An elliptic curve on the quadric

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The curve of genus 21 at the centre of the construction lies on the quadric surface $\Sigma=\mathbb P^1\times\mathbb P^1$, where it has contact of order four with an elliptic curve $\Gamma$ at six points. This lesson builds $\Gamma$ and the data carried by the six points. Section 1 collects the facts about curves on $\Sigma$ used throughout the course: cohomology, degrees, and the residue description of the canonical bundle. Section 2 constructs an explicit smooth curve $\Gamma$ of bidegree $(2,2)$ on which the two rulings differ by a line bundle of exact order four. Section 3 attaches to $\Gamma$ a nondegenerate quadratic form on a two-dimensional space of sections. Section 4 chooses six points of $\Gamma$ and six vectors in that quadratic space with two normalizations that the gluing of vector fields in a later lesson requires.

Prerequisites. Derivations and the Lipman–Zariski problem for the overall plan. From the course Cohomology of quasi-coherent sheaves (AG-QC): Cohomology of projective space, Exercise 7.5 with its solution (cohomology of $\mathcal O(a,b)$ on $\mathbb P^1\times\mathbb P^1$ over any ring); Dualizing sheaves and Serre duality for projective schemes, Theorem 4.2 and Proposition 8.2. From the course The abstract theory of ideals (NOE-ABS): Algebraic functions of one variable: conductors and adjoints, Theorem 4.3 (Riemann–Roch). From Commutative algebra for geometry (AG-CA): Kähler differentials, Theorem 5.1; Smooth algebras over a field and the Jacobian criterion, Corollary 2.2.

Throughout, the ground field is $\mathbb C$. A curve is a smooth connected projective curve. Line bundles on a curve are written additively: $B+D$ means $B\otimes\mathcal O(D)$ for a divisor $D$, and $mB$ means $B^{\otimes m}$. For a line bundle $B$, $h^i(B)=\dim H^i(B)$. The canonical bundle of a curve $X$ is $K_X=\Omega^1_X$; by Proposition 8.2 of the lesson on dualizing sheaves it is the dualizing sheaf, so Theorem 4.2 there gives $h^1(B)=h^0(K_X-B)$, and Riemann–Roch reads $\begin{gathered} h^0(B)-h^0(K_X-B)\\ =\deg B+1-g(X),\\ g(X)=h^1(\mathcal O_X)=h^0(K_X). \end{gathered}$ Taking $B=K_X$ gives $\deg K_X=2g(X)-2$.

1. Curves on the quadric

Write $\Sigma=\mathbb P^1\times\mathbb P^1$ with homogeneous coordinates $[X_0:X_1]$ and $[Y_0:Y_1]$, affine coordinates $x=X_1/X_0$, $y=Y_1/Y_0$, and $\mathcal O(a,b)=\mathrm{pr}_1^*\mathcal O(a)\otimes\mathrm{pr}_2^*\mathcal O(b)$. For $a,b\ge0$, the global sections of $\mathcal O(a,b)$ are the bihomogeneous polynomials of bidegree $(a,b)$. In the chart $X_0Y_0\ne0$ a section $S$ is represented by its dehomogenization $s(x,y)=S(1,x;1,y)$, a polynomial of degree at most $a$ in $x$ and at most $b$ in $y$.

Put $V_0(d)=H^0(\mathbb P^1,\mathcal O(d))$ and $V_1(d)=H^1(\mathbb P^1,\mathcal O(d))$, of dimensions $\max(d+1,0)$ and $\max(-d-1,0)$. Exercise 7.5 of the lesson on cohomology of projective space proves $\begin{aligned} &H^i(\Sigma,\mathcal O(a,b))\\ &\quad\cong\bigoplus_{p+q=i}V_p(a)\otimes V_q(b). \end{aligned} \tag{1.1}$

Lemma 1.1 (vanishing on the quadric). (a) $H^0(\Sigma,\mathcal O(a,b))=0$ if $a<0$ or $b<0$. (b) $H^1(\Sigma,\mathcal O(a,b))=0$ unless one of $a,b$ is at least $0$ and the other at most $-2$; in particular it vanishes when $a,b\ge-1$ and when $a,b\le-1$. (c) $H^2(\Sigma,\mathcal O(a,b))$ has dimension $(a+1)(b+1)$ if $a,b\le-2$, and vanishes otherwise.

Proof. Read off (1.1): $V_0(d)\ne0$ exactly when $d\ge0$, and $V_1(d)\ne0$ exactly when $d\le-2$. $\square$

Lemma 1.2 (curves on the quadric). Let $K$ be a section of $\mathcal O(c,d)$, $c,d\ge1$, whose zero scheme $C=V(K)$ is smooth of dimension one. Then:

$H^0(C,\mathcal O_C)=\mathbb C$ and $h^1(\mathcal O_C)=(c-1)(d-1)$; in particular $C$ is connected, hence irreducible, and $g(C)=(c-1)(d-1)$.
$\deg\bigl(\mathcal O(a,b)|_C\bigr)=ad+bc$.

Proof. (1) Multiplication by $K$ gives the exact sequence $0\to\mathcal O(-c,-d)\to\mathcal O_\Sigma\to\mathcal O_C\to0$. By Lemma 1.1, $\mathcal O(-c,-d)$ has $H^0=H^1=0$ and $h^2=(c-1)(d-1)$, while $\mathcal O_\Sigma$ has $h^0=1$, $H^1=H^2=0$. The cohomology sequence gives $H^0(\mathcal O_C)\cong H^0(\mathcal O_\Sigma)=\mathbb C$ and $H^1(\mathcal O_C)\cong H^2(\mathcal O(-c,-d))$. A decomposition of $C$ into two disjoint nonempty closed subsets would give a nonconstant idempotent in $H^0(\mathcal O_C)$, so $C$ is connected. Its local rings are regular, hence domains, so two irreducible components cannot meet; being connected, $C$ is irreducible.

