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.,
$$\chi^{-1}(P\_1) + \chi^{-1}(P\_2) + \chi^{-1}(P\_3) + \chi^{-1}(P\_4) = (0,0)?$$


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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.




How can one parametrize a real elliptic normal curve such that four points are coplanar iff their parameters sum to zero?
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.


