---
abstract: 'We investigate conformality of the differential of a mapping between Riemannian manifolds if the tangent bundles are equipped with a generalized metric of Cheeger-Gromoll type.'
author:
- 'W. Kozłowski'
- 'K. Niedziałomski'
title: 'Conformality of a differential with respect to Cheeger-Gromoll type metrics'
---

Introduction and Preliminaries
==============================

Generalized metrics of Cheeger-Gromoll type or $(p,q)$-metrics $h_{p,q}$, being a generalization of Sasaki metric $h_S$ [@Kw] and Cheeger-Gromoll metric $h_{CG}$ [@GK], have been recently introduced by M. Benyounes, E. Loubeau and C. M. Wood in [@BLW] in the context of harmonic sections. In [@BLW1], the same authors showed that the geometry of the tangent bundle equipped with this kind of metric is of independent interest. It is worth noticing that M. I. Munteanu in [@M] investigated independently the geometry of tangent bundle equipped with a certain deformation of Cheeger-Gromoll metric other that in [@BLW]. Yet in [@KW], Sz. Walczak and the first named author considered $(p,q)$-metrics in the context of Riemannian submersions and Gromov-Hausdorff topology.

In this paper we introduced $(p,q,\alpha)$-metrics which are more general that $(p,q)$-metrics. (In contrast to [@BLW] we do not assume that $p,q$ and $\alpha$ are constant). We investigate relations between conformality of a map $\varphi:(M,g)\to (M',g')$ between Riemannian manifolds and its differential $\Phi=\varphi_\ast: (TM, h)\to 
(TM',h')$ between their tangent bundles equipped with $(p,q,\alpha)$-metric $h$ and $(r,s,\beta)$-metric $h'$, respectively.

Interesting enough, there is essential difference between the cases $\dim M=2$ and $\dim M\geq 3$.

We prove that in the second case (Theorem \[MR\_T1\]) $\Phi$ is conformal if and only if $\varphi$ is a homothety and totally geodesic immersion and some special relations between triples $(p,q,\alpha)$ and $(r,s,\beta)$ hold. In this case $\Phi$ is also a homothety with the same dilatation as $\varphi$.

However, in the first case it may happen that $\Phi $ is conformal, although $\varphi$ is not a totally geodesic immersion (Theorem \[MR\_T2\]). Then $\Phi$ is no longer a homothety. An example of such a map is given.

Cheeger-Gromoll type metrics {#S_CGM}
----------------------------

Consider a Riemannian manifold $(M,g)$, and let $\pi :TM \to M$ be its tangent bundle. The Levi-Civita connection $\nabla$ of $g$, gives a natural splitting $T(TM) = \mathcal H \oplus \mathcal V$ of the second tangent bundle $\pi_\ast :T(TM)\to TM$, where the [*vertical*]{} distribution $\mathcal V$ is the kernel of $\pi_\ast$, and the [ *horizontal*]{} distribution is the kernel of, so called, connection map $K$. If $X,Z\in T_x M$ then by $X^v_Z$ we denote the vertical lift of $X$ to the point $Z$, i.e., $X^v_Z$ is a tangent vector to the curve $t \mapsto Z + tX$ at $t=0$. Every $A\in T_Z(TM)$ can be uniquely written as $A=\mathcal H A+ \mathcal V A$, where $\mathcal H A\in \mathcal H_Z$ and $\mathcal V A\in \mathcal V_Z$ denote its horizontal and vertical part respectively. The vertical part of $A$ is given by $(KA)^v_Z$.

Recall that $K$ is a smooth $\mathbb R$-linear bundle morphism determined by the conditions:

-   For every $Z\in TM$, $K: T_Z(TM)\to T_{\pi(Z)} M$ is the canonical isomorphism, i.e., $K(X^v_Z)=X$.

-   For every vector field $X$ on $M$ and every $v\in T_xM$, $K(X_\ast v) = \nabla_v X$.

Notice that (K1) and (K2) imply the following properties

-   For every Riemannian manifold $(M', g')$ and every $X,Z\in T_x M$ and every map $\varphi: M\to M'$, $\varphi_{\ast \ast} X^v_Z  = 
    (\varphi_\ast X)^v_{\varphi_\ast (Z)}$.

-   For every curve $\gamma$ in $M$ and every vector field $\xi$ along $\gamma$, $K(\dot \xi)=\nabla_{\dot \gamma} \xi$.

Let $p,q,\alpha$ be smooth functions on $M$. Assume $q$ is non-negative and $\alpha$ is positive. Define $(p,q,\alpha)$-metric $h=h_{p,q,\alpha}$ on $TM$ as follows: For every $A,B\in T_Z(TM)$, $Z\in T_x M$, $$h(A,B) = g(\pi_\ast A,\pi_\ast B) +
\omega_{\alpha}(Z)^{p}
\big( g(KA,KB) + qg(KA,Z)g(KB,Z)\big),$$ where $\omega_\alpha(Z)=(1+\alpha g(Z,Z))^{-1}$. Here all functions $p,q,\alpha$ are evaluated at $x$. For any $p,q,\alpha$, the Riemannian metric $h_{p,q,\alpha}$ is a special case of a metric considered in [@M]. Notice that if $p,q,\alpha$ are constants and $\alpha =1$ then $h_{p,q,\alpha}$ becomes a metric from [@BLW]. In particular, $h_{0,0,1}$ (resp. $h_{1,1,1}$) is Sasaki metric $h_S$ [@Kw] (resp. Cheeger-Gromoll metric $h_{CG}$ [@GK]).\

Conformal mappings and metrics {#SS_CM}
------------------------------

Recall that a map $\varphi: (M,g)\to (M',g')$ between Riemannian manifolds is [*conformal*]{} if $\varphi^\ast g' = \lambda g$ for some positive function $\lambda$ on $M$. The function $\lambda$ is called a [*dilatation*]{}. A conformal mapping with constant dilatation is called a [*homothety*]{}. If $\dim M < \dim M'$ a conformal mapping $\varphi$ is often called [*weakly conformal*]{}.