(2) Both sides are additive in $(a,b)$, so it suffices to treat $(1,0)$ and $(0,1)$. The line $\ell=V(X_0)\cong\mathbb P^1$ is not contained in $C$: otherwise the irreducible curve $C$ would equal $\ell$, and $K$ would be a unit multiple of a power of $X_0$, of bidegree $(m,0)$, while $d\ge1$. So $K|_\ell$ is a nonzero binary form of degree $d$ in $Y_0,Y_1$, and $X_0|_C$ is a nonzero section of $\mathcal O(1,0)|_C$. At each point the order of vanishing of a section of a line bundle on the curve $C$ is the length of the quotient of the local ring by it, so the degree of the divisor of $X_0|_C$ is the length of $\mathcal O_C/(X_0)=\mathcal O_\Sigma/(K,X_0)=\mathcal O_\ell/(K|_\ell),$ which is the number $d$ of zeros of $K|_\ell$ counted with multiplicity. The same argument with $V(Y_0)$ gives $\deg\mathcal O(0,1)|_C=c$. $\square$

Residues. For sections $S$ of $\mathcal O(c-2,d-2)$ and $K\neq0$ of $\mathcal O(c,d)$, let $S\varepsilon/K$ denote the rational two-form on $\Sigma$ equal to $s\,dx\wedge dy/k$ in the chart $X_0Y_0\ne0$, where $s,k$ are the dehomogenizations. In the chart with coordinates $x'=1/x$ and $y$, the dehomogenizations are $s'=x'^{c-2}s$ and $k'=x'^{c}k$, and $dx=-x'^{-2}dx'$, so $\frac{s\,dx\wedge dy}{k}=-\frac{s'\,dx'\wedge dy}{k'} ;$ the charts with $y'=1/y$ behave in the same way. Hence the poles of $S\varepsilon/K$ lie on $V(K)$. The notation records that the two-form $\varepsilon$, equal to $dx\wedge dy$ in the first chart, has no zeros and has poles of order two along $V(X_0)$ and $V(Y_0)$; in particular $K_\Sigma\cong\mathcal O(-2,-2)$.

Let $C$ be a smooth curve on a smooth surface and $\psi$ a rational two-form near a point of $C$ such that $k\psi$ is regular, where $k$ is a local equation of $C$ with $dk\ne0$ along $C$. Writing $k\psi=dk\wedge\alpha+k\beta$ with regular forms $\alpha,\beta$ (possible because $dk$ is part of a local basis of one-forms), the residue $\operatorname{Res}_C\psi$ is the restriction of $\alpha$ to $C$. The normal differential stands first: $\operatorname{Res}_{y=0}\bigl((dy/y)\wedge dz\bigr)=dz$.

The residue is well defined. If $dk\wedge\alpha=k\beta'$ for regular $\alpha,\beta'$, then at each point of $C$ the two-form $dk\wedge\alpha$ vanishes, so $\alpha$ is a multiple of $dk$ pointwise along $C$, and its restriction to $C$ is zero. If $k$ is replaced by $uk$ with $u$ a unit, then $\begin{aligned} uk\psi&=u\,dk\wedge\alpha+uk\beta\\ &=d(uk)\wedge\alpha\\ &\quad+uk\bigl(\beta-u^{-1}du\wedge\alpha\bigr), \end{aligned}$ so the same $\alpha$ serves for the new equation.

Lemma 1.3 (adjunction). In the situation of Lemma 1.2, the map $S|_C\mapsto\operatorname{Res}_C(S\varepsilon/K)$ is an isomorphism of line bundles $\mathcal O(c-2,d-2)|_C\xrightarrow{\ \sim\ }K_C .$ In the chart $X_0Y_0\ne0$ it is given by $\begin{aligned} &\operatorname{Res}_C\Bigl(\frac{s\,dx\wedge dy}{k}\Bigr)\\ &\quad=-\frac{s\,dx}{k_y}=\frac{s\,dy}{k_x}, \end{aligned} \tag{1.2}$ the first expression where $k_y\ne0$ and the second where $k_x\ne0$. In particular $\deg K_C=(c-2)d+(d-2)c$, which agrees with $2g(C)-2$.

Proof. If $S$ vanishes on $C$, then $S=KS'$ locally and $S\varepsilon/K=S'\varepsilon$ is regular, with residue zero; so the map is defined on $\mathcal O(c-2,d-2)|_C$. In the chart, $dk\wedge dx=k_y\,dy\wedge dx$, so where $k_y\ne0$, $\frac{s\,dx\wedge dy}{k}=\frac{dk}{k}\wedge\Bigl(-\frac{s\,dx}{k_y}\Bigr),$ which gives the first expression; the second follows in the same way from $dk\wedge dy=k_x\,dx\wedge dy$. On $C$ the relation $k_x\,dx+k_y\,dy=0$ holds in $\Omega^1_C$ (Theorem 5.1 of the lesson on Kähler differentials), so the two expressions agree where both are defined. By the Jacobian criterion (Corollary 2.2 of the lesson on smooth algebras), $k_x$ and $k_y$ have no common zero on the smooth curve $C$. Where $k_y\ne0$, the same Jacobian presentation shows that $\Omega^1_C$ is free on $dx$, so $-s\,dx/k_y$ generates $\Omega^1_C$ exactly where $s\ne0$; similarly where $k_x\ne0$. In the other charts the same computation applies to $\pm s'\,dx'\wedge dy/k'$. Hence the map is an isomorphism of line bundles. The degree follows from Lemma 1.2(2). $\square$

