Differential as a harmonic morphism with respect to Cheeger--Gromoll type metrics

TL;DR

This paper studies the conditions under which the differential of a map between Riemannian manifolds with Cheeger--Gromoll type metrics acts as a harmonic morphism, focusing on horizontal conformality and related properties.

Contribution

It provides a characterization of the differential as a harmonic morphism in the context of Cheeger--Gromoll type metrics on tangent bundles.

Findings

Characterization of harmonic morphisms with Cheeger--Gromoll metrics

Conditions for horizontal conformality of differentials

Insights into the geometry of tangent bundle maps

Abstract

We investigate horizontal conformality of a differential of a map between Riemannian manifolds where the tangent bundles are equipped with Cheeger--Gromoll type metrics. As a corollary, we characterize the differential of a map as a harmonic morphism.

Equations

$$h(A,B)=g(\pi_{\ast}A,\pi_{\ast}B)+\omega_{\alpha}(\xi)^{p}\big{(}g(KA,KB)+qg(KA,\xi)g(KB,\xi)\big{)},$$

$${\nabla }_X^{\phi }(Y\circ \phi )={\nabla }_{{\phi }_{\ast }X}^{N}Y,X\in T_{x}M,Y\in \Gamma (TN).$$

$${\nabla }_X^{\phi }{\phi }_{\ast }Y-{\nabla }_Y^{\phi }{\phi }_{\ast }X={\phi }_{\ast }[X,Y],X,Y\in \Gamma (TM).$$

$$B(X,Y)={\nabla }_X^{\phi }{\phi }_{\ast }Y-{\phi }_{\ast }({\nabla }_X^{M}Y),X,Y\in \Gamma (TM).$$

$${\phi }_{\ast \ast }{X}_{\xi }^v$$

$${\phi }_{\ast \ast }{X}_{\xi }^h$$

$$X=X^{\top }+X^{\perp }$$

$$⟨{\phi }_{\ast x}X,{\phi }_{\ast x}Y{⟩}_N=\lambda (x)⟨X,Y{⟩}_M,X,Y\in {\mathcal{H}}_x^{\phi },$$

$${\mathcal{V}}_{\xi }^{\Phi }=\mathrm{Span}\{{\eta }_{\xi }^h+({\nabla }_{\xi }^M\eta {)}_{\xi }^v\mid \eta \in \Gamma ({\mathcal{V}}^{\phi })\},$$

$${\mathcal{H}}_{\xi }^{\Phi }=\mathrm{Span}\{X_{\xi }^v-q{\omega }_q(\xi )⟨X,\xi ⟩{\xi }_{\xi }^v+{\omega }_{\alpha }(\xi {)}^p({\nabla }_{\xi }^{M}X{)}_{\xi }^h\mid X\in \Gamma ({\mathcal{H}}^{\phi })\}.$$

$${\Phi }_{\ast }({\eta }_{\xi }^h+({\nabla }_{\xi }^M\eta {)}_{\xi }^v)=({\phi }_{\ast }\eta {)}_{{\phi }_{\ast }\xi }^h+B(\eta ,\xi {)}_{{\phi }_{\ast }\xi }^v+({\phi }_{\ast }{\nabla }_{\xi }^M\eta {)}_{{\phi }_{\ast }\xi }^v=0.$$

$${\mathcal{V}}_{\xi }^{\Phi }\ni (f\eta {)}_{\xi }^h+({\nabla }_{\xi }^{M}f\eta {)}_{\xi }^v={\eta }_{\xi }^v.$$

$$\displaystyle\begin{split}\langle A,B\rangle_{TM}&=\omega_{\alpha}(\xi)^{p}\bigg{(}\langle X,\nabla^{M}_{\xi}\eta\rangle_{M}+q\langle X,\xi\rangle_{M}\langle\xi,\nabla^{M}_{\xi}\eta\rangle_{M}\\ &-q\omega_{q}(\xi)\langle X,\xi\rangle_{M}\langle\nabla^{M}_{\xi}\eta,\xi\rangle_{M}\\ &-q^{2}\omega_{q}(\xi)\langle X,\xi\rangle_{M}\langle\nabla^{M}_{\xi}\eta,\xi\rangle_{M}|\xi|^{2}_{M}+\langle\nabla^{M}_{\xi}X,\eta\rangle_{M}\bigg{)}\\ &=q\omega_{\alpha}(\xi)^{p}\langle X,\xi\rangle_{M}\langle\nabla^{M}_{\xi}\eta,\xi\rangle_{M}(1-\omega_{q}(\xi)-q\omega_{q}(\xi)|\xi|^{2}_{M})\\ &=0.\end{split}$$

