Surfaces containing two circles through each point

Abstract

We find all analytic surfaces in 3-dimensional Euclidean space such that through each point of the surface one can draw two transversal circular arcs fully contained in the surface (and analytically depending on the point). The search for such surfaces traces back to the works of Darboux from XIXth century. We prove that such a surface is an image of a subset of one of the following sets under some composition of inversions:

• the set $\{p+q:p\in \alpha ,q\in \beta \}$, where $\alpha ,\beta$ are two circles in $\mathbb {R}^3$;

• the set $\{2\frac{p \times q^{}}{|p+q|^2}:p\in \alpha ,q\in \beta ,p+q\ne 0\}$, where $\alpha ,\beta$ are circles in the unit sphere ${S}^2$;

• the set $\{(x,y,z): Q(x,y,z,x^2+y^2+z^2)=0\}$, where $Q\in \mathbb {R}[x,y,z,t]$ has degree 2 or 1.

  The proof uses a new factorization technique for quaternionic polynomials.

Appendix. A rationality result

In this appendix we give a direct proof of Lemma 4.8 going along the lines of [29] and exposed for nonspecialists. We use notation of Sect. 4, in particular, from the statements of lemmas there.

A projective algebraic surface $\Psi$ is unirationally ruled (or simply uniruled), if there are an algebraic curve $\eta$ and a rational map $\eta \times P^1\dasharrow \Psi$ whose image is the whole $\Psi$ possibly besides a 1-dimensional set. A surface $\Psi$ is birationally ruled (or simply ruled), if there is a birational map $\eta \times P^1\dasharrow \Psi$. A curve $\eta$ is rational, if there is a birational map $P^1\dasharrow \eta$.

Lemma 5.15

The surface ${\hat{\Phi }}$ is unirationally ruled.

Proof

Let ${\hat{\gamma }}$ be the irreducible curve in the variety of conics given by Lemma 4.5. Consider the algebraic set ${\hat{\Psi }} := \{(P, \alpha )\in {\hat{\Phi }}\times {\hat{\gamma }} : P\in \alpha \}.$ The second projection ${\hat{\Psi }}\rightarrow {\hat{\gamma }}$ is a rational map such that a generic fiber is a conic (hence a rational curve). By the Noether–Enriques theorem [2, Theorem III.4] ${\hat{\Psi }}$ is birationally ruled. In particular there is an algebraic curve $\eta$ and a rational map $\eta \times P^1\dasharrow {\hat{\Psi }}$, whose image is the whole ${\hat{\Psi }}$ besides a 1-dimensional set. Compose the map with the first projection ${\hat{\Psi }}\rightarrow {\hat{\Phi }}$. This projection is surjective by [30, Theorem 2 in §I.5.2] because the image contains the 2-dimensional subset $\bigcup _v \alpha _v\cap \Omega =\Phi \cap \Omega$ by Lemma 4.2. We get a rational map $\eta \times P^1\dasharrow {\hat{\Phi }}$, whose image is ${\hat{\Phi }}$ besides a 1-dimensional set, i.e., ${\hat{\Phi }}$ is unirationally ruled. $\square$

Lemma 5.16

Each unirationally ruled surface is birationally ruled.

Proof

Let ${\hat{\Phi }}$ be a unirationally ruled surface and $\eta \times P^1\dasharrow {\hat{\Phi }}$ be a rational map whose image is the whole ${\hat{\Phi }}$ besides a 1-dimensional set. Take a desingularization $d:{\tilde{\Phi }}\rightarrow {\hat{\Phi }}$ defined in the 1st paragraph of the proof of Lemma 4.9. Let ${\hat{\Phi }}\dasharrow {\tilde{\Phi }}$ be the inverse rational map of the desingularization.

Consider the rational map $\eta \times P^1\dasharrow {\hat{\Phi }}\dasharrow {\tilde{\Phi }}$. By the theorem on eliminating indeterminacy [2, Theorem II.7] this rational map equals a composition $\eta \times P^1\dasharrow {\tilde{\Psi }}\rightarrow {\tilde{\Phi }}$, where ${\tilde{\Psi }}$ is a smooth projective algebraic surface, the first map is birational, and the second map is rational and defined everywhere. In particular, ${\tilde{\Psi }}$ is birationally ruled. Since the image $\eta \times P^1\dasharrow {\hat{\Phi }}$ is the whole ${\hat{\Phi }}$ besides a 1-dimensional set and the surfaces are compact, by [30, Theorem 2 in §I.5.2] it follows that ${\tilde{\Psi }}\rightarrow {\tilde{\Phi }}$ is surjective.

By the Enriques theorem [2, Theorem VI.17 and Proposition III.21], a smooth projective algebraic surface is birationally ruled if and only if for each $k>0$ the k-th tensor power of the exterior square of the cotangent bundle has no sections except identical zero (in other terminology, the surface has Kodaira dimension$-\infty$, or all plurigeni vanish). Assume, to the contrary, that ${\tilde{\Phi }}$ has such a section (pluricanonical section). Then the pullback under the surjective rational map ${\tilde{\Psi }}\rightarrow {\tilde{\Phi }}$ is such a section for the birationally ruled surface ${\tilde{\Psi }}$, a contradiction. Thus ${\tilde{\Phi }}$, and hence ${\hat{\Phi }}$, is birationally ruled. $\square$

Proof of Lemma 4.8

By Lemmas 5.15 and 5.16 the surface ${\hat{\Phi }}$ is birationally ruled. Thus there is a birational map ${\hat{\Phi }}\dasharrow \eta \times P^1$. Consider the two conics through a generic point of the surface ${\hat{\Phi }}$. Their images are two distinct rational curves through a point of $\eta \times P^1$. Since there is only one $P^1$-fiber through each point, at least one of the rational curves is nonconstantly projected to the curve $\eta$. By the Lüroth theorem [2, Theorem V.4] the curve $\eta$ must be rational, and hence ${\hat{\Phi }}$ is rational. $\square$

Remark 5.17

We conjecture that the following generalization of Lemma 4.8 is true: an algebraic surface containing two rational curves through almost every point is rational. See [17, Definition IV.3.2, Theorems  IV.5.4, IV.3.10.3, IV.2.10, Corollary IV.5.2.1, Exercise IV.3.12.2] for a sketch of the proof. $\square$