Proof. Taking the derivative with respect to t of $ {\langle}f_{t*}X_i^t,f_{t*}X_j^t{\rangle}=\delta_{ij} $ at t = 0 and using Lemma (5), we obtain,

\begin{align*} 0= & \frac{\partial}{\partial t}|_{t=0}{\langle}f_{t*}X_i^t,f_{t*}X_j^t{\rangle}\\ =&{\langle}\tilde\nabla_{X_i}{\cal T}+f_*X_i',f_*X_j{\rangle}+{\langle}f_*X_i,\tilde\nabla_{X_j}{\cal T}+f_*X_j'{\rangle}\\ =& {\langle}X_i',X_j{\rangle}+{\langle}\tilde\nabla_{X_i}{\cal T},f_*X_j{\rangle}+{\langle}X_i,X_j'{\rangle}+{\langle}f_*X_i,\tilde\nabla_{X_j}{\cal T}{\rangle}. \end{align*}

Combining the preceding equation with (4) yields:

\begin{equation*} {\langle}X_i',X_j{\rangle}+{\langle}X_i,X_j'{\rangle}+2\rho{\langle}X_i,X_j{\rangle}=0,\,\,\,1\leq i,j\leq n. \end{equation*}

Proof. Let us first compute $\partial/\partial t|_{t=0}\|\alpha^t\|^2$. Let $\{X_i^t\}_{1\leq i\leq n}$ be a one-parameter family of frames such that, for each fixed $t\in (-\epsilon, \epsilon)$, $\{X_i^t\}_{1\leq i\leq n}$ is orthonormal with respect to the metric induced by f_t. By (12) we have,

(16)\begin{equation} {\langle}\frac{\partial}{\partial t}|_{t=0}\alpha^t(X^t_i,X^t_j), \eta{\rangle} ={\langle}\beta(X_i,X_j)+\rho\alpha(X_i,X_j)+\alpha(X_i',X_j)+\alpha(X_i,X_j'),\eta{\rangle}, \end{equation}

hence

\begin{align*} \frac{1}{2}\frac{\partial}{\partial t}|_{t=0}\|\alpha^t(X^t_i,X^t_j)\|^2=&{\langle}\frac{\partial}{\partial t}|_{t=0}\alpha^t(X^t_i,X^t_j),\alpha(X_i,X_j){\rangle}\\ =&{\langle}\beta(X_i,X_j)+\rho\alpha(X_i,X_j),\alpha(X_i,X_j){\rangle}\\ &+{\langle}\alpha(X_i',X_j)+\alpha(X_i,X_j'),\alpha(X_i,X_j){\rangle}. \end{align*}

Thus

\begin{align*} \frac{\partial}{\partial t}|_{t=0}\|\alpha^t\|^2=&2\rho\|\alpha\|^2+2\sum_{i,j=1}^n{\langle}\beta(X_i,X_j),\alpha(X_i,X_j){\rangle} \\ &+2\sum_{i,j=1}^n{\langle}\alpha(X_i',X_j)+\alpha(X_i,X_j'),\alpha(X_i,X_j){\rangle}\\ =&2\rho\|\alpha\|^2+2\sum_{i,j=1}^n{\langle}\beta(X_i,X_j),\alpha(X_i,X_j){\rangle} \\ &+4\sum_{i,j=1}^n{\langle}\alpha(X_i',X_j),\alpha(X_i,X_j){\rangle}. \end{align*}

It follows from (9) that:

\begin{align*} 2{\langle}\beta(X_i,X_j),\alpha(X_i,X_j){\rangle}=& {\langle}\beta(X_i,X_i),\alpha(X_j,X_j){\rangle}+{\langle}\beta(X_j,X_j),\alpha(X_i,X_i){\rangle}\\ &+{\langle}X_j,X_j{\rangle}\mbox{Hess}\,\rho(X_i,X_i)+{\langle}X_i,X_i{\rangle}\mbox{Hess}\,\rho(X_j,X_j), \end{align*}

for all $1\leq i, j\leq n$ with i ≠ j. Therefore

\begin{equation*} 2\sum_{i,j=1}^n{\langle}\beta(X_i,X_j),\alpha(X_i,X_j){\rangle}= 2(n-1)\Delta\rho +2\sum_{i, j=1}^n{\langle}\beta(X_i,X_i),\alpha(X_j,X_j){\rangle}, \end{equation*}

where Δ denotes the Laplacian. On the other hand, from the Gauss equation,

\begin{equation*} {\langle}\alpha(X_i',X_j),\alpha(X_i,X_j){\rangle}={\langle}\alpha(X_i',X_i),\alpha(X_j,X_j){\rangle}+{\langle}R(X_i',X_j)X_i,X_j{\rangle}, \end{equation*}

where R denotes the Riemann curvature tensor of Mⁿ, we obtain,

\begin{equation*} \sum_{i\neq j}{\langle}\alpha(X_i',X_j),\alpha(X_i,X_j){\rangle}= \sum_{i\neq j}{\langle}\alpha(X_i',X_i),\alpha(X_j,X_j){\rangle}-\sum_{i=1}^n \mbox{Ric }(X_i',X_i). \end{equation*}

Thus

\begin{align*} \frac{\partial}{\partial t}|_{t=0}\|\alpha^t\|^2=&2\rho\|\alpha\|^2+2(n-1)\Delta\rho +2\sum_{i, j=1}^n{\langle}\beta(X_i,X_i),\alpha(X_j,X_j){\rangle}\\ &+4\sum_{i,j=1}^n{\langle}\alpha(X_i',X_i),\alpha(X_j,X_j){\rangle}-4\sum_{i=1}^n \mbox{Ric }(X_i',X_i)\\ =&2\rho\|\alpha\|^2+2(n-1)\Delta\rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}\\ &+4n\sum_{i=1}^n{\langle}\alpha(X_i',X_i),\mathcal{H}{\rangle}-4\sum_{i=1}^n\mbox{Ric }(X_i',X_i). \end{align*}

By (13), we may write $X_i'=-\rho X_i+\sum_{i\neq k}{\langle}X_i',X_k{\rangle}X_k$; hence

\begin{align*} \sum_{i=1}^n{\langle}\alpha(X_i',X_i),\mathcal{H}{\rangle}=&-\rho n\|\mathcal{H}\|^2+\sum_{i\neq k}{\langle}X_i',X_k{\rangle}{\langle}\alpha(X_k,X_i),\mathcal{H}{\rangle}\\ =&-\rho n\|\mathcal{H}\|^2, \end{align*}

where the last equality follows from (14). Similarly,

\begin{equation*} \sum_{i=1}^n\mbox{Ric }(X_i',X_i)=-\rho n(n-1)s, \end{equation*}

where $s=\frac1{n(n-1)}\sum_{i=1}^n{\textstyle\text{Ric}}(X_i,X_i)$ is the scalar curvature of Mⁿ. Thus

\begin{align*} \frac{\partial}{\partial t}|_{t=0}\|\alpha^t\|^2=&2\rho\|\alpha\|^2+2(n-1)\Delta\rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}\\ &-4n^2\rho\|\mathcal{H}\|^2+4\rho n(n-1)s. \end{align*}

Using that

\begin{equation*} s=\frac{n}{n-1}\|\mathcal{H}\|^2-\frac{1}{n(n-1)}\|\alpha\|^2, \end{equation*}

we obtain,

(17)\begin{equation} \frac{\partial}{\partial t}|_{t=0}\|\alpha^t\|^2=2(n-1)\Delta \rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}-2\rho\|\alpha\|^2. \end{equation}

