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A degree-zero complex analytic divisor $E$ on a compact Riemann surface is principal exactly when its Abel-Jacobi map of a compact Riemann surface is zero. Normalize a differential of the third kind with residues equal to the coefficients of $E$. The Riemann bilinear relations for a compact surface express its $b$-periods as $2\pi i$ times the Abel integrals. If those integrals are in the period lattice, subtract a suitable integral multiple of the normalized holomorphic differential forms; all periods then lie in $2\pi i\mathbb{Z}$, so exponentiation gives the required meromorphic function. Conversely, the logarithmic differential of a meromorphic function has integer residues and periods in $2\pi i\mathbb{Z}$, giving a zero Abel class by the same calculation.

On a compact Riemann surface, the degree of a holomorphic vector bundle is the degree of a line bundle of its determinant line bundle. A Hermitian metric on a holomorphic vector bundle and its Chern connection express this degree by the trace-curvature integral. The value is independent of the metric because the normalized trace curvature represents the First Chern class.

A meromorphic one-form on a compact Riemann surface with only simple poles is a differential of the third kind. Prescribed residues can be realized exactly when their sum is zero: in the residue exact sequence for $K_C(S)$, the obstruction in $H^1(C,K_C)\cong \mathbb{C}$ is the sum of residues. Subtracting a holomorphic differential form normalizes its $a$-periods. Integer residues and periods in $2\pi i\mathbb{Z}$ make the exponential of its integral a single-valued meromorphic function with those orders of zeros and poles.

Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. For a closed complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

For a nonzero holomorphic vector bundle $F$ on the curve, its degree of a holomorphic vector bundle and slope of a holomorphic vector bundle areWe obtain a bound independent of the particular holomorphic subbundle by fixing a Hermitian metric on a holomorphic vector bundle $E$ once and for all. Let $D_E$ be its Chern connection, $F_E=D_E^2$ its Chern curvature, and $\omega$ any positive area form on the compact Riemann surface. WriteHere $K$ is a smooth Hermitian endomorphism of $E$. By compactness, its largest eigenvalue is bounded above by some finite constant $C$ on the whole curve.

Let $F\subseteq E$ be a holomorphic subbundle of positive rank $r$, with the induced metric. Use the smooth orthogonal splitting of a vector subbundle $E=F\oplus F^{\perp }$. In this splitting the connection has block formwhere $D_F$ is the projected Chern connection and $B$ is the second fundamental form of a holomorphic subbundle. The latter has type $(1,0)$: the $(0,1)$ part of $D_E$ preserves the holomorphic subbundle. The adjoint $B^{†}$ includes conjugation of the form factor and has type $(0,1)$. Squaring this block matrix givesThis derives the curvature formula for a holomorphic subbundle in the column-vector convention used here. Locally write $B=bdz$. ThenThe degree of a holomorphic vector bundle is the trace-curvature integral, since the trace is the curvature of the determinant line bundle. HenceThe last integrand is a nonnegative real two-form and is independent of the local unitary frame and coordinate. Since $K\le C{I}_E$, the compressed trace is at most $rC$ pointwise. ThereforeThis is the subbundle slope bound from ambient Chern curvature. The right side depends only on $E$ and the initially chosen metrics, and is finite. It proves the requested uniform upper bound for every positive-rank subbundle, without assuming any stability property of $E$ or using a bound that depends on $F$.

Take the compact Riemann surface $X={\mathbb{P}}^1$ and the holomorphic vector bundle $E={\mathcal{O}}_X\oplus {\mathcal{O}}_X$. For every positive integer $m$, use homogeneous coordinates $[Z_0:Z_1]$ and the mapThe two sections $Z_0^m,Z_1^m$ never vanish together on ${\mathbb{P}}^1$. Thus this map has rank one at every point; locally one of its two components is invertible in suitable line-bundle frames, so its image is a genuine holomorphic subbundle and its quotient is a locally free sheaf. More explicitly, there is an exact sequenceExactness follows fibrewise from the absence of a common zero and hence also locally as a sequence of vector bundles. The image $F_m$ is isomorphic to ${\mathcal{O}}_X(-m)$ and has rank one, soThus there need not be a lower bound for subbundle slopes, even for a fixed trivial rank-two bundle. Merely multiplying a section of one fixed subline by a function with zeros would not suffice: that could give only a subsheaf with a torsion quotient. The nowhere-simultaneously-zero pair above avoids that problem and realizes unbounded negative degrees of line subbundles.

