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# Polyharmonic surfaces in $3$-dimensional homogeneous spaces

S. Montaldo  Università degli Studi di Cagliari
Dipartimento di Matematica e Informatica
Via Ospedale 72
09124 Cagliari, Italia<EMAIL_ADDRESS>,  C. Oniciuc  Faculty of
Mathematics
“Al.I. Cuza” University of Iasi
Bd. Carol I no. 11
700506 Iasi, ROMANIA<EMAIL_ADDRESS>and  A. Ratto  Università degli Studi
di Cagliari
Dipartimento di Matematica e Informatica
Via Ospedale 72
09124 Cagliari, Italia<EMAIL_ADDRESS>
###### Abstract.

In the first part of this paper we shall classify proper triharmonic
isoparametric surfaces in $3$-dimensional homogeneous spaces (Bianchi-Cartan-
Vranceanu spaces, shortly BCV-spaces). We shall also prove that triharmonic
Hopf cylinders are necessarily CMC. In the last section we shall determine a
complete classification of CMC $r$-harmonic Hopf cylinders in BCV-spaces,
$r\geq 3$. This result ensures the existence, for suitable values of $r$, of
an ample family of new examples of $r$-harmonic surfaces in BCV-spaces.

###### Key words and phrases:

Triharmonic maps, polyharmonic maps, Bianchi-Cartan-Vranceanu spaces,
$3$-dimensional homogeneous spaces

###### 2010 Mathematics Subject Classification:

Primary: 58E20; Secondary: 53C42, 53C43.

The authors S.M. and A.R. are members of the Italian National Group
G.N.S.A.G.A. of INdAM. The author C.O. was supported by a project funded by
the Ministry of Research and Innovation within Program 1 - Development of the
national RD system, Subprogram 1.2 - Institutional Performance - RDI
excellence funding projects, Contract no. 34PFE/19.10.2018.

##  1\. Introduction

In order to introduce the geometrical setting of this paper we recall that
harmonic maps are the critical points of the energy functional

(1.1) $E(\varphi)=\frac{1}{2}\int_{M}\,|d\varphi|^{2}\,dV\,,$

where $\varphi:M\to N$ is a smooth map between two Riemannian manifolds
$(M,g)$ and $(N,h)$. A map $\varphi$ is harmonic if it is a solution of the
Euler-Lagrange system of equations associated to (1.1), i.e.,

(1.2) $-d^{*}d\varphi={\operatorname{trace}}\,\nabla d\varphi=0\,.$

The left member of (1.2) is a vector field along the map $\varphi$ or,
equivalently, a section of the pull-back bundle $\varphi^{-1}TN$: it is called
tension field and denoted $\tau(\varphi)$. In addition, we recall that if
$\varphi$ is an isometric immersion, then $\varphi$ is a harmonic map if and
only if the immersion $\varphi$ defines a minimal submanifold of $N$ (see [8,
9] for background). Let us denote $\nabla^{M}$, $\nabla^{N}$ and
$\nabla^{\varphi}$ the induced connections on the bundles $TM$, $TN$ and
$\varphi^{-1}TN$ respectively. The rough Laplacian on sections of
$\varphi^{-1}TN$, denoted $\overline{\Delta}$, is defined by

(1.3)
$\overline{\Delta}=d^{*}d=-\sum_{i=1}^{m}\left(\nabla^{\varphi}_{e_{i}}\nabla^{\varphi}_{e_{i}}-\nabla^{\varphi}_{\nabla^{M}_{e_{i}}e_{i}}\right)\,,$

where $\\{e_{i}\\}_{i=1}^{m}$ is a local orthonormal frame field tangent to
$M$.

Now, in order to define the notion of an $r$-harmonic map, we consider the
following family of functionals which represent a version of order $r$ of the
classical energy (1.1).

If $r=2s$, $s\geq 1$:

(1.4) $\displaystyle E_{2s}(\varphi)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{M}\,\langle\,\underbrace{(d^{*}d)\ldots(d^{*}d)}_{s\,{\rm
times}}\varphi,\,\underbrace{(d^{*}d)\ldots(d^{*}d)}_{s\,{\rm
times}}\varphi\,\rangle_{{}_{N}}\,\,dV$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{M}\,\langle\,\overline{\Delta}^{s-1}\tau(\varphi),\,\overline{\Delta}^{s-1}\tau(\varphi)\,\rangle_{{}_{N}}\,\,dV\,.$

In the case that $r=2s+1$, $s\geq 0$:

(1.5) $\displaystyle E_{2s+1}(\varphi)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{M}\,\langle\,d\underbrace{(d^{*}d)\ldots(d^{*}d)}_{s\,{\rm
times}}\varphi,\,d\underbrace{(d^{*}d)\ldots(d^{*}d)}_{s\,{\rm
times}}\varphi\,\rangle_{{}_{N}}\,\,dV$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{M}\,\sum_{j=1}^{m}\langle\,\nabla^{\varphi}_{e_{j}}\,\overline{\Delta}^{s-1}\tau(\varphi),\,\nabla^{\varphi}_{e_{j}}\,\overline{\Delta}^{s-1}\tau(\varphi)\,\rangle_{{}_{N}}\,\,dV\,.$

We say that a map $\varphi$ is $r$-harmonic if, for all variations
$\varphi_{t}$,

$\left.\frac{d}{dt}\,E_{r}(\varphi_{t})\,\right|_{t=0}\,=\,0\,\,.$

This condition is equivalent to the vanishing of the $r$-tension field
$\tau_{r}(\varphi)$. We recall that the expressions which describe the
$r$-tension field of a general map $\varphi:M\to N$ between two Riemannian
manifolds were computed by Maeta (see [10]) and are the following:

$\displaystyle\tau_{2s}(\varphi)$ $\displaystyle=$
$\displaystyle\overline{\Delta}^{2s-1}\tau(\varphi)-R^{N}\left(\overline{\Delta}^{2s-2}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})$
$\displaystyle-\sum_{\ell=1}^{s-1}\,\left\\{R^{N}\left(\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s+\ell-2}\tau(\varphi),\overline{\Delta}^{s-\ell-1}\tau(\varphi)\right)d\varphi(e_{i})\right.$
$\displaystyle\qquad\qquad-\,\left.R^{N}\left(\overline{\Delta}^{s+\ell-2}\tau(\varphi),\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s-\ell-1}\tau(\varphi)\right)d\varphi(e_{i})\right\\}\,\,,$

where $\overline{\Delta}^{-1}=0$ and $\\{e_{i}\\}_{i=1}^{m}$ is a local
orthonormal frame field tangent to $M$ (the sum over $i$ is not written but
understood). Similarly,

$\displaystyle\tau_{2s+1}(\varphi)$ $\displaystyle=$
$\displaystyle\overline{\Delta}^{2s}\tau(\varphi)-R^{N}\left(\overline{\Delta}^{2s-1}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})$
$\displaystyle-\sum_{\ell=1}^{s-1}\,\left\\{R^{N}\left(\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s+\ell-1}\tau(\varphi),\overline{\Delta}^{s-\ell-1}\tau(\varphi)\right)d\varphi(e_{i})\right.$
$\displaystyle\qquad\qquad-\,\left.R^{N}\left(\overline{\Delta}^{s+\ell-1}\tau(\varphi),\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s-\ell-1}\tau(\varphi)\right)d\varphi(e_{i})\right\\}$
$\displaystyle\,-\,R^{N}\Big{(}\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s-1}\tau(\varphi),\overline{\Delta}^{s-1}\tau(\varphi)\Big{)}d\varphi(e_{i})\,\,.$

If $r=1$, the functional (1.5) is just the energy. In the case that $r=2$, the
functional (1.4) is called bienergy and its critical points are the so-called
biharmonic maps. At present, a very ample literature on biharmonic maps is
available and, again, we refer to [17] and references therein for an
introduction to this topic.

More generally, the $r$-energy functionals $E_{r}(\varphi)$ defined in (1.4),
(1.5) have been intensively studied (see [2, 10, 11, 12, 13, 14, 15], for
instance). Inspection of the Euler-Lagrange equations for $E_{r}(\varphi)$
shows that a harmonic map is always $r$-harmonic for any $r\geq 2$. When the
target manifold is nonflat, we use to call an $r$-harmonic map proper if it is
not harmonic (similarly, an $r$-harmonic submanifold, i.e., an $r$-harmonic
isometric immersion, is proper if it is not minimal). As a general fact, when
the ambient space has nonpositive sectional curvature there are several
results which assert that, under suitable conditions, an $r$-harmonic
submanifold is minimal (see [11] and [12], for instance).

Things drastically change when the ambient space is positively curved. Let us
denote by ${\mathbb{S}}^{m+1}$ the sphere ${\mathbb{S}}^{m+1}(1)$ of radius
$1$. Moreover, let $A$ be the shape operator of $M^{m}$ into
${\mathbb{S}}^{m+1}$ and ${\mathbf{H}}=f\eta$ the mean curvature vector field,
where $\eta$ is the unit normal vector field and $f$ is the mean curvature
function. Throughout the whole paper, when we write that $M^{m}$ is a CMC
hypersurface we mean that $f$ is a constant which will be denoted by $\alpha$.

In [16] Ou derived the equation for biharmonic hypersurfaces in a generic
Riemannian manifold. More precisely, he proved:

######  Theorem 1.1.

[16] Let $\varphi:M^{m}\to N^{m+1}$ be an isometric immersion of codimension-
one with mean curvature vector ${\mathbf{H}}=f\eta$. Then $\varphi$ is
biharmonic if and only if:

(1.8) $\begin{cases}\Delta f+f|A|^{2}-f{\rm Ric}^{N}(\eta,\eta)=0,\\\
2A\,({\rm grad}\,f)+mf{\rm grad}\,f-2\,f\,({\rm
Ric}^{N}\,(\eta))^{\top}=0,\end{cases}$

where ${\rm Ric}^{N}:T_{q}N\longrightarrow T_{q}N$ denotes the Ricci operator
of the ambient space defined by $\langle{\rm Ric}^{N}\,(Z),W\rangle={\rm
Ric}^{N}(Z,W)$ and $A$ is the shape operator of the hypersurface with respect
to the unit normal vector $\eta$.

We point out that, contrary to [16], the sign convention for $\Delta$ in this
paper is such that $\Delta f=-f^{\prime\prime}$ on ${\mathbb{R}}$. If the mean
curvature $f$ is constant, say $f\equiv\alpha$, then the biharmonic equation
reduces to

(1.9) $\begin{cases}-\alpha|A|^{2}+\alpha{\rm Ric}^{N}(\eta,\eta)=0\\\
\,\alpha\,({\rm Ric}^{N}\,(\eta))^{\top}=0\,,\end{cases}$

from which we deduce that a non minimal CMC hypersurface $M^{m}$ is proper
biharmonic if and only if

(1.10) ${\rm Ric}^{N}(\eta)=|A|^{2}\eta\,.$

In the instance that $M^{m}$ is a hypersurface of ${\mathbb{S}}^{m+1}$ the
biharmonic condition (1.10) reduces to

(1.11) $|A|^{2}-m=0\,.$

As for the $r$-harmonic case, condition (1.11) was generalized in [13]:

######  Theorem 1.2.

Let $M^{m}$ be a non-minimal CMC hypersurface in ${\mathbb{S}}^{m+1}$ and
assume that $|A|^{2}$ is constant. Then $M^{m}$ is proper $r$-harmonic ($r\geq
3$) if and only if

(1.12) $|A|^{4}-m|A|^{2}-(r-2)m^{2}\alpha^{2}=0\,.$

As an application of Theorem 1.2, several new examples of isoparametric
$r$-harmonic hypersurfaces were illustrated in [13], where it was stressed
that the value of $r$, $r\geq 2$, plays a crucial role when the ambient is
positively curved. By contrast, when the target space form has nonpositive
curvature, generally non-existence results are confirmed for all values of
$r$, $r\geq 2$.

