Proof. In an open neighbourhood of any point $p\in F^{-1}(c)$ , the vector field $\mathbf N$ can be extended to an orthonormal frame $E_1,\dots ,E_{n-1}, E_n=\mathbf N$ , where n is the dimension of the manifold. Then using the equations

$$\begin{align*}\langle \nabla_{E_n}\mathbf N,\mathbf N\rangle=0;\quad \nabla_{E_i}F=0\text{ for }1\leq i<n;\quad \text{and}\quad \nabla_{E_n(p)}F=\langle \nabla F(p), E_n(p)\rangle=\sqrt{b(c)},\end{align*}$$

we obtain

$$ \begin{align*} h(p)&=\sum_{i=1}^{n-1}\langle -\nabla_{E_i(p)}\mathbf N, E_i(p)\rangle=\sum_{i=1}^{n}\langle -\nabla_{E_i(p)}\mathbf N, E_i(p)\rangle =\sum_{i=1}^{n}\left\langle -\nabla_{E_i(p)} \left(\frac{\nabla F}{\sqrt{b\circ F}}\right), E_i(p)\right\rangle \\ &=-\frac{\Delta F(p)}{\sqrt{b(c)}}-\nabla_{E_n(p)} \left(\frac{1}{\sqrt{b\circ F}}\right)\langle \nabla F(p), E_n(p)\rangle=- \frac{ a(c)}{\sqrt{b(c)}}+\frac{b'(c)}{2\sqrt{b(c)}}\\[-41pt] \end{align*} $$

Proof. The formula follows from $\langle J_\alpha (E_i),E_k\rangle =\langle [E_i,E_k],F_{\alpha }\rangle $ .

Proof. The first equation follows from the skew-symmetry of the Lie bracket, the second identity can be obtained by evaluating the identity $J_Z^2=-\|Z\|^2 \mathrm {Id}_{\mathfrak v}$ on the basis vectors $Z=F_{\alpha }$ .

Proof. Polarising the identity $J_Z^2=-\|Z\|^2\mathrm {Id}_{\mathfrak v}$ , we obtain

(5) $$ \begin{align} J_X J_Y+J_Y J_X = -2\langle X,Y\rangle \mathrm{Id}_{\mathfrak v} \end{align} $$

If $Y\in \mathfrak z$ is an arbitrary element, then this gives

$$\begin{align*}\langle [J_XU,V]-[U,J_XV],Y\rangle&=\langle J_YJ_XU,V\rangle-\langle J_YU,J_XV\rangle\\&=\langle J_YJ_XU,V\rangle+\langle J_XJ_YU,V\rangle=-2\langle U,V\rangle \langle X,Y\rangle \end{align*}$$

and this implies the statement.

Proof. The statement is true for $V=0$ . If $V\neq 0$ , then $\left \{\frac {1}{\|V\|}J_{\alpha }V : 1\leq \alpha \leq m\right \}$ is an orthonormal basis of $J_{\mathfrak z}V$ by (6), and

$$\begin{align*}\langle [V,V_1], [V,V_2]\rangle &=\sum_{\alpha=1}^m\langle F_{\alpha},[V,V_1]\rangle\cdot\langle F_{\alpha}, [V,V_2]\rangle\\& =\sum_{\alpha=1}^m\langle J_{\alpha}V,V_1\rangle\cdot \langle J_{\alpha}V,V_2\rangle =\|V\|^2\langle P_V(V_1),P_V(V_2)\rangle.\\[-43pt] \end{align*}$$

Proof. Write V and W as a linear combination $V=\sum _{j=1}^nV_jE_j$ and $W=\sum _{k=1}^nW_kE_k$ . Then

$$ \begin{align*} \sum_{i=1}^n\langle[E_i,V],[E_i,W]\rangle&= \sum_{i,j,k=1}^n\langle[E_i,E_j],[E_i,E_k]\rangle V_jW_k= \sum_{i,j,k=1}^n\sum_{\alpha=1}^m C_{ij\alpha}C_{ik\alpha} V_jW_k \\&= \sum_{\alpha=1}^m\sum_{j,k=1}^n \left(\sum_{i=1}^n -C_{ji\alpha}C_{ik\alpha}\right) V_jW_k=\sum_{\alpha=1}^m\sum_{j,k=1}^n \delta_{jk} V_jW_k =m \langle V,W\rangle. \end{align*} $$

Proof. It is clear that $v\in \mathbb R v\oplus J_{\mathfrak z}v\leq \mathrm {Cl}(\mathfrak z,q)v$ for any v, so we need to show that $\mathbb R v\oplus J_{\mathfrak z}v$ is a $\mathrm {Cl}(\mathfrak z,q)$ -module if and only if v satisfies the $J^2$ -condition.