Let $X$ be a connected compact Riemann surface of geometric genus $g$, equivalently a smooth projective curve over $\mathbb{C}$, and put $V=H^0(X,K_X)$. Choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$ with $a_i\cdot b_j={\delta }_{ij}$. For a closed differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The underlying bilinear identity isIt holds for any two smooth closed one-forms. In particular, the Riemann bilinear relations for holomorphic differential forms arefor $\omega ,\eta \in V$ and $\omega \ne 0$ in the inequality. The inequality fixes the orientation and the sign convention.

To prove the identity, cut $X$ along the symplectic basis to a fundamental polygon $P$. On $P$, a closed one-form $\alpha$ has a primitive $F$. The Stokes theorem givesOpposite copies of each cut have opposite orientations, while their values of $F$ differ by the corresponding period of $\alpha$. Pairing those edges therefore leaves $A_i(\alpha ){\int }_{b_i}\beta -B_i(\alpha ){\int }_{a_i}\beta$ for the $i$th handle. Summing proves the identity. The same calculation can be made with a small disk about the polygon's vertices removed; its extra boundary terms tend to zero, so the vertices cause no additional term.

For holomorphic differential forms $\omega ,\eta$, their wedge product of differential forms vanishes: locally both are multiples of $dz$. This proves the first relation. If $\omega =f(z)dz$ and $z=x+iy$, thenA nonzero holomorphic differential form is nonzero on an open set, so the integral is strictly positive. Applying the identity to $\omega ,\overline{\omega }$ proves the second relation.

These relations first prove that the map $V\to {\mathbb{C}}^g$ taking $a$-periods is injective: if every $A_i(\omega )$ vanishes, the positive integral just computed would vanish. Since $\dim V=g$ by the Riemann-Roch theorem, it is an isomorphism. Consequently there is a unique basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$. Set ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. The first relation gives ${\tau }_{ij}={\tau }_{ji}$, and the second gives, for any nonzero column $c$,Thus the normalized period matrix of a complex torus is symmetric and has positive-definite imaginary part:

The integration map sends a cycle $\gamma$ to the functional $\omega \mapsto {\int }_{\gamma }\omega$ in $V^{\ast }$. In the normalized basis its image is the period latticeIndeed, the real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: its imaginary part is $Yy$, and $Y$ is invertible. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and has compact quotient. This proves thatis a well-defined $g$-dimensional complex torus, rather than merely a quotient by an arbitrary subgroup.

The relations also provide its canonical principal polarization of a complex torus. With the convention that a Hermitian form is conjugate-linear in its first argument, put $H(z,w)={\overline{z}}^{,t}{Y}^{-1}w$ and $E=\mathrm{Im}H$. Symmetry of $\tau$ givesIn particular $E$ is integral and unimodular on $\Lambda$, and it is compatible with the complex structure. These are exactly the positivity and integrality conditions for a Riemann form on a complex torus; they make the complex torus a Jacobian variety with a principal polarization of a complex torus. If instead one uses a first-argument-linear Hermitian form, the same alternating form is $-\mathrm{Im}H$. Changing the symplectic basis changes the coordinates but not this polarized quotient. For $g=0$, $V=0$ and the Jacobian variety is a point.

Fix $p_0\in X$ on a connected compact Riemann surface, and construct its Jacobian variety by periods as in the first solution. For a divisor on an algebraic curve $E={\sum }_p{n}_{p}p$ of degree zero, defineChanging the paths adds a period lattice element; changing $p_0$ adds a common integral multiplied by $\sum n_p=0$. The Abel theorem for divisors saysEquivalently, effective divisors on an algebraic curve of the same degree have the same Abelian sum exactly when they are equivalent under linear equivalence of divisors.

We need a meromorphic version of the Riemann bilinear relations. Let $\eta$ be a differential of the third kind, with residues $n_p$ at its simple poles, and let $\omega$ be a holomorphic differential form. Choose representatives of the cycles and paths avoiding the poles in the same cut polygon, and let $F(p)={\int }_{p_0}^p\omega$ there. ThenHere the equality is an equality of the chosen complex numbers, before taking any quotient. To prove it, remove small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ on the remaining polygon, its boundary integral is zero. Pairing opposite outer edges gives the left side, just as in the first solution. The clockwise boundary about $p$ contributes $-2\pi i{n}_{p}F(p)$ by the residue theorem. Moving these terms to the other side proves the formula and its sign.

