Minimal conformally flat hypersurfaces

TL;DR

This paper classifies minimal conformally flat hypersurfaces with three distinct principal curvatures in space forms, showing they are generalized cones over Clifford tori or belong to a unique family in Euclidean space.

Contribution

It extends classification results for conformally flat hypersurfaces, especially characterizing minimal cases with three principal curvatures in space forms.

Findings

No such hypersurfaces exist if mean curvature H is non-zero.

Minimal conformally flat hypersurfaces with three principal curvatures are generalized cones over Clifford tori when c≠0.

In Euclidean space, a unique one-parameter family of such hypersurfaces exists.

Abstract

We study conformally flat hypersurfaces $f\colon M^{3} \to \Q^{4}(c)$ with three distinct principal curvatures and constant mean curvature $H$ in a space form with constant sectional curvature $c$. First we extend a theorem due to Defever when $c=0$ and show that there is no such hypersurface if $H\neq 0$. Our main results are for the minimal case $H=0$. If $c\neq 0$, we prove that if $f\colon M^{3} \to \Q^{4}(c)$ is a minimal conformally flat hypersurface with three distinct principal curvatures then $f(M^3)$ is an open subset of a generalized cone over a Clifford torus in an umbilical hypersurface $\Q^{3}(\tilde c)\subset \Q^4(c)$, $\tilde c>0$, with $\tilde c\geq c$ if $c>0$. For $c=0$, we show that, besides the cone over the Clifford torus in $\Sf^3\subset \R^4$, there exists precisely a one-parameter family of (congruence classes of) minimal isometric immersions $f\colon M^3 \to \R^4$ with three distinct principal curvatures of simply-connected conformally flat Riemannian manifolds.

Equations

$$G(x,t)={\exp }_{g(x)}(t\xi (g(x)))$$

$$I=\sum _{i=1}^n{v}_i^{2}d{u}_i^2\text{mbox}andII=\sum _{i=1}^n{V}_i{v}_{i}d{u}_i^2.$$

$$\displaystyle\left\{\begin{array}[]{l}(i)\ \dfrac{\partial v_{i}}{\partial u_{j}}=h_{ji}v_{j},\,\,\,\,\,\,\,\,\,(ii)\ \dfrac{\partial h_{ik}}{\partial u_{j}}=h_{ij}h_{jk},\vspace{1ex}\\ (iii)\ \dfrac{\partial h_{ij}}{\partial u_{i}}+\dfrac{\partial h_{ji}}{\partial u_{j}}+\sum_{k\neq i,j}h_{ki}h_{kj}+V_{i}V_{j}+cv_{i}v_{j}=0,\vspace{1ex}\\ (iv)\ \dfrac{\partial V_{i}}{\partial u_{j}}=h_{ji}V_{j},\,\,\,\,1\leq i\neq j\neq k\neq i\leq n.\end{array}\right.$$

$$\sum _{i=1}^3{\delta }_i{v}_i^2=0,\sum _{i=1}^3{\delta }_i{v}_i{V}_i=0\text{mbox}and\sum _{i=1}^3{\delta }_i{V}_i^2=1,$$

$$\begin{array}[]{lcl}v_{2}^{2}=v_{1}^{2}+v_{3}^{2},&&\displaystyle{V_{2}=-\frac{1}{3}\Big{(}\frac{v_{1}}{v_{3}}-\frac{v_{3}}{v_{1}}\Big{)}+\frac{v_{2}}{3}H},\vspace{.2cm}\\ \displaystyle{V_{1}=-\frac{1}{3}\Big{(}\frac{v_{2}}{v_{3}}+\frac{v_{3}}{v_{2}}\Big{)}+\frac{v_{1}}{3}H},&&\displaystyle{V_{3}=\frac{1}{3}\Big{(}\frac{v_{1}}{v_{2}}+\frac{v_{2}}{v_{1}}\Big{)}+\frac{v_{3}}{3}H},\end{array}$$

$$\begin{array}[]{l}H=\sum_{i=1}^{3}{V_{i}v_{i}^{-1}}\end{array}$$

$$⟨(x_1,x_2,x_3),(y_1,y_2,y_3)⟩=x_1{y}_1-x_2{y}_2+x_3{y}_3.$$

$$1=⟨V,V⟩=⟨av+bw,av+bw⟩=b^2⟨w,w⟩=b^2{v}_2^2.$$

$$\begin{array}[]{lll}\displaystyle{V_{1}=\frac{1}{v_{2}}(V_{2}v_{1}-\lambda v_{3})}\qquad\mbox{and}\qquad\displaystyle{V_{3}=\frac{1}{v_{2}}(V_{2}v_{3}+\lambda v_{1})}.\end{array}$$

$$\begin{array}[]{l}\displaystyle{V_{2}=-\frac{\lambda}{3}\Big{(}\frac{v_{1}}{v_{3}}-\frac{v_{3}}{v_{1}}\Big{)}+\frac{v_{2}}{3}H}.\end{array}$$

$$V_{1}=-\frac{\lambda}{3}\Big{(}\frac{v_{2}}{v_{3}}+\frac{v_{3}}{v_{2}}\Big{)}+\frac{v_{1}}{3}H\,\,\,\,\,\,\mbox{and}\,\,\,\,\,V_{3}=\frac{\lambda}{3}\Big{(}\frac{v_{1}}{v_{2}}+\frac{v_{2}}{v_{1}}\Big{)}+\frac{v_{3}}{3}H,$$

$$g(x_1,x_2)=\frac{1}{\sqrt{2}}(\cos (\sqrt{2}{x}_1),\sin (\sqrt{2}{x}_1),\cos (\sqrt{2}{x}_2),\sin (\sqrt{2}{x}_2)),$$

$$F(s,x)=sg(x),x=(x_1,x_2),$$

$$\eta (s,x_1,x_2)=\frac{1}{\sqrt{2}}(\cos (\sqrt{2}{x}_1),\sin (\sqrt{2}{x}_1),-\cos (\sqrt{2}{x}_2),-\sin (\sqrt{2}{x}_2))$$

$$\begin{array}[]{lll}I=ds^{2}+s^{2}(dx_{1}^{2}+dx_{2}^{2})\qquad\text{and}\qquad I\!I=s(-dx^{2}_{1}+dx^{2}_{2}).\end{array}$$

$$u_2=\log s,u_1=\sqrt{2}{x}_1\text{mbox}and{u}_3=\sqrt{2}{x}_2,$$

$$I=\frac{e^{2u_2}}{2}(d{u}_1^2+2d{u}_2^2+d{u}_3^2)\text{mbox}andII=\frac{e^{u_2}}{2}(-d{u}_1^2+d{u}_3^2),$$

$$F(s,x)=\frac{1}{\sqrt{c}}(\cos (\sqrt{c}s)e_5+\sin (\sqrt{c}s)g(x)),$$

$$N_{s_0}=F_{\ast }\frac{\partial }{\partial s}{|}_{s=s_0}=-\sin (\sqrt{c}{s}_0)e_5+\cos (\sqrt{c}{s}_0)g$$

$$F(s+s_0,x)=\cos (\sqrt{c}s)F_{s_0}(x)+\sin (\sqrt{c}s)N_{s_0}(x),$$

$$\eta (s,x_1,x_2)=\frac{1}{\sqrt{2}}(\cos (\sqrt{2}{x}_1),\sin (\sqrt{2}{x}_1),-\cos (\sqrt{2}{x}_2),-\sin (\sqrt{2}{x}_2),0)$$