2. A curve of bidegree (2,2) with a torsion line bundle

Lemma 2.1 (smoothness of a (2,2) curve). Let $G$ be a section of $\mathcal O(2,2)$, written $G=\alpha(X)\,Y_0^2+\beta(X)\,Y_0Y_1+\gamma(X)\,Y_1^2$ with binary quadratic forms $\alpha,\beta,\gamma$ in $X=(X_0,X_1)$, and let $\delta=\beta^2-4\alpha\gamma$, a binary quartic form. If $\delta$ has four distinct zeros on $\mathbb P^1$, then $V(G)$ is smooth. Moreover, $V(G)$ meets the fibre $\{x\}\times\mathbb P^1$ in one point if $\delta(x)=0$ and in two points otherwise.

Proof. Fix $x\in\mathbb P^1$ and let $q_x(Y)=\alpha(x)Y_0^2+\beta(x)Y_0Y_1+\gamma(x)Y_1^2$, defined up to a nonzero factor. If $q_x$ were identically zero, then $\alpha,\beta,\gamma$ would all vanish at $x$, and $\delta$ would vanish to order at least two there; so $q_x\ne0$, and $V(G)\cap(\{x\}\times\mathbb P^1)$ is the zero set of the binary quadratic $q_x$, with discriminant $\delta(x)$.

If $\delta(x)\ne0$, $q_x$ has two simple zeros, and at each of them the derivative of $G$ along the fibre is nonzero, so $V(G)$ is smooth there.

If $\delta(x)=0$, $q_x$ has a double zero $y_0$. Choose a local coordinate $t$ on the first factor with $t(x)=0$, and an affine coordinate $s$ on the second factor with $s(y_0)=0$ and $s=\infty$ not a zero of $q_x$. Near $(x,y_0)$, in a trivialization, $G=a(t)+b(t)s+c(t)s^2$ with $a(0)=b(0)=0$ and $c(0)\ne0$. The discriminant $b^2-4ac$ equals $\delta$ times a nowhere vanishing function, because changing the fibre coordinate by a Möbius transformation and the trivialization by a unit multiplies the discriminant of a binary quadratic by a nonzero factor. Differentiating at $t=0$, $\frac{d}{dt}\bigl(b^2-4ac\bigr)\Big|_{t=0}=-4a'(0)c(0).$ Since $x$ is a simple zero of $\delta$, $a'(0)\ne0$, so $\partial G/\partial t\ne0$ at $(x,y_0)$, and $V(G)$ is smooth there. $\square$

The fibres over the four zeros of $\delta$ are the four points where the projection $V(G)\to\mathbb P^1$ to the first factor is ramified: at such a point, $s$ is a local coordinate on the curve and $t=-c(0)s^2/a'(0)+O(s^3)$.

The zigzag curve. Consider the eight points $\begin{gathered} (0,0),\ (0,1),\ (1,1),\ (1,2),\\ (2,2),\ (2,3),\ (3,3),\ (3,0) \end{gathered}$ of the chart $X_0Y_0\ne0$: a closed path on the grid $\{0,1,2,3\}^2$ that alternates vertical and horizontal steps. They lie on the curve $\Gamma=V(G)$, where $G$ has dehomogenization $\begin{aligned} g(x,y)={}&x^2y^2-5x^2y\\ &+3x^2-xy^2\\ &+11xy-9x\\ &-3y^2+3y . \end{aligned} \tag{2.1}$ On the grid lines, $\begin{aligned} g(0,y)&=-3y(y-1),\\ g(1,y)&=-3(y-1)(y-2),\\ g(2,y)&=-(y-2)(y-3),\\ g(3,y)&=3y(y-3),\\ g(x,0)&=3x(x-3),\\ g(x,1)&=-x(x-1),\\ g(x,2)&=-3(x-1)(x-2),\\ g(x,3)&=-3(x-2)(x-3). \end{aligned}$ Each of these has degree exactly two, so it has no zero at infinity, and each grid line meets $\Gamma$ exactly in the two marked points on it. Writing $g=A(x)y^2+B(x)y+C(x)$ with $A=x^2-x-3$, $B=-5x^2+11x+3$, $C=3x^2-9x$, the quartic $\delta$ of Lemma 2.1 has dehomogenization $\begin{gathered} \delta(x)=B^2-4AC\\ =13x^4-62x^3+91x^2-42x+9 . \end{gathered}$ Its leading coefficient is $13\ne0$, so $\delta$ does not vanish at $x=\infty$, and its discriminant is $2^8\cdot3^8\cdot7\ne0$, so its four zeros in $\mathbb C$ are distinct. By Lemma 2.1, $\Gamma$ is smooth. By Lemma 1.2, $\Gamma$ is irreducible of genus $(2-1)(2-1)=1$. (The factorizations and the discriminant are finite exact computations; the check script of this course verifies them and, independently, that $g,g_x,g_y$ generate the unit ideal in each of the four charts.)