Such an $\eta$ exists whenever $\sum n_p=0$. Here is the required existence argument. Let $S$ be the reduced support of $E$, and considerLocally the last map takes $cdz/z$ to $c$. By Serre duality, the obstruction in $H^1(K_X)\cong \mathbb{C}$ is dual to restriction of constant functions, so it is the sum of the residues. Thus the kernel of this sum is exactly the image of global meromorphic differentials. Alternatively, the Riemann-Roch theorem gives $h^0(K_X(S))-h^0(K_X)=|S|-1$, and the residue theorem identifies that image with the same codimension-one subspace. Subtracting a linear combination of the normalized holomorphic differential forms makes $A_i(\eta )=0$ for every $i$.

Suppose now that $\mathrm{AJ}(E)=0$. Let $u_i={\sum }_p{n}_p{\int }_{p_0}^p{\omega }_i$ with the paths used above. There are integer columns $m,n$ such that $u=m+\tau n$. The normalized meromorphic bilinear formula gives $B_i(\eta )=2\pi i{u}_i$. Thereforehas $a$-periods $-2\pi i{n}_i$ and $b$-periods $2\pi i{m}_i$. Its integrals around the poles are also in $2\pi i\mathbb{Z}$. These cycles generate the homology of the punctured surface, sois a single-valued, nowhere-zero holomorphic function away from the support of $E$. Near $p$, write ${\eta }^{\prime }=n_{p}dz/z+h(z)dz$. Then $f=z^{n_p}\exp (H(z))$ times a nonzero constant, with $H^{\prime }=h$. It extends meromorphically with order exactly $n_p$. Hence $\mathrm{div}(f)=E$.

Conversely, if $E=\mathrm{div}(f)$, take $\eta =df/f$. It has residues $n_p$ and every period lies in $2\pi i\mathbb{Z}$: continuation of a local logarithm of $f$ changes it by an integral multiple of $2\pi i$. Write $A_i(\eta )=2\pi i{r}_i$ and $B_i(\eta )=2\pi i{s}_i$. Applying the meromorphic bilinear formula to ${\omega }_i$ givesThus $u\in \Lambda$ and $\mathrm{AJ}(E)=0$, completing both directions without assuming the conclusion as a property of the Jacobian variety.

For completeness, the effective degree-$d$ version lives on the symmetric product of a curve $X^{(d)}=X^d/{\mathfrak{S}}_d$. The sum of integrals on $X^d$ is invariant under the finite symmetric group and descends to the Abelian sum map $u_d$. The quotient is locally described by elementary symmetric coordinates, so repeated points are included. The proof above identifies its fibre over $u_d(D)$ withExplicitly, a nonzero global section gives its effective zero divisor, and two sections give the same divisor precisely when their quotient is constant. These maps are holomorphic in local symmetric coordinates. More intrinsically, a family of effective divisors on an algebraic curve in this fixed class determines a line of sections of ${\mathcal{O}}_X(D)$, after locally trivializing any line bundle pulled back from the parameter space. Conversely such a line of sections gives its family of zero divisors on an algebraic curve. These operations are inverse and unaffected by that trivialization, proving the identification with the projective space in families as well as on points.

Finally this description identifies the analytic Jacobian variety with ${\mathrm{Pic}}^0(X)$. Every degree-zero line bundle has a meromorphic section: after twisting by a sufficiently large effective divisor, the Riemann-Roch theorem supplies a nonzero global section. Its divisor represents the original line bundle. The correspondence to the period quotient is well-defined and injective by the theorem. It is surjective because one may choose $g$ distinct points with independent evaluations of holomorphic differential forms; the derivative of the sum of their Abel integrals is then an isomorphism. Hence the subgroup generated by point differences contains an open subset of the connected complex torus, and must be the whole group. Local integration coordinates, or the exponential sequence for ${\mathcal{O}}_X$, give the same holomorphic identification. This justifies the subsequent use of degree-$d$ line bundles as points of a translate of the Jacobian variety.

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.

