Harmonic and biharmonic Riemannain submersions from Sol space

TL;DR

This paper classifies harmonic and biharmonic Riemannian submersions from Sol space to surfaces, proving their non-existence except when the base is a hyperbolic space form.

Contribution

It provides a complete classification and non-existence results for harmonic and biharmonic submersions from Sol space, highlighting the special case of hyperbolic base spaces.

Findings

No harmonic or biharmonic submersions from Sol space to any surface.

Existence of submersions only when the base is a hyperbolic space form.

Classification results for Riemannian submersions from Sol space.

Abstract

In this paper, we give a complete classification of harmonic and biharmonic Riemannian submersions $\pi:(R^3,g_{Sol})\to (N^2,h)$ from Sol space into a surface by proving that there is neither harmonic nor biharmonic Riemannian submersion $\pi:(R^3,g_{Sol})\to (N^2,h)$ from Sol space no matter what the base space $(N^2,h)$ is. We also prove that a Riemannian submersion $\pi:(R,g_{Sol})\to (N^2,h)$ from Sol space exists only when the base space is a hyperbolic space form.

Equations

$$E\left(\phi ,\Omega \right)=\frac{1}{2}{\int }_{\Omega }{\left|\mathrm{d}\phi \right|}^2\mathrm{d}x.$$

$$E^2\left(\phi ,\Omega \right)=\frac{1}{2}{\int }_{\Omega }{\left|\tau (\phi )\right|}^2\mathrm{d}x$$

$${\tau }^2(\phi ):={\mathrm{Trace}}_g({\nabla }^{\phi }{\nabla }^{\phi }-{\nabla }_{{\nabla }^M}^{\phi })\tau (\phi )-{\mathrm{Trace}}_g{R}^N(\mathrm{d}\phi ,\tau (\phi ))\mathrm{d}\phi =0,$$

$$R^N(X,Y)Z=[{\nabla }_X^N,{\nabla }_Y^N]Z-{\nabla }_{[X,Y]}^{N}Z.$$

$$E_1=e^{-z}\frac{\partial }{\partial x},E_2=e^z\frac{\partial }{\partial y},E_3=\frac{\partial }{\partial z}.$$

$$[E_1,E_2]=0,[E_2,E_3]=-E_2,[E_1,E_3]=E_1,$$

$$\begin{array}[]{lll}\nabla_{E_{1}}E_{1}=-E_{3},\;\nabla_{E_{1}}E_{3}=E_{1},\;\nabla_{E_{2}}E_{2}=E_{3},\;\nabla_{E_{2}}E_{3}=-E_{2},\\ {\rm all\;other}\;\nabla_{E_{i}}E_{j}=0,\;i,j=1,2,3.\\ \end{array}$$

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

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

$$\begin{array}[]{lll}R_{1212}=g(R(E_{1},E_{2})E_{2},E_{1})=1,\;R_{1313}=g(R(E_{1},E_{3})E_{3},E_{1})=-1,\\ R_{2323}=g(R(E_{2},E_{3})E_{3},E_{2})=-1,\;\;{\rm all\;other}\;R_{ijkl}=0,i,j,k,l=1,2,3.\\ \end{array}$$

$$[e_1,e_3]=f_3{e}_2+{\kappa }_1{e}_3,[e_2,e_3]=-f_3{e}_1+{\kappa }_2{e}_3,[e_1,e_2]=f_1{e}_1+f_2{e}_2-2\sigma e_3.$$

$${\nabla }_{e_1}{e}_1=-f_1{e}_2,{\nabla }_{e_1}{e}_2=f_1{e}_1-\sigma e_3,{\nabla }_{e_1}{e}_3=\sigma e_2,$$

$${\nabla }_{e_2}{e}_1=-f_2{e}_2+\sigma e_3,{\nabla }_{e_2}{e}_2=f_2{e}_1,{\nabla }_{e_2}{e}_3=-\sigma e_1,$$

$${\nabla }_{e_3}{e}_1=-{\kappa }_1{e}_3+(\sigma -f_3)e_2,{\nabla }_{e_3}{e}_2=-(\sigma -f_3)e_1-{\kappa }_2{e}_3,{\nabla }_{e_3}{e}_3={\kappa }_1{e}_1+{\kappa }_2{e}_2.$$

$$\{\begin{array}{l}{e}_3(f_1)+({\kappa }_1+f_2)f_3-e_1(f_3)=0,\\ e_3(f_2)+({\kappa }_2-f_1)f_3-e_2(f_3)=0,\\ 2e_3(\sigma )+{\kappa }_1{f}_1+{\kappa }_2{f}_2+e_2({\kappa }_1)-e_1({\kappa }_2)=0,\end{array}$$

