Plectic points and the algebraic rank of quadratic twists

Let A/FA/F be a modular elliptic curve, let E/FE/F be a quadratic CM extension of class number one, let AE/FA^E/F be the quadratic twist of AA by E/FE/F, and let ψ\psi be an optimal embedding of conductor 11. Write ralg(AE/F)r_{\mathrm{alg}}(A^E/F) for the algebraic rank. Plectic rank conjecture. There exists a non-torsion plectic point Qψ,j\mathbbm1\mathrm{Q}_{\psi,j}^{\mathbbm{1}} if and only if

ralg(AE/F)=r.r_{\mathrm{alg}}(A^E/F)=r.

Moreover, if the cited Kol\operatorname{Kol}-type condition and the algebraicity conjecture hold, then the points wψ,1\mathbbm1,,wψ,r\mathbbm1w_{\psi,1}^{\mathbbm{1}},\dots,w_{\psi,r}^{\mathbbm{1}} generate AE(F)A^E(F). This connects the non-torsion of the distinguished plectic points with the full algebraic rank of the quadratic twist and predicts generators of its rational points.

Sources & referencesView supporting material

Primary source

Michele Fornea, “Plectic Jacobians”, arXiv:2305.01130 (2023).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.