We now compute $\partial/\partial t|_{t=0}\|\mathcal{H}^t\|^2$. With $\{X_i^t\}_{1\leq i\leq n}$ as above, we have,

\begin{align*} \frac{\partial}{\partial t}|_{t=0}\|\mathcal{H}^t\|^2=&2{\langle}\frac{\partial}{\partial t}|_{t=0}\mathcal{H}^t,\mathcal{H}{\rangle}\\ =&2{\langle}\frac{1}{n}\sum_{i=1}^n\frac{\partial}{\partial t}|_{t=0}\alpha^t(X^t_i,X^t_i),\mathcal{H}{\rangle}\\ =&2\rho\|\mathcal{H}\|^2+2{\langle}\mathcal{L},\mathcal{H}{\rangle}+\frac{4}{n}\sum_{i=1}^n{\langle}\alpha(X_i',X_i),\mathcal{H}{\rangle}, \end{align*}

where the last step follows from (16). Using (13) and (14) as before, we obtain:

(18)\begin{equation} \frac{\partial}{\partial t}|_{t=0}\|\mathcal{H}^t\|^2=2{\langle}\mathcal{L},\mathcal{H}{\rangle}-2\rho\|\mathcal{H}\|^2. \end{equation}

It follows from (17) and (18) that:

\begin{align*} \frac{\partial}{\partial t}|_{t=0}\phi_t^2=&\frac{n}{n-1}\left(2(n-1)\Delta \rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}-2\rho\|\alpha\|^2\right.\\ &\left.-2n{\langle}\mathcal{L},\mathcal{H}{\rangle}+2n\rho\|\mathcal{H}\|^2\right)\\ =&\frac{n}{n-1}(2(n-1)\Delta\rho+2n(n-1){\langle}\mathcal{L},\mathcal{H}{\rangle}\\ &-2\rho\|\alpha\|^2+2\rho n\|\mathcal{H}\|^2)\\ =&2n\Delta\rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}-2\phi^2\rho, \end{align*}

where we have used (1) in the last equality.

Proof of Theorem (2)

If ${\cal T}$ is the variational vector field of an infinitesimal Moebius variation $F\colon (-\epsilon, \epsilon)\times M^n\to \mathbb{R}^m$ of f, then the corresponding conformal factor ρ is given by (5). Thus, (15) yields:

\begin{equation*} -2\phi^2\rho=2n\Delta\rho-2\phi^2\rho+2n^2{\langle}\mathcal{L},\mathcal{H}{\rangle}, \end{equation*}

and hence (7) holds.

For the converse, assume that ${\cal T}$ is a conformal infinitesimal bending of f whose conformal factor ρ satisfies (7). The variation ${\cal F}\colon\mathbb{R}\times M^n\to\mathbb{R}^m$ given by: $ {\cal F}(t,x)=f(x)+t{\cal T}(x), $ is a conformal infinitesimal variation with variational vector field ${\cal T}$. Let $f_t={\cal F}(t,\cdot)$ and let ϕ_t be given by (1) for each f_t, $t\in \mathbb{R}$. We claim that ${\cal F}$ is an infinitesimal Moebius variation of f. Indeed, we have

\begin{equation*}\frac\partial{\partial t}\vert_{t=0}(\phi_t^2\langle f_{t\ast}X,f_{t\ast}Y\rangle)=\frac\partial{\partial t}\vert_{t=0}(\phi_t^2)\langle X,Y\rangle+\phi^2(\langle{\widetilde\nabla}_X\mathcal T,f_\ast Y\rangle+\langle f_\ast X,{\widetilde\nabla}_Y\mathcal T\rangle)\end{equation*}

for all $X,Y\in\mathfrak{X}(M)$, hence,

\begin{equation*} \frac{\partial}{\partial t}|_{t=0}(\phi_t^2{\langle}f_{t*X},f_{t*}Y{\rangle})=\left(\frac{\partial}{\partial t}|_{t=0}(\phi_t^2)+2\phi^2\rho\right){\langle}X,Y{\rangle} \end{equation*}

by (4). On the other hand, from (15) and (7) we have:

\begin{equation*} \frac{\partial}{\partial t}|_{t=0}(\phi_t^2)+2\phi^2\rho=0, \end{equation*}

which proves the claim and completes the proof.

Proof. It follows from (4) and Lemma 5 that:

\begin{eqnarray*} \partial/\partial t|_{t=0}{\langle}\alpha^t(X,Y),N_t{\rangle}&=&\partial/\partial t|_{t=0}{\langle}{f_t}_*A_tX, {f_t}_*Y{\rangle}\\ &=&{\langle}A'X,Y{\rangle}+2\rho{\langle}AX,Y{\rangle} \end{eqnarray*}

for all $X,Y\in\mathfrak{X}(M)$. On the other hand, from (12) we obtain,

\begin{equation*} \partial/\partial t|_{t=0}{\langle}\alpha^t(X,Y),N_t{\rangle}={\langle}\mathcal{B}X,Y{\rangle}+\rho{\langle}AX,Y{\rangle}. \end{equation*}

Comparing the two preceding equations yields (19).

Proof. Since f is infinitesimally Moebius bendable, it admits, in particular, a conformal infinitesimal bending ${\cal T}$ that is non-trivial on any open subset. By Lemma 11, there exists an open and dense subset $\mathcal{U}$ of Mⁿ where $\mathcal{B}$, H and A share a common eigenbundle Δ of constant dimension $\dim \Delta\geq n-2$. In the proof of Proposition 15 in [Reference Dajczer, Jimenez and Vlachos7] (see Eq. (35) therein), it was shown that:

(20)\begin{equation} bA+\lambda(\mathcal{B}-bI)+\mbox{Hess}\,\rho=0, \end{equation}

on $\mathcal{U}$. Let $\mathcal{H}$ and $\mathcal{L}$ be given by $\mbox{tr}\, A=n\mathcal{H}$ and $\mbox{tr}\, \mathcal{B}=n\mathcal{L}$. Taking traces in (20) yields

\begin{equation*} nb\mathcal{H}+n\lambda\mathcal{L}-n\lambda b+\Delta\rho=0. \end{equation*}

We write the preceding equation as:

\begin{equation*} n(\mathcal{H}-\lambda)(\mathcal{L}-b)=\Delta\rho+n\mathcal{L}\mathcal{H}, \end{equation*}

which is also equivalent to:

(21)\begin{equation} \mbox{tr}\,(A-\lambda I)\mbox{tr}\,(\mathcal{B}-bI)=n(\Delta\rho+n\mathcal{L}\mathcal{H}). \end{equation}

Taking into account that (7) reduces to $\Delta\rho+n\mathcal{L}\mathcal{H}=0$, it follows from (21) and Theorem (2) that $\mbox{tr}\,(A-\lambda I)\mbox{tr}\,(\mathcal{B}-bI)~=~0$.

Finally, notice that the preceding condition cannot occur on any open subset where $\dim \Delta= n-1$. Indeed, if $\dim \Delta= n-1$, then the condition $\mbox{tr}\,(A-\lambda I)=0$ would imply that $A=\lambda I$, whereas $\mbox{tr}\,(\mathcal{B}-bI)=0$ would yield $\mathcal{B}=bI$, in contradiction with the assumptions that f is free of umbilic points and that the infinitesimal bending ${\cal T}$ is non-trivial, respectively.