$$\{\begin{array}{l}{R}^M(e_1,e_3,e_1,e_2)=-e_1(\sigma )+2{\kappa }_1\sigma =-2a_2^3{a}_3^3,\\ R^M(e_1,e_3,e_1,e_3)=e_1({\kappa }_1)+{\sigma }^2-{\kappa }_1^2+{\kappa }_2{f}_1=2(a_2^3{)}^2-1,\\ R^M(e_1,e_3,e_2,e_3)=e_1({\kappa }_2)-e_3(\sigma )-{\kappa }_1{f}_1-{\kappa }_1{\kappa }_2=-2a_1^3{a}_2^3,\\ R^M(e_1,e_2,e_1,e_2)=e_1(f_2)-e_2(f_1)-f_1^2-f_2^2+2f_3\sigma -3{\sigma }^2=2(a_3^3{)}^2-1,\\ R^M(e_1,e_2,e_2,e_3)=-e_2(\sigma )+2{\kappa }_2\sigma =2a_1^3{a}_3^3,\\ R^M(e_2,e_3,e_1,e_3)=e_2({\kappa }_1)+e_3(\sigma )+{\kappa }_2{f}_2-{\kappa }_1{\kappa }_2=-2a_1^3{a}_2^3,\\ R^M(e_2,e_3,e_2,e_3)={\sigma }^2+e_2({\kappa }_2)-{\kappa }_1{f}_2-{\kappa }_2^2=2(a_1^3{)}^2-1.\end{array}$$

$$K^N=e_1(f_2)-e_2(f_1)-f_1^2-f_2^2+2f_3\sigma ,$$

$$e_3(K^N)=e_3[e_1(f_2)-e_2(f_1)-f_1^2-f_2^2+2f_3\sigma ]=0.$$

$$K^N=e_1(f_2)-e_2(f_1)-f_1^2-f_2^2.$$

$$\{\begin{array}{l}-e_1(\sigma )=-2a_2^3{a}_3^3,\\ {\sigma }^2=2(a_2^3{)}^2-1,\\ -e_3(\sigma )=-2a_1^3{a}_2^3=0,\\ K^N=3{\sigma }^2+2(a_3^3{)}^2-1,\\ -e_2(\sigma )=2a_1^3{a}_3^3,\\ e_3(\sigma )=-2a_1^3{a}_2^3=0,\\ {\sigma }^2=2(a_1^3{)}^2-1.\end{array}$$

$$\{\begin{array}{l}-{\Delta }^M{\kappa }_1-2\sum _{i=1}^2{f}_i{e}_i({\kappa }_2)-{\kappa }_2\sum _{i=1}^2\left(e_i(f_i)-{\kappa }_i{f}_i\right)+{\kappa }_1\left(-K^N+\sum _{i=1}^2{f}_i^2\right)=0,\\ -{\Delta }^M{\kappa }_2+2\sum _{i=1}^2{f}_i{e}_i({\kappa }_1)+{\kappa }_1\sum _{i=1}^2(e_i(f_i)-{\kappa }_i{f}_i)+{\kappa }_2\left(-K^N+\sum _{i=1}^2{f}_i^2\right)=0\end{array}$$

$$\begin{array}[]{lll}a_{1}^{1}=0,\;a_{3}^{1}\neq\pm 1\;{\rm and}\;a_{2}^{1}\neq 0.\end{array}$$

$$f_1=0,{\nabla }_{e_1}{e}_1=0.$$

$$\begin{array}[]{lll}\nabla_{e_{1}}e_{1}=\nabla_{e_{1}}(\sum\limits_{i=1}^{3}a_{1}^{i}E_{i})=\sum\limits_{i=1}^{3}e_{1}(a_{1}^{i})E_{i}+\sum\limits_{i,j=1}^{3}a_{1}^{j}a_{1}^{i}\nabla_{E_{j}}E_{i}.\end{array}$$

$${\nabla }_{e_1}{e}_1=-f_1{e}_2=-f_1\sum _{i=1}^3{a}_2^i{E}_i.$$

$$\begin{array}[]{lll}-f_{1}a_{2}^{1}=\langle-f_{1}\sum\limits_{i=1}^{3}a_{1}^{i}E_{i},E_{1}\rangle=\langle\nabla_{e_{1}}e_{1},E_{1}\rangle\\ =\langle\sum\limits_{i=1}^{3}e_{1}(a_{1}^{i})E_{i}+\sum\limits_{i,j=1}^{3}a_{1}^{j}a_{1}^{i}\nabla_{E_{j}}E_{i},E_{1}\rangle=e_{1}(a_{1}^{1})=0,\end{array}$$

$$\{\begin{array}{l}{e}_1(a_1^2)=a_1^2{a}_1^3,\\ e_1(a_1^3)=-(a_1^2{)}^2,\\ e_1(a_2^1)=-\sigma a_3^1,\\ e_1(a_2^2)=a_1^2{a}_2^3-\sigma a_3^2,\\ e_1(a_2^3)=-a_1^2{a}_2^2-\sigma a_3^3,\\ e_1(a_3^1)=\sigma a_2^1,\\ e_1(a_3^2)=a_1^2{a}_3^3+\sigma a_2^2,\\ e_1(a_3^3)=-a_1^2{a}_3^2+\sigma a_2^3,\\ f_2{a}_2^1=-a_1^3{a}_2^1+\sigma a_3^1=-a_3^2+\sigma a_3^1,\\ e_2(a_3^1)=-a_2^1{a}_3^3,\\ e_2(a_2^1)=-a_2^1{a}_2^3,\\ {\kappa }_1{a}_3^1=(\sigma -f_3)a_2^1-a_1^3{a}_3^1=(\sigma -f_3)a_2^1+a_2^2.\text{par}\end{array}$$