Assume first that v satisfies the $J^2$ -condition and show that $\mathbb R v\oplus J_{\mathfrak z}v$ is a $\mathrm {Cl}(\mathfrak z,q)$ -module. Since $\mathrm {Cl}(\mathfrak z,q)$ is generated by the elements of $\mathfrak z\subset \mathrm {Cl}(\mathfrak z,q)$ , it suffices to prove that $J_{z_1}(\mathbb R v\oplus J_{\mathfrak z}v)\subseteq \mathbb R v\oplus J_{\mathfrak z}v$ for all $z_1\in \mathfrak z$ . Choose an arbitrary element $\lambda v+J_zv$ of $\mathbb R v\oplus J_{\mathfrak z}v$ and decompose z as $z=\mu z_1+z_2$ , where $z_1\perp z_2$ . Then there is an element $z_3\in \mathfrak z$ such that $J_{z_1}J_{z_2}v=J_{z_3}v$ , thus $J_{z_1}(\lambda v+J_zv)=-\mu \|z_1\|^2 v+J_{\lambda z_1+z_3}v\in \mathbb R v\oplus J_{\mathfrak z}v$ , as we wanted to show.

Conversely, if $\mathbb R v\oplus J_{\mathfrak z}v$ is a $\mathrm {Cl}(\mathfrak z,q)$ -module, and $z_1\perp z_2$ are two elements of $\mathfrak z$ , then $J_{z_2}v\in \mathbb R v\oplus J_{\mathfrak z}v$ implies $J_{z_1}J_{z_2}v\in \mathbb R v\oplus J_{\mathfrak z}v$ , so there exist $\lambda \in \mathbb R$ and $z_3$ such that ${J_{z_1}J_{z_2}v=\lambda v+J_{z_3}v}$ . Since

$$\begin{align*}\langle v, J_{z_3}v\rangle=\langle z_3,[v,v]\rangle=0 \quad\text{and}\quad \langle v, J_{z_1}J_{z_2}v\rangle=-\langle J_{z_1}v, J_{z_2}v\rangle=-\langle z_1, z_2\rangle \|v\|^2=0, \end{align*}$$

$\lambda $ must vanish, therefore $J_{z_1}J_{z_2}v=J_{z_3}v$ .

Proof. The first part follows from the fact that the orthogonal complement of a $\mathrm {Cl}(\mathfrak z,q)$ -submodule of $\mathfrak v$ is also a $\mathrm {Cl}(\mathfrak z,q)$ -submodule of $\mathfrak v$ as the operators $J_z$ are skew adjoint. To show the second part, it is enough to check that if $w\in \ker _{\mathfrak v} (\operatorname {\mathrm {ad}} v)$ , then $J_zw\perp v$ for any $z\in \mathfrak z$ . However, this follows from $\langle J_zw,v\rangle =\langle [w,v],z\rangle =0$ .

Proof. Recall the classification of Clifford modules over $\mathrm {Cl}(\mathfrak z,q)$ (see [Reference Berndt, Tricerri and Vanhecke3, Sec. 3.1.2] or [Reference Lawson and Michelsohn29, Ch. I, §. 5]).

1. (a) If $m \not \equiv 3$ (mod $4$ ), then there exists a unique (up to isomorphism) irreducible $\mathrm {Cl}(\mathfrak z,q)$ -module $\mathfrak d$ . Every $\mathrm {Cl}(\mathfrak z,q)$ -module $\mathfrak v$ is isomorphic to a k-fold direct sum of $\mathfrak d$ , that is, $\mathfrak v\cong \bigoplus ^k \mathfrak d$ .

