Periods, Jacobians, and Abel--Jacobi Theory: Examples and Counterexamples

The eight corners of the octagon lie in one vertex class: side pairings give $v_0\sim v_3\sim v_2\sim v_1\sim v_4\sim v_7\sim v_6\sim v_5\sim v_0$, and the corresponding corner sectors form one link cycle. There are four paired edges and one face. Every side pair has opposite exponents, so the face orientation descends to an orientation of the closed connected surface. The word has two commutator blocks, hence is the genus-two normal form by [F1].

Applying [F2] to this model gives $H_1({\mathrm{\Sigma }}_2;\mathbb{Z})={\mathbb{Z}}^4$ with ordered basis $e_1,e_2,e_3,e_4$.

Let $J=\mathrm{diag}(J_2,J_2)$ and let $x_1,x_2,x_3,x_4$ be the evaluation-dual cohomology basis. By [F3], $⟨x_p⌣x_q,[{\mathrm{\Sigma }}_2]⟩=J_{pq}$.

By the adjunction in [F4], $⟨x_q,D(x_p)⟩=⟨x_p⌣x_q,[{\mathrm{\Sigma }}_2]⟩=J_{pq}$, so $D(x_p)={\sum }_q{J}_{pq}{e}_q$. Put $z_q={\sum }_p{J}_{pq}{x}_p$. Since $⟨x_r,D(z_q)⟩={\sum }_p{J}_{pq}{J}_{pr}=(J^{\mathsf{T}}J{)}_{qr}={\delta }_{qr}$, we have $D(z_q)=e_q$ and $D^{-1}(e_q)=z_q$.

Substituting these coordinates into [F4] gives $⟨e_p,e_q⟩={\sum }_{r,s}{J}_{rp}{J}_{sq}{J}_{rs}=(J^{\mathsf{T}}JJ{)}_{pq}=J_{pq}$. Thus the displayed matrix is $\mathrm{diag}(J_2,J_2)$; its determinant is $\det (J_2{)}^2=1$, proving the symplectic and unimodular claims.

The same A-page suppliers [F2–F5] give the endpoint cases: for genus zero, $H_1=0$ and the unique form on the zero group has empty matrix and determinant $1$ by convention; for genus one, the standard square has $H_1\cong {\mathbb{Z}}^2$ with matrix $J_2$ and determinant $1$. These computations use the cellular, cup-pairing, and intersection-form suppliers rather than importing an examples-page result.

By the addition rule of [F1] applied with base point $q_0$, $u_{q_0}(p)=u_{q_0}(p_0)+[\omega \mapsto {\int }_{p_0}^p\omega ]=u_{q_0}(p_0)+u_{p_0}(p)$; hence $u_{p_0}(p)-u_{q_0}(p)=-u_{q_0}(p_0)$ for every $p$, the displayed degree-one formula. Summing with coefficients $n_p$ gives ${\sum }_p{n}_p{u}_{p_0}(p)-{\sum }_p{n}_p{u}_{q_0}(p)=-\left({\sum }_p{n}_p\right)u_{q_0}(p_0)=0$ because $\mathrm{deg}D={\sum }_p{n}_p=0$.

The formula $u((q)-(p))=[\omega \mapsto {\int }_{\gamma }\omega ]$ is the addition rule of [F1], read for the difference of two points; it is independent of $\gamma$ by the well-definedness lemma [F1]. Adding the two classes for the pairs $(q,p)$ and $(r,q)$ and using additivity of the integral under concatenation gives $u((q)-(p))+u((r)-(q))=u((r)-(p))$.

When $g\ge 1$, [F3] makes $u_{q_0}$ nonconstant, while $u_{q_0}(q_0)=0$ by [F1]. Hence there exists $p_0\mathrm{\ne }{q}_0$ with $u_{q_0}(p_0)\mathrm{\ne }0$. Step 1.1 then makes the degree-one difference $u_{p_0}(p)-u_{q_0}(p)$ nonzero for every $p$. This proves the claimed base-point dependence in degree one without assuming a torus model.

Claims: base-point independence on ${\mathrm{Div}}^0(X)$ by step 1.1, the path-integral formula and three-point additivity by step 1.2, and the degree-one dependence by step 2.1; all under the inherited AC of [F4].