The argument gives the useful additional descriptionFor $r=1$, its nonzero linear equation is $⟨\xi ,st⟩=0$. This agrees with the derivative of the Abelian sum map: if $\mathrm{div}(s)=D$, then $st$ is the unique holomorphic differential form vanishing along $D$. For $r\ge 2$, the local equation has no linear term and the point is singular. When $g=1$, $W_0$ is the single trivial line bundle in ${\mathrm{Pic}}^0(X)$, with multiplicity one; the argument includes this case using the empty effective divisor. In genus zero there is no $W_{g-1}$ of effective divisors on an algebraic curve, so that hypothesis is necessary.

For a nonconstant holomorphic map between compact Riemann surfaces, choose local coordinates making the map $z\mapsto z^{e_p}$. Its local degree is $e_p$, and its ramification excess is $e_p-1$; branching order conventions use either of these quantities, so both are specified below. The degree $d$ is the number of inverse images of any regular value, or the sum of local degrees in any fiber. The Riemann-Hurwitz formula is $2g_X-2=d(2g_Y-2)+{\sum }_p(e_p-1)$.

The projective closure is $S^m-T^m-Z^m=0$. Its three partial derivatives cannot vanish simultaneously at any projective point, so it is nonsingular and inherits the complex structure of a smooth projective curve. The affine polynomial is irreducible by Eisenstein applied to $t^m-(s^m-1)$ at the prime $s-1$ in $\mathbb{C}[s]$, so the smooth projective curve is connected. At $Z=0$ its points are $[\zeta :1:0]$, where ${\zeta }^m=1$; these are exactly the $m$ points added at infinity.

The extension of the coordinate projection is $F=[S:Z]$. There is no point on the curve with $S=Z=0$, so this gives a holomorphic map everywhere. More explicitly, in the chart $T=1$ put $\sigma =S/T$, $\tau =Z/T$. Near an infinity point, ${\sigma }^m=1+{\tau }^m$ and $\sigma (0)=\zeta \ne 0$. Thus $\tau$ is a local coordinate and $1/F=\tau /\sigma$ has nonzero derivative $1/\zeta$ at zero. Each infinity point is unramified, with local degree one.

For a generic finite $s$, there are $m$ distinct solutions of $t^m=s^m-1$, so $\mathrm{deg}F=m$. Ramification occurs exactly at $t=0$, $s^m=1$. At each such point, $t$ is a local coordinate and $s-s_0=t^m/(m{s}_0^{m-1})+O(t^{2m})$, giving local degree $m$ and ramification excess $m-1$. There are no other critical points, since away from $t=0$ the projection $s$ is a local coordinate. Therefore

Adjoin ${\mathbb{P}}^1(\mathbb{Q})$ to the complex upper half-plane. Ordinary points retain their usual neighborhoods. At a rational boundary point $a=\sigma \infty$, a neighborhood basis consists of $a$ together with $\sigma \{z\in \mathbb{H}:\mathrm{Im}z>Y\}$, with $Y$ arbitrarily large. Different choices of $\sigma \in S{L}_2(\mathbb{Z})$ change it by a translation and sign, so give the same neighborhood system. Give $X(1)$ the quotient topology. The modular group acts properly discontinuously on $\mathbb{H}$; its modular cusp neighborhoods descend to punctured neighborhoods with their modular cusp point filled in. This quotient is Hausdorff: interior orbits can be separated by proper discontinuity, and a sufficiently high modular cusp neighborhood avoids any prescribed compact interior neighborhood.

Away from fixed points of the effective group $PS{L}_2(\mathbb{Z})$, a small disk is disjoint from its nonidentity translates, so its coordinate gives a holomorphic quotient coordinate. At an elliptic fixed point ${\tau }_0$, useIts effective stabilizer rotates $u$ by an $e$th root of unity, where $e=2$ at $i$ and $e=3$ at $\rho$. The quotient coordinate is $w=u^e$, not $u$. All rational boundary points lie in one modular cusp class. At infinity the stabilizer acts by $\tau \mapsto \tau +1$, andis a punctured-disk coordinate; adjoining $q=0$ fills in the modular cusp. These charts and their holomorphic transition maps define the complex structure of the compactified modular curve.

For compactness use the standard regionIts part with $\mathrm{Im}\tau \le Y$, $Y>1$, is compact: the imaginary part is at least $\sqrt{3}/2$. The quotient of its upper tail, with its two vertical sides identified and infinity added, is the closed disk $|q|\le e^{-2\pi Y}$. These two compact images cover $X(1)$, including the boundary-edge identifications. Thus $X(1)$ is a compact Riemann surface.

