Periods, Jacobians, and Abel--Jacobi Theory

By [F1], it is enough to compute on the indicated polygonal model; the homeomorphism transfers the resulting singular homology groups to $X$. In the commutator polygon, let $v_0,\dots ,v_{4g}=v_0$ be the successive corners. For each handle block $a_i{b}_i{a}_i^{-1}{b}_i^{-1}$, its side identifications join the four corners in that block successively: the $a_i$ pair identifies the first with the fourth and the second with the third, while the $b_i$ pair identifies the second with the next block's first and the third with the fourth. Thus every corner lies in one class. There are consequently one $0$-cell, $2g$ paired $1$-cells, and one $2$-cell.

For $g=0$, the sphere digon has two corner classes: its paired sides identify each corner with itself, leaving the two distinct corners $v_0,v_1$. Orient the single paired edge from $v_0$ to $v_1$. The $1$-cell incidence formula gives ${\mathrm{\partial }}_{1}e=v_1-v_0$. The attaching word $a{a}^{-1}$ has total exponent zero, so its $2$-cell has ${\mathrm{\partial }}_2=0$. Therefore $\ker {\mathrm{\partial }}_1=0$ and $\mathrm{coker}{\mathrm{\partial }}_1\cong \mathbb{Z}$.

When $g\ge 1$, every $1$-cell is a loop at the unique vertex, so ${\mathrm{\partial }}_1=0$ by [F3]. The coefficient of each $1$-cell in ${\mathrm{\partial }}_2$ is the degree of the attaching word after the other edges are collapsed; each label occurs once with each exponent, so that degree is $1-1=0$. Hence ${\mathrm{\partial }}_2=0$. The chain groups are $C_2=\mathbb{Z}$, $C_1={\mathbb{Z}}^{2g}$, $C_0=\mathbb{Z}$, and $C_q=0$ for $q\ge 3$.

Taking kernels modulo images in the chain complexes of steps 2.1 and 1.2 gives the stated cellular homology groups in both cases. By [F4], these are the singular homology groups, and the cellular generators in the commutator model are exactly its side-loop classes. Counting cells gives $1-2g+1=2-2g$ for $g\ge 1$ and $2-1+1=2$ for the digon, proving the Euler-characteristic claim by [F5].

To verify the subdivision claim, let the new vertex be $w$ and orient the two replacement edges from the old vertex $v$ to $w$ and from $w$ to $v$. Their boundaries are $w-v$ and $v-w$; all other loop edges still have zero boundary. In the attaching word the subdivided side occurs once in each direction, so each new edge also has total exponent zero and ${\mathrm{\partial }}_2=0$. The kernel of ${\mathrm{\partial }}_1$ is generated by the sum of the two replacement edges and the other $2g-1$ loop edges. This gives the asserted basis, with the original side class represented by that sum.

If $a<b$, around each point of the compact trace $\gamma ([a,b])$ choose a coordinate disk on which [F2] gives a local primitive of the coefficient in [F1]. The preimages of these disks cover $[a,b]$; [F5] supplies a Lebesgue number, so a finite subdivision can be chosen with each subpath contained in one primitive disk. There are only finitely many disk and primitive choices, so [F10] suffices and no unrestricted choice is used. If $a=b$, the empty subdivision gives the stated zero convention.

On a fixed coordinate disk, two local primitives have the same derivative $h$ in its coordinate; their difference has zero derivative and is constant by [F3]. Therefore replacing a chosen primitive on one subinterval does not change its endpoint increment.

Suppose a connected subpath lies in two coordinate disks $U,V$, with coordinates $z,w$ and local primitives $H_U,H_V$. Its image lies in one connected component of $U\cap V$ by [F4]. Put $G_U=H_U\circ z^{-1}$ and $G_V=H_V\circ w^{-1}$. In the $z$ coordinate, [F1] and [F9] give $\frac{d}{dz}(G_V(w(z))-G_U(z))=h_V(w(z))w^{\mathrm{\prime }}(z)-h_U(z)=0$, so [F3] makes this difference constant on that component. The two endpoint increments are equal.

Compare any two admissible subdivisions by taking the finite common refinement of their breakpoints. On each refined subinterval, the two original primitive disks both contain the path image, so step 1.3 identifies their increments; splitting an increment inside one disk changes nothing because the same primitive values telescope. Thus the two defining sums agree, proving independence of subdivision, chart, and local primitive.

Splitting the defining sum at a join proves additivity under concatenation; reversing each subinterval swaps the two endpoint values and negates the sum; for a constant path every endpoint increment is zero. If $\lambda ,\mu \in \mathbb{C}$, local primitives for $\lambda {\omega }_1+\mu {\omega }_2$ are $\lambda H_1+\mu H_2$, so the formula is complex-linear.