Switches. Put $F=\mathcal O(1,0)|_\Gamma$ and $S=\mathcal O(0,1)|_\Gamma$, both of degree two by Lemma 1.2(2). For $p\in\Gamma$, the vertical line through $p$ is the zero set of a section of $\mathcal O(1,0)$; its restriction to $\Gamma$ has a divisor of degree two containing $p$, which we write $p+\iota_F(p)$. This defines a map $\iota_F$ on the points of $\Gamma$ with $\iota_F^2=\mathrm{id}$; its fixed points are the four ramification points of the first projection. Define $\iota_S$ in the same way with horizontal lines, and put $T=\iota_S\circ\iota_F .$ By construction $p+\iota_F(p)$ is a divisor of $F$ and $p+\iota_S(p)$ a divisor of $S$ for every $p$.

Lemma 2.2. On a curve $X$ of genus one, $K_X\cong\mathcal O_X$, and for points $p,q$ one has $\mathcal O_X(p)\cong\mathcal O_X(q)$ only if $p=q$. A line bundle of degree $e>0$ has $h^0=e$; a line bundle of degree $0$ has $h^0=1$ if it is trivial and $h^0=0$ otherwise; a line bundle of negative degree has $h^0=0$.

Proof. $\deg K_X=0$ and $h^0(K_X)=1$; a nonzero section of a degree-zero line bundle has no zeros, so it trivializes the bundle. Hence $K_X\cong\mathcal O_X$. A line bundle of negative degree has no nonzero section, since the divisor of a section is effective. For $\deg B=e>0$, Riemann–Roch gives $h^0(B)=e+h^0(-B)=e$. For $\deg B=0$, a nonzero section trivializes $B$. Finally $h^0(\mathcal O_X(p))=1$, so if $p\ne q$ and $\mathcal O_X(p)\cong\mathcal O_X(q)$, a rational function with divisor $q-p$ would be a nonconstant section of $\mathcal O_X(p)$, a contradiction. $\square$

Lemma 2.3 (switches and torsion). Let $D_0=S-F$. For every $p\in\Gamma$:

$\mathcal O_\Gamma(T(p)-p)\cong D_0$, and $\mathcal O_\Gamma(T^k(p)-p)\cong kD_0$ for $k\ge1$.
$kD_0\cong\mathcal O_\Gamma$ if and only if $T^k(p)=p$; in particular, whether $T^k(p)=p$ does not depend on $p$.
$\iota_FT\iota_F=T^{-1}$.

Proof. (1) $p+\iota_F(p)$ is a divisor of $F$ and $\iota_F(p)+\iota_S(\iota_F(p))$ a divisor of $S$; subtracting, $T(p)-p$ is a divisor of $S-F$. Summing over $T^i(p)$, $0\le i<k$, gives the second statement. (2) If $T^k(p)=p$, then $kD_0\cong\mathcal O_\Gamma(0)$. If $T^k(p)\ne p$ and $kD_0$ were trivial, then $\mathcal O_\Gamma(T^k(p))\cong\mathcal O_\Gamma(p)$, contradicting Lemma 2.2. (3) Since $\iota_F^2=\iota_S^2=\mathrm{id}$, $\iota_FT\iota_F=\iota_F\iota_S=T^{-1}$. $\square$

Proposition 2.4. On the zigzag curve, $D_0=S-F$ has exact order four. Explicitly, the sections $\begin{gathered} P_X=\prod_{i=0}^3(X_1-iX_0),\\ P_Y=\prod_{i=0}^3(Y_1-iY_0) \end{gathered}$ of $\mathcal O(4,0)$ and $\mathcal O(0,4)$ restrict to sections of $4F$ and $4S$ with the same divisor, the eight zigzag points, each simple, and $\zeta:4F\xrightarrow{\ \sim\ }4S,\qquad \zeta(P_X|_\Gamma)=P_Y|_\Gamma,$ is an isomorphism.

Proof. From the factorizations on the grid lines, $\begin{gathered} \iota_F(0,0)=(0,1),\ \iota_S(0,1)=(1,1),\\ \iota_F(1,1)=(1,2),\ \iota_S(1,2)=(2,2), \end{gathered}$ $\begin{gathered} \iota_F(2,2)=(2,3),\ \iota_S(2,3)=(3,3),\\ \iota_F(3,3)=(3,0),\ \iota_S(3,0)=(0,0). \end{gathered}$ So $T$ moves $(0,0)$ to $(1,1)$, $(2,2)$, $(3,3)$ and back to $(0,0)$: $T^4(0,0)=(0,0)$ and $T^2(0,0)=(2,2)\ne(0,0)$. By Lemma 2.3(2), $4D_0$ is trivial and $2D_0$ is not, so $D_0$ has order exactly four. The section $P_X|_\Gamma$ vanishes on the four vertical grid lines, hence at the eight zigzag points; since $\deg4F=8$, these are all its zeros and each is simple. The same holds for $P_Y|_\Gamma$. Two sections of line bundles with the same divisor determine an isomorphism carrying one to the other. $\square$

