3.1. Definition, contact structure

The Heisenberg group $\mathfrak{H}$ is the set ${\mathbb{C}}\times {\mathbb{R}}$ with multiplication $*$ given by

\begin{equation*} (z,t)*(w,s)=(z+w,t+s+2\Im (z\overline {w})). \end{equation*}

The Heisenberg group $\mathfrak{H}$ is a 2-step nilpotent Lie group. Consider the left-invariant vector fields

\begin{eqnarray*} X=\frac {\partial }{\partial x}+2y\frac {\partial }{\partial t},\quad Y=\frac {\partial }{\partial y}-2x\frac {\partial }{\partial t}, \quad T=\frac {\partial }{\partial t}, \end{eqnarray*}

satisfying $[X,Y]=-4T.$ We also use complex fields

\begin{equation*} Z=\frac {1}{2}(X-iY)=\frac {\partial }{\partial z}+i\overline {z}\frac {\partial }{\partial t},\quad \overline {Z}=\frac {1}{2}(X+iY)=\frac {\partial }{\partial \overline {z}}-iz\frac {\partial }{\partial t}. \end{equation*}

The vector fields $X,Y,T$ form a basis for the Lie algebra of left-invariant vector fields of $\mathfrak{H}$ . The Lie algebra $\mathfrak{h}$ of $\mathfrak{H}$ has a grading $\mathfrak{h} = \mathfrak{v}_1\oplus \mathfrak{v}_2$ with

\begin{align*} \mathfrak{v}_1 = \mathrm{span}_{{\mathbb{R}}}\{X, Y\}\quad \text{and}\quad \mathfrak{v}_2=\mathrm{span}_{{\mathbb{R}}}\{T\}. \end{align*}

In $\mathfrak{H}$ we consider the 1-form

(2) \begin{equation} \omega =dt+2xdy-2ydx=dt+2\Im (\overline {z}dz). \end{equation}

The following proposition holds; it summarises well-known facts about $\mathfrak{H}$ :

The sub-Riemannian structure of $\mathcal{H}$ is defined by the relations

\begin{equation*} g_{cc}(X,X)=g_{cc}(Y,Y)=1,\quad g_{cc}(X,Y)=0. \end{equation*}

The sub-Riemannian metric is then given by

\begin{equation*} g_{cc}=ds_{cc}^2=dx^2+dy^2. \end{equation*}

The isometry group $\textrm {Isom}_{cc}(\mathfrak{H})$ of the sub-Riemannian metric $g_{cc}$ comprises compositions of:

1. (1) Left-translations $T_{(\zeta ,s)}$ , $(\zeta ,s)\in \mathfrak{H}$ , defined by $ T_{(\zeta ,s)}(z,t)=(\zeta ,s)*(z,t).$ The group of left-translations is isomorphic to $\mathfrak{H}$ .

2. (2) Conjugation $j$ , defined by $ j(z,t)=(\overline {z},-t).$

3. (3) Rotations $R_\theta$ , $\theta \in {\mathbb{R}}$ , defined by $ R_\theta (z,t)=(ze^{i\theta },t)$ for every $(z,t)\in \mathfrak{H}$ . The group of rotations is isomorphic to $\textrm {U}(1)$ .

Left-translations and rotations are CR maps which preserve $\omega$ whereas conjugation is anti-CR, which skew-preserves $\omega$ : $j^*(\omega )=-\omega$ . The isometry group of $g_{cc}$ comprises of composites of the above mappings:

\begin{equation*} \textrm {Isom}(\mathfrak{H}, g_{cc})\simeq \mathfrak{H}\rtimes \textrm {U}(1)\times \mathbb{Z}_2. \end{equation*}

The dilations $ D_\delta$ ( $\delta \gt 0$ ) which are defined by

\begin{equation*} D_\delta (z,t)=(\delta z, \delta ^2 t) \end{equation*}

for every $(z,t)\in \mathfrak{H}$ are homotheties for the metric $g_{cc}$ and they are also $\textrm {CR}$ maps. The group of dilations is isomorphic to the multiplicative group ${\mathbb{R}}_{\gt 0}$ .

3.2. From Riemannian approximants to contact Riemannian structure

Using the sub-Riemannian metric of $\mathfrak{H}$ , we construct a contact Riemannian structure in the Heisenberg group as follows: for $L\gt 0$ , we consider a Riemannian metric in $\mathfrak{H}$ such that the frame

\begin{equation*} \{X,Y, T_L=T/\sqrt {L}\} \end{equation*}

is orthonormal. Explicitly,

\begin{equation*} g_L=ds_L^2=dx^2+dy^2+L\omega ^2=ds^2_{cc}+L\omega ^2. \end{equation*}

Note here that $g_L$ is the metric defined in [Reference Capogna, Danielli, Pauls and Tyson5] (mind only the different notation of the definition of the Heisenberg group). Let $\phi$ be the extension of $J$ in the whole tangent bundle ${T}(\mathfrak{H})$ by setting $\phi (T_L)=0.$ Then $(\mathfrak{H};\,\sqrt {L}\omega ,T_L,$ $ \phi ,g_L)$ is a contact Riemannian structure if and only if $L=1/4$ by checking the equations equation (1): the equation

\begin{equation*} (1/2)\,d(\sqrt {L}\omega )(X, Y)=g_L(\phi X, Y), \end{equation*}

is equivalent to

\begin{equation*}(\sqrt {L}/2)\,d\omega (X, Y)=2\sqrt {L}=g_L(Y, Y)=1,\end{equation*}

i.e., $L=1/4.$ From now, we will write $g$ instead of $g_{1/4}$ , $\widetilde {T}$ instead of $2T$ and $\widetilde {\omega }$ instead of $(1/2)\omega$ .

Let $\nabla$ be the Levi-Civita connection. Using Koszul’s formula (in the case of orthonormal frames)

(3) \begin{equation} g(\nabla _VU,W)=-(1/2)\left (g([U,W],V)+g([V,W],U)+g([U,V],W)\right )\!, \end{equation}

we find

\begin{eqnarray*} && \nabla _{X}X=0,\quad \nabla _{X}Y=-\widetilde {T},\quad \nabla _{X}\widetilde {T}=Y,\\ && \nabla _{Y}X=\widetilde {T},\quad \nabla _{Y}Y=0,\quad \nabla _{Y}\widetilde {T}=-X,\\ && \nabla _{\widetilde {T}}X=Y,\quad \nabla _{\widetilde {T}}Y=-X,\quad \nabla _{\widetilde {T}}\widetilde {T}=0. \end{eqnarray*}

Let

(4) \begin{equation} R(U,V)W=\nabla _{V}\nabla _{U}W-\nabla _{U}\nabla _{V}W +\nabla _{[U,V]}W, \end{equation}

be the Riemannian curvature tensor. We have

\begin{eqnarray*} && R(X,\widetilde {T})X=\widetilde {T},\quad R(X,\widetilde {T})Y=0,\quad R(X,\widetilde {T})\widetilde {T}=-X,\\ && R(Y,\widetilde {T})X=0,\quad R(Y,\widetilde {T})Y=\widetilde {T},\quad R(Y,\widetilde {T})\widetilde {T}=-Y\\ && R(X,Y)X=-3Y,\quad R(X,Y)Y=3X,\quad R(X,Y)\widetilde {T}=0. \end{eqnarray*}

Using the relation $ K(U,V)=g(R(U,V)U,V)$ for sectional curvature of planes spanned by unit vectors $U,V$ we obtain the following:

Corollary 3.2. The sectional curvatures of the distinguished planes spanned by a) $X,Y$ , b) $X,\widetilde {T}$ and c) $Y,\widetilde {T}$ are, respectively:

\begin{eqnarray*} && K(X,Y)= -3,\quad K(X,\widetilde {T})= 1,\quad K(Y,\widetilde {T})= 1. \end{eqnarray*}

3.3. Sasakian structure

Recall the definition given in Section 2. In our case, we have first that the Reeb vector field $\widetilde {T}$ is by definition unit; moreover, it is in fact Killing.