2. (b) If $m\equiv 3$ (mod $4$ ), then there exists exactly two non-isomorphic irreducible $\mathrm {Cl}(\mathfrak z,q)$ -modules $\mathfrak d_1$ and $\mathfrak d_2$ . Every $\mathrm {Cl}(\mathfrak z,q)$ -module $\mathfrak v$ is isomorphic to the direct sum $\mathfrak v\cong \left (\bigoplus ^{k_1} \mathfrak d_1\right )\oplus \left (\bigoplus ^{k_2} \mathfrak d_2\right )$ for some $k_1$ and $k_2$ . The modules $\mathfrak d_1$ and $\mathfrak d_2$ have the same dimension.

The formula for the dimension $n_0$ of the modules $\mathfrak d$ , $\mathfrak d_1$ , and $\mathfrak d_2$ depends on the modulo 8 residue class of m and is given in the following table.

$$\begin{align*}\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline m & 8p & 8p+1 & 8p+2 & 8p+3 & 8p+4 & 8p+5 & 8p+6 & 8p+7 \\[3pt] \hline &&&&&&&&\\[-9pt] n_0 & 2^{4p} & 2^{4p+1} & 2^{4p+2}& 2^{4p+2} & 2^{4p+3}&2^{4p+3}&2^{4p+3}&2^{4p+3} \\ \hline \end{array} \end{align*}$$

If there exists a non-zero vector $v\in \mathfrak v$ which satisfies the $J^2$ -condition, then $\mathrm {Cl}(\mathfrak z,q)v=\mathbb R v\oplus J_{\mathfrak z}v$ . The dimension of $\mathrm {Cl}(\mathfrak z,q)v$ is a multiple of $n_0$ , the dimension of $\mathbb R v\oplus J_{\mathfrak z}v$ is $m+1$ , so we must have $n_0\leq m+1$ . As $8p+7<2^{4p}$ if $p>0$ , inequality $n_0\leq m+1$ can hold only if $p=0$ . Among the eight values of m corresponding to $p=0$ , exactly the values $0,1,3,7$ satisfy the inequality $n_0\leq m+1$ . Since in these four cases we have $n_0=m+1$ in fact, $\mathbb R v\oplus J_{\mathfrak z}v$ is a non-trivial irreducible $\mathrm {Cl}(\mathfrak z,q)$ -module.

Proof. We consider all cases simultaneously, writing $\mathfrak v=\mathfrak D_1\oplus \dots \oplus \mathfrak D_K$ , where $K=k$ and $\mathfrak D_i=\mathfrak d$ for all i in case (i), while in cases (ii) and (iii), we set $K= k_1+k_2$ , $\mathfrak D_i= \mathfrak d_1$ for $1\leq i\leq k_1$ and $\mathfrak D_i= \mathfrak d_2$ for $k_1< i\leq k_1+k_2$ .

Assume that v satisfies the $J^2$ -condition. Let $\pi _i\colon \mathfrak v\to \mathfrak D_i$ be the projection onto the ith component and $v_i=\pi _i(v)$ . The restriction $\pi _i^v$ of $\pi _i$ onto the submodule $\mathrm {Cl}(\mathfrak z,q)v$ is a module homomorphism between two irreducible modules, hence it is either the $0$ -homomorphism or an isomorphism of modules. This implies that if $v_i\neq 0$ , then ${\mathfrak D_i\cong \mathrm {Cl}(\mathfrak z,q)v}$ . Hence v must be isotypic.

Furthermore, if $i<j$ are two indices for which $v_i\neq 0$ and $v_j\neq 0$ , then ${\pi _j\circ (\pi _i^v)^{-1}\colon \mathfrak D_i\to \mathfrak D_j}$ is a module isomorphism mapping $v_i$ to $v_j$ . Actually, the condition that for any pair of indices $i<j$ for which $v_i\neq 0$ and $v_j\neq 0$ , there exists a module isomorphism $\mathfrak D_i\to \mathfrak D_j$ mapping $v_i$ to $v_j$ is also sufficient for an isotypic vector to satisfy the $J^2$ -condition.