$${\eta }_{\xi }^v$$

$$X_{\xi }^h$$

$$T(X,Y)=\frac{1}{2}[X,Y{]}^{\top }.$$

$$-⟨({\nabla }_{\xi }^M\hat{Z}{)}^{\perp },\hat{W}{⟩}_M=⟨\xi ,T(\hat{Z},\hat{W}){⟩}_M.$$

$${\nabla }_X^{\lambda }Y={\nabla }_X^{M}Y+S(X,Y),$$

$$S(X,Y)=\frac{1}{2\lambda }((X\lambda )Y+(Y\lambda )X-⟨X,Y{⟩}_M\mathrm{grad}\lambda ).$$

$$P(X,\xi )=\sum _{i=1}^n⟨\xi ,T(X,e_i){⟩}_M{\phi }_{\ast }{e}_i,\xi \in T_{x}M,X\in {\mathcal{H}}_x^{\phi },$$

$${\Phi }_{\ast }(X_{\xi }^h)=({\phi }_{\ast }X{)}_{{\phi }_{\ast }\xi }^h+({\phi }_{\ast }S(X,\xi ){)}_{{\phi }_{\ast }\xi }^v+P(X,\xi {)}_{{\phi }_{\ast }\xi }^v.$$

$${\nabla }_{{\phi }_{\ast }\dot{\gamma }}^N{\phi }_{\ast }\eta ={\nabla }_{{\phi }_{\ast }\dot{\gamma }}^N{\phi }_{\ast }{\eta }^{\perp }={\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\perp })$$

$${\nabla }_{\dot{\gamma }}^{\lambda }\eta =S(\dot{\gamma },\eta ).$$

$${\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\perp }=S(\dot{\gamma },\eta )-{\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top }.$$

$$⟨{\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top }),Y{⟩}_N=\lambda ⟨{\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top },\hat{Y}{⟩}_M=\lambda ⟨{\nabla }_{{\eta }^{\top }}^{\lambda }\dot{\gamma },\hat{Y}{⟩}_M=-\lambda ⟨{\eta }^{\top },T(\dot{\gamma },\hat{Y}){⟩}_M.$$

$${\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top })=\frac{1}{\lambda }\sum _i⟨{\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top }),{\phi }_{\ast }{e}_i{⟩}_N{\phi }_{\ast }{e}_i=-\lambda \sum _i⟨{\eta }^{\top },T(\dot{\gamma },e_i){⟩}_M{\phi }_{\ast }{e}_i.$$

$$\begin{array}{rl}K({\Phi }_{\ast }{X}_{\xi }^h)& ={\nabla }_{{\phi }_{\ast }X}^N{\phi }_{\ast }\eta \\ & ={\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\perp })\\ & ={\phi }_{\ast }S(X,\eta )-{\phi }_{\ast }({\nabla }_{\dot{\gamma }}^{\lambda }{\eta }^{\top })\\ & ={\phi }_{\ast }S(X,\eta )+\sum _i⟨{\eta }^{\top },T(X,e_i){⟩}_M{\phi }_{\ast }{e}_i\\ & ={\phi }_{\ast }S(X,\xi )+\sum _i⟨\xi ,T(X,e_i){⟩}_M{\phi }_{\ast }{e}_i\end{array}$$

$${\pi }_{\ast }({\Phi }_{\ast }{X}_{\xi }^h)={\phi }_{\ast }({\pi }_{\ast }{X}_{\xi }^h)={\phi }_{\ast }X,$$

$$B(X,\xi )$$

$$⟨B(X,\xi ),{\phi }_{\ast }Y{⟩}_N$$

$$⟨{\phi }_{\ast }(S(X,\xi )),P(X,\xi ){⟩}_N$$

$$|B(X,\xi {)}_0^v{|}_{TN}^2$$

Full text

1. Introduction and preliminary results

1.1. Introduction

In [2], M. Benyounes, E. Loubeau and C. M. Wood intoduced a new class of Riemannian metrics in the tangent bundle over a Riemanian manifold. These metrics $h_{p,q}$, depending on two constants $p,q$, generalize Sasaki and Cheeger–Gromoll metrics. Wider class of this type of metrics was investigated by M. I. Munteanu [8] and by the authors [6]. Now, there are several articles concerning geometry of a tangent bundle equipped with Cheeger–Gromoll type metric. For example, M. Benyounes, E. Loubeau and C. M. Wood [3] considered $h_{p,q}$ metrics in the context of harmonic maps, whereas the first named author and Sz. M. Walczak [7] in the context of Riemanian submersions and Gromov–Hausdorff topology.

