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The Nevanlinna ramification index is the normalized lower limiting integrated excess multiplicity over a target value, with pole orders used for infinity. It is an asymptotic value-distribution quantity, distinct from the integer ramification index of a holomorphic map at one source point. It can be positive even when the target has positive Nevanlinna deficiency.

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.

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

Homogenizing the affine equation gives the projective closureThis is the Klein quartic. It is smooth: if all three projective partial derivatives vanished, thenMultiplying them gives $28X^3{Y}^3{Z}^3=0$. If one coordinate is zero, the displayed equations force all three to vanish, which is impossible in projective space. Thus $C$ is a smooth compactification of $X^{\prime }$. The smooth compactification compatible with the extended coordinate functions is its normalization, and since $C$ is already smooth it is biholomorphic to $C$.

For a direct genus computation, consider the extended coordinate map ${\pi }_x:X\to {\mathbb{C}}_{\infty }$. A generic $x$ leaves the cubic equationso $\mathrm{deg}{\pi }_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven pointsAt the origin, $x=-y^3+O(y^6)$, so the ramification index of a holomorphic map is three and the contribution is two. At each of the other seven points, $F_x=-7/2\ne 0$, $F_y=0$, and $F_{yy}=6y\ne 0$, so the ramification index is two and each contributes one.

There are two points at infinity. Near $P=[1:0:0]$, in coordinates $u=Y/X$, $v=Z/X$, the equation isand $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation isThus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is thereforeThe Riemann-Hurwitz formula for the degree-three map to the sphere givesand henceThis also agrees with the genus of a smooth plane curve of degree four.

The formula $[X:Y:Z]\mapsto [X:Z]$ is initially undefined at the unique point $P_{\infty }=[0:1:0]$ of $X$ at infinity. Since $X$ is a smooth projective curve, the extension of a rational map from a smooth projective curve extends it uniquely to a morphismOn the affine chart it is the rational function $x$, and the equation $y^3=x^4+1$ shows that $\phi$ has degree three.

For each of the four roots $\alpha$ of $x^4+1$, let $P_{\alpha }=(\alpha ,0)$. Since the root is simple, $t=y$ is a uniformizer at $P_{\alpha }$, and the equation givesThus the ramification index of a holomorphic map at every $P_{\alpha }$ is three.

At $P_{\infty }$, use the chart $Y=1$ and put $u=X/Y$, $v=Z/Y$. The equation becomesThe implicit function theorem makes $u$ a uniformizer and gives $v=u^4$ times a local unit. Henceso $P_{\infty }$ also has ramification index three. These are all the ramification points of a holomorphic map because the resulting ramification degree is $5(3-1)=10$. Indeed, the Riemann-Hurwitz formula givesTherefore $g_X=3$, in agreement with the genus of a smooth plane curve, and the Euler characteristic of the underlying compact orientable surface is

Let $F:X\to Y$ be a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces. At $p\in X$, let $e_p$ be the ramification index of a holomorphic map. Choose a triangulation of $Y$ containing every branch value among its vertices, and lift it to $X$.

If the target triangulation has $V$ vertices, $E$ edges, and $T$ triangles, then the lifted triangulation has $dE$ edges and $dT$ triangles, because no branch value lies in an edge or face interior. For a target vertex $v$, local degree counting givesConsequently the number of vertices above $v$ isWriting $R={\sum }_{p\in X}(e_p-1)$, the Euler characteristic is thereforeSince a compact orientable surface of genus $g$ has Euler characteristic $2-2g$, this Triangulation proof of the Riemann-Hurwitz formula givesor equivalently

Now extend a cubic polynomial $f:\mathbb{C}\to \mathbb{C}$ to a degree-three holomorphic mapof the Riemann sphere. Both genera are zero, so Riemann--Hurwitz givesThe point at infinity is totally ramified with $e_{\infty }=3$, contributing two. Thus the finite points contribute exactly two. There are consequently two cases.

If there is one finite ramification point $\alpha$, it has index three. Choose an affine source map $h$ sending zero to $\alpha$. Thenfor some $c\ne 0$, and integration givesAn affine target map subtracts $d$ and rescales by $3/c$, yielding $z^3$.

Otherwise there are two distinct finite ramification points, each of index two. Choose an affine source map $h$ sending $-1$ and $1$ to them. The derivative of $f\circ h$ is then a nonzero multiple of $z^2-1$:HenceThe affine target map $g(w)=(w-d)/c$ givesTherefore the affine normal forms of a complex cubic polynomial are precisely

In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the formThe point $p$ is a ramification point of a holomorphic map when its ramification index of a holomorphic map satisfies $e_p>1$. A branch value of a holomorphic map, also called a branch point in the target, is a value $f(p)$ of some ramification point.

Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the holomorphic inverse function theorem supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Henceis an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.

The monodromy theorem says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the path lifting theorem lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma ]\in {\pi }_1({\mathbb{C}}_{\infty }\setminus B,w)$, giving the monodromy group of a covering.

For the stated function, make the Möbius change of coordinateThenThe critical points are $t=0,\pm 1/\sqrt{2},\infty$, with branch valuesA loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these areThey generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group isthe dihedral group in its action on the four vertices of a square.