To understand what this characterisation of the $J^2$ -condition means in different cases, we need a description of the module automorphisms of the irreducible Clifford modules. Let $\mathbb K$ denote $\mathbb R$ , $\mathbb C$ , $\mathbb H$ , $\mathbb O$ corresponding to the cases $m=0,1,3,7$ , respectively. In each case, we can construct a realisation of the modules $\mathfrak d$ or $\mathfrak d_1$ and $\mathfrak d_2$ on the linear space $\mathbb K$ thinking of $\mathfrak z$ as the linear space of purely imaginary elements $\mathbb K$ . Then the operator $J_z\colon \mathbb K\to \mathbb K$ for $z\in \mathfrak z$ is the multiplication by z in the commutative cases $\mathbb K\in \{\mathbb R,\mathbb C\}$ . When $\mathbb K$ is not commutative, $J_z$ can act on $\mathbb K$ both by left and by right multiplications by z, providing the two non-isomorphic Clifford module structures $\mathfrak d_1$ and $\mathfrak d_2$ of $\mathbb K$ . In each case when $J_z$ equals the left multiplication by z, a module automorphism $\phi \colon \mathbb K\to \mathbb K$ is a right multiplication by a non-zero element of the nucleus of $\mathbb K$ . Recall that the nucleus of an alternative algebra consists of elements x satisfying the associativity identity $(ab)x=a(bx)$ for every a, b. Similarly, when $J_z$ is the right multiplication by z, a module automorphism $\phi \colon \mathbb K\to \mathbb K$ is a left multiplication by a non-zero element of the nucleus. It is clear that the nucleus of an associative algebra equals the whole algebra, and it is known that the nucleus of the algebra of octonions is $\mathbb R$ . This means that in the associative cases $\mathbb K\in \{\mathbb R,\mathbb C,\mathbb H\}$ , the automorphism groups of the irreducible Clifford modules act transitively on non-zero vectors, while in the case of $\mathbb O$ , two non-zero vectors belong to the same orbit of the automorphism group if and only if they are real multiples of one another. This completes the proof.

Proof. By the above remark, the equivalence (i) $\iff $ (ii) is proved in [Reference Cowling, Dooley, Korányi and Ricci12] for Damek–Ricci spaces and it also follows from Theorem 4.6.

Implication (ii) $\Longrightarrow $ (iii) follows from the fact that if $v_1, v_2\in \mathfrak v\setminus \{0\}$ satisfy the $J^2$ -condition, then ${J_{\mathfrak z} v_1\oplus \mathbb R v_1}$ and $J_{\mathfrak z} v_2\oplus \mathbb R v_2$ are irreducible Clifford modules, so their intersection, which is a submodule of both, is either $0$ or equal to both.

Now we prove (iii) $\Longrightarrow $ (ii). By Lemma 4.3, we need to show that for any non-zero vector $v\in \mathfrak v$ , $J_{\mathfrak z}v\oplus \mathbb Rv$ is a $\mathrm {Cl}(\mathfrak z,q)$ -submodule. Choose an arbitrary non-zero element $w\in J_{\mathfrak z}v\oplus \mathbb Rv$ . Then w is in the intersection $(J_{\mathfrak z}v\oplus \mathbb Rv)\cap (J_{\mathfrak z}w\oplus \mathbb Rw)$ , hence $J_{\mathfrak z}v\oplus \mathbb Rv=J_{\mathfrak z}w\oplus \mathbb Rw$ , therefore $J_{\mathfrak z}w\subseteq J_{\mathfrak z}v\oplus \mathbb Rv$ .

Proof. Observe that the geodesic is contained in the linear subspace generated by the pairwise orthogonal vectors $v, J_zv, z,A$ . Choose an orthonormal system of vectors $E_v,E_{J},E_z\in \mathfrak n$ such that

$$\begin{align*}v=\|v\|E_v,\quad J_zv=\|J_zv\|E_J, \quad z=\|z\|E_z, \end{align*}$$

and denote by $X(\theta ),Y(\theta ),Z(\theta ),W(\theta )$ the coefficients of $\gamma (\theta )$ in the decomposition

$$\begin{align*}\gamma(\theta)=X(\theta)E_v+Y(\theta)E_J+Z(\theta)E_z+W(\theta) A. \end{align*}$$