In [6], the authors studied the conformality of a differential $\Phi=\varphi_{\ast}:(TM,\tilde{h})\to(TN,h)$ of a map $\varphi:M\to N$ between Riemanian manifolds, where $\tilde{h}=h_{p,q,\alpha}$ and $h=h_{r,s,\beta}$ are metrics of Cheeger–Gromoll type. In this paper, we continue considerations concerning the differential of a map. We give necessary and sufficient conditions for a differential to be horizontally conformal and as a corollary we characterize the differential as a harmonic morphisms.

The idea is the following. The vertical part of Cheeger–Gromoll type metrics is nonlinear with respect to the base point, excluding Sasaki metric. Hence, conformal change of a metric on $M$ does not effect conformal change of Cheeger–Gromoll type metric. Scaling the base point, the condition of horizontal conformality reduces to vanishing of a certain polynomial, which gives restrictions to coefficients $p,q,r,s$ and $\alpha,\beta$. That is why, for horizontal conformality of $\Phi$ there are not many natural choices of $\tilde{h}$ and $h$ (Theorem 2.1), whereas in the context of a harmonic morphism the only possibility is Sasaki metric (Theorem 2.2).

1.2. Preliminary results

Let $(M,g)$ be a Riemannian manifold and $\pi:TM\to M$ its tangent bundle. The Levi-Civita connection and the projection $\pi$ gives the natural splitting $TTM=\mathcal{H}\oplus\mathcal{V}$ of the second tangent bundle $\pi_{\ast}:TTM\to TM$, where the vertical distribution $\mathcal{V}$ is the kernel of $\pi_{\ast}$ and the horizontal distribution $\mathcal{H}$ is the kernel of the conection map $K$. If $X,\xi\in T_{x}M$ then there is a unique vertical vector $X^{v}_{\xi}$ and a unique horizontal vector $X^{h}_{\xi}$ in $T_{\xi}TM$ such that $\pi_{\ast}X^{h}_{\xi}=X$ and $KX^{v}_{\xi}=X$. Moreover, any $A\in T_{\xi}TM$ has a unique decomposition into horozontal and vertical part, $A=\mathcal{V}A+\mathcal{H}A$. For more details on decomposition of the tangent bundle see [5].

Let $p,q,\alpha$ be constants, $q$ non–negative, $\alpha$ positive. Define $(p,q,\alpha)$-metric $h=h_{p,q,\alpha}$ on $TM$ as follows. For every $A,B\in T_{\xi}(TM)$,

[TABLE]

where $\omega_{\alpha}(\xi)=(1+\alpha g(\xi,\xi))^{-1}$. The Riemannian metric $h_{p,q,\alpha}$ is a generalization of the metric considered in [2, 3] and is a special case of a metric considered in [8]. In particular, $h_{0,0,\alpha}$ (or $h_{p,0,0}$) is Sasaki metric, $h_{1,1,1}$ Cheeger-Gromoll metric.

We will often write $\langle\cdot,\cdot\rangle_{M}$ for $g$ and $\langle\cdot,\cdot\rangle_{TM}$ for $h=h_{p,q,\alpha}$. The lenght of a vector will be denoted by $|\cdot|_{M}$ and $|\cdot|_{TM}$, respecitively.

Consider now a smooth map $\varphi:M\to N$ between Riemannian manifolds $M$ and $N$. Let $\varphi^{-1}TN\to M$ be a pull–back bundle. There is a unique connection $\nabla^{\varphi}$ in this bundle characterised by the property [1]

[TABLE]

Then we easily obtain

[TABLE]

The second fundamental form of $\varphi$ is $B=\nabla\varphi_{\ast}$,

[TABLE]

By (1.1), we get that $B$ is symmetric and hence tensorial in both variables. If $B=0$ we say that $\varphi$ is totally geodesic. We will need the following lemma.

Lemma 1.1**.**

For $X,\xi\in T_{x}M$

[TABLE]

Proof.

Follows by the definition of a connection map and vertical and horizontal distributions. Details are left to the reader. ∎

Let $\varphi:(M,\langle\cdot,\cdot\rangle_{M})\to(N,\langle\cdot,\cdot\rangle_{N})$ be a smooth map between Riemannian manifolds, $\dim M>\dim N$. Let $\mathcal{V}^{\varphi}={\rm ker}\varphi_{\ast}$ be the vertical distribution and $\mathcal{H}^{\varphi}=(\mathcal{V}^{\varphi})^{\bot}$, the orthogonal complement of $\mathcal{H}^{\varphi}$ with respect to $\langle\cdot,\cdot\rangle_{M}$, the horozontal distribution, $TM=\mathcal{V}^{\varphi}\oplus\mathcal{H}^{\varphi}$. Each $X\in T_{x}M$ has therefore a unique decomposition

[TABLE]

into vertical and horizontal part.

