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arXiv · 2609.14206

The two cubic separable minimal surfaces: $\mathcal D_1\mathcal D_2\mathcal D_3=1$ and $\mathcal C_1+\mathcal C_2+\mathcal C_3=3\,\mathcal C_1\mathcal C_2\mathcal C_3$

Abstract

A minimal surface in $\mathbb{R}^3$ is $\textit{separable}$ if it is the zero set of $f(x)+g(y)+h(z)$, and $\textit{isotropic}$ if moreover $f=g=h$. We announce that there are exactly two non-planar isotropic separable minimal surfaces up to homothety and rigid motion, that they are conjugate, and that they are Schwarz's diamond surface $\mathrm{D}$ and Schwarz's primitive surface $\mathrm{P}$. In lattice-normalized coordinates their implicit equations are \[\mathrm{D}:\; \mathcal{D}(x)\mathcal{D}(y)\mathcal{D}(z)=1,\] \[\mathrm{P}:\; \mathcal{C}(x)+\mathcal{C}(y)+\mathcal{C}(z) =3\,\mathcal{C}(x)\mathcal{C}(y)\mathcal{C}(z),\] where $\mathcal{D}(t)=\frac{\operatorname{sn}\operatorname{dn}}{\operatorname{cn}}(K[\frac{1}{4}]t,\frac{1}{4})$ and $\mathcal{C}(t)=\operatorname{cn}(2K[\frac{3}{4}]t,\frac{3}{4})$. The first equation is classical -- it is due to Schwarz, later Cayley, and appears in Nitsche's $\textit{Lectures}$ -- we identify Nitsche's transcendental function exactly as $\operatorname{sn}\operatorname{dn}/\operatorname{cn}$ at parameter $\frac{1}{4}$. The second appears to be new; it is the symmetric member of the two-parameter family of Kim and Ogata (2024), and it proves their assertion that the family contains $\mathrm{P}$. Both surfaces are governed by one differential equation with two signs, \[φ'^{\,2}=1+2\cosh 2φ \ (\mathrm{D}),\] \[φ'^{\,2}=1+2\cos 2φ \ (\mathrm{P}),\] the two signs of the separation constant of the classical reduction; and in that common normalization the two lattice periods are $2K[\frac{1}{4}]$ and $2K[\frac{3}{4}]$, whose ratio is Schwarz's 1866 constant $K'[\frac{1}{4}]/K[\frac{1}{4}]=1.2792615...$, the necessary scale ratio for a conjugate pair. Proofs will appear in [D, P].

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Steven Finch. 2026-09-13. The two cubic separable minimal surfaces: $\mathcal D_1\mathcal D_2\mathcal D_3=1$ and $\mathcal C_1+\mathcal C_2+\mathcal C_3=3\,\mathcal C_1\mathcal C_2\mathcal C_3$. https://arxiv.org/abs/2609.14206

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