The bundle identities. Define on $\Gamma$ $\Delta=2D_0,\quad \Phi=F+\Delta,\quad M=3\Phi .$ Thus $\Delta$ is nontrivial with $2\Delta\cong\mathcal O_\Gamma$. Every line bundle in this list, and every restriction $\mathcal O(a,b)|_\Gamma=aF+bS$, is of the form $aF+bS$ with $a,b\in\mathbb Z$: $\Delta=-2F+2S$, $\Phi=-F+2S$, $M=-3F+6S$. We fix once and for all the isomorphisms $\begin{aligned} &\zeta^m:\ aF+bS\\ &\quad\xrightarrow{\ \sim\ }(a-4m)F\\ &\quad+(b+4m)S\quad(m\in\mathbb Z) \end{aligned} \tag{2.2}$ induced by $\zeta$. They are compatible with tensor products and with composition, $\zeta^m\zeta^{m'}=\zeta^{m+m'}$. Every identification below between bundles of this form is one of them. With $\Gamma|_\Gamma=\mathcal O(2,2)|_\Gamma=2F+2S$, they give $\begin{gathered} \Gamma|_\Gamma=4F+\Delta,\qquad M=3F+\Delta,\\ 2M=6F,\qquad 2\Phi=2F, \end{gathered}$ $\begin{gathered} 4M=(2\Gamma+4F)|_\Gamma,\\ 2M-(\Gamma+F)|_\Gamma=\Phi,\\ (\Gamma+2F)|_\Gamma-2M=\Delta . \end{gathered} \tag{2.3}$ For instance $4M=-12F+24S$ and $(2\Gamma+4F)|_\Gamma=8F+4S$ are identified by $\zeta^{-5}$. In (2.3), $\Gamma$ and $F$ also denote the line bundles $\mathcal O(2,2)$ and $\mathcal O(1,0)$ on $\Sigma$, a notation used from here on.

3. The quadratic form

The logarithmic form. Put $\begin{gathered} \Omega=\frac{\varepsilon}{G},\qquad \Omega=\frac{dx\wedge dy}{g}\\ \text{in the chart }X_0Y_0\ne0 . \end{gathered}$ By Section 1, $\Omega$ is a rational two-form on $\Sigma$ without zeros and with a simple pole exactly along $\Gamma$ (the case $(c,d)=(2,2)$, $S=1$ of the discussion before Lemma 1.3). Its residue $\omega=\operatorname{Res}_\Gamma\Omega=-\frac{dx}{g_y}=\frac{dy}{g_x}$ is a nowhere vanishing differential on $\Gamma$, by Lemma 1.3 with $\mathcal O(0,0)|_\Gamma=\mathcal O_\Gamma$.

The Wronskian section. Let $u=(u_1,u_2)$ be a basis of $H^0(\Sigma,\mathcal O(1,0))$ obtained from $(X_0,X_1)$ by a matrix of determinant one; the next lesson fixes a convenient choice. Restricting to $\Gamma$, put $\begin{aligned} \eta_\Gamma&=u_1\,du_2-u_2\,du_1\\ &\in H^0(\Gamma,K_\Gamma+2F). \end{aligned}$ This means: in a local frame $e$ of $F$, with $u_i=f_ie$, the section is $(f_1\,df_2-f_2\,df_1)\,e^2$. It does not depend on the frame, since replacing $e$ by $he$ replaces $f_i$ by $f_i/h$, and $\begin{aligned} &(f_1/h)\,d(f_2/h)-(f_2/h)\,d(f_1/h)\\ &\qquad=h^{-2}(f_1\,df_2-f_2\,df_1). \end{aligned}$ Replacing $u$ by $Au$, $A\in GL_2(\mathbb C)$, multiplies $\eta_\Gamma$ by $\det A$, so $\eta_\Gamma$ is the same for all admissible $u$ and may be computed with $u=(X_0,X_1)$: in the frame $X_0$, $f=(1,x)$ and $\eta_\Gamma=dx\cdot X_0^2$.

Lemma 3.1. The section $\eta_\Gamma$ vanishes exactly at the four ramification points $r_1,\ldots,r_4$ of the first projection $\Gamma\to\mathbb P^1$, each simply. In the chart $X_0Y_0\ne0$, $-\frac{\eta_\Gamma}{\omega}=g_y\cdot X_0^2 .$

Proof. By Lemma 1.3, $\omega=-dx/g_y$ where $g_y\ne0$, so $\eta_\Gamma/\omega=-g_yX_0^2$ on that set, and by continuity on the whole chart. Its zeros in the chart are the points of $\Gamma$ with $g_y=0$, which are the ramification points of the first projection lying in the chart. At such a point, in the coordinates of the proof of Lemma 2.1, $g$ is a unit times $G=a(t)+b(t)s+c(t)s^2$ and $s$ is a Möbius function of $y$, so along $\Gamma$, where $G=0$, the derivative $g_y$ is a unit times $\partial G/\partial s$. There $s$ is a local coordinate on $\Gamma$, $t=O(s^2)$ along $\Gamma$, and $\partial G/\partial s=b(t)+2c(t)s$, which is $2c(0)s+O(s^2)$, a simple zero. The section $\eta_\Gamma$ does not depend on a chart, and the same computation in the charts with $x'=1/x$ or $y'=1/y$, where $\eta_\Gamma/\omega$ is $\pm$ the derivative of the local equation along the fibre times a frame, covers the remaining ramification points. The total number of zeros is $\deg(K_\Gamma+2F)=4$, so these are all. $\square$

The Φ-map. By Lemma 2.2, $h^0(\Gamma,\Phi)=2$, and for each $q\in\Gamma$ the sections of $\Phi$ vanishing at $q$ form a line, since $h^0(\Phi-q)=1$. The divisor of such a section is $q+\iota_\Phi(q)$ for a unique point $\iota_\Phi(q)$. In particular $\Phi$ has no base point, and it defines a morphism of degree two $\begin{gathered} \varphi_\Phi:\Gamma\to\mathbb P\bigl(H^0(\Gamma,\Phi)^*\bigr)\cong\mathbb P^1,\\ q\mapsto[\mathrm{ev}_q], \end{gathered}$ where $\mathrm{ev}_q(\sigma)=\sigma(q)/e(q)$ for a local frame $e$ of $\Phi$ at $q$; changing $e$ rescales $\mathrm{ev}_q$. The fibre through $q$ is $\{q,\iota_\Phi(q)\}$. The ramification points of $\varphi_\Phi$, where $\iota_\Phi(q)=q$, are finite in number, being the zeros of the nonzero differential of $\varphi_\Phi$; their images are the branch values.

