Biharmonic Riemannian Submersions from a Three-Dimensional Non-Flat Torus

Abstract

In this paper, we study Riemannian submersions from a three-dimensional non-flat torus $T^2\times S^1$ to a surface and their biharmonicity. In local coordinates, a complete characterization of such Riemannian submersions is provided. Based on this result, it is proven that a Riemannian submersion from this torus to a surface is biharmonic if and only if it is harmonic, and such a map is, up to an isometry, the projection onto the first factor $T^2$ followed by a Riemannian covering map. As a by-product, we also prove that the Riemannian submersion $\varphi :([T^2\times S^1]\setminus \{t=\pi \},{r^2(1+cost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)\to (N^2,h)$ from a cuspidal torus to a surface is proper biharmonic if and only if, up to an isometry, it is the projection map $[T^2\times S^1]\setminus \{t=\pi \}\to S^1\setminus \{t=\pi \}\times S^1$.

1. Introduction and Preliminaries

Let $(M,g)$ and $(N,h)$ be two Riemannian manifolds. A smooth map $\phi :(M,g)\to (N,h)$ is called a harmonic map if its tension field vanishes identically (see e.g., [1,2]):

$$\tau \left(\phi \right):={\mathrm{Trace}}_g\nabla \mathrm{d}\phi \equiv 0.$$

Variationally, a harmonic map is a critical point of the energy functional:

$$E\left(\phi \right)={\displaystyle \frac{1}{2}}{\int }_M{\left|\mathrm{d}\phi \right|}^2\mathrm{d}{v}_g.$$

A natural generalization is the concept of biharmonic maps, which arise as the critical points of the bienergy functional:

$$E_2\left(\phi \right)={\displaystyle \frac{1}{2}}{\int }_M{\left|\tau \left(\phi \right)\right|}^2\mathrm{d}{v}_g.$$

The corresponding Euler–Lagrange equation gives the biharmonic map equation (see e.g., [3]):

$${\tau }_2\left(\phi \right):={\mathrm{Trace}}_g({\nabla }^{\phi }{\nabla }^{\phi }-{\nabla }_{{\nabla }^M}^{\phi })\tau \left(\phi \right)-{\mathrm{Trace}}_g{R}^N(\mathrm{d}\phi ,\tau \left(\phi \right))\mathrm{d}\phi =0,$$

(1)

where $R^N$ denotes the curvature operator of $(N,h)$, defined by

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

This implies that $\phi$ is biharmonic if and only if its bitension field ${\tau }_2\left(\phi \right)$ vanishes identically. Since $\tau \left(\phi \right)\equiv 0$ implies ${\tau }_2\left(\phi \right)\equiv 0$, any harmonic map is automatically biharmonic. Those with ${\tau }_2\left(\phi \right)\equiv 0$ but $\tau \left(\phi \right)\ne 0$ are called proper biharmonic maps.

The geometric study of biharmonic maps primarily examines the biharmonicity of geometrically significant maps, particularly isometric immersions and Riemannian submersions. For some results and recent progress on biharmonic isometric immersions (i.e., biharmonic submanifolds), see [4,5,6,7,8,9,10,11] and the extensive references therein.

Riemannian submersions serve as the natural dual to isometric immersions (equivalently, submanifolds). The investigation of their biharmonicity was initiated in [12]. In [13], the so-called integrability data were introduced as a useful tool to study biharmonic Riemannian submersions from generic 3-manifolds. Later in [14], this was generalized to higher dimensions with one-dimensional fibers. Complete classifications of biharmonic Riemannian submersions from three-dimensional space forms and three-dimensional BCV spaces into a surface were established in [13,15]. Recently, in [16], some local characterizations of biharmonic Riemannian submersions from the product space $M^2\times \mathbb{R}$ into a surface were obtained, where the authors first constructed a special adapted frame for the Riemannian submersion and then studied biharmonic Riemannian submersions from $M^2\times \mathbb{R}$.

One of the aims of this paper is to investigate biharmonic Riemannian submersions from a three-dimensional non-flat torus $T^2\times S^1$. We emphasize that our approach first characterizes Riemannian submersions from this space and then examines their biharmonicity, which differs methodologically from that in [16].

Recall that a three-dimensional non-flat torus is topologically equivalent to the Cartesian product of three circles $S^1\times S^1\times S^1$. In local coordinates, a three-dimensional non-flat torus $T^2\times S^1$ can be represented as an isometric immersion into a Euclidean space ${\mathbb{R}}^5$:

$$\left\{\begin{array}{c}{x}_1=(R+rcost)cosu,\hfill \\ x_2=(R+rcost)sinu,\hfill \\ x_3=rsint,\hfill \\ x_4=r^{\prime }cosv,\hfill \\ x_5=r^{\prime }sinv,\hfill \end{array}\right.$$

where $u,t,v\in [0,2\pi )$ are parameters, and the positive constants $R,r$ and $r^{\prime }$ are radii with $R>r$. By a direct calculation, the induced metric g on this torus can be derived as

$$g={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2.$$

Here, the radii R, r and $r^{\prime }$ determine the size and shape of the torus. We note that the three-dimensional non-flat torus can be used for the study of curvature theory in differential geometry and has important applications in theoretical physics, particularly in cosmology (see e.g., [17,18,19]).

From now on, when no confusion arises, we denote the three-dimensional torus as $(T^2\times S^1,g)$ or simply $T^2\times S^1$. Similarly, we denote the two-dimensional torus $T^2$ by $(T^2,g_{T^2})=\left(T^2,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2\right)$.