Proof. It suffices to show that for every vector fields $U,V$ we have

\begin{equation*} g(\nabla _V\widetilde {T},U)+g(\nabla _U\widetilde {T}, V)=0. \end{equation*}

Set $U=a_1X+b_1Y+c_1\widetilde {T}$ , $V=a_2X+b_2Y+c_2\widetilde {T}$ . Then

\begin{eqnarray*} && \nabla _U\widetilde {T}=a_1Y-b_1X,\\ && \nabla _V\widetilde {T}=a_2Y-b_2X \end{eqnarray*}

and

\begin{equation*} g(\nabla _V\widetilde {T},U)+g(\nabla _U\widetilde {T}, V)=a_2b_1-b_2a_1+a_1b_2-b_1a_2=0. \end{equation*}

Now, if $U$ and $V$ are as above then

\begin{equation*} R(U,\widetilde {T})V=-a_1c_2X-b_1c_2Y+(a_1a_2+b_1b_2)\widetilde {T}. \end{equation*}

On the other hand, a direct calculation yields

\begin{equation*} g(U,V)\widetilde {T}-g(\widetilde {T},V)U=-a_1c_2X-b_1c_2Y+(a_1a_2+b_1b_2)\widetilde {T}. \end{equation*}

From Lemma 3.3 and the above relations, we have

3.4. $\textrm {CR}$ and Sasakian automorphisms

Let $\mathfrak{CR}(\mathfrak{H})$ be the group of $\textrm {CR}$ maps of $\mathfrak{H}$ and let also $\mathfrak{Aut}(\mathfrak{H})$ be the group of Sasakian automorphisms of $\mathfrak{H}$ : that is, $g$ -isometries $f$ which also satisfy $f^*\widetilde {\omega }=\widetilde {\omega }$ . Denote by $(\mathfrak{CR}(\mathfrak{H}))_0$ and $(\mathfrak{Aut}(\mathfrak{H}))_0$ their respective connected components of identity. We have the following theorem (Theorem 5.3 of [Reference Boyer2]):

4.1. Kähler form I

Let $\mathcal{C}(M)$ be the cone over an $n$ -dimensional Riemannian manifold $(M, g)$ . It is well-known that the holonomy of $\mathcal{C}(M)$ encodes significant geometric information about the manifold $M.$ In particular, the manifold $(M, g)$ is Sasakian if and only if the cone $(\mathcal{C}(M), dr^2+r^2g)$ is Kähler (see, for instance, [Reference Boyer and Galicki3]). Now Proposition 3.4 immediately implies that $ \mathcal{C}(\mathfrak{H})=(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{J},g_r,\Omega _r)$ is Kähler. We describe below the features of $\mathcal{C}(\mathfrak{H})$ : first, we consider the orthonormal basis $\{X_r,Y_r,T_r,\partial _r\}$ for the metric $g_r$ , that is,

\begin{equation*} X_r=(1/r)X,\quad Y_r=(1/r)Y,\quad T_r=(1/r)\widetilde {T},\quad \partial _r=d/dr. \end{equation*}

We note that all Lie brackets vanish besides

\begin{equation*} [X_r,Y_r]=-(2/r)T_r,\quad [X_r,\partial _r]=(1/r)X_r, \quad [Y_r,\partial _r]=(1/r)Y_r,\quad [T_r,\partial _r]=(1/r)T_r. \end{equation*}

The action of the complex structure $\mathbb{J}$ is given by

\begin{equation*} \mathbb{J} X_r=Y_r,\quad \mathbb{J} Y_r=-X_r,\quad \mathbb{J} T_r=-\partial _r,\quad \mathbb{J} \partial _r=T_r. \end{equation*}

Let also $ \phi =dx$ and $ \psi =dy.$ The following hold:

\begin{equation*} r\phi =X_r^*,\quad r\psi =Y_r^*,\quad r\widetilde {\omega }=T_r^*,\quad dr=\partial _r^*. \end{equation*}

We write $\phi _1=r(\phi +i\psi )$ , $\phi _2=dr+ir\widetilde {\omega }$ and $Z_r=(1/2)(X_r-iY_r)$ , $V_r=(1/2)(\partial _r-iT_r)$ , so that $\phi _1=Z_r^*$ and $\phi _2=V_r^*$ . The Kähler metric $g_r$ and the Kähler form $\Omega _r$ are given respectively by

\begin{eqnarray*} && g_r=dr^2+r^2g=dr^2+r^2(\phi ^2+\psi ^2+(\widetilde \omega )^2)=|\phi _1|^2+|\phi _2|^2,\\ && \Omega _r=d\left (\frac {r^2}{2}\widetilde \omega \right )=rdr\wedge \widetilde \omega +r^2\;\phi \wedge \psi =\frac {i}{2}(\phi _1\wedge \overline {\phi _1}+\phi _2\wedge \overline {\phi _2}). \end{eqnarray*}

4.1.1. Curvature

If $\nabla ^r$ is the Riemannian connection, we obtain by Koszul’s formula that

\begin{eqnarray*} && \nabla ^r_{X_r}X_r=-(1/r)\partial _r,\quad \nabla ^r_{Y_r}X_r=(1/r)T_r,\quad \nabla ^r_{T_r}X_r=(1/r)Y_r,\quad \nabla ^r_{\partial _r}X_r=0,\\ && \nabla ^r_{X_r}Y_r=-(1/r)T_r,\quad \nabla ^r_{Y_r}Y_r=-(1/r)\partial _r,\quad \nabla ^r_{T_r}Y_r=-(1/r)X_r,\quad \nabla ^r_{\partial _r}Y_r=0,\\ && \nabla ^r_{X_r}T_r=(1/r)Y_r,\quad \nabla ^r_{Y_r}T_r=-(1/r)X_r,\quad \nabla ^r_{T_r}T_r=-(1/r)\partial _r,\quad \nabla ^r_{\partial _r}T_r=0,\\ && \nabla ^r_{X_r}\partial _r=(1/r)X_r,\quad \nabla ^r_{Y_r}\partial _r=(1/r)Y_r,\quad \nabla ^r_{T_r}\partial _r=(1/r)T_r,\quad \nabla ^r_{\partial _r}\partial _r=0. \end{eqnarray*}

In the next proposition, we compute the sectional curvatures of distinguished planes.