If $\gamma$ is piecewise $C^1$, refine the partition so each coordinate subpath is piecewise $C^1$. By [F6] it is a rectifiable contour, and [F7] identifies its usual contour integral with the increment of the local primitive. Summing the finitely many chart segments gives exactly the path integral defined above.

Let $K\subseteq Y$ be compact. By [F3], $K$ is closed in $Y$, and by [F2] its preimage $f^{-1}(K)$ is closed in $X$. Since $X$ is compact by [F1], $f^{-1}(K)$ is compact. This holds for every compact $K$, so $f$ is proper and the degree in [F4] is defined.

Suppose $d=1$. For every $y\in Y$, [F4] gives a nonempty finite fibre with ${\sum }_{x\in f^{-1}(y)}{e}_x(f)=1$. Each summand is a positive integer by [F5], so the fibre has exactly one point, with ramification index $1$. Thus $f$ is bijective and is unramified at every point.

For each $x\in X$, [F6] gives charts in which $f$ is $z\mapsto z$ near $x$, so it has a holomorphic local inverse near $f(x)$. The set-theoretic inverse $g$ from step 1.2 agrees with each such local inverse on its domain, hence is holomorphic on all of $Y$. Its defining identities $g\circ f={\mathrm{i}\mathrm{d}}_X$ and $f\circ g={\mathrm{i}\mathrm{d}}_Y$ show that $f$ is an isomorphism.

For a path contained in a coordinate disk $V$ identified with the unit disk, write its endpoints as $a,b$ and choose $0<r<r^{\mathrm{\prime }}<1$ with the entire compact path image in $\{\mathrm{\mid }z\mathrm{\mid }<r\}$. Choose $\psi$ equal to $1$ on $\{\mathrm{\mid }z\mathrm{\mid }\le r\}$ and supported in $\{\mathrm{\mid }z\mathrm{\mid }<r^{\mathrm{\prime }}\}$ by [F3]. If $a=b$, set $f_0=1$; the integral of a closed form on this path is zero by [F1]. If $a\mathrm{\ne }b$, put $R(z)=(z-b)\mathrm{/}(z-a)$ and use $L=\mathrm{Log}R$ outside the segment $[a,b]$. Indeed $R(z)$ belongs to the nonpositive real ray only on $[a,b]$, which lies inside $\{\mathrm{\mid }z\mathrm{\mid }<r\}$, so [F2] defines $L$ throughout the annulus $r<\mathrm{\mid }z\mathrm{\mid }<1$. Set $f_0=R$ for $\mathrm{\mid }z\mathrm{\mid }\le r$ and $f_0=\exp (\psi L)$ for $r<\mathrm{\mid }z\mathrm{\mid }<1$. These formulas agree smoothly across $\mathrm{\mid }z\mathrm{\mid }=r$: where $L$ is defined, $f_0=R\exp ((\psi -1)L)$, and $\psi -1$ vanishes smoothly on the inner disk. Since $f_0=1$ near $\mathrm{\partial }V$, extend it by $1$ off $V$. This gives a weak solution of $Q-P$ with a simple zero at $b$, a simple pole at $a$, and no other zeros or poles.

For claim 3, let $f$ and $g$ be weak solutions of $D$. Near a support point both have the local form $z^{n_p}{h}_f$ and $z^{n_p}{h}_g$ with the same integer $n_p=D(p)$ and smooth nowhere-vanishing factors, so $g\mathrm{/}f=h_g\mathrm{/}{h}_f$ is smooth and nowhere vanishing near $p$; away from $\mathrm{\mid }D\mathrm{\mid }$ both functions are smooth and nowhere vanishing. Hence $g\mathrm{/}f$ is smooth and nowhere vanishing on all of $X$.

The endpoint integral of [F1] is invariant under fixed-endpoint homotopy: subdivide the compact homotopy square into sufficiently small triangles in primitive neighborhoods, where every boundary sum telescopes, and cancel the interior edges. A coordinate disk contracts to a point, so this makes integration from a fixed point path independent on the disk. The resulting function is a local primitive plus a constant near each point, hence a smooth global primitive of any closed form there. Let $\omega =dg$ near the closed disk $\mathrm{\mid }z\mathrm{\mid }\le r^{\mathrm{\prime }}$, and multiply the local primitive by a cutoff supported in $V$ and equal to $1$ near that disk, obtaining a global smooth $g$ without changing $dg$ where $d{f}_0\mathrm{/}{f}_0$ is supported. Put $\alpha =d{f}_0\mathrm{/}{f}_0$ away from $a,b$. It is closed there, since $d(f_0^{-1}d{f}_0)=-f_0^{-2}d{f}_0\wedge d{f}_0=0$. The local form in [F4] shows that $\alpha \wedge dg$ is absolutely integrable: its coefficient is $O(1\mathrm{/}\mathrm{\mid }z-a\mathrm{\mid })+O(1\mathrm{/}\mathrm{\mid }z-b\mathrm{\mid })$. On a compact coordinate disk containing its support with small disks about $a,b$ deleted, $d(g\alpha )=dg\wedge \alpha =-\alpha \wedge dg$. The outer boundary contributes zero since $\alpha =0$ there. On the two inner circles the boundary orientation is clockwise; [F4] therefore gives $\int \alpha \wedge dg=2\pi i(g(b)-g(a))$ as their radii tend to zero. Thus $\frac{1}{2\pi i}{\int }_X(d{f}_0\mathrm{/}{f}_0)\wedge \omega =g(b)-g(a)={\int }_{\gamma }\omega$ by [F1]. For $a=b$ both sides vanish.

