Divisors, Riemann--Roch, and Duality: Examples and Counterexamples

The zero function is a polynomial. For nonzero $f\in L(D)$, [F3] writes $f=P\mathrm{/}Q$ with $P,Q$ coprime. If $Q$ were nonconstant, [F4] would give a root $a\in \mathbb{C}$; if $P(a)=0$, the same factorization result would make $z-a$ divide both $P$ and $Q$, contrary to coprimeness. Thus $P(a)\mathrm{\ne }0$ and $f$ has a finite pole at $a$, impossible for $f\in L(d[\mathrm{\infty }])$. Hence $Q$ is constant. If $P$ has degree $m$, its expression in $u=1\mathrm{/}z$ has leading term $c{u}^{-m}$, so its pole order at infinity is $m$; membership in $L(D)$ forces $m\le d$. Conversely every polynomial of degree at most $d$ has no finite poles and pole order at infinity at most $d$, so belongs to $L(D)$. The monomials are linearly independent as polynomials and span this space, proving the dimension and basis claims.

By [F6], $s_j=z^j{s}_D$ has divisor $(z^j)+D=j[0]+(d-j)[\mathrm{\infty }]$, since $z$ has a simple zero at $0$ and a simple pole at $\mathrm{\infty }$. At every finite point $s_0=s_D$ is nonzero because its divisor is $d[\mathrm{\infty }]$; at infinity $s_d$ is nonzero because its divisor is $d[0]$. Thus the basis sections have no common zero. Since $d+1\ge 2$, [F7] applies to $V=H^0(X,E)$ and gives the asserted linear-system map and pullback isomorphism.

On the source chart $z_0\mathrm{\ne }0$, the section $s_D$ is a local frame and the coefficients of $s_j$ are $z^j$, so the map has coordinates $[1:z:\cdots :z^d]$. On the chart $z_1\mathrm{\ne }0$, the coordinate is $u=z_0\mathrm{/}{z}_1=1\mathrm{/}z$ and $s_d=z^d{s}_D$ is a local frame because its divisor is $d[0]$; the coefficients of $s_0,\dots ,s_d$ in this frame are $u^d,u^{d-1},\dots ,1$. Hence the map is $[u^d:u^{d-1}:\cdots :1]$ there. On the overlap, multiplying $[1:z:\cdots :z^d]$ by $u^d$ gives the second tuple, so the formulas agree and are holomorphic in both charts; together they give the stated homogeneous formula on all of ${\mathbb{P}}^1$.

In the finite chart, the target chart coordinates are $(z,z^2,\dots ,z^d)$, whose first coordinate recovers $z$ and whose derivative has first component $1$. In the chart at infinity, the target coordinates are $(u^d,u^{d-1},\dots ,u)$, whose last coordinate recovers $u$ and whose derivative has last component $1$. The point at infinity maps to $[0:\cdots :0:1]$, outside the target chart with first coordinate nonzero, while every finite point lies in that chart. Thus the map is globally injective and has nonzero differential at every point. For any nonzero hyperplane form $L(W)={\sum }_{j=0}^d{a}_j{W}_j$, monomial independence from step 1.1 shows its pullback ${\sum }_{j=0}^d{a}_j{z}_0^{d-j}{z}_1^j$ is a nonzero homogeneous polynomial of degree $d$. If its dehomogenization on $z_0\mathrm{\ne }0$ has degree $m\le d$, [F4] gives $m$ finite roots counted with multiplicity, and in the infinity coordinate it has a zero of order $d-m$ when $m<d$; hence every hyperplane section has total multiplicity $d$. Under the hyperplane-section definition of degree, the image is the rational normal curve of degree $d$.

The chart formulas show that ${\phi }_d$ is continuous. If $C\subseteq {\mathbb{P}}^1$ is closed, [F9] makes $C$ compact; pulling any open cover of ${\phi }_d(C)$ back along ${\phi }_d$ gives an open cover of $C$, so compactness gives a finite subcover and ${\phi }_d(C)$ is compact. The target ${\mathbb{P}}^d$ is Hausdorff by [F8], so [F9] makes ${\phi }_d(C)$ closed. Hence the continuous bijection from ${\mathbb{P}}^1$ onto its image is closed and has continuous inverse. Together with the nonzero differential from step 4.1, this proves that ${\phi }_d$ is a holomorphic embedding, including the case $d=1$.