Proposition 4.1. The sectional curvatures at all other pairs of distinguished planes vanish besides that of the distinguished plane spanned by $X_r,Y_r$ :

\begin{equation*} K_r(X_r,Y_r)=-\frac {4}{r^2}\lt 0. \end{equation*}

Proof. If $R_r$ is the Riemannian curvature tensor, we have

\begin{eqnarray*} && R_r(X_r,Y_r)X_r=-(4/r^2)Y_r, \end{eqnarray*}

and

\begin{eqnarray*} && R_r(X_r,T_r)X_r= R_r(X_r,\partial _r)X_r= R_r(Y_r,T_r)Y_r\\ && = R_r(Y_r,\partial _r)Y_r= R_r(T_r,\partial _r)T_r=0. \end{eqnarray*}

Thus, the holomorphic sectional curvature of the plane spanned by $X_r$ and $Y_r$ is $ K_r(X_r,Y_r)=-4/r^2\lt 0.$ All other sectional curvatures of distinguished planes vanish.

Corollary 4.2. The Ricci curvatures of $g_r$ in the directions of $X_r, Y_r, T_r$ and $\partial _r$ are respectively

\begin{eqnarray*} && \textrm {Ric}(X_r)=\textrm {Ric}(Y_r)=-\frac {4}{3r^2},\quad \textrm {Ric}(T_r)=\textrm {Ric}(\partial _r)=0, \end{eqnarray*}

and the scalar curvature is

\begin{equation*} K=-\frac {2}{3r^2}. \end{equation*}

4.2. Kähler form II

On $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ we consider the basis of the tangent space comprising

\begin{equation*} X,Y,T,\partial _r \end{equation*}

and we define an almost complex structure $\mathbb{I}$ on $\mathcal{C}(\mathfrak{H})$ by the relations

\begin{equation*} \mathbb{I} X=Y,\quad \mathbb{I} Y=-X,\quad \mathbb{I} T=\partial _r,\quad \mathbb{I}\partial _r=-T. \end{equation*}

For positive continuous functions $a=a(r)$ and $b=b(r)$ we consider the Riemannian metrics $g_{a.b}$ on $\mathcal{C}(\mathfrak{H})$ defined by

\begin{equation*} g_{a,b}=\frac {dx^2+dy^2}{a^2}+\frac {\omega ^2+dr^2}{b^2}. \end{equation*}

Recall that $\omega =dt+2xdy-2ydx$ and ${\ker \omega }=\textrm {Span}\{X, Y\}$ (see Proposition 3.1). An orthonormal basis for this metric comprises the vector fields

\begin{equation*} X^{\prime}=aX,\quad Y^{\prime}=aY \quad T^{\prime}=bT,\quad R^{\prime}=b\partial _r, \end{equation*}

which satisfy the following bracket relations:

\begin{eqnarray*} && [X^{\prime},Y^{\prime}]=-(4a^2/b)T^{\prime},\quad [X^{\prime},T^{\prime}]=0,\quad [X^{\prime},R^{\prime}]=-(b\dot a/a)X^{\prime},\\ && [Y^{\prime},T^{\prime}]=0,\quad [Y^{\prime},R^{\prime}]=-(b\dot a/a)Y^{\prime},\\ && [T^{\prime},R^{\prime}]=-\dot b T^{\prime}. \end{eqnarray*}

The corresponding fundamental form for $g_{a,b}$ is

\begin{equation*} \Omega _{a,b}=\frac {dx\wedge dy}{a^2}+\frac {\omega \wedge dr}{b^2}. \end{equation*}

The triple $(\mathbb{I}, g_{a,b}, \Omega _{a,b})$ is a Hermitian structure on $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ : we verify that $\mathbb{I}$ is integrable since the Nijenhuis tensor vanishes. Moreover,

\begin{equation*} g_{a, b}(\mathbb{I} U, \mathbb{I} V)=g_{a, b}(U, V),\quad \Omega _{a, b}(U, V)=g_{a, b}(\mathbb{I} U, V) \end{equation*}

for any vector fields $U, V.$

Proof. The Hermitian manifold $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g_{a,b}, \Omega _{a,b})$ is Kähler if and only if $\Omega _{a,b}$ is closed. We check at once that

\begin{eqnarray*} d\Omega _{a,b}&=&-2(\dot a/a^3)dr\wedge dx\wedge dy+(4/b^2)dx\wedge dy\wedge dr\\ &=&\left (4/b^2-2\dot a/a^3\right )dx \wedge dy\wedge dr. \end{eqnarray*}

If $\nabla$ is the Riemannian connection, then

\begin{eqnarray*} && \nabla _{X^{\prime}}X^{\prime}=(b\dot a/a)R^{\prime},\quad \nabla _{Y^{\prime}}X^{\prime}=(2a^2/b)T^{\prime},\quad \nabla _{T^{\prime}}X^{\prime}=(2a^2/b)Y^{\prime},\quad \nabla _{R^{\prime}}X^{\prime}=0,\\[2pt] && \nabla _{X^{\prime}}Y^{\prime}=-(2a^2/b)T^{\prime},\quad \nabla _{Y^{\prime}}Y^{\prime}=(b\dot a/a)R^{\prime},\quad \nabla _{T^{\prime}}Y^{\prime}=-(2a^2/b)X^{\prime},\quad \nabla _{R^{\prime}}Y^{\prime}=0,\\[2pt] && \nabla _{X^{\prime}}T^{\prime}=(2a^2/b)Y^{\prime}, \quad \nabla _{Y^{\prime}}T^{\prime}=-(2a^2/b)X^{\prime},\quad \nabla _{T^{\prime}}T^{\prime}=\dot bR^{\prime},\quad \nabla _{R^{\prime}}T^{\prime}=0,\\[2pt] && \nabla _{X^{\prime}}R^{\prime}=-(b\dot a/a)X^{\prime},\quad \nabla _{Y^{\prime}}R^{\prime}=-(b\dot a/a)Y^{\prime},\quad \nabla _{T^{\prime}}R^{\prime}=-\dot bT^{\prime},\quad \nabla _{R^{\prime}}R^{\prime}=0. \end{eqnarray*}

By requiring $\Omega _{a,b}$ to be closed, we obtain that

\begin{eqnarray*} && \nabla _{X^{\prime}}X^{\prime}=\nabla _{Y^{\prime}}Y^{\prime}=(2a^2/b)R^{\prime},\\[2pt] && \nabla _{X^{\prime}}R^{\prime}=-(2a^2/b)X^{\prime},\quad \nabla _{Y^{\prime}}R^{\prime}=-(2a^2/b)Y^{\prime}. \end{eqnarray*}

By calculating the Riemannian curvature tensor $R,$ we find that

\begin{eqnarray*} && R(X^{\prime},Y^{\prime})X^{\prime}=-(16a^4/b^2)Y^{\prime},\quad R(T^{\prime},R^{\prime})T^{\prime}=(b\ddot b-\dot b^2)R^{\prime}\\[2pt] && R(X^{\prime},T^{\prime})X^{\prime}= ((2a^2/b)^2-(2a^2/b)\dot b)T^{\prime},\quad R(X^{\prime},R^{\prime})X^{\prime}=(b(d/dr)(2a^2/b)-(2a^2/b)^2)R^{\prime},\\[2pt] && R(Y^{\prime},T^{\prime})Y^{\prime}=((2a^2/b)^2-(2a^2/b)\dot b)T^{\prime},\quad R(Y^{\prime},R^{\prime})Y^{\prime}=(b(d/dr)(2a^2/b)-(2a^2/b)^2)R^{\prime} \end{eqnarray*}