Lemma 3.2. $\iota_\Phi=T^2\circ\iota_F$. The four ramification points of the first projection form two fibres $\{r,T^2(r)\}$ of $\varphi_\Phi$, and $\varphi_\Phi$ is unramified at them.

Proof. By Lemma 2.3(1), $p+T^2\iota_F(p)$ is a divisor of $F+2D_0=\Phi$, since $p+\iota_F(p)$ is a divisor of $F$. By uniqueness of the residual point, $\iota_\Phi(p)=T^2\iota_F(p)$. For a fixed point $r$ of $\iota_F$, $\iota_\Phi(r)=T^2(r)$, which differs from $r$ because $2D_0$ is nontrivial (Lemma 2.3(2)). By Lemma 2.3(3), $\iota_FT^2\iota_F=T^{-2}$, and $T^{-2}=T^2$ because $T^4=\mathrm{id}$; hence $\iota_F(T^2r)=T^2\iota_F(r)=T^2(r)$, so $T^2(r)$ is again a fixed point of $\iota_F$. $\square$

An element $\sigma=\sum_i\sigma_i\otimes\tau_i$ of $\operatorname{Sym}^2H^0(\Gamma,\Phi)$ defines the quadratic form $Q_\sigma(\lambda)=\sum_i\lambda(\sigma_i)\lambda(\tau_i)$ on $H^0(\Gamma,\Phi)^*$, and has the image $m(\sigma)=\sum_i\sigma_i\tau_i\in H^0(\Gamma,2\Phi)$ under multiplication. For every $q$ and every local frame $e$ of $\Phi$ at $q$, $Q_\sigma(\mathrm{ev}_q)=\frac{m(\sigma)(q)}{e(q)^2}. \tag{3.1}$

Proposition 3.3 (the quadratic form). There is a unique $\sigma\in\operatorname{Sym}^2H^0(\Gamma,\Phi)$ with $\begin{aligned} m(\sigma)&=-\frac{\eta_\Gamma}{\omega}\\ &\in H^0(\Gamma,2F)\\ &=H^0(\Gamma,2\Phi), \end{aligned} \tag{3.2}$ the equality of section spaces being $\zeta$ from (2.3). The quadratic form $Q=Q_\sigma$ on $H^0(\Gamma,\Phi)^*$ is nondegenerate, and a direction $[\mathrm{ev}_q]$ is isotropic exactly when $q$ is one of the ramification points $r_1,\ldots,r_4$ of the first projection.

Proof. The map $m$ is injective: identifying $\operatorname{Sym}^2H^0(\Gamma,\Phi)$ with the binary quadratic forms on $H^0(\Gamma,\Phi)^*$, $m(\sigma)$ is the pullback of $Q_\sigma$ along $\varphi_\Phi$, and $\varphi_\Phi$ is surjective onto $\mathbb P^1$, being a nonconstant morphism from an irreducible projective curve. By Lemma 3.2, the divisor $r_1+r_2+r_3+r_4$ of $-\eta_\Gamma/\omega$ (Lemma 3.1) is the pullback along $\varphi_\Phi$ of $a+b$, where $a\ne b$ are the images of the two fibres, and each fibre is reduced. Let $\sigma_0$ correspond to a binary quadratic form with simple zeros at $a$ and $b$. Then $m(\sigma_0)$ and $-\eta_\Gamma/\omega$ are sections of $2\Phi$ with the same divisor, so their ratio is a nowhere vanishing regular function on the connected projective curve $\Gamma$, a nonzero constant $c$. Then $\sigma=c\sigma_0$ satisfies (3.2), and it is unique by injectivity. Its binary quadratic form has two distinct zeros, so $Q$ is nondegenerate. By (3.1), $Q(\mathrm{ev}_q)=0$ exactly when $-\eta_\Gamma/\omega$ vanishes at $q$. $\square$

The polar form of $Q$ is written $\begin{gathered} v\cdot w=\tfrac12\bigl(Q(v+w)-Q(v)-Q(w)\bigr),\\ v^2=Q(v). \end{gathered}$

4. Six points with prescribed data

Suppose six vectors $w_1,\ldots,w_6$ and their sum $w$ satisfy $w_j^2+w_j\cdot w=1$ for each $j$. Summing over $j$, $\sum_{j=1}^6\bigl(w_j^2-1\bigr)=-\sum_jw_j\cdot w=-w^2 .$ The curve of the next lesson carries a line bundle $L$ of degree $10$; the choice $w^2=90=9\deg L$ will make this total equal to $-9\deg L$, the value that a later obstruction computation requires, while all six pointwise normalizations are kept.