As a natural further step, in this paper we shall focus on the study of
$r$-harmonic surfaces into $3$-dimensional homogeneous spaces with group of
isometries of dimension $4$.

It is well-known (see, e.g., [1], [3], [6]) that $3$-dimensional homogeneous
spaces with group of isometries of dimension $4$ admit, as a canonical model,
the so called Bianchi-Cartan-Vranceanu spaces (shortly, BCV-spaces)

(1.13)
$M^{3}_{m,\ell}=\left(\bar{M}\times{\mathbb{R}},g=\frac{dx^{2}+dy^{2}}{[1+m(x^{2}+y^{2})]^{2}}+\Big{[}dz+\frac{\ell}{2}\,\frac{ydx-
xdy}{1+m(x^{2}+y^{2})}\Big{]}^{2}\right)\,,$

where $\bar{M}=\\{(x,y)\in{\mathbb{R}}^{2}\colon 1+m(x^{2}+y^{2})>0\\}$.

The space $M^{3}_{m,\ell}$ is the total space of the following Riemannian
submersion over a simply connected complete surface $M^{2}(4m)$ of constant
curvature $4m$, see [6]:

(1.14) $\displaystyle\pi:M^{3}_{m,\ell}\longrightarrow
M^{2}(4m)=\left(\bar{M},h=\frac{dx^{2}+dy^{2}}{[1+m(x^{2}+y^{2})]^{2}}\right),\;\;\;\pi(x,y,z)=(x,y).$

We point out that while in ${\mathbb{S}}^{n}(\rho)$ the letter $\rho$
indicates the radius, in $M^{2}(4m),\,{\mathbb{H}}^{2}(4m)$ the real number
within the brackets represents the sectional curvature.

These BCV-spaces are also a model for Thurston’s eight $3$-dimensional
geometries with the exception of the hyperbolic space ${\mathbb{H}}^{3}$ and
$\rm Sol$. More precisely, they include the $3$-dimensional space forms
${\mathbb{R}}^{3}$ ($\ell=m=0$), ${\mathbb{S}}^{3}(1/\sqrt{m})$
($\ell^{2}=4m$), the product spaces $M^{2}(4m)\times{\mathbb{R}}$ ($\ell=0$),
${\rm Nil_{3}}$ ($m=0$), $\widetilde{SL}(2,{\mathbb{R}})$ ($\ell\neq 0,\,m<0$)
and $SU(2)$ ($\ell\neq 0,\,m>0$, $\ell^{2}\neq 4m$). See Figure 1 for a
representation of the BCV-spaces with respect to the values of the parameters
$\ell$ and $m$.

Figure 1. Distribution of BCV-spaces w.r.t. to the values of $\ell$ and
$m$.$\ell$$m$${\rm Nil_{3}}$${\rm
Nil_{3}}$${\mathbb{S}}^{3}(\frac{1}{\sqrt{m}})$${\mathbb{S}}^{2}(\frac{1}{2\sqrt{m}})\times{\mathbb{R}}$${\mathbb{H}}^{2}(4m)\times{\mathbb{R}}$$SU(2)$$SU(2)$$SU(2)$$SU(2)$${\mathbb{R}}^{3}$$\widetilde{SL}(2,{\mathbb{R}})$$\widetilde{SL}(2,{\mathbb{R}})$

In his paper [18], Ou used equation (1.8) to study biharmonic surfaces in BCV-
spaces. He first showed that a totally umbilical biharmonic surface in any
$3$-dimensional Riemannian manifold has constant mean curvature. Then he used
this to show that the only totally umbilical proper biharmonic surface in
$3$-dimensional geometries is a part of ${\mathbb{S}}^{2}(1/\sqrt{2m})$ in
${\mathbb{S}}^{3}(1/\sqrt{m})$.

Moreover, he proved the following characterization of CMC biharmonic surfaces:

######  Theorem 1.3.

(see [18]) A CMC surface in a $3$-dimensional Bianchi-Cartan-Vranceanu space
is proper biharmonic if and only if it is a part of one of the following:

  1. (i)

${\mathbb{S}}^{2}(\frac{1}{\sqrt{2m}})$ in
${\mathbb{S}}^{3}(\frac{1}{\sqrt{m}})$,

  2. (ii)

${\mathbb{S}}^{1}(\frac{1}{2\sqrt{2m}})\times\mathbb{R}$ in
${\mathbb{S}}^{2}(\frac{1}{2\sqrt{2m}})\times\mathbb{R}$,

  3. (iii)

a Hopf cylinder in $SU(2)$ with $4m-\ell^{2}>0$ over a circle of radius
$R=1/\sqrt{8m-\ell^{2}\>}$ in the base sphere
$M^{2}(4m)={\mathbb{S}}^{2}(\frac{1}{2\sqrt{m}})$.

The main aim of our paper is to investigate the existence of triharmonic and,
more generally, $r$-harmonic surfaces in this geometric setting $(r\geq 3)$.

Our paper is organised as follows. In Section 2 we state our main results on
triharmonic surfaces in BCV-spaces. These results will be proved in Sections 4
and 5. Finally, in Section 6, we shall determine a complete classification of
proper CMC $r$-harmonic Hopf cylinders in Bianchi-Cartan-Vranceanu spaces,
$r\geq 3$. As an application, we shall be able to describe, for suitable
values of $r$, an ample family of new examples of $r$-harmonic surfaces in
BCV-spaces.

##  2\. Statement of the results on triharmonic surfaces in BCV-spaces

In order to state our results, it is convenient to recall first some basic
facts and terminology.

For a Bianchi-Cartan-Vranceanu $3$-space given in (1.13), one can easily check
that the vector fields

$E_{1}=F\frac{\partial}{\partial x}-\frac{\ell y}{2}\frac{\partial}{\partial
z},\quad E_{2}=F\frac{\partial}{\partial y}+\frac{\ell
x}{2}\frac{\partial}{\partial z},\quad E_{3}=\frac{\partial}{\partial z},$

where $F=1+m(x^{2}+y^{2})$, form a global orthonormal frame field (see [3,
18]).

Now, let $\gamma(s)=(x(s),y(s)),\,s\in I$ be a smooth curve in the base space
$M^{2}(4m)$ of the Riemannian submersion (1.14). Then the Hopf cylinder
$\Sigma_{\gamma}=\Sigma$ over the curve $\gamma$ is defined as

(2.1) $\Sigma=\cup_{s\in I}\pi^{-1}(\gamma(s))\,.$

Then the surface $\Sigma$ can be parametrized as $r(s,t)=(x(s),y(s),t)$ since
the fiber of $\pi$ over a point $(x_{0},y_{0})$ is
$\pi^{-1}(x_{0},y_{0})=\\{(x_{0},y_{0},t):t\in{\mathbb{R}}\\}$.

It is convenient to assume that the base curve $\gamma$ is parametrized by arc
length, i.e.,

$\frac{\dot{x}^{2}+\dot{y}^{2}}{F^{2}}=1\,.$

Next, we define:

(2.2)
$X=\displaystyle{\frac{\dot{x}}{F}E_{1}+\frac{\dot{y}}{F}E_{2}}\,,\quad\eta=\displaystyle{\frac{\dot{y}}{F}E_{1}-\frac{\dot{x}}{F}E_{2}}\,.$

Then the unit vector field $\eta$ is normal to the Hopf cylinder $\Sigma$ and
$\\{X,E_{3},\eta\\}$ is a global orthonormal frame field adapted to $\Sigma$.

Our first result is:

######  Theorem 2.1.

Let $\Sigma$ be a triharmonic Hopf cylinder in a BCV-space $M^{3}_{m,\ell}$.
Then $\Sigma$ is CMC.

The analysis in the proof of Theorem 2.1 shows that, if $\Sigma$ is a Hopf
cylinder, then its tension field is $\tau=-\kappa_{g}\eta$, where $\kappa_{g}$
denotes the geodesic curvature of its base curve.

Our second result is:

######  Theorem 2.2.

Let $\Sigma$ be a CMC Hopf cylinder in a BCV-space $M^{3}_{m,\ell}$.

  * (i)

If $4m\leq\ell^{2}$ and $\Sigma$ is triharmonic, then $\Sigma$ is minimal.

  * (ii)

If $4m>\ell^{2}$ and the geodesic curvature $\kappa_{g}$ of its base curve
verifies

(2.3) $\kappa_{g}^{2}=2(4m-\ell^{2})\,,$

then $\Sigma$ is proper triharmonic.

######  Remark 2.3.

The base curve of a proper triharmonic Hopf cylinder in $SU(2)$ as in Theorem
2.2 (ii) is a circle of radius $R=1/\sqrt{12m-2\ell^{2}}$ in
${\mathbb{S}}^{2}(1/(2\sqrt{m}))$.

The analysis of the Hopf cylinders fits naturally into the context of the
study of isoparametric surfaces. We recall that, in a general Riemannian
manifold, a hypersurface is said to be isoparametric if itself and its locally
defined nearby equidistant hypersurfaces have constant mean curvature. In the
$30$’s, Cartan characterized isoparametric hypersurfaces in space forms as
those with constant principal curvatures and achieved their classification in
hyperbolic spaces ${\mathbb{H}}^{n}$. Segre obtained a similar result for
Euclidean spaces ${\mathbb{R}}^{n}$. In both cases, isoparametric
hypersurfaces are also open parts of extrinsically homogeneous hypersurfaces,
that is, codimension one orbits of isometric actions on the ambient space. By
contrast, the classification problem in spheres ${\mathbb{S}}^{n}$ is much
more complicated and rich, and there are inhomogeneous examples (see [5] and
references therein, for instance).

In spaces of nonconstant curvature, very few classification results are known.
In the case of interest for us we have the following important result:

######  Theorem 2.4.

[7] Let $\Sigma$ be an immersed surface in $M^{3}_{m,\ell}$, $4m-\ell^{2}\neq
0$. Then the following assertions are equivalent:

  * (i)

$\Sigma$ is an open subset of a homogeneous surface.

  * (ii)

$\Sigma$ is isoparametric.

  * (iii)

$\Sigma$ has constant principal curvatures.

  * (iv)

$\Sigma$ is an open subset of one of the following complete surfaces:

(a) a Hopf cylinder over a complete curve of constant curvature in
$M^{2}(4m)$;

(b) a horizontal slice $M^{2}(4m)\times t_{0}$ with $\ell=0$;

(c) a parabolic helicoid $P_{\alpha,m,\ell}$ with $\alpha^{2}+m<0$.

######  Remark 2.5.

We point out that in references [4, 6, 7] the authors use parameters
$\kappa,\tau$ instead of $m,\ell$. The relationship between these parameters
is $\kappa=4m,\,\tau=\ell/2$. Also, in these papers our constant $\alpha$ is
denoted by $H$.

The parabolic helicoids $P_{\alpha,m,\ell}$ will be described explicitly in
Section 4.

Our main result in the context of isoparametric surfaces is the following:

######  Theorem 2.6.

Let $\Sigma$ be an isoparametric immersed surface in $M^{3}_{m,\ell}$,
$4m-\ell^{2}\neq 0$. If $\Sigma$ is proper triharmonic, then it is an open
part of a Hopf cylinder as in (ii) of Theorem 2.2.