The $j$ function also extends over the modular cusp with one simple pole, since $j=q^{-1}+O(1)$. Its local expressions at elliptic points descend to holomorphic functions of $u^e$. By Question 1 it maps the remaining quotient bijectively onto $\mathbb{C}$, so its extension is a degree-one holomorphic map $X(1)\to {\mathbb{P}}^1$. A degree-one map of compact Riemann surfaces is biholomorphic: any local ramification would already give degree at least two. Therefore $X(1)$ is the Riemann sphere; this will also identify the base genus in Question 5.

A weight-$2k$ modular form gives an invariant tensor differentialbecause $d(\gamma \tau )=(c\tau +d{)}^{-2}d\tau$ cancels its automorphy factor. Assume first $k\ge 0$. At an ordinary quotient point this differential is holomorphic. At an elliptic point of order $e$, use $u$ and $w=u^e$ from part (a), and write the pullback as $A(u)(du{)}^k$, with $A$ holomorphic. Its invariance says $A(\zeta u){\zeta }^k=A(u)$, so a nonzero term $a_r{u}^r$ has $r+k\equiv 0(\mathrm{mod}e)$. Sincethe exponent in the descended coefficient is the integer $(r-k(e-1))/e$. With $r\ge 0$, it is at least $-\lfloor k(e-1)/e\rfloor$. At the modular cusp, $d\tau =dq/(2\pi iq)$ and $f$ has a power series with nonnegative exponents, so ${\omega }_f$ has a pole of order at most $k$.

Consequently the linear map $f\mapsto {\omega }_f$ injects $M_{2k}$ into the meromorphic sections of $K^{\otimes k}$ with poles bounded byThis section space is $H^0(X(1),K^{\otimes k}(D))$. It is finite-dimensional on a compact Riemann surface by the Riemann-Roch theorem. Concretely choose one nonzero meromorphic section of the line bundle; dividing by it identifies these sections with a Riemann-Roch space with a fixed finite divisor bound. This proves the requested finite-dimensionality by compact-surface theory.

In fact the injection is an isomorphism. If a descended coefficient has order at least $-\lfloor k(e-1)/e\rfloor$, its pulled-back exponent $ed+k(e-1)$ is nonnegative; the modular cusp pole bound likewise gives a holomorphic Fourier expansion of a modular form. These pullbacks therefore recover modular forms. Since $X(1)$ is the Riemann sphere, $\mathrm{deg}K=-2$, and a line bundle of degree $d$ on the sphere has $\max (0,d+1)$ sections. HenceFor negative $k$, the valence formula excludes a nonzero holomorphic modular form because all its orders would be nonnegative but their sum would be $k/6<0$; the space is zero. The mechanism behind finite-dimensionality is modular forms as meromorphic differentials on X(1), with explicit elliptic and modular cusp pole bounds.

Let $V=H^0(C,{\Omega }_C^1)$, of complex dimension $g$, and choose a symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(C,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. Integration gives a homomorphismWe first prove both injectivity and discreteness, rather than treating periods as a lattice by definition.

Cut the compact Riemann surface along this basis to a polygon. For closed one-forms $\omega ,\eta$, choose a primitive $F$ of $\omega$ on the cut polygon and apply Stokes theorem to $d(F\eta )=\omega \wedge \eta$. Pairing its boundary edges records the jumps of $F$ across the cuts and gives the Riemann bilinear relations for a compact surface:If a holomorphic differential form $\omega$ has all $a$-periods zero, take $\eta =\overline{\omega }$. The right side is zero, while in a holomorphic coordinate, for $\omega =f(z)dz$,Its integral is strictly positive unless $\omega =0$. Thus the $a$-period map $V\to {\mathbb{C}}^g$ is injective, hence an isomorphism. Choose a basis ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_j}{\omega }_i={\delta }_{ij}$, and put ${\tau }_{ij}={\int }_{b_j}{\omega }_i$. Applying the same boundary identity to ${\omega }_i,{\omega }_j$, whose wedge product is zero, gives ${\tau }_{ij}={\tau }_{ji}$. Applying it to $\omega =\sum c_i{\omega }_i$ and its conjugate givesHence $\mathrm{Im}\tau$ is positive definite.

