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For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree isThe valency theorem says that this integer is independent of $y\in Y$.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree iswhere $e_x$ is the ramification index of a holomorphic map at $x$. The valency theorem makes this sum independent of $y\in Y$. The Riemann-Hurwitz formula is

Write $E=\mathbb{C}/\Lambda$. The map $\psi (z+\Lambda )=-z+\Lambda$ is well-defined and biholomorphic, with ${\psi }^2=\mathrm{id}$. Its fixed points satisfyThey are therefore indexed by $\frac{1}{2}\Lambda /\Lambda \cong (\mathbb{Z}/2\mathbb{Z}{)}^2$, so there are four, as in the negation map on a one-dimensional complex torus.

For each $x\in E^{\prime }$ choose a small open disc $U_x$ for which $U_x\cap \psi (U_x)=\varnothing$. The projection restricts to a homeomorphismUse $(p{|}_{U_x}{)}^{-1}$ followed by a holomorphic coordinate on $U_x$ as a chart on $p(U_x)$. Choosing $\psi (U_x)$ instead changes this coordinate by the holomorphic map induced by $\psi$, so the transition functions are holomorphic. These charts make $S^{\prime }$ a Riemann surface and $p:E^{\prime }\to S^{\prime }$ a holomorphic double covering map.

At each omitted fixed point, a centred local coordinate changes by $z\mapsto -z$, so the invariant quotient coordinate is $w=z^2$. Hence the extension has degree two and exactly four ramification points, each with $e_x=2$. Since the complex torus $E$ has genus one, Riemann--Hurwitz givesThus $\boxed{g_S=0}$, in agreement with the general quotient of a one-dimensional complex torus by negation.

If $F:X\to Y$ is a nonconstant holomorphic map between compact connected Riemann surfaces, its local degree at $p$ is the integer $m_F(p)\ge 1$ for which suitable local coordinates giveThe valency theorem states thatis independent of $y\in Y$. This common value is the degree of a holomorphic map, denoted $\mathrm{deg}F$.

Now let $f$ be a nonconstant rational function of degree $d$. If its distinct finite poles have orders $m_1,\dots ,m_r$, and its pole order at infinity is $m_{\infty }\ge 0$, thenThe derivative $f^{\prime }$ has a pole of order $m_j+1$ at each finite pole. When $m_{\infty }>0$, the expansion $f(z)\sim c{z}^{m_{\infty }}$ shows that $f^{\prime }(z)\sim c{m}_{\infty }{z}^{m_{\infty }-1}$ at infinity. Hence the degree of the derivative of a rational function isIn the first case $r\ge 0$, while in the second $1\le r\le d$; thereforeFor every $d\ge 1$, the lower bound is attained by $f(z)=z^d$, whose derivative has degree $d-1$ (with a constant assigned degree zero). For distinct $a_1,\dots ,a_d$, the functionhas $d$ simple poles, degree $d$, and a derivative with $d$ double poles, so $\mathrm{deg}{f}^{\prime }=2d$. Thus both rational bounds are sharp for every $d$.

Let next $g$ be a nonconstant elliptic function for the period lattice $\Lambda$. Its degree is the total order of its poles in a fundamental parallelogram; by the valency theorem, this is also the degree of the induced map $\mathbb{C}/\Lambda \to \hat{\mathbb{C}}$. If these poles have orders $n_1,\dots ,n_s$, then $d=\mathrm{deg}g={\sum }_j{n}_j$, and $g^{\prime }$ has poles of orders $n_j+1$. The degree of the derivative of an elliptic function is consequentlySince every nonconstant elliptic function has at least one pole and $1\le s\le d$,

Let $d\ge 3$ be odd. The Weierstrass elliptic function supplies the lower-bound exampleIt has one pole modulo $\Lambda$, of order $d$, so its derivative has one pole of order $d+1$. For the upper bound, choose distinct points $a_1,\dots ,a_d$ modulo $\Lambda$ and nonzero constants $c_j$ with ${\sum }_j{c}_j=0$. The quasi-periodicity of the Weierstrass zeta function makeselliptic. It has exactly $d$ simple poles, while $g^{\prime }$ has $d$ double poles. Therefore the two bounds are attained for every required odd degree.

The valency theorem states that for a nonconstant analytic map between compact connected Riemann surfaces,for every target point $w$.

For a rational map $f=p/q$ on the Riemann sphere, first cancel common factors; then the degree of a rational map of the Riemann sphere isThe analytic isomorphisms are exactly the degree-one maps, namely the Möbius transformations.

The required transformation isIt sends $\infty \leftrightarrow 1$ and $0\leftrightarrow -1$. Its fixed points satisfy

The octahedral rotation orbits on the Riemann sphere have possible sizescorresponding respectively to vertices, face centres, edge centres, and generic points.

In the displayed $F$, no numerator factor vanishes at $0$ or at a fourth root of one. Hence $0,\pm 1,\pm i$ are poles of order four. The numerator and denominator have degrees $24$ and $20$, so $F(z)\sim z^4$ at infinity; infinity is also a pole of order four. Thusand

Finally let $z$ have stabilizer of order $e$. A local coordinate turns its stabilizer action into rotation by $e$th roots of unity. Since $F$ is invariant, the first nonconstant term of its local expansion has exponent divisible by $e$, so $m_F(z)\ge e$. The orbit has $24/e$ points and therefore contributes at least $24$ to the fibre through $F(z)$. The valency theorem and $\mathrm{deg}F=24$ leave no room for any further point. The degree-sized invariant separates finite-group orbits, so $F(z)=F(w)$ implies that $z$ and $w$ lie in the same orbit.