Remark 1. 

Consider a hypersurface into a Euclidean space ${\mathbb{R}}^5$ parametrized by

$$\left\{\begin{array}{c}{x}_1=r(1+cost)cosu,\hfill \\ x_2=r(1+cost)sinu,\hfill \\ x_3=rsint,\hfill \\ x_4=r^{\prime }cosv,\hfill \\ x_5=r^{\prime }sinv,\hfill \end{array}\right.$$

where $u,t,v\in [0,2\pi )$ and $t\ne \pi$ are parameters, and the positive constants r and $r^{\prime }$ are radii. The induced metric g on this hypersurface is given by

$$g_{\pi }=r^2{(1+cost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2.$$

We call the hypersurface a cuspidal torus, denoted by $\left([T^2\times S^1]\setminus \{t=\pi \},g_{\pi }\right)$ or $[T^2\times S^1]\setminus \{t=\pi \}$. It can also be viewed as the standard torus $(T^2\times S^1,g)$ with radius $R=r$ minus the singular point $\{t=\pi \}$. Herein, the non-flat torus refers to a smooth and non-singular torus, excluding the cuspidal torus.

In this paper, we study Riemannian submersions from a three-dimensional non-flat torus $(T^2\times S^1,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)$ to a surface and their biharmonicity. In local coordinates, a complete classification and characterization of such Riemannian submersions is provided (Theorem 1). Based on this result, it is proven that a Riemannian submersion from this torus to a surface is biharmonic if and only if it is harmonic, and such a map is, up to an isometry, the projection onto the first factor $(T^2,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2)$ followed by a Riemannian covering map (Theorem 2). As a by-product, we also prove that the Riemannian submersion $\varphi :([T^2\times S^1]\setminus \{t=\pi \},{r^2(1+cost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)\to (N^2,h)$ from a cuspidal torus to a surface is proper biharmonic if and only if, up to an isometry, it is the projection map $\varphi :([T^2\times S^1]\setminus \{t=\pi \},g_{\pi })\to (S^1\setminus \{t=\pi \}\times S^1,r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)$, $\varphi (u,t,v)=(t,v)$ (Proposition 2).

2. Riemannian Submersions from T² × S¹

Consider the three-dimensional non-flat torus $T^2\times S^1$ with the metric

$$g={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2,$$

where $u,t,v\in [0,2\pi )$ are parameters, and the constants $R,r$, $r^{\prime }$ are radii satisfying $R>r>0$ and $r^{\prime }>0$.

The orthonormal frame on $T^2\times S^1$ is given by

$$E_1={\displaystyle \frac{1}{R+rcost}}{\displaystyle \frac{\partial }{\partial u}},E_2={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial t}},E_3={\displaystyle \frac{1}{r^{\prime }}}{\displaystyle \frac{\partial }{\partial v}}.$$

(2)

A direct computation yields the Lie bracket products and the Levi-Civita connection of $T^2\times S^1$:

$$\begin{array}{c}[E_1,E_2]={\displaystyle \frac{-sint}{R+rcost}}{E}_1,{\nabla }_{E_1}{E}_1={\displaystyle \frac{sint}{R+rcost}}{E}_2,{\nabla }_{E_1}{E}_2=-{\displaystyle \frac{sint}{R+rcost}}{E}_1,\hfill \\ \mathrm{all}\mathrm{other}[E_i,E_j]=0,{\nabla }_{E_i}{E}_j=0,i,j=1,2,3.\hfill \end{array}$$

(3)

The only possible nonzero components of the Riemannian curvature of $T^3$ are

$$\begin{array}{c}{R}_{1212}=R(E_1,E_2,E_1,E_2)=-E_2\left(f\right)-f^2={\displaystyle \frac{cost}{r(R+rcost)}},\hfill \end{array}$$

(4)

where $f=-{\displaystyle \frac{sint}{R+rcost}}$.

For the two-dimensional non-flat torus $(T^2,g_0={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2)$, the Gauss curvature is

$$\begin{array}{c}{K}^{T^2}=R(E_1,E_2,E_1,E_2)={\displaystyle \frac{cost}{r(R+rcost)}}.\hfill \end{array}$$

(5)

Hereafter, throughout the paper, the frame $\{E_1,E_2,E_3\}$ on $T^2\times S^1$ will be consistently used to denote that defined by (2).

Let $\pi :(T^2\times S^1,g)\to (N^2,h)$ be a Riemannian submersion equipped with a local orthonormal frame $\{e_1,e_2,e_3\}$, where $e_3$ is vertical (i.e., $d\pi \left(e_3\right)=0$).

By Proposition 2.2 in [13], the Lie brackets of $T^2\times S^1$ in terms of the orthonormal frame $\{e_1,e_2,e_3\}$ are given by

$$[e_1,e_3]=f_3{e}_2+k_1{e}_3,[e_2,e_3]=-f_3{e}_1+k_2{e}_3,[e_1,e_2]=f_1{e}_1+f_2{e}_2-2\sigma e_3.$$

Here, $f_1,f_2,f_3,k_1,k_2$ and $\sigma$ are called the generalized integrability data associated with this frame. The frame is adapted to $\pi$ if and only if $f_3=0$, and the components of the tension field of the Riemannian submersion with respect to the frame $\{e_1,e_2,e_3\}$ are ${\kappa }_1$ and ${\kappa }_2$ (see [13], Proposition 2.2, for details).

