Rationality of meromorphic functions between real algebraic sets in the plane

In 1907, Poincaré [22] proved that if a biholomorphic map defined in an open set in $\mathbb{C}^{2}$ maps an open piece of a three dimensional sphere into another, it is necessarily a rational map. This result was extended by Tanaka [26] and then Alexander [1] (for holomorphic maps) to real spheres in higher dimensions and Forstnerič [11] gave a uniform bound on the degree of the rational maps. Webster [27] then discovered a general algebraicity phenomenon for holomorphic mappings between open pieces of algebraic Levi nondegenerate hypersurfaces in the complex Euclidean $m$-space $\mathbb{C}^{m}$, $m\geq 2$. Webster’s result leads to the following general problem:

Under what conditions must a holomorphic mapping $f:\mathbb{C}^{m}\to\mathbb{C}^{n}$ sending a real algebraic set $A\subset\mathbb{C}^{m}$ onto another real algebraic set $A^{\prime}\subset\mathbb{C}^{n}$ be algebraic ?

Here, a subset $A\subset\mathbb{C}^{m}$ is a real algebraic set if it is defined by the vanishing of real-valued polynomials in $2m$ real variables and we shall always assume that $A$ is irreducible. The definition of algebraicity of holomorphic mappings can be found for example in [2].

When $m,n\geq 2$, there is a substantial literature related to the above problem, see for example, [1],[21], [11], [13],[2], [14],[16] and [18]. In this paper, we will investigate this problem for the case $m=n=1$ which to the best of our knowledge, has not been studied in the literature, may be due to the fundamental difference between the geometric complex analysis in dimension $>1$ and the one-variable theory (see the comments in page 826 of [7]). To be more precise, suppose $f$ is a meromorphic function in the complex plane $\mathbb{C}$ and $\mathcal{C}$ is a real algebraic curve in $\mathbb{R}^{2}\cong\mathbb{C}$. We would like to know, for a real algebraic set $A\subset\mathcal{C}$, when $f(A)$ can lie in a real algebraic curve in $\mathbb{R}^{2}$. For the case that $f$ is a polynomial or rational function, it is known that $f(\mathcal{C})$ is lying in some real algebraic curve in $\mathbb{R}^{2}$. The following rationality result shows that $f$ has to be a rational function if $f(\mathcal{C})$ is assumed to be real algebraic and $\mathcal{C}$ is the unit circle $\mathbb{S}^{1}=\{z\in\mathbb{C}:|z|=1\}$. Moreover, the second part of the result can be considered as a one dimensional analog of Poincaré and Tanaka’s rationality results ([22] and [26]).

Notice that in Poincaré, Tanaka and Alexander’s rationality results mentioned above, the mappings involved are only required to be biholomorphic or simply holomorphic in an open set containing a piece of a $2m-1$ dimensional sphere in $\mathbb{C}^{m}$. On the other hand, Theorem 1.1 requires the map to be meromorphic in the whole complex plane. This is because the rationality result is not true if the map is defined only in an open set containing a piece of the circle $\mathbb{S}^{1}$ as can be seen from the following

The main tool of proving Theorem 1.1 and other results in this paper are the Schwarz functions coined by Davis in [9] (or the Schwarz reflection functions by Davis and Pollak [8]). One can also find a generalization of Schwarz function to higher dimensions in [25].

From Chapter 6 of [9], we know that if $\Gamma$ is an arc of a circle with center at $z_{0}$ and radius $r$, then $S_{\Gamma}(z)=\overline{z_{0}}+\frac{r^{2}}{z-z_{0}}$. When $\Gamma$ is an arc of a straight line passing through $z_{1}$ and $z_{2}$ in $\mathbb{C}$, then $S_{\Gamma}(z)=\frac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}(z-z_{2})+\overline{z_{2}}$. These are the only possibilities for $S_{\Gamma}(z)$ to be a rational function ([8] or pages 104-106 of [9]).

We remark that Schwarz functions may have a branch point at infinity. If there exists no branch point at infinity, Shapiro [24] and Millar [17] have proven that if $\Gamma$ is a simple analytic closed curve such that $S_{\Gamma}$ is analytic outside $\Gamma$ (which is also analytic at infinity), then:

1. (i)

   when $S_{\Gamma}(z)=O(z)$ as $z\to\infty$, $\Gamma$ is an ellipse;

2. (ii)

   when $S_{\Gamma}(z)=O(1)$ as $z\to\infty$, $\Gamma$ is a circle;

3. (iii)

   when $S_{\Gamma}(z)=O(\frac{1}{z})$ as $z\to\infty$, $\Gamma$ is a circle centred at the origin.