Let $\lambda$ be a strictly positive $C^\infty$-function on $M$, and $g^\lambda= \lambda g$. The Levi-Civita connections $\nabla$ and $\nabla^\lambda$ of $g$ and $g^\lambda$ are related as follows: $\nabla^\lambda = \nabla + S_M^{g,\lambda}$ where $S=S_M^{g,\lambda}$ is a symmetric $(1,2)$-tensor field given by (compare [@Ch], page 64): $$S(X,Y) = \frac{1}{2\lambda}\big( (X\lambda)Y + (Y\lambda)X - g(X,Y) {\rm grad\,} \lambda \big),
\quad X,Y \in \Gamma(M, TM).$$

Suppose $\varphi:M\to M'$ is an immersion, e.g., conformal mapping. Then for every $x\in M$ we may choose an open neighbourhood $U_x$ of $x$ such that $L_{x'}=\varphi(U_x)$ ($x'=\varphi(x)$) is a regular submanifold of $M'$. Let $j:L_{x'}\to M'$ be the inclusion map, and let $\bar g =j^\ast g'$ be the induced metric tensor on $L_{x'}$. Moreover, let $\Pi$ denote the second fundamental form of $L_{x'}$. We say that the immersion $\varphi:(M,g)\to (M',g')$ is [*totally geodesic*]{} if for every $x\in M$, $L_{x'}$ is a totally geodesic submanifold of $(M',g')$. One can prove the following

\[CM\_nable\] Suppose $\varphi: (M,g)\to (M',g')$ is a conformal mapping with a dilatation $\lambda$. Choose $x\in M$ and put $x'=\varphi(x)$. Let $\gamma:(-\epsilon,\epsilon)\to L_{x'}$ be a curve in $M$ and $\xi$ be a vector field along $\gamma$. Put $\gamma'=\varphi\circ \gamma$ and $\xi'=\varphi_\ast \xi$. Then $$\bar \nabla_{\dot \gamma'}\xi' = \varphi_\ast\nabla_{\dot \gamma}\xi + \varphi_\ast S(\dot \gamma, \xi),$$ where $S=S_M^{g,\lambda}$, and $\nabla$ and $\bar \nabla$ are the Levi-Civita connections of $g$ and $\bar g$ respectively.

Adopt the notations from Lemma \[CM\_nable\]. Put $Z= \xi(0)$, $Z'=\xi'(0)$, $v= \dot\gamma (0)$ and $v'=\dot\gamma'(0)$. Suppose that a vector $A\in 
T_Z(TM)$ is tangent to the curve $\xi$ (it is convenient to think of vector fields along curves as of curves in the tangent bundle), i.e., $A=\dot \xi(0)$. Next, let $K$ and $K'$ denote connection maps induced from $\nabla$ and $\nabla'$ respectively. Moreover put $\Phi=\varphi_\ast:TM\to TM'$. As a direct consequence of Lemma \[CM\_nable\], the equation $\nabla'=\nabla+\Pi$ and properties of connection map we get

\[CM\_KK\] The vectors $K(A)$ and $K'(\Phi_\ast A)$ are related as follows: $$K'(\Phi_\ast A) = \varphi_\ast K(A) + \varphi_\ast S(v,Z)+\Pi(v',Z').$$ In particular, if $A$ is horizontal then $ K'(\Phi_\ast A) = \varphi_\ast S(v,Z)+\Pi(v',Z').$

If $\dim M=\dim M'$ then the term $\Pi$ is omitted.

Let $U=U_x$ and $L'=L_{x'}$ and let $\pi':TL'\to L'$ be a natural projection. Since $\varphi: U\to L'$ is a conformal diffeomorphism, so is its inverse. In particular $\Phi: TU\to TL'$ is a diffeomorphism. Therefore we have

\[CM\_cor\_-1\] Take $A'\in T_{Z'}(TL')$ such that $v'=\pi'_\ast A$. Then $$K(\Phi_\ast^{-1} A')= \varphi_\ast^{-1} K'(A') + \varphi_\ast^{-1} S'(v',Z')-\varphi_\ast^{-1}\Pi(v',Z'),$$ or equivalently $$K(\Phi_\ast^{-1} A')= \varphi_\ast^{-1} \bar K(A') + \varphi_\ast^{-1} S'(v',Z'),$$ where $S'=S_{L'}^{\bar g, \mu}$ with $\mu= (1/\lambda)\circ \varphi^{-1}$, and $\bar K$ is the connection map induced from $\bar \nabla$.

\[CM\_co\] Suppose $\varphi:(M,g)\to (M',g')$ is a conformal mapping and $M$ is connected. $\Phi_\ast$ maps horizontal vectors onto horizontal vectors if and only if $\varphi$ is a totally geodesic homothety.