By Remark 1 in [13], the Levi-Civita connection of $T^2\times S^1$ in terms of the orthonormal frame $\{e_1,e_2,e_3\}$ is given by

$$\left\{\begin{array}{c}{\nabla }_{e_1}{e}_1=-f_1{e}_2,{\nabla }_{e_1}{e}_2=f_1{e}_1-\sigma e_3,{\nabla }_{e_1}{e}_3=\sigma e_2,\hfill \\ {\nabla }_{e_2}{e}_1=-f_2{e}_2+\sigma e_3,{\nabla }_{e_2}{e}_2=f_2{e}_1,{\nabla }_{e_2}{e}_3=-\sigma e_1,\hfill \\ {\nabla }_{e_3}{e}_1=-{\kappa }_1{e}_3+(\sigma -f_3)e_2,{\nabla }_{e_3}{e}_2=-(\sigma -f_3)e_1-{\kappa }_2{e}_3,{\nabla }_{e_3}{e}_3=-{\kappa }_1{e}_1+{\kappa }_2{e}_2.\hfill \end{array}\right.$$

(6)

Let $e_i={\displaystyle \sum _{j=1}^3}{a}_i^j{E}_j$ (for $i=1,2,3$) be the transformation relations between the frames $\left\{e_i\right\}$ and $\left\{E_i\right\}$. Here, the frame $\{E_1,E_2,E_3\}$ of $T^2\times S^1$ is defined by (2), and hence, $\left(a_i^j\right)\in SO\left(3\right)$. We use Equation (5) and Equation (12) in [13] to obtain the only possible nonzero components of the Riemannian curvature R:

$$\left\{\begin{array}{c}R(e_1,e_3,e_1,e_2)=-e_1\left(\sigma \right)+2{\kappa }_1\sigma =-a_2^3{a}_3^3{K}^{T^2},\hfill \\ R(e_1,e_3,e_1,e_3)=e_1\left({\kappa }_1\right)+{\sigma }^2-{\kappa }_1^2+{\kappa }_2{f}_1={\left(a_2^3\right)}^2{K}^{T^2},\hfill \\ R(e_1,e_3,e_2,e_3)=e_1\left({\kappa }_2\right)-e_3\left(\sigma \right)-{\kappa }_1{f}_1-{\kappa }_1{\kappa }_2=-a_1^3{a}_2^3{K}^{T^2},\hfill \\ R(e_1,e_2,e_1,e_2)=e_1\left(f_2\right)-e_2\left(f_1\right)-f_1^2-f_2^2+2f_3\sigma -3{\sigma }^2={\left(a_3^3\right)}^2{K}^{T^2},\hfill \\ R(e_1,e_2,e_2,e_3)=-e_2\left(\sigma \right)+2{\kappa }_2\sigma =a_1^3{a}_3^3{K}^{T^2},\hfill \\ R(e_2,e_3,e_1,e_3)=e_2\left({\kappa }_1\right)+e_3\left(\sigma \right)+{\kappa }_2{f}_2-{\kappa }_1{\kappa }_2=-a_1^3{a}_2^3{K}^{T^2},\hfill \\ R(e_2,e_3,e_2,e_3)={\sigma }^2+e_2\left({\kappa }_2\right)-{\kappa }_1{f}_2-{\kappa }_2^2={\left(a_1^3\right)}^2{K}^{T^2},\hfill \end{array}\right.$$

(7)

where $K^{T^2}={\displaystyle \frac{cost}{r(R+rcost)}}$ is the Gauss curvature of $(T^2,g_{T^2})$.

Finally, by the 4th equation of (7) and the O’Neill Curvature Formula for a Riemannian submersion [20], we conclude that the Gauss curvature of the base space is given by (see also [13], Remark 1, for details)

$$K^N=R_{1212}^N\circ \pi =e_1\left(f_2\right)-e_2\left(f_1\right)-f_1^2-f_2^2+2f_3\sigma .$$

(8)

Remark 2. 

Recall that $f_3=0$ if and only if the orthonormal frame $\left\{e_1,e_2,e_3\right\}$ on $(T^2\times S^1,g)$ is adapted to the Riemannian submersion. Moreover, when $f_3=0$, it follows that $e_3\left(f_1\right)=e_3\left(f_2\right)=0$ (see [15]). We note that $K^N$ is constant along each fiber of π, which is equivalent to $e_3\left(K^N\right)=e_3(e_1\left(f_2\right)-e_2\left(f_1\right)-f_1^2-f_2^2+2f_3\sigma )=0$ (see [13]).

By Proposition 2.5 and Theorem 2.6 in [16], one can easily see that there exists a special frame adapted to a Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ as follows.

Lemma 1. 

Let $\pi :(T^2\times S^1,g)\to (N^2,h)$ be a Riemannian submersion from a three-dimensional non-flat torus to a surface.Then, there exists an orthonormal frame adapted to this Riemannian submersion as

$$\left\{\begin{array}{cc}\hfill & e_1=cos\theta E_1+sin\theta E_2,\hfill \\ \hfill & e_2=-cos\alpha (sin\theta E_1-cos\theta E_2)+sin\alpha E_3,\hfill \\ \hfill & e_3=sin\alpha (sin\theta E_1-cos\theta E_2)+cos\alpha E_3\hfill \end{array}\right.$$

(9)

such that

$$\left\{\begin{array}{cc}\hfill & f_1=0,f_2=-\overline{f}{cos}^2\alpha ,f_3=-E_3\left(\theta \right)=0,\sigma =-\overline{f}sin\alpha cos\alpha ,\hfill \\ \hfill & {\kappa }_1=-\overline{f}{sin}^2\alpha ,{\kappa }_2=-e_3\left(\alpha \right),\hfill \end{array}\right.$$

(10)

where $cos\theta =⟨e_1,E_1⟩$, $cos\alpha =⟨e_3,E_3⟩$, $\overline{f}=-sin\theta E_1\left(\theta \right)+cos\theta E_2\left(\theta \right)+fsin\theta$, $f=-{\displaystyle \frac{sint}{R+rcost}}$. Furthermore, θ depends only on u and t, i.e., $\theta =\theta (u,t)$.

