(4) From (3) it is clear that a nonempty sum (finite) of functions of the form (a) $r_kz^{4k+1}$ or (b) $ r_kz^{4k+3}\ (k \in \mathbb{N_0}, r_k \neq 0)$ is ok assuming that all $r_k$ are positive or all are negative (all the summons are decreasing or all are increasing). In that case a divergent series means a pole (all the summons are of the same sign), and we know that there are no poles, so one series can be used for the whole line.

(5) Can we use in our series $z^n$ where $n$ is even? No. In that case $f(yi) \in \hat{\mathbb{I}}$ would imply, that the part of the (finite or infinite) series where $z$ comes in even powers is constantly equal to zero. Such a series is a zero-series.

(6) Can we use non-real $r_k$? No. Since $h(x)$ is a real function all its derivatives are real, so all $r_k$ are real.

The remaining part is the question, wheather all the functions are of the form (a) or (b) defined in (4). Unfortunately not. A simple example is $z - (2 z^3)/3 + z^5/5$ (its derivative $z^4 - 2z^2 + 1$ is nonnegative for real $z$ and at least one for $z \in \hat{\mathbb{I}}$ so $g$ and $h$ are monotone). The assumption that $r_k$ in (a) or (b) are of a constant sign is also too strong: $z - z^5/5 + z^9/9$ (its derivative is $1 - z^4 + z^8$).

It looks like the condition for $g,h$ to be bijective is just that their derivatives are of a constant sign. The derivatives of $g$ and $h$ are in fact equal to the derivative of $f$, so our condition (with (2) in mind) is: $f(0) = 0$ and (finite) $f'$ has constant sign on $\mathbb{R}$ and on $\mathbb{I}$ (it must be real, so the condition that we do not use even powers is hidden here except for the power zero which is written explicitly). I can't think of something more simple. Maybe a diffrent approach is needed.

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