The translation action of $\mathrm{\Lambda }$ on $\mathbb{C}$ leaves $dz$ invariant, so by [F2] $dz$ descends to a holomorphic differential on $X$, and it is nowhere vanishing because its local expressions are the constant coefficient $1$. Hence $\mathrm{\Omega }(X)\mathrm{\ne }0$, so $g\ge 1$ by [F4], and since a nonzero holomorphic differential has exactly $2g-2$ zeros counted with multiplicity while $dz$ has none, $2g-2=0$ and $g=1$; then [F4] gives ${\dim }_{\mathbb{C}}\mathrm{\Omega }(X)=1$ and, since $dz\mathrm{\ne }0$, $\mathrm{\Omega }(X)=\mathbb{C}\cdot dz$ with $(dz)=0$ of degree $0$; in particular $\mathrm{\ell }(K)=1$ and $\mathrm{deg}K=0$.

The loops ${\pi }_1,{\pi }_2$ lift to the paths $t\mapsto t{\omega }_1$ and $t\mapsto t{\omega }_2$ on $[0,1]$; by [F5] the path integral of $dz$ along ${\pi }_1$ is ${\omega }_1-0={\omega }_1$ and along ${\pi }_2$ is ${\omega }_2$. By [F3] the period lattice is exactly $e(H_1(X;\mathbb{Z}))=\mathrm{\Lambda }$; under $\mathrm{\Omega }(X{)}^{\ast }\cong \mathbb{C}$, $\alpha \mapsto \alpha (dz)$, the periods ${\omega }_1,{\omega }_2$ of the two standard loops are therefore the two generators of $\mathrm{\Lambda }$. Hence $\mathrm{Jac}(X)=\mathbb{C}\mathrm{/}\mathrm{\Lambda }=X$.

By [F5] the point map sends $p=[z]$ to the class of the functional $\omega \mapsto {\int }_{p_0}^p\omega$; under the identification $\mathrm{\Omega }(X)=\mathbb{C}dz$ of step 1.1 and $\mathrm{Jac}(X)=\mathbb{C}\mathrm{/}\mathrm{\Lambda }$ of step 1.2 this is the class of $z-z_0$, that is, $u_{p_0}(p)=p-p_0$ in the group $X$. Hence $u_{p_0}$ is the translation by $-p_0$, a biholomorphism. Consequently $u(D)=0$ for $D\in {\mathrm{Div}}^0(X)$ exactly when ${\sum }_p{n}_p(p-p_0)=0$ in $\mathbb{C}\mathrm{/}\mathrm{\Lambda }$, i.e. when the group sum of $D$ vanishes; by [F6] (Abel's criterion) this is exactly the condition that $D$ is principal, and ${\mathrm{Pic}}^0(X)\cong X$. For $D=(q)-(p)$ the class is $q-p$.

The four displayed claims are steps 1.1, 1.2 and 2.1, under the inherited AC of [F7].

The homogeneous equations define a closed subset of compact Hausdorff projective space by [F9]. On $Z_0\mathrm{\ne }0$, their points are $[1:x:x^2:x^3:y]$ with $y^2=x^5-1$. This affine equation is smooth: either $x\mathrm{\ne }0$, when the local inverse of $x\mapsto x^5$ expresses $x$ holomorphically in $y$, or $y\mathrm{\ne }0$, when the local inverse of $y\mapsto y^2$ expresses $y$ in $x$; both cannot vanish on the equation. If $Z_0=0$, the equations force $Z_1=Z_2=Z_4=0$, giving just $p_{\mathrm{\infty }}=[0:0:0:1:0]$. In the $Z_3=1$ chart put $u=Z_2,v=Z_4$; then $Z_1=u^2,Z_0=u^3$ and $v^2=u-u^6$. The inverse theorem for $u\mapsto u-u^6$ gives $u=v^{2}b(v)$, with $b(0)=1$ a holomorphic unit, so $v$ is a local coordinate. Thus $X_0$ is a smooth compact projective curve. The transformation $T_0(x,y)=(\zeta x,y)$ preserves the equations and has order five; it sends $(u,v)$ to $({\zeta }^{-1}u,{\zeta }^{2}v)$. The function $y=v\mathrm{/}{u}^3$ has a pole of order five at $p_{\mathrm{\infty }}$. For $y\mathrm{\ne }\pm i,\mathrm{\infty }$, its five distinct $x$-roots form one $T_0$-orbit. At $(0,\pm i)$ the $x$ chart gives local degree five, and the infinity chart gives the same degree. Hence $y$ is the degree-five orbit quotient to the sphere with precisely those three branch values; its positive local monodromies relative to $T_0$ are $1,1,3$, since the local multipliers are $\zeta ,\zeta ,{\zeta }^2$.