As $\chi $ is a quadratic polynomial of $\theta $ , the functions $X,Y,Z,W$ are rational functions of $\theta $ :

$$\begin{align*}X(\theta)=\frac{2\theta(1-s\theta)}{\chi(\theta)}\|v\|,\quad Y(\theta)=\frac{2\theta^2}{\chi(\theta)}\|z\|\|v\|,\quad Z(\theta)=\frac{2\theta}{\chi(\theta)}\|z\|,\quad W(\theta)=\frac{1-\theta^2}{\chi(\theta)}. \end{align*}$$

Using the relation $\|\xi \|^2=\|v\|^2+\|z\|^2+s^2=1$ , a simple algebraic computation shows that the functions $X,Y,Z,W$ satisfy the equations

(10) $$ \begin{align} \|z\|X+sY-\|v\| Z & =0, \end{align} $$

(11) $$ \begin{align} \left(1-\frac{\|v\|^2}{2}\right)Y-s\|v\| Z+\|v\| \|z\| W&=\|v\| \|z\|, \end{align} $$

(12) $$ \begin{align} \|z\| (X^2 + Y^2) - 2 \|v\| Y&=0. \end{align} $$

Case (i): If $z\neq 0$ , we distinguish two cases depending on v.

If $v\neq 0$ , then we can express Z and W as affine functions of the vector $XE_v+YE_J$ from the equations (10) and (11); therefore, the image of $\gamma $ is contained in an affine image of the linear subspace spanned by $E_v$ and $E_J$ . Since (12) defines a circle in this linear subspace, the image of $\gamma $ is an ellipse. In this case, we can express $\theta $ as the rational function $\frac {Y}{Z\|v\|}$ of the coordinates of $\gamma (\theta )$ , thus $\gamma $ is a birational equivalence.

If $v=0$ , then $X=Y=0$ and $\gamma $ is in the linear subspace spanned by $E_z$ and A. It can be verified that in this case, the coordinate functions Z and W satisfy the quadratic equation

$$\begin{align*}Z^2 + W^2 - \frac{2 s}{\|z\|} Z= 1, \end{align*}$$

which defines a circle. The parameter $\theta $ can be expressed as the rational function $\frac {Z}{s Z+\|z\|W+\|z\|}$ of the coordinates of $\gamma (\theta )$ , hence $\gamma $ is a birational equivalence.

Case (ii): If $z=0$ , but $v\neq 0$ , then $Y=Z=0$ , thus, the image of $\gamma $ is in the linear subspace spanned by $E_v$ and A, and the coordinates of $\gamma (\theta )$ satisfy equation

$$ \begin{align*} 4\|v\|^2 W + X^2 - (2 \|v\| + s X)^2=0, \end{align*} $$

which defines a parabola with axis parallel to $\mathfrak a$ . As $\theta =\frac {X(\theta )}{2\|v\|+sX(\theta )}$ , $\gamma $ is a birational equivalence.

Case (iii): If $z=0$ and $v=0$ , then $s=1$ and $\gamma (\theta )=\big (0,0,\frac {1+\theta }{1-\theta }\big )$ , so $\gamma $ parameterises the projective line containing $\mathfrak a$ .

Proof. It suffices to show that there exist smooth functions a, b such that $\Delta D_{x_0}=a\circ D_{x_0}$ and $\|\nabla D_{x_0}\|^2=b \circ D_{x_0}$ . We refer to [Reference Berndt, Tricerri and Vanhecke3, Sec. 4.4, Lemma] for the computation of the Laplace operator $\Delta $ of S. We note that there is an unnecessary coefficient $\frac 12$ in the formula in [Reference Berndt, Tricerri and Vanhecke3]. The correct formula and its transcription to the half-space model are