In the dual coordinates, $I(\sum m_i{a}_i+n_i{b}_i)=m+\tau n$. The real-linear map $(x,y)\mapsto x+\tau y$ from ${\mathbb{R}}^{2g}$ to ${\mathbb{C}}^g$ is an isomorphism: taking imaginary parts of a zero gives $(\mathrm{Im}\tau )y=0$, then $y=0$ and $x=0$. Its image of ${\mathbb{Z}}^{2g}$ is therefore discrete and cocompact, and no nonzero integral cycle has zero image. Thus

The Abel-Jacobi map of a compact Riemann surface sends an effective complex analytic divisor $D=\sum Q_i$ to ${\sum }_i{\int }_{P_0}^{Q_i}({\omega }_1,\dots ,{\omega }_g)$ modulo $\Lambda$. Changing a path adds an integral period, so the map is well-defined. For two degree-$d$ complex analytic divisors, let $E=D-D^{\prime }={\sum }_P{n}_{P}P$, with $\sum n_P=0$, and putWe prove that $E$ is principal exactly when $v\in \Lambda$, including the construction of a meromorphic function.

First construct a differential of the third kind $\eta$ with residues $n_P$ at the support of $E$ and no other poles. For the reduced support $S$, the residue exact sequence isOn sheaf cohomology, the obstruction to prescribing residues belongs to $H^1(C,K_C)\cong \mathbb{C}$ by Serre duality; its pairing with the constant section of ${\mathcal{O}}_C$ is the sum of residues. This follows directly by integrating the local simple-pole terms around the boundary circles. Thus the obstruction is zero because $\sum n_P=0$, and the required $\eta$ exists. Subtract a linear combination of the normalized holomorphic differential forms so that all its $a$-periods are zero.

Take its poles and the integration paths inside the same cut polygon. Apply the boundary calculation above to a primitive $F_i(P)={\int }_{P_0}^P{\omega }_i$, removing small circles around the poles of $\eta$. Since $d(F_i\eta )=0$ away from the poles, the paired outer edges contribute $B_i(\eta )$, while the small circles contribute $2\pi i{\sum }_P{n}_P{F}_i(P)$ by the residue theorem. ThereforeThe same argument without normalizing $\eta$ gives the useful reciprocity identityThese identities use compatible cuts and paths; the final conclusions are unchanged by integral periods.

Suppose $v=m+\tau n$ with $m,n\in {\mathbb{Z}}^g$. DefineIts $a$-periods are $-2\pi i{n}_i$, and its $b$-periods are $2\pi i{m}_i$. Around a pole its integral is $2\pi i{n}_P$. These cycles generate the first homology group of the punctured surface, so every period of ${\eta }^{\prime }$ belongs to $2\pi i\mathbb{Z}$. Choosing an auxiliary point $Q$ away from the poles, setIt is single-valued away from the poles. Locally ${\eta }^{\prime }=n_{P}dt/t+\text{a holomorphic one-form}$, so $f=t^{n_P}$ times a nonvanishing holomorphic function. It extends meromorphically with $\mathrm{div}(f)=E$. This proves sufficiency and explicitly produces the required principal complex analytic divisor.

Conversely, if $E=\mathrm{div}(f)$, the meromorphic differential $df/f$ has residues $n_P$ and every period in $2\pi i\mathbb{Z}$, since each integral counts the winding number of $f$ about zero. Write its $a$-period vector as $2\pi ia$ and its $b$-period vector as $2\pi ib$, with $a,b\in {\mathbb{Z}}^g$. The reciprocity identity gives $v=b-\tau a\in \Lambda$. Since $E$ has degree zero, changing the base point cancels in its Abel sum; a base point at the support of $E$ causes no problem, because the Abel integrals use holomorphic forms. We have provedThis establishes the Abel theorem for divisors directly from the period and residue calculations.

For a compact Riemann surface with a symplectic basis $a_i,b_i$ of its first homology group, cutting along that basis and integrating a primitive givesfor closed smooth one-forms. For a holomorphic differential form $\omega$ and a differential of the third kind $\eta$, the corresponding boundary calculation givesThe paths and representatives of the cycles are chosen in the same cut surface. Normalized holomorphic differential forms consequently have a symmetric period matrix with positive definite imaginary part.