$$a_1^1=f_1=f_2=0.$$

$$\{\begin{array}{l}-e_1(\sigma )+2{\kappa }_1\sigma =-2a_2^3{a}_3^3,\\ e_1({\kappa }_1)+{\sigma }^2-{\kappa }_1^2=2(a_2^3{)}^2-1,\\ e_1({\kappa }_2)-e_3(\sigma )-{\kappa }_1{\kappa }_2=-2a_1^3{a}_2^3,\\ 2f_3\sigma -3{\sigma }^2=2(a_3^3{)}^2-1,\\ -e_2(\sigma )+2{\kappa }_2\sigma =2a_1^3{a}_3^3,\\ e_2({\kappa }_1)+e_3(\sigma )-{\kappa }_1{\kappa }_2=-2a_1^3{a}_2^3,\\ {\sigma }^2+e_2({\kappa }_2)-{\kappa }_2^2=(2a_1^3{)}^2-1.\end{array}$$

$$\begin{array}[]{lll}a_{3}^{2}=\sigma a_{3}^{1}.\end{array}$$

$$\begin{array}[]{lll}e_{1}(\kappa_{1})a_{3}^{1}+\kappa_{1}e_{1}(a_{3}^{1})=e_{1}(\sigma)a_{2}^{1}+\sigma e_{1}(a_{2}^{1})-e_{1}(f_{3})a_{2}^{1}-f_{3}e_{1}(a_{2}^{1})+e_{1}(a_{2}^{2}),\end{array}$$

Full text

*Supported by the Natural Science Foundation of China (No. 11861022).

** Supported by a grant from the Simons Foundation ( 427231, Ye-Lin Ou).

Abstract

In this paper, we give a complete classification of harmonic and biharmonic Riemannian submersions \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space into a surface by proving that there is neither harmonic nor biharmonic Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space no matter what the base space $(N^{2},h)$ is. We also prove that a Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space exists only when the base space is a hyperbolic space form.

1. Introduction and Preliminaries

All manifolds, maps, tensor fields studied in this paper are assumed to be smooth unless there is an otherwise statement.

Recall that a harmonic map $\varphi:(M,g)\to(N,h)$ between Riemannian manifolds is a critical point of the energy functional

[TABLE]

The Euler-Lagrange equation is given by the vanishing of the tension filed $\tau(\varphi)={\rm Trace}_{g}\nabla{\rm d}\varphi$ (see [7]). Clearly, the map $\varphi$ is harmonic if and only if $\tau(\varphi)={\rm Trace}_{g}\nabla{\rm d}\varphi=0$ holds identically.

The study of biharmonic maps as a special case of $k$-polyharmonic maps were first proposed by J. Eells and L. Lemaire in [7]. A biharmonic map $\varphi:(M,g)\to(N,h)$ between Riemannian manifolds is a critical point of the bienergy

[TABLE]

for every compact subset $\Omega$ of $M$, where $\tau(\varphi)={\rm Trace}_{g}\nabla{\rm d}\varphi$ is the tension field of $\varphi$. Jiang [9] first computed the first variation of the functional to see that $\varphi$ is biharmonic if and only if its bitension field vanishes identically, i.e.,

[TABLE]

where $R^{N}$ is the curvature operator of $(N,h)$ defined by

[TABLE]

We call a submanifold that is a biharmonic submanifold if the isometric immersion that defines the submanifold is a biharmonic map. Analogously, a Riemannian submersion is called a biharmonic (respectively, harmonic) Riemannian submersion if the Riemannian submersion is a biharmonic (respectively, harmonic) map. Obviously, any harmonic map is always biharmonic whilst biharmonic maps include harmonic maps as special cases. We use proper biharmonic maps (respectively, submanifolds, Riemannian submersion) to call those biharmonic maps that are not harmonic maps.

For harmonicity of Riemannian submersions, one of our motivations is that the definition of Riemannian submersions, in a sense, are considered as the dual notion of isometric immersions (i.e., submanifolds). There are many interesting examples of harmonic isometric immersions of a surface (i.e., minimal surfaces) into 3-manifolds, such as planes or catenoid in \mbox{{\mathbb{R}}}^{3} or harmonic embedding of $S^{2}$ into $S^{3}$ [16]. On the other hand, there exist many interesting examples and classification results of harmonic Riemannian submersions from 3-dimensional Riemannian manifolds into a surface: Hopf fibration $\pi:S^{3}\to S^{2}(4)$ and the orthogonal projection \pi:\mbox{{\mathbb{R}}}^{3}\to\mbox{{\mathbb{R}}}^{2} are harmonic Riemannian submersion; there is no harmonic Riemannian submersion $\pi:H^{3}\to(N^{2},h)$ no matter what $(N^{2},h)$ is (see [20, 24]); harmonic Riemannian submersions from Thurston’s 3-dimensional geometries, 3-dimensional BCV spaces and a Berger sphere $S_{\varepsilon}^{3}$ have been completely classified and many explicit constructions of harmonic Riemannian submersions were given (see [24] for details).