We say that $\varphi$ is horizontally conformal if for every $x\in M$ either $\varphi_{\ast x}=0$ or $\varphi_{\ast x}:\mathcal{H}^{\varphi}_{x}\to T_{\varphi(x)}N$ is surjective and

[TABLE]

where $\lambda(x)$ is positive. We call $\lambda$ the dilatation of $\varphi$. Let $C_{\varphi}$ denote the set of critical points i.e., points $x\in M$ such that $\varphi_{\ast x}=0$. One can easily prove

Lemma 1.2**.**

For $\Phi=\varphi_{\ast}:TM\to TN$ we have $\pi(C_{\Phi})\subset C_{\varphi}$.

Moreover, by Lemma 1.1 we have

Lemma 1.3**.**

If a map $\varphi:M\to N$ is a submersion, i.e., surjective map of maximal rank, then so is $\Phi=\varphi_{\ast}:TM\to TN$.

Remark 1.4**.**

By Lemmas 1.2 and 1.3, if $\varphi:M\to N$ and $\Phi=\varphi_{\ast}:TM\to TN$ are horizontally conformal, then the differential of horizontally conformal submersion $\varphi:M\setminus C_{\varphi}\to N$ is a horizontaly conformal submersion $\Phi:TM\setminus\pi^{-1}(C_{\varphi})\to TN$. Hence, throughout the paper, without loss of generality, we may assume that horizontaly conformal map $\varphi$ is a submersion i.e. the set of critical points is empty.

Theorem 1.5**.**

Assume $\varphi:M\to N$ is a submersion. Let $\Phi=\varphi_{\ast}:TM\to TN$ and equip tangent bundles $TM$ and $TN$ with Cheeger–Gromoll type metrics $h_{p,q,\alpha}$ and $h_{r,s,\beta}$, respectively. Then, with respect to submersion $\Phi$, the second tangent bundle splits into orthogonal sum $T_{\xi}TM=\mathcal{V}^{\Phi}_{\xi}\oplus\mathcal{H}^{\Phi}_{\xi}$, where

[TABLE]

and

[TABLE]

Proof.

Let $\varphi:(M,g_{M})\to(N,g_{N})$ be a submersion. Fix $\xi\in T_{x}M$, $x\in M$. Denote the right hand side of (1.5) by $\mathbb{V}$ and the right hand side of (1.6) by $\mathbb{H}$. By Lemma 1.1, for $\eta\in\Gamma(\mathcal{V}^{\varphi})$

[TABLE]

Hence, $\mathbb{V}\subset\mathcal{V}^{\Phi}_{\xi}$. Let $f$ be a smooth function such that $\xi f=1$ and $f(x)=0$. Then

[TABLE]

Therefore, we see that $\dim\mathbb{V}=2(m-n)$, where $m=\dim M$ and $n=\dim N$. Thus $\mathcal{V}^{\Phi}_{\xi}=\mathbb{V}$, so (1.5) holds. Now, let $\eta\in\Gamma(\mathcal{V}^{\varphi})$ and $X\in\Gamma(\mathcal{H}^{\varphi})$. For $A=\eta^{h}_{\xi}+(\nabla^{M}_{\xi}\eta)^{v}_{\xi}$ and $B=X^{v}_{\xi}-q\omega_{q}(\xi)\langle X,\xi\rangle\xi^{v}_{\xi}+\omega_{\alpha}(\xi)^{p}(\nabla^{M}_{\xi}X)^{h}_{\xi}$

[TABLE]

Thus $\mathbb{H}$ is orthogonal to $\mathcal{V}^{\Phi}_{\xi}$. Again, as above, taking a function $f$ such that $\xi f=1$ and $f(x)=0$, we get $X^{h}_{\xi}\in\mathbb{H}$ for $X\in\Gamma(\mathcal{H}^{\varphi})$. Hence, $\dim\mathbb{H}=2m$. Therefore, $\mathbb{H}=\mathcal{H}^{\Phi}_{\xi}$ and (1.6) holds. ∎

The proof of Theorem 1.5 also implies

Corollary 1.6**.**

Assume $\varphi:M\to N$ is a submersion. Let $\Phi=\varphi_{\ast}:TM\to TN$ and equip tangent bundles $TM$ and $TN$ with Cheeger–Gromoll type metrics $h_{p,q,\alpha}$ and $h_{r,s,\beta}$, respectively. Then

[TABLE]