Proof. If $\Delta=\ker (A-\lambda I)$, then it is the eigenbundle corresponding to the principal curvature λ, and hence umbilical. Thus we only need to consider the case in which Δ coincides with $\ker (B-bI)$ and is a proper subspace of $\ker(A-\lambda I)$. Equation (11) can be written as:

\begin{equation*} (\nabla_X(\mathcal{B}-bI))Y-(\nabla_Y(\mathcal{B}-bI))X+(X\wedge Y)(A\nabla\rho-\nabla b)=0, \end{equation*}

for any $X,Y\in\mathfrak{X}(M)$, where $\nabla b$ and $\nabla\rho$ are the gradients of b and of the conformal factor ρ, respectively. Let $T,S\in\Gamma(\Delta)$ be orthogonal and take $X\in\Gamma(\Delta^\perp)$. Evaluating the preceding equation in X and T and taking the inner product of both sides with S gives $ {\langle}\nabla_T(\mathcal{B}-bI)X,S{\rangle}=0. $ Since we are assuming that $\mbox{rank }(\mathcal{B}-bI)=2$, the above equation gives:

\begin{equation*} (\nabla_TS)_{\Delta^\perp}=0, \end{equation*}

for all $T,S\in\Gamma(\Delta)$ with ${\langle}T,S{\rangle}=0$. Thus Δ is an umbilical distribution.

Proof. It follows from (20) that, at each $x\in \mathcal{U}$, Eq. (10) can be written as:

(22)\begin{equation} \bar{\mathcal{B}}X \wedge \bar A Y= \bar{\mathcal{B}}Y \wedge \bar A X, \end{equation}

or equivalently,

\begin{equation*}{\langle}\bar{A}Y,X{\rangle}{\langle}\bar{\mathcal{B}}X, Y{\rangle}-{\langle}\bar{\mathcal{B}}X,X{\rangle}{\langle}\bar{A}Y, Y{\rangle}={\langle}\bar{A}X,X{\rangle}{\langle}\bar{\mathcal{B}}Y,Y{\rangle}-{\langle}\bar{\mathcal{B}}Y,X{\rangle}{\langle}\bar{A}X,Y{\rangle},\end{equation*}

for all $X, Y\in T_x \mathcal{U}$. Applying the preceding equation to orthogonal unit eigenvectors X and Y of $\bar A|_{\Delta^\perp}$, with $\bar A X=\mu_1X$ and $\bar A Y=\mu_2Y$, gives

\begin{equation*} -\mu_2{\langle}\bar{\mathcal{B}}X,X{\rangle}=\mu_1{\langle}\bar{\mathcal{B}}Y,Y{\rangle}. \end{equation*}

Therefore, at each point of $\mathcal{U}_1$, either $\mu_1=\mu_2:=\mu\neq 0$, and hence $\bar A|_{\Delta^\perp}=\mu I$, or ${\langle}\bar{\mathcal{B}}X,X{\rangle}=0={\langle}\bar{\mathcal{B}}Y,Y{\rangle}$. In the latter case, denoting $\theta={\langle}\bar{\mathcal{B}}X,Y{\rangle}$, we have $\bar{\mathcal{B}}X=\theta Y$ and $\bar{\mathcal{B}}Y=\theta X$.

Proof. Denoting by λ the principal curvature of f with constant multiplicity n − 2 with respect to a unit normal vector field N, the map

\begin{equation*}x\in M^n\mapsto f(x)+\frac{1}{\lambda(x)}N\end{equation*}

determines a two-parameter congruence of hyperspheres that is enveloped by f. As explained in the introduction, this congruence of hyperspheres is determined by a space-like surface $s\colon L^2\to\mathbb{S}_1^{n+2}$.

Since f is infinitesimally Moebius bendable, it admits, in particular, a conformal infinitesimal bending that is non-trivial on any open subset. By Proposition 15 in [Reference Dajczer, Jimenez and Vlachos7], the hypersurface f is either elliptic, hyperbolic or parabolic with respect to $J\in\Gamma(\mbox{End}(\Delta^\perp))$ satisfying $J^2=-I$, $J^2=I$ or $J^2=0$, respectively, with J ≠ I if $J^2=I$ and J ≠ 0 if $J^2=0$. Moreover, the tensor $\mathcal{B}$ associated with ${\cal T}$ satisfies:

(23)\begin{equation} \bar{\mathcal{B}}=\mu \bar{A} J, \end{equation}

where $0\neq\mu\in C^\infty(M)$ is constant along the leaves of Δ.

It was also shown in [Reference Dajczer, Jimenez and Vlachos7] that, in the hyperbolic and elliptic cases, the tensor J is projectable with respect to the quotient map $\pi\colon M^n\to L^2$ onto the spaces of leaves of the eigenbundle Δ of λ, that is, there exists $\bar{J} \in \mbox{End}(TL)$ such that $\bar{J}\circ \pi_*=\pi_*\circ J$. Moreover, the surface $s\colon L^2\to\mathbb{S}_1^{n+2}$ is either a special elliptic or special hyperbolic surface with respect to $\bar J$. This means that,

(24)\begin{equation} \alpha^s(\bar{J}\bar{X},\bar{Y})=\alpha^s(\bar{X},\bar{J}\bar{Y}), \end{equation}

for all $\bar{X},\bar{Y}\in \mathfrak{X}(L)$, and that there exists $\mu\in C^\infty(L)$ such that $\mu \bar J$ is a Codazzi tensor on L ².

In the sequel we will show that, under the assumptions of the proposition, the tensors J and $\bar{J}$ act as a rotation of angle $\pi/2$ on $\Delta^\perp$ and on each tangent space of L ², respectively, that is, both J and $\bar{J}$ are orthogonal tensors satisfying $J^2=-I$ and $\bar{J}^2=-I$. From the orthogonality of J and the symmetry of $\bar{B}$ it will follow that the tensor $\bar{A}=A-\lambda I$ is traceless by (23). This implies that λ is the mean curvature function of f, and hence the congruence of hyperspheres determined by s is its central sphere congruence. On the other hand, the orthogonality of $\bar{J}$ and the fact that $\bar{J}^2=-I$ implies the minimality of s by (24), and this will conclude the proof.

First we rule out the parabolic case. So, assume that there exists $J\in\Gamma(\Delta^\perp)$ such that $J^2=0$, J ≠ 0, $\nabla^h_T J=0$ for all $T\in\Gamma(\Delta)$, and such that $C_T\in\mbox{span}\{I,J\}$ for all $T\in\Gamma(\Delta)$. By Proposition 16 of [Reference Dajczer, Jimenez and Vlachos7], f is conformally ruled, with the leaves of the distribution $\Delta\oplus\ker(J)$ as the rulings of f. Let $X,Y\in\Gamma(\Delta^\perp)$ be an orthonormal basis of $\Delta^\perp$ such that JX = Y and JY = 0, let $\lambda_1, \lambda'\in C^{\infty}(M)$ be such that $\bar A X=\lambda_1X+\lambda' Y$ and $\bar A Y=\lambda' X$. From (23) we see that $\bar{\mathcal B}X=\mu\lambda' X$ and $\bar{\mathcal B} Y=0$. Since $\mathcal{B}$ is not a multiple of the identity endomorphism, then $\lambda'\neq 0$, and hence $\mbox{tr}\, \bar{\mathcal B}\neq 0$. It follows from Proposition (12) that $\lambda_1=\mbox{tr}\, \bar A=0$.

It follows from the Codazzi equation that:

(25)\begin{equation} \nabla^h_T \bar A=\bar{A}C_T \end{equation}

for any $T\in\Gamma(\Delta)$.

For a fixed $T\in\Gamma(\Delta)$, write $ C_T=dI+eJ, $ for some smooth functions d and e. On one hand, $ \nabla^h_T\bar{A}X=\nabla^h_T\lambda'Y=T(\lambda')Y, $ where we have used that $\nabla^h_TY=0$, for Y is tangent to the rulings. On the other hand,

\begin{equation*} \bar{A}C_TX=\bar{A}(dX+eY)=d\lambda'Y+e\lambda'X.\end{equation*}

Therefore e = 0 by (25) and, since $T\in\Gamma(\Delta)$ was chosen arbitrarily, it follows that $C_T\in \mbox{span}\{I\}$ for any $T\in\Gamma(\Delta)$, a contradiction with our assumption.