Each factor $z-a_j$ has divisor $[a_j]-[\mathrm{\infty }]$ by the coordinates in [F3], so [F2] gives $(u)=-D+s[\mathrm{\infty }]$. For any nonzero $f$, $(f\mathrm{/}u)+s[\mathrm{\infty }]=(f)+D$, so division by $u$ identifies $L(D)$ with $L(s[\mathrm{\infty }])$. A rational function with no finite poles is a polynomial: in a reduced quotient of polynomials, a nonconstant denominator would have a finite root by [F4], hence a pole, contrary to the divisor bound. A polynomial of degree $d$ has pole order $d$ at infinity in the coordinate $w=1\mathrm{/}z$, so for $s\ge 0$ this space consists precisely of the polynomials of degree at most $s$. Its basis $1,z,\dots ,z^s$ yields the displayed basis of $L(D)$ and its divisor formula. If $s<0$, [F2] gives $L(D)=0$.

By [F8], different local lifts of a torus point differ by a lattice translation, whose derivative is $1$, so their differentials $dz$ agree; by [F5] they define a nowhere-vanishing holomorphic differential with divisor $0$. Thus [F6] identifies the canonical bundle with ${\mathcal{O}}_X(0)$. By [F7], $\mathrm{\ell }(0)=1$ and $i(0)=\mathrm{\ell }(K)=1$; applying Riemann–Roch at $0$ gives $0=1-g$, hence $g=1$. For $n\ge 1$, the divisor $K-n[o]=-n[o]$ has negative degree, so [F2] gives $\mathrm{\ell }(K-n[o])=0$. Riemann–Roch and duality [F7] now give $\mathrm{\ell }(n[o])=n$ and $L(0)=\mathbb{C}$.

On the finite chart $dz$ has no zeros or poles; at infinity $dz=-w^{-2}dw$, so [F5] gives $(dz)=-2[\mathrm{\infty }]=K$. Step 1.1 applied to $K$ gives $\mathrm{\ell }(K)=0$, and [F6] identifies this zero space with the holomorphic differentials. Applied to $K-D$, the same calculation gives $\mathrm{\ell }(K-D)=\max (0,-s-1)$. If $s\ge -1$, subtracting it from $\mathrm{\ell }(D)$ gives $s+1$; if $s\le -2$, the difference is $0-(-s-1)=s+1$. This is Riemann–Roch for the genus-zero sphere, and [F7] identifies the second term with $i(D)$.

By [F9], ${\mathrm{\wp }}^j({\mathrm{\wp }}^{\mathrm{\prime }}{)}^{\epsilon }$ descends to the torus, has no poles away from $o$, and has pole order exactly $2j+3\epsilon$ at $o$, with nonzero leading coefficient $(-2{)}^{\epsilon }$. The chosen representatives have distinct orders $2,3,\dots ,n$, so each belongs to $L(n[o])$; together with $1$ they are linearly independent, because in a nontrivial linear combination the term of largest pole order has a principal coefficient that no other term can cancel. There are $n$ such functions for $n\ge 1$ (only $1$ when $n=1$), so step 1.2 makes them a basis. This proves the gap and the first nonconstant-function claims.

Embed $C$ projectively by [F3]. Its complex points $X$ are closed in compact projective space because its homogeneous defining equations are continuous, so $X$ is compact, Hausdorff and second countable. At each point [F4] gives a standard smooth chart with one free coordinate $z$; the implicit-function lemma supplies a holomorphic graph chart, and the transitions are holomorphic. Thus each component of $X$ is a compact Riemann surface, with finitely many components by [F10]. An algebraic morphism is holomorphic in these charts: its coordinate functions are regular fractions whose denominators are nonzero near the point, and compositions with the graph chart are holomorphic. In particular $\varphi$ induces a holomorphic map $X\to \widehat{\mathbb{C}}$.

In a standard smooth chart the invertible Jacobian minor lets [F5] eliminate all dependent coordinate differentials, leaving $dz$ as a regular algebraic frame of ${\omega }_C$ and as the analytic differential frame. The cotangent isomorphism in [F5] says $z-z(p)$ has nonzero class in the one-dimensional ${\mathfrak{m}}_p\mathrm{/}{\mathfrak{m}}_p^2$; since the local ring is a DVR by [F6], it is an algebraic uniformizer. Any rational coefficient is therefore $(z-z(p){)}^{a}u$ with $a\in \mathbb{Z}$ and $u$ a regular unit; analytically $u$ is holomorphic with nonzero value at $p$, so its algebraic and analytic orders are equal. This also proves that a nonzero rational coefficient cannot vanish identically on an analytic neighbourhood. Consequently algebraic rational differentials become nonzero meromorphic differentials with precisely the same divisor, and regular differentials become holomorphic; the restriction of their section spaces is injective. Pullbacks and line-bundle isomorphisms have the same local regular transition formulas and therefore induce the corresponding holomorphic bundle maps.

