Let $E \subset \mathbb{P}^3_{\mathbb{R}}$ be a real elliptic normal curve with two non-null-homotopic connected components. Is there a parametrization $$ \chi: (\mathbb{R}/\mathbb{Z})\times (\mathbb{Z}/2\mathbb{Z}) \longrightarrow E$$ such that any four points $P_1,P_2,P_3,P_4$ on $E$ are coplanar if and only if their parameters sum to zero, i.e., $$\chi^{-1}(P_1) + \chi^{-1}(P_2) + \chi^{-1}(P_3) + \chi^{-1}(P_4) = (0,0)?$$

Equip $\mathbb{P}^3_{\mathbb{R}}$ with homogeneous coordinates $[W:X:Y:Z]$. Any elliptic normal curve $E\subset \mathbb{P}^3_{\mathbb{R}}$ is the complete intersection of two quadrics $$ Q_i: [W,X,Y,Z] M_i [W,X,Y,Z]^t = 0,\qquad i = 1,2,$$ for some symmetric matrices $M_1, M_2$. The pair $(Q_1, Q_2)$ has discriminant $\Delta = \det (s M_1 + t M_2)$, which is homogeneous of degree four.

For a rectangular lattice $\Lambda = \omega_1 \mathbb{Z} + \omega_2 \mathbb{Z} i \subset \mathbb{C}$ with associated Weierstrass function $\wp$, the map $$\mathbb{C}/\Lambda \longrightarrow \mathbb{P}^3_{\mathbb{C}},\qquad z\longmapsto [1: \wp(z): \wp'(z): \wp''(z)],$$ embeds the torus as a complex elliptic normal curve. A real elliptic normal curve $E$ is obtained as a restriction of this parametrization. The Frobenius-Stickelberger addition formula shows that this parametrization has the required property. More generally, with $O$ the distinguished point of the elliptic curve, one can use any basis of the Riemann-Roch space $\mathcal{L}(4[O])$ as the components of the parametrization; see [1, Ex. 3.11]. However, the differential equation of $\wp$ shows that the discriminant $\Delta$ of this curve always has two or four linear factors over the real numbers.

Can one find a parametrization with the required property for the remaining case that $\Delta$ has no linear factors over the real numbers? This corresponds to a real elliptic normal curve with two non-null-homotopic connected components. A concrete example of such a curve is the intersection of the quadrics $$Q_1 : XY + ZW = 0,\qquad Q_2 : -X^2 + Y^2 - 2Z^2 + ZW + 2W^2 = 0.$$

[1] Silverman, The arithmetic of elliptic curves, 2nd ed., Graduate Texts in Mathematics, vol. 106, Springer, Dordrecht, 2009.


1 Answer 1


It looks like "no", even if you do not assume any special kind of regularity. The image of $0$ would have to be a point of $4$-fold intersection with a plane (some kind of maximal "flex"). On the other hand, since each component is not null-homologous, each plane intersects each component, hence, there's no real points of $4$-fold intersection.


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