######  Remark 2.7.

If $4m-\ell^{2}=0$, then $M^{3}_{m,\ell}$ is a space form with nonnegative
sectional curvature and in this case the only proper triharmonic isoparametric
surface is ${\mathbb{S}}^{2}(\frac{1}{\sqrt{3m}})$ in
${\mathbb{S}}^{3}(\frac{1}{\sqrt{m}})$ (see [11]).

##  3\. Preliminaries

In order to prepare the ground for our proofs we need to carry out some
preliminary work. Generally, the use of a bar over a symbol indicates that we
refer to an object of the ambient space. We adopt the following notation and
sign convention for the Riemannian curvature tensor field:

$\overline{R}(X,Y)Z=\overline{\nabla}_{X}\overline{\nabla}_{Y}Z-\overline{\nabla}_{Y}\overline{\nabla}_{X}Z-\overline{\nabla}_{[X,Y]}Z\,.$

Moreover,

$\begin{array}[]{lll}&&\overline{R}(X,Y,Z,W)=\langle\overline{R}(X,Y)W,Z\rangle,\\\
&&\\\
&&{\rm\overline{Ric}}(X,Y)=\sum_{i=1}^{3}\overline{R}(X,e_{i},Y,e_{i})=\sum_{i=1}^{3}\langle\overline{R}(X,e_{i})e_{i},Y\rangle.\end{array}$

A straightforward computation shows that

(3.1) $[E_{1},E_{2}]=-[E_{2},E_{1}]=2mxE_{2}-2myE_{1}+\ell E_{3}\,,\;\;{\rm
all\;\;others}\;\;[E_{i},E_{j}]=0,\;\;i,j=1,2,3.$

Then, using the Koszul formula

$\displaystyle 2\langle Z,\overline{\nabla}_{Y}X\rangle$ $\displaystyle=$
$\displaystyle X\langle Y,Z\rangle+Y\langle Z,X\rangle-Z\langle X,Y\rangle$
$\displaystyle-\langle[X,Z],Y\rangle-\langle[Y,Z],X\rangle-\langle[X,Y],Z\rangle\,,$

it is easy to compute:

(3.3)
$\begin{array}[]{ll}\overline{\nabla}_{E_{1}}E_{1}=2myE_{2}\,,&\overline{\nabla}_{E_{2}}E_{2}=2mxE_{1}\,,\vspace{2mm}\\\
\overline{\nabla}_{E_{1}}E_{2}=-2myE_{1}+\dfrac{\ell}{2}E_{3}\,,&\overline{\nabla}_{E_{2}}E_{1}=-2mxE_{2}-\dfrac{\ell}{2}E_{3}\,,\vspace{2mm}\\\
\overline{\nabla}_{E_{3}}E_{1}=\overline{\nabla}_{E_{1}}E_{3}=-\dfrac{\ell}{2}E_{2}\,,&\overline{\nabla}_{E_{3}}E_{2}=\overline{\nabla}_{E_{2}}E_{3}=\dfrac{\ell}{2}E_{1}\,,\vspace{2mm}\\\
{\rm
all\;\;others\;\;}\overline{\nabla}_{E_{i}}E_{j}=0,\;i,j=1,2,3.&\end{array}$

Similarly, a further computation gives the possible nonzero values of the
sectional curvatures:

(3.4)
$\begin{split}\overline{R}_{1212}&=\langle\overline{R}(E_{1},E_{2})E_{2},E_{1}\rangle=4m-\frac{3\ell^{2}}{4},\\\
\overline{R}_{1313}&=\langle\overline{R}(E_{1},E_{3})E_{3},E_{1}\rangle=\frac{\ell^{2}}{4},\\\
\overline{R}_{2323}&=\langle\overline{R}(E_{2},E_{3})E_{3},E_{2}\rangle=\frac{\ell^{2}}{4}\,.\end{split}$

The Riemannian curvature tensor field $\overline{R}$ of $M^{3}_{m,\ell}$ can
be described as follows (see [6], where the opposite sign convention for the
curvature tensor is adopted).

(3.5) $\displaystyle\overline{R}(X,Y)Z$ $\displaystyle=$
$\displaystyle\left(4m-\frac{3\ell^{2}}{4}\right)\big{(}-\langle X,Z\rangle
Y+\langle Y,Z\rangle X\big{)}-(4m-\ell^{2})$ $\displaystyle\big{(}\langle
Y,E_{3}\rangle\langle Z,E_{3}\rangle X+\langle Y,Z\rangle\langle
X,E_{3}\rangle E_{3}-\langle X,E_{3}\rangle\langle Z,E_{3}\rangle Y-\langle
X,Z\rangle\langle Y,E_{3}\rangle E_{3}\big{)}\,.$

Another useful formula is the following (see [6]):

(3.6) $\overline{\nabla}_{X}\,E_{3}=\frac{\ell}{2}\,X\times E_{3}\,,$

where $\times$ here has the following meaning:

$\langle X\times Y,Z\rangle=\det{}_{(E_{1},E_{2},E_{3})}(X,Y,Z)\,.$

We shall study oriented immersed surfaces $\varphi:M^{2}\to M^{3}_{m,\ell}$
and denote by $\eta$ the unit normal vector field.

The vector field $E_{3}^{\top}=E_{3}-\nu\,\eta$, where we have set

(3.7) $\nu=\langle E_{3},\eta\rangle\,,$

will play a basic role in our analysis. This vector field plays an important
part also in the previous literature on this subject. For our purposes, it is
useful to recall (see Proposition 3.3 of [6]):

(3.8)
$\overline{\nabla}_{X}\,E_{3}^{\top}=\nabla_{X}\,E_{3}^{\top}+B(X,E_{3}^{\top})=\nu\,\left(A(X)-\frac{\ell}{2}\,J(X)\right)+\langle
A(X),E_{3}^{\top}\rangle\,\eta\,,$

where $J$ denotes the $\pi/2$ rotation on $TM^{2}$.

If $p$ is an arbitrarily fixed point of $M^{2}$, then, as $\nabla J=0$, we can
consider a geodesic frame field $\\{X_{1},X_{2}\\}$ such that, in a small
neighbourhood of $p$, $J(X_{1})=X_{2}$, $J(X_{2})=-X_{1}$.

Moreover, taking into account the definition (3.7), we compute:

(3.9) $X(\nu)=-\langle A(X)-\frac{\ell}{2}\,J(X),E_{3}\rangle=-\langle
A(X)-\frac{\ell}{2}\,J(X),E_{3}^{\top}\rangle\,.$

##  4\. Proof of Theorems 2.1 and 2.2

Let $\\{T,N\\}$ denote the canonically oriented unit tangent and normal fields
to $\gamma$ in the base space $M^{2}(4m)=(\bar{M},h)$, i.e.,

$T=(\dot{x},\dot{y})\,,\quad N=(-\dot{y},\dot{x})\,,$

so that the geodesic curvature $\kappa_{g}$ of $\gamma$ is defined in the base
space by means of

$\nabla_{T}T=\kappa_{g}N\,,\quad\nabla_{T}N=-\kappa_{g}T\,.$

Note that we denote $\overline{\nabla}=\nabla^{g},\,\nabla=\nabla^{\Sigma}$. A
computation shows that

$\kappa_{g}=\frac{2m}{F}(\dot{x}y-\dot{y}x)+\frac{\ddot{y}\dot{x}-\ddot{x}\dot{y}}{F^{2}}\,.$

Our first lemma is:

######  Lemma 4.1.

Let $X,\eta$ be the vector fields defined in (2.2). Then

(4.1)
$\begin{array}[]{llllll}{\rm(i)}&\overline{\nabla}_{X}X=-\kappa_{g}\eta&{\rm(ii)}&\overline{\nabla}_{X}\eta=\kappa_{g}X-\dfrac{\ell}{2}E_{3}&{\rm(iii)}&\overline{\nabla}_{E_{3}}X=\dfrac{\ell}{2}\eta\vspace{2mm}\\\
{\rm(iv)}&\overline{\nabla}_{E_{3}}\eta=-\dfrac{\ell}{2}X&{\rm(v)}&\overline{\nabla}_{E_{3}}E_{3}=0&{\rm(vi)}&\overline{\nabla}_{X}E_{3}=\dfrac{\ell}{2}\eta\end{array}$

###### Proof.

We note that $X=T^{\mathcal{H}}$, i.e., $X$ is the horizontal lift of $T$ and,
similarly, $\eta=-N^{\mathcal{H}}$. Because $\pi$ is a Riemannian submersion,
we know that

(4.2)
$d\pi\left(\overline{\nabla}_{V^{\mathcal{H}}}W^{\mathcal{H}}\right)=\nabla_{d\pi(V^{\mathcal{H}})}d\pi(W^{\mathcal{H}})\,.$

By way of example, we prove (ii):

$d\pi\left(\overline{\nabla}_{X}\eta\right)=-\nabla_{T}N=\kappa_{g}T=d\pi(\kappa_{g}X)\,.$

It follows that

$\overline{\nabla}_{X}\eta=\kappa_{g}X+cE_{3}\,,$

where

$c=\langle\overline{\nabla}_{X}\eta,E_{3}\rangle=-\langle\eta,\overline{\nabla}_{X}E_{3}\rangle\,.$

Next, using (3.6), we deduce that

$c=-\frac{\ell}{2}\langle\eta,X\times E_{3}\rangle=-\frac{\ell}{2}\,.$

The other computations of this lemma are similar and so we omit them. ∎

It is easy to deduce from Lemma 4.1 that the tension field of the Hopf
cylinder is

$\tau=-\kappa_{g}\eta\,.$

In particular, the Hopf cylinder $\Sigma$ is CMC ($\alpha=-\kappa_{g}/2$) if
and only if its base curve $\gamma$ has constant geodesic curvature. Next, we
need:

######  Lemma 4.2.

(4.3) $\overline{\Delta}\tau=AX+BE_{3}+C\eta\,,$

where

$A=3\kappa_{g}\dot{\kappa}_{g}\,,\quad B=-\ell\dot{\kappa}_{g}\,,\quad
C=\ddot{\kappa}_{g}-\frac{\ell^{2}}{2}\kappa_{g}-\kappa_{g}^{3}\,.$

###### Proof.

We compute:

$\overline{\Delta}\tau=-\Big{\\{}\overline{\nabla}_{X}\overline{\nabla}_{X}\tau-\overline{\nabla}_{\nabla_{X}X}\tau+\overline{\nabla}_{E_{3}}\overline{\nabla}_{E_{3}}\tau-\overline{\nabla}_{\nabla_{E_{3}}E_{3}}\tau\Big{\\}}\,.$

Since $\tau$ is orthogonal to $\Sigma$, we deduce from Lemma 4.1(i) that
$\nabla_{X}X=0$. Also, $\nabla_{E_{3}}E_{3}=0$ and so, using again Lemma 4.1,
we compute:

$\displaystyle\overline{\Delta}\tau$ $\displaystyle=$
$\displaystyle-\Big{\\{}\overline{\nabla}_{X}\overline{\nabla}_{X}\tau+\overline{\nabla}_{E_{3}}\overline{\nabla}_{E_{3}}\tau\Big{\\}}$
$\displaystyle=$
$\displaystyle\overline{\nabla}_{X}\big{[}\dot{\kappa}_{g}\eta+\kappa_{g}(\kappa_{g}X-\frac{\ell}{2}E_{3})\big{]}-\overline{\nabla}_{E_{3}}\big{[}\kappa_{g}\frac{\ell}{2}X\big{]}$
$\displaystyle=$
$\displaystyle\ddot{\kappa}_{g}\eta+\dot{\kappa}_{g}(\kappa_{g}X-\frac{\ell}{2}E_{3})+2\kappa_{g}\dot{\kappa}_{g}X-\kappa_{g}^{3}\eta$
$\displaystyle-\frac{\ell}{2}\dot{\kappa}_{g}E_{3}-\frac{\ell^{2}}{4}\kappa_{g}\eta-\frac{\ell^{2}}{4}\kappa_{g}\eta$

and the conclusion follows readily. ∎

Now, we compute:

######  Lemma 4.3.

Let $A,B,C$ be the function defined in Lemma 4.1. Then

$\displaystyle\overline{\Delta}^{2}\tau$ $\displaystyle=$
$\displaystyle\Big{[}\frac{\ell^{2}}{4}A+A\kappa_{g}^{2}-\ddot{A}-\frac{\ell}{2}B\kappa_{g}-2\dot{C}\kappa_{g}-C\dot{\kappa}_{g}\Big{]}X$
$\displaystyle+\Big{[}-\frac{\ell}{2}A\kappa_{g}+\frac{\ell^{2}}{4}B-\ddot{B}+\ell\dot{C}\Big{]}E_{3}$
$\displaystyle+\Big{[}2\dot{A}\kappa_{g}+A\dot{\kappa}_{g}-\ell\dot{B}-\ddot{C}+C\kappa_{g}^{2}+\frac{\ell^{2}}{2}C\Big{]}\eta\,.$

###### Proof.

Obviously,

$\overline{\Delta}^{2}\tau=\overline{\Delta}(AX)+\overline{\Delta}(BE_{3})+\overline{\Delta}(C\eta)\,.$

Computing as in Lemma 4.1 we find:

$\displaystyle\overline{\Delta}(AX)$ $\displaystyle=$
$\displaystyle\Big{[}\frac{\ell^{2}}{4}A+A\kappa_{g}^{2}-\ddot{A}\Big{]}X-\Big{[}\frac{\ell}{2}A\kappa_{g}\Big{]}E_{3}+\Big{[}2\dot{A}\kappa_{g}+A\dot{k}_{g}\Big{]}\eta$
$\displaystyle\overline{\Delta}(BE_{3})$ $\displaystyle=$
$\displaystyle-\Big{[}\frac{\ell}{2}B\kappa_{g}\Big{]}X+\Big{[}\frac{\ell^{2}}{4}B-\ddot{B}\Big{]}E_{3}-\Big{[}\ell\dot{B}\Big{]}\eta$
$\displaystyle\overline{\Delta}(C\eta)$ $\displaystyle=$
$\displaystyle-\Big{[}2\dot{C}\kappa_{g}+C\dot{\kappa}_{g}\Big{]}X+\Big{[}\ell\dot{C}\Big{]}E_{3}+\Big{[}\frac{\ell^{2}}{2}C+C\kappa_{g}^{2}-\ddot{C}\Big{]}\eta\,.$

Adding up these three terms we obtain (4.3). ∎

Now, using again Lemma 4.1 and the explicit expression (3.5) of the Riemannian
curvature tensor field, we can compute the two curvature terms of the
$3$-tension field (1) and we find:

######  Lemma 4.4.

$\displaystyle-\overline{R}\left(\overline{\Delta}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})$
$\displaystyle=$
$\displaystyle-\frac{\ell^{2}}{4}AX-\frac{\ell^{2}}{4}BE_{3}+\left(-4m+\frac{\ell^{2}}{2}\right)C\eta\,;$
$\displaystyle-\,\overline{R}\Big{(}\nabla^{\varphi}_{e_{i}}\,\tau(\varphi),\tau(\varphi)\Big{)}d\varphi(e_{i})$
$\displaystyle=$
$\displaystyle\left(4m-\frac{3\ell^{2}}{4}\right)\kappa_{g}^{3}\eta\,.$

Finally, adding up the terms computed in Lemmata 4.3–4.4 and simplifying using
the explicit expression of the functions $A,B,C$ defined in Lemma 4.1, we
obtain the explicit expression of the $3$-tension field of a Hopf cylinder.
This is summarized in the following

######  Proposition 4.5.

As in (2.1), let $\Sigma$ be a Hopf cylinder in a BCV-space $M^{3}_{m,\ell}$.
Then its $3$-tension field is given by

$\displaystyle\tau_{3}$ $\displaystyle=$
$\displaystyle\Big{[}2\dot{\kappa}_{g}(\ell^{2}\kappa_{g}+5\kappa_{g}^{3}-5\ddot{\kappa}_{g})-5\kappa_{g}\kappa_{g}^{(3)}\Big{]}X$
$\displaystyle+\frac{\ell}{2}\Big{[}-(\ell^{2}+9\kappa_{g}^{2})\dot{\kappa}_{g}+4\kappa_{g}^{(3)}\Big{]}E_{3}$
$\displaystyle+\frac{1}{4}\Big{[}\kappa_{g}\left(60\dot{\kappa}_{g}^{2}-(\ell^{2}+4\kappa_{g}^{2})(2\ell^{2}-8m+\kappa_{g}^{2})\right)+2(5\ell^{2}-8m+20\kappa_{g}^{2})\ddot{\kappa}_{g}-4\kappa_{g}^{(4)}\Big{]}\eta\,.$

We deduce from (4.5) that a Hopf cylinder $\Sigma$ is $3$-harmonic if and only
if the geodesic curvature $\kappa_{g}$ of its base curve $\gamma$ verifies:

(4.6)
$\begin{cases}2\dot{\kappa}_{g}(\ell^{2}\kappa_{g}+5\kappa_{g}^{3}-5\ddot{\kappa}_{g})-5\kappa_{g}\kappa_{g}^{(3)}=0\\\
-\ell(\ell^{2}+9\kappa_{g}^{2})\dot{\kappa}_{g}+4\ell\kappa_{g}^{(3)}=0\\\
\kappa_{g}\left(60\dot{\kappa}_{g}^{2}-(\ell^{2}+4\kappa_{g}^{2})(2\ell^{2}-8m+\kappa_{g}^{2})\right)+2(5\ell^{2}-8m+20\kappa_{g}^{2})\ddot{\kappa}_{g}-4\kappa_{g}^{(4)}=0\,.\end{cases}$

###### Proof of Theorem 2.1.

Because $\tau=-\kappa_{g}\eta$, the proof amounts to showing that the base
curve $\gamma$ of $\Sigma$ has constant geodesic curvature $\kappa_{g}$. We
denote by $K1,K2,K3$ respectively the left-hand sides of the three equations
in (4.6).

Case $\ell\neq 0$

We argue by contradiction. So, let us suppose that (4.6) admits a nonconstant
solution. It follows that there exists an open interval $I$ such that
$\kappa_{g}(s),\dot{\kappa}_{g}(s)$ are both different from zero on $I$. We
work on $I$. First, from

$4\ell K1+5\kappa_{g}K2=0$

we deduce that

(4.7) $\ddot{\kappa}_{g}=\frac{\kappa_{g}}{40}(3\ell^{2}-5\kappa_{g}^{2})\,.$

Next, replacing the derivative of (4.7) into $K2$ gives

$-\frac{7}{10}\ell(\ell^{2}+15\kappa_{g}^{2})\dot{\kappa}_{g}=0\,,$

Thus, since $\ell\neq 0$, we must have that $\dot{\kappa}_{g}=0$ obtaining a
contradiction.

Case $\ell=0$

In this case (4.6) is equivalent to

(4.8)
$\left\\{\begin{array}[]{ll}{\rm(i)}&2\dot{\kappa}_{g}(\kappa_{g}^{3}-\ddot{\kappa}_{g})-\kappa_{g}\kappa_{g}^{(3)}=0\\\
{\rm(ii)}&15\kappa_{g}\dot{\kappa}_{g}^{2}-\kappa_{g}^{3}(\kappa_{g}^{2}-8m)+2(5\kappa_{g}^{2}-2m)\ddot{\kappa}_{g}-\kappa_{g}^{(4)}=0\,.\end{array}\right.$

Explicit integration of (4.8) (i) yields

(4.9) $2\kappa_{g}\ddot{\kappa}_{g}=\kappa^{4}_{g}-\dot{\kappa}^{2}_{g}+c,$

where $c$ is a real constant. Taking the first and the second derivative of
(4.9) we obtain the expression of $\ddot{\kappa}_{g}$, $\kappa^{(3)}_{g}$ and
$\kappa^{(4)}_{g}$ in terms of $\kappa_{g}$, $\dot{\kappa}_{g}$ and $c$.
Substituting these expressions in (4.8) (ii) we obtain

(4.10)
$c^{2}+12m\kappa_{g}^{6}+7\kappa_{g}^{8}-8c\dot{\kappa}_{g}^{2}+7\dot{\kappa}_{g}^{4}+4m\kappa_{g}^{2}(\dot{\kappa}_{g}^{2}-c)+2\kappa_{g}^{4}(5c+7\dot{\kappa}_{g}^{2})=0\,.$

If $m=c=0$, then (4.10) becomes

$7(\kappa_{g}^{2}+\dot{\kappa}_{g}^{2})^{2}=0$

from which we obtain a contradiction. Now, let $(m,c)\neq(0,0)$. From (4.10)
we obtain

(4.11)
$\dot{\kappa}_{g}^{2}=\frac{1}{7}\left(4c-2m\kappa_{g}^{2}-7\kappa_{g}^{4}\pm\sqrt{9c^{2}+12cm\kappa_{g}^{2}+(4m^{2}-126c)\kappa_{g}^{4}-56m\kappa_{g}^{6}}\right)\,.$

We can restrict our attention to an open subinterval of $I$ where

$9c^{2}+12cm\kappa_{g}^{2}+(4m^{2}-126c)\kappa_{g}^{4}-56m\kappa_{g}^{6}$

is positive. Taking the derivative of (4.11) (we choose the solution with the
sign $+$ before the square root, the other case is similar) we obtain an
expression of $\ddot{\kappa}_{g}$ as a function of $\kappa_{g}$ and $c$.
Substituting this expression and (4.11) into (4.9) we easily obtain that
$\kappa_{g}$ is a root of the following polynomial with constant coefficients

$7056m\kappa_{g}^{10}+196(81c+26m^{2})\kappa_{g}^{8}+384(42cm-m^{3})\kappa_{g}^{6}+12c(945c-121m^{2})\kappa_{g}^{4}-1764c^{2}m\kappa_{g}^{2}-567c^{3}\,.$

Since $(m,c)\neq(0,0)$, it follows again that $\kappa_{g}$ is constant, a
contradiction. Thus the proof is ended. ∎

###### Proof of Theorem 2.2.

The proof follows easily from Proposition 4.5 using the assumption that
$\kappa_{g}$ is a constant. ∎

##  5\. Proof of Theorem 2.6

Our first goal is to compute all the terms which occur in the expression of
the $3$-tension field

(5.1)
$\tau_{3}(\varphi)=\overline{\Delta}^{2}\tau(\varphi)-R\left(\overline{\Delta}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})\,-\,R\Big{(}\nabla^{\varphi}_{e_{i}}\,\tau(\varphi),\tau(\varphi)\Big{)}d\varphi(e_{i})\,.$

For this purpose, we now establish a series of useful preliminary lemmata.

######  Lemma 5.1.

Let $\varphi:M^{2}\to M^{3}_{m,\ell}$ be an oriented surface. Let $A$ denote
the shape operator, $f=(1/2)\operatorname{trace}A$ the mean curvature function
and $\eta$ the unit normal. Then

  * (a)

$(\nabla A)(X,Y)=(\nabla A)(Y,X)-(R(X,Y)\eta)^{\top}$;

  * (b)

$\langle(\nabla A)(X,Y),Z\rangle=\langle(\nabla A)(X,Z),Y\rangle$;

  * (c)

$\operatorname{trace}(\nabla
A)(\cdot,\cdot)=2\operatorname{grad}f+(4m-\ell^{2})\,\nu\,E_{3}^{\top}$.