On each component $X_j$, the map $\varphi$ is nonconstant: if locally constant at a point over $t$, a local coordinate of the target vanishing at $t$ would pull back to an identically zero germ, contrary to the finite DVR order of that nonzero rational pullback in step 2.1 and [F7]. Compactness makes each restriction proper. By [F10], it has a positive integer analytic degree $d_j$. Equality of local orders in step 2.1 and the algebraic fibre formula [F7] give ${\sum }_j{d}_j=2$. If $X$ had two components, both degrees would be one. The fibre formula would then make each restriction bijective and unramified, and the local inverses in [F10] would make each component biholomorphic to the sphere. The sphere has no nonzero holomorphic differential: the meromorphic differential $dz$ has divisor $-2[\mathrm{\infty }]$, since $dz=-w^{-2}dw$ in $w=1\mathrm{/}z$. Dividing a holomorphic differential by $dz$ would give an element of $L(-2[\mathrm{\infty }])$, which is zero by [F9]. But [F8] supplies a nonzero regular algebraic differential, whose restriction is nonzero by step 2.1 and holomorphic on every component; if every component were a sphere it would vanish everywhere, contradicting this injectivity. Thus $X$ is connected and $\varphi :X\to \widehat{\mathbb{C}}$ has degree two.

Choose a nonzero regular algebraic differential by [F8], and let $K_C$ be its divisor. Step 2.1 identifies its algebraic divisor with the analytic canonical divisor $K$ on connected $X$, coefficient by coefficient. All residue degrees are one by [F7], so their degrees agree. If $h$ is the topological genus of $X$, [F8] and [F9] give $2g-2=\mathrm{deg}{K}_C=\mathrm{deg}K=2h-2$, hence $h=g$. Restriction of regular algebraic differentials is injective by step 2.1; its source has dimension $g$ by [F8] and its target has dimension $h=g$ by [F9], so it is an isomorphism. This proves the required canonical-section and genus comparison without assuming a general algebraic/analytic cohomology comparison.

A general fibre of the analytic degree-two map consists of two distinct points $P,Q$: [F10] makes the branch-value set finite. It is the same algebraic fibre by step 2.1. The pullback of the standard section of $\mathcal{O}(1)$ vanishing at $t$ has divisor $P+Q$ on $X$, so $L$ induces ${\mathcal{O}}_X(P+Q)$. The bundle isomorphism in [F2] and step 2.1 therefore give $K\sim D=(g-1)(P+Q)$. By [F9], $\mathrm{\ell }(K)=g$ and linear equivalence gives $\mathrm{\ell }(D)=g$; equivalently RR gives $\mathrm{\ell }(D)-\mathrm{\ell }(K-D)=g-1$ with $K-D\sim 0$ and $\mathrm{\ell }(0)=1$. The $g$ pulled-back monomial sections in [F2] are now a basis of both algebraic and analytic canonical spaces by step 4.1, so their coordinate map is precisely the Veronese factorization, up to a projective basis change. Since $P\mathrm{\ne }Q$ have the same image under $\varphi$, their canonical images agree; hence the canonical map is not an embedding and has degree two onto the rational normal curve. Finally [F8], [F9] and step 4.1 give all displayed RR and duality dimensions.

By [F3] at $A=0$, $\mathrm{\ell }(K)=i(0)=g$. At $A=K$, duality gives $i(K)=\mathrm{\ell }(0)=1$, so Riemann–Roch gives $g-1=\mathrm{deg}K+1-g$, hence $\mathrm{deg}K=2g-2$. If $\mathrm{deg}D<0$, [F2] gives $\mathrm{\ell }(D)=0$ and [F3] gives $\chi ({\mathcal{O}}_X(D))=\mathrm{\ell }(D)-i(D)=-\mathrm{\ell }(K-D)$. If $\mathrm{deg}D>2g-2$, then $\mathrm{deg}(K-D)<0$ and [F2] gives $\mathrm{\ell }(K-D)=0$; [F3] therefore gives $i(D)=0$ and $\mathrm{\ell }(D)=\mathrm{deg}D+1-g$.

Suppose $\mathrm{deg}D=0$ and $0\mathrm{\ne }f\in L(D)$. The effective divisor $(f)+D$ has degree zero by [F2], so all its nonnegative coefficients vanish; thus $(f)=-D$ and $D\sim 0$. Conversely, if $D\sim 0$, [F2] and [F3] identify $L(D)$ with the one-dimensional space $L(0)=\mathbb{C}$. This proves the degree-zero equivalence and the dimension bound.