$$\begin{array}[]{lll}\displaystyle{I=ds^{2}+\frac{1}{c}\sin^{2}({\sqrt{c}}s)(dx_{1}^{2}+dx_{2}^{2})}\;\;\mbox{and}\;\;\displaystyle{I\!I=\frac{\sin({\sqrt{c}}s)}{\sqrt{c}}(-dx^{2}_{1}+dx^{2}_{2})}.\end{array}$$

$$\frac{d{u}_2}{ds}=\frac{\sqrt{c}}{\sin (\sqrt{c}s)},u_1=\sqrt{2}{x}_1\text{mbox}and{u}_3=\sqrt{2}{x}_2,$$

$$I=\frac{{\sin }^2\theta }{2c}(d{u}_1^2+2d{u}_2^2+d{u}_3^2)\text{mbox}andII=\frac{\sin \theta }{2\sqrt{c}}(-d{u}_1^2+d{u}_3^2),$$

$$\frac{d\theta }{d{u}_2}=\sin \theta .$$

$$F(s,x)=\frac{1}{\sqrt{-c}}(\cosh (\sqrt{-c}s)e_5+\sinh (\sqrt{-c}s)g(x)),$$

$$I=d{s}^2+\frac{1}{-c}{\sinh }^2(\sqrt{-c}s)(d{x}_1^2+d{x}_2^2)$$

$$II=\frac{\sinh (\sqrt{-c}s)}{\sqrt{-c}}(-d{x}_1^2+d{x}_2^2).$$

$$\frac{d{u}_2}{ds}=\frac{\sqrt{-c}}{\sinh (\sqrt{-c}s)},u_1=\sqrt{2}{x}_1\text{mbox}and{u}_3=\sqrt{2}{x}_2,$$

$$I=\frac{{\sinh }^2\theta }{-2c}(d{u}_1^2+2d{u}_2^2+d{u}_3^2)\text{mbox}andII=\frac{\sinh \theta }{2\sqrt{-c}}(-d{u}_1^2+d{u}_3^2),$$

Full text

It was shown by E. Cartan [3] that if $f\colon M^{n}\to\mathbb{Q}^{n+1}(c)$ is a hypersurface of dimension $n\geq 4$ of a space form with constant sectional curvature $c$ and dimension $n+1$, then $M^{n}$ is conformally flat if and only if $f$ has a principal curvature of multiplicity at least $n-1$. Recall that a Riemannian manifold $M^{n}$ is conformally flat if each point of $M^{n}$ has an open neighborhood that is conformally diffeomorphic to an open subset of Euclidean space ${\mathbb{R}}^{n}$.

Cartan also proved that any hypersurface $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ with a principal curvature of multiplicity two is conformally flat, and that the converse is no longer true in this case. The study of conformally flat hypersurfaces by Cartan was taken up by Hertrich-Jeromin [6], who showed that a conformally flat hypersurface $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures admits locally principal coordinates $(u_{1},u_{2},u_{3})$ such that the induced metric $ds^{2}=\sum_{i=1}^{3}v_{i}^{2}du_{i}^{2}$ satisfies, say, $v_{2}^{2}=v_{1}^{2}+v_{3}^{2}$. A recent improvement by Canevari and the second author [2] (see Theorem 5 below) of Hertrich-Jeromin’s theorem is the starting point for the results of this paper.

Conformally flat hypersurfaces $f\colon M^{n}\to\mathbb{Q}^{n+1}(c)$ of dimension $n\geq 4$ with constant mean curvature were shown by do Carmo and Dajczer [4] to be rotation hypersurfaces whose profile curves satisfy a certain ODE. In particular, they concluded that minimal conformally flat hypersurfaces with dimension $n\geq 4$ of $\mathbb{Q}^{n+1}(c)$ are generalized catenoids, extending a previous result for $c=0$ by Blair [1]. The same conclusions apply for hypersurfaces $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ that have a principal curvature of multiplicity two. The case in which $f$ has three distinct principal curvatures was studied by Defever [5], who proved that no such hypersurface exists if $c=0$. Our first theorem extends this result for $c\neq 0$.

Examples of minimal conformally flat hypersurfaces $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures can be constructed for any $c\in{\mathbb{R}}$ as follows. Take a Clifford torus $g\colon{\mathbb{R}}^{2}\to\mathbb{Q}^{3}(\tilde{c})$, $\tilde{c}>0$, with $\tilde{c}\geq c$ if $c>0$, in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset\mathbb{Q}^{4}(c)$, and consider the generalized cone over $g$ in $\mathbb{Q}^{4}(c)$, that is, the hypersurface parametrized, on the open subset of regular points, by the map $G\colon{\mathbb{R}}^{2}\times{\mathbb{R}}\to\mathbb{Q}^{4}({c})$ given by

[TABLE]

where $\xi$ is a unit normal vector field to the inclusion $i\colon\mathbb{Q}^{3}(\tilde{c})\to\mathbb{Q}^{4}({c})$ and $\exp$ is the exponential map of $\mathbb{Q}^{4}({c})$. If $c=0$, the map $G$ parametrizes a standard cone over a Clifford torus in the sphere $\mathbb{Q}^{3}(\tilde{c})\subset{\mathbb{R}}^{4}$.

The next result states that there are no further examples of minimal conformally flat hypersurfaces $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures if $c\neq 0$.

Our last and main result shows that the preceding statement is not true in Euclidean space ${\mathbb{R}}^{4}$. In fact, we show that, besides the cone over a Clifford torus in ${\mathbb{S}}^{3}$, there exists precisely a one-parameter family of further examples.

More precisely, we show that there exist an algebraic variety $M^{4}\subset{\mathbb{R}}^{6}$, which contains a pair of straight lines $\ell_{-}$ and $\ell_{+}$ whose complement $\tilde{M}^{4}=M^{4}\setminus(\ell_{-}\cup\ell_{+})$ is a regular submanifold of ${\mathbb{R}}^{6}$, an involutive distribution ${\cal D}$ of codimension one on $\tilde{M}^{4}$ and a finite group $G$ of involutions of $\tilde{M}^{4}$ isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ such that the following assertions hold:

• (i)

To each leaf $\sigma$ of ${\cal D}$ one can associate a minimal immersion $f_{\sigma}\colon U_{\sigma}\to{\mathbb{R}}^{4}$ with three distinct principal curvatures of a simply connected open subset $U_{\sigma}\subset{\mathbb{R}}^{3}$, which is conformally flat with the metric induced by $f_{\sigma}$, and a covering map $\phi_{\sigma}\colon U_{\sigma}\to\sigma$. The singular set $\ell_{-}\cup\ell_{+}$ of $M^{4}$ corresponds to the cone over a Clifford torus in ${\mathbb{S}}^{3}$.

• (ii)

If $\sigma$ and $\tilde{\sigma}$ are distinct leaves of ${\cal D}$ then $f_{\tilde{\sigma}}$ is congruent to $f_{\sigma}$ if and only if there exist a diffeomorphism $\psi\colon U_{\sigma}\to U_{\tilde{\sigma}}$ and $\Theta\in G$ such that $\phi_{\tilde{\sigma}}\circ\psi=\Theta\circ\phi_{\sigma}$. In particular, $\tilde{\sigma}=\Theta(\sigma)$.