Set $\kappa =4^{1\mathrm{/}5}{e}^{\pi i\mathrm{/}5}$, so ${\kappa }^5=-4$, and on $\mathrm{\mid }z\mathrm{\mid }<1$ use the branch of $(1-z^5{)}^a$ normalized to one at zero. It exists by [F10] because $\mathrm{Re}(1-z^5)>0$. For $\epsilon =\pm 1$ put $x(z)=\kappa z(1-z^5{)}^{-2\mathrm{/}5}$ and $y_{\epsilon }(z)=\epsilon i(1+z^5)\mathrm{/}(1-z^5)$. Direct substitution gives $x^5=y_{\epsilon }^2+1$. The Cayley coordinate $r=(y-\epsilon i)\mathrm{/}(y+\epsilon i)$ lies in the unit disk in the corresponding upper or lower half-plane; its five roots $z$ give exactly the five points over $y$, and at $y=\epsilon i$ the parameter $x=\kappa z+O(z^6)$ is regular. Thus each formula parametrizes the full half-plane preimage biholomorphically by a disk. At a boundary root $z^5=1$, $u=1\mathrm{/}x$ and $v=y\mathrm{/}{x}^3$ tend to zero with orders $2\mathrm{/}5$ and $1\mathrm{/}5$, so the parametrization extends continuously to $p_{\mathrm{\infty }}$. Between successive boundary roots, $y_{+}=-\cot (5\theta \mathrm{/}2)$ runs from $-\mathrm{\infty }$ to $+\mathrm{\infty }$, while $y_{-}$ runs in reverse. The five open arcs cover the five real-$y$ sheets $x={\zeta }^j(1+y^2{)}^{1\mathrm{/}5}$.

Let $H(0)=0$ and $H^{\mathrm{\prime }}(z)=(1-z^5{)}^{-2\mathrm{/}5}$, using the primitive in [F10]. Rotation gives $H(\zeta z)=\zeta H(z)$. At a fifth root $a$ factor $1-z^5=(a-z)q(z)$, $q(a)=5a^4\mathrm{\ne }0$. A compatible local branch and the Taylor expansion of the holomorphic unit $q^{-2\mathrm{/}5}$ integrate to a constant plus a convergent series in $(a-z{)}^{n+3\mathrm{/}5}$; thus $H$ extends continuously there with difference $O(\mathrm{\mid }z-a{\mathrm{\mid }}^{3\mathrm{/}5})$. Away from these roots it extends holomorphically across each boundary arc. Consequently it is continuous on the closed disk. Its value $L=H(1)={\int }_0^1(1-r^5{)}^{-2\mathrm{/}5}dr$ is finite and positive. For $0<\theta <2\pi \mathrm{/}5$, $\arg (1-e^{5i\theta })=5\theta \mathrm{/}2-\pi \mathrm{/}2$, so $\arg (dH(e^{i\theta })\mathrm{/}d\theta )=7\pi \mathrm{/}10$ and its magnitude is positive. This arc therefore traces the straight segment from $L$ to $\zeta L$ monotonically; rotation gives the other four segments. The boundary map is a bijection onto the positively oriented convex regular pentagon with vertices $L{\zeta }^k$.