For a real algebraic planar curve $\mathcal{C}$, by construction $S_{\mathcal{C}}$ is an algebraic function (see Proposition 1). Then there exists a slit or simple curve $\gamma=\gamma(t):[0,\infty)\rightarrow\mathbb{C}$ tending to $\infty$ as $t$ goes to infinity and each analytic branch of the Schwarz function $S_{\mathcal{C}}$ exists in $\mathbb{C}\backslash\gamma$. Here, we slightly abuse the notation and we still denote each branch of $S_{\mathcal{C}}$ in $\mathbb{C}\backslash\gamma$ by $S_{\mathcal{C}}$. Moreover, by the theory of Puiseux series, we have

for some $c\in\mathbb{C}\cup\{\infty\}$.

From the expression of the Schwarz function for a circle, it is then clear that the following result generalizes the first part of Theorem 1.1 from the unit circle $\mathbb{S}^{1}$ to a more general real planar algebraic curve $\mathcal{C}$.

Recall that the Schwarz function for a line passing through $z_{1}$ and $z_{2}$ in $\mathbb{C}$ is given by $S(z)=\frac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}(z-z_{2})+\overline{z_{2}}$ which does not satisfy condition (1.1). The following example shows that condition (1.1) in Theorem 1.3 is essential.

The functions in Example 1.5 belong to the so-called class $W$ (like Weierstrass), which consists of elliptic functions, rational functions of one exponential $\exp(\alpha z),\alpha\in\mathbb{C}\backslash\{0\}$ and rational functions in $z$. These are the only meromorphic functions which satisfy the addition formulae. They are also the only meromorphic solutions of certain non-linear complex differential equations (see [10],[4], [5], [6], [19] and [12]).

From Examples 1.5, one may ask if there exists some non-rational functions in class $W$ that can map real irreducible planar algebraic sets (other than the straight lines) to another real planar algebraic set. The following result shows that this is impossible.

Theorem 1.6 shows that for any transcendental $f$ in class $W$, $f(\mathbb{S}^{1})$ cannot be part of a real planar algebraic curve.

Let $S_{L}$ and $S_{B}$ be the Schwarz functions associated with $L$ and $B$ respectively. Then from (9.3) of Davis and Pollak’s paper [8], we have

in a neighborhood $N$ of $L$. Notice that as $L\subset\mathbb{S}^{1}$, its Schawrz function is $S_{L}(z)=\frac{1}{z}$. Hence we have in $N$,

Since $f$ is meromorphic in $\mathbb{C}$, $g(z):=\overline{f(\frac{1}{\overline{z}})}$ is meromorphic in $\mathbb{C}\backslash\{0\}$ and it follows that $S_{B}(f(z))$ is meromorphic in $N$.

To get the Schwarz function of $B$, we rewrite $P(x,y)=0$ as $Q(z,\overline{z})=0$ for some $Q\in\mathbb{C}[z,\overline{z}]$. Then $f(L)\subset\mathcal{C}^{\prime}$ implies that

in $N$. Thus we have $Q(f(z),g(z))\equiv 0$ in $N$ and hence in the (connected) subset of $\mathbb{C}\backslash\{0\}$ where both $f$ and $g$ are analytic.

If $Q$ depends on $z$ only, then $f$ will be a single valued algebraic function in $\mathbb{C}$ and hence a constant which is impossible. So we may let $Q(f,g)=q(f)g^{n}+\cdots$ where $n\geq 1$ and $q$ is a polynomial. We first consider the case $q(f(0))\neq 0$. Then we have $0<c<|q(f(z))|<d$ in some closed disk $\overline{\mathbb{D}(0;r)}$. By choosing $r$ sufficiently small, we also have $Q(f(z),g(z))\equiv 0$ in $\overline{\mathbb{D}(0;r)}\backslash\{0\}$. Then it follows from Lemma 2.1 below that $|g(z)|$ is bounded above in $\overline{\mathbb{D}(0;r)}\backslash\{0\}$ and hence $g$ has a removable singularity at $0$. This will force $f$ to have a removable singularity at infinity so that $f$ must be rational.