Subdivide each curve into finitely many subpaths contained in coordinate disks, using compactness and the Lebesgue-number argument of [F1], and construct the factor $f_{j\mathrm{\ell }}$ for each subpath by step 1.1. Each construction depends only on its subpath and disk, not on $\omega$; every closed form has a primitive on such a disk, since local primitive endpoint integrals are invariant under fixed-endpoint homotopy by subdividing a compact homotopy square into primitive neighborhoods, whose boundary increments telescope. Form the finite product $f={\prod }_{j,\mathrm{\ell }}{f}_{j\mathrm{\ell }}$. At any endpoint of the subdivided curves, each factor has an integer coordinate power times a smooth unit; changing a centered holomorphic coordinate multiplies that power by a holomorphic unit. Summing the endpoint exponents gives exactly the boundary of the original chain: intermediate endpoint contributions cancel, even when endpoints repeat or a subpath is closed. Therefore $f=z^{D(p)}h$ with $h$ smooth and nowhere zero at support points, and $f$ extends smoothly and nonvanishingly at all canceled intermediate endpoints. Away from the endpoints it is smooth and nonzero. On their complement, the finite product rule gives $df\mathrm{/}f={\sum }_{j,\mathrm{\ell }}d{f}_{j\mathrm{\ell }}\mathrm{/}{f}_{j\mathrm{\ell }}$.

Sum the absolutely convergent model identities from step 2.1 and use the product rule of step 2.2 and finite additivity of the path integral in [F1]. This gives $\frac{1}{2\pi i}{\int }_X(df\mathrm{/}f)\wedge \omega ={\sum }_{j,\mathrm{\ell }}{\int }_{{\gamma }_{j\mathrm{\ell }}}\omega ={\int }_c\omega$ for every closed smooth complex $1$-form. For holomorphic $\omega$, the $(1,0)$ term wedges to zero by [F7], while the local form $f=z^{D(p)}h$ gives $\bar{\mathrm{\partial }}f\mathrm{/}f=\bar{\mathrm{\partial }}h\mathrm{/}h$, a smooth form across every support point. Hence the displayed holomorphic identity has an ordinary smooth top-form integral.

The construction proves the weak-solution and integral claims in steps 2.2 and 3.1, and step 1.2 proves uniqueness up to a smooth nowhere-vanishing factor. The zero chain gives the empty product $f=1$; a nonzero closed chain can instead give a nonconstant nowhere-zero $f$, as its period identity requires. All selections in the construction are finite, and full AC covers the inherited Stokes hypothesis in [F9].

Choose the one-polygon normal form in [F1] with its orientation matching the canonical orientation of $X$. If this orientation gives the inverse commutator word, use the finite relabeling $a_i:=b_{g+1-i}$ and $b_i:=a_{g+1-i}$; since $[a,b{]}^{-1}=[b,a]$, this changes the reversed word back to ${\prod }_i{a}_i{b}_i{a}_i^{-1}{b}_i^{-1}$ without changing the quotient orientation. By [F2], for $g\ge 1$ the ordered side loops are a basis of $H_1(X;\mathbb{Z})$, and for $g=0$ the group is zero.

Suppose $g\ge 1$. The universal coefficient sequence [F3] has kernel ${\mathrm{Ext}}_{\mathbb{Z}}^1(H_0(X;\mathbb{Z}),\mathbb{Z})={\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Z},\mathbb{Z})=0$, since $\mathbb{Z}$ is free. Thus evaluation $\beta :H^1(X;\mathbb{Z})\to \mathrm{Hom}(H_1(X;\mathbb{Z}),\mathbb{Z})$ is an isomorphism; let $e_p^{\ast }$ be the coordinate functional with $e_p^{\ast }(e_q)={\delta }_{pq}$ and set $x_p={\beta }^{-1}(e_p^{\ast })$. These classes form the cohomology basis dual to the side-loop basis.