Now assume that f is hyperbolic, that is, that there exists $J\in\Gamma(\mbox{End}(\Delta^\perp))$ such that $J^2=I$, with J ≠ I, $\nabla^h_T J=0$ and such that $C_T\in\mbox{span}\{I,J\}$ for all $T\in\Gamma(\Delta)$. Let $\{X,Y\}$ be a frame of $\Delta^\perp$ of unit eigenvectors of J, with JX = X and $JY=-Y$. Since $\nabla^h_T J=0$ for all $T\in\Gamma(\Delta)$, it follows that $\nabla^h_T X=0=\nabla^h_T Y$. The symmetry of $\bar{\mathcal{B}}=\mu\bar{A}J$ yields ${\langle}\bar{A}X,Y{\rangle}=0$. Write $\bar{A}X=\alpha X+\beta Y$ and $\bar{A}Y=\gamma X+ \delta Y$ for some smooth functions α, β, γ, δ. Then

(26)\begin{equation} {\langle}\bar{A}X,X{\rangle}=\alpha+\beta{\langle}Y,X{\rangle},\,\,\,\,\,{\langle}\bar{A}Y,Y{\rangle}=\gamma{\langle}Y,X{\rangle}+\delta,\end{equation}

and from ${\langle}\bar{A}X,Y{\rangle}=0={\langle}X, \bar{A}Y{\rangle}$ we obtain,

(27)\begin{equation} \alpha{\langle}X,Y{\rangle}+\beta=0=\gamma+\delta{\langle}X,Y{\rangle}. \end{equation}

On the other hand, writing as before $C_T=dI+eJ$ for some smooth functions d and e, Eq. (25) gives,

(28)\begin{equation} {\langle}\nabla^h_T\bar{A}X, X{\rangle}={\langle}\bar{A}C_TX, X{\rangle}=(d+e){\langle}\bar{A}X, X{\rangle}, \end{equation}

and similarly,

(29)\begin{equation} {\langle}\nabla^h_T\bar{A}Y, Y{\rangle}={\langle}\bar{A}C_TY, Y{\rangle}=(d-e){\langle}\bar{A}Y, Y{\rangle}. \end{equation}

Suppose that $\mbox{tr}\,\bar{A}=0$. Then $\alpha=-\delta$, hence (27) implies that $\beta=-\gamma$. Thus ${\langle}\bar{A}X,X{\rangle}=-{\langle}\bar{A}Y,Y{\rangle}$ by (26), and hence

\begin{equation*}{\langle}\nabla^h_T\bar{A}X, X{\rangle}=T{\langle}\bar{A}X, X{\rangle}=-T{\langle}\bar{A}Y, Y{\rangle}=-{\langle}\nabla^h_T\bar{A}Y, Y{\rangle}. \end{equation*}

Comparing with (28) and (29) gives $d+e=d-e$, for ${\langle}\bar{A}X, X{\rangle}\neq 0$ by the assumption that $\mbox{rank }\bar A=2$. Hence e = 0.

If $\mbox{tr}\, \bar{\mathcal{B}}=0$, then (23) gives $\alpha=\delta$, and hence $\gamma=\beta$ by (27). Therefore ${\langle}\bar{A}X,X{\rangle}={\langle}\bar{A}Y,Y{\rangle}$ by (26), and hence

\begin{equation*}{\langle}\nabla^h_T\bar{A}X, X{\rangle}=T{\langle}\bar{A}X, X{\rangle}=T{\langle}\bar{A}Y, Y{\rangle}={\langle}\nabla^h_T\bar{A}Y, Y{\rangle}. \end{equation*}

Then, we obtain as before that e = 0 by comparing with (28) and (29), and we conclude as in the parabolic case that $C_T\in \mbox{span}\{I\}$ for any $T\in\Gamma(\Delta)$, a contradiction.

Finally, suppose that f is elliptic, that is, that there exists $J\in\Gamma(\mbox{End}(\Delta^\perp))$ such that $J^2=-I$, $\nabla^h_T J=0$, and such that $C_T\in\mbox{span}\{I,J\}$ for all $T\in\Gamma(\Delta)$. Let $\{X,Y\}$ be a frame of $\Delta^\perp$ such that JX = Y and $JY=-X$. This is equivalent to asking the complex vector fields X − iY and X + iY to be pointwise eigenvectors of the $\mathbb{C}$-linear extension of J, also denoted by J, associated with the eigenvalues i and −i, respectively. Thus $z(X-iY)=(sX+tY)+i(tX-sY)$ and $z(X+iY)=(sX-tY)+i(tX+sY)$ are also eigenvectors of J associated to i and −i, respectively, for any $z=s+it\in \mathbb{C}$, that is, $\bar{X}=sX+tY$ and $\bar{Y}=-tX+sY$ form a new frame of $\Delta^\perp$ such that $J\bar X=\bar Y$ and $J\bar Y=-\bar X$. It is easily seen that s and t can be chosen so that $\bar{X}$ and $\bar{Y}$ are unit vector fields. In summary, we can always choose a frame $\{X,Y\}$ of unit vector fields such that JX = Y and $JY=-X$.

Since $\nabla^h_T J=0$ for all $T\in\Gamma(\Delta)$, then $J\nabla^h_T X=\nabla^h_T Y$ and $J\nabla^h_T Y=-\nabla^h_T X$. Denoting $\hat X=\nabla^h_T X$ and $\hat Y=\nabla^h_T Y$, it follows that $\hat X-i\hat Y=(s+it)(X-iY)$ for some $s+it\in \mathbb{C}$, that is,

\begin{equation*}\hat X=sX+tY\,\,\,\,\,\mbox{and}\,\,\,\,\,\hat Y=-tX+sY.\end{equation*}

Since X and Y have unit length, then ${\langle}\hat X,X{\rangle}=0={\langle}\hat Y, Y{\rangle}$. Thus

\begin{equation*}s+t{\langle}X,Y{\rangle}=0=s-t{\langle}X,Y{\rangle},\end{equation*}

and hence $t{\langle}X,Y{\rangle}=0=s$.

Assume that J is not an orthogonal tensor, that is, that ${\langle}X,Y{\rangle}\neq 0$. Then $s=0=t$, that is, $\nabla^h_T X=0=\nabla^h_T Y$ for all $T\in \Gamma(\Delta)$.

Write $\bar{A}X=\alpha X+\beta Y$ and $\bar{A}Y=\gamma X+ \delta Y$ for some smooth functions α, β, γ and δ. The symmetry of $\bar{\mathcal{B}}$ gives:

(30)\begin{equation} {\langle}\bar{A}X,X{\rangle}+{\langle}\bar{A}Y,Y{\rangle}=0. \end{equation}

Then

(31)\begin{equation} {\langle}\bar{A}X,X{\rangle}=\alpha+\beta{\langle}Y,X{\rangle},\,\,\,\,\,{\langle}\bar{A}Y,Y{\rangle}=\gamma{\langle}Y,X{\rangle}+\delta,\end{equation}

and from (30) and ${\langle}\bar{A}X,Y{\rangle}={\langle}X, \bar{A}Y{\rangle}$ we obtain, respectively,

(32)\begin{equation} (\alpha+\delta) +(\beta+\gamma){\langle}X,Y{\rangle}=0, \end{equation}

and

(33)\begin{equation} \alpha{\langle}X,Y{\rangle}+\beta=\gamma+\delta{\langle}X,Y{\rangle}. \end{equation}

On the other hand, writing as before $ C_T=dI+eJ $ for some smooth functions d and e, Eq. (25) gives:

(34)\begin{equation} {\langle}\nabla^h_T\bar{A}X, Y{\rangle}={\langle}\bar{A}C_TX, Y{\rangle}=d{\langle}\bar{A}X, Y{\rangle} +e{\langle}\bar{A}Y, Y{\rangle}. \end{equation}

Now assume that $\mbox{tr}\,\bar{A}=0$. Then $\alpha=-\delta$, hence $\beta=-\gamma$ by (32). Thus ${\langle}\bar{A}X,Y{\rangle}=0$ by (33), and hence ${\langle}\nabla^h_T\bar{A}X, Y{\rangle}=T{\langle}\bar{A}X, Y{\rangle}=0$. It follows from (34) that e = 0, for ${\langle}\bar{A}Y, Y{\rangle}\neq 0$ by the assumption that $\mbox{rank }\bar A=2$.