For the frame determined by (9), we have the following.

Lemma 2. 

Let $\pi :(T^2\times S^1,g)\to (N^2,h)$ be a Riemannian submersion with the orthonormal frame determined by (9). Then, we have

$$\begin{array}{c}{e}_1\left(\alpha \right)=-\sigma =\overline{f}sin\alpha cos\alpha ,e_2\left(\alpha \right)=0,\hfill \\ e_1\left(\theta \right)=fcos\theta ,\theta =arccos\left({\displaystyle \frac{c}{R+rcost}}\right),\hfill \\ \overline{f}=-sin\theta E_1\left(\theta \right)+cos\theta E_2\left(\theta \right)+fsin\theta ,f=-{\displaystyle \frac{sint}{R+rcost}},\hfill \end{array}$$

(11)

where c is a constant along the vector field $e_1$ with $\left|c\right|<R-r$.

Proof. 

First, a straightforward computation using (11) gives

$$-f_1{e}_2={\nabla }_{e_1}{e}_1={\nabla }_{e_1}\left(\sum _{i=1}^3{a}_1^i{E}_i\right)=\sum _{i=1}^3{e}_1\left(a_1^i\right)E_i+\sum _{i,j=1}^3{a}_1^j{a}_1^i{\nabla }_{E_j}{E}_i,$$

which, together with $f_1=0$, implies

$$e_1\left(a_1^1\right)=-a_1^1{a}_1^{2}f.$$

Recall that $f=-{\displaystyle \frac{sint}{R+rcost}}$, $a_1^1=cos\theta$ and $sin\theta =a_1^2$. Therefore, we get

$$e_1\left(\theta \right)=fcos\theta =-{\displaystyle \frac{sint}{R+rcost}}cos\theta .$$

(12)

Similarly, the first line of (11) can be obtained.

On the other hand, we observe that applying the Liouville formula (see e.g., [21]) to the surface $(T^2,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2)$, Equation (13) coincides with the geodesic equation on the surface $T^2$ given by

$${\displaystyle \frac{d\theta }{ds}}=-{\displaystyle \frac{sint}{R+rcost}}cos\theta ,$$

or equivalently (see e.g., [21]),

$${\displaystyle \frac{sin\theta d\theta }{cos\theta dt}}=-{\displaystyle \frac{rsint}{R+rcost}}.$$

(13)

Solving the ordinary differential Equation (13), we obtain

$$\theta =arccos{\displaystyle \frac{c}{R+rcost}},$$

where c is a constant along $e_1$ with $\left|c\right|<R-r$.

Since $\overline{f}=-sin\theta E_1\left(\theta \right)+cos\theta E_2\left(\theta \right)+fsin\theta$, we have the remaining terms of (11).

Combining all results above, we obtain the lemma. □

We now state one of our main results.

Theorem 1. 

Let $\pi :(T^2\times S^1,g={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)\to (N^2,h)$ be a Riemannian submersion from the standard torus. Then,

(a) The base space is non-flat. Up to an isometry, either

(i) π is the projection $T^2\times S^1\to T^2$ followed by a Riemannian covering;

(ii) one can choose local coordinates $(\psi ,t)$ on $(N^2,h)$ such that the map can be represented as

$$\begin{array}{c}\pi :\left(T^2\times S^1,g\right)\to \left(N^2,{\displaystyle \frac{c_1^2{(R+rcost)}^2}{{c_1^2(R+rcost)}^2+1}}d{\psi }^2+r^{2}d{t}^2\right),\pi (u,t,v)=(r^{\prime }v-{\displaystyle \frac{u}{c_1}},t),\hfill \end{array}$$

where $c_1\in \mathbb{R}\setminus \left\{0\right\}$.

(b) The base space is flat, up to an isometry, one can choose local coordinates $(t,v)$ on $(N^2,h)$ such that the map can be represented as

$$\begin{array}{c}\pi :(T^2\times S^1,g)\to (S^1\times S^1,r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2),\pi (u,t,v)=(t,v).\hfill \end{array}$$

Proof. 

We prove the theorem by discussing the following three cases:

Case I: $cos\alpha =a_3^3\equiv \pm 1$. In this case, we observe that $E_3$ is vertical, and $e_1=E_2$, $e_2=E_1$ are basic. It follows from Theorem 2.2 in [16] that locally, the Riemannian submersion $\pi$ is, up to an isometry, the projection defined by

$$(T^2\times S^1,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)\to (T^2,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2)$$

followed by a Riemannian covering map, and it is harmonic.

Case II: $a_3^3=cos\alpha \equiv 0$.

In this case, by (7), (9) and (10), we have $f_1=f_2=f_3={\kappa }_2=\sigma =0$ and ${\kappa }_1=-\overline{f}\ne 0$. The adapted frame takes the form $\{e_1=cos\theta E_1+sin\theta E_2,e_2=E_3,e_3=sin\theta E_1-cos\theta E_2\}$, where $\theta =arccos\left({\displaystyle \frac{c}{R+rcost}}\right)$ and c is a constant with $\left|c\right|<R-r$ (by Lemma 2).