Since biharmonic maps are considered as the generalizations of harmonic maps and include harmonic maps as a subset, it would be very interesting to study biharmonicity of Riemannian submersions. Based on this, we will study biharmonicity of Riemannian submersions from 3-dimensional Sol space into a surface in the second part of the paper. Biharmonic Riemannian submersions were first studied by Oniciuc in [13]. In [20], the authors first introduced so-called integrability data and then used the main tool to obtain a complete classification of biharmonic Riemannian submersions from a 3-dimensional space form into a surface. In [1], the authors studied biharmonicity of a general Riemannian submersion and obtained biharmonic equations for Riemannian submersions with one-dimensional fibers and Riemannian submersions with basic mean curvature vector fields of fibers, and they first used the so-called integrability data to study biharmonic Riemannian submersions from $(n+1)$-dimensional spaces with one-dimensional fibers. In [18], the author studied biharmonicity a more general setting of Riemannian submersion with a $S^{1}$ fiber over a compact Riemannian manifold. In [8] , the authors studied generalized harmonic morphisms and obtained many examples of biharmonic Riemannian submersions which are maps between Riemannian manifolds that pull back local harmonic functions to local biharmonic functions.

In addition to these, we refer the readers to the following classification results. In 2023, the authors [21] classified all proper biharmonic Riemannian submersions from BCV 3-diemnsional spaces into a surface. In a recent paper [22], the authors also gave complete classifications of biharmonic Riemannian submersions from 3-dimensional Berger sphere. And also, biharmonic Riemannian submersions from product spaces M^{2}\times\mbox{{\mathbb{R}}} to a surface have been completely classified in [23].

Recall that Sol space is one of Thurston’s eight models of 3-dimensional geometry. It is the Riemannian manifold (\mbox{{\mathbb{R}}}^{3},g_{Sol}), where the metric can be described by $g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2}$ with respect to Euclidean coordinates on \mbox{{\mathbb{R}}}^{3}.

First of all, one observes that it is easy to find Riemannian submersions from Sol space. For example, the projections

\pi_{1}:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(\mbox{{\mathbb{R}}}^{2},e^{2z}{\rm d}x^{2}+{\rm d}z^{2}),\;\pi_{1}(x,y,z)=(x,z), and

\pi_{2}:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(\mbox{{\mathbb{R}}}^{2},e^{2z}{\rm d}y^{2}+{\rm d}z^{2}),\;\pi_{2}(x,y,z)=(y,z) are both Riemannian submersions.

One may wonder whether these are harmonic or biharmonic, whether there is any harmonic or biharmonic Riemannian submersions from Sol space. In this paper, we prove that there is neither harmonic nor biharmonic Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space no matter what the base space $(N^{2},h)$ is. We also prove that a Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space exists only when the base space is a hyperbolic space form.

2. Harmonic Riemannian submersions from Sol space

In this section, we obtain a nonexistence classification results for harmonic Riemannian submersions from Sol space to a surface.

Let (\mbox{{\mathbb{R}}}^{3},g_{Sol}) denote Sol space, where the metric $g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2}$ with respect to local coordinates $(x,y,z)$ in \mbox{{\mathbb{R}}}^{3}. We have a defined orthonormal basis as

[TABLE]

With respect to this orthonormal frame, the Lie brackets and the Levi-Civita connection are given by:

[TABLE]

[TABLE]

One adopts the following notation and sign convention for Riemannian curvature operator.

[TABLE]

the Riemannian and the Ricci curvature tensors are given by

[TABLE]

A straightforward computation gives

[TABLE]

Let \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) be a Riemannian submersion from Sol space with an orthonormal frame $\{e_{1},\;e_{2},\;e_{3}\}$ on (\mbox{{\mathbb{R}}}^{3},g_{Sol}) and $e_{3}$ being vertical. By a treatment similar to to those used treating Remark 1 in [21], we then have the following (7)–(13) (see [21] for details)

[TABLE]

where $f_{1},f_{2},f_{3},\;\kappa_{1},\;\kappa_{2}\;{\rm and}\;\sigma$ are the (generalizd) integrability data of the Riemannian submersion $\pi$. When $f_{3}=0$, the horizontal distribution $\{e_{1},\;e_{2}\}$ are basic and $\{f_{1},f_{2},\kappa_{1},\;\kappa_{2},\;\sigma\}$ is the integrability data of the adapted frame.

The Levi-Civita connection is given by

[TABLE]

Denoting by $e_{i}=\sum\limits_{j=1}^{3}a_{i}^{j}E_{j},\;i=1,2,3$, using (3), (6) and (8), then the Jacobi identity applied to the frame $\{e_{1},e_{2},e_{3}\}$ gives

[TABLE]

and the terms of the curvature tension as follows

[TABLE]

Gauss curvature of the base space as

[TABLE]

[TABLE]

When $f_{3}=0$, then Gauss curvature of the base space becomes

[TABLE]

Now we are ready to give the following classification of harmonic Riemannian submersions from Sol space.

Proposition 2.1**.**

(see[21]) A Riemannian submersion $\pi:(M^{3},g)\to(N^{2},h)$ is harmonic if and only if  $\nabla^{M}_{e_{3}}e_{3}=0$, i.e., $\kappa_{1}=\kappa_{2}=0$.