• (iii)

If $f\colon M^{3}\to{\mathbb{R}}^{4}$ is a minimal isometric immersion with three distinct principal curvatures of a simply connected conformally flat Riemannian manifold, then either $f(M^{3})$ is an open subset of the cone over a Clifford torus in ${\mathbb{S}}^{3}$ or there exist a leaf $\sigma$ of ${\cal D}$ and a local diffeomorphism $\rho\colon M^{3}\to V$ onto an open subset $V\subset U_{\sigma}$ such that $f$ is congruent to $f_{\sigma}\circ\rho$.

1 Preliminaries

In this section we discuss a local characterization of conformally flat hypersurfaces $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures and present the examples of minimal conformally flat hypersurfaces in $\mathbb{Q}^{4}(c)$ with three distinct principal curvatures given by generalized cones over Clifford tori.

1.1 Characterization of conformally flat hypersurfaces

First we recall the notion of holonomic hypersurfaces. One says that a hypersurface $f\colon M^{n}\to\mathbb{Q}^{n+1}(c)$ is holonomic if $M^{n}$ carries global orthogonal coordinates $(u_{1},\ldots,u_{n})$ such that the coordinate vector fields $\partial_{j}=\dfrac{\partial}{\partial u_{j}}$ diagonalize the second fundamental form $I\!I$ of $f$.

Set $v_{j}=\|\partial_{j}\|$ and define $V_{j}\in C^{\infty}(M)$, $1\leq j\leq n$, by $I\!I(\partial_{j},\partial_{j})=V_{j}v_{j}$, $1\leq j\leq n$. Then the first and second fundamental forms of $f$ are

[TABLE]

Denote $v=(v_{1},\ldots,v_{n})$ and $V=(V_{1},\ldots,V_{n})$. We call $(v,V)$ the pair associated to $f$. The next result is well known.

Proposition 4**.**

The triple $(v,h,V)$, where $h_{ij}=\frac{1}{v_{i}}\frac{\partial v_{j}}{\partial u_{i}},$ satisfies the system of PDE’s

[TABLE]

Conversely, if $(v,h,V)$ is a solution of $(\ref{sistema-hol})$ on a simply connected open subset $U\subset{\mathbb{R}}^{n}$, with $v_{i}>0$ everywhere for all $1\leq i\leq n$, then there exists a holonomic hypersurface $f\colon U\to\mathbb{Q}^{n+1}(c)$ whose first and second fundamental forms are given by $(\ref{fundforms}).$

The following characterization of conformally flat hypersurfaces $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures was given in [2], improving a theorem due to Hertrich-Jeromin [6].

Theorem 5**.**

Let $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ be a holonomic hypersurface whose associated pair $(v,V)$ satisfies

[TABLE]

where $(\delta_{1},\delta_{2},\delta_{3})=(1,-1,1)$. Then $M^{3}$ is conformally flat and $f$ has three distinct principal curvatures.

Conversely, any conformally flat hypersurface $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures is locally a holonomic hypersurface whose associated pair $(v,V)$ satisfies $(\ref{holo.flat}).$

It will be convenient to use the following equivalent version of Theorem 5.

Corollary 6**.**

Let $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ be a holonomic hypersurface whose associated pair $(v,V)$ satisfies

[TABLE]

where $H$ is the mean curvature function of $f$. Then $M^{3}$ is conformally flat and $f$ has three distinct principal curvatures.

Conversely, any conformally flat hypersurface $f\colon\,M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures is locally a holonomic hypersurface whose associated pair $(v,V)$ satisfies $(\ref{holo.flat.med.2}).$

Proof.

It suffices to show that equations (6) together with

[TABLE]

are equivalent to (7). For that, consider the Minkowski space ${\mathbb{L}}^{3}$ endowed with the Lorentz inner product

[TABLE]

Then the conditions in (6) say that $v=(v_{1},v_{2},v_{3})$ and $V=(V_{1},V_{2},V_{3})$ are orthogonal with respect to such inner product, $v$ is light-like and $V$ is a unit space-like vector. Since $w=(-v_{3},0,v_{1})\in{\mathbb{L}}^{3}$ is orthogonal to $v,$ we have $v^{\perp}=\mbox{span}\{v,w\}.$ As $V\in v^{\perp},$ we can write $V=av+bw$ for some $a,b\in C^{\infty}(M^{3})$. Note that $V_{2}=av_{2}.$ Using $(\ref{holo.flat})$ we obtain

[TABLE]

Thus $V=\frac{V_{2}}{v_{2}}v+\frac{\lambda}{v_{2}}w,$ with $\lambda=\pm 1.$ Therefore

[TABLE]

Substituting (9) in (8) we obtain

[TABLE]

Substituting (10) in (9) yields

[TABLE]

and changing the orientation, if necessary, we may assume that $\lambda=1$.∎

1.2 Generalized cones over Clifford tori

First we show that, if $g\colon{\mathbb{R}}^{2}\to{\mathbb{S}}^{3}\subset{\mathbb{R}}^{4}$ is the Clifford torus parametrized by

[TABLE]

then the standard cone $F\colon(0,\infty)\times{\mathbb{R}}^{2}\to{\mathbb{R}}^{4}$ over $g$ given by

[TABLE]

is a minimal conformally flat hypersurface.

The first and second fundamental forms of $F$ with respect to the unit normal vector field

[TABLE]

are

[TABLE]

In terms of the new coordinates $u_{1},u_{2},u_{3}$, related to $s,x_{1},x_{2}$ by

[TABLE]

the first and second fundamental forms of $F$ become

[TABLE]

hence $F$ is a minimal conformally flat hypersurface with three distinct principal curvatures, one of which being zero.

The preceding example can be extended to the case in which the ambient space is any space form, yielding examples of minimal conformally flat hypersurfaces $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures also for $c\neq 0$.

Start with the Clifford torus $g\colon{\mathbb{R}}^{2}\to{\mathbb{S}}^{3}\subset{\mathbb{R}}^{4}$ parametrized by (11). If $c>0$, define $F\colon(0,\pi/\sqrt{c})\times{\mathbb{R}}^{2}\to{\mathbb{S}}^{4}(c)\subset{\mathbb{R}}^{5}={\mathbb{R}}^{4}\times{\mathbb{R}}$ by

[TABLE]

where $x=(x_{1},x_{2})$ and $e_{5}$ is a unit vector spanning the factor ${\mathbb{R}}$ in the orthogonal decomposition ${\mathbb{R}}^{5}={\mathbb{R}}^{4}\times{\mathbb{R}}$.

Notice that, for each fixed $s=s_{0}$, the map $F_{s_{0}}\colon{\mathbb{R}}^{2}\to{\mathbb{S}}^{4}(c)$, given by $F_{s_{0}}(x)=F(s_{0},x)$, is also a Clifford torus in an umbilical hypersurface ${\mathbb{S}}^{3}(\tilde{c})\subset{\mathbb{S}}^{4}(c)$ with curvature $\tilde{c}=c/\sin^{2}(\sqrt{c}s_{0})$, which has

[TABLE]

as a unit normal vector field along $F_{s_{0}}$. Notice also that

[TABLE]

thus $s\mapsto F(s,x)$ parametrizes the geodesic in ${\mathbb{S}}^{4}(c)$ through $F_{s_{0}}(x)$ tangent to $N_{s_{0}}(x)$ at $F_{s_{0}}(x)$. Hence $F$ is a generalized cone over $F_{s_{0}}$.