Therefore, the Gauss curvature of the base surface $K^N=R(e_1,e_2,e_1,e_2)=0$ (by (8)), and $e_1$ is a geodesic vector field on $(T^2,g_{T^2}={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2)$ since ${\nabla }_{e_1}{e}_1=f_1{e}_1=0$.

Let $\gamma :I\to (T^2,g_{T^2})$ be an arbitrary integral curve of the vector field $e_1$ parameterized by arc length $y_1$. Thus, $\gamma$ is a geodesic in $T^2$. It is a well-established result that one can choose local semi-geodesic coordinates $(x_1,y_1)$ on the 2-torus $T^2$ such that the coordinate frame satisfies ${\overline{E}}_1=e^{-p(x_1,y_1)}{\partial }_{x_1}=e_3=sin\theta E_1-cos\theta E_2,{\overline{E}}_2={\partial }_{y_1}=e_1=cos\theta E_1+sin\theta E_2,$ where the metric on $T^2$ admits the expression $g_{T^2}=e^{2p(x_1,y_1)}d{x}_1^2+d{y}_1^2$ (see e.g., [21], Proposition 4; [16], Lemma 2.1). By Lemma 2.1 in [16] and Lemma 2, we have $p_{y_1}={\displaystyle \frac{\partial p}{\partial y_1}}=-{\kappa }_1=\overline{f}$ and the Gaussian curvature of $T^2$ is $K^{T^2}=-p_{y_1{y}_1}-p_{y_1}^2={\displaystyle \frac{cost}{r(R+rcost)}},$ with $y_1$ related to t via the transformation $y_1=\int {\displaystyle \frac{r(R+rcost)}{\sqrt{{(R+rcost)}^2-c^2}}}dt$ (i.e., $d{y}_1={\displaystyle \frac{r(R+rcost)}{\sqrt{{(R+rcost)}^2-c^2}}}dt$), where c is a constant along $e_1={\partial }_{y_1}$ with $\left|c\right|<R-r$. Further computations show that $p(x_1,y_1)=ln\sqrt{{(R+rcost)}^2-c^2}$ and c is a constant, where t is understood as a function of $y_1$ induced by the inverse of the above integral transformation.

Since the target surface is flat (by $K^N=R(e_1,e_2,e_1,e_2)=0$) and $f_3=0$, the frame $\{e_1={\overline{E}}_2={\partial }_{y_1},e_2=E_3={\displaystyle \frac{1}{r^{\prime }}}{\partial }_v,e_3={\overline{E}}_1=e^{-p(x_1,y_1)}{\partial }_{x_1}\}$ is an adapted frame to the Riemannian submersion $\pi$ with the integrability data $\{f_1=f_2={\kappa }_2=\sigma =0,{\kappa }_1=-\overline{f}\ne 0\}$.

Note that since $\sigma =0$, the horizontal distribution of the Riemannian submersion $\pi$ is integrable with flat integral submanifolds. Therefore, locally, up to an isometry of the domain and/or the codomain manifold, the Riemannian submersion $\pi$ is the projection along the fibers which are the integral curves of ${\overline{E}}_1=e^{-p(x_1,y_1)}{\partial }_{x_1}$ onto the integral submanifold and hence can be described by

$$(T^2\times S^1,e^{2p(x_1,y_1)}d{x}_1^2+d{y}_1^2+r^{\prime 2}d{v}^2)$$

$$\to (S^1\times S^1,d{y}_1^2+r^{\prime 2}d{v}^2),\pi (x_1,y_1,v)=(y_1,v).$$

Note also that $g_{T^2}=e^{2p(x_1,y_1)}d{x}_1^2+d{y}_1^2={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2$. Then, the Riemannian submersion can also be described by

$$\pi :(T^2\times S^1,{(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+r^{\prime 2}d{v}^2)\to$$

$$(S^1\times S^1,{\displaystyle \frac{r^2{(R+rcost)}^2}{{(R+rcost)}^2-c^2}}d{t}^2+r^{\prime 2}d{v}^2),\pi (u,t,v)=(t,v).$$

Since $d\pi \left(e_3\right)=0$ (where $e_3$ is vertical), it follows that $c=0$.

Case III: $a_3^3=cos\alpha \ne 0,\pm 1$ and $f_2\ne 0$. This, together with (10), implies $sin\alpha cos\alpha \ne 0$ and ${\kappa }_1=-\overline{f}{sin}^2\alpha \ne 0$. From the 5th and 7th equations of (7) and (10), we obtain

$$e_2\left({\kappa }_2\right)={\kappa }_2^2, e_2\left(\sigma \right)=2{\kappa }_2\sigma .$$

Using $e_2\left(\alpha \right)=0$ and $\overline{f}=-{\displaystyle \frac{\sigma }{sin\alpha cos\alpha }}={\displaystyle \frac{e_1\left(\alpha \right)}{sin\alpha cos\alpha }}$, we derive $e_2\left(f_2\right)=2{\kappa }_2{f}_2$ and $e_2\left(\overline{f}\right)=2{\kappa }_2\overline{f}$. This shows that $\alpha$ and ${\kappa }_2$ depend only on the variable t. Therefore, applying the identity ${\kappa }_2{e}_3\left(f_2\right)=[e_2,e_3]\left(f_2\right)=e_2{e}_3\left(f_2\right)-e_3{e}_2\left(f_2\right)$ and the condition $f_3=0$, we deduce that $e_3\left(f_2\right)=0$ (by Remark 2) and $e_3{e}_2\left(f_2\right)=e_3\left(2{\kappa }_2{f}_2\right)=0$, and hence $e_3\left({\kappa }_2\right)=0$.