Let $\varphi:M\to N,$ be a Riemannian submersion, $\Phi=\varphi_{\ast}:TM\to TN$. A vector field $\hat{Z}\in\Gamma(\mathcal{H}^{\varphi})$ is called basic if there is a vector field $Z\in\Gamma(TN)$ such that $\varphi_{\ast}\hat{Z}=Z\circ\varphi$. The correspondence $Z\mapsto\hat{Z}$ is one–to–one, since $\varphi_{\ast x}:\mathcal{H}^{\varphi}_{x}\to T_{\varphi(x)}N$ is an isomorphisms. Let $T:\Gamma(\mathcal{H}^{\varphi})\times\Gamma(\mathcal{H}^{\varphi})\to\Gamma(\mathcal{V}^{\varphi})$ be an itegrability tensor of $\varphi$,

[TABLE]

Let $\nabla^{M}$ and $\nabla^{N}$ be the Levi–Civita conections of $\langle\cdot,\cdot\rangle_{M}$ and $\langle\cdot,\cdot\rangle_{N}$, respectively. Then one can prove that for basic vector fields $\hat{Z},\hat{W}$ and vertical vector field $\xi$

[TABLE]

For more details on Riemannian submersions see [4, Chapter 9].

Assume now, $\varphi:(M,\langle\cdot,\cdot\rangle_{M})\to(N,\langle\cdot,\cdot\rangle_{N})$ is a horizontally conformal map with a dilatation $\lambda$. If we put $\langle\cdot,\cdot\rangle_{\lambda}=\lambda\langle\cdot,\cdot\rangle_{M}$, then $\varphi:(M,\langle\cdot,\cdot\rangle_{\lambda})\to(N,\langle\cdot,\cdot\rangle_{N})$ is a Riemannian submersion. Clearly, horizontal and vertical distrbutions with respect to $\langle\cdot,\cdot\rangle_{M}$ and $\langle\cdot,\cdot\rangle_{\lambda}$ coincide. The Levi–Civita connections $\nabla^{M}$ and $\nabla^{\lambda}$ of $\langle\cdot,\cdot\rangle_{M}$ and $\langle\cdot,\cdot\rangle_{\lambda}$ satisfy

[TABLE]

where $S$ is a symmetric tensor field given by

[TABLE]

Let $\Phi=\varphi_{\ast}:TM\to TN$ and equip $TM$ and $TN$ with Cheeger–Gromoll type metrics $h_{p,q}$ and $h_{r,s}$, respectively. For simplicity, put

[TABLE]

where $e_{1},\ldots,e_{n}$ is an orthonormal basis of $\mathcal{H}^{\varphi}_{x}$.

Lemma 1.7**.**

Let $X\in\mathcal{H}^{\varphi}_{x}$, $\xi\in T_{x}M$, $x\in M$. Then

[TABLE]

Proof.

We may assume that $X^{h}_{\hat{\zeta}}=(\gamma^{h})^{\cdot}(0)$, where $\gamma$ is a curve on $M$ such that $\gamma(0)=x$, $\dot{\gamma}(t)\in\mathcal{H}^{\varphi}$ and $\dot{\gamma}(0)=X$, and $\gamma^{h}$ is a horizontal lift of $\gamma$ to $TM$ such that $\gamma^{h}(0)=\xi$. Put $\eta=\gamma^{h}$ for simplicity. Then

[TABLE]

and by (1.10) and the fact that $\eta$ is parallel with respect to $\nabla^{M}$,

[TABLE]

Therefore,

[TABLE]

Next, for a vector field $Y$ on $N$, by (1.9)

[TABLE]

Thus

[TABLE]

Since

[TABLE]

and

[TABLE]

the equality (1.11) holds. ∎

By Lemma 1.7 we have (see also [1, Lemma 4.5.1]).

Corollary 1.8**.**

1. $(1)$

Let $X,Y\in\mathcal{H}^{\varphi}_{x}$, $\xi\in T_{x}M$. Then

[TABLE] 2. $(2)$

Let $X\in\mathcal{H}^{\varphi}_{x}$, $\xi\in\mathcal{V}^{\varphi}_{x}$. Then

[TABLE] 3. $(3)$

$B(X,\xi)=0$* for all $X\in\Gamma(\mathcal{H}^{\varphi})$, $\xi\in\Gamma(TM)$ if and only if $T=0$ and $S=0$.*

Proof.

(1.12) follows by (1.11) and (1.4). For $X,Y\in\mathcal{H}^{\varphi}_{x}$ and $\xi\in T_{x}M$ we have

[TABLE]

which proves (1.13). Now, let $X\in\mathcal{H}^{\varphi}_{x}$ and $\xi\in\mathcal{V}^{\varphi}_{x}$. Then

[TABLE]

Thus

[TABLE]

since $T$ is skew–symmetric. Hence (1.14) holds, which implies (1.15). $(3)$ is a consequence of (1.12) and (1.14). ∎

Let $X\in\mathcal{H}^{\varphi}_{x}$ and $\xi\in T_{x}M$. Decompose $S(X,\xi)^{v}_{\xi}$ into horizontal and vertical part with respect to $\mathcal{V}^{\Phi}$ and $\mathcal{H}^{\Phi}$,

[TABLE]

Equip tangent bundles $TM$ and $TN$ with Cheeger–Gromoll type metrics $h_{p,q,\alpha}$ and $h_{r,s,\beta}$, respectively.

