# $f^3 + g^3=1$ for two meromorphic functions

Can you find two non-constant meromorphic functions $f,g$ such that $f^3 +g^3=1$?

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## 3 Answers

Yes. According to remark 2 on p.236 of Remmert's book, Classical topics in complex function theory, Springer GTM 172, a (more or less) explicit example can be given by using the Weierstrass $\wp$-function associated to the triangular lattice $\{m + n e^{2\pi i /3}\,:\,m,n \in \mathbb{Z}\}$. This function satisfies the differential equation $$(\wp')^2 = 4 \wp^3 - g_2 \wp - g_3 \qquad \text{with}\; g_2 = 0 \; \text{and} \; g_3 = \pm \frac{\Gamma\big(\frac{1}{3}\big)^{18}}{(2\pi)^6}.$$

Putting $a = \frac{1}{2} \sqrt[3]{\phantom{|}g_{3}}$ and $b = \dfrac{1}{\sqrt{24a}}$ the functions $$f = \frac{a + b\, \wp'}{\wp} \qquad \text{and} \qquad g = \frac{a - b\,\wp'}{\wp}$$ satisfy $f^3 + g^3 = 1$ by the above differential equation (the values of $a$ and $b$ are found using the ansatz $\big(a + b\,\wp'\big)^3 + \big(a - b\,\wp'\big)^3 = \wp^3$ and the differential equation).

Observe however that the zeroes of $\wp$ and the poles of $\wp'$ are poles of both $f$ and $g$ and indeed, according to the proposition on the same page this is necessarily the case (by a beautiful application of Picard's theorem). In fact, this proposition states: If $f$ and $g$ are meromorphic functions on $\mathbb{C}$ satisfying $\,f^n + g^n = 1$ for some $n \geq 3$ then $f$ and $g$ are both either constant or they have common poles. Added: In fact, as user8268 points out in his comment below, one can even show that there are actually no non-constant such functions for $n \geq 4$ (thanks a lot for that!). This is also alluded to by Remmert in his remark 1 on the same page 236.

See Remmert's book for further details.

Added Later.

Following the links on the Wikipedia page I stumbled over Hans Lundmark's beautiful page on elliptic functions where you can find some domain coloring plots of some $\wp$-functions towards the end of the page.

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for $f^n+g^n=1$ (assuming they are meromorphic functions defined on all $\mathbb{C}$): such functions would give you a holomorphic map from $\mathbb{C}$ to the (compactification of) the Riemann surface $x^n+y^n=1$, which is of genus $>1$ if $n>3$. The universal cover of the Riemann surface is thus the unit disc $D$, and the map would lift to a holomorphic map $\mathbb{C}\to D$, which must be constant by Liuoville thm. Hence no non-constant $f,g$ for $n>3$. – user8268 Mar 30 '11 at 14:45
@user8268: Very nice, thanks for this remark (in fact, this is remark 1 in Remmert's book which I skipped, unfortunately). @joriki: thanks for the correction! – t.b. Mar 30 '11 at 14:54
(to include some numbers to my previous comment:) the genus of $x^n+y^n=1$ is (as computed e.g. by Riemann-Hurwitz formula) $g=(n-1)(n-2)/2$. For $n=1$ we get $g=1$, so the universal covering of the surface is $\mathbb{C}$, hence $x$ and $y$ pull-back to meromorphic functions satisfying $f^3+g^3=1$ - which is as non-explicit proof of their existence as it gets :) – user8268 Mar 30 '11 at 15:04
...sorry I meant "for $n=3$" – user8268 Mar 30 '11 at 15:12
New address: users.mai.liu.se/hanlu09/complex – Hans Lundmark Aug 14 '15 at 10:24

As I mentioned in this other post, A.C. Dixon studied in 1890 a class of elliptic functions based on inverting the Abelian integral

$$\int\frac{\mathrm dt}{\left(1-t^3\right)^{2/3}}$$

(actually, he studied a more general class of elliptic functions based on the cubic $x^3+y^3-3\alpha xy=1$, but I'll confine myself to the case $\alpha=0$ and refer the interested to Dixon's paper for the general case.)

In particular, there are the two Dixon elliptic functions $\mathrm{sm}(z,0)=\mathrm{sm}(z)$ and $\mathrm{cm}(z)$ that satisfy $\mathrm{sm}^3(z)+\mathrm{cm}^3(z)=1$ and a bunch of other identities, and can be expressed in terms of Weierstrass $\wp$:

$$\mathrm{sm}(z)=\frac{6\wp\left(z;0,\frac1{27}\right)}{1-3\wp^\prime\left(z;0,\frac1{27}\right)}$$

$$\mathrm{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime\left(z;0,\frac1{27}\right)-1}$$

Both have the independent periods $\pi_3=B\left(\frac13,\frac13\right)$ (where $B(a,b)$ is the beta function) and $\pi_3\exp(2i\pi/3)$, and have poles at the points $z=\frac23\pi_3+k\pi_3,\;k\in\mathbb Z$ in the real line, as well as having a total of seven poles and three zeros in each "fundamental hexagon". As can be surmised, these are related to the general solutions in Theo's answer, due to the fact that the Weierstrass function satisfies homogeneity relations.

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More general "hyprgoniometric functions" were studied by Lundberg in 1879(!) that satisfy relations of the form $s^n+c^n=1$; see this and this for instance. – J. M. May 9 '11 at 16:00

As an addendum of sorts to Theo's answer, the two functions $f$ and $g$ in the answer can be simplified through the use of the homogeneity relations

$$\wp\left(cz;\frac{g_2}{c^4},\frac{g_3}{c^6}\right)=\frac1{c^2}\wp\left(z;g_2,g_3\right),\quad \wp^\prime\left(cz;\frac{g_2}{c^4},\frac{g_3}{c^6}\right)=\frac1{c^3}\wp^\prime\left(z;g_2,g_3\right)$$

Focusing on the case of positive $g_3$ (the treatment for negative $g_3$ is similar), we have

$$f(z)=\frac{3+\sqrt{3}\wp^\prime\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}{6\wp\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)},\quad g(z)=\frac{3-\sqrt{3}\wp^\prime\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}{6\wp\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}$$

where now only "equianharmonic case" Weierstrass elliptic functions are involved.

If these functions are plotted in the complex plane, a hexagonal structure similar to what is observed for the Dixon elliptic functions can be seen. This suggests that there might be a relationship between these functions and the Dixon elliptic functions.

In particular, using the homogeneity relations, one can show that

$$f(z)=-\frac{\mathrm{cm}\left(\frac{\Gamma(1/3)^3}{2\pi}\sqrt{3}z\right)}{\mathrm{sm}\left(\frac{\Gamma(1/3)^3}{2\pi}\sqrt{3}z\right)}$$

As promised, here are plots of $f(z)$ on the real line:

a contour plot of the real and imaginary parts of $f(z)$:

and plots of a single "hexagonal tile" of $f(z)$:

Plots of $g(z)$ are similar since $g(z)=f(-z)$.

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I'll edit this answer later to include pictures and show the equivalence with the Dixon elliptic functions. – J. M. Jul 21 '11 at 9:09
I hope you haven't forgotten... I'd love to see those pictures! – t.b. Jul 23 '11 at 0:59
Don't worry @Theo, I haven't; once I get to a Mathematica machine I'll post those pictures post-haste. – J. M. Jul 23 '11 at 1:01
Very nice! Thanks a lot! – t.b. Jul 27 '11 at 13:15