Note that $e_3=sin\alpha (sin\theta E_1-cos\theta E_2)+cos\alpha E_3$, $e_2\left({\kappa }_2\right)={\kappa }_2^2$ and ${\kappa }_2$ depends only on t. Together with (9), (10) and (11), it follows that ${\kappa }_2=0$ and $\theta ={\displaystyle \frac{\pi }{2}}$, which implies $c=0$. Combining these, we can conclude that

$$e_2\left(\overline{f}\right)=e_2\left({\kappa }_1\right)=e_2\left(\sigma \right)=e_3\left(\alpha \right)=e_2\left(f_2\right)=0,$$

$$e_2\left(K^{T^2}\right)=e_3\left(K^{T^2}\right)=e_3\left(\overline{f}\right)=e_3\left({\kappa }_1\right)=e_3\left(\sigma \right)=0$$

and the frame adapted to the Riemannian submersion has the form

$$e_1=E_2, e_2=-cos\alpha E_1+sin\alpha E_3, e_3=sin\alpha E_1+cos\alpha E_3.$$

Moreover, by (11), we have $e_1\left(\alpha \right)={\displaystyle \frac{{\alpha }^{\prime }}{r}}=-{\displaystyle \frac{sint}{R+rcost}}sin\alpha cos\alpha$, which yields

$$tan\alpha \left(t\right)=c_1(R+rcost),$$

(14)

where $c_1$ is a nonzero constant.

Since $e_1=E_2={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial t}}$ is a basic vector field with ${\nabla }_{e_1}{e}_1=0$, there exists a tangent vector field ${\epsilon }_1$ on $(N^2,h)$ whose integral curves are geodesics such that $d\pi \left(e_1\right)={\epsilon }_1$. According to the differential geometry theory of surfaces (see e.g., [21]), we can take a local semi-geodesic coordinate system $(t,\varphi )$ on $N^2$ such that the metric has the form $h=r^{2}d{t}^2+e^{2\lambda (t,\varphi )}d{\varphi }^2$ with the orthonormal frame ${\epsilon }_1={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial t}}$, ${\epsilon }_2=e^{-\lambda }{\displaystyle \frac{\partial }{\partial \varphi }}$ in the base space satisfying $d\pi \left(e_1\right)={\epsilon }_1$ and $d\pi \left(e_2\right)={\epsilon }_2=e^{-\lambda }{\displaystyle \frac{\partial }{\partial \varphi }}$.

A direct computation gives $F_1\circ \pi {\epsilon }_1+F_2\circ \pi {\epsilon }_2=[{\epsilon }_1,{\epsilon }_2]=-{\displaystyle \frac{1}{r}}{\lambda }_t{\epsilon }_2$, yielding

$${\displaystyle \frac{1}{r}}{\lambda }_t=-F_2\circ \pi =-f_2={\displaystyle \frac{cos\alpha e_1\left(\alpha \right)}{sin\alpha }}={\displaystyle \frac{cos\alpha {\alpha }^{\prime }\left(t\right)}{rsin\alpha }}.$$

This implies that $\lambda (t,\varphi )=ln\left|sin\alpha \left(t\right)\right|+w\left(\varphi \right)$, where $w=w\left(\varphi \right)$ is a function of $\varphi$. Under the coordinate transformation $t=t,\psi =\int e^{w\left(\varphi \right)}d\varphi$ in the base space (where $\psi$ is independent of t), the metric of the base space takes the form $h=r^{2}d{t}^2+e^{2u}d{\psi }^2$ with $u\left(t\right)=ln\left|sin\alpha \left(t\right)\right|$. So locally, up to an isometry, the Riemannian submersion $\pi$ can be expressed as

$$\pi :(T^2\times S^1,g)\to (N^2,h=r^{2}d{t}^2+{sin}^2\alpha \left(t\right)d{\psi }^2), \pi (u,t,v)=(t,\psi (u,v\left)\right)$$

where $\psi =\psi (u,v)$ is a function to be determined.

Using Equation (14), the frame $e_1={\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial t}}$, $e_2=-cos\alpha E_1+sin\alpha E_3$, $e_3=sin\alpha E_1+cos\alpha E_3$, and the relations $d\pi (e_1$$)={\epsilon }_1$, $d\pi \left(e_2\right)={\epsilon }_2$, $d\pi \left(e_3\right)=0$, we obtain the PDE system:

$$\begin{array}{c}-c_1{cos}^2\alpha {\psi }_u+{\displaystyle \frac{{sin}^2\alpha }{r^{\prime }}}{\psi }_v=1,c_{1}cos\alpha {\psi }_u+{\displaystyle \frac{cos\alpha }{r^{\prime }}}{\psi }_v=0.\hfill \end{array}$$

(15)

Solving (15) via the method of first integrals gives

$$\begin{array}{c}\psi (u,v)=\mathsf{ϝ}(c_1{r}^{\prime }v-u),\hfill \end{array}$$

(16)

where $\mathsf{ϝ}$ is a nonconstant differentiable function.