If $\mbox{tr}\,\bar{\mathcal{B}}=0$, then (23) gives $\gamma=\beta$, hence $\alpha=\delta$ by (33). Therefore ${\langle}\bar{A}X,X{\rangle}=0={\langle}\bar{A}Y,Y{\rangle}$ by (32) and (31). Then ${\langle}\nabla^h_T\bar{A}X, X{\rangle}=T{\langle}\bar{A}X, X{\rangle}=0$, and, on the other hand,

\begin{eqnarray*} {\langle}\nabla^h_T\bar{A}X, X{\rangle}&=&{\langle}\bar{A}C_TX,X{\rangle}\\ &=&{\langle}\bar{A}(dX+eY),X{\rangle}\\ &=& d{\langle}\bar{A}X,X{\rangle}+e{\langle}\bar{A}Y,X{\rangle}\\ &=& e{\langle}\bar{A}Y,X{\rangle}. \end{eqnarray*}

It follows that e = 0, for ${\langle}\bar{A}Y, X{\rangle}\neq 0$ by the assumption that f is free of points with a principal curvature of multiplicity at least n − 1. We conclude as in the previous cases that $C_T\in \mbox{span}\{I\}$ for any $T\in \Gamma(\Delta)$, a contradiction.

It follows that J must be an orthogonal tensor, that is, ${\langle}X,Y{\rangle}=0$. It remains to show that the tensor $\bar{J} \in \mbox{End}(TL)$ given by $\bar{J}\circ \pi_*=\pi_*\circ J$ is also orthogonal. For this, we use the fact that the metric ${\langle}\cdot,\cdot{\rangle}'$ on L ² induced by s is related to the metric of Mⁿ by:

(35)\begin{equation} {\langle}\bar{Z},\bar{W}{\rangle}'={\langle}\bar{A}Z, \bar{A}W{\rangle}, \end{equation}

for all $\bar{Z},\bar{W}\in\mathfrak{X}(L)$, where Z, W are the horizontal lifts of $\bar{Z}$ and $\bar{W}$, respectively. Let $\bar{X}\in\mathfrak{X}(L)$ and denote by $X\in\Gamma(\Delta^\perp)$ its horizontal lift. Using the symmetry of $\bar{A}J$, we have:

\begin{align*} {\langle}\bar{X},\bar{J}\bar{X}{\rangle}'&={\langle}\bar{A}X, \bar{A}JX{\rangle}\\ &={\langle}\bar{A}J\bar{A}X,X{\rangle}\\ &={\langle}J\bar{A}X,\bar{A}X{\rangle}\\ &=0, \end{align*}

where in the last step we have used that J acts as a rotation of angle $\pi/2$ on $\Delta^\perp$. Using again the symmetry of $\bar{A}J$, the proof of the orthogonality of $\bar{J}$ is completed by noticing that:

\begin{align*} {\langle}\bar{J}\bar{X},\bar{J}\bar{X}{\rangle}'&={\langle}\bar{A}JX,\bar{A}JX{\rangle}\\ &={\langle}J\bar{A}JX,\bar{A}X{\rangle}\\ &={\langle}JJ^t\bar{A}X,\bar{A}X{\rangle}\\ &=-{\langle}J^2\bar{A}X,\bar{A}X{\rangle}\\ &={\langle}\bar{X},\bar{X}{\rangle}'. \end{align*}

Proof of Theorem 1

Let $\mathcal{U}$ and Δ be, respectively, the open and dense subset of Mⁿ and the distribution of rank n − 2 given by Proposition 12. By that result, $\mathcal{U}$ splits as $\mathcal{U}=\mathcal{U}_1\cup \mathcal{U}_2$, with $\mbox{tr}\, \bar{\mathcal{B}}=0$ on $\mathcal{U}_1$ and $\mbox{tr}\, \bar{A}=0$ on $\mathcal{U}_2$.

We also consider the decompositions $\mathcal{U}=\mathcal{V}_1\cup \mathcal{V}_2$ and $\mathcal{U}=\mathcal{W}_1\cup \mathcal{W}_2$, where $\mathcal{V}_1$ and $\mathcal{V}_2$ are the subsets where the dimension of $\ker \bar{A}$ is either n − 2 or n − 1, respectively, $\mathcal{W}_2$ is the subset where $C(\Gamma(\Delta))\subset \mbox{span}\{I\}$ and $\mathcal{W}_1=\mathcal{U}\setminus \mathcal{W}_2$.

In the following we denote by S ⁰ the interior of the subset S. We will show that the direct statement holds on the open and dense subset:

\begin{equation*}\mathcal{U}^*=\mathcal{V}^0_2\cup (\mathcal{V}_1\cap \mathcal{W}_1)\cup (\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{U}^0_2)\cup (\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_2)\cup (\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_1),\end{equation*}

where $\mathcal{Y}_1$ and $\mathcal{Y}_2$ are the subsets of $\mathcal{U}_1$ given by Proposition 14.

It follows from Proposition 15 that the central sphere congruence of $f|_{\mathcal{V}_1\cap \mathcal{W}_1}$ is determined by a minimal space-like surface $s\colon L^2\to \mathbb{S}_1^{n+2}$.

The proof of the direct statement will be completed once we prove that, for each connected component $\mathcal{W}$ of the subsets $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_2$, $\mathcal{V}^0_2$, $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{U}^0_2$ and $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_1$, respectively, $f|_{\mathcal{W}}$ is a conformally surface-like hypersurface determined by a surface $h\colon L^2\to \mathbb{Q}_\epsilon$, $\epsilon \in \{-1,0,1\}$ of one of the following types:

1. (i) an isothermic surface;

2. (ii) a generalized cone over a unit-speed curve $\gamma\colon J\to \mathbb{Q}_{c}^2$ in an umbilical surface $\mathbb{Q}_{c}^2\subset \mathbb{Q}_{\epsilon}^3$, $c\geq \epsilon$;

3. (iii) a minimal surface;

4. (iv) an umbilical surface.