The first and second fundamental forms of $F$ with respect to the unit normal vector field

[TABLE]

are

[TABLE]

In terms of the new coordinates $u_{1},u_{2},u_{3}$, related to $s,x_{1},x_{2}$ by

[TABLE]

the first and second fundamental forms of $F$ become

[TABLE]

where $\theta=\sqrt{c}s$, which, in view of the first equation in (12), satisfies

[TABLE]

It follows from (13) that $F$ is a minimal conformally flat hypersurface.

If $c<0$, define $F\colon(0,\infty)\times{\mathbb{R}}^{2}\to\mathbb{H}^{4}(c)\subset{\mathbb{L}}^{5}$ by

[TABLE]

where $x=(x_{1},x_{2})$, $e_{5}$ is a unit time-like vector in ${\mathbb{L}}^{5}$ and $e_{5}^{\perp}$ is identified with ${\mathbb{R}}^{4}$.

As in the previous case, for each fixed $s=s_{0}$ the map $F_{s_{0}}\colon{\mathbb{R}}^{2}\to\mathbb{H}^{4}(c)$, given by $F_{s_{0}}(x)=F(s_{0},x)$, is also a Clifford torus in an umbilical hypersurface ${\mathbb{S}}^{3}(\tilde{c})\subset\mathbb{H}^{4}(c)$ with curvature $\tilde{c}=-c/\sinh^{2}(\sqrt{-c}s_{0})$, and $F$ is a generalized cone over $F_{s_{0}}$.

Now the first and second fundamental forms of $F$ are

[TABLE]

and

[TABLE]

In terms of the new coordinates $u_{1},u_{2},u_{3}$, related to $s,x_{1},x_{2}$ by

[TABLE]

they become

[TABLE]

where $\theta(s)=\sqrt{-c}s$ satisfies

[TABLE]

It follows from (14) that $F$ is a minimal conformally flat hypersurface with three distinct principal curvatures, one of which being zero.

2 The proofs of Theorems 1 and 2

First we derive a system of PDE’s for new unknown functions associated to a conformally flat hypersurface $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ with three distinct principal curvatures under the assumption that $f$ has constant mean curvature.

Proposition 7**.**

Let $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ be a holonomic hypersurface with constant mean curvature $H$ whose associated pair $(v,V)$ satisfies $(\ref{holo.flat.med.2})$. Set

[TABLE]

Then $v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3}$ satisfy the differential equations

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

as well as the algebraic relations

[TABLE]

where

[TABLE]

Proof.

The triple $(v,h,V),$ where $h_{ij}=\frac{1}{v_{i}}\frac{\partial v_{j}}{\partial u_{i}},$ satisfies the system of PDE’s

[TABLE]

Equations $(i)$, $(ii)$, $(iii)$ and $(iv)$ are due to the fact that $f$ is a holonomic hypersurface with $(v,V)$ as its associated pair, and $(v)$ and $(vi)$ follow by differentiating (6).

Using (7) and equations $(i)$, $(iv)$, $(v)$ and $(vi)$ in (25) one can show that

[TABLE]

From $(i)$ and $(v)$ of (25), together with (26), one obtains the formulae for the derivatives $\frac{\partial v_{i}}{\partial u_{j}},$ $1\leq i,j\leq 3$. In a similar way, using $(i)$, $(iii)$ and $(v)$, together with (26), one finds the derivatives $\frac{\partial\alpha_{i}}{\partial u_{j}},$ $1\leq i\neq j\leq 3$. In order to compute $\frac{\partial\alpha_{i}}{\partial u_{i}},$ $1\leq i\leq 3$, we note that equation $(ii)$, together with (26) and the remaining equations in (25), determines the system of linear equations

[TABLE]

in the variables $\frac{\partial\alpha_{i}}{\partial u_{i}},$ $1\leq i\leq 3$, where

[TABLE]

and

[TABLE]

One can check that such system has a unique solution given by (18), (20) and (23).

Finally, computing the mixed derivatives $\frac{\partial^{2}\alpha_{i}}{\partial u_{j}\partial u_{k}}=\frac{\partial^{2}\alpha_{i}}{\partial u_{k}\partial u_{j}},$ $1\leq i,k,j\leq 3$, from (25) we obtain

[TABLE]

[TABLE]

and

[TABLE]

In the lemmata that follows we assume the hypotheses of Proposition 7 to be satisfied and use the notations therein.

Lemma 8**.**

If $v_{1}=v_{3}$ everywhere then $H=0$.

Proof.

By the assumption and the first equation in (7) we have $v_{2}=\sqrt{2}v_{1}$. We obtain from any two of the equations in (15) that $\alpha_{1}=0$, whereas any two of the equations in (17) imply that $\alpha_{3}=0$. Then (18) and (23) give

[TABLE]

which imply that $H=0$. ∎

Lemma 9**.**

The functions $\alpha_{1},\alpha_{2},\alpha_{3}$ can not vanish symultaneously on any open subset of $M^{3}$.

Proof.

If $\alpha_{1},\alpha_{2},\alpha_{3}$ all vanish on the open subset $U\subset M^{3}$, then (18), (20) and (23) become

[TABLE]

[TABLE]

[TABLE]

Comparying (27) with (28) and (29) yields, respectively,

[TABLE]

which is a contradiction. ∎

Lemma 10**.**

There does not exist any open subset of $M^{3}$ where $v_{1}-v_{3}$ is nowhere vanishing and $\alpha_{1}=0=\alpha_{3}$.

Proof.

Assume that $\alpha_{1}=0=\alpha_{3}$ and that $v_{1}-v_{3}$ does not vanish on the open subset $U\subset M^{3}$. By Lemma 9, $\alpha_{2}$ must be nonvanishing on an open dense subset $V\subset U$. Then equations (15), (16), (17), (19), (21) and (22) reduce to the following on $V$:

[TABLE]

[TABLE]

and, since $\alpha_{1}=\alpha_{3}=0,$ equations (18), (20) and (23) become, respectively,

[TABLE]

[TABLE]

[TABLE]

Multiplying (32) and (34) by $2v_{3}^{8}$ and $3v_{1}^{2}v_{2}^{4}v_{3}^{2}$, respectively, and subtracting one from the other, yield

[TABLE]

On one hand, substituting (35) in (33) we obtain

[TABLE]

On the other hand, differentiating (35) with respect to $u_{2}$ and using (31) gives

[TABLE]

Using that $\alpha_{2}\neq 0$ and $v_{1}-v_{3}\neq 0$ on $V$, we obtain from (36) and (37) that

[TABLE]

Differentiating (38) with respect to $u_{2},$ and using (31) we obtain

[TABLE]

Since $v_{1}-v_{3}\neq 0,$ we must have $(22v_{1}^{6}-11v_{1}^{4}v_{3}^{2}-4v_{1}^{2}v_{3}^{4}+8v_{3}^{6})\neq 0$ on $V.$ Therefore, (39) implies that

[TABLE]

Finally, differentiating (40) with respect to $u_{2}$ and using (31) we obtain

[TABLE]

which is a contradiction, for the right-hand-side of (41) is nonzero.∎

Lemma 11**.**

There does not exist any open subset of $M^{3}$ where $\alpha_{2}=0=\alpha_{j}$ for some $j\in\{1,3\}$.