The boundary calculation alone is not a univalence assumption. If $w$ is off that polygon boundary, uniform convergence of $H(r{e}^{i\theta })$ to $H(e^{i\theta })$ makes the quotient $(H(r{e}^{i\theta })-w)\mathrm{/}(H(e^{i\theta })-w)$ lie in a disk of radius less than one about $1$ for $r$ sufficiently close to one. Its continuous argument returns to its initial value, so the loops have the same argument increment. The polygon increment is $2\pi$ for an interior $w$, since every ray from $w$ meets its convex boundary once; it is zero for exterior $w$, since a separating line puts the polygon in a half-plane with one continuous argument. By [F11] applied to $H-w$ on $\mathrm{\mid }z\mathrm{\mid }=r$, there is exactly one preimage, counted with multiplicity, for every interior $w$ and none for exterior $w$. Letting $r$ tend to one proves this on the entire disk. An interior point cannot map to the polygon boundary, because openness [F11] would then give exterior image points. Since $H^{\mathrm{\prime }}\mathrm{\ne }0$, [F9] gives a holomorphic inverse. Thus $H$ maps the open disk biholomorphically to the regular pentagon interior and its closed-disk extension is a homeomorphism to the closed pentagon.

Put $\alpha =dx\mathrm{/}y$. In the two disk parameters, differentiation gives $\alpha =\kappa \mathrm{/}(\epsilon i)(1-z^5{)}^{-2\mathrm{/}5}dz$. The primitive coordinates are therefore $W_{+}=cH$ and $W_{-}=-cH$, $c=\kappa \mathrm{/}i$, on two opposite regular pentagons. On real sheet $j$, oriented from $y=-\mathrm{\infty }$ to $+\mathrm{\infty }$, one has $\alpha =(2\mathrm{/}5){\zeta }^j(1+y^2{)}^{-4\mathrm{/}5}dy$, also at $y=0$ by continuity. Both face primitives have this derivative along the same seam, so their seam identification is a translation, with reversed boundary orientation. If $V_k=cL{\zeta }^k$, the upper arc between ${\zeta }^k,{\zeta }^{k+1}$ lies on sheet $j=k+1$; the paired opposite side is glued by $W\mapsto W-V_k-V_{k+1}$, sending $V_k$ to $-V_{k+1}$ and $V_{k+1}$ to $-V_k$. All ten vertices map to $p_{\mathrm{\infty }}$. The translation quotient maps continuously and bijectively to $X_0$ by the two disk parametrizations and these five seams, so its compact source and Hausdorff target satisfy the homeomorphism criterion in [F9]. Interior and seam coordinates are holomorphic primitives with nonzero derivative. At the common vertex the $v$ chart gives $\alpha =-v^{2}b(v)(2b(v)+v{b}^{\mathrm{\prime }}(v))dv$; a primitive is $v^3$ times a holomorphic unit. Its cube-root coordinate is $v$ times a holomorphic unit, and is locally invertible by [F9,F10]. This is exactly the holomorphic cone chart of the ten angles $3\pi \mathrm{/}5$ totaling $6\pi$. Hence the identification is analytic, including the vertex, and $T_0$ rotates both pentagons by $\zeta$.

The two closed-disk maps give an actual finite cell structure: one vertex $p_{\mathrm{\infty }}$, five oriented loop edges $e_j$ on the real-$y$ sheets, and two disk faces. Each upper boundary traverses every $e_j$ once positively, and the lower boundary traverses each once negatively by step 2.1. Thus [F12] gives $C_2={\mathbb{Z}}^2,C_1={\mathbb{Z}}^5,C_0=\mathbb{Z}$, $d_1=0$, and $d_2(a,b)=(a-b)(1,1,1,1,1)$. The singular-homology comparison yields $H_1(X_0;\mathbb{Z})={\mathbb{Z}}^5\mathrm{/}⟨e_0+e_1+e_2+e_3+e_4⟩$. The relation eliminates $e_4$ with coefficient one and imposes no relation among the first four, so they form an integral basis. With $C_0=[e_0]$, $T_0{e}_j=e_{j+1}$ gives the asserted four orbit basis and the five-orbit relation. Consequently $\mathbb{Z}[t]\mathrm{/}(1+t+\cdots +t^4)\to H_1$, $p\mapsto p(T_0)C_0$, is an isomorphism, not merely a full-rank submodule. The two disk closures meet along the seams, proving connectedness; $\chi =1-5+2=-2$ and [F12] give genus two.