Lemma 4.1 (six vectors). In a two-dimensional nondegenerate quadratic space over $\mathbb C$ one can choose three distinct nonisotropic directions, avoiding any prescribed finite set of directions, and nonzero vectors $w_1,\ldots,w_6$, using each of the three directions twice with the two vectors in each pair equal, such that $\begin{gathered} w=\sum_{j=1}^6w_j,\quad w^2=90,\\ w_j^2+w_j\cdot w=1\\ (1\le j\le6). \end{gathered} \tag{4.1}$

Proof. Choose coordinates with $Q(x,y)=xy$, so that $(x,y)\cdot(x',y')=(xy'+yx')/2$. Put $\begin{gathered} h=\sqrt{90},\quad R=\sqrt{1+h^2/4},\quad b=\frac{2h}{R},\\ d(z)=\Bigl(Rz-\frac h2,\ \frac Rz-\frac h2\Bigr)\quad(z\ne0). \end{gathered}$ Expanding, $\begin{aligned} &d(z)^2+d(z)\cdot(h,h)\\ &=\Bigl(R^2-\frac{hR}2\bigl(z+z^{-1}\bigr)+\frac{h^2}4\Bigr)\\ &\quad+\Bigl(\frac{hR}2\bigl(z+z^{-1}\bigr)-\frac{h^2}2\Bigr)\\ &=R^2-\frac{h^2}4=1 . \end{aligned}$ For a nonzero complex number $\lambda$, let $z_1,z_2,z_3$ be the roots of $X^3-bX^2+b\lambda X-\lambda=0 .$ By Vieta's formulas, $\sum_iz_i=b$, $\sum_{i<j}z_iz_j=b\lambda$ and $z_1z_2z_3=\lambda$, so $\sum_iz_i^{-1}=b\lambda/\lambda=b$. Hence $\begin{aligned} \sum_{i=1}^3d(z_i)&=\Bigl(Rb-\frac{3h}2,\ Rb-\frac{3h}2\Bigr)\\ &=\Bigl(\frac h2,\ \frac h2\Bigr), \end{aligned}$ because $Rb=2h$. Take $w_{2i-1}=w_{2i}=d(z_i)$. Then $w=(h,h)$, $w^2=h^2=90$, and $w_j^2+w_j\cdot w=1$ by the expansion.

It remains to choose $\lambda$ so that the three directions are distinct and nonisotropic. As $\lambda\to0$ the cubic tends to $X^2(X-b)$: two roots tend to $0$ and one to $b$. The discriminant of the cubic, a polynomial in $\lambda$, has no constant term and linear coefficient $-4b^3\ne0$, so the roots are distinct for all small $\lambda\ne0$. The direction of $d(z)$ is given by the ratio of its entries, $\rho(z)=\frac{Rz-h/2}{R/z-h/2}=\frac{z(Rz-h/2)}{R-hz/2},$ which is holomorphic near $0$ with $\rho(0)=0$ and $\rho'(0)=-h/(2R)\ne0$. So $\rho$ is injective near $0$, and the two small roots give distinct directions, both near the direction $\rho=0$. At $z=b$ the entries of $d(b)$ are $Rb-h/2=3h/2$ and $R/b-h/2=(R^2-h^2)/(2h)$, that is $(1-3h^2/4)/(2h)$, both nonzero, so the third direction tends to a direction with $\rho\ne0$, distinct from the other two for small $\lambda$. A vector $(x,y)$ is isotropic exactly when $x=0$ or $y=0$; the entries of $d(z_i)$ tend to nonzero limits or to infinity, so the three vectors are nonisotropic for small $\lambda\ne0$.

Finally, for $\nu\in\mathbb C^*$ the map $(x,y)\mapsto(\nu x,\nu^{-1}y)$ is an isometry; applied to all six vectors it preserves (4.1) and multiplies every ratio $\rho$ by $\nu^2$. A prescribed finite set of directions excludes only finitely many values of $\nu^2$ for each of the three directions, so a suitable $\nu$ avoids it. $\square$

The marked points. Apply Lemma 4.1 to $(H^0(\Gamma,\Phi)^*,Q)$, excluding the branch values of $\varphi_\Phi$. Each of the three directions is $[\mathrm{ev}_q]$ for exactly two distinct points of $\Gamma$, a fibre $\{q,\iota_\Phi(q)\}$ of $\varphi_\Phi$. Label the six points $q_1,\ldots,q_6$ so that $[w_j]=[\mathrm{ev}_{q_j}]$, and put $Z=q_1+\cdots+q_6 . \tag{4.2}$ Each pair $q_{2i-1}+q_{2i}$ is a divisor of $\Phi$, so $Z$ is a divisor of $3\Phi=M$. Choose a section $\beta$ of $M$ with divisor $Z$. The directions are nonisotropic, so by Proposition 3.3 no $q_j$ is a ramification point of the first projection. The first coordinates $x(q_j)$ are not all equal, because a vertical line meets $\Gamma$ in at most two points.

Summary. The smooth curve $\Gamma\subset\Sigma$ of bidegree $(2,2)$ and genus one carries: the line bundles $F,S,D_0,\Delta,\Phi,M$ with the identifications (2.2) and (2.3); the logarithmic form $\Omega$ and its residue $\omega$; the nondegenerate quadratic form $Q$ of Proposition 3.3; six distinct points $q_1,\ldots,q_6$, none a ramification point of the first projection, with $Z$ a divisor of $M$ cut out by $\beta$; and nonzero vectors $w_j\in H^0(\Gamma,\Phi)^*$ in the directions $[\mathrm{ev}_{q_j}]$, using three distinct directions, satisfying (4.1).