Proof.

We argue for the case in which $j=1$, the other case being similar. So, assume that $\alpha_{1}$ and $\alpha_{2}$ vanish on an open subset $U\subset M^{3}$. By Lemma 9, $\alpha_{3}$ is nonzero on an open dense subset $V\subset U$. Equations (18) and (20) can be rewritten as follows on $V$:

[TABLE]

Eliminating $\alpha_{3}^{2}$ from the equations in (42) yields

[TABLE]

Differentiating (43) with respect to $u_{3}$ and using that $\alpha_{3}\neq 0$ we obtain

[TABLE]

Finally, differentiating (44) with respect to $u_{3}$ and using the fact that $H$ is constant we obtain

[TABLE]

which is a contradiction. ∎

Lemma 12**.**

If there exist $p\in M^{3}$ and $1\leq i\neq j\leq 3$ such that $\alpha_{i}(p)\neq 0\neq\alpha_{j}(p)$ then $H=0=c$.

Proof.

We give the proof for the case in which $i=1$ and $j=2$, the remaining ones being similar. Let $U\subset M^{3}$ be an open neighborhood of $p$ where $\alpha_{1}$ and $\alpha_{2}$ are nowhere vanishing. Then (24) gives

[TABLE]

and

[TABLE]

or equivalently,

[TABLE]

Subtracting one of the equations in (45) from the other we obtain

[TABLE]

Differentiating (46) with respect to $u_{1}$ we obtain

[TABLE]

Since $\alpha_{1}\neq 0$, equation (47) implies that $H=0$, and hence $c=0$ by (46). ∎

Lemma 13**.**

If $v_{1}\neq v_{3}$ at some point of $M^{3}$ then $H=0=c$.

Proof.

Assume that $v_{1}(p_{0})\neq v_{3}(p_{0})$ for some $p_{0}\in M^{3}$, and hence that $v_{1}\neq v_{3}$ on some open neighborhood $U\subset M^{3}$ of $p_{0}$. By Lemma 9, there exist an open subset $U^{\prime}\subset U$ and $i\in\{1,2,3\}$ such that $\alpha_{i}(p)\neq 0$ for all $p\in U^{\prime}$. It follows from Lemma 10 and Lemma 11 that there exist $q\in U^{\prime}$ and $j\in\{1,2,3\},\ j\neq i,$ such that $\alpha_{j}(q)\neq 0$. Thus there exists $q\in M^{3}$ such that $\alpha_{i}(q)\neq 0$ and $\alpha_{j}(q)\neq 0,\ i\neq j,$ and the conclusion follows from Lemma 12. ∎

Proof of Theorem 1: Follows immediately from Lemma 8 and Lemma 13.∎

Proof of Theorem 2: Given $p\in M^{3}$, let $u_{1},u_{2},u_{3}$ be $U$ be local principal coordinates on an open neighborhood $U$ of $p$ as in Corollary 6. It follows from Lemma 13 that the associated pair $(v,V)$ satisfies $v_{1}=v_{3}$ on $U$. Thus $\lambda_{2}$ vanishes on $U$, and hence everywhere on $M^{3}$ by analyticity. The statement is now a consequence of the next proposition. ∎

Proposition 14**.**

Let $f\colon M^{3}\to\mathbb{Q}^{4}(c)$ be a conformally flat hypersurface with three distinct principal curvatures. If one of the principal curvatures is everywhere zero, then either $c=0$ and $f$ is locally a cylinder over a surface $g\colon M^{2}(\bar{c})\to{\mathbb{R}}^{3}$ with constant Gauss curvature $\bar{c}\neq 0$ or $f$ is locally a generalized cone over a surface $g\colon M^{2}(\bar{c})\to\mathbb{Q}^{3}(\tilde{c})$ with constant Gauss curvature $\bar{c}\neq\tilde{c}$ in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset\mathbb{Q}^{4}(c)$, $\tilde{c}\geq c$, with $\tilde{c}>0$ if $c=0$. If, in addition, $f$ is minimal, then $f(M^{3})$ is an open subset of a generalized cone over a Clifford torus in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset\mathbb{Q}^{4}(c)$, $\tilde{c}>0$, with $\tilde{c}\geq c$ if $c>0$.

Proof.

Let $e_{1},e_{2},e_{3}$ denote local unit vector fields which are principal directions correspondent to the distinct principal curvatures $\lambda_{1},\lambda_{2},\lambda_{3}$, respectively. Then conformal flatness of $M^{3}$ is equivalent to the relations

[TABLE]

and

[TABLE]

for all distinct indices $i,j,k$ (see [7], p. 84). It follows from Codazzi’s equation and (48) that

[TABLE]

If, say, $\lambda_{2}=0$, then equation (49) yields

[TABLE]

hence the distribution $\{e_{2}\}^{\perp}$ spanned by $e_{1}$ and $e_{3}$ is umbilical in $M^{3}$ by (50).

If $\varphi$ is identically zero on $M^{3}$, then $\{e_{2}\}^{\perp}$ is a totally geodesic distribution, and hence $M^{3}$ is locally isometric to a Riemannian product $I\times M^{2}$ by the local de Rham theorem. Since $M^{3}$ is conformally flat, it follows that $M^{2}$ must have constant Gauss curvature. Moreover, by Molzan’s theorem (see Corollary $17$ in [8]), $f$ is locally an extrinsic product of isometric immersions of the factors, which is not possible if $c\neq 0$ because $f$ has three distinct principal curvatures. Therefore $c=0$ and $f$ is locally a cylinder over a surface with constant Gauss curvature in ${\mathbb{R}}^{3}$.

If $\varphi$ is not identically zero on $M^{3}$, given $x\in M^{3}$ let $\sigma$ be the leaf of $\{e_{2}\}^{\perp}$ containing $x$ and let $j\colon\sigma\to M^{3}$ be the inclusion of $\sigma$ into $M^{3}$. Denote $\tilde{g}=f\circ j$. Then the normal bundle $N_{\tilde{g}}\sigma$ of $\tilde{g}$ splits as

[TABLE]

and

[TABLE]

for all $X\in\mathfrak{X}(\sigma)$, where $\tilde{\nabla}$ is the induced connection on ${\tilde{g}}^{*}T\mathbb{Q}^{4}(c)$. It follows that the normal vector field $\eta=f_{*}e_{2}$ of $\tilde{g}$ is parallel with respect to the normal connection of $\tilde{g}$, and that the shape operator of $\tilde{g}$ with respect to $\eta$ is given by $A^{\tilde{g}}_{\eta}=\varphi I$. It is a standard fact that this implies $\tilde{g}(\sigma)$ to be contained in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset\mathbb{Q}^{4}(c)$, $\tilde{c}\geq c$, that is, there exist an umbilical hypersurface $i\colon\mathbb{Q}^{3}(\tilde{c})\to\mathbb{Q}^{4}(c)$ and an isometric immersion $g\colon M^{2}=\sigma\to\mathbb{Q}^{3}({\tilde{c}})$ such that $\tilde{g}=i\circ g$. Moreover, since at any $y\in\sigma$ the fiber $L(y)\!=\!\mbox{span}\{\eta(y)\}$ coincides with the normal space of $i$ at $g(y)$, it follows that $f$ coincides with the generalized cone over $g$ in a neighborhood of $x$.