Therefore, it follows that

\begin{eqnarray*} && K(X^{\prime},Y^{\prime})=-16a^4/b^2,\quad K(T^{\prime},R^{\prime})=b\ddot b-\dot b^2,\\[2pt] && K(X^{\prime},T^{\prime})=(2a^2/b)^2-(2a^2/b)\dot b,\quad K(X^{\prime},R^{\prime})=b(d/dr)(2a^2/b)-(2a^2/b)^2,\\[2pt] && K(Y^{\prime},T^{\prime})=(2a^2/b)^2-(2a^2/b)\dot b,\quad K(Y^{\prime},R^{\prime})=b(d/dr)(2a^2/b)-(2a^2/b)^2. \end{eqnarray*}

At this point, we need the following lemma, which is verified after straightforward calculations:

Lemma 4.4. If $C\gt 0$ , equations

\begin{equation*} -16a^4/b^2=-C^2, \quad b\ddot b-\dot b^2=-C^2, \end{equation*}

have positive solutions

\begin{equation*} a=\sqrt {Cb}/2; \quad b(r)=Cr \quad {or} \quad b(r)=(C/c)\sinh (cr),\quad c\gt 0. \end{equation*}

If $b(r)=Cr$ , we have

\begin{eqnarray*} && K(X^{\prime},Y^{\prime})=K(T^{\prime},R^{\prime})=-C^2,\\[2pt] && K(X^{\prime},R^{\prime})=K(Y^{\prime},R^{\prime})=-C^2/4,\\[2pt] && K(X^{\prime},T^{\prime})=K(Y^{\prime},T^{\prime})=-C^2/4. \end{eqnarray*}

On the other hand, if $b(r)=(C/c)\sinh (cr)$ we have

\begin{eqnarray*} && K(X^{\prime},Y^{\prime})=K(T^{\prime},R^{\prime})=-C^2,\\[2pt] && K(X^{\prime},R^{\prime})=K(Y^{\prime},R^{\prime})=-C^2/4,\\[2pt] && K(X^{\prime},T^{\prime})=K(Y^{\prime},T^{\prime})=-(C^2/4)(2\cosh (cr)-1). \end{eqnarray*}

Observe that the sectional curvature is always negative but not bounded below in the second case. The following proposition now follows immediately.

For $a=\sqrt {b}/2$ and $b=r$ as above we shall write $g^{\prime}$ and $\Omega ^{\prime}$ instead of $g_{a,b}$ and $\Omega _{a,b}$ , respectively. When $a=\sqrt {b}/2$ , $b=(1/c)\sinh (cr)$ , $c\gt 0$ , we shall write $g^c$ and $\Omega ^c$ , respectively.

In what follows, we consistently refer to the two Kähler manifolds mentioned in Proposition 4.5 for the sake of convenience.

5.1. PCR mappings

The definitions stated below are based on definitions given in [Reference Platis10] as well as in [Reference Bejancu1].

We comment that the condition of the definition implies $F_*(JX)=\mathbb{J}(F_*X)$ for every vector field $X$ in the underlying real subbundle of $\mathcal{H}$ . We also wish to comment that this is a weaker version of the definition of CR submanifold, as it is given by Bejancu in [Reference Bejancu1], p. 20.

Again, compare Definition 5.2 with Bejancu’s definition. We now focus on a particular case, which is described in the next proposition.

Proof. Let $\mathcal{H}=\ker (\eta )$ be the strictly pseudoconvex CR structure of $M$ , and let $\iota \,:\,M\to \mathcal{C}(M)$ be the mapping $p\mapsto (p,1)$ . Then $\iota$ is clearly an embedding and if $Z_p\in \mathcal{H}_p$ , $p\in M$ , then

\begin{equation*} \iota _{*,p}(Z_p)=Z_{(p,1)}(=Z_p+\textbf {0}_1)\in (\mathcal{H}_r)_{(p,1)}. \end{equation*}

Hence $\iota$ is $(\mathcal{H},\mathcal{H}_r)$ -PCR. On the other hand, $(g_r)|_{\mathcal{H}_r\times \mathcal{H}_r}=r^2g_{cc}$ and thus

\begin{equation*} \iota ^{*}(g_r|_{\mathcal{H}_r\times \mathcal{H}_r})=g_{cc}=g\mid _{\mathcal{H}\times \mathcal{H}}. \end{equation*}

From Example iv) of Section 7.1 and Proposition 5.3, we obtain the following:

5.2. PCR-Kähler equivalence

Note that condition ii) is equivalent to

1. ii’) $(\mathcal{H},\mathcal{H}^{\prime})$ -Kähler if $G^*(\Omega ^{\prime}\mid _{\mathcal{H}^{\prime}\times \mathcal{H}^{\prime}})=\Omega \mid _{\mathcal{H}\times \mathcal{H}}$ .

When the above hold, the manifolds $N$ and $N^{\prime}$ are called PCR-Kähler equivalent (with respect to prescribed bundles $\mathcal{H}$ and $\mathcal{H}^{\prime}$ ); we will write

\begin{equation*} (N,\mathbb{J},g,\mathcal{H})\sim _G(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime}). \end{equation*}

The above is indeed an equivalence relation:

1. (1) Reflexive: by taking the identity map $G=id_N$ we have

   \begin{equation*} (N,\mathbb{J},g,\mathcal{H})\sim _G(N,\mathbb{J},g,\mathcal{H}). \end{equation*}

2. (2) Symmetric: if $ (N,\mathbb{J},g,\mathcal{H})\sim _G(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime})$ then $ (N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime})\sim _{G^{-1}}(N,\mathbb{J},g,\mathcal{H}).$

3. (3) Transitive: if $ (N,\mathbb{J},g,\mathcal{H})\sim _G(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime})$ and if $ (N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime})\sim _{G^{\prime}}(N^{{\prime\prime}},\mathbb{J}^{{\prime\prime}},g^{{\prime\prime}},\mathcal{H}^{{\prime\prime}})$ then if $ (N,\mathbb{J},g,\mathcal{H})\sim _{G^{\prime}\circ G}(N^{{\prime\prime}},\mathbb{J}^{{\prime\prime}},g^{{\prime\prime}},\mathcal{H}^{{\prime\prime}}).$

Suppose now that $(N,\mathbb{J},g,\Omega )$ is as above and $\mathcal{H}$ is a prescribed subbundle of ${T}^{1,0}N$ . The following holds:

1. (1) If $G\,:\,N\to N$ lies in the subgroup of the holomorphic isometry group of $N$ comprising maps that preserve $\mathcal{H}$ , then $(N,\mathbb{J},g,\mathcal{H})\sim _G(N,\mathbb{J},g,\mathcal{H})$ .