Notice that any surface as in (ii), (iii) and (iv) is also isothermic (for h as in (ii) see Corollary 12 in [Reference Jimenez and Tojeiro11]). Notice also that $f|_{\mathcal{W}}$ being a conformally surface-like hypersurface determined by a surface $h\colon L^2\to \mathbb{Q}_\epsilon$, $\epsilon \in \{-1,0,1\}$, is equivalent to ${\mathcal{W}}$ being (isometric to) a Riemannian product $L^2\times N^{n-2}$ and to $f|_{\mathcal{W}}$ being given by $f|_{\mathcal{W}}=\mathcal{I}\circ\Phi\circ (h\times i)$, where i is the inclusion map of an open subset $N^{n-2}$ of either $\mathbb{R}^{n-2}$ or $\mathbb{Q}_{-\epsilon}^{n-2}$, according to whether ϵ is zero or not, $\mathcal{I}$ is a Moebius transformation of $\mathbb{R}^{n+1}$, and Φ is the standard isometry $\Phi\colon \mathbb{R}^3\times \mathbb{R}^{n-2}\to \mathbb{R}^{n+1}$ if ϵ = 0 and, if $\epsilon =-1$ or 1, respectively, the conformal diffeomorphism:

• • $\Phi\colon \mathbb{H}^3\times \mathbb{S}^{n-2}\subset \mathbb{L}^{4}\times \mathbb{R}^{n-1}\to \mathbb{R}^{n+1}\setminus \mathbb{R}^{2}$, $\Phi(x, y)=\frac{1}{x_0}(x_1, x_2,y) $ for all $x=x_0e_0+x_1e_1+x_2e_2+x_3e_3\in \mathbb{L}^{4}$ and $y=(y_1, \ldots, y_{n-1})\in \mathbb{S}^{n-2}\subset \mathbb{R}^{n-1}$, where $\{e_0, e_1, e_2, e_3\}$ is a pseudo-orthonormal basis of the Lorentzian space $\mathbb{L}^{k+1}$ with ${\langle}e_0, e_0{\rangle}=0={\langle}e_3, e_3{\rangle}$ and ${\langle}e_0, e_3{\rangle}=-1/2$.

• • $\Phi\colon \mathbb{S}^{3}\times \mathbb{H}^{n-2}\subset \mathbb{R}^{4}\times \mathbb{L}^{n-1}\to \mathbb{R}^{n+1}\setminus \mathbb{R}^{n-3}$, $\Phi(x, y)=\frac{1}{y_0}(x, y_0, \ldots, y_{n-3})$ for all $x=(x_1, \ldots, x_4)\in \mathbb{S}^3\subset \mathbb{R}^4$ and $y=y_0e_0+\ldots y_{n-3}e_{n-2}\in \mathbb{H}^{n-2}\subset\mathbb{L}^{n-1}$, where $\{e_0, \ldots, e_{n-2}\}$ is a pseudo-orthonormal basis of $\mathbb{L}^{n-1}$ with ${\langle}e_0, e_0{\rangle}=0={\langle}e_{n-2}, e_{n-2}{\rangle}$ and ${\langle}e_0, e_{n-2}{\rangle}=~-1/2$.

Case (i): Let $\mathcal{W}$ be a connected component of $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_2$. Since, in particular, $\mathcal{W}\subset \mathcal{W}_2$, then $C(\Gamma(\Delta))\subset \mbox{span}\{I\}$ on $\mathcal{W}$. By Lemma 16, $f|_{\mathcal{W}}$ is a conformally surface-like hypersurface determined by a surface $h\colon L^2\to \mathbb{Q}_\epsilon$, $\epsilon \in \{-1,0,1\}$. Thus, with notations as in the preceding paragraph, we may write $\mathcal{W}= L^2\times N^{n-2}$ and $f|_{\mathcal{W}}=\mathcal{I}\circ\Phi\circ (h\times i)$. In particular, the distributions Δ and ${\Delta}^\perp$ are given by the tangent spaces to $N^{n-2}$ and L ², respectively.

Denote by g ₁ the product metric of $\mathbb{Q}_\epsilon^3\times \mathbb{Q}_{-\epsilon}^{n-2}$ and let g ₂ be the metric on $\mathbb{Q}_\epsilon^3\times \mathbb{Q}_{-\epsilon}^{n-2}$ induced from the metric of $\mathbb{R}^{n+1}$ by the conformal diffeomorphism $\mathcal{I}\circ \Phi$. Let $\varphi \in C^{\infty}(\mathbb{Q}_\epsilon^3\times \mathbb{Q}_{-\epsilon}^{n-2})$ be the conformal factor of g ₂ with respect to g ₁, that is, $g_2=\varphi^2 g_1$. Then the shape operators of $F_1=h\times i\colon L^2\times N^{n-2}\to \mathbb{Q}_\epsilon^3\times \mathbb{Q}_{-\epsilon}^{n-2}$ and of $F_2=F_1\colon (L^2\times N^{n-2}, F_1^*g_2)\to (\mathbb{Q}_\epsilon^3\times \mathbb{Q}_{-\epsilon}^{n-2}, g_2)$ with respect to unit normal vector fields N ₁ and $N_2=N_1/\varphi$, respectively, are related by:

(36)\begin{equation} A_{N_2}^{F_2}=\frac{1}{\varphi\circ F_1}A^{F_1}_{N_1}-\frac{g_1(\nabla^1 \varphi, N_1)}{(\varphi\circ F_1)^2} I. \end{equation}

We recall that the Levi-Civita connections $\bar\nabla$ and $\nabla$ of the metrics $\bar g$ and $g={\langle}\,,\,{\rangle}$ on $M^n=L^2\times N^{n-2}$ induced by F ₁ and F ₂, respectively, satisfy:

(37)\begin{equation} \nabla_XY=\bar\nabla_XY+\frac{1}{\bar\varphi}(X(\bar\varphi)Y+Y(\bar\varphi)X-\bar g(X,Y)\bar\nabla\bar\varphi), \end{equation}

where $\bar\varphi=\varphi\circ F_1$. It follows from (37) that the mean curvature vector field $\delta\in\Gamma(\Delta^\perp)$ of Δ (with respect to the metric g) is:

(38)\begin{equation} \delta=-\bar{\varphi}^{-3}(\bar\nabla\bar\varphi)_{\Delta^\perp}. \end{equation}

Now, we can write (11) as:

(39)\begin{equation} (\nabla_X \bar{\mathcal{B}})Y-(\nabla_Y\bar{\mathcal{B}})X+(X\wedge Y)(A\nabla\rho-\nabla b)=0, \end{equation}

for all $X,Y\in\mathfrak{X}(M)$. The Δ-component of (39) evaluated in unit vector fields $Z\in\Gamma(\Delta^\perp)$ and $T\in\Gamma(\Delta)$ gives: $ {\langle}\bar{\mathcal{B}}Z,\nabla_TT{\rangle}={\langle}Z,A\nabla\rho-\nabla b{\rangle}, $ or equivalently,

(40)\begin{equation} \bar{\mathcal{B}}\delta=A\nabla\rho-\nabla b. \end{equation}

Since $\mathcal{W}\subset \mathcal{Y}^0_2$, there exists locally a smooth function θ and an orthonormal frame $\{X, Y\}$ of $\Delta^\perp$ given by principal directions of f such that $\bar{\mathcal{B}} X=\theta Y$ and $\bar{\mathcal{B}} Y=\theta X$. From (37) we have ${\langle}\nabla_XT,X{\rangle}={\langle}\nabla_YT,Y{\rangle}=T(\log\circ\bar{\varphi})$. Evaluating (39) in T and X (or Y) gives:

(41)\begin{equation} T(\theta)=-T(\log\circ\bar{\varphi})\theta, \end{equation}

whereas (39) evaluated in X and Y yields:

(42)\begin{equation} X(\theta)=2\theta{\langle}\nabla_YY,X{\rangle}-{\langle}Y,A\nabla\rho-\nabla b{\rangle}, \end{equation}

and

(43)\begin{equation} Y(\theta)=2\theta{\langle}\nabla_XX,Y{\rangle}-{\langle}X,A\nabla\rho-\nabla b{\rangle}. \end{equation}