In particular, $M^{3}$ is a warped product $I\times_{\rho}M^{2}$, and since $M^{3}$ is conformally flat, $M^{2}$ must have constant Gauss curvature. If, in addition, $f$ is minimal, then $g$ must be a Clifford torus in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset\mathbb{Q}^{4}(c)$, $\tilde{c}>0$, with $\tilde{c}\geq c$ if $c>0$, and the preceding argument shows that $f(M^{3})$ is an open subset of a generalized cone over $g$.∎

3 Proof of Theorem 3

First we rewrite Proposition (7) when $H=0=c$ and state a converse to it.

Proposition 15**.**

Let $f\colon M^{3}\to{\mathbb{R}}^{4}$ be a holonomic hypersurface whose associated pair $(v,V)$ satisfies

[TABLE]

and

[TABLE]

Set

[TABLE]

Then $\phi=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$ satisfies the system of PDE’s

[TABLE]

where

[TABLE]

as well as the algebraic equation

[TABLE]

Conversely, if $\phi=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$ is a solution of $(\ref{flat.min.reduz.})$ satisfying (51) on an open simply-connected subset $U\subset{\mathbb{R}}^{3}$, then $\phi$ satisfies (54) and the triple $(v,h,V)$, where $v=(v_{1},v_{2},v_{3})$, $V=(V_{1},V_{2},V_{3})$ is given by $(\ref{holo.flat.min.2})$ and $h=(h_{ij})$, with $h_{ij}=\frac{1}{v_{i}}\frac{\partial v_{j}}{\partial u_{i}}$, $1\leq i\neq j\leq 3$, satisfies (5), and hence gives rise to a holonomic hypersurface $f\colon U\to{\mathbb{R}}^{4}$ whose associated pair $(v,V)$ satisfies (51) and (52).

In view of Corollary 6 and Proposition 15, minimal conformally flat hypersurfaces of ${\mathbb{R}}^{4}$ are in correspondence with solutions $\phi=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$ of (53) satisfying (51) and (54). We shall prove that such solutions are, in turn, in correspondence with the leaves of a foliation of codimension one on the algebraic variety constructed in the next result.

Proposition 16**.**

Define $G,F:{\mathbb{R}}^{6}={\mathbb{R}}^{3}\times{\mathbb{R}}^{3}\to{\mathbb{R}}$ by

[TABLE]

and

[TABLE]

Let $M^{4}:=F^{-1}(0)\cap G^{-1}(0)\cap\{(x,y)\in{\mathbb{R}}^{6};x_{1}>0,x_{2}>0,x_{3}>0\,\,\mbox{and}\,\,y\neq 0\}$ and let $\ell_{\pm}$ be the half lines in $M^{4}$ given by

[TABLE]

Then $\tilde{M}^{4}=M^{4}\setminus(\ell_{-}\cup\ell_{+})$ is a regular submanifold of ${\mathbb{R}}^{6}$ and $\ell_{-}\cup\ell_{+}$ is the singular set of $M^{4}$.

Proof.

If $p\in M^{4}$ we have \nabla G(p)=\big{(}-2x_{1},2x_{2},-2x_{3},0,0,0\big{)}, while the components of $\nabla F(p)$ are given by

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

That $M^{4}\setminus(\ell_{+}\cup\ell_{-})$ is a smooth submanifold of ${\mathbb{R}}^{6}$ and $\ell_{-}\cup\ell_{+}$ is the singular set of $M^{4}$ is a consequence of the next two facts.

Fact 1: $\nabla F(p)\neq 0$ for all $p\in M^{4}.$

Proof.

If $\nabla F(p)=0$ at $p=(x_{1},x_{2},x_{3},y_{1},y_{2},y_{3})$ then from $\frac{\partial F}{\partial y_{1}}(p)=0$ it follows that $y_{1}=0$, whereas $\frac{\partial F}{\partial y_{3}}(p)=0$ implies that $y_{3}=0$. Thus $y_{2}\neq 0$, and hence $x_{3}=x_{1}$ from $\frac{\partial F}{\partial y_{2}}(p)=0$. Therefore $x_{2}=\sqrt{2}x_{1},$ and then $0=\frac{\partial F}{\partial x_{2}}(p)=-20\sqrt{2}x_{1}^{11},$ which contradicts the fact that $x_{1}>0$.

Fact 2: The subset $\{p\in M^{4}\,:\,\nabla F(p)=a\nabla G(p)\,\,\,\mbox{for some}\,\,\,a\in{\mathbb{R}}-\{0\}\}$ coincides with $\ell_{-}\cup\ell_{+}.$

Proof.

Assume that

[TABLE]

for some $a\in{\mathbb{R}}-\{0\}$. Equation (55) gives us six equations, the last three of which yield $y_{1}=y_{3}=0$ and $x_{3}=x_{1}$. Since $x_{2}^{2}=x_{1}^{2}+x_{3}^{2},$ we obtain that $x_{2}=\sqrt{2}x_{1}.$ Using this and the second of such equations we obtain that $a=-10x_{1}^{10}.$ Finally, the first one implies that $y_{2}^{2}=1$. ∎

Proposition 17**.**

Let $X_{1},X_{2},X_{3}\colon M^{4}\to{\mathbb{R}}^{6}$ be defined by

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

Then the following assertions hold:

• (i)

$X_{1}(p),X_{2}(p),X_{3}(p)$* are linearly independent for all $p\in~{}\tilde{M}^{4}$.*

• (ii)

$\{p\in M^{4};X_{1}(p)=0\}=\ell_{-}\cup\ell_{+}=\{p\in M^{4};X_{3}(p)=0\}$.

• (iii)

The vector fields $X_{1},X_{2},X_{3}$ are everywhere tangent to $\tilde{M}^{4}$ and the curves $\gamma_{\pm}\colon{\mathbb{R}}\to{\mathbb{R}}^{6}$ given by $\gamma_{\pm}(t)=(e^{t},\sqrt{2}e^{t},e^{t},0,\pm 1,0)$, are integral curves of $X_{2}$ with $\gamma_{\pm}({\mathbb{R}})=\ell_{\pm}$.

• (iv)

$[X_{i},X_{j}]=0$* on $\tilde{M}^{4}$ for all $1\leq i\neq j\leq 3$.*

Proof.

First notice that $X_{2}(p)=0$ if and only if $y_{2}=0$ and $A_{2}(p)=0$. Since $A_{2}(p)<0$ whenever $y_{2}=0$, it follows that $X_{2}(p)\neq 0$ for all $p~{}\in~{}M^{4}$.

Now observe that $X_{1}(p)=0$ if and only if $y_{1}=0$ and $A_{1}(p)=0$. Thus, if $p=(x_{1},x_{2},x_{3},y_{1},y_{2},y_{3})$ is such that $X_{1}(p)=0$ then

[TABLE]

where

[TABLE]

[TABLE]

and

[TABLE]

Such system has a unique solution given by

[TABLE]

Thus we must have $y_{3}=0$, and hence $x_{1}=x_{3}$ and $y_{2}=\pm 1.$ It follows that the subset $\{p\in M^{4};X_{1}(p)=0\}$ coincides with $\ell_{-}\cup\ell_{+}$.