($\Rightarrow$) If $\Phi_\ast$ maps horizontal vectors onto horizontal vectors then by Lemma \[CM\_KK\], $\varphi_\ast S + \Pi$ vanishes identically. Since $\varphi_\ast S$ and $\Pi$ are always orthogonal and a conformal mapping is an immersion it follows that $S$ and $\Pi$ vanish identically. Applying the definition of $S$ with $X=Y={\rm grad\,} \lambda$, we get that ${\rm grad\,} \lambda$ is the zero vector field. Consequently, $\lambda$ is constant and therefore $\varphi$ is a homothety. Since $\Pi$ vanishes, $\varphi$ is totally geodesic.

($\Leftarrow$) Obvious.

 Algebraic lemmas
-----------------

Suppose two finite dimensional spaces $V$ and $W$ equipped with inner products ${\langle},{\rangle}_V $ and ${\langle},{\rangle}_W $ are given. Let $B:V\times V\to W$ be a symmetric, bilinear form on $V$. Moreover, let $C\geq 0$. Consider a condition $$\label{SS_AL_E1}
{\langle}B(X,Z),B(Y,Z){\rangle}_W = C{\langle}X,Y{\rangle}_V {\langle}Z,Z{\rangle}_V.$$ for every $X,Y,Z\in V$. Then, if $X,Y$ are orthogonal $$\label{SS_AL_E01}
{\langle}B(X,X),B(Y,Y){\rangle}_W =
 -C{\langle}X,X{\rangle}_V {\langle}Y,Y{\rangle}_V.$$

\[Lem\_AL\] Assume that $B$ satisfies the condition . If $\dim V \geq 3$ then $C=0$. In particular, $B$ vanishes.

Suppose that $C\ne 0$. Take an orthonormal pair $X,Y$. Let $\xi = B(X,X)$ and $\zeta = B(Y,Y)$. By we have $$\label{SS_AL_E2}
{\langle}\xi,\xi{\rangle}= {\langle}\zeta,\zeta{\rangle}= C>0.$$ In particular, $\xi\ne 0$ and $\zeta\ne 0$. Applying we see that $${\langle}\xi, \zeta {\rangle}= - C.$$ Using above and one can obtain that $\xi=-\zeta$. Next, since $\dim V\geq 3$ we may find $Z\in V$ such that $X,Y,Z$ is an orthonormal triple. Let $\eta =  B(Z,Z)$. Then, by above $\xi=-\zeta=\eta=-\xi$, which contradicts the fact that $\xi\ne 0$.

Notice that the assumption $\dim V\geq 3$ is essential. Namely we have

\[SS\_AL\_PRP\] $($a$)$ A symmetric bilinear form $B:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}^2$ satisfies if and only if there exists an angle $\theta$ such that $$\label{AL_E}
B(X,Y)=\pm\sqrt{C}e^{i\theta}XY\quad {\rm or}\quad B(X,Y)=\pm\sqrt{C}e^{i\theta}\bar{X}\bar{Y},$$ where we identify $\mathbb{R}^2$ with $\mathbb{C}$.\
$($b$)$ If $\dim V=2$ and a non-zero symmetric bilinear form $B:V\times V\to W$ satisfies the condition then there exists a $2$-dimensional subspace $U$ of $W$ such that the image of $B$ is equal to $U$ and with respect to orthonormal bases of $V$ and $U$, $B$ is of the form .

\(a) Elementary exercise. (b) Take an orthonormal basis $X,Y$ of $V$. Since $B\ne 0$, we have $C\ne 0$. Consequently, $\xi = B(X,X)$, $\zeta = B(Y,Y)$ and $\eta = B(X,Y)$ are nonzero vectors of $W$ of length $\sqrt{C}$. By we see that ${\langle}\xi,\eta{\rangle}= {\langle}\zeta, \eta{\rangle}= 0$. Moreover, ${\langle}\xi, \zeta{\rangle}= -C$, by . It follows that $\xi = -\zeta$. Consequently, the image $U$ of $B$ is a two-dimensional subspace spanned by $\xi,\eta$.

Now taking orthonormal bases of $V$ and $U$, e.g., $X,Y$ and $\xi/\sqrt{C}, \eta/\sqrt{C}$, we reduce (b) to (a).

Conformality of a differential
==============================

In this section all manifolds are connected. Let $(M,g)$ and $(M',g')$ be Riemannian manifolds of dimensions $m$ and $m'$, respectively. We assume that $m,m'\geq 2$. Denote by $\nabla$ and $\nabla'$ the Levi-Civita connections of $g$ and $g'$, respectively. Equip their tangent bundles $\pi:TM\to M$ and $\pi':TM'\to M'$ with $(p,q,\alpha)$-metric $h$ and $(r,s,\beta)$-metric $h'$, respectively.

Let $\varphi: M\to M'$ and $\Phi = \varphi_\ast: TM\to TM'$. Put $g={\langle},{\rangle}$ and $g'={\langle},{\rangle}'$. Denote by $|\cdot|$ and $|\cdot|'$ the norms induced by $g$ and $g'$, respectively. Moreover, denote by $\|\cdot\|$ and $\|\cdot\|'$ the norms induced by $h$ and $h'$, respectively.