Substituting this into the 2nd PDE of (15), we deduce that

$$\begin{array}{c}\psi (u,v)=r^{\prime }v-{\displaystyle \frac{u}{c_1}}.\hfill \end{array}$$

(17)

By Equation (14), we also have $sin\alpha ={\displaystyle \frac{c_1(R+rcost)}{\sqrt{{c_1^2(R+rcost)}^2+1}}}$.

Combining all cases, the theorem follows. □

Remark 3. 

(1) Although local coordinates are used in the proof, Theorem 1 is globally valid: the Riemannian submersion π and metric g are well-defined on the entire non-flat torus $T^2\times S^1$.

(2) Note that the base space $(N^2,h={\displaystyle \frac{c_1^2{(R+rcost)}^2}{{c_1^2(R+rcost)}^2+1}}d{\psi }^2+r^{2}d{t}^2)$ (with $c_1\in \mathbb{R}\setminus \left\{0\right\}$) is a non-standard torus. By Statement (a)(ii) of Theorem 1, the Riemannian submersion $\pi :\left(T^2\times S^1,g\right)\to \left(N^2,h\right)$ with $\pi (u,t,v)=(r^{\prime }v-{\displaystyle \frac{u}{c_1}},t)$ is a map from the three-dimensional non-flat torus onto this non-standard torus.

By an argument analogous to the proof of Theorem 1, we obtain the following:

Proposition 1. 

Let $\varphi :([T^2\times S^1]\setminus \{t=\pi \}, g_{\pi }={r^2(1+cost)}^{2}d{u}^2+r^{2}d{t}^2+r^{\prime 2}d{v}^2)\to (N^2,h)$ be a Riemannian submersion from the cuspidal torus. Then,

(a) The base space is non-flat. Up to an isometry, either

(i) the map is the projection onto the first factor $[T^2\times S^1]\setminus \{t=\pi \}\to T^2\setminus \{t=\pi \}$ followed by a Riemannian covering map;

(ii) one can take local coordinates $(\psi ,t)$ on $(N^2,h)$ such that the map can be expressed as

$$\begin{array}{c}\varphi :\left([T^2\times S^1]\setminus \{t=\pi \},g_{\pi }\right)\to \left(N^2\setminus \{t=\pi \},{\displaystyle \frac{c_1^2{r}^2{(1+cost)}^2}{1+c_1^2{r}^2{(1+cost)}^2}}d{\psi }^2+r^{2}d{t}^2\right),\hfill \\ \varphi (u,t,v)=(r^{\prime }v-{\displaystyle \frac{u}{c_1}},t),\mathrm{where}{c}_1\in \mathbb{R}\setminus \left\{0\right\}.\hfill \end{array}$$

(b) The base space is flat, and up to an isometry, one can take local coordinates $(t,v)$ on $(N^2,h)$ such that the projection can be expressed as

$$\begin{array}{c}\varphi :([T^2\times S^1]\setminus \{t=\pi \},g_{\pi })\to (S^1\setminus \{t=\pi \}\times S^1,r^{2}d{t}^2+r^{\prime 2}d{v}^2),\varphi (u,t,v)=(t,v).\hfill \end{array}$$

Proof. 

For the statement (a), by analogy with Theorem 1 (a) and setting $R=r$, the conclusion follows immediately.

For the statement (b), analogous to Theorem 1 (b) with $R=r$, the submersion structure is preserved. Due to the global smoothness of $[T^2\times S^1]\setminus \{t=\pi \}$ on $t\in [0,2\pi )$$\setminus \left\{\pi \right\}$, we must have $c=0$. Thus, the base metric simplifies to $r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2$, and the conclusion follows. □

3. Biharmonic Riemannian Submersions from ${\mathit{T}}^{\mathbf{2}}\times {\mathit{S}}^{\mathbf{1}}$

In this section, we investigate biharmonicity of the Riemannian Submersions $(T^2\times S^1,g)\to (N^2,h)$ from a three-dimensional non-flat torus onto a surface by applying Theorem 1 obtained in Section 2.

Let $\pi :(T^2\times S^1,g)\to (N^2,h)$ be a Riemannian submersion with an orthonormal frame $\left\{e_1,e_2,e_3\right\}$, where $e_3$ is vertical. Let $\{f_1,f_2,f_3,k_1,k_2,\sigma \}$ denote the generalized integrability data associated to this frame.

Remark 4. 

(i) It is well known from [15] that π is harmonic if and only if the integrability data ${\kappa }_1={\kappa }_2=0$.

(ii) For a Riemannian submersion $\pi :(M^3,g)\to (N^2,h)$ with the adapted frame $\{e_1,e_2,e_3\}$ and the integrability data $\{f_1,f_2,{\kappa }_1,{\kappa }_2\sigma \}$, Proposition 2.1 in [15] states that π is biharmonic if and only if

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

(18)

where $K^N=R_{1212}^N\circ \pi =e_1\left(f_2\right)-e_2\left(f_1\right)-f_1^2-f_2^2$ denotes the Gauss curvature of Riemannian manifold $(N^2,h)$.

We are ready to state another main result.

Theorem 2. 

Let $\pi :(T^2\times S^1,g={(R+rcost)}^{2}d{u}^2+r^{2}d{t}^2+{r^{\prime }}^{2}d{v}^2)\to (N^2,h)$ be a Riemannian submersion from a three-dimensional non-flat torus to a surface. Then, π is biharmonic if and only if it is harmonic, which holds if and only if π is the projection onto the first factor $T^2\times S^1\to T^2$ followed by a Riemannian covering map.