Lemma 1.9**.**

Assume $\varphi$ and $\Phi$ are both horizontally conformal with dilatations $\lambda$ and $\Lambda$, respectively. Let $X\in\mathcal{H}^{\varphi}_{x}$, $\xi\in T_{x}M$. Then

[TABLE]

Proof.

(1.16) follows directly by (1.4) (and Corollary 1.6). To prove (1.17), define $V=X^{h}_{\xi}-S(X,\xi)^{v}_{\varphi_{\ast}\xi}$. Then

[TABLE]

Therefore,

[TABLE]

since $\langle X^{h}_{\xi},S^{\bot}_{\Phi}(X,\xi)\rangle_{TM}=\langle X^{h}_{\xi},S(X,\xi)^{v}_{\xi}\rangle=0$, and, by (1.11),

[TABLE]

Hence (1.17) holds. ∎

Lemma 1.10**.**

If $\Phi$ and $\varphi$ are both horizontally conformal, then $\lambda$ is constant.

Proof.

Let $X\in\mathcal{H}^{\varphi}$ and $\xi\in\mathcal{V}^{\varphi}$. Comparing (1.16) and (1.17) and using (1.15) we get

[TABLE]

which implies, $\varphi_{\ast}(S(X,\xi))=0$, so $\xi\lambda=0$. It follows that ${\rm grad}\,\lambda$ is horizontal.

Assume now, $\xi\in\mathcal{H}^{\varphi}$. Then $P(X,\xi)=0$. Again, comparing (1.16) and (1.17) we get

[TABLE]

Thus

[TABLE]

Hence, as before, $\varphi_{\ast}S(X,\xi)=0$. Since ${\rm grad}\,\lambda$ is horizontal, we may put $X=\xi={\rm grad}\,\lambda$. Then

[TABLE]

Therefore, ${\rm grad}\,\lambda$ is vertical. Finally, $\lambda$ is constant. ∎

Lemma 1.11**.**

Assume $\dim N\geq 2$. Let $\Phi$ be a horizontally conformal map with dilatation $\Lambda$. Then $\varphi$ is horizontally conformal with constant dilatation $\lambda=\Lambda(0)$. Moreover, $\Lambda$ is constant and the horizontal distribution $\mathcal{H}^{\varphi}$ is integrable.

Proof.

Let $X,Y\in\mathcal{H}^{\varphi}_{x}$, $0=0_{x}\in T_{x}M$. Then by Corollary 1.6, $X^{h}_{0},Y^{h}_{0}\in\mathcal{H}^{\Phi}_{0}$. By (1.4), $\Phi_{\ast}(X^{h}_{0})=(\varphi_{\ast}X)^{h}_{0}$ and $\Phi_{\ast}(Y^{h}_{0})=(\varphi_{\ast}Y)^{h}_{0}$. Thus

[TABLE]

Hence $\varphi$ is conformal with dilatation $\lambda(x)=\Lambda(0_{x})$. By Lemma 1.10 $\lambda$ is constant. Thus $S=0$.

Let $X,Y\in\Gamma(\mathcal{H}^{\varphi})$, $\xi\in\mathcal{V}^{\varphi}$. Then $X^{v}_{\xi}+\omega_{\alpha}(\xi)^{p}(\nabla_{\xi}X)^{h}_{\xi},Y^{h}_{\xi}\in\mathcal{H}^{\Phi}_{\xi}$ by Theorem 1.5 and Corollary 1.6. Moreover, by Corollary 1.8

[TABLE]

Hence, by horizontal conformality of $\Phi$ and (1.9) we get

[TABLE]

Put $A(\xi)=\Lambda(\xi)-\lambda$. Then by (1.16), $A(t\xi)=t^{2}A(\xi)$. Hence, replacing $\xi$ by $t\xi$ and assuming for simplicity $|\xi|_{M}=1$, we get

[TABLE]

Therefore, $\langle\xi,T(Y,X)\rangle_{M}=0$ and, since $\xi\in\Gamma(\mathcal{V}^{\varphi})$ was taken arbitrary, $T=0$. This implies, together with the fact that $S=0$ and Corollary 1.8 that

[TABLE]

Hence, by (1.16), $\Lambda(\xi)=\lambda$ for any $\xi$. ∎

2. Main results

2.1. Differential as a horizontally confomal map

Consider a smooth map $\varphi:(M,\langle\cdot,\cdot\rangle_{M})\to(N,\langle\cdot,\cdot\rangle_{N})$, $\dim M>\dim N\geq 2$, between Riemanian manifolds. Let $\Phi=\varphi_{\ast}:(TM,\tilde{h})\to(TN,h)$, where $\tilde{h}=h_{p,q,\alpha}$ and $h=h_{r,s,\beta}$ are Cheeger–Gromoll type metrics.