A degree-one proper nonconstant holomorphic map to the sphere is a biholomorphism: [F5] makes every fibre a single point of multiplicity $1$, so the map is bijective; [F6] provides local holomorphic inverses, which agree with its unique inverse and hence glue to a holomorphic inverse. It would force $g=0$ by [F7]. Now let $\mathrm{deg}D=1$, $g\ge 1$, and suppose $f_0,f_1\in L(D)$ are independent. Put $E_j=(f_j)+D$, effective of degree one by [F2], and $r=f_1\mathrm{/}{f}_0$. Independence makes $r$ nonconstant, and its pole divisor is bounded by $E_0$ since $(r)=E_1-E_0$. By [F4] and [F5], it is a proper map of degree at most one and at least one, contradicting the preceding genus consequence. Thus $\mathrm{\ell }(D)\le 1$. A nonzero section gives an effective degree-one divisor $(f)+D=[p]$, so $D\sim [p]$. Conversely $D\sim [p]$ identifies $L(D)$ with $L([p])$, which contains constants and has dimension at most one; hence equality holds. For effective degree-one $D=[p]$, this also proves $L(D)=\mathbb{C}$.

Let $\mathrm{deg}D=2$ and $g\ge 2$. Choosing any point $p$, [F8] and step 1.3 give $\mathrm{\ell }(D)\le \mathrm{\ell }(D-[p])+1\le 2$. Suppose $\mathrm{\ell }(D)=2$ and choose independent $f_0,f_1\in L(D)$, representing sections as in [F9]. Their effective zero divisors $E_j=(f_j)+D$ each have degree two. Let $B$ be their common effective divisor, with coefficient $B(q)=\min \{E_0(q),E_1(q)\}$. If $B\mathrm{\ne }0$, choose a point $q$ in its support; then both $f_j$ belong to $L(D-[q])$, contrary to the degree-one bound in step 1.3. Hence $B=0$. The nonconstant ratio $r=f_1\mathrm{/}{f}_0$ has divisor $E_1-E_0$; because the two effective divisors have disjoint support, its pole divisor is exactly $E_0$, of degree two. By [F4] and [F5], $r:X\to \widehat{\mathbb{C}}$ has degree two, proving hyperellipticity in the stated analytic sense. Its contrapositive and the bound $\mathrm{\ell }(D)\le 2$ give $\mathrm{\ell }(D)\le 1$ for every degree-two divisor on a nonhyperelliptic surface.

By [F2], local lift coordinates differ by translations, so their differentials glue to a nowhere-zero holomorphic differential $\omega =dz$ with $K=(\omega )=0$. By [F4], $i(0)=\mathrm{\ell }(K)=\mathrm{\ell }(0)=1$, so $g=1$ and $h^0(X,K_X)=1$. Thus every holomorphic differential is $cdz$. The prescribed principal part is nonzero by [F3] and would force a simple pole at $o$ with no other poles.

If a meromorphic function $f$ realized the data, [F5] would give $(f)=E-[o]$ with $E$ effective of degree one. Hence $E=[q]$ and $q\mathrm{\ne }o$. The weighted fibre over infinity has one simple point, so [F6] gives degree one; every fibre is then one point of multiplicity one. The local holomorphic inverses in [F6] glue to a global inverse, making $X$ biholomorphic to the sphere, contrary to $g=1$ in step 1.1. Independently, $f\omega$ would have residue $1$ at $o$ and zero elsewhere by [F3], contradicting [F7]. This directly proves nonsolvability without a good-cover hypothesis.

Choose $0<r<\delta \mathrm{/}4$ using [F2]. The quotient images of radius-$r$ plane balls cover $X$ and are disc chart domains; compactness gives a finite subcover. In the lift of any one member, another member that intersects it has at most one relevant translated radius-$r$ ball: two such centres would be within $4r<\delta$ of each other, contrary to [F2]. Thus every nonempty finite intersection lifts injectively to an intersection of finitely many plane balls. It is bounded, open and convex; straight-line contraction to an interior point makes it contractible, and [F8] makes it disc-biholomorphic. This is a finite good cover subordinate to the global holomorphic frame $1$ of ${\mathcal{O}}_X(0)$ from [F8]. Supply compatible metrics by [F9]; the canonical comparison in [F10] supplies the comparison map used by its residue criterion. By [F3], pairing the data with $cdz$ gives ${\mathrm{Res}}_o(cdz\mathrm{/}z)=c$, since all other terms vanish; replacing the representative $1\mathrm{/}z$ by a holomorphic perturbation leaves this residue unchanged. Step 1.1 identifies the entire differential space, so this is the complete residue functional, and [F10] is the exact necessary-and-sufficient obstruction criterion. Its value $1$ at $dz$ proves the claimed failure.