Theorem 2.2**.**

There exists no harmonic Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space no matter what $(N^{2},h)$ is.

Proof.

Let \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) be a Riemannian submersion with an orthonormal frame $\{e_{1},\;e_{2},\;e_{3}\}$, $e_{3}$ being vertical, and the (generalized) integrability data $\{f_{1},f_{2},f_{3},\kappa_{1},\kappa_{2},\sigma\}$. By Proposition 2.1, the Riemannian submersion $\pi$ is harmonic if and only if $\kappa_{1}=\kappa_{2}=0$. Using (10) and Proposition 2.2 in [24], we obtain

[TABLE]

where $K^{N}=e_{1}(f_{2})-e_{2}(f_{1})-f_{1}^{2}-f_{2}^{2}+2f_{3}\sigma$.

Comparing the 2nd equation with the 7th equation of (14), we have $(a_{1}^{3})^{2}=(a_{2}^{3})^{2}$. However, using the 3rd equation of (14), we get $a_{1}^{3}a_{2}^{3}=0$ and hence $a_{1}^{3}=a_{2}^{3}=0$. We substitute this into the 2nd equation of (14) to have $\sigma^{2}=-1$, a contradiction.

From which we obtain the theorem.

∎

3. Biharmonic Riemannian submersions from Sol space

We state the following proposition ([20]) which will be later used in the rest of the paper.

Proposition 3.1**.**

(see [20]) Let $\pi:(M^{3},g)\to(N^{2},h)$ be a Riemannaian submersion with the adapted frame $\{e_{1},\;e_{2},\;e_{3}\}$ and the integrability data $f_{1},f_{2},\kappa_{1},\;\kappa_{2}\;{\rm and}\;\sigma$. Then, the Riemannaian submersion $\pi$ is biharmonic if and only if

[TABLE]

where $K^{N}=R^{N}_{1212}\circ\pi=e_{1}(f_{2})-e_{2}(f_{1})-f_{1}^{2}-f_{2}^{2}$ is Gauss curvature of Riemannian manifold $(N^{2},h)$.

The following proposition was found in [21].

Proposition 3.2**.**

(see [21]) Let $\pi:(M^{3},g)\longrightarrow(N^{2},h)$ be a Riemannian submersion from 3-manifolds with an orthonormal frame $\{e_{1},e_{2},e_{3}\}$ and $e_{3}$ being vertical. If $\nabla_{e_{1}}e_{1}=0$, then either $\nabla_{e_{2}}e_{2}=0$,  or  $\nabla_{e_{2}}e_{2}\not\equiv 0$, and the frame $\{e_{1},e_{2},e_{3}\}$ is an adapted frame.

We will prove the important conclusion used proving our main theorem

Theorem 3.3**.**

Let \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2})\to(N^{2},h) be a Riemannian submersion. Then, we have such an adapted frame $\{e_{1}=a_{1}^{2}E_{2}+a_{1}^{3}E_{3},\;e_{2},\;e_{3}\}$ of the Riemannian submersion $\pi$ with $e_{3}$ being vertical. Moreover, if $E_{1}$ is not vertical, then $\nabla_{e_{2}}e_{2}\neq 0$, i.e., $f_{2}\neq 0$.

Proof.

Obviously, if $E_{1}$ is tangent to the fiber of the Riemannian submersion $\pi$ , then any basic field is of the form $e=a^{2}E_{2}+b^{2}E_{3},\;{\rm and}\;a^{2}+b^{2}=1$.

From this time on, we only need to suppose that $E_{1}$ is not vertical, i.e., $e_{3}\neq\pm E_{1}$. Then, the vector filed $e_{1}=e_{3}\times E_{1}$ is horizontal and hence $\langle e_{1},E_{1}\rangle=0$. From this, we have a defined orthonormal frame $\{e_{1},\;e_{2}=e_{3}\times e_{1},\;e_{3}\}$ on $M^{3}$. If denoting by $e_{i}=\sum\limits_{j=1}^{3}a_{i}^{j}E_{j},i=1,2,3$, together with $\langle e_{1},E_{1}\rangle=0$, then $e_{1}$ is expressed as the form $e_{1}=a_{1}^{2}E_{2}+a_{1}^{3}E_{3}$ and hence  $(a_{1}^{2})^{2}+(a_{1}^{3})^{2}=1$. From these, we have the following

[TABLE]

One can further check the following equalities as

[TABLE]

By a direct computation, we get

[TABLE]

However, using (8), the above has another expression as

[TABLE]

By equating (18) and (19) and comparing the coefficient of $E_{1}$, we obatin

[TABLE]

which has been used (3) and $a_{1}^{1}=0$. This leads to $f_{1}=0$ for $a_{2}^{1}\neq 0$, and hence (17) holds.