Set $\bar{\theta}=\bar{\varphi}\theta$ . It follows from (41) that $T(\bar{\theta})=0$ for any $T\in\Gamma(\Delta)$. Thus, $\bar{\theta}$ induces a function on L ², which we also denote by $\bar \theta$. By (36), the vector fields $\bar{X}=\bar{\varphi}X$ and $\bar{Y}=\bar{\varphi}Y$ form an orthonormal frame of principal directions of h. Using (37), (38), (40) and (42) we obtain:

(44)\begin{align} \bar{X}(\bar{\theta})&=\bar{X}(\bar{\varphi})\theta+\bar{\varphi}^2X(\theta)\nonumber\\ &=\bar{X}(\bar{\varphi})\theta+\bar{\varphi}^2(2\theta{\langle}\nabla_YY,X{\rangle}-{\langle}Y,A\nabla\rho-\nabla b{\rangle})\nonumber\\ &=\bar{X}(\bar{\varphi})\theta+\bar{\varphi}^2(2\theta(\bar{\varphi}^{-1}\bar{g}(\bar{\nabla}_{\bar{Y}}\bar{Y},\bar{X})-\bar{\varphi}^{-2}\bar{X}(\bar{\varphi}))-{\langle}\bar{B}Y,\delta{\rangle})\nonumber\\ &=\bar{X}(\bar{\varphi})\theta+2\bar{\theta}\bar{g}(\bar{\nabla}_{\bar{Y}}\bar{Y},\bar{X})-2\theta\bar{X}(\bar{\varphi})-\bar{\varphi}^2\theta{\langle}X,\delta{\rangle}\nonumber\\ &=\bar{X}(\bar{\varphi})\theta+2\bar{\theta}\bar{g}(\bar{\nabla}_{\bar{Y}}\bar{Y},\bar{X})-2\theta\bar{X}(\bar{\varphi})+\theta\bar{\varphi}^{-1}{\langle}X,\bar{\nabla}\bar{\varphi}{\rangle}\nonumber\\ &=\bar{X}(\bar{\varphi})\theta+2\bar{\theta}\bar{g}(\bar{\nabla}_{\bar{Y}}\bar{Y},\bar{X})-2\theta\bar{X}(\bar{\varphi})+\theta\bar{X}(\bar{\varphi})\nonumber\\ &=2\bar{\theta}\bar{g}(\bar{\nabla}_{\bar{Y}}\bar{Y},\bar{X}). \end{align}

A similar computation using (43) instead of (42) gives:

(45)\begin{equation} \bar{Y}(\bar{\theta})=2\bar{\theta}\bar{g}(\bar{\nabla}_{\bar{X}}\bar{X},\bar{Y}). \end{equation}

Let $\mathcal{B}^*\in\Gamma(\mbox{End}(TL))$ be defined by $\mathcal{B}^* X=\bar{\theta}\bar{Y}$ and $\mathcal{B}^* Y=\bar{\theta}\bar{X}$. Then (44) and (45) are equivalent to $\mathcal{B}^*$ being a Codazzi tensor on L ². By (36), the shape operator A^h of h is a multiple of $\bar{A}|_{\Delta^\perp}$, which has rank two, for $\mathcal{W}\subset \mathcal{V}_1$, and $\mathcal{B}^*$ is a multiple of $\bar{\mathcal{B}}|_{\Delta^\perp}$. Therefore, the fact that $\bar{A}$ and $\bar{\mathcal{B}}$ satisfy (22) implies that $\mathcal{B}^*$ and A^h also satisfy (22) with respect to $\bar{g}$. Since, in addition, $\mbox{tr}\, \mathcal{B}^*=0$, it follows from Proposition 8 of [Reference Jimenez and Tojeiro11], together with Theorem 4.7 in [Reference Dajczer and Jimenez6] (also stated in [Reference Jimenez and Tojeiro11] as Theorem 2), that h is locally infinitesimally Bonnet bendable, hence isothermic by Proposition 9 of [Reference Jimenez and Tojeiro11].

Case (ii): First we show that the interior $\mathcal{V}^0_2$ of the subset $\mathcal{V}_2$ where $\dim \ker \bar{A}=n-1$ is contained in $\mathcal{W}^0_2\cap \mathcal{Y}^0_2$. Clearly, $\mathcal{V}_2\subset \mathcal{U}_1$, the subset where $\mbox{tr}\, \bar{B}=0$, for if $\mbox{tr}\, \bar{A}(x)=0$, that is, if $x\in \mathcal{V}_2\cap \mathcal{U}_2$, then we would have $\bar{A}(x)=0$, in contradiction with the assumption that f is free of umbilic points. Also, since $\dim \ker \bar{A}=n-1$ on $\mathcal{V}_2$, then $\bar{A}|_{\Delta^\perp}$ cannot be a multiple of the identity endomorphism $I\in \Gamma(\mbox{End}(\Delta^\perp))$ at any point of $\mathcal{V}_2$. Thus $\mathcal{V}_2\subset \mathcal{Y}_2$.

We now show that, on any connected component $\mathcal{W}$ of $\mathcal{V}^0_2$, the splitting tensor $C\colon\Gamma(\Delta)\to\Gamma(\mbox{End}(\Delta^\perp))$ of Δ satisfies $C(\Gamma(\Delta))\subset \mbox{span}\{I\}$, which will yield the inclusion $\mathcal{V}^0_2\subset \mathcal{W}^0_2$. So let $\mathcal{W}$ be such a connected component. Since we already know that $\mathcal{W}\subset \mathcal{Y}^0_2$, there exist locally smooth functions $\theta, \mu$ and unit vector fields $Y\in \Gamma(\Delta^\perp\cap \ker \bar{A})$ and $X\in \Gamma((\ker{\bar A})^\perp)$ such that $\bar{A}X=\mu X$, $\bar{\mathcal{B}}X=\theta Y$ and $\bar{\mathcal{B}}Y=\theta X$.

Applying (39) to T and Y gives:

(46)\begin{equation} \begin{array}{l}T(\theta)X+\theta\nabla_TX-\bar{\mathcal{B}}\nabla_TY+\bar{\mathcal{B}}\nabla_YT+{\langle}\lambda\nabla\rho-\nabla b, Y{\rangle}T\\ \hspace{20ex}-{\langle}\lambda\nabla\rho-\nabla b, T{\rangle}Y=0. \end{array} \end{equation}

Taking the X-component of (46) we obtain:

(47)\begin{equation} T(\theta)=\theta{\langle}\nabla_YY, T{\rangle}, \end{equation}

whereas the Y-component and the T-component give, respectively,

(48)\begin{equation} {\langle}\lambda\nabla\rho-\nabla b, T{\rangle}=0, \end{equation}

and

\begin{equation*} {\langle}\lambda\nabla\rho-\nabla b, Y{\rangle}=\theta{\langle}\nabla_TT, X{\rangle}. \end{equation*}

Applying (39) to T and X and using (48) give:

(49)\begin{equation} \begin{array}{l}T(\theta)Y+\theta\nabla_TY-\bar{\mathcal{B}}\nabla_TX +\bar{\mathcal{B}}\nabla_XT+{\langle}\mu\nabla\rho-\nabla b, X{\rangle}T=0. \end{array} \end{equation}

Taking the S-component of (49) for $S\in \Gamma(\Delta)$ with ${\langle}S, T{\rangle}=0$ gives:

\begin{equation*}{\langle}\nabla_T S, Y{\rangle}=0,\end{equation*}

and taking its T-component yields:

\begin{equation*} \theta{\langle}\nabla_TT, Y{\rangle}={\langle}\mu\nabla\rho-\nabla b, X{\rangle}. \end{equation*}

Using that $\ker\bar{A}=\{X\}^\perp$ is an umbilical distribution, it follows that the same holds for Δ.