In the paper we use the following notation: If $\varphi$ (resp. $\Phi$) is conformal mapping then its dilatation will be always denoted by $\lambda$ (resp. $\Lambda$).

Technical lemmas
----------------

\[C1\] Suppose that $\varphi$ and $\Phi$ are conformal mappings. Then for any $Z\in T_x M$ and $x'=\varphi(x)$, $$\begin{aligned}
\Lambda(Z) &=& \lambda(x) \frac{(1 + \alpha(x) |Z|^2)^{p(x)}}{(1 + \lambda(x) \beta(x') |Z|^2)^{r(x')}},\label{C11}\\ 
   q (x) &=& \lambda (x) s (x').\label{C12}\end{aligned}$$

Let $X,Z\in T_xM$. Applying conformality of $\Phi$ and (K3) we have: $    \| (\varphi_\ast X)^v_{\varphi_\ast Z} \|'^2=\| \Phi_\ast X_Z^v \|'^2= \Lambda(Z) \|X^v_Z\|^2$. Using now the definitions of $h$ and $h'$ and conformality of $\varphi$, one can easily get $$\label{C13}
 \lambda(x) \frac{|X|^2 + \lambda(x) s(x') {\langle}X,Z{\rangle}^2}{(1 + \lambda(x) \beta(x') |Z|^2)^{r(x')}} = \Lambda(Z) \frac{|X|^2 + q(x) {\langle}X,Z{\rangle}^2}{(1+\alpha(x) 
|Z|^2)^{p(x)}}.$$

Taking nonzero vector $X$ orthogonal to $Z$, becomes . Next, let $Z\ne 0$. Putting $X=Z$ in and comparing the result with we get .

\[C3\] If $\Phi$ is conformal then so is $\varphi$. Moreover $\lambda(x) = \Lambda(0_x)$, $x\in M$.

Let $x\in M$ and $Z=0_x\in T_xM$. By (K3) and conformality of $\Phi$, for every $X,Y\in T_xM$ $$\begin{aligned}
{\langle}\varphi_\ast X,\varphi_\ast Y{\rangle}' &=& h'((\varphi_\ast X)^v_{\varphi_\ast Z},(\varphi_\ast Y)^v_{\varphi_\ast Z})\\
&=& \Lambda(Z) h(X^v_Z, Y^v_Z)=\Lambda(Z){\langle}X,Y{\rangle}.\end{aligned}$$

\[C2\] Adopt the notation from Section \[SS\_CM\]. Suppose that $\varphi$ and $\Phi$ are conformal mappings. Then

-   $\varphi$ is a homothety.

-   For every $x\in M$ one of the following conditions holds: $$\begin{aligned}
    &&p(x)=r(x')=0,\label{C2E1}\\
    &&p(x)=r(x')\ne 0\quad {\rm  and}\quad \lambda \beta(x') = \alpha(x),\label{C2E2}\\
    &&p(x)=r(x')=1 \quad {\rm and} \quad \lambda \beta(x') \ne \alpha(x),\label{C2E3} \\
    &&p(x)=1\quad {\rm  and}\quad r(x')=0.\label{C2E4}\end{aligned}$$

-   If for every $x\in M$ either or holds then $\Phi$ is also a homothety with the dilatation $\Lambda=\lambda$. Moreover, $\varphi$ is totally geodesic.

-   If or holds globally then for every $v,w,Z\in T_x M$ $$\label{C2E5} 
    {\langle}\Pi(\varphi_\ast v,\varphi_\ast Z), \Pi(\varphi_\ast w,\varphi_\ast Z){\rangle}'  = C{\langle}v,w{\rangle}{\langle}Z,Z{\rangle},$$ where $C=\lambda (\alpha(x)-\lambda \beta(x'))\ne 0$ in the case , and $C=\lambda\alpha(x)\ne 0$ in the case . In particular, $\varphi$ is not totally geodesic.

Assume that $v\ne 0$. Take vectors $A\in T_Z(TM)$ and $A'\in T_{Z'}(TL')$ as in Lemma \[CM\_nable\] and Corollary \[CM\_cor\_-1\]. Moreover we may assume that $A$ and $A'$ are horizontal with respect to $\nabla$ and $\bar \nabla$, respectively, i.e., $K(A)=0$ and $\bar K(A')=0$.

Put $U=U_x$ and $L'=L_{x'}$. Let $J:TL'\to TM'$ be the inclusion map. Put $\bar h=J^\ast h'$. Since $\varphi: (U, g)\to (L',\bar g)$ and $\Phi:(TU,h)\to (TL,\bar h)$ are conformal diffeomorphisms, so are $\varphi^{-1}:(L',\bar g)\to (U, g)$ and $
\Phi^{-1}:(TL',\bar h)\to (TU, h)$. Dilatations of $\varphi^{-1}$ and $\Phi^{-1}$ are equal to $\mu = (1/\lambda)\circ \varphi^{-1}$ and $\hat \mu = (1/\Lambda)\circ \Phi^{-1}$, respectively. Thus we have $$\begin{aligned}
\|\Phi_\ast A\|'^2 &=& \Lambda(Z)\|A\|^2,\\
\|\Phi^{-1}_\ast A'\|^2 &=& \hat \mu(Z')\|A'\|'^2.\end{aligned}$$