Applying (3), (8) and $a_{1}^{1}=f_{1}=0$ and a further computation similar to those used calculating (18)–(20) gives

[TABLE]

Since $\nabla_{e_{1}}e_{1}=0$, we conclude from Proposition 3.2 to have either $\nabla_{e_{2}}e_{2}\neq 0$, and the frame $\{e_{1},e_{2},e_{3}\}$ is adapted to the Riemannian submersion $\pi$; or  $\nabla_{e_{2}}e_{2}=0$. Now, we just need to consider the latter case $\nabla_{e_{2}}e_{2}=0$, i.e., $f_{2}=0$. From these, one has the following

[TABLE]

Then, (10) turns into

[TABLE]

We now show that the latter case (i.e., $a_{1}^{1}=f_{1}=f_{2}=0$, $a_{3}^{1}\neq\pm 1$ and $a_{2}^{1}\neq 0$) can not happen by considering the following two cases:

Case I: $a_{3}^{1}=0,\;f_{2}=0$. In this case, since $a_{1}^{1}=0$, we have $a_{2}^{1}=\pm 1$ and hence $a_{2}^{3}=0$. By the 9th equation of (21), one easily sees that $a_{3}^{2}=a_{3}^{1}=0$ and hence $a_{3}^{3}=\pm 1$. This leads to $a_{1}^{3}=0$ and $a_{1}^{2}=\pm 1$. Substituting this into the 6th equation of (21), we have $\sigma=0$. However, we substitute $\sigma=0$ and $a_{3}^{3}=\pm 1$ into the 4th equation of (23) to find $0=1$, a contradiction.

Case II: $a_{3}^{1}\neq 0,\;\pm 1$ and $f_{2}=0$. In this case, since $a_{1}^{1}=0$, we then have $a_{2}^{1}\neq 0,\pm 1$. Substitute $f_{2}=0$ into the 9th equation of (21) to have

[TABLE]

Applying $e_{1}$ to both sides the 12th equation of (21), we get

[TABLE]

which can be rewritten as

[TABLE]

Using the 3rd, the 4th, the 6th equation of (21), the 1st, the 2nd equation of (23) and the 1st equation of (9), a straightforward computation gives

[TABLE]

One substitutes the 12th equation of (21) and (24) into (26), together with $a_{2}^{1}a_{2}^{3}+a_{3}^{1}a_{3}^{3}=0$ and $a_{2}^{2}=a_{1}^{3}a_{3}^{1}$, to compute the following

[TABLE]

the last equality holds by using the fact $a_{2}^{1}a_{1}^{3}=a_{3}^{2}$, $a_{3}^{2}=\sigma a_{3}^{1}$ and $a_{3}^{1}=a_{1}^{2}a_{2}^{3}-a_{1}^{3}a_{2}^{2}$.

Since $a_{3}^{1}\neq 0$, then (28) becomes

[TABLE]

Substituting the 4th equation of (23) into (29), together with $(a_{1}^{3})^{2}+(a_{2}^{3})^{2}+(a_{3}^{3})^{2}=1$, and simplifying the resulting equation, we get

[TABLE]

This implies

[TABLE]

Since $\sigma a_{3}^{1}=a_{3}^{2}$, the above equation is equivalent to

[TABLE]

or,

[TABLE]

On the other hand, let $\theta$, $\alpha$ denote angles between $e_{1}$ and $E_{2}$, between $e_{3}$ and $E_{1}$, respectively, since $a_{1}^{1}=0$, we have

[TABLE]

where $a_{1}^{2}=\cos\theta,\;a_{1}^{3}=\sin\theta,\;a_{2}^{1}=\sin\alpha,\;a_{2}^{2}=-\sin\theta\cos\alpha,\;a_{2}^{3}=\cos\theta\cos\alpha,\;a_{3}^{1}=\cos\alpha,\;a_{3}^{2}=\sin\theta\sin\alpha$ and $a_{3}^{3}=-\cos\theta\sin\alpha$.

Using the 1st and the 3rd equation of (21), it is not difficult to check the following

[TABLE]

Since $a_{2}^{3}=\cos\theta\cos\alpha,\;a_{3}^{1}=\cos\alpha$ and $\;a_{3}^{3}=-\cos\theta\sin\alpha$, then (33) turns into

[TABLE]

Note that $\theta$ is angle between $e_{1}$ and $E_{2}$, but $\alpha$ angle between $e_{3}$ and $E_{1}$, then the two functions: $\cos^{2}\theta$,  $2\cos^{4}\alpha+\sin^{2}\alpha$  are linearly independent. Then, Eq. (36) implies taht $\theta$ and $\alpha$ have to be constants, and hence $\cos\theta=0$ since (35). Substituting this into Eq. (36). we have $-1=0$, a contradiction.

Summarizing all results in the above cases, the theorem follows.

∎

Remark 1*.*

Let \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2})\to(N^{2},h) be a Riemannian submersion with $e_{3}$ being vertical. If $e_{3}\neq\pm E_{1}$, i.e., $a_{3}^{1}\neq\pm 1$, by Theorem 3.3, one can choose such an adapted frame $\{e_{1}=a_{1}^{2}E_{2}+a_{1}^{3}E_{3},\;e_{2},\;e_{3}\}$ to $\pi$ and $f_{2}\neq 0$. From these, the case corresponding to $a_{1}^{1}=f_{1}=f_{3}=0$, $f_{2}\neq 0$, $a_{3}^{1}\neq\pm 1$ and $a_{2}^{1}\neq 0$. Clearly, this implies that the case $a_{3}^{1}\neq\pm 1$, and $a_{1}^{1}=f_{1}=f_{2}=0$ can not happen.