Theorem 2.1**.**

$\Phi$* is horizontally conformal if and only if $\varphi$ is totally geodesic and horizontally conformal with constant dilatation $\lambda$, $p\alpha=r\beta=0$ and $q=\lambda s$. Then, the dilatation $\Lambda$ of $\Phi$ is constant and equal to $\lambda$, horizontal distribution $\mathcal{H}^{\varphi}$ is integrable and vertical distribution $\mathcal{V}^{\varphi}$ is totally geodesic.*

Proof.

Assume $\Phi$ is horizontally conformal. Then by Lemma 1.11, $\varphi$ is horizontally conformal with constant dilatation $\lambda$, the dilatation $\Lambda$ of $\Phi$ is constant and equal to $\lambda$ and by Corollary 1.8, $B(X,\xi)=0$ for $X\in\mathcal{H}^{\varphi}$ and $\xi\in TM$.

Let $X\in\Gamma(\mathcal{H}^{\varphi})$ and $\xi\in\Gamma(\mathcal{V}^{\varphi})$. Then $A=X^{v}_{\xi}+\omega_{\alpha}(\xi)^{p}(\nabla^{M}_{\xi}X)^{h}_{\xi}\in\mathcal{H}^{\Phi}_{\xi}$ by Theorem 1.5. Since $T=0$, by (1.9) $\nabla^{M}_{\xi}X\in\mathcal{V}^{\varphi}$. Then $|\Phi_{\ast}A|^{2}_{TN}=\lambda|A|^{2}_{TM}$ implies

[TABLE]

Replacing $\xi$ by $t\xi$, assuming $|\xi|^{2}_{M}=1$ and computing the second derivative at $t=0$, we get

[TABLE]

Since $\langle B(\nabla_{\xi}X,\xi),\varphi_{\ast}X\rangle_{N}=\lambda|\nabla^{M}_{\xi}X|^{2}_{M}$, we obtain $-p\alpha|X|^{2}_{M}=|\nabla^{M}_{\xi}X|^{2}_{M}$. Therefore, $p\alpha=0$ and $\nabla^{M}_{\xi}X=0$. This implies $B=0$ globally, so $\varphi$ is totally geodesic. In particular, horizontal distrubution is integrable and vertical distribution is totally geodesic.

Moreover, for $X\in\Gamma(\mathcal{H}^{\varphi})$ basic and $\xi\in\mathcal{H}^{\varphi}_{x}$ such that $\xi$ and $X_{x}$ are orthogonal, by Theorem 1.5 and the fact that $\alpha p=0$ we have $A=X^{v}_{\xi}+(\nabla^{M}_{\xi}X)^{h}_{\xi}\in\mathcal{H}^{\Phi}_{\xi}$. Since $\nabla^{M}_{\xi}X\in\mathcal{H}^{\varphi}_{x}$, the condition $|\Phi_{\ast}A|^{2}_{TN}=\lambda|A|^{2}_{TM}$ simplfies to

[TABLE]

This implies $r\beta=0$. Now, for arbitrary $X,\xi\in\Gamma(\mathcal{H}^{\varphi})$, $A=X^{v}_{\xi}-q\omega_{q}(\xi)\langle X,\xi\rangle\xi^{v}_{\xi}+(\nabla^{M}_{\xi}X)^{h}_{\xi}\in\mathcal{H}^{\Phi}$. Then, as above by horizontal conformality of $\Phi$, after some computations

[TABLE]

Hence $q=\lambda s$.