5. Exercises

5.1 (easy). Verify that the eight zigzag points lie on $g=0$, and derive the factorizations of $g$ on the eight grid lines.

5.2 (medium). Prove the converse in Lemma 2.1: if $\delta$ has a multiple zero, or vanishes identically, then $V(G)$ is singular.

5.3 (medium). Let $\Gamma'\subset\Sigma$ be a smooth curve of bidegree $(2,2)$ passing through the six points $(x_0,y_0),(x_0,y_1),(x_1,y_1)$, $(x_1,y_2),(x_2,y_2),(x_2,y_0)$, with $x_0,x_1,x_2$ distinct and $y_0,y_1,y_2$ distinct. Show that $\mathcal O(-1,1)|_{\Gamma'}$ has exact order three.

5.4 (medium). Show that $\iota_\Phi$ has exactly four fixed points, and that a point $p$ is fixed by $\iota_\Phi$ if and only if $\iota_F(p)=T^2(p)$.

5.5 (medium). Check directly that the residue $\omega$ of $\Omega$ has no zero at the points of $\Gamma$ with $x=\infty$, using the chart $(x',y)$, $x'=1/x$.

6. Solutions

5.1. Substitution gives $g=0$ at each point. For example, $\begin{aligned} g(0,y)&=-3y^2+3y=-3y(y-1),\\ g(x,1)&=(1-5+3)x^2\\ &\quad+(-1+11-9)x+(-3+3)\\ &=-x^2+x=-x(x-1),\\ g(3,y)&=(9-3-3)y^2\\ &\quad+(-45+33+3)y+(27-27)\\ &=3y^2-9y=3y(y-3); \end{aligned}$ the others are similar.

5.2. Suppose first that $q_x\equiv0$ for some $x$. Then $\alpha,\beta,\gamma$ vanish at $x$, so near the fibre $G=tH$, where $t$ is a local coordinate on the first factor vanishing at $x$ and $H$ restricts to the fibre as a binary quadratic form $H_0$, possibly zero. Along the fibre, $\partial G/\partial s=tH_s=0$ and $\partial G/\partial t=H$. Every binary quadratic form over $\mathbb C$, including zero, has a zero on $\mathbb P^1$; at such a point of the fibre both partial derivatives vanish, so $V(G)$ is singular there.

Otherwise $q_x\ne0$ for every $x$. If $\delta$ has a multiple zero at $x$, which happens at every $x$ when $\delta\equiv0$, then $q_x$ has a double zero $y_0$. In the coordinates of the proof of Lemma 2.1, $b^2-4ac$ is a unit times $\delta$, so its derivative at $t=0$ vanishes: $-4a'(0)c(0)=0$. As $c(0)\ne0$, $a'(0)=0$, and both partial derivatives $a'(0)$ and $b(0)$ of $G$ vanish at $(x,y_0)$.

5.3. The vertical and horizontal lines through the six points meet $\Gamma'$ exactly in these points, two on each line, because a line of either ruling meets $\Gamma'$ in a divisor of degree two. With $T'=\iota_S\iota_F$ on $\Gamma'$, one gets $T'(x_0,y_0)=(x_1,y_1)$, $T'^2(x_0,y_0)=(x_2,y_2)$, $T'^3(x_0,y_0)=(x_0,y_0)$. Lemma 2.3 holds for any smooth curve of bidegree $(2,2)$ (its proof uses only Lemma 2.2 and genus one), so $3(S-F)$ is trivial and $S-F$ is not; the order is three.

5.4. By Lemma 3.2, $\iota_\Phi(p)=T^2\iota_F(p)$, so $\iota_\Phi(p)=p$ if and only if $\iota_F(p)=T^{-2}(p)$, and $T^{-2}=T^2$. To count the fixed points, which are the ramification points of $\varphi_\Phi$, choose an affine coordinate $\tau$ on the target $\mathbb P^1$ such that $\tau=\infty$ is not a branch value, and let $\nu=d\tau$, a meromorphic differential without zeros and with a double pole at $\infty$. Where $\varphi_\Phi$ is unramified, $\varphi_\Phi^*\nu$ has the same order as $\nu$ at the image point. At a ramification point $p$, $\tau-\tau(p)$ is a unit times the square of a local coordinate, so $\varphi_\Phi^*\nu$ has a simple zero. Hence the divisor of $\varphi_\Phi^*\nu$ is the sum of the ramification points minus twice the two points over $\infty$. Its degree is $\deg K_\Gamma=0$, so there are exactly four ramification points.

5.5. In the chart $(x',y)$, $\Omega=-dx'\wedge dy/g'$ with $g'(x',y)=x'^2g(1/x',y)$. Where $g'_{y}\ne0$, $\omega=dx'/g'_y$, and where $g'_{x'}\ne0$, $\omega=-dy/g'_{x'}$. These are the expressions of Lemma 1.3 with the sign of $\varepsilon$ in this chart; by smoothness one of $g'_{x'},g'_y$ is nonzero at each point, and the corresponding differential $dx'$ or $dy$ generates $\Omega^1_\Gamma$ there. So $\omega$ has no zero.

Sources

[OpenAI-048] OpenAI, A singular normal affine surface with free tangent sheaf (manuscript, 23 September 2026), in the release github.com/openai/math, family 048. Section 2: the elliptic model, the quadratic form and the six-vector lemma, for an arbitrary elliptic curve with a line bundle of order four, using its group law. The zigzag curve of Section 2 is a concrete instance, and the switches of Lemma 2.3 serve for the group law.