Theorem 3.4**.**

A Riemannian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2})\to(N^{2},h) from Sol space exists only in (\mbox{{\mathbb{R}}}^{3},g_{Sol})\to H^{2} with Gauss curvature of the base space $K^{N}=-1$.

Proof.

Let \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2})\to(N^{2},h) be a Riemannian submersion with $e_{3}$ being vertical. For the above notations and signs, we just need to consider the two cases $a_{3}^{1}=0$ or $a_{3}^{3}=\pm 1$. We use the proof by contradiction to obtain the theorem. We now assume $a_{3}^{1}\neq 0$ and $a_{3}^{3}\neq\pm 1$. It follows from Theorem 3.3 and Remark 1 that there exists such an adapted frame $\{e_{1}=a_{1}^{2}E_{2}+a_{1}^{3}E_{3},\;e_{2},\;e_{3}\}$ to $\pi$, and hence the following hold

[TABLE]

Then, (10) becomes as

[TABLE]

We apply $e_{1}$ to both sides the 12th equation of (21), together with $f_{3}=0$, to get

[TABLE]

Using the 3rd, the 4th, the 6th equation of (21), the 1st, the 2nd equation of (38) and the 1st equation of (9), a straightforward computation gives

[TABLE]

Substituting the 12th equation of (21) into (39), together with $a_{2}^{1}a_{2}^{3}+a_{3}^{1}a_{3}^{3}=0$, $f_{3}=0$ and $a_{2}^{2}=a_{1}^{3}a_{3}^{1}$, a direct computation gives

[TABLE]

the last equality holds for using the fact $a_{2}^{1}a_{1}^{3}=a_{3}^{2}$ and $a_{3}^{1}=a_{1}^{2}a_{2}^{3}-a_{1}^{3}a_{2}^{2}$.

Since $a_{1}^{1}=0$, one can assume that

[TABLE]

where $a_{1}^{2}=\cos\theta,\;a_{1}^{3}=\sin\theta,\;a_{2}^{1}=\sin\alpha,\;a_{2}^{2}=-\sin\theta\cos\alpha,\;a_{2}^{3}=\cos\theta\cos\alpha,\;a_{3}^{1}=\cos\alpha,\;a_{3}^{2}=\sin\theta\sin\alpha$ and $a_{3}^{3}=-\cos\theta\sin\alpha$.

We applying the 1st and the 3rd equation of (21) to see that

[TABLE]

Since $a_{1}^{3}=\sin\theta,\;a_{3}^{1}=\cos\alpha$ and $\;a_{3}^{2}=\sin\theta\sin\alpha$, then (41) becomes

[TABLE]

One solves the above equation to obtain

[TABLE]

where $\varphi(\alpha)=\frac{-\sin\alpha\pm\sqrt{1+7\cos^{2}\alpha}}{\cos\alpha}.$

Substituting $a_{2}^{1}=\sin\alpha,\;a_{2}^{2}=-\sin\theta\cos\alpha$, $a_{3}^{1}=\cos\alpha$, $f_{3}=0$ and (45) into the 12th equation of (21), we have

[TABLE]

where denote by $\psi(\alpha)=-\cos\alpha+\sin\alpha\varphi(\alpha)$ and $\varphi(\alpha)=\frac{-\sin\alpha\pm\sqrt{1+7\cos^{2}\alpha}}{\cos\alpha}.$

We substitute (45) and (46) into the 1st equation of (38) to have

[TABLE]

where $\psi(\alpha)=-\cos\alpha+\sin\alpha\varphi(\alpha)(\alpha)$ and $\varphi(\alpha)=\frac{-\sin\alpha\pm\sqrt{1+7\cos^{2}\alpha}}{\cos\alpha}.$

On the other hand, substitute (45) into the left-hand side of (47) to compute as

[TABLE]

the last holds by using (43).

Comparing (47) with (48), we deduce

[TABLE]

Solving the above equation, we obtain a contradiction. Indeed, if $\theta$ is a constant, by the 2nd equation of (43), we have $\cos\theta=0$ and $\sin\theta=\pm 1$. Substituting this into the above equation we have

[TABLE]

this implies $\alpha$ being constant and hence $\sigma=0$ together with the 1st equation of (43). Hence, using (44), we get $a_{3}^{1}=\cos\alpha=0$  since  $\sin\theta=\pm 1$, a contradiction.

If $\theta\neq{\rm constant}$, but the two functions $\sin^{2}\theta$,  $\cos^{2}\theta$ are linearly independent, then (49) means

[TABLE]

where $\psi(\alpha)=-\cos\alpha+\sin\alpha\varphi(\alpha)$ and $\varphi(\alpha)=\frac{-\sin\alpha\pm\sqrt{1+7\cos^{2}\alpha}}{\cos\alpha}.$

Clearly, the second equation of the above equation implies $\alpha$ being a constant and hence $\sigma=0$ together with the 1st equation of (43). Moreover, by (44) and $a_{3}^{1}=\cos\alpha\neq 0$, one finds that $\sin\theta=0$ and hence $\theta$ is a constant contradicting the assumption $\theta\neq{\rm constant}$. From these, when $a_{3}^{1}\neq\pm 1,0$, there is no a Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space no matter what $(N^{2},h)$ is.