Now suppose $q(f(0))=0$, then we can write $q(f(z))=z^{m}f_{0}(z)$ where $m\in\mathbb{N}$, $f_{0}$ is meromorphic in $\mathbb{C}$ with $f_{0}(0)\neq 0$. Then we may assume that $0<c_{0}<|f_{0}(z)|<d_{0}$ in some closed disk $\overline{\mathbb{D}(0;r_{0})}$ with $r_{0}<1$. We will also let $f(z)=z^{k}f_{1}(z)$ where $k\geq 0$ and $f_{1}$ is meromorphic in $\mathbb{C}$ with $f_{1}(0)\neq 0$. By Lemma 2.1, we have for $z\in\overline{\mathbb{D}(0;r_{0})}$,

for some positive constant $C$. Since $g(z)=\frac{1}{z^{k}}\overline{f_{1}(\frac{1}{\overline{z}})}$, it follows that

in $\overline{\mathbb{D}(0;r_{0})}$. It follows that $|\frac{f_{1}(w)}{w^{m-k}}|\leq C$ for all large $w$. Then the meromorphic function $\frac{f_{1}(w)}{w^{m-k}}$ has a removable singularity at infinity and hence $\frac{f_{1}(w)}{w^{m-k}}$ is rational. Therefore $f_{1}$ and hence $f$ must be rational.

Now if $\mathcal{C}^{\prime}=\mathbb{S}^{1}$, then we need to show that the rational function $f$ must be a quotient of two finite Blaschke products. This will follow from the following result of Pakovich and Shparlinski [20, Theorem 2.2].

As $\mathcal{C}^{\prime}=\mathbb{S}^{1}$, We can take $p_{1}(z)=z$, $p_{2}=f$ and since the infinite set $A$ is a subset of $\{z\in\mathbb{C}:|p_{1}(z)|=|p_{2}(z)|=1\}$, we must have $z=b_{1}\circ q$ and hence the degree of the rational function $q$ is one and thus $q$ is a Möbius transformation. From $z=b_{1}\circ q$, we deduce that $q^{-1}=b_{1}$ and hence $|q^{-1}(z)|=1$ whenever $|z|=1$ as $b_{1}$ is a quotient of finite Blaschke products. Hence $q$ must be a degree one finite Blaschke product. Notice that $f=b_{2}\circ q=\frac{B_{1}}{B_{2}}\circ q$ where $B_{1}$ and $B_{2}$ are finite Blaschke products. Since a composition of two finite Blaschke products is still a finte Blaschke product, $f$ will be a quotient of finite Blaschke products. $\Box$

To prove Theorem 1.3, we need the following proposition.

We first consider the case that $f$ is analytic in some neighborhood of $\overline{c}$. By the assumption (1.1) and the fact that $S_{\mathcal{C}}\equiv S_{L}$ for some branch of $S_{L}$ in $\mathbb{C}\backslash\gamma$ (by Proposition 1), we have $\lim_{z\rightarrow\infty,z\notin\gamma}S_{L}(z)=c$.Thus, we have $\lim\limits_{n\rightarrow\infty}G(z_{n})=\overline{f(\overline{c})}$ and

Since there are infinitely many $a$ satisfying $Q(a,\overline{f(\overline{c})})=0$, we have $Q(z,\overline{f(\overline{c})})\equiv 0$ in $\mathbb{C}$. Hence $Q(u,v)=(v-\overline{f(\overline{c})})Q^{\prime}(u,v)$ for some non-constant $Q^{\prime}\in\mathbb{C}[u,v]$, which contradicts the assumption that $Q$ is irreducible in $\mathbb{C}[u,v]$.

It remains to consider the case that $\overline{c}$ is a pole of $f$. Assume that $Q(F,G)=\sum_{k=0}^{n}p_{k}(F)G^{k}$, where $p_{k}\in\mathbb{C}[z]$ for $1\leq k\leq n$ and $p_{n}\not\equiv 0$. We may also assume that $p_{n}(a)\neq 0$ as there are infinitely many non-Picard exceptional values of $F$. Then by Lemma 2.1, $\{G(z_{n})\}_{n=1}^{\infty}$ will be a bounded sequence. This contradicts to the fact that

and we conclude that $f$ must be rational. $\Box$

Like what we did in the proofs of Theorem 1.1 and 1.3, we can assume that $A$ is an analytic arc and there exists an analytic branch of $S_{A}$ in $\mathbb{C}\backslash\gamma$ and an irreducible polynomial $Q$ in $\mathbb{C}[u,v]$ such that

in the complement of the slit $\gamma$ in $\mathbb{C}$. As $f$ is non-constant elliptic, $n=\deg_{v}Q(u,v)\geq 1$. Let $g(z)=\overline{f(\overline{S_{A}(z)})}$. As $\overline{f(\overline{z})}$ is meromorphic in $\mathbb{C}$ and $S_{A}$ is analytic in $\mathbb{C}\backslash\gamma$, $g$ is meromorphic in $\mathbb{C}\backslash\gamma$.

