For a divisor on an algebraic curve $D$ on a smooth projective curve $X$ of genus $g$, the Riemann-Roch theorem stateswhere $K_X$ is a canonical divisor. Taking $D=0$ gives $1-\ell (K_X)=1-g$, so $\ell (K_X)=g$. Taking $D=K_X$ then gives

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch givesThe first two equalities show that the complete linear system of a divisor $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a very ample divisor, in accordance with the general fact that a high-degree divisor is very ample on a smooth projective curve, and its sections define a closed embeddingThe ambient dimension therefore depends only on $g$.

The Riemann-Hurwitz formula for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ isChoose a smooth plane quartic $C$, so $g(C)=3$, and form the product of projective varieties $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its normalization $\tilde{B}$. At least one coordinate projection $\tilde{B}\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz givesso the geometric genus of $B$ is at least three. Hence $S$ is the required surface; this is the product surface without low-genus curves construction.

Finally let $X\subset {\mathbb{P}}^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ isIts two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in ${\mathbb{P}}^2$ is $p$. The criterion for a morphism of algebraic varieties therefore shows that the restrictionis a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the projection of a plane curve from an exterior point has degree $d$.

By the genus of a smooth plane curve, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its ramification divisor $R_{\pi }$ givesEvery ramification point contributes at least one to this degree, so

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.

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.