Taking the X-component of (49) yields:

(50)\begin{equation} {\langle}\nabla_XY, T{\rangle}=0, \end{equation}

whereas the Y-component gives:

(51)\begin{equation} T(\theta)=\theta{\langle}\nabla_XX, T{\rangle}. \end{equation}

It follows from (47), (50) and (51), taking into account that one also has ${\langle}\nabla_YX, T{\rangle}~=~0$, that the distribution $\Delta^\perp$ is umbilical with mean curvature vector field $\zeta=(\nabla \log \theta)|_{\Delta}$, which is equivalent to the splitting tensor $C\colon\Gamma(\Delta)\to\Gamma(\mbox{End}(\Delta^\perp))$ of Δ satisfying $C_T={\langle}\zeta, T{\rangle} I$ for all $T\in\Gamma(\Delta)$.

Now that we know that $\mathcal{V}^0_2\subset \mathcal{W}^0_2\cap \mathcal{Y}^0_2$, the argument used in case (i) shows that $f|_{\mathcal{W}}$ is a conformally surface-like hypersurface determined by an isothermic surface $h\colon L^2\to \mathbb{Q}^3_\epsilon$, $\epsilon \in \{-1,0,1\}$. But since $\mbox{rank } \ker \bar{A}=n-1$ on $\mathcal{V}_2$, then h has index of relative nullity equal to one at any point. By Corollary 12 in [Reference Jimenez and Tojeiro11], h is a generalized cone over a unit-speed curve $\gamma\colon J\to \mathbb{Q}_{c}^2$ in an umbilical surface $\mathbb{Q}_{c}^2\subset \mathbb{Q}_{\epsilon}^3$, $c\geq \epsilon$.

Cases (iii) and (iv): Let $\mathcal{W}$ be a connected component of $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{U}_2^0$ (respectively, $\mathcal{V}_1\cap \mathcal{W}^0_2\cap \mathcal{Y}^0_1$). As in cases (i) and (ii), $f|_{\mathcal{W}}$ is a conformally surface-like hypersurface determined by a surface $h\colon L^2\to \mathbb{Q}_\epsilon$, $\epsilon \in \{-1,0,1\}$, by Lemma 16. Since $\mbox{tr}\, \bar{A}=0$ on $\mathcal{U}_2$ (respectively, $\bar{A}|_{\Delta^\perp}$ is a multiple of the identity endomorphism of $\Delta^\perp$ on $\mathcal{Y}_1$), it follows from (36) that also $\mbox{tr}\, A^h=0$ (respectively, A^h is a multiple of the identity endomorphism of TL), hence h is a minimal surface (respectively, h is umbilical).

We now prove the converse. Assume first that $f\colon M^n\to\mathbb{R}^{n+1}$ is a simply connected hypersurface whose central sphere congruence is determined by a minimal space-like surface $s\colon L^2\to \mathbb{S}_{1,1}^{n+2}$. Let $\bar{J}\in\Gamma(\mbox{End}(TL))$ represent a rotation of angle $\pi/2$, and let $\bar{X},\bar{Y}$ be an orthonormal frame satisfying $\bar{J}\bar{X}=\bar{Y}$ and $\bar{J}\bar{Y}=-\bar{X}$. Then $\bar{J}$ is parallel with respect to the Levi-Civita connection $\nabla'$ on L ², hence it is, in particular, a Codazzi tensor on L ². Since s is minimal, then $ \alpha'(\bar{X},\bar{X})+\alpha'(\bar{Y},\bar{Y})=0, $ hence s is a special elliptic surface by Proposition 11 in [Reference Dajczer, Jimenez and Vlachos7].

By Theorem 1 in [Reference Dajczer, Jimenez and Vlachos7], f admits a non-trivial conformal infinitesimal bending ${\cal T}$. We now show that ${\cal T}$ is also an infinitesimal Moebius bending. Let $X,Y\in\mathfrak{X}(M)$ be the lifts of $\bar{X}$ and $\bar{Y}$. From (35) we see that $\bar{A}X$ and $\bar{A}Y$ form an orthonormal frame of $\Delta^\perp$. Let $J\in\Gamma(\mbox{End}(\Delta^\perp))$ be the lift of $\bar{J}$. It was shown in the proof of the converse of Theorem 1 in [Reference Dajczer, Jimenez and Vlachos7] that $\bar{A}J$ is symmetric. Thus

\begin{equation*}{\langle}J\bar{A}X,\bar{A}X{\rangle}={\langle}\bar{A}J\bar{A}X,X{\rangle}={\langle}\bar{A}X,\bar{A}JX{\rangle}={\langle}\bar{X},\bar{J}\bar{X}{\rangle}'=0.\end{equation*}

Similarly, ${\langle}J\bar{A}Y,\bar{A}Y{\rangle}=0$, ${\langle}J\bar{A}X,\bar{A}Y{\rangle}=-1$ and ${\langle}J\bar{A}Y,\bar{A}X{\rangle}=1$. Hence J is an orthogonal tensor, and the symmetry of $\bar{A}J$ implies that $\mbox{tr}\,\bar{A}=0$. By Proposition 12, ${\cal T}$ is an infinitesimal Moebius bending.

Now let $f\colon M^n\to \mathbb{R}^{n+1}$ be a conformally surface-like hypersurface determined by an isothermic surface $h\colon L^2\to \mathbb{Q}_\epsilon$, $\epsilon \in \{-1,0,1\}$. Then h is locally infinitesimally Bonnet bendable (see, e.g., Proposition 9 and Remark 11 of [Reference Jimenez and Tojeiro11]), that is, it admits locally a non-trivial infinitesimal variation $h_t\colon L^2\to \mathbb{Q}_\epsilon$ such that the metrics $\bar{g}_t$ induced by the ${h_t}'s$ and their mean curvatures $\mathcal{H}_{t}$ satisfy: $ \partial/\partial t|_{t=0}\bar{g}_t=0=\partial/\partial t|_{t=0}\mathcal{H}_t. $

Thus also $\partial/\partial t|_{t=0}K_t=0$, where K_t is the Gauss curvature of $\bar{g}_t$.

Let f_t be the variation of f given by the conformally surface-like hypersurfaces determined by h_t. The Moebius metric of f_t is (see Remark 3.7 in [Reference Li, Ma and Wang13]):

\begin{equation*} \left(4\mathcal{H}_t^2-\frac{2n}{n-1}(K_t-\epsilon)\right)(\bar{g}_t+g_{-\epsilon}), \end{equation*}

where g _−ϵ is the metric of $\mathbb{Q}_{-\epsilon}^{n-2}$ and $\bar{g}_t+g_{-\epsilon}$ denotes the product metric on $L^2\times \mathbb{Q}_{-\epsilon}^{n-2}$. Therefore, the immersions f_t determine an infinitesimal Moebius variation of f. It remains to argue that the latter is non-trivial.

From Proposition 10 we know that the associated tensor $\mathcal{B}$ satisfies (19). On the other hand, by (36) the shape operator A_t of f_t has the form:

(52)\begin{equation} A_t=\delta_{1}(t)\bar{A}_t+\delta_2(t)I, \end{equation}

for some smooth functions δ ₁ and δ ₂, with $\delta_1(0)\neq 0$. Here $\bar{A}_t$ denotes the second fundamental form of h_t extended to TM by defining $\bar{A}_tT=0$ for any T tangent to $\mathbb{Q}_{-\epsilon}^{n-2}$. Since $\partial/\partial t|_{t=0} \bar{A}_t\neq 0$, for h_t determine a non-trivial infinitesimal Bonnet variation of h, it follows from (19) and (52) that $\mathcal{B}$ is not a multiple of the identity endomorphism. Hence the infinitesimal Moebius variation of f determined by f_t is non-trivial (see Remarks 4–2)).