In addition, if $a_{3}^{1}=0$, we have $a_{2}^{1}=\pm 1$ since $a_{1}^{1}=0$, in this case, a straightforward computation similar to those used computing Case II in Theorem 3.5 gives $f_{1}=f_{3}=0$, $f_{2}=-1$ and hence Gauss curvature of the base space $K^{N}=e_{1}(f_{2})-f_{2}^{2}=-1$; if $a_{3}^{1}=\pm 1$, we have $a_{2}^{1}=0$ since $a_{1}^{1}=0$, in this case, a direct calculation similar to those used calculating Case I in Theorem 3.5 gives $f_{1}=0$, $f_{2}=1$ and hence Gauss curvature of the base space $K^{N}=e_{1}(f_{2})-f_{2}^{2}=-1$. Clearly, This implies that the a Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) exists only in (\mbox{{\mathbb{R}}}^{3},g_{Sol})\to H^{2} with Gauss curvature of the base space $K^{N}=-1$.

From which we obtain the theorem.

∎

Theorem 3.5**.**

There exists no biharmonic Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) no matter what $(N^{2},h)$ is.

Proof.

Let $\nabla$ denote the Levi-Civita connection on Sol space (\mbox{{\mathbb{R}}}^{3},g_{Sol}) with an orthonormal frame $\{e_{1},\;e_{2},\;e_{3}\}$ and $e_{3}$ being vertical. We denote by $e_{i}=\sum\limits_{j=1}^{3}a_{i}^{j}E_{j},\;i=1,2,3$. To complete the proof of the theorem, from Theorem 3.4, we only discuss biharmonicity of a Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to H^{2}. Furthermore, from the proof of Theorem 3.4, we only need to consider the two cases $a_{3}^{1}=\pm 1$ or $a_{3}^{1}=0$.

Case I: $a_{3}^{1}=\pm 1$. In this case, one sees that $e_{3}=\pm E_{1}$ and hence take an orthogonal frame $\{e_{1}=E_{3},\;e_{2}=E_{2},\;e_{3}=-E_{1}\}$ on (\mbox{{\mathbb{R}}}^{3},g_{Sol}) with $e_{3}$ being vertical. A direct computation using (2) and (3) gives

[TABLE]

It follows that the ( generalized) integrability data $f_{1}=f_{3}=\kappa_{2}=\sigma=0,\kappa_{1}=-f_{2}=-1$ and hence $\{e_{1}=E_{3},\;e_{2}=E_{2},\;e_{3}=E_{1}\}$ is actually adapted to $\pi$ with $e_{3}$ being vertical. Then, biharmonic equation (15) reduces to

[TABLE]

However, the left-hand term of (52) can be computed as

[TABLE]

Therefore, the Riemannian submersion $\pi$ is not biharmonic in this case.

Case II: $a_{3}^{1}=0.$ In this case, we have $a_{2}^{1}=\pm 1$ since $a_{1}^{1}=0$. Then, we can take an orthonormal frame $\{e_{1}=E_{3},\;e_{2}=E_{1},,\;e_{3}=E_{2}\}$ with $e_{3}$ being vertical. A direct computation using (2) and (3) gives

[TABLE]

This follows that the ( generalized) integrability data $f_{1}=f_{3}=\kappa_{2}=\sigma=0,\kappa_{1}=-f_{2}=1$ and hence $\{e_{1}=E_{3},\;e_{2}=E_{1},,\;e_{3}=E_{2}\}$ becomes adapted to $\pi$ with $e_{3}$ being vertical. Substituting this into biharmonic equation (15) and a direct computation, we have

[TABLE]

which is a contradiction. Thus, the Riemannian submersion $\pi$ is not biharmonic.

Summarizing all results in the above cases we obtain the theorem. ∎

Remark 2*.*

We would like to point out that, with respect to local coordinates, a Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to H^{2} can be locally expressed as the following (up to equivalence):

$(a)$: the Riemannaian submersion

[TABLE]

or,

$(b)$: the Riemannaian submersion

[TABLE]

By Theorem 2.2 and Theorem 3.5, these Riemannaian submersions are neither harmonic nor biharmonic.

As a consequence of Theorem 2.2 and Theorem 3.5, we state the following fact

Corollary 3.6**.**

Any Riemannaian submersion \pi:(\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) from Sol space to a surface is neither harmonic nor biharmonic.

Although there is no (harmonic) biharmonic Riemannaian submersion from Sol space to a surface, there exist many (harmonic) biharmonic maps (\mbox{{\mathbb{R}}}^{3},g_{Sol})\to(N^{2},h) which are not Riemannaian submersions.

Example 1*.*

The maps \phi:(\mbox{{\mathbb{R}}}^{3},g_{Sol}=e^{2z}{\rm d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm d}z^{2})\to(\mbox{{\mathbb{R}}}^{2},du^{2}+dv^{2}),

$\phi(x,y,z)=(u,v)=(y,Az^{3}+Bz^{2}+Cz+D)$ are biharmonic, where $A,B,C,D$ are constants. In particular, when $A^{2}+B^{2}>0$, this family of maps are proper biharmonic. Note that these maps are not Riemannaian submersions.