2. (2) If $N^{\prime}$ is a manifold and $G\,:\,N\to N^{\prime}$ is a diffeomorphism, then $G$ defines a point in the PCR-Kähler equivalence class of $N$ . Indeed, we may turn $N^{\prime}$ into a Kähler manifold by defining a complex structure $\mathbb{J}^{\prime}=G_{*}\circ \mathbb{J}\circ G^{-1}_{*}$ and a metric $g^{\prime}=(G^{-1})^*g$ . Then $\mathcal{H}^{\prime}=G_*(\mathcal{H})\in {T}^{1,0}N^{\prime}$ and

   \begin{equation*} (N,\mathbb{J},g,\mathcal{H})\sim _G(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime}). \end{equation*}

3. (3) In order to detect whether another given Kähler manifold $(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\Omega ^{\prime})$ with no prescribed subbundle $\mathcal{H}^{\prime}$ lies within the PCR-Kähler equivalence class of $N$ , one has to construct a diffeomorphism $G\,:\,N\to N^{\prime}$ and to check if

   \begin{equation*} (N,\mathbb{J},g,\mathcal{H})\sim _G(N^{\prime},\mathbb{J}^{\prime},g^{\prime},\mathcal{H}^{\prime}=G_*(\mathcal{H})). \end{equation*}

Proof. Recall that $Z= (X-iY)/2$ and $\dot a=2a^3/b^2.$ By Lemma 4.3, we know that $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g_{a,b},\Omega _{a, b})$ is Kähler when $\dot a=2a^3/b^2.$ Now consider the map

\begin{equation*}G\,:\,(\mathcal{C}(\mathfrak{H}),\mathbb{J}, g_r,\Omega _r)\to (\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g_{a,b},\Omega _{a, b})\end{equation*}

with the formula

\begin{equation*} G(z,t,r)=(z,t, a^{-1}(1/r)). \end{equation*}

If $\mathcal{H}=\langle Z\rangle$ , then we have $G_*(\mathcal{H})=\mathcal{H}$ , which shows that $G$ is $(\mathcal{H},\mathcal{H})$ -PCR. To avoid confusion with $r,$ we write

\begin{equation*}g_{a,b}{\mid _{\mathcal{H}\times \mathcal{H}}}=\frac {1}{a^2}(dx^2+dy^2),\end{equation*}

where $a=a(d)$ in accordance with the coordinate $(z, t, d)\in \mathfrak{H}\times {\mathbb{R}}_{\gt 0}.$ Then,

\begin{equation*} G^*g_{a,b}{\mid _{\mathcal{H}\times \mathcal{H}}}=\frac {1}{(a(a^{-1}(1/r)))^2}(dx^2+dy^2)=r^2(dx^2+dy^2)=g_r{\mid _{\mathcal{H}\times \mathcal{H}}} \end{equation*}

and thus $G$ is also $(\mathcal{H},\mathcal{H})$ -Kähler. This completes the proof.

There is a partial inverse to Theorem5.6:

Proposition 5.7. Let $(g,\mathbb{I},\Omega )$ be a Kähler structure in $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ and let also $G\,:\,(\mathcal{C}(\mathfrak{H}),\mathbb{J}, g_r,\Omega _r)\to (\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g, \Omega )$ be an $(\mathcal{H},\mathcal{H})$ - PCR Kähler diffeomorphism such that $G_*Z=Z$ . Then

\begin{equation*} g\mid _{\mathcal{H}\times \mathcal{H}}=\frac {dx^2+dy^2}{a^2(r)}, \end{equation*}

for some strictly positive function $a(r)$ .

Proof. Suppose $G(z,t,r)=(w,s,q)$ and $G_*Z=Z$ . Since

\begin{equation*} G_*Z=Zw\frac {\partial }{\partial z}+Z\overline {w}\frac {\partial }{\partial \overline {z}}+Zs\frac {\partial }{\partial t}+Zq\frac {\partial }{\partial r}, \end{equation*}

we must have

\begin{equation*} Zw=1,\quad Z\overline {w}=0,\quad Zs=i\overline {z},\quad Zq=0. \end{equation*}

It follows that

\begin{equation*} w=z+c_1(r)+ic_2(r),\quad s=t+c_3(r),\quad q=c_4(r), \end{equation*}

where $c_4$ is a strictly positive 1-1 function of $r$ and $c_i$ , $i=1,2,3,$ are real functions of $r$ . Set $c_4^{-1}(r)=1/a(r)$ . Then

\begin{eqnarray*} (G^*g)\mid _{\mathcal{H}\times \mathcal{H}}&=&(g_{11}\circ G)(dx+c_1^{\prime}(r)dr)^2+2(g_{12}\circ G)(dx+c_1^{\prime}(r)dr)(dy+c_2^{\prime}(r)dr)\\ &&+(g_{22}\circ G)(dy+c_2^{\prime}(r)dr)^2\\ &=&r^2(dx^2+dy^2). \end{eqnarray*}

Therefore, $c_i(r)=c_i$ , where $c_i$ , $i=1,2$ are constants and

\begin{equation*} g\mid _{\mathcal{H}\times \mathcal{H}}=\frac {dx^2+dy^2}{a^2(r)}. \end{equation*}

6.1. Complex hyperbolic plane

The Heisenberg group $\mathfrak{H}$ appears naturally within the context of complex hyperbolic geometry as the boundary of the complex hyperbolic plane. In the concept of this paper, the complex hyperbolic plane $\textbf {H}^2_{\mathbb{C}}$ is considered as the Siegel domain $ \mathcal{S}=\{(z_1,z_2)\in {\mathbb{C}}^2\;|\;\rho (z_1,z_2)\gt 0\},$ where $ \rho (z_1,z_2)=-2\Re (z_1)-|z_2|^2.$ The complex hyperbolic plane $\textbf {H}^2_{\mathbb{C}}$ is a complex manifold; there is a natural Kähler structure defined on $\textbf {H}^2_{\mathbb{C}}$ coming from the Bergman metric:

(5) \begin{equation} ds^2=\frac {4}{\rho }|dz_2|^2+\frac {4}{\rho ^2}|\partial \rho |^2=\frac {4}{\rho }|dz_2|^2+\frac {4}{\rho ^2}|dz_1+\overline {z_2}dz_2|^2. \end{equation}

The Kähler form is then

(6) \begin{equation} \Omega =-4i\partial \overline {\partial }(\log \rho )=-4i\left (\frac {1}{\rho }dz_2\wedge d\overline {z_2}-\frac {1}{\rho ^2}\partial \rho \wedge \overline {\partial }\rho \right )\!. \end{equation}

The group of holomorphic isometries is $\textrm {PU}(2,1)$ .