$$ \begin{align*} \Delta&=e^{\tau}\sum_{i=1}^n\partial_{v_i}^2+e^{\tau}\left(e^{\tau}+\frac{1}{4}\sum_{i=1}^nv_i^2\right)\sum_{\alpha=1}^m\partial_{z_{\alpha}}^2 +\partial_{\tau}^2-\left(m+\frac{n}{2}\right)\partial_{\tau}+e^{\tau}\sum_{i,j=1}^n\sum_{\alpha=1}^mC_{ij\alpha}v_i\partial_{v_j}\partial_{z_{\alpha}}\\ &=t\sum_{i=1}^n\partial_{v_i}^2+t\left(t+\frac{1}{4}\sum_{i=1}^nv_i^2\right)\sum_{\alpha=1}^m\partial_{z_{\alpha}}^2 +t^2\partial_{t}^2-\left(m+\frac{n}{2}-1\right)t\partial_{t}+t\sum_{i,j=1}^n\sum_{\alpha=1}^mC_{ij\alpha}v_i\partial_{v_j}\partial_{z_{\alpha}}, \end{align*} $$

where the second line is obtained from the preceding line using $\partial _{\tau }=t\partial _t$ and $\partial _{\tau }^2=t^2\partial _t^2+t\partial _t$ .

Evaluating $\Delta D_{x_0}$ at $x=(V,Z,t)\in \mathfrak v\oplus \mathfrak z\times \mathbb R_+$ , where $V=\sum _{i=1}^n V^iE_i$ , we get

(14) $$ \begin{align} \begin{aligned} \Delta D_{x_0}(x) ={}&\sum_{i=1}^n\left(t+t_0 +\frac{\|V-V_0\|^2}{4}+\frac{1}{2}\langle E_i,V-V_0\rangle^2+\frac12\| [E_i,V_0]\|^2\right)\\ &+\left(t+\frac{1}{4}\|V\|^2\right)\left(\sum_{\alpha=1}^m2\right)\\ &+\frac{2}{t}\left(\left(t_0+\frac{\|V-V_0\|^2}{4}\right)^2+ \left\|Z-Z_0+\frac12[V,V_0]\right\|^2\right)\\ &-\left( m+\frac{n}{2}-1\right)\left(t -\frac{1}{t}\left(\left(t_0+\frac{\|V-V_0\|^2}{4}\right)^2+ \left\|Z-Z_0+\frac12[V,V_0]\right\|^2\right)\right)\\ &+\sum_{i,j=1}^n\sum_{\alpha=1}^mC_{ij\alpha}V^i\langle F_{\alpha},[E_j,V_0]\rangle. \end{aligned} \end{align} $$

It is clear that

$$\begin{align*}\sum_{i=1}^n \langle E_i,V-V_0\rangle^2 =\|V-V_0\|^2. \end{align*}$$

Furthermore, Lemma 3.6 yields

$$\begin{align*}\sum_{i=1}^n \| [E_i,V_0]\|^2=m\|V_0\|^2, \end{align*}$$

and from Lemma 3.2, we obtain

$$ \begin{align*} \sum_{i,j=1}^n\sum_{\alpha=1}^mC_{ij\alpha}V^i\langle F_{\alpha},[E_j,V_0]\rangle &= \sum_{i,j,k=1}^n \sum_{\alpha=1}^mC_{ij\alpha}V^iC_{jk\alpha}\langle V_0,E_k \rangle\\&= \sum_{i,k=1}^n \sum_{\alpha=1}^m -\delta_{ik}V^i\langle V_0,E_k\rangle=-m\langle V,V_0\rangle. \end{align*} $$

Plugging these equations into (14), a simple algebraic rearrangement gives

$$\begin{align*}\Delta D_{x_0}(x)=\left( m+\frac{n}{2}+1\right)D_{x_0}(x)-2(m+1)t_0.\end{align*}$$

Consider now the squared norm of the gradient of $D_{x_0}$ . Denote by $\mathbf E_1,\dots ,\mathbf E_n;\mathbf F_1,\dots ,\mathbf F_m; \mathbf A$ the left-invariant vector fields corresponding to the orthonormal basis $E_1,\dots ,E_n;F_1,\dots ,F_m;A\in \mathfrak s$ . These vector fields are computed in [Reference Berndt, Tricerri and Vanhecke3, Sec. 4.1.5]. The derivative of $D_{x_0}$ with respect to these vector fields are