Put for a while $S=S(v,Z)$, $S'=S'(v',Z')$ and $\Pi'=\Pi(v',Z')$. Since $\pi_\ast A= v$, $\pi'_\ast A'=v'$, $K(A)=0$, $K'(A')=\bar K(A')+\Pi'=\Pi'$ and $Z'$ is orthogonal to $\Pi'$, we have $$\begin{aligned}
\|A\|^2 &=& |v|^2,\\
\|A'\|'^2 &=& |v'|'^2 + \omega_\beta(Z')^r|\Pi'|'^2\end{aligned}$$ Next applying Lemma \[CM\_nable\] and Corollary \[CM\_cor\_-1\], the equalities $\pi'_\ast \Phi_\ast A = v'$ and $\pi_\ast \Phi^{-1}_\ast A'=v$, and the fact that $\varphi_\ast S$ is orthogonal to $\Pi'$ we get $$\begin{aligned}
\|\Phi_\ast A \|'^2 &= &|v'|'^2 + \omega_\beta(Z')^r\big(\lambda(x)|S|^2 + s\lambda^2(x){\langle}S,Z{\rangle}^2 + |\Pi'|'^2\big)\\ 
\|\Phi^{-1}_\ast A' \|^2 &=& |v|^2 + \omega_\alpha(Z)^p\big(\mu(x') 
|S'|'^2 + q\mu^2(x') {\langle}S',Z'{\rangle}'^2\big)\end{aligned}$$

Combining now above equalities and using the definitions of $\mu$ and $\hat \mu$ we get $$\begin{aligned}
&&\Lambda |v|^2 = |v'|'^2 + \omega_\beta(Z')^r\big(\lambda|S|^2 + \lambda^2{\langle}S,Z{\rangle}^2 + |\Pi'|'^2\big),\label{TL_E1}\\
&&\frac{1}{\Lambda} \big(|v'|'^2 + \omega_\beta(Z')^r|\Pi'|'^2\big) = |v|^2 + \omega_\alpha(Z)^p\big(\frac{1}{\lambda} 
|S'|'^2 + \frac{1}{\lambda^2}  {\langle}S',Z'{\rangle}'^2\big)\label{TL_E2},\end{aligned}$$ where $\lambda=\lambda(x)$ and $\Lambda=\Lambda(Z)$. Multiplying equations and side by side we conclude that $$0 = \omega_\beta(Z')^r\lambda|v|^2|S|^2 + \text{non-negative expression}.$$ Since $v\ne 0$, $S(v,Z)=0$. Since $x\in M$, $v,Z\in T_x M$ were arbitrary the tensor field $S$ vanishes identically. Therefore, $\lambda$ is a constant function and thus $\varphi$ is a homothety. Hence (a) is proved.

Substituting $S=0$ in we get $$|\Pi(v',Z')|'^2 = \frac{\Lambda(Z)-\lambda}{ \omega_\beta(Z')^r} |v|^2.$$ Applying Lemma \[C1\] we get $$\label{C2EP}
|\Pi(\varphi_\ast v,\varphi_\ast Z)|'^2 = \lambda ((1+\alpha|Z|^2)^p-(1+\lambda \beta |Z|^2)^r)|v|^2.$$ Using the facts that the map $(v,Z)\mapsto |\Pi(\varphi_\ast v,\varphi_\ast Z)|'^2$ is non negative and symmetric with respect to $v,Z$, we conclude (b).

If or holds then by it follows that $\Lambda=\lambda$. Moreover, in these cases becomes $|\Pi(\varphi_\ast v,\varphi_\ast Z)|'^2=0$. This proves (C).

If or holds then it is an elementary computation to check that $\Pi$ satisfies , proving (d).

\[Lem\_CC2\] Suppose that $\dim M\geq 3$ or $\dim M'\leq \dim M+ 1$. Then under the assumptions of Lemma \[C2\] we have: $\varphi $ is totally geodesic homothety, $\Phi $ is a homothety and its dilatation $\Lambda$ is equal to $\lambda$.

It suffices to show that under the assumptions the conditions and cannot hold. Then the assertion follows from Lemma \[C2\] (a) and (c). Suppose that or holds. Then by Lemma \[C2\] (d) it follows that the symmetric bilinear form $B:T_{x}M \times T_x M \to T_{x'}M'$ given by $B(v,w)=\Pi(\varphi_\ast v,\varphi_\ast w)$ satisfies the condition with $C\ne 0$. If $\dim M\geq 3$ then we have a contradiction with Lemma \[Lem\_AL\], if $\dim M'\leq \dim M +1$ then we have a contradiction with Lemma \[SS\_AL\_PRP\].