6.2. Horospherical map

The boundary $\partial \mathcal{S}$ of $\mathcal{S}$ ( $\rho (z_1,z_2)=0)$ admits a strictly pseudoconvex CR structure with contact form $\omega ^{\prime}=-\Im (\partial \rho )$ , and with this CR structure $\partial \mathcal{S}$ and the Heisenberg group $\mathfrak{H}$ are CR equivalent; the CR diffeomorphism between them is given by

(7) \begin{equation} h\,:\, \mathfrak{H}\ni (z, t)\longmapsto ((-|z|^2+it)/2,z)\in \partial \mathcal{S}, \end{equation}

which identifies $\mathfrak{H}$ to $\partial \textbf {H}^2_{\mathbb{C}}$ in a CR manner. To see this, we calculate

\begin{eqnarray*} h_*(Z)&=&(1/2)Z(-|z|^2+it)(\partial /\partial z_1)+(1/2)Z(-|z|^2-it)(\partial /\partial \overline {z_1})\\ &&+Z(z)(\partial /\partial z_2)+Z(\overline {z})(\partial /\partial \overline {z_2})\\ &=&-\overline {z_2}(\partial /\partial z_1)+(\partial /\partial z_2)\in {T}^{1,0}\mathcal{S}. \end{eqnarray*}

The map $h$ is the boundary map of the horospherical map which describes the horospherical model for the complex hyperbolic plane:

Definition 6.1. The horospherical model for $\textbf {H}^2_{\mathbb{C}}$ is given by the horospherical map defined by

\begin{equation*} H\,:\, \mathfrak{H}\times {\mathbb{R}}_{\gt 0}\ni (z, t,r)\longmapsto ((-|z|^2-r+it)/2,\;z )\in \mathcal{S}. \end{equation*}

Proof. By Proposition 4.5, one can see that a Kähler manifold $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g_{a, b},\Omega _{a,b})$ has constant holomorphic sectional curvature only when it is holomorphic isometric to either $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g^{\prime},\Omega ^{\prime})$ or $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g^c,\Omega ^c).$ However, the manifold $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g^c,\Omega ^c)$ has unbounded negative real sectional curvature. Therefore, it is enough to show that $H$ is holomorphic isometric mapping between Kähler manifold $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I}, g^{\prime},\Omega ^{\prime})$ and $\textbf {H}^2_{\mathbb{C}}.$

We firstly show that $H_*(Z)$ and $H_*(W)$ are in ${T}^{1,0}\textbf {H}^2_{\mathbb{C}}.$ Here, $Z=\partial _z+i\overline {z}\partial _t$ and $W=(1/2)(\partial _t-i\partial _r)$ . Indeed,

\begin{eqnarray*} H_*(Z)&=&(1/2)Z(-|z|^2-r+it)(\partial /\partial z_1)+(1/2)Z(-|z|^2-r-it)(\partial /\partial \overline {z_1})\\[2pt] &&+Z(z)(\partial /\partial z_2)+Z(\overline {z})(\partial /\partial \overline {z_2})\\[2pt] &=&-\overline {z_2}(\partial /\partial z_1)+(\partial /\partial z_2)\in {T}^{1,0}\textbf {H}^2_{\mathbb{C}}. \end{eqnarray*}

Also,

\begin{eqnarray*} H_*(W)&=&(1/4)(\partial _t-i\partial _r)(-|z|^2-r+it)(\partial /\partial z_1)\\[2pt] &&+(1/4)(\partial _t-i\partial _r)(-|z|^2-r-it)(\partial /\partial \overline {z_1})\\[2pt] &&+(1/2)(\partial _t-i\partial _r)(z)(\partial /\partial z_2)+(1/2)(\partial _t-i\partial _r)(\overline {z})(\partial /\partial \overline {z_2})\\[2pt] &=&(i/2)(\partial /\partial z_1)\in {T}^{1,0}\textbf {H}^2_{\mathbb{C}}. \end{eqnarray*}

On the other hand, from equation (5) it follows immediately that $ H^*ds^2=g^{\prime}.$ This completes the proof.

7.1. Submanifolds of $\mathcal{C}(\mathfrak{H})=(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{J},g_r,\Omega _r)$

(i) Half-plane. We embed $\mathcal{U}=\{(t,r)\;:\,t\in {\mathbb{R}}, r\gt 0\}$ into $\mathcal{C}(\mathfrak{H})$ by setting

\begin{equation*} \iota _{\mathcal{U}}(t,r)=(0,2t,r). \end{equation*}

The pullback metric is then

\begin{equation*} g_{\mathcal{U}}=\iota _{\mathcal{U}}^{\ast }g_r=dr^2+r^2dt^2 \end{equation*}

and is a flat Kähler metric on $\mathcal{U}$ and the submanifold $\mathcal{U}$ is a totally geodesic submanifold of $\mathcal{C}(\mathfrak{H})$ (the second fundamental form vanishes). To see this, let $\{\partial _r,\;(1/r)\partial _t\}$ be an (orthonormal) basis for $\mathcal{U}$ and let $\{\partial _r,\; (1/2)T_r\}$ be a local extension to the tangent bundle of $\mathcal{C}(\mathfrak{H})$ . If $B$ is the second fundamental form, then

\begin{eqnarray*} && B(\partial _r,\partial _r)=\left (\nabla ^r_{\partial _r}\partial _r\right )^N=0^N=0,\\[3pt] && B(\partial _r,(1/r)\partial _t)=\left ((1/2)\nabla ^r_{\partial _r}T_r\right )^N=0^N=0,\\[3pt] && B((1/r)\partial _t,(1/r)\partial _t)=\left ((1/4)\nabla ^r_{T_r}T_r\right )^N=\left (-1/(4r)\partial _r\right )^N=0. \end{eqnarray*}

Therefore, we obtain that the second fundamental form is identically zero and hence $(\mathcal{U},g)$ is totally geodesic. As for the curvature, it follows from Proposition 4.1, the Gauss equation, and the vanishing of the second fundamental form that the sectional curvature is zero.

(ii) Complex plane. We embed $\mathbb{C}$ into $\mathcal{C}(\mathfrak{H})$ by setting

\begin{equation*} \iota _{{\mathbb{C}}}(z)=(z,0,1). \end{equation*}

The pullback metric is

(8) \begin{equation} g_{\mathbb{C}}=\iota _{{\mathbb{C}}}^{\ast }g_r=|dz|^2+\Im ^2({\overline z}dz)=(1+y^2)dx^2-2xydxdy+(1+x^2)dy^2, \end{equation}

which is again Kähler. It is not totally geodesic: to see this, we consider vector fields $\partial _x$ and $\partial _y$ on $\mathbb{C}$ and their respective local extensions to $\mathcal{C}(\mathfrak{H})$ . The normal space to $\mathbb{C}$ is spanned by the orthonormal vector fields

\begin{equation*} N_1=\frac {y\partial _x-x\partial _y+2(1+x^2+y^2)\partial _t}{\sqrt {1+x^2+y^2}},\quad N_2=\partial _r. \end{equation*}

If $B$ is the second fundamental form, then

\begin{eqnarray*} && B(\partial _x,\partial _x)=\frac {2xy}{\sqrt {1+x^2+y^2}} N_1-(1+y^2)N_2,\\ && B(\partial _x,\partial _y)=\frac {y^2-x^2}{\sqrt {1+x^2+y^2}}N_1+xyN_2,\\ && B(\partial _y,\partial _y)=-\frac {2xy}{\sqrt {1+x^2+y^2}}N_1-(1+x^2)N_2. \end{eqnarray*}

Thus, submanifold $\mathbb{C}$ is not a totally geodesic submanifold of $\mathcal{C}(\mathfrak{H})$ . Now, $ R_r(\partial _x,\partial _y)\partial _x=R_r(X,Y)X=-4\partial _y+8x\partial _t$ and $g_r(R_r(\partial _x,\partial _y)\partial _x,\;\partial _y)=-4$ and, by the Gauss equation, we obtain

\begin{equation*} K(\partial _x,\partial _y)=-\frac {3+2x^2+2y^2}{(1+x^2+y^2)^2}. \end{equation*}

In this way, $\mathbb{C}$ is a submanifold with unbounded negative (holomorphic) sectional curvature.