Conversely, assume $\varphi$ is totally geodesic and horizontally conformal with constant dilatation $\lambda$, $p\alpha=r\beta=0$ and $q=\lambda s$. By Theorem 1.5 $\Phi:TM\to TN$ is a submersion and vertical and horizontal distributions $\mathcal{V}^{\Phi}$ and $\mathcal{H}^{\Phi}$ are given by (1.5) and (1.6), respectively. It remains to show that $\Phi_{\ast}:\mathcal{H}^{\Phi}\to T(TN)$ is conformal, but this follows immediately by simple calculations. ∎

2.2. Differential as a harmonic morphism

Let $\varphi:(M,\langle\cdot,\cdot\rangle_{M})\to(N,\langle\cdot,\cdot\rangle_{N})$. Consider notation from the first section. We say that $\varphi$ is harmonic if its tension field $\tau(\varphi)={\rm tr}B$ vanishes. If $e_{1},\ldots,e_{n}$ is an orthonormal frame on $M$ then

[TABLE]

A map $\varphi:M\to N$ is said to be a harmonic morphism if for every harmonic function $f:G\to\mathbb{R}$ defined on an open subset $G$ of $N$ with $\varphi^{-1}(G)$ nonempty, the composition $f\circ\varphi$ is a harmonic map on $\varphi^{-1}(G)$. There is a useful characterisation of harmonic morphisms due to Fuglede and Ishihara. Namely, see [1], $\varphi$ is a harmonic morphisms if and only if it is horizontally conformal and harmonic. Moreover, if $\varphi$ is horizontally conformal, then its tension field simplifies to [1]

[TABLE]

where $m=\dim M$, $n=\dim N$ and $\kappa_{\varphi}$ is the mean curvature vector field of the fibres of $\varphi$,

[TABLE]

where $e_{1},\ldots,e_{m-n}$ is an orthonormal frame of $\mathcal{V}^{\varphi}$.

Equip tangent bundle $TM$ with Cheeger–Gromoll type metric $h_{p,q,\alpha}$. Let $\omega_{q}(\xi)=(1+q|\xi|^{2}_{M})^{-1}$. The Levi–Civita connection $\nabla^{TM}$ corresponding to $h_{p,q,\alpha}$ evaluated at $\xi$ is given by [8]

[TABLE]

Let $\varphi:(M,\langle\cdot,\cdot\rangle_{M})\to(N,\langle\cdot,\cdot\rangle_{N})$, $\dim M>\dim N\geq 2$, be a smooth map between Riemanian manifolds, $\Phi=\varphi_{\ast}:(TM,\tilde{h})\to(TN,h)$, where $\tilde{h}=h_{p,q,\alpha}$ and $h=h_{r,s,\beta}$ are Cheeger–Gromoll type metrics.

Theorem 2.2**.**

$\Phi$* is a harmonic morphism if and only if $\varphi$ is a totally geodesic harmonic morphism, the dilatation of $\varphi$ is constant and $\tilde{h},h$ are Sasaki metrics.*

Proof.

Assume $\Phi$ is a harmonic morphism. In particular, $\Phi$ is horizontally conformal. Hence, by Theorem 2.1, $\varphi$ is horizontally conformal, the dilatations of $\Phi$ and $\varphi$ are constant, $\Lambda=\lambda$, and $\varphi$ is totally geodesic. Therefore $\varphi$ is harmonic, so $\varphi$ is a harmonic morphism. Thus, by (2.2), $\kappa_{\varphi}=0$, $\kappa_{\Phi}=0$. Moreover, by Theorem 2.1, $p\alpha=r\beta=0$ and $q=\lambda s$. Let $\xi\in\mathcal{H}^{\varphi}$ and let $e_{1},\ldots,e_{m-n}$ be an orthonormal frame for $\mathcal{V}^{\varphi}_{\xi}$, $m=\dim M$, $n=\dim N$. Since vertical distribution $\mathcal{V}^{\varphi}$ is, by Theorem 2.1 totally geodesic, by Theorem 1.5 and Corollary 1.6

[TABLE]

is an orthonormal frame for $\mathcal{V}^{\Phi}_{\xi}$. Then

[TABLE]

Therefore,

[TABLE]

Since $\Phi_{\ast}(\kappa_{\Phi})=0$ and $\varphi_{\ast}(\kappa_{\varphi})=0$, it follows that $q=0$. Hence, $s=0$ and $\tilde{h}$, $h$ are Sasaki metrics.

Conversely, assume $\varphi$ is a totally geodesic harmonic morphism, the dilatation of $\varphi$ is constant and $\tilde{h}$, $h$ are Sasaki metrics. Then, by Theorem 2.1 $\Phi$ is horizontally conformal with constant dilatation. Moreover, $\varphi_{\ast}(\kappa_{\varphi})=0$. Hence, similarly as above,

[TABLE]

and by (2.2), $\Phi$ is a harmonic morphism. ∎

Remark 2.3**.**

Condition $p\alpha=0$ for Cheeger–Gromoll metric $h_{p,q,\alpha}$ is equavalent to condition $p=0$ for the metric $h_{p,q}$. It follows that in the case of horizontal conformality of a differential, and in consequence for a differential to be a harmonic morphism, it is sufficient to consider only $h_{p,q}$ metrics introduced in [3].