###### Proof.

(a) This is just the Codazzi equation.

(b)

$\displaystyle\langle(\nabla A)(X,Y),Z\rangle$ $\displaystyle=$
$\displaystyle\langle\nabla_{X}A(Y)-A(\nabla_{X}Y),Z\rangle$ $\displaystyle=$
$\displaystyle\langle\nabla_{X}A(Y),Z\rangle-\langle A(\nabla_{X}Y),Z\rangle$
$\displaystyle=$ $\displaystyle X\langle A(Y),Z\rangle-\langle
A(Y),\nabla_{X}Z\rangle-\langle\nabla_{X}Y,A(Z)\rangle\,.$

Because the above expression is symmetric with respect to $Y,Z$ the conclusion
follows.

(c) Let $p\in M^{2}$ be an arbitrarily fixed point and consider a geodesic
frame field $\\{X_{i}\\}_{i=1}^{2}$ around $p$. At $p$ we have:

$\displaystyle{\rm Trace}(\nabla A)(\cdot,\cdot)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{2}(\nabla
A)(X_{i},X_{i})=\sum_{i=1}^{2}(\nabla_{X_{i}}A)(X_{i})=\sum_{i=1}^{2}\nabla_{X_{i}}A(X_{i})$
$\displaystyle=$ $\displaystyle\sum_{i,j=1}^{2}\nabla_{X_{i}}(\langle
A(X_{i}),X_{j}\rangle X_{j})=\sum_{i,j=1}^{2}\nabla_{X_{i}}(\langle
X_{i},A(X_{j})\rangle X_{j})$ $\displaystyle=$
$\displaystyle\sum_{i,j=1}^{2}\langle X_{i},\nabla_{X_{i}}A(X_{j})\rangle
X_{j}=\sum_{i,j=1}^{2}\langle X_{i},(\nabla A)(X_{i},X_{j})\rangle X_{j}\,.$

Now, first using (a) and then the explicit expression of the curvature tensor
field (3.5), we continue the previous sequence of equalities as follows:

$\displaystyle{\rm Trace}(\nabla A)(\cdot,\cdot)$ $\displaystyle=$
$\displaystyle\ldots=\sum_{i,j=1}^{2}\langle
X_{i},\nabla_{X_{j}}A(X_{i})-\overline{R}(X_{i},X_{j})\eta\rangle X_{j}$
$\displaystyle=$ $\displaystyle\sum_{i,j=1}^{2}\langle
X_{i},\nabla_{X_{j}}A(X_{i})\rangle X_{j}-\langle
X_{i},\overline{R}(X_{i},X_{j})\eta\rangle X_{j}$ $\displaystyle=$
$\displaystyle\sum_{i,j=1}^{2}\Big{(}X_{j}\langle
X_{i},A(X_{i})\rangle\Big{)}X_{j}$
$\displaystyle+\sum_{i,j=1}^{2}(4m-\ell^{2})\Big{[}\langle
X_{j},E_{3}\rangle\nu-\langle X_{i},E_{3}\rangle\nu\delta_{ij}\Big{]}X_{j}$
$\displaystyle=$ $\displaystyle
2\sum_{j=1}^{2}X_{j}(f)X_{j}+(4m-\ell^{2})\Big{[}2\nu E_{3}^{\top}-\nu
E_{3}^{\top}\Big{]}$ $\displaystyle=$ $\displaystyle
2\operatorname{grad}f+(4m-\ell^{2})\nu E_{3}^{\top}$

and so the proof of the lemma is ended. ∎

Next, we compute:

######  Lemma 5.2.

Let $\varphi:M^{2}\to M^{3}_{m,\ell}$ be an oriented surface and denote by
${\mathbf{H}}=f\eta$ its mean curvature vector field. Then

(5.2) $\overline{\Delta}{\mathbf{H}}=(\Delta
f+f|A|^{2})\eta+2A(\operatorname{grad}f)+2f\operatorname{grad}f+(4m-\ell^{2})\,f\,\nu\,E_{3}^{\top}\,.$

###### Proof.

We work with a geodesic frame field $\left\\{X_{i}\right\\}_{i=1}^{2}$ around
an arbitrarily fixed point $p\in M^{2}$. Again, we simplify the notation
writing $\nabla$ for $\nabla^{M^{2}}$. Since ${\mathbf{H}}=f\eta$, around $p$
we have:

$\nabla_{X_{i}}^{\varphi}{\mathbf{H}}=\nabla_{X_{i}}^{\perp}{\mathbf{H}}-A_{{\mathbf{H}}}(X_{i})=\left(X_{i}f\right)\eta-
fA\left(X_{i}\right)\,.$

Then, denoting by $B$ the second fundamental form, at $p$ we have:

$\displaystyle\nabla_{X_{i}}^{\varphi}\nabla_{X_{i}}^{\varphi}{\mathbf{H}}$
$\displaystyle=$
$\displaystyle\left(X_{i}X_{i}f\right)\eta-\left(X_{i}f\right)A\left(X_{i}\right)-\left(X_{i}f\right)A\left(X_{i}\right)-f\big{(}\nabla_{X_{i}}A\left(X_{i}\right)+B\left(X_{i},A\left(X_{i}\right)\right)\big{)}$
$\displaystyle=$
$\displaystyle\left(X_{i}X_{i}f\right)\eta-2\left(X_{i}f\right)A\left(X_{i}\right)-f(\nabla
A)(X_{i},X_{i})-f|A\left(X_{i}\right)|^{2}\eta\,.$

Now, taking the sum over $i$ and using Lemma 5.1, we obtain (5.2) (note that
the sign convention for $\Delta$ and $\overline{\Delta}$ is as in (1.3)). ∎

In the special case that the surface is CMC, say $f=\alpha$, then (5.2)
becomes:

(5.3)
$\overline{\Delta}{\mathbf{H}}=\alpha\,|A|^{2}\eta+\alpha\,(4m-\ell^{2})\,\nu\,E_{3}^{\top}\,.$

Since the mean curvature vector field and the tension field are related by
$\tau(\varphi)=2{\mathbf{H}}$, setting for convenience of notation
$\tau(\varphi)=\tau$ and

(5.4) $V=\nu\,E_{3}^{\top}\,,$

we rewrite (5.3) as

(5.5)
$\overline{\Delta}\tau=2\alpha\,|A|^{2}\eta+2\alpha\,(4m-\ell^{2})\,V\,.$

Next, from (5.5) and performing a computation as in Lemma 5.2 we find:

######  Lemma 5.3.

Let $\varphi:M^{2}\to M^{3}_{m,\ell}$ be an oriented surface and assume that
$f\equiv\alpha$. Then

(5.6) $\overline{\Delta}^{2}\tau=2\alpha(|A|^{4}+\Delta|A|^{2})\,\eta+4\alpha
A(\operatorname{grad}|A|^{2})+2\alpha|A|^{2}(4m-\ell^{2})\,V+2\alpha(4m-\ell^{2})\,\overline{\Delta}\,V\,.$

The first difficulty is to compute in a convenient way $\overline{\Delta}\,V$.
We have

######  Lemma 5.4.

$\displaystyle\overline{\Delta}\,V$ $\displaystyle=$ $\displaystyle
2\nu\Big{(}2A^{2}(E_{3}^{\top})+\frac{3\ell}{4}\big{(}A(J(E_{3}^{\top}))-J(A(E_{3}^{\top})\big{)}\Big{)}$
$\displaystyle+\Big{\\{}(4m-\ell^{2})(|E_{3}^{\top}|^{2}-\nu^{2})+|A|^{2}+\frac{5\ell^{2}}{4}\Big{\\}}\nu
E_{3}^{\top}$ $\displaystyle+\Big{\\{}2|A(E_{3}^{\top})|^{2}+\ell\langle
JE_{3}^{\top},A(E_{3}^{\top})\rangle-2\nu^{2}|A|^{2}-(4m-\ell^{2})\nu^{2}|E_{3}^{\top}|^{2}\Big{\\}}\eta\,.$

###### Proof.

We can work with a geodesic frame field $\left\\{X_{i}\right\\}_{i=1}^{2}$
such that $J(X_{1})=X_{2}$, $J(X_{2})=-X_{1}$ around an arbitrarily fixed
point $p\in M^{2}$. Taking into account (3.8) and (3.9) we obtain around $p$:

$\nabla_{X_{i}}^{\varphi}V=\nu^{2}\left(A(X_{i})-\frac{\ell}{2}J(X_{i})\right)-\langle
A(X_{i})-\frac{\ell}{2}J(X_{i}),E_{3}^{\top}\rangle E_{3}^{\top}+\nu\langle
A(X_{i}),E_{3}^{\top}\rangle\eta\,.$

Next, in a similar fashion, we compute at $p$:

$\displaystyle\nabla_{X_{i}}^{\varphi}\big{(}\nabla_{X_{i}}^{\varphi}V\big{)}$
$\displaystyle=$ $\displaystyle
2\nu\,X_{i}(\nu)\big{(}A(X_{i})-\frac{\ell}{2}J(X_{i})\big{)}$
$\displaystyle+\nu^{2}\Big{[}(\nabla
A)(X_{i},X_{i})+|A(X_{i})|^{2}\eta-\frac{\ell}{2}\langle
A(X_{i}),J(X_{i})\rangle\eta\Big{]}$ $\displaystyle+\Big{[}\langle-(\nabla
A)(X_{i},X_{i}),E_{3}^{\top}\rangle-\nu\big{|}A(X_{i})-\frac{\ell}{2}J(X_{i})\big{|}^{2}\Big{]}E_{3}^{\top}$
$\displaystyle-\langle
A(X_{i})-\frac{\ell}{2}J(X_{i}),E_{3}^{\top}\rangle\Big{[}\nu\big{(}A(X_{i})-\frac{\ell}{2}J(X_{i})\big{)}+\langle
A(X_{i}),E_{3}^{\top}\rangle\eta\Big{]}$ $\displaystyle+X_{i}(\nu)\langle
A(X_{i}),E_{3}^{\top}\rangle\eta+\nu\langle(\nabla
A)(X_{i},X_{i}),E_{3}^{\top}\rangle\eta$ $\displaystyle+\nu^{2}\langle
A(X_{i}),A(X_{i})-\frac{\ell}{2}J(X_{i})\rangle\eta-\nu\langle
A(X_{i}),E_{3}^{\top}\rangle A(X_{i})\,.$

Now, since $\left\\{X_{i}\right\\}_{i=1}^{2}$ is a geodesic frame field, we
have at $p$:

$\overline{\Delta}V=-\sum_{i=1}^{2}\nabla_{X_{i}}^{\varphi}\big{(}\nabla_{X_{i}}^{\varphi}V\big{)}\,.$

Therefore,

$\displaystyle\overline{\Delta}V$ $\displaystyle=$
$\displaystyle-2\nu\big{(}A(\operatorname{grad}\nu)-\frac{\ell}{2}J(\operatorname{grad}\nu)\big{)}$
$\displaystyle-\nu^{2}\Big{[}(4m-\ell^{2})\nu
E_{3}^{\top}+|A|^{2}\eta-\frac{\ell}{2}\sum_{i=1}^{2}\langle
A(X_{i}),J(X_{i})\rangle\eta\Big{]}$
$\displaystyle+\Big{[}(4m-\ell^{2})\nu|E_{3}^{\top}|^{2}+\sum_{i=1}^{2}\nu\big{|}A(X_{i})-\frac{\ell}{2}J(X_{i})\big{|}^{2}\Big{]}E_{3}^{\top}$
$\displaystyle+\sum_{i=1}^{2}\langle
A(X_{i})-\frac{\ell}{2}J(X_{i}),E_{3}^{\top}\rangle\Big{[}\nu\big{(}A(X_{i})-\frac{\ell}{2}J(X_{i})\big{)}+\langle
A(X_{i}),E_{3}^{\top}\rangle\eta\Big{]}$ $\displaystyle-\langle
A(\operatorname{grad}\nu),E_{3}^{\top}\rangle\eta-(4m-\ell^{2})\nu^{2}|E_{3}^{\top}|^{2}\eta$
$\displaystyle-\nu^{2}\sum_{i=1}^{2}\langle
A(X_{i}),A(X_{i})-\frac{\ell}{2}J(X_{i})\rangle\eta+\nu\sum_{i=1}^{2}\langle
A(X_{i}),E_{3}^{\top}\rangle A(X_{i})\,.$