(iii) Vertical plane. We continue to consider the immersion vertical plane $\mathbb{V}$ into $\mathcal{C}(\mathfrak{H})$ by setting

\begin{equation*} \iota _{\mathbb{V}}(x+it)=(x,2t,1). \end{equation*}

The pullback metric is

(9) \begin{equation} g_{\mathbb{V}}=\iota _{\mathbb{V}}^{\ast }g_r=dx^2+dt^2, \end{equation}

which is the Euclidean metric. We consider vector fields $\partial _x$ and $\partial _t$ on $\mathbb{C}$ and their respective local extensions to $\mathcal{C}(\mathfrak{H})$ . The normal space to the vertical plane is spanned by the orthonormal vector fields

\begin{equation*} N_1=\partial _y-2x\partial _t,\quad N_2=\partial _r. \end{equation*}

If $B$ is the second fundamental form, then

\begin{equation*}B(\partial _x, \partial _x)=B(\partial _t, \partial _t)=-N_2,\quad B(\partial _x, \partial _t)=N_1.\end{equation*}

Thus, submanifold $\mathbb{V}$ is not a totally geodesic submanifold of $\mathcal{C}(\mathfrak{H})$ . Now, by $g_r(R_r(\partial _x,\partial _t)\partial _x,\;\partial _t)=0$ and Gauss equation we obtain

\begin{equation*} K(\partial _x,\partial _t)=0, \end{equation*}

which yields that the vertical plane $\mathbb{V}$ is a flat submanifold in $\mathcal{C}(\mathfrak{H})$ .

(iv) Half-space. We embed the half space $\mathcal{U}_3$ into $\mathcal{C}(\mathfrak{H})$ by setting

\begin{equation*} \iota _{\mathcal{U}_3}(x+it,r)=(x,2t,r) \end{equation*}

and the induced metric is

\begin{equation*} g_{\mathcal{U}_3}=r^2(dx^2+dt^2)+dr^2; \end{equation*}

thus, $\mathcal{U}_3$ is indeed the warped product ${\mathbb{R}}_{\gt 0}\times _r\mathbb{V}$ and $\mathbb{V}$ here is considered with the Euclidean metric as shown in equation (9). The tangent space of $\mathcal{U}_3$ is generated by the orthonormal vector fields $\{(1/r) \partial _x$ , $(1/r)\partial _t$ , $\partial r\},$ which are all normal to the unit vector field

\begin{equation*} N=Y_r=\frac {1}{r}(\partial _y-2x\partial _t). \end{equation*}

If $B$ is the second fundamental form, then simple calculations deduce

\begin{equation*} B((1/r)\partial _x,(1/r)\partial _t)=\frac {1}{2r}N, \end{equation*}

whereas all other components of $B$ vanish. Therefore, $\mathcal{U}_3$ is not totally geodesic. By the Gauss equation, we obtain for the sectional curvatures of distinguished planes that

\begin{equation*} K((1/r)\partial _x,(1/r)\partial _t)=-\frac {1}{4r^2} \end{equation*}

whereas all other sectional curvatures vanish, i.e., the half-space $\mathcal{U}_3$ is a submanifold with unbounded negative sectional curvature.

(v) Heisenberg group. We embed the Heisenberg group $\mathfrak{H}$ into $\mathcal{C}(\mathfrak{H})$ by setting

\begin{equation*} \iota _{\mathfrak{H}}(z,t)=(z,t,1) \end{equation*}

and the induced metric is of course $g$ as in Section 3.2. The tangent space of $\mathfrak{H}$ is generated by the vector fields $X,Y,\widetilde {T}$ , which are all normal to the unit field $ N=\partial _r.$ If $B$ is the second fundamental form, then

\begin{eqnarray*} && B(X,X)=B(Y,Y)=B(\widetilde T,\widetilde T)=-N, \end{eqnarray*}

whereas all other components of $B$ vanish. Therefore, $\mathfrak{H}$ is not totally geodesic. Now, by the Gauss equation, we subsequently recover the formulas for the sectional curvatures of distinguished planes as in Corollary 3.2.

7.2. Submanifolds of $(\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g^c,\Omega ^c)$

We study the submanifolds of $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ with respect to the metric $g^c$ in a similar manner as above. In particular, it is only necessary to consider the case when $c=1$ . Indeed, if we consider the mapping $F\,:\, (\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g^1,\Omega ^1) \to (\mathfrak{H}\times {\mathbb{R}}_{\gt 0},\mathbb{I},g^c,\Omega ^c)$ by

\begin{equation*}F(z, t, r)=(z/\sqrt {c}, t/c, r/c),\end{equation*}

then it shows that $F^{*}g^c=(4/\!\sinh r)(dx^2+dy^2)+(1/\!\sinh ^2r)(\omega ^2+dr^2)=g^1,$ from which one can easily know that $F$ is holomorphic isometric mapping. In what follows, we let $b_1=\sinh r,$ and $a_1=\sqrt {b_1}/2.$

(i) Half-plane. Let $\mathcal{U}=\{(t,r)\;:\,t\in {\mathbb{R}}, r\gt 0\}$ embed into $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ by setting

\begin{equation*} \tau _{\mathcal{U}}(t,r)=(0,t,r). \end{equation*}

The pullback metric is then

\begin{equation*} g^1_{\mathcal{U}}=\tau _{\mathcal{U}}^{\ast }g_1=(1/\!\sinh ^2 r)(dt^2+dr^2) \end{equation*}

and is a hyperbolic metric on $\mathcal{U}$ and the submanifold $\mathcal{U}$ is a totally geodesic submanifold. Indeed, we let $\{b_1\partial _t,\;b_1\partial _r\}$ be an (orthonormal) basis for $\mathcal{U}$ and consider a local extension to the tangent bundle of $\mathcal{C}(\mathfrak{H})$ . If $B$ is the second fundamental form, by direct computations, we obtain that

\begin{equation*} B(b_1\partial _t, b_1\partial _t)=B(b_1\partial _t, b_1\partial _r) =B(b_1\partial _r, b_1\partial _r)=0. \end{equation*}

Therefore, we obtain that the second fundamental form is identically zero and hence $(\mathcal{U},g^1_{\mathcal{U}})$ is totally geodesic. It follows from Proposition 4.5 and Gauss’s equation that the sectional curvature is $-1.$

(ii) Complex plane. In what follows, we always let $c=\sinh 1.$ We embed $\mathbb{C}$ into $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ by setting

\begin{equation*} \tau _{{\mathbb{C}}}(z)=((c/\sqrt {2})z,0,1). \end{equation*}

The pullback metric is

\begin{equation*} g^1_{\mathbb{C}}=\tau _{{\mathbb{C}}}^{\ast }g^1=2c(dx^2+dy^2)+(xdy-ydx)^2=(2c+y^2)dx^2-2xydxdy+(2c+x^2)dy^2, \end{equation*}

which is again Kähler. We consider vector fields $\partial _x$ and $\partial _y$ on $\mathbb{C}$ and their respective local extensions to $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ . The normal space to $\mathbb{C}$ is spanned by the orthonormal vector fields

\begin{equation*} N_1=\frac {\sqrt {c} (y\partial _x-x\partial _y+2(x^2+y^2+c)\partial _t)}{2\sqrt {x^2+y^2+c}},\quad N_2=c\partial _r. \end{equation*}