Both $\alpha =dx\mathrm{/}y$ and $\beta =xdx\mathrm{/}y$ are holomorphic: away from $y=0$ this is immediate; at $y=0$, where $x^5=1$, their local expressions are $2\mathrm{/}(5x^4)dy$ and $2\mathrm{/}(5x^3)dy$. At infinity, with $u=v^{2}b(v)$ from step 1.1, $\alpha =-udu\mathrm{/}v=-v^{2}b(v)(2b(v)+v{b}^{\mathrm{\prime }}(v))dv$ and $\beta =-du\mathrm{/}v=-(2b(v)+v{b}^{\mathrm{\prime }}(v))dv$; at $x=0,y=\pm i$ the $x$ chart is regular. Their ratio is the nonconstant function $x$, so they are independent, and the dimension supplier [F1] applied after step 6.1 makes them a basis of $\mathrm{\Omega }(X_0)$. Direct pullback gives $T_0^{\ast }\alpha =\zeta \alpha$ and $T_0^{\ast }\beta ={\zeta }^2\beta$.

Their continuous closed-path periods on $C_0=e_0$ are the strictly positive numbers $I_1=(2\mathrm{/}5){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(1+y^2{)}^{-4\mathrm{/}5}dy$ and $I_2=(2\mathrm{/}5){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(1+y^2{)}^{-3\mathrm{/}5}dy$. These converge because the tails are $O(\mathrm{\mid }y{\mathrm{\mid }}^{-8\mathrm{/}5})$ and $O(\mathrm{\mid }y{\mathrm{\mid }}^{-6\mathrm{/}5})$. The holomorphic infinity chart and local primitive endpoint differences in [F5] identify these improper limits with the full continuous-path integrals, so neither normalization is zero. With ${\omega }_1=\alpha \mathrm{/}{I}_1,{\omega }_2=\beta \mathrm{/}{I}_2$, the real-sheet formulas give $P(T_0^k{C}_0,{\omega }_i)={\zeta }^{ki}$. Combined with the integral basis in step 6.1, this establishes the named witness for all hypotheses and its actual normalized period vectors.

The preceding construction proves the hypotheses for $X_0,T_0,C_0$. For any $X,T,C$ satisfying the general hypotheses, put $V=\mathrm{\Omega }(X)$, of complex dimension two by [F1], and $A=T^{\ast }{\mathrm{\mid }}_V$. Since $A^5=1$, the projections $E_j=\frac{1}{5}{\sum }_{k=0}^4{\zeta }^{-jk}{A}^k$ satisfy ${\sum }_{j=0}^4{E}_j=1$ and $A{E}_j={\zeta }^j{E}_j$, so $V$ is a sum of eigenspaces; [F3] excludes eigenvalue $1$. By [F4], the real-linear extension of the period map $H_1(X;\mathbb{Z}){\otimes }_{\mathbb{Z}}\mathbb{R}\to V^{\ast }$ is an isomorphism: it maps the real basis of cycles to the real basis given by their full-lattice period vectors. Naturality [F5] intertwines $T_{\ast }$ with the dual operator $\xi \mapsto \xi \circ A$. The cyclic basis of [F2] gives characteristic polynomial $t^4+t^3+t^2+t+1$ for $T_{\ast }$, with each of $\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4$ occurring once. On the underlying real space of $V^{\ast }$, its complexification has eigenvalues ${\lambda }_1,{\lambda }_2,{\bar{\lambda }}_1,{\bar{\lambda }}_2$, where ${\lambda }_1,{\lambda }_2$ are those of $A$. Consequently the exponents $k,l$ of ${\lambda }_1={\zeta }^k,{\lambda }_2={\zeta }^l$ are distinct and nonopposite modulo five. Their ratio $l\mathrm{/}k$ is $2$ or $3$ modulo five; by exchanging them if necessary it is $2$. Replace $T$ by $T^r$ with $rk=1$ modulo five. Its eigenvalues on $V$ are then $\zeta ,{\zeta }^2$, and $T^r$ generates the same ring as $T$, so $C$ remains a cyclic generator by [F2]. Choose corresponding eigenvectors ${\omega }_1,{\omega }_2$. This proves the required eigenvalue normalization without a separate Hodge-decomposition premise.