Riemann-Roch theorem

For a smooth projective curve of genus $g$ and a canonical divisor $K$,

On a smooth projective curve, the exact sequence $0\to \mathcal{O}(D-P)\to \mathcal{O}(D)\to k(P)\to 0$ increases the Euler characteristic by one. Iteration for positive and negative coefficients gives $\chi (\mathcal{O}(D))=\mathrm{deg}D+1-g$. Serre duality identifies $h^1(\mathcal{O}(D))$ with $h^0(\mathcal{O}(K-D))$, proving $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$.

If a line bundle on a smooth projective curve has both $h^0(\mathcal{L})>0$ and $h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$, the displayed inequality holds. Write these dimensions as $r,s$. The product dimension bound for sections on a curve gives $r+s-1\le g$, and the Riemann-Roch theorem gives $r-s=\mathrm{deg}\mathcal{L}+1-g$. Adding gives $2r\le \mathrm{deg}\mathcal{L}+2$.

At a point $p$ of a smooth projective curve, a positive integer $n$ is a Weierstrass gap when no rational function has a pole of order exactly $n$ at $p$ and no other poles. Equivalently,

At every point of a smooth projective curve of genus $g$, exactly $g$ positive integers are Weierstrass gap. Indeed, each quotient $L(np)/L((n-1)p)$ has dimension zero or one, while Riemann--Roch gives $\ell (np)=n+1-g$ for all $n>2g-2$.

On a genus-two smooth projective curve, the gap set at any point is either $\{1,2\}$ or $\{1,3\}$. The integer $1$ is always a gap on a positive-genus curve. If $2$ is not a gap, then $3$ must be a gap, since pole orders are closed under addition and $2,3$ would otherwise generate every integer at least two.

A canonical divisor is the divisor of any nonzero rational differential. Its divisor class is independent of the differential and has degree $2g-2$.

A rational differential on a smooth projective curve is a rational section of its canonical bundle. In a local coordinate $t$ it has the form $f(t)dt$ for a rational function $f$.

If $t$ is a uniformizer at a point $p$ of a smooth projective curve and a nonzero rational differential is $\omega =f(t)dt$, then ${\nu }_p(\omega )={\nu }_p(f)$. This integer is independent of the chosen uniformizer. Its positive and negative values are respectively the orders of the zero and pole of $\omega$ at $p$.

If $K_X=(\omega )$ is represented by a nonzero rational differential, multiplication by $\omega$ identifies the Riemann-Roch space $L(K_X)$ with the vector space of holomorphic differential forms. The Riemann-Roch theorem gives $\ell (K_X)=g$ for a smooth projective curve of genus $g$.

For the affine coordinate $t$ on ${\mathbb{P}}^1$, put $s=t^{-1}$ at infinity. Since $dt=-s^{-2}ds$,so $K_{{\mathbb{P}}^1}\sim -2\infty$. More generally,because $L(d\infty )$ consists of polynomials of degree at most $d$ when $d\ge 0$ and only the zero function when $d<0$.

A smooth plane curve of degree $d$ has genus

If a smooth degree-$d$ projective plane curve had genus two, the genus-degree formula would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

A smooth plane quartic is a smooth degree-four curve in ${\mathbb{P}}^2$ and has genus three.

The projective closure of the superelliptic curve $y^3=x^4+1$ is the smooth plane quarticIts unique point at infinity is $P_{\infty }=[0:1:0]$. The rational function $x=X/Z$ defines a degree-three morphism to ${\mathbb{P}}^1$ ramified with index three at $P_{\infty }$ and at the four affine points $(\alpha ,0)$ with ${\alpha }^4=-1$. Its genus is three, and

The Klein quartic has the smooth projective plane modelIts affine chart $Z=1$ is $x^{3}y+y^3+x=0$. If $\zeta$ is a primitive seventh root of unity, the projective transformationis an automorphism of order seven.

The projective completion ofis the smooth plane quarticThe coordinate relabelling $[A:B:C]=[Y:X:Z]$ turns this equation into the standard Klein quartic equation $A^{3}B+B^{3}C+C^{3}A=0$.

The coordinate function $x$ has degree three on the Klein quartic. It has ramification index three at $(0,0)$, index two at the seven affine points satisfying $y^7=-3/8$ and $x=2y^3$, and index two at $[0:1:0]$. The total ramification is ten, so the Riemann-Hurwitz formula gives genus three.

For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the Adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

Projection from a point of a smooth plane quartic gives a degree-three map to the projective line. A degree-two map would make the curve hyperelliptic, but its canonical map is its plane embedding, whereas a hyperelliptic canonical map is not an embedding. Its gonality is therefore three.

For $g\ge 2$, a basis of the space of regular differentials defines the canonical map $X\to {\mathbb{P}}^{g-1}$. Equivalently, after choosing a differential with divisor $K$, one may use a basis of $L(K)$.

For a nonhyperelliptic smooth projective curve of geometric genus four, the canonical map is an embedding of degree six. Dimension counts $10>h^0(2K)=9$ and $20>h^0(3K)=15$ supply a quadric and a cubic not divisible by it. The quadric is irreducible because the curve is nondegenerate. Degree six and unmixedness of a complete intersection then identify the intersection with the curve as a scheme.