If $B$ is the second fundamental form, then

\begin{eqnarray*} && B(\partial _x,\partial _x)=\frac {4xy}{c^{3/2}\sqrt {x^2+y^2+c}}N_1+\frac {2\sqrt {1+c^2}(c+2y^2)}{c^2}N_2,\\[3pt] && B(\partial _x,\partial _y)=\frac {2(y^2-x^2)}{c^{3/2}\sqrt {x^2+y^2+c}}N_1-\frac {4\sqrt {1+c^2}}{c^2}xyN_2,\\[3pt] && B(\partial _y,\partial _y)=-\frac {4xy}{c^{3/2}\sqrt {x^2+y^2+c}}N_1+\frac {2\sqrt {1+c^2}(c+2x^2)}{c^2}N_2. \end{eqnarray*}

Therefore, submanifold $\mathbb{C}$ is not a totally geodesic submanifold of $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ . A direct calculation shows that

\begin{equation*}g^1(R(\partial _x,\partial _y)\partial _x,\;\partial _y) =-\frac {4((6+c^2)c+(2+c^2)y^2)(x^2+c)}{c^4}. \end{equation*}

Now, by the Gauss equation, we obtain

\begin{eqnarray*} K(\partial _x,\partial _y)= 2\frac {c(-2c^2+1)f_{xy}^2+(((c^2+2)y^2+c^3+6c)(x^2+c)+c^2(c^2-1))f_{xy}-c^3} {c^5(f_{xy}+c)f_{xy}} \end{eqnarray*}

where $f_{xy}=x^2+y^2+c.$ Thus, $\mathbb{C}$ is a submanifold with unbounded negative holomorphic sectional curvature.

(iii) Vertical plane. We embed the vertical plane $\mathbb{V}$ into $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ by setting

\begin{equation*} \tau _{\mathbb{V}}\,:\,(x, t)\mapsto ((\sqrt {c}/2)x,ct,1). \end{equation*}

The pullback metric is

(10) \begin{equation} g^1_{\mathbb{V}}=\tau _{\mathbb{V}}^{\ast }g^1=dx^2+dt^2, \end{equation}

which is again Euclidean metric. We consider vector fields $\partial _x$ and $\partial _t$ in $\mathbb{V}$ and their respective local extensions to $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ . The normal space to $\mathbb{V}$ is spanned by the orthonormal vector fields

\begin{equation*} N_1=\frac {\sqrt {c}}{2}(\partial _y-2x\partial _t),\quad N_2=c\partial _r. \end{equation*}

Then we have

\begin{eqnarray*} && B(\partial _x,\partial _x)=\frac {2}{c} N_2,\\ && B(\partial _x,\partial _t)=(c)^{-3/2}N_1,\\ && B(\partial _t,\partial _t)=\frac {\sqrt {1+c^2}}{c^2 }N_2. \end{eqnarray*}

Thus, submanifold $\mathbb{V}$ is not a totally geodesic submanifold of $\mathcal{C}(\mathfrak{H})$ . It is not difficult to get that

\begin{equation*}g^1(R(\partial _x,\partial _t)\partial _x,\;\partial _t)=-\frac {1+2c^2 }{c^3}.\end{equation*}

Now, by the Gauss equation, we obtain

\begin{equation*} K(\partial _x,\partial _t)=-\frac {2(1+c^2-\sqrt {1+c^2})}{c^3},\end{equation*}

i.e., $\mathbb{V}$ is a submanifold with constant negative sectional curvature.

(iv) Half-space. We embed the half space $\mathcal{U}_3$ into $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ by setting

\begin{equation*} \tau _{\mathcal{U}_3}\,:\, (x+it, r)\mapsto (x,t,r) \end{equation*}

and the induced metric is

\begin{equation*} g^1_{\mathcal{U}_3}=(4/\!\sinh r)dx^2+(1/\!\sinh ^2 r)(dt^2+dr^2). \end{equation*}

The tangent space of $\mathcal{U}_3$ is generated by the orthonormal vector fields $\{a_1 \partial _x$ , $b_1 \partial _t$ , $b_1\partial r\}.$ Their respective local extensions to $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ are all normal to the unit vector field $Y^{\prime}=a_1(\partial _y-2x\partial _t).$ Direct calculations deduce

\begin{equation*}B(a_1 \partial _x,b_1 \partial _t)=\frac {1}{2}Y^{\prime},\end{equation*}

and all other components of the second fundamental form vanish. Therefore, $\mathcal{U}_3$ is not totally geodesic. By $g^{1}(R(a_1\partial _x,b_1\partial _t)a_1\partial _x,b_1\partial _t)=(1-2\cosh ^2r)/4$ and Gauss equation, we obtain for the sectional curvatures of distinguished planes that

\begin{equation*}K(a_1\partial _x,b_1\partial _t)=-\frac {1}{2}\cosh ^2 r,\end{equation*}

whereas all other sectional curvatures vanish. Thus, the half-space $\mathcal{U}_3$ is a submanifold with unbounded negative sectional curvature.

(v) Heisenberg group. We embed the Heisenberg group $\mathfrak{H}$ into $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ by setting

\begin{equation*} \tau _{\mathfrak{H}}\,:\,(z,t)\mapsto ((\sqrt {c}/2)z,(c/4)t,1) \end{equation*}

and the induced metric is $g^1_{\mathfrak{H}}=dx^2+dy^2+(1/16)\omega ^2$ , which is a Riemannian approximation metric. We consider the vector fields $\{X,Y,T\}$ on $\mathfrak{H}$ and their respective extension to $\mathcal{C}(\mathfrak{H}).$ They are all normal to the unit field $ N=\sinh 1\partial _r.$ We obtain

\begin{eqnarray*} && B(X,X)=B(Y,Y)=\frac {2\sqrt {1+c^2}}{c}N, \quad B(T,T)=\frac {\sqrt {1+c^2}}{c^2 }N, \end{eqnarray*}

whereas all other components of $B$ vanish. Therefore, $\mathfrak{H}$ is not totally geodesic. By direct calculations we find

\begin{eqnarray*} &&g^1(R(X,Y)X,Y)=-\frac {4(4+3c^2)}{c^2},\\ &&g^1(R(X,T)X,T)=g^1(R(Y,T)Y,T)=\frac {1-2c(1+c^2)}{c^3}. \end{eqnarray*}

Finally, by Gauss theorem we subsequently obtain the sectional curvatures of the distinguished planes

\begin{equation*}K(X, Y)=-\frac {4(3+2c^2)}{c^2},\quad K(X, T)=K(Y, T)=\cfrac {16(1-2(1+c^2)(c-1))}{c^3}.\end{equation*}

Thus, the Heisenberg group is a submanifold of $\mathfrak{H}\times {\mathbb{R}}_{\gt 0}$ with constant negative sectional curvature.