Main results
------------

We begin with some definitions. Suppose $\bar M$ is a submanifold of a Riemannian manifold $(M',g')$. Suppose that a real-valued non-negative function $C$ on $\bar M$ is given. We say that $\bar M$ is [*optimal with a coefficient $C$*]{} if for every $x'\in \bar M$ the second fundamental form $\Pi$ of $\bar M$ at $x'$ satisfies with the constant $C(x')$ that is $${\langle}\Pi(u,w),\Pi(v,w){\rangle}=C(x'){\langle}u, v{\rangle}{\langle}w,w{\rangle},\quad u,v,w\in T_{x'}\bar M.$$

In particular, every totally geodesic submanifold is optimal with the coefficient $0$. By Lemma \[Lem\_AL\] and Lemma \[SS\_AL\_PRP\] it follows that if $\dim \bar M\geq 3$ or ${\rm codim\,} \bar M\leq 1$ then each optimal submanifold is totally geodesic.

\[MR\_R1\] Observe that if $\varphi:M\to \bar M$ is a conformal diffeomorphism such that holds then $\bar M$ is optimal with the coefficient $C/(\lambda\circ\varphi^{-1})^2$.

\[PRP\_OPT\] Suppose $\dim \bar M=2$. Denote by $\kappa'$ and $\bar \kappa$ the sectional curvatures of $M'$ and $\bar M$, and let $\sigma = T_{x'} \bar M$. If $\bar M$ is optimal submanifold of $M'$ with a coefficient $C$ then $\bar M$ is minimal submanifold of $M'$ and $\bar \kappa(\sigma) = \kappa'(\sigma) - 2C(x')$. In particular, if $C$ is constant and $M'$ is a space of constant curvature then so is $\bar M$.

The fact that $\bar M$ is minimal follows immediately from Lemma \[SS\_AL\_PRP\]: it suffices to calculate the trace of the bilinear form given by . The second statement follows from , and the Gauss Equation.

Suppose now that two Riemannian manifolds $(M,g)$ and $(M',g')$ are given and $\dim M\leq \dim M'$. Equip their tangent bundles $TM$ and $TM'$ with $(p,q,\alpha)$-metric $h$ and $(r,s,\beta)$-metric $h'$ respectively. Suppose next that the functions $p,q,r,s,\alpha,\beta$ are constant, and $\varphi:M\to M'$ is an imbedding (injective immersion). Let $\bar M=\varphi(M)$.

\[MR\_T1\] Let $\dim M\geq 3$ or $\dim M'\leq \dim M +1$.

-   Suppose that $\varphi$ is a conformal mapping with a dilatation $\lambda$. Then $\Phi=\varphi_\ast:TM\to TM'$ is conformal if and only if $q=\lambda (s\circ\varphi^{-1})$, $\varphi$ is a homothety, $\bar M$ is totally geodesic and for every $x\in M$ $(x'=\varphi(x))$ one of the conditions or holds.

-   If $\Phi$ is a conformal mapping then $\varphi$ and $\Phi$ are homotheties and $\Lambda = \lambda$.

\[MR\_T2\] Let $\dim M = 2$ and $\dim M'\geq \dim M +2$.

-   Suppose that $\varphi$ is a conformal mapping with a dilatation $\lambda$. Then $\Phi=\varphi_\ast:TM\to TM'$ is conformal if and only if $\varphi$ is a homothety, $q=\lambda (s\circ\varphi^{-1})$, $\bar M$ is optimal with the coefficient $$C=\frac{1}{\lambda}((p\alpha)\circ\varphi^{-1}-\lambda \beta r)$$ and for every $x\in M$ $(x'=\varphi(x))$ one of the properties - is satisfied.

-   Suppose $\Phi$ is a conformal mapping. Then $\varphi$ a homothety and

    -   If for every $x\in M$ one of the conditions or holds then $\Phi$ is also a homothety and $\Lambda = \lambda$.

    -   If one of the conditions or holds globally then $\varphi$ is a minimal immersion and for every plane $\sigma=\varphi_\ast (T_x M)$, $x\in M$, the Gauss curvature $\kappa(\sigma)$ of $M$ is $$\label{MR_T2_cur} 
        \kappa(\sigma) =\lambda \kappa'(\sigma) - 2 C(\varphi(x))\lambda,$$ where $\kappa'(\sigma)$ is the Gauss curvature of $M'$. Moreover, $\Phi$ is not a homothety. Its dilatation $\Lambda$ is $$\Lambda(Z)= \lambda \frac{1+\alpha(x) g(Z,Z)}{1+\lambda \beta(x')r(x') g(Z,Z)}, \quad Z\in T_x M, x'=\varphi(x)\label{MR_T2_E1}.$$

(I, III $\Rightarrow$) Suppose $\varphi$ and $\Phi$ are conformal mappings. By Lemma \[C2\], $\varphi$ is a homothety and for every $x\in M$ ($x'=\varphi(x)$) one of the conditions - holds. By Lemma \[C1\], $q=\lambda (s\circ\varphi^{-1})$. If $\dim M\geq 3$ or $\dim M'\leq \dim M +1$, by Lemma \[C2\] (d) and Lemma \[Lem\_CC2\] conditons and cannot hold. Therefore Lemma \[C2\] (c) implies that $\bar M$ is totally geodesic. Moreover, if or is satisfied then by Lemma \[C2\] (d) and Remark \[MR\_R1\], $\bar M$ is optimal with the coefficient $C=(1/\lambda)((p\alpha)\circ\varphi^{-1}-\lambda \beta r)$.

(I, III $\Leftarrow$) Taking horizontal (resp. vertical) $A\in T_ZTM$, computing $h'(\Phi_\ast A,\Phi_\ast A)$ and applying relations between $p,q,r,s,\alpha,\beta$ and $\lambda$ one can conclude that $\Phi$ is conformal.

