Period matrix and Jacobian of the pentagon curve

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.

For each eigenvector ${\omega }_i$, if $P(C,{\omega }_i)=0$, naturality [F5] gives $P(T^{k}C,{\omega }_i)={\zeta }^{ki}P(C,{\omega }_i)=0$ for every $k$. The cyclic basis [F2] then makes every period vanish on ${\omega }_i$. But [F4] says the real span of period functionals is all of $V^{\ast }$, so every complex-linear functional vanishes on ${\omega }_i$, forcing ${\omega }_i=0$, a contradiction. Thus rescaling by $P(C,{\omega }_i{)}^{-1}$ preserves the eigenvector and normalizes its period to $1$.

By naturality [F5] and the eigenvalue relations, $P(T^{k}C,{\omega }_i)=P(C,(T^k{)}^{\ast }{\omega }_i)={\zeta }^{ki}P(C,{\omega }_i)={\zeta }^{ki}$ for every $k\ge 0$; the case $k=4$ is consistent because $T^{4}C=-C-TC-T^{2}C-T^{3}C$ and ${\zeta }^{4i}=-{\sum }_{k=0}^3{\zeta }^{ki}$ for $i=1,2$.

Every class in $H_1(X;\mathbb{Z})$ is ${\sum }_k{a}_k{T}^{k}C$ with integers $a_k$ by [F2]; by additivity of $P$ and step 11.1 its period vector is $({\sum }_k{a}_k{\zeta }^k,{\sum }_k{a}_k{\zeta }^{2k})=(p(\zeta ),p({\zeta }^2))$ for the polynomial $p={\sum }_k{a}_k{t}^k$ reduced in $A$. Hence $\mathrm{\Lambda }$ is the displayed lattice; the identification with $\{(a,\sigma (a)):a\in \mathbb{Z}[\zeta ]\}$ uses $\sigma (\zeta )={\zeta }^2$.

The four period vectors of $C,TC,T^{2}C,T^{3}C$ are $(1,1),(\zeta ,{\zeta }^2),({\zeta }^2,{\zeta }^4),({\zeta }^3,\zeta )$ by step 11.1, and generate $\mathrm{\Lambda }$ by step 12.1. If ${\sum }_{k=0}^3{x}_k({\zeta }^k,{\zeta }^{2k})=0$ with real $x_k$, then $p(t):={\sum }_{k=0}^3{x}_k{t}^k$ vanishes at $\zeta ,{\zeta }^2$ and, by conjugation, at ${\zeta }^4,{\zeta }^3$. These four complex numbers are distinct, so [F7] forces $p=0$ and every $x_k=0$. The real coordinate matrix $M$ defined in clause 3 therefore has trivial kernel and is invertible by [F7]. Its columns generate $\mathrm{\Lambda }$, so the full-rank lattice definition gives $\mathrm{covol}(\mathrm{\Lambda })=\mathrm{\mid }\det M\mathrm{\mid }>0$, in agreement with [F4].

By [F6] the Jacobian is ${\mathbb{C}}^2\mathrm{/}\mathrm{\Lambda }$ with $\mathrm{\Lambda }$ as in step 12.1, every point of it is represented by a degree-zero divisor, and ${\mathrm{Pic}}^0(X)\cong \mathrm{Jac}(X)$; this is the last claim.

The concrete witness is established in steps 1.1–8.1, and claims 1-4 are steps 10.1, 12.1, 13.1 and 13.2, under the inherited AC of [F8].