Now (5.4) can be computed using (3.9) and performing some simplifications such
as:

$\begin{split}\langle A(X_{1}),J(X_{1})\rangle+\langle
A(X_{2}),J(X_{2})\rangle&=0\\\
\sum_{i=1}^{2}|A(X_{i})-\frac{\ell}{2}J(X_{i})\big{|}^{2}&=|A|^{2}+\frac{\ell^{2}}{2}\\\
\sum_{i=1}^{2}\langle X_{i},J(E_{3}^{\top})\rangle\langle
X_{i},A(E_{3}^{\top})\rangle\eta&=\langle
J(E_{3}^{\top}),A(E_{3}^{\top})\rangle\eta\\\
\operatorname{grad}\nu=-\sum_{i=1}^{2}\langle
A(X_{i})-\frac{\ell}{2}J(X_{i}),E_{3}^{\top}\rangle
X_{i}&=-A(E_{3}^{\top})-\frac{\ell}{2}J(E_{3}^{\top})\,.\end{split}$

Indeed,

$\displaystyle\overline{\Delta}V$ $\displaystyle=$ $\displaystyle 2\nu
A^{2}(E_{3}^{\top})+\ell\nu A(J(E_{3}^{\top}))-\ell\nu
J(A(E_{3}^{\top}))+\nu\frac{\ell^{2}}{2}E_{3}^{\top}$
$\displaystyle-\nu^{2}\Big{[}(4m-\ell^{2})\nu E_{3}^{\top}+|A|^{2}\eta\Big{]}$
$\displaystyle+\Big{[}(4m-\ell^{2})\nu|E_{3}^{\top}|^{2}+\nu\left(|A|^{2}+\frac{\ell^{2}}{2}\right)\Big{]}E_{3}^{\top}$
$\displaystyle+\nu
A^{2}(E_{3}^{\top})+\nu\frac{\ell}{2}A(J(E_{3}^{\top}))-\nu\frac{\ell}{2}J(A(E_{3}^{\top}))+\nu\frac{\ell^{2}}{4}E_{3}^{\top}$
$\displaystyle+\Big{[}|A(E_{3}^{\top})|^{2}+\frac{\ell}{2}\langle
J(E_{3}^{\top}),A(E_{3}^{\top})\rangle\Big{]}\eta$
$\displaystyle+|A(E_{3}^{\top})|^{2}\eta+\frac{\ell}{2}\langle
J(E_{3}^{\top}),A(E_{3}^{\top})\rangle\eta-(4m-\ell^{2})\nu^{2}|E_{3}^{\top}|^{2}\eta$
$\displaystyle-\nu^{2}|A|^{2}\eta+\nu A^{2}(E_{3}^{\top})$

from which (5.4) follows immediately (note that each line of (5) is equal to
the corresponding line of (5)). ∎

Now, we can state the main result which is of independent interest and
summarizes the preliminary work which we have carried in this section.

######  Proposition 5.5.

Let $M^{2}$ be an oriented surface in $M^{3}_{m,\ell}$. Assume that $M^{2}$
has CMC equal to $\alpha$. Then its $3$-tension field is

$\displaystyle\tau_{3}$ $\displaystyle=$
$\displaystyle\alpha\Big{\\{}2\Delta|A|^{2}+2|A|^{4}+|A|^{2}\big{[}\ell^{2}(1+2\nu^{2})-8m(1+\nu^{2})\big{]}$
$\displaystyle+4(4m-\ell^{2})|A(E_{3}^{\top})|^{2}+2\ell(4m-\ell^{2})\langle
J(E_{3}^{\top}),A(E_{3}^{\top})\rangle-4\alpha(4m-\ell^{2})\langle
A(E_{3}^{\top}),E_{3}^{\top}\rangle$
$\displaystyle+2\alpha^{2}\ell^{2}(3+4\nu^{2})-32m\alpha^{2}(1+\nu^{2})\Big{\\}}\eta$
$\displaystyle+2\alpha(4m-\ell^{2})\nu\Big{\\{}3|A|^{2}+4\alpha^{2}+3\ell^{2}\nu^{2}+m(4-12\nu^{2})\Big{\\}}E_{3}^{\top}$
$\displaystyle+4\alpha(4m-\ell^{2})\nu\Big{\\{}2A^{2}(E_{3}^{\top})-\frac{3\ell}{4}J(A(E_{3}^{\top}))+\frac{3\ell}{4}(A(J(E_{3}^{\top}))\Big{\\}}-4\alpha^{2}(4m-\ell^{2})\nu
A(E_{3}^{\top})$ $\displaystyle+4\alpha A(\operatorname{grad}|A|^{2})\,,$

where $\nu$ is defined in (3.7).

###### Proof.

The explicit expression of the curvature tensor field is given in (3.5) and so
we have all the ingredients to compute the $3$-tension field (5.1). More in
detail:

(I) $\overline{\Delta}^{2}\tau$ is given in Lemmata 5.3 and 5.4.

(II) Here we compute the first curvature term

$\overline{R}\left(\overline{\Delta}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})\,,$

which we rewrite as

(5.11)
$\sum_{i=1}^{2}\overline{R}\left((\overline{\Delta}\tau)^{\perp},X_{i}\right)X_{i}+\sum_{i=1}^{2}\overline{R}\left((\overline{\Delta}\tau)^{\top},X_{i}\right)X_{i}\,,$

where, according to (5.5), we have

(5.12)
$(\overline{\Delta}\tau)^{\perp}=2\alpha\,|A|^{2}\eta\,,\quad(\overline{\Delta}\tau)^{\top}=2\alpha\,(4m-\ell^{2})\,V\,.$

Then, using the general expression (3.5) for the curvature tensor field, we
find:

$\displaystyle\sum_{i=1}^{2}\overline{R}\left((\overline{\Delta}\tau)^{\perp},X_{i}\right)X_{i}$
$\displaystyle=$ $\displaystyle
2\alpha|A|^{2}\Big{\\{}2\left(4m-\frac{3\ell^{2}}{4}\right)\eta$
$\displaystyle\qquad\qquad-(4m-\ell^{2})\big{[}(|E_{3}^{\top}|^{2}+2\nu^{2})\eta+\nu
E_{3}^{\top}\big{]}\Big{\\}}\,.$
$\displaystyle\sum_{i=1}^{2}\overline{R}\left((\overline{\Delta}\tau)^{\top},X_{i}\right)X_{i}$
$\displaystyle=$ $\displaystyle
2\alpha(4m-\ell^{2})\nu\Big{\\{}\left(4m-\frac{3\ell^{2}}{4}\right)\,E_{3}^{\top}$
$\displaystyle\qquad\qquad\qquad-(4m-\ell^{2})\big{[}|E_{3}^{\top}|^{2}E_{3}^{\top}+|E_{3}^{\top}|^{2}\nu\,\eta\big{]}\Big{\\}}\,.$

Adding up the results in (5) and (5) we can easily handle the term in (II).

(III) Here we deal with the other curvature term, i.e.,

$\overline{R}\Big{(}\nabla^{\varphi}_{e_{i}}\,\tau(\varphi),\tau(\varphi)\Big{)}d\varphi(e_{i})\,.$

Using again the general expression (3.5) for the curvature tensor field and
observing that

$\nabla^{\varphi}_{X_{i}}\tau=-2\alpha A(X_{i})$

we find (slight abuse of notation: we identify $X_{i}$ and $d\varphi(X_{i})$):

$\displaystyle\sum_{i=1}^{2}\overline{R}\left(\nabla_{X_{i}}^{\varphi}\tau,\tau\right)X_{i}$
$\displaystyle=$
$\displaystyle-4\alpha^{2}\sum_{i=1}^{2}\overline{R}\left(A(X_{i}),\eta\right)X_{i}$
$\displaystyle=$
$\displaystyle-4\alpha^{2}\Big{\\{}-2\alpha\left(4m-\frac{3\ell^{2}}{4}\right)\eta$
$\displaystyle+(4m-\ell^{2})\Big{[}\nu A(E_{3}^{\top})-\langle
A(E_{3}^{\top}),E_{3}^{\top}\rangle\eta-2\alpha\nu
E_{3}^{\top}-2\alpha\nu^{2}\eta\Big{]}\Big{\\}}$

Using the results obtained in the three cases (I), (II), (III) we can easily
compute the $3$-tension field described explicitly in (5.1). More precisely,
after some routine simplifications we obtain (5.5) (where we have used that
$|E_{3}^{\top}|^{2}=1-\nu^{2}$) and so the proof of Proposition 5.5 is
completed. ∎

###### Proof of Theorem 2.6.

According to Theorem 2.4, we just have to study the three possible cases
(iv)(a),(b),(c). Case (iv)(a) is analysed in detail in Theorem 2.2 and
provides examples of proper triharmonic surfaces. By contrast, horizontal
slices of the type (iv)(b) are totally geodesic and therefore there exists no
proper triharmonic surface of this type. By way of summary, the proof will be
complete if we show that any parabolic helicoid $P_{\alpha,m,\ell}$ as in
Theorem 2.4 (iv)(c) cannot be proper triharmonic. Despite the simplicity of
this plan for the proof, the involved computations are quite long and will be
carried out using the half-space model for $M^{3}_{m,\ell}$. More precisely,
recalling that in the case of parabolic helicoids $m<0$ by assumption, we
shall work in

$\tilde{M}^{3}_{m,\ell}=\left(\\{(\tilde{x},\tilde{y},\tilde{z})\in{\mathbb{R}}^{3}\,:\,\tilde{y}>0\\},\,\frac{d\tilde{x}^{2}+d\tilde{y}^{2}}{-4m\tilde{y}^{2}}+\left(d\tilde{z}+\frac{\ell}{4m\tilde{y}}d\tilde{x}\right)^{2}\right)\,.$

Now, we recall (see [4]) some basic facts about the half-space model. We have
an explicit isometry $\Theta:M^{3}_{m,\ell}\to\tilde{M}^{3}_{m,\ell}$ given by

$\Theta(x,y,z)=\left(\frac{2y}{\sqrt{-m}G^{2}},\frac{1+m(x^{2}+y^{2})}{-mG^{2}},z+\frac{\ell}{2m}\arccos\left(\frac{y}{G}\right)\right)\,,$

where we have set

$G=\sqrt{\left(\frac{1}{\sqrt{-m}}+x\right)^{2}+y^{2}}\,.$

Moreover, an explicit positively oriented global orthonormal frame field on
the half-space model is:

$\tilde{E}_{1}=2\sqrt{-m}\,\tilde{y}\,\partial_{\tilde{x}}+\frac{\ell}{2\sqrt{-m}}\,\partial_{\tilde{z}}\,,\quad\tilde{E}_{2}=2\sqrt{-m}\,\tilde{y}\,\partial_{\tilde{y}}\,,\quad\tilde{E}_{3}=\partial_{\tilde{z}}\,.$