\(II) Suppose $\Phi$ is conformal. By Lemma \[C3\], $\varphi$ is also conformal. Therefore (I) and Lemma \[C1\] imply that $\Phi$ and $\varphi$ are homotheties and $\Lambda=\lambda$.

\(IV) As above we conclude that $\varphi$ is conformal.

(IV1) It is a consequence of Lemma \[C2\] (c).

(IV2) Suppose for every $x\in M$ one of the conditions or holds. By Proposition \[PRP\_OPT\], (I) and the fact that the curvature under the action of a homothety with dilatation $\lambda$ is scaled by $1/\lambda$, $\varphi$ is a minimal immersion and holds. Conditions , and equation imply .

As a direct consequence of of Theorem \[MR\_T1\] we obtain

\[Cor\_MRT1\] Suppose $\dim M\geq 3$ or $\dim M'\leq \dim M+1$. Let $\varphi:(M,g)\to (M',g')$ be an imbedding. Then we have:

-   $\Phi:(TM,h_S)\to (TM',h'_S)$ is conformal if and only if $\varphi$ is totally geodesic homothety.

-   $\Phi:(TM,h_{CG})\to (TM',h'_{CG})$ is conformal if and only if $\varphi$ is totally geodesic isometric imbedding.

-   $\Phi:(TM,h_{CG})\to (TM',h'_S)$ is never conformal.

An example to Theorem \[MR\_T2\] {#SS_EX}
--------------------------------

It is important to show that there is essential difference between Theorems \[MR\_T1\] and \[MR\_T2\]. To do this we give an example of $2$-dimensional manifold $M$, $4$-dimensional manifold $M'$ and an immersion $\varphi:M\to M'$ such that $\bar M= \varphi (M)$ is optimal but not totally geodesic.

Let $\Sigma^d(\rho)$ denote Euclidean $d$-dimensional sphere of radius $\rho$ centred at the origin in $\mathbb{R}^{d+1}$. Recall (see [@Ch], Chapter 4 §5 page 139) that the [*second standard immersion*]{} of $\Sigma^2(1)$ it is a map $\varphi:\Sigma^2(1)\to \Sigma^4(1/\sqrt{3})$ defined as follows: Consider harmonic homogeneous polynomials $u_i$, $i=1,\dots,5$, in $\mathbb{R}^3$ given by $$\begin{aligned}
u_1&=&x_2 x_3, \quad u_2=x_1x_3,\quad u_3=x_1x_2\\
u_4&=& \frac12(x_1^2-x_2^2),\quad u_5=\frac{\sqrt 3}{6}(x_1^2+x_2^2-2x_3^2)\end{aligned}$$ and let $u=(u_1,\dots,u_5)$. We define $\varphi$ to be the restriction $u|\Sigma^2(1)$. Then $\varphi:\Sigma^2(1)\to \Sigma^4(1/\sqrt{3})$ is an isometric immersion (but not imbedding). Nevertheless, $\bar M = \varphi (\Sigma^2(1))$ is a minimal submanifold of $\Sigma^4(1/\sqrt{3})$. We show that $\bar M$ is optimal with the constant coefficient $C=1$.

\[Lem\_EXS\] Suppose $(M,g)$ is a Riemannian manifold and $u:M\to \mathbb{R}^{d+1}$. Assume that the image $\bar M=u(M)$ is contained in $\Sigma=\Sigma^d(\rho)$ and $u: M\to \Sigma$ is an imbedding. Denote by $\Pi$ and $\bar \Pi$ the second fundamental form of $\bar M$ in $\Sigma$ and $\bar M$ in $\mathbb{R}^{d+1}$, respectively. Then for every $x'\in \Sigma$ and for every basis $(e_i)$ of $T_{x'} \bar M$ $${\langle}\Pi(e_i,e_k),\Pi(e_j,e_l){\rangle}=  {\langle}\bar \Pi(e_i,e_k),\bar \Pi(e_j,e_l){\rangle}- \frac{1}{\rho^2}{\langle}e_i,e_k{\rangle}{\langle}e_j, e_l {\rangle},$$ where ${\langle},{\rangle}$ is the canonical inner product in $\mathbb{R}^{d+1}$.

Elementary exercise.

If $\varphi$ is the second standard immersion then the submanifold $\bar M=\varphi(\Sigma^2(1))$ is optimal with the constant coefficient $C=1$.

Let $M=\Sigma^2(1)$ and $\Sigma= \Sigma^4(1/\sqrt{3})$. Adopt the notations from Lemma \[Lem\_EXS\]. Denote by $\tilde\varphi$ the restriction of $\varphi$ to lower half sphere $\Sigma^2_-(1)$. Since $\varphi(\Sigma^2(1))$ coincides with the closure of $\tilde \varphi(\Sigma^2_-(1))$, it suffices to prove that $\bar M_-=\tilde \varphi(\Sigma^2_-(1))$ is optimal with the coefficient one.