In a similar way one shows that the subset $\{p\in M^{4};X_{3}(p)=0\}$ coincides with $\ell_{-}\cup\ell_{+}$, and the proof of $(ii)$ is completed.

To prove $(i)$, first notice that $X_{1}(p),X_{2}(p),X_{3}(p)$ are pairwise linearly independent. This already implies that if $\lambda_{1},\lambda_{2},\lambda_{3}\in{\mathbb{R}}$ are such that

[TABLE]

then either $\lambda_{1}=\lambda_{2}=\lambda_{3}=0$ ou $\lambda_{1}\neq 0,\lambda_{2}\neq 0$ e $\lambda_{3}\neq 0.$ We will show that the last possibility can not occur.

Equation (56) gives the system of equations

[TABLE]

Thus, it suffices to prove that the system of equations

[TABLE]

has no solutions for $p=(x_{1},x_{2},x_{3},y_{1},y_{2},y_{3})\in\tilde{M}^{4}$. We write the preceding system as a linear system

[TABLE]

in the variables $y_{1}^{2},y_{2}^{2}$ and $y_{3}^{2}$, where

[TABLE]

and

[TABLE]

Since

[TABLE]

such system has no solutions. Thus $(i)$ is proved.

Now, for $p\in M^{4}$ we have

[TABLE]

hence $X_{1},X_{2},X_{3}$ are everywhere tangent to $\tilde{M}^{4}$. That $\gamma_{\pm}$ is an integral curve of $X_{2}$ follows by checking that $X_{2}(\gamma_{\pm}(t))=(e^{t},\sqrt{2}e^{t},e^{t},0,0,0)=\gamma^{\prime}_{\pm}(t)$. Finally, a straightforward computation gives

[TABLE]

The proof of the next proposition is straightforward.

Proposition 18**.**

$(i)$* For each $\epsilon=(\epsilon_{1},\epsilon_{2},\epsilon_{3})$, $\epsilon_{j}\in\{-1,1\}$ for $1\leq j\leq 3$, the map $\Phi^{\epsilon}\colon\tilde{M}^{4}\to\tilde{M}^{4}$ given by*

[TABLE]

*satisfies $\Phi^{\epsilon}_{*}X_{j}(p)=\epsilon_{j}X_{j}(\Phi^{\epsilon}(p))$ for all $p\in\tilde{M}^{4}$.

$(ii)$ The map $\Psi\colon\tilde{M}^{4}\to\tilde{M}^{4}$ given by*

[TABLE]

satisfies

[TABLE]

*for all $p=(x_{1},x_{2},x_{3},y_{1},y_{2},y_{3})\in\tilde{M}^{4}$.

$(iii)$ The maps $\Psi$ and $\Phi^{\epsilon}$, $\epsilon\in\{-1,1\}\times\{-1,1\}\times\{-1,1\}$, generate a group of involutions of $\tilde{M}^{4}$ isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ that preserves the distribution ${\cal D}$ spanned by the vector fields $X_{1},X_{2}$ and $X_{3}$.*

Proof of Theorem 3: First we associate to each leaf $\sigma$ of ${\cal D}$ a covering map $\phi_{\sigma}\colon U_{\sigma}\to\sigma$ from a simply-connected open subset $U_{\sigma}\subset{\mathbb{R}}^{3}$ and a minimal immersion $f_{\sigma}\colon U_{\sigma}\to{\mathbb{R}}^{4}$ with three distinct principal curvatures whose induced metric is conformally flat.

For any $q\in\tilde{M}^{4}$ and for $1\leq i\leq 3$ denote by $\tau_{q}^{i}\colon J_{q}^{i}\to\tilde{M}^{4}$ the maximal integral curve of $X_{i}$ through $q$, that is, $0\in J_{q}^{i}$, $\tau^{i}_{q}(0)=q$, $(\tau^{i}_{q})^{\prime}(t)=X_{i}(\tau^{i}_{q}(t))$ for all $t\in J_{q}^{i}$, and $J_{q}^{i}$ is maximal with these properties. Let

[TABLE]

and let $\varphi^{i}\colon{\cal D}(X_{i})\to\tilde{M}^{4}$ be the flow of $X_{i}$, given by $\varphi^{i}(t,q)=\tau^{i}_{q}(t)$. For a fixed $p\in\sigma$ define $U_{\sigma}=U_{\sigma}(p)$ by

[TABLE]

and $\phi_{\sigma}=\phi^{p}_{\sigma}$ by

[TABLE]

Then $0\in U_{\sigma}$, $\phi_{\sigma}(0)=p$, and for all $u\in U_{\sigma}$ we have

[TABLE]

We claim that $\phi_{\sigma}$ is a covering map onto $\sigma$. Given $x\in\sigma$, let $\tilde{B}_{2\epsilon}(0)$ be an open ball of radius $2\epsilon$ centered at the origin such that $\phi_{\sigma}^{x}|_{\tilde{B}_{2\epsilon}(0)}$ is a diffeomorphism onto $B_{2\epsilon}(x)=\phi_{\sigma}^{x}(\tilde{B}_{2\epsilon}(0))$. Since

[TABLE]

whenever both sides are defined, where $t=(t_{1},t_{2},t_{3})$ and $s=(s_{1},s_{2},s_{3})$, if $x=\phi_{\sigma}^{p}(s)$, $s=(s_{1},s_{2},s_{3})\in U_{\sigma}$, then for any $y=\phi_{\sigma}^{x}(t)\in B_{2\epsilon}(x)$, $t=(t_{1},t_{2},t_{3})\in\tilde{B}_{2\epsilon}(0)$, we have

[TABLE]

This shows that $B_{2\epsilon}(x)\subset\phi_{\sigma}^{p}(U_{\sigma})$ if $x\in\phi_{\sigma}^{p}(U_{\sigma})$, hence $\phi_{\sigma}^{p}(U_{\sigma})$ is open in $\sigma$. But since $y=\phi_{\sigma}^{x}(t)$ if and only if $x=\phi_{\sigma}^{y}(-t)$, as follows from (57), the same argument shows that $x\in\phi_{\sigma}^{p}(U_{\sigma})$ if $y\in\phi_{\sigma}^{p}(U_{\sigma})$ for some $y\in B_{2\epsilon}(x)$, and hence $\sigma\setminus\phi_{\sigma}^{p}(U_{\sigma})$ is also open. It follows that $\phi_{\sigma}^{p}$ is onto $\sigma$.