For future use, we observe that $d\Theta(E_{3})=\tilde{E}_{3}$. Next, we
compute

$[\tilde{E}_{1},\tilde{E}_{2}]=-2\sqrt{-m}\tilde{E}_{1}+\ell\tilde{E}_{3}\,;\quad[\tilde{E}_{1},\tilde{E}_{3}]=[\tilde{E}_{2},\tilde{E}_{3}]=0\,.$

Then, using the Koszul identity (3), it is easy to verify that the version of
(3.3) in this context is:

(5.16)
$\begin{array}[]{lll}\overline{\nabla}_{\tilde{E}_{1}}\tilde{E}_{1}=2\sqrt{-m}\tilde{E}_{2}\,,&\overline{\nabla}_{\tilde{E}_{1}}\tilde{E}_{2}=-2\sqrt{-m}\tilde{E}_{1}+\dfrac{\ell}{2}\tilde{E}_{3}\,,&\overline{\nabla}_{\tilde{E}_{1}}\tilde{E}_{3}=-\dfrac{\ell}{2}\tilde{E}_{2}\,,\vspace{2mm}\\\
\overline{\nabla}_{\tilde{E}_{2}}\tilde{E}_{1}=-\dfrac{\ell}{2}\tilde{E}_{3}\,,&\overline{\nabla}_{\tilde{E}_{2}}\tilde{E}_{2}=0&\overline{\nabla}_{\tilde{E}_{2}}\tilde{E}_{3}=\dfrac{\ell}{2}\tilde{E}_{1}\,,\vspace{2mm}\\\
\overline{\nabla}_{\tilde{E}_{3}}\tilde{E}_{1}=-\dfrac{\ell}{2}\tilde{E}_{2}\,,&\overline{\nabla}_{\tilde{E}_{3}}\tilde{E}_{2}=\dfrac{\ell}{2}\tilde{E}_{1}\,,&\overline{\nabla}_{\tilde{E}_{3}}\tilde{E}_{3}=0\,.\end{array}$

The parabolic helicoids $P_{\alpha,m,\ell}$ (see [7]) are the CMC surfaces in
$\tilde{M}^{3}_{m,\ell}$ parametrized by

$X(u,v)=\left(u,v,a\log v\right)\,,\quad v>0\,,$

where $a$ is non-vanishing real constant whose relation with the geometrical
parameters $\alpha,m,\ell$ will be made explicit in (5.21) below.

We have to verify that a parabolic helicoid $P_{\alpha,m,\ell}$, $\alpha\neq
0$, cannot be triharmonic. For this purpose, we apply Proposition 5.5. In
order to compute all the terms which appear in the expression of the
$3$-tension field $\tau_{3}$ (see (5.5) it is convenient to express and
compute all the relevant quantities with respect to the global orthonormal
frame field $\\{\tilde{E}_{1},\tilde{E}_{2},\tilde{E}_{3}\\}$. Writing
$[\tilde{x}_{1},\tilde{x}_{2},\tilde{x}_{3}]$ for
$\tilde{x}_{1}\tilde{E}_{1}+\tilde{x}_{2}\tilde{E}_{2}+\tilde{x}_{3}\tilde{E}_{3}$,
we compute:

(5.17) $\displaystyle X_{u}$ $\displaystyle=$
$\displaystyle\left[\frac{1}{2\sqrt{-m}\,v},0,\frac{\ell}{4mv}\right]$
$\displaystyle X_{v}$ $\displaystyle=$
$\displaystyle\left[0,\frac{1}{2\sqrt{-m}\,v},\frac{a}{v}\right]$
$\displaystyle\eta$ $\displaystyle=$
$\displaystyle\frac{1}{L}\Big{[}\ell,4am,2\sqrt{-m}\Big{]}\,,$

where we have set

(5.18) $L=\sqrt{\ell^{2}+4m(4a^{2}m-1)}$

Note that

$\nu=\eta\cdot\tilde{E}_{3}=\frac{2\sqrt{-m}}{L}$

and so we can compute $E_{3}^{\top}=\tilde{E}_{3}-\nu\eta$. We find

$E_{3}^{\top}=\left[-\frac{2\ell\sqrt{-m}}{L^{2}},\frac{8a(-m)^{3/2}}{L^{2}},\frac{4m}{L^{2}}+1\right]\,.$

Moreover,

$J(E_{3}^{\top})=\eta\times
E_{3}^{\top}=\left[\frac{4am}{L},-\frac{\ell}{L},0\right]\,.$

Now, we recall that $A(X)=-\overline{\nabla}_{X}\eta$ and $J(X)=\eta\times X$
for all $X$ tangent to $P_{\alpha,m,\ell}$. Next, we observe that all the
coefficients of $\eta,E_{3}^{\top},J(E_{3}^{\top})$ are constant. Therefore,
using (5.16) it is not difficult to compute:

(5.19) $\displaystyle A(E_{3}^{\top})$ $\displaystyle=\left[-\frac{2a\ell
m}{L},\frac{\ell^{2}}{2L},0\right]$ $\displaystyle A(J(E_{3}^{\top}))$
$\displaystyle=\left[\frac{\sqrt{-m}\left(32a^{2}m^{2}+\ell^{2}\right)}{L^{2}},\frac{4a\ell(-m)^{3/2}}{L^{2}},-\frac{16a^{2}\ell
m^{2}+\ell^{3}}{2L^{2}}\right]$ $\displaystyle A^{2}(E_{3}^{\top})$
$\displaystyle=\left[-\frac{\ell^{2}\sqrt{-m}}{L^{2}},\frac{4a\ell(-m)^{3/2}}{L^{2}},\frac{16a^{2}\ell
m^{2}+\ell^{3}}{2L^{2}}\right]$ $\displaystyle J(A(E_{3}^{\top}))$
$\displaystyle=\left[-\frac{\ell\sqrt{-m}\left(32a^{2}m^{2}+\ell^{2}\right)}{2L^{2}},-\frac{2a\ell^{2}(-m)^{3/2}}{L^{2}},\frac{16a^{2}\ell^{2}m^{2}+\ell^{4}}{4L^{2}}\right]$

Using (5.19), we are in the right position to compute the mean curvature
$\alpha$ and $|A|^{2}$. We have

(5.20) $2\alpha=\frac{1}{|E_{3}^{\top}|^{2}}\big{(}\langle
A(E_{3}^{\top}),E_{3}^{\top}\rangle+\langle
A(J(E_{3}^{\top})),J(E_{3}^{\top})\rangle\big{)}=\frac{8a(-m)^{3/2}}{L}.$

Inverting (5.20), together with (5.18), we deduce that the relationship
between $a$ and the geometrical parameters $\alpha,\ell,m$ is:

(5.21) $a=\frac{\alpha}{4m}\sqrt{\frac{4m-\ell^{2}}{m+\alpha^{2}}}.$

As for $|A|^{2}$, using (5.21) and (5.18) a straight computation returns

(5.22) $|A|^{2}=\frac{1}{|E_{3}^{\top}|^{2}}\big{(}\langle
A(E_{3}^{\top}),A(E_{3}^{\top})\rangle+\langle
A(J(E_{3}^{\top})),A(J(E_{3}^{\top}))\rangle\big{)}=\frac{1}{2}(8\alpha^{2}+\ell^{2}).$

Next, using (5.19), (5.21) and (5.22) all the terms in (5.5) can be easily
computed and, after adding up and simplifying, we find

$\tau_{3}=\alpha\,\left[\frac{1}{\sqrt{-m}}\sqrt{\frac{\alpha^{2}+m}{4m-\ell^{2}}}\,T1,\frac{\alpha}{\sqrt{-m}}\,T2,-2\sqrt{\frac{\alpha^{2}+m}{4m-\ell^{2}}}\,T3\right]\,,$

where

$\displaystyle T1$ $\displaystyle=$
$\displaystyle\ell\Big{[}\ell^{4}+\ell^{2}(-16m+6\alpha^{2})+32(2m^{2}-3m\alpha^{2}-2\alpha^{4})\Big{]}\,;$
$\displaystyle T2$ $\displaystyle=$
$\displaystyle\ell^{4}+\ell^{2}(-16m+6\alpha^{2})+32(2m^{2}-3m\alpha^{2}-2\alpha^{4})\,;$
$\displaystyle T3$ $\displaystyle=$
$\displaystyle\ell^{4}+\ell^{2}(-8m+2\alpha^{2})+64\,\alpha^{2}(m+\alpha^{2})\,.$

By way of summary, a parabolic helicoid $P_{\alpha,m,\ell}$ is proper
triharmonic if and only if

(5.23) $\left\\{\begin{array}[]{l}T1=0\\\ T2=0\\\ T3=0\,.\end{array}\right.$

Now, if $\ell=0$, the third equation of system (5.23) becomes

$64\,\alpha^{2}(m+\alpha^{2})=0\,,$

which has not relevant solutions since in our construction $m+\alpha^{2}<0$ by
hypothesis.

Next, we handle the case $\ell\neq 0$. First, we observe that the condition
$T2-T3=0$ implies

(5.24)
$\ell^{2}=\frac{-8(-2m^{2}+5m\alpha^{2}+4\alpha^{4})}{2m-\alpha^{2}}\,.$

Now, replacing this value of $\ell^{2}$ into $T2$ and simplifying we find that
necessarily

$(2m+3\alpha^{2})(-2m^{2}+9m\alpha^{2}+8\alpha^{4})=0\,.$

Because $m<0$, the only possibilities are:

(5.25) $m=-\frac{3\alpha^{2}}{2}\quad{\rm or}\quad
m=\frac{\alpha^{2}}{4}(9-\sqrt{145})\,.$

The first value for $m$ is not acceptable because, if we replace it into
(5.24), we find

$\ell^{2}=-16\alpha^{2}\,,$

a contradiction. As for the second value of $m$ into (5.25), it suffices to
observe that it would imply $m+\alpha^{2}>0$, a fact which contradicts our
assumption.

Therefore, there exists no proper triharmonic parabolic helicoid
$P_{\alpha,m,\ell}$ and so the proof of Theorem 2.6 is completed. ∎

##  6\. Proper CMC $r$-harmonic Hopf cylinders

In this section we focus on the study of proper CMC $r$-harmonic Hopf
cylinders. We shall show that the existence of such submanifolds depends not
only on the curvature of the ambient space, but also on the value of $r$. This
will be illustrated in Corollary 6.3 which is a consequence of the following
theorem.

######  Theorem 6.1.

Assume that $r\geq 2$. Let $\Sigma$ be a non minimal CMC Hopf cylinder in a
BCV-space $M^{3}_{m,\ell}$. Then $\Sigma$ is proper $r$-harmonic if and only
if the geodesic curvature $\kappa_{g}$ of its base curve $\gamma$ is a non
zero constant which verifies

(6.1)
$\kappa_{g}^{4}+\left[-4m(r-1)+\frac{3\ell^{2}r}{4}\right]\kappa_{g}^{2}-\frac{\ell^{2}}{2}(4m-\ell^{2})=0\,.$

######  Remark 6.2.

From (4.1) (ii) and (iv), the norm $|A|^{2}$ of the shape operator of $\Sigma$
is (see also [18]):

(6.2) $|A|^{2}=\kappa_{g}^{2}+\frac{\ell^{2}}{2}\,.$

Using (6.2), condition (6.1) is equivalent to

(6.3)
$|A|^{4}-\left(4m-\frac{\ell^{2}}{2}\right)|A|^{2}-(r-2)\left(4m-\frac{3\ell^{2}}{4}\right)\kappa_{g}^{2}=0\,.$

Thus, setting $r=2$ into (6.3), it is immediate to recover the result of Ou
[18] concerning CMC biharmonic Hopf cylinders. Also, in the special case $r=3$
in (6.1), it is easy to recover the statement of Theorem 2.2.