Let $f$ be the stereographic projection $\Sigma^2(1)\to \mathbb{R}^2$ from the north pole. Put $\psi = f\circ \tilde\varphi^{-1}$. Fix $x'=\psi^{-1}(t)$, where $t=(t_1;t_2) \in \mathbb{R}^2$, $|t|<1$. Let $e_i=(\partial / \partial 
\psi_i)(x')$. Put $t^2=t_1^2+t_2^2$ and $t^4 = (t^2)^2$. Since $\varphi$ is an isometric immersion and $f$ is the stereographic projection we have $${\langle}e_i,e_j{\rangle}=
 \frac{4\delta_{ij}}{(t^2+1)^2},\quad i,j=1,2,$$ where $\delta_{ij}$ is the Kronecker symbol. In the light of Lemma \[Lem\_EXS\] (with $\rho = 1/\sqrt 3$), to finish the proof it suffices to show that $$\begin{aligned}
{\langle}\bar \Pi(e_i,e_k), \bar \Pi(e_j,e_k) {\rangle}&=& (3\delta_{ik}+1)\frac{16 \delta_{ij}}{(t^2+1)^4}\label{SS_EX_PRPE1}\\
& =&(3\delta_{ik}+1) {\langle}e_i,e_j{\rangle}^2,\quad i,j,k = 1,2.\nonumber 
$$

After elementary but laborious calculations we get $$\begin{gathered}
\bar \Pi (e_1,e_1)=\frac{4}{(t^2+1)^4}\Big(4t_2(1-t^2-2t_1^2);8t_1(1-t_1^2);4t_1t_2(t_1^2-t_2^2-3);\\
t^4-8t_1^2t_2^2+6t_2^2-6t_1^2+1;\sqrt{3}(t^4-2t^2-4t_1^2+1)\Big)\end{gathered}$$ and $$\begin{gathered}
\bar \Pi (e_2,e_2)=\frac{4}{(t^2+1)^4} \Big( 8t_2(1-t^2_2);4t_1(1-t^2-2t_2^2);4t_1t_2(t_2^2-t_1^2-3);\\
6t_2^2-6t_1^2 +8t_1^2t_2^2-t^4-1;\sqrt{3}(t^4-2t^2-4t_2^2+1)\Big)\end{gathered}$$ and $$\begin{gathered}
\bar \Pi (e_1,e_2)=\frac{4}{(t^2+1)^4}\Big( 2t_1(t^2-4t_2^2+1);2t_2( t^2-4t_1^2+1);\\
8t_1^2t_2^2-t^4+1;
4t_1t_2(t_1^2-t_2^2);-4\sqrt{3}t_1t_2\Big).\end{gathered}$$ Now one can check that holds.

Now let $\mathbb{R}P^2$ denote the real $2$-dimensional projective space. We treat $\mathbb{R}P^2$ as a Riemannian manifold whose metric $g$ is given by the standard two-sheeted covering map $\hat \pi:\Sigma^2(1)\to \mathbb{R}P^2$. Put $\hat \varphi(\hat x) = \varphi(x)$ if $\hat x = \hat \pi(x)$. Since $\varphi(x)=\varphi(-x)$, the map $\hat \varphi$ is well defined. Moreover, $\hat \varphi :\mathbb{R}P^2 \to \Sigma^4(1/\sqrt 3)$ is an imbedding. It is called [*the first standard imbedding* ]{} of $\mathbb{R}P^2$ into $\Sigma^4(1/\sqrt 3)$.

Take constants $q,\alpha >0$. Suppose Cheeger-Gromoll type metrics $h$ and $h'$ on $T(\mathbb{R}P^2)$ and $T(\Sigma^4(1/\sqrt 3))$ are given. If

\(1) $h=h_{1,q,\alpha+1}$ and $h'=h'_{1,q,\alpha}$, or

\(2) $h=h_{1,q,1}$ and $h'=h'_{0,q,1}$\
then $\hat \varphi_\ast$ is a conformal mapping, but not a homothety. Its dilatation is $\Lambda(Z)= (1+ (\alpha+1)g(Z,Z))/(1 + \alpha g(Z,Z))$ in the case of (1) and $\Lambda(Z) = 1+g(Z,Z)$ in the case of (2).

[99]{} M. Benyounes, E. Loubeau, C. M. Wood, [*Harmonic sections of Riemannian bundles and metrics of Cheeger-Gromoll type*]{}, Diff. Geom. Appl. [**25**]{} (2007), 322-334. M. Benyounes, E. Loubeau, C. M. Wood, [*The geometry of generalized Cheeger-Gromoll metrics*]{}, Preprint (arXiv: math.DG/0703059v1). B. Chen, [*Total mean curvature and submanifolds of finite type*]{}, World Scientific, Singapore (1984). S. Gudmundsson, E. Kappos, [*On the geometry of the tangent bundle with the Cheeger-Gromoll metric*]{}, Tokyo J. Math. [**25**]{} (2002), 75-83. O. Kowalski, [*Curvature of the induced Riemannian metric on the tangent bundle of a Riemannian manifold*]{}, J. reine anegew. Math. [**250**]{} (1971) pp. 124-129. W. Kozłowski, Sz. M. Walczak, [*Collapse of unit horizontal bundles equipped with a metric of Cheeger-Gromoll type*]{}, Diff. Geom. Appl., to appear. M. I. Munteanu, [*Some aspects on the geometry of the tangent bundles and tangent sphere bundles of a Riemannian manifold*]{}, Mediterr. J. Math. [**5**]{} (2008), no. 1, 43–59.

Wojciech Kozłowski: wojciech@math.uni.lodz.pl\
Kamil Niedziałomski: kamiln@math.uni.lodz.pl\
Faculty of Mathematics and Computer Science\
Łódź University\
Banacha 22, 90-238 Łódź\
Poland