Now, for any $x\in\sigma$ write

[TABLE]

and for each $\alpha\in A$ let $\tilde{B}_{2\epsilon}(\tilde{x}_{\alpha})$ denote the open ball of radius $2\epsilon$ centered at $\tilde{x}_{\alpha}$. Define a map $\psi_{\alpha}\colon B_{2\epsilon}(x)\to\tilde{B}_{2\epsilon}(\tilde{x}_{\alpha})$ by

[TABLE]

By (57) we have

[TABLE]

for all $y\in B_{2\epsilon}(x)$. Thus $\phi_{\sigma}^{p}$ is a diffeomorphism from $\tilde{B}_{2\epsilon}(\tilde{x}_{\alpha})$ onto $B_{2\epsilon}(x)$ having $\psi_{\alpha}$ as its inverse. In particular, this implies that $\tilde{B}_{\epsilon}(\tilde{x}_{\alpha})$ and $\tilde{B}_{\epsilon}(\tilde{x}_{\beta})$ are disjoint if $\alpha$ and $\beta$ are distinct indices in $A$. Finally, it remains to check that if $\tilde{y}\in(\phi_{\sigma}^{p})^{-1}(B_{\epsilon}(x))$ then $\tilde{y}\in\tilde{B}_{\epsilon}(\tilde{x}_{\alpha})$ for some $\alpha\in A$. This follows from the fact that

[TABLE]

For the last equality, observe from (57) that for all $x,y\in\sigma$ we have that $\phi_{\sigma}^{x}(t)=y$ if and only if $\phi_{\sigma}^{y}(-t)=x$.

Writing $\phi_{\sigma}=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$, it follows that $\phi_{\sigma}$ satisfies (53), as well as (51) and (54). Defining $h_{ij}=\frac{1}{v_{i}}\frac{\partial v_{j}}{\partial u_{i}}$, $1\leq i\neq j\leq 3$, and $V=(V_{1},V_{2},V_{3})$ by (52), it follows from Proposition 15 that the triple $(v,h,V)$, where $v=(v_{1},v_{2},v_{3})$ and $h=(h_{ij})$, satisfies (5), and hence gives rise to a minimal conformally flat hypersurface $f_{\sigma}\colon U_{\sigma}\to{\mathbb{R}}^{4}$ with three distinct principal curvatures by Corollary 6.

Given two distinct leaves $\sigma$ and $\tilde{\sigma}$ of ${\cal D}$, the corresponding immersions $f_{\sigma}$ and $f_{\tilde{\sigma}}$ are congruent if and only if there exists a diffeomorphism $\psi\colon U_{\sigma}\to U_{\tilde{\sigma}}$ such that

[TABLE]

where $I_{\sigma}$ and $I\!I_{\sigma}$ are the first and second fundamental formulae of $f_{\sigma}$, respectively, and $I_{\tilde{\sigma}}$, $I\!I_{\tilde{\sigma}}$ are those of $f_{\tilde{\sigma}}$. A long but straightforward computation shows that, up to a translation, either $\psi$ coincides with the map given by

[TABLE]

for some $\epsilon=(\epsilon_{1},\epsilon_{2},\epsilon_{3})$ with $\epsilon_{i}\in\{-1,1\}$ for $1\leq i\leq 3$, or it is the composition of such a map with the map given by

[TABLE]

It is easy to check that this is the case if and only if there exists $\Theta\in G$ such that $\phi_{\tilde{\sigma}}\circ\psi=\Theta\circ\phi_{\sigma}$.

Now let $f\colon M^{3}\to{\mathbb{R}}^{4}$ be a minimal isometric immersion with three distinct principal curvatures of a simply connected conformally flat Riemannian manifold. We shall prove that either $f(M^{3})$ is an open subset of the cone over a Clifford torus in ${\mathbb{S}}^{3}$ or there exist a leaf $\sigma$ of ${\cal D}$ and a local diffeomorphism $\rho\colon M^{3}\to V$ onto an open subset $V\subset U_{\sigma}$ such that $f$ is congruent to $f_{\sigma}\circ\rho$.

First we associate to $f$ a map $\phi_{f}\colon M^{3}\to M^{4}\subset{\mathbb{R}}^{6}$ as follows. Fix a unit normal vector field $N$ along $f$ and denote by $\lambda_{1}<\lambda_{2}<\lambda_{3}$ the distinct principal curvatures of $f$ with respect to $N$. For each $1\leq j\leq 3$, the eigenspaces $E_{\lambda_{j}}=\ker(A-\lambda_{j}I)$ associated to $\lambda_{j}$, where $A$ is the shape operator with respect to $N$ and $I$ is the identity endomorphism, form a field of directions along $M^{3}$, and since $M^{3}$ is simply connected we can find a smooth global unit vector field $Y_{j}$ along $M^{3}$ such that $\mbox{span}\{Y_{j}\}=E_{\lambda_{j}}$. Let the functions $v_{1},v_{2},v_{3}$ be defined on $M^{3}$ by

[TABLE]

and let $\alpha_{1},\alpha_{2},\alpha_{3}$ be given by

[TABLE]

Define $\phi_{f}\colon M^{3}\to{\mathbb{R}}^{6}$ by $\phi_{f}=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$.

Now, it follows from Theorem 5 that each point $p\in M^{3}$ has a connected open neighborhood $U\subset M^{3}$ endowed with principal coordinates $u_{1},u_{2},u_{3}$ such that the pair $(v,V)$ associated to $f$ satisfies (51) and (52). Let $\phi_{U}\colon U\to{\mathbb{R}}^{6}$ be given by $\phi_{U}=(v_{1},v_{2},v_{3},\alpha_{1},\alpha_{2},\alpha_{3})$, with (\alpha_{1},\alpha_{2},\alpha_{3})=\big{(}\frac{1}{v_{2}}\frac{\partial v_{2}}{\partial u_{1}},\frac{1}{v_{3}}\frac{\partial v_{3}}{\partial u_{2}},\frac{1}{v_{1}}\frac{\partial v_{1}}{\partial u_{3}}\big{)}. It is easy to check that $\phi_{f}|_{U}=\Theta\circ\phi_{U}$ for some $\Theta\in G$. On the other hand, by Proposition 15 we have that $\phi_{U}$ satisfies (53), as well as the algebraic equation (54). It follows that $\phi_{U}(U)\subset M^{4}$ and that

[TABLE]

for all $u\in U$. Therefore either $\phi_{U}(U)$ is an open subset of a leaf $\sigma_{U}$ of the distribution ${\cal D}$ on $\tilde{M}^{4}$ spanned by $X_{1},X_{2},X_{3}$, or $\phi_{U}(U)$ is an open segment of either $\ell_{+}$ or $\ell_{-}$. If the latter possibility holds for some open subset $U\subset M^{3}$, then $v_{1}=v_{3}$ on $U$, hence $\lambda_{2}=0$ on $U$. By analyticity, $\lambda_{2}=0$ on $M^{3}$, and hence Proposition 14 implies that $f(M^{3})$ is an open subset of a cone over a Clifford torus in an umbilical hypersurface $\mathbb{Q}^{3}(\tilde{c})\subset{\mathbb{R}}^{4}$, $\tilde{c}>0$.

Otherwise we have that each point $p\in M^{3}$ has an open neighborhood $U\subset M^{3}$ such that $\phi_{f}(U)$ is an open subset of a leaf $\sigma_{U}$ of ${\cal D}$. It follows that $\phi_{f}(M^{3})$ is an open subset of a leaf $\sigma$ of ${\cal D}$. If $\rho\colon M^{3}\to U_{\sigma}$ is a lift of $\phi_{f}$ with respect to $\phi_{\sigma}$, that is, $\phi_{f}=\phi_{\sigma}\circ\rho$, then $\rho$ is a local diffeomorphism such that $f$ and $f_{\sigma}\circ\rho$ have the same first and second fundamental forms. Therefore $f$ is congruent to $f_{\sigma}\circ\rho$.∎