In the following corollary we shall indicate for which values of the
parameters $\ell$, $m$ and $r$ there are acceptable solutions of (6.1). We
suggest that the reader keeps in mind the geometrical counterpart of the cases
(i), (ii), (iii) of Corollary 6.3 referring to the diagram in Figure 1.

######  Corollary 6.3.

Let $M^{3}_{m,\ell}$ be a BCV-space. Then there exists a proper CMC
$r$-harmonic Hopf cylinder if and only if one of the following holds:

  1. (i)

$4m-\ell^{2}>0$ and $r\geq 2$;

  2. (ii)

$4m-\ell^{2}=0$, $\ell\neq 0$ and $r\geq 5$;

  3. (iii)

$4m-\ell^{2}<0$, $\overline{R}_{1212}=4m-{3\ell^{2}}/{4}>0$ and

(6.4)
$r\geq\frac{4\left(\sqrt{2}\sqrt{\ell^{4}-4\ell^{2}m}+4m\right)}{16m-3\ell^{2}}\,.$

######  Remark 6.4.

The assumptions $4m-\ell^{2}<0$ and $\overline{R}_{1212}=4m-{3\ell^{2}}/{4}>0$
in Case (iii) of Corollary 6.3 have a geometrical meaning because they state
that the ambient space is $SU(2)$ endowed with a metric with positive
sectional curvature. Here we point out that from the analytical view point
these hypotheses are equivalent just to the condition
$(3/4)\ell^{2}<4m<\ell^{2}$ which corresponds to the region between the two
parabolas $4m=(3/4)\ell^{2}$ and $4m=\ell^{2}$ (see Figure 2). It is
convenient to describe this region as the union
$\cup_{a}\gamma_{a},\,3/4<a<1$, where $\gamma_{a}$ is the parabola
$4m=a\ell^{2}$. Now, on $\gamma_{a}$ the lower bound for $r$ in (6.4) becomes

$r_{a}=\frac{4\left(a+\sqrt{2-2a}\right)}{4a-3}\,.$

We observe that on $(3/4,1)$ $r_{a}$ is a strictly decreasing function of $a$
with $\lim_{a\to 3/4^{+}}r_{a}=+\infty$ and $\lim_{a\to 1^{-}}r_{a}=4$.
Therefore, the more we approach $\gamma_{3/4}$, the larger $r$ must be in
order to have a proper CMC $r$-harmonic Hopf cylinder.

Figure 2. Distribution of proper CMC $r$-harmonic Hopf cylinders in BCV-spaces
w.r.t. to the values of $\ell$, $m$ and $r$.$\ell$$m$$4m=\ell^{2}$ – solutions
if $r\geq 5$$4m=\frac{3}{4}\ell^{2}$ – no solutions$4m=a\ell^{2}$ – solutions
if $r\geq r_{a}$solutions if $r\geq 2$solutions if $r\geq 2$no solutionsno
solutionsno solutionsno solutionsno solutions

Now, we prove the results of this section.

###### Proof of Theorem 6.1.

We know that $\tau=-\kappa_{g}\eta$ and $\kappa_{g}$ is constant. To simplify
the notation, we set

$c=\kappa_{g}^{2}+\frac{\ell^{2}}{2}\quad(=|A|^{2})\,.$

We know from Lemma 4.2 that $\overline{\Delta}\tau=-\kappa_{g}c\eta$ and then
we deduce that

(6.5) $\overline{\Delta}^{r}\tau=-\kappa_{g}c^{r}\eta\,.$

Next, using Lemma 4.1, we obtain:

(6.6)
$\overline{\nabla}_{X}(\overline{\Delta}^{r}\tau)=-\kappa_{g}^{2}c^{r}X+\frac{\ell}{2}\kappa_{g}c^{r}E_{3}\,;\qquad\overline{\nabla}_{E_{3}}(\overline{\Delta}^{r}\tau)=\frac{\ell}{2}\kappa_{g}c^{r}X\,.$

We shall also need the following equalities which can easily by derived from
(3.5):

(6.7)
$\overline{R}(\eta,X)X=\left(4m-\frac{3\ell^{2}}{4}\right)\eta\,;\quad\overline{R}(\eta,E_{3})E_{3}=\frac{\ell^{2}}{4}\eta\,;\quad\overline{R}(X,\eta)E_{3}=0\,.$

Now, using (6.5), (6.6) and (6.7), we can compute the $r$-tension field whose
expression is given in (1), (1). We have:

$\displaystyle\tau_{2s}(\varphi)$ $\displaystyle=$
$\displaystyle\overline{\Delta}^{2s-1}\tau(\varphi)-R^{N}\left(\overline{\Delta}^{2s-2}\tau(\varphi),d\varphi(e_{i})\right)d\varphi(e_{i})$
$\displaystyle-\sum_{p=1}^{s-1}\,\left\\{R^{N}\left(\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s+p-2}\tau(\varphi),\overline{\Delta}^{s-p-1}\tau(\varphi)\right)d\varphi(e_{i})\right.$
$\displaystyle\qquad\qquad-\,\left.R^{N}\left(\overline{\Delta}^{s+p-2}\tau(\varphi),\nabla^{\varphi}_{e_{i}}\,\overline{\Delta}^{s-p-1}\tau(\varphi)\right)d\varphi(e_{i})\right\\}$
$\displaystyle=$
$\displaystyle-\kappa_{g}c^{2s-1}\eta+\kappa_{g}c^{2s-2}\Big{[}\overline{R}(\eta,X)X+\overline{R}(\eta,E_{3})E_{3}\Big{]}$
$\displaystyle-\sum_{p=1}^{s-1}\,\Big{\\{}\kappa_{g}^{3}c^{2s-3}\overline{R}(X,\eta)X-\frac{\ell}{2}\kappa_{g}^{2}c^{2s-3}\overline{R}(E_{3},\eta)X$
$\displaystyle\qquad\quad\quad-\frac{\ell}{2}\kappa_{g}^{2}c^{2s-3}\overline{R}(X,\eta)E_{3}-\kappa_{g}^{3}c^{2s-3}\overline{R}(\eta,X)X$
$\displaystyle\qquad\quad\quad+\frac{\ell}{2}\kappa_{g}^{2}c^{2s-3}\overline{R}(\eta,E_{3})X-\frac{\ell}{2}\kappa_{g}^{2}c^{2s-3}\overline{R}(\eta,X)E_{3}\Big{\\}}$
$\displaystyle=$
$\displaystyle-\kappa_{g}c^{2s-1}\eta+\kappa_{g}c^{2s-2}\Big{[}\left(4m-\frac{3\ell^{2}}{4}\right)\eta+\frac{\ell^{2}}{4}\eta\Big{]}$
$\displaystyle-c^{2s-3}\sum_{p=1}^{s-1}\,\Big{\\{}-\kappa_{g}^{3}\left(4m-\frac{3\ell^{2}}{4}\right)\eta-\kappa_{g}^{3}\left(4m-\frac{3\ell^{2}}{4}\right)\eta\Big{\\}}$
$\displaystyle=$
$\displaystyle-\kappa_{g}c^{2s-3}\Big{[}c^{2}-\left(4m-\frac{\ell^{2}}{2}\right)c-(2s-2)\left(4m-\frac{3\ell^{2}}{4}\right)\kappa_{g}^{2}\Big{]}\eta\,.$

It follows from the last equality that the Hopf cylinder $\Sigma$ is proper
$r$-harmonic ($r=2s$) if and only if

(6.8)
$c^{2}-\left(4m-\frac{\ell^{2}}{2}\right)c-(r-2)\left(4m-\frac{3\ell^{2}}{4}\right)\kappa_{g}^{2}=0\,.$

Finally, using $c=\kappa_{g}^{2}+(\ell^{2}/2)$ in (6.8) we obtain (6.1).

The computations for $\tau_{2s+1}$ are analogous and so we omit further
details. ∎

###### Proof of Corollary 6.3.

We will proceed by an accurate analysis of the dependence of the roots of
equation (6.1) on the values of the parameters $\ell$, $m$ and $r$. Putting
$x=k_{g}^{2}$ in (6.1), the existence of a proper $r$-harmonic CMC Hopf
cylinder is equivalent to the existence of a positive solution of the equation

(6.9)
$x^{2}+\left[-4m(r-1)+\frac{3\ell^{2}r}{4}\right]x-\frac{\ell^{2}}{2}(4m-\ell^{2})=0\,.$

We divide the analysis in a series of cases.

If $m\leq 0$, then the coefficients of (6.9) are nonnegative and thus there
exists no positive solution.

Thus, from now on, we assume that $m>0$. Then we have the following subcases:

  * (i)

If $4m-\ell^{2}>0$, then there exists a unique positive solution
$\kappa_{g}^{2}$ of (6.9) for all $r\geq 2$.

  * (ii)

If $4m-\ell^{2}=0$, then equation (6.9) admits the positive solution

$\kappa_{g}^{2}=\frac{r-4}{4}\,$

if and only if $r\geq 5$. We point out that this result is in accordance with
the discussion in Remark 1.3 of [14].

  * (iii)

If $4m-\ell^{2}<0$, then replacing $\overline{R}_{1212}=4m-{3\ell^{2}}/{4}$ in
(6.9) we obtain

(6.10)
$x^{2}+\left[4m-r\overline{R}_{1212}\right]x-\frac{\ell^{2}}{2}\overline{R}_{1212}+\frac{\ell^{4}}{8}=0\,.$

Thus, if $\overline{R}_{1212}\leq 0$ there exists no acceptable solution.
Finally, if $\overline{R}_{1212}=4m-{3\ell^{2}}/{4}>0$, then a straightforward
check shows that (6.9) admits two distinct positive solutions if and only if

(6.11)
$r>\frac{4\left(\sqrt{2}\sqrt{\ell^{4}-4\ell^{2}m}+4m\right)}{16m-3\ell^{2}}\,,$

and only one positive solution when

(6.12)
$r=\frac{4\left(\sqrt{2}\sqrt{\ell^{4}-4\ell^{2}m}+4m\right)}{16m-3\ell^{2}}\,.$

∎

######  Remark 6.5.

(i) We observe that, for all $r\geq 5$, there always exist suitable couples
$m,\ell$ such that (6.11), or (6.12), is verified. In the case that (6.11)
holds the corresponding two solutions give rise to two non-congruent
$r$-harmonic Hopf cylinders.

(ii) Concerning the case $4m-\ell^{2}>0$, if $r=2$ we find
$\kappa_{g}^{2}=4m-\ell^{2}$. On the other hand, Ou and Wang (see Theorem 1.3
(iii)) find that $\gamma$ must be a circle of radius

(6.13) $R=\frac{1}{\sqrt{8m-\ell^{2}}}$

in the base space
${\mathbb{S}}^{2}\left(\frac{1}{2\sqrt{m}}\right)=M^{2}(4m)$. In order to
check that our result is coherent with that of Ou and Wang, we observe that a
curve of constant curvature $\kappa_{g}$ in ${\mathbb{S}}^{2}(\rho)$ is a
plane curve in ${\mathbb{R}}^{3}$ with constant curvature $\kappa$ and radius

(6.14) $R=\frac{1}{\kappa}=\frac{\rho}{\sqrt{\kappa_{g}^{2}\rho^{2}+1}}\,.$

Substituting $\rho=1/(2\sqrt{m})$ and $\kappa_{g}^{2}=4m-\ell^{2}$ in (6.14)
we recover (6.13).

## References

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