Proof. 

By Lemma 1, we consider a Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ such that the adapted frame $\{e_1,e_2,e_3\}$ takes the form (9) and the integrability data $\{f_1,f_2,{\kappa }_1,{\kappa }_2\sigma \}$ is given by (10).

To prove the theorem, we examine three cases that correspond to those in the proof of Theorem 1.

We note first that by Remark 4, the Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ is harmonic if and only if ${\kappa }_1={\kappa }_2=0$. We verify that ${\kappa }_1={\kappa }_2=0$ if and only if Case I in Theorem 1 occurs, which establishes the statement $\left(i\right)$.

We now study biharmonicity of the Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ by considering the remaining two cases (Case II and Case III) in the proof of Theorem 1.

By Lemma 1, Lemma 2 and Case II in the proof of Theorem 1, we have

$$cos\alpha =0,f_1=f_2=f_3={\kappa }_2=\sigma =K^N=c=0,{\kappa }_1={\displaystyle \frac{sint}{R+rcost}}.$$

(19)

Substituting this into (18) and simplifying the resulting equation, we conclude that the Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ is biharmonic if and only if

$$\Delta {\kappa }_1=\Delta \left({\displaystyle \frac{sint}{R+rcost}}\right)=0.$$

(20)

However, a straightforward computation gives

$$\begin{array}{c}\Delta \left({\displaystyle \frac{sint}{R+rcost}}\right)={\displaystyle \frac{1}{r^2}}{\left({\displaystyle \frac{sint}{R+rcost}}\right)}^{″}-{\displaystyle \frac{1}{r}}\left({\displaystyle \frac{sint}{R+rcost}}\right){\left({\displaystyle \frac{sint}{R+rcost}}\right)}^{\prime }\hfill \\ ={\displaystyle \frac{(r^2-R^2)sint}{r^2{(R+rcost)}^3}}\not\equiv 0(\mathrm{since}R>r>0),\hfill \end{array}$$

(21)

contradicting the requirement, where $sin\theta ={\displaystyle \frac{\sqrt{{(R+rcost)}^2-c^2}}{(R+rcost)}}=1$ (with $c=0$). This shows that the case cannot occur.

Similarly, applying Lemma 1, Lemma 2 and Case III in the proof of Theorem 1 yields

$$\begin{array}{c}\alpha =arctan\left[c_1(R+rcost)\right],sin\alpha ={\displaystyle \frac{c_1(R+rcost)}{\sqrt{{c_1^2(R+rcost)}^2+1}}},cos\alpha ={\displaystyle \frac{1}{\sqrt{{c_1^2(R+rcost)}^2+1}}},\hfill \\ cos\theta =f_1=f_3={\kappa }_2=0,f_2=-{\displaystyle \frac{{\alpha }^{\prime }cos\alpha }{rsin\alpha }},\sigma =-{\displaystyle \frac{{\alpha }^{\prime }}{r}},\hfill \\ K^N={\displaystyle \frac{1}{r}}{\left(f_2\right)}^{\prime }-f_2^2,{\kappa }_1=-{\displaystyle \frac{{\alpha }^{\prime }sin\alpha }{rcos\alpha }}.\hfill \end{array}$$

(22)

By Remark 4 and (22), we deduce that the Riemannian submersion $\pi :(T^2\times S^1,g)\to (N^2,h)$ is biharmonic if and only if

$$\Delta {\kappa }_1-{\kappa }_1\left\{-K^N+f_2^2\right\}=0,$$

(23)

which is equivalent to

$$\begin{array}{c}{\alpha }^{‴}sin\alpha {cos}^2\alpha +cos\alpha ({sin}^2\alpha +3){\alpha }^{\prime }{\alpha }^{″}+sin\alpha (2{cos}^2\alpha +3){{\alpha }^{\prime }}^3=0.\hfill \end{array}$$

(24)

For $\alpha =arctan\left[c_1(R+rcost)\right]$ (with $c_1\in \mathbb{R}\setminus \left\{0\right\}$), a straightforward computation shows it does not satisfy Equation (24), yielding a contradiction. Thus, the case can not occur.

Summarizing all the above results, we prove the theorem. □

We follow the proof of Theorem 2 replacing R with r to have the following.

Proposition 2. 

Let $\varphi :([T^2\times S^1]\setminus \{t=\pi \},g_{\pi }={r^2(1+cost)}^{2}d{u}^2+r^{2}d{t}^2+r^{\prime 2}d{v}^2)\to (N^2,h)$ be Riemannian submersion from a cuspidal torus to a surface. Then,

(i) The Riemannian submersion ϕ is harmonic if and only if it is the projection onto the first factor $[T^2\times S^1]\setminus \{t=\pi \}\to T^2\setminus \{t=\pi \}$ followed by a Riemannian covering map.

(ii) The Riemannian submersion ϕ is proper biharmonic if and only if up to an isometry, it is the projection

$$\varphi :([T^2\times S^1]\setminus \{t=\pi \},g_{\pi })\to (S^1\setminus \{t=\pi \}\times S^1,r^{2}d{t}^2+r^{\prime 2}d{v}^2), \varphi (u,t,v)=(t,v).$$

Remark 5. 

From the proof of Theorem 2, it is seen that the integrable data ${\kappa }_1$ (cf. Equation (19)) and the differential equation for α (cf. Equation (24)), corresponding to flat and non-flat base spaces $(N^2,h)$, respectively, can also be obtained by applying Theorem 2.9 in [16]. Our approach, however, differs from that in [16].