To proceed, we will make use of the finite map property of $f$ and a counting argument (see page 553 of [3]). Now choose any complex number $z_{0}$ in $\mathbb{C}\backslash\gamma$ such that $\deg_{v}Q(f(z_{0}),v)=n$ and the orbit $z_{0}+\mathbb{Z}\omega_{1}$ is a subset of $\mathbb{C}\backslash\gamma$ and does not contain a pole of $f$ or $g$. Then for all $n\in\mathbb{Z}$,

As $Q(f(z_{0}),v)$ has at most $n$ distinct zeros, there must be a subset $M_{z_{0}}$ of $\{1,\ldots,nd+1\}$ with cardinality at least $d+1$ such that $g(z_{0}+m_{k}\omega_{1})=g(z_{0}+m_{l}\omega_{1})$ for any $m_{k},m_{l}\in M_{z_{0}}$. As each $M_{z_{0}}$ is a subset of $\{1,\ldots,nd+1\}$, there are only finitely many distinct $M_{z_{0}}$. Since there is an uncountable number of choices of $z_{0}$, there must be an uncountable set $U$ of $z$ for which the set $M$ is independent of $z_{0}$ in $U$. Since any uncountable subset of $\mathbb{C}$ has an uncountable set of accumulation points, we deduce that for any $m_{k}$ and $m_{l}$ in $M$, $g(z+m_{k}\omega_{1})\equiv g(z+m_{l}\omega_{1})$ in $\mathbb{C}\backslash\gamma$ and hence we have

in $\mathbb{C}\backslash\gamma$. As $f:\mathbb{C}/\Lambda\to\mathbb{C}\cup\{\infty\}$ is of degree $d$ and $M$ contains at least $d+1$ points, it follows from (2.6) and a similar counting argument that there exist distinct $n_{1},n_{2}\in M$ such that

for $z\in\mathbb{C}\backslash\gamma$. As $\Lambda$ is a discrete set and $S_{A}(z+n_{1}\omega_{1})-S_{A}(z+n_{2}\omega_{1})$ is analytic in the connected set $\mathbb{C}\backslash\{(\gamma-n_{1}\omega_{1})\cup(\gamma-n_{2}\omega_{1})\}$, we must have

in $\mathbb{C}\backslash\{(\gamma-n_{1}\omega_{1})\cup(\gamma-n_{2}\omega_{1})\}$ for some $l_{1},l_{2}\in\mathbb{Z}$. Since $S_{A}(z+n_{2}\omega_{1})$ is analytic in $\gamma-n_{1}\omega_{1}$, it follows that $S_{A}(z+n_{1}\omega_{1})$ can be extended to an entire function by the above identity. Since $S_{A}$ is algebraic and entire, it must be a polynomial. Notice that the polynomial $S_{A}^{\prime}$ is periodic. This will reduce $S_{A}^{\prime}$ to a constant and hence $S_{A}(z)=az+b$. Then from (6.11) of Davis’ book [9], we have $\overline{S_{A}(\overline{S_{A}(z)})}=z$, and we can deduce that $|a|=1$. To see that $A$ is a straight line segment, consider any two points $z_{1},z_{2}$ in $A$. Then $\overline{z_{i}}=S_{A}(z_{i})=az_{i}+b$ for $i=1,2$. It then follows that $\frac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}=a$ and we are done.

Now consider that case that $f(z)=R(e^{\alpha z})$ for some rational function $R$ and $\alpha\in\mathbb{C}\backslash\{0\}$. Without loss of generality, we may assume that $\alpha=2\pi i$ so that $f$ has period $1$. Let $D=\{x+iy:0\leq x<1\}$. Then $f:D\to\mathbb{C}\cup\{\infty\}$ is a finite map. One can apply a similar argument to show that $A$ is a straight line segment. $\Box$