Suppose $g\ge 1$ and let $J=\mathrm{diag}(J_2,\dots ,J_2)$. The polygon cup-pairing computation [F4] gives $⟨x_p⌣x_q,[X]⟩=J_{pq}$, with the positive generator $\omega$ normalized by $⟨\omega ,[X]⟩=1$. Thus the cup-pairing matrix on this evaluation-dual basis is $J$, so it is unimodular.

Suppose $g\ge 1$. By the adjunction identity in [F5], $⟨x_q,D_X(x_p)⟩=⟨x_p⌣x_q,[X]⟩=J_{pq}$. Since $x_q$ is evaluation-dual to $e_q$ by step 1.2, these are the coordinates of $D_X(x_p)$, so $D_X(x_p)={\sum }_q{J}_{pq}{e}_q$. For each $q$ put $z_q={\sum }_p{J}_{pq}{x}_p$. Then $⟨x_r,D_X(z_q)⟩={\sum }_p{J}_{pq}{J}_{pr}=(J^{\mathsf{T}}J{)}_{qr}={\delta }_{qr}$, hence $D_X(z_q)=e_q$ and $D_X^{-1}(e_q)=z_q$.

Suppose $g\ge 1$. Substituting the inverse-duality coordinates from step 3.1 into the definition in [F5] gives $⟨e_p,e_q{⟩}_X={\sum }_{r,s}{J}_{rp}{J}_{sq}{J}_{rs}=(J^{\mathsf{T}}JJ{)}_{pq}=J_{pq},$ because $J^{\mathsf{T}}J=I$. Thus the matrix is $J$, each $2\times 2$ block has determinant $1$, and the form is unimodular for $g\ge 1$.

If $g=0$, [F2] gives $H_1(X;\mathbb{Z})=0$, and [F3] then gives $H^1(X;\mathbb{Z})=0$ because $H_0(X;\mathbb{Z})\cong \mathbb{Z}$ is free. Both pairings are on the zero group, whose empty matrix has determinant $1$ by convention, so both are unimodular. For $g\ge 1$, whenever smooth transverse embedded-curve representatives of the side-loop classes are given, [F6,F7] identifies their signed counts with the same intersection form. Reversing orientation changes $[X]$ to $-[X]$ and negates the intersection form; replacing each $b_i$ by $-b_i$ restores the displayed symplectic basis convention.

By definition, $[\theta ]=0$ in ${\mathrm{\Omega }}^{0,1}(X)\mathrm{/}\bar{\mathrm{\partial }}{\mathrm{\Omega }}^{0,0}(X)$ exactly when $\theta$ belongs to the image of the global operator, which is exactly the existence of a smooth $g$ with $\theta =\bar{\mathrm{\partial }}g$. Every $(0,1)$-form is closed because the next bidegree is $(0,2)=0$, so this quotient is the Dolbeault group stated above.

If $\theta$ is replaced by $\theta +\bar{\mathrm{\partial }}g$ and $\omega$ is holomorphic, then $d(g\omega )=\bar{\mathrm{\partial }}g\wedge \omega$: the term $\mathrm{\partial }g\wedge \omega$ has type $(2,0)$ and vanishes on a curve, while $d\omega =0$ by [F2]. Thus Stokes [F3] gives ${\int }_X(\bar{\mathrm{\partial }}g)\wedge \omega ={\int }_{X}d(g\omega )=0$. The integral therefore depends only on $[\theta ]$. It is complex-bilinear, since wedge product and integration are complex-linear in each factor. If $\theta =\bar{\mathrm{\partial }}g$, the same identity gives ${\int }_X\theta \wedge \omega =0$ for every $\omega$, proving (1)$\Rightarrow$(3).

Choose a compatible Hermitian metric on $X$ and a Hermitian metric on the trivial line bundle, as supplied by [F4]; the full Axiom of Choice implies the countable-choice assumption used for this metric existence. Apply [F4] to $E={\mathcal{O}}_X$, so $E^{\ast }\cong {\mathcal{O}}_X$ and $K\otimes E^{\ast }\cong K$. Its integration pairing is exactly $B$, hence the induced map $[\theta ]\mapsto B([\theta ],\cdot )$ is an isomorphism, in particular injective. If (3) holds, this functional is zero, so injectivity gives $[\theta ]=0$ and (2) follows; then (1) follows from step 1.1. This also covers the zero class and the case $H^0(X,K)=0$: in the latter case the isomorphism forces $H^{0,1}(X,{\mathcal{O}}_X)=0$, so the vacuous orthogonality condition still implies solvability. Full AC supplies the assumptions of harmonic-star duality and the countable choice required by the metric-existence and Stokes interfaces. The quotient calculation in step 1.1 uses no choice.