$$ \begin{align*} \mathbf{E}_iD_{x_0}(x)={}&\sqrt t\partial_{v_i}D_{x_0}(x)-\frac12\sqrt t\sum_{\alpha=1}^m\sum_{j=1}^nC_{ij\alpha}v_j\partial_{z_\alpha} D_{x_0}(x)\\ ={}&\frac1{\sqrt t}\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)\langle V-V_0,E_i\rangle+\frac1{\sqrt t}\left\langle Z-Z_0+\frac1{2} [V,V_0],[E_i,V_0]\right\rangle\\ &-\sum_{\alpha=1}^m\sum_{j=1}^nC_{ij\alpha}v_j\frac1{\sqrt t}\left\langle Z-Z_0+\frac12[V,V_0],F_\alpha\right\rangle\\ ={}&\frac1{\sqrt t}\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)\langle V-V_0,E_i\rangle+\frac1{\sqrt t}\left\langle Z-Z_0+\frac1{2} [V,V_0],[E_i,V_0]\right\rangle\\ &-\frac1{\sqrt t}\left\langle Z-Z_0+\frac12[V,V_0],[E_i,V]\right\rangle \\ ={}&\frac1{\sqrt t}\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)\langle V-V_0,E_i\rangle+\frac1{\sqrt t}\left\langle J_{Z-Z_0+\frac1{2} [V,V_0]}(V-V_0),E_i\right\rangle, \\ \mathbf F_\alpha D_{x_0}(x)={}&t\partial_{z_\alpha} D_{x_0}(x)=2\left\langle Z-Z_0+\frac12[V,V_0],F_\alpha\right\rangle, \\ \mathbf A D_{x_0}(x)={}& \partial_\tau D_{x_0}(x)=t\partial_t D_{x_0}(x)=-D_{x_0}(x)+2\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right). \end{align*} $$

The squared norm of the gradient is

$$ \begin{align*} \big\|\nabla D_{x_0}(x)\big\|^2={}&\sum_{i=1}^n\big(\mathbf E_iD_{x_0}(x)\big)^2+\sum_{\alpha=1}^m \big(\mathbf F_\alpha D_{x_0}(x)\big)^2+\big(\mathbf A D_{x_0}(x)\big)^2\\ ={}&\frac1t\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)^2\|V-V_0\|^2+\frac1t\left\|Z-Z_0+\frac12[V,V_0]\right\|^2\left\|V-V_0\right\|^2\\ &+4\left\|Z-Z_0+\frac12[V,V_0]\right\|^2\\ &+D^2_{x_0}(x)-4D_{x_0}(x)\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)+4\left(t+t_0+\left\|\frac{V-V_0}{2}\right\|^2\right)^2\\ ={}&D^2_{x_0}(x)-4t_0D_{x_0}(x).\\[-37pt] \end{align*} $$

Proof. The coefficients of the polynomial function $tD_{\eta (\theta )}((V,Z,t))$ diverge as $\theta $ tends to $1/s$ , (we set $1/s=\infty $ if $s=0$ ), but if we multiply $D_{\eta (\theta )}$ with the constant $(1/\theta -s)^2$ to slow down the increase of the coefficients, the normalised polynomials will converge to a non-zero polynomial. Using the relation $s^2+\|v\|^2=1$ , one can bring $(1/\theta -s)^2t D_{\eta (\theta )}((V,Z,t))$ into the form

$$\begin{align*}\!\left(\!\left(\frac{1}{\theta}-s\right)\!\!\left( t+{\bar t}+\left\|\frac{\bar V-V}{2}\right\|^2\right)+2 s {\bar t}-\sqrt{\bar t}\langle V-\bar V,v\rangle \right)^2\!+ \left\| \left(\frac{1}{\theta}-s\right)\!\!\left(Z-\bar Z+\frac{1}{2}[V,\bar V]\right)+\sqrt{\bar t}[V-\bar V,v]\right\|^2\!\!. \end{align*}$$

Thus, we have

$$\begin{align*}\lim_{\theta\to 1/s}(1/\theta-s)^2 D_{\eta(\theta)}((V,Z,t))=\frac{\bar t}{t}\left(\left(2s\sqrt{\bar t}-\langle V-\bar V,v\rangle \right)^2+\big\|[V-\bar V,v] \big\|^2\right), \end{align*}$$

which coincides with $D_{\circledast }^{\eta }(V,Z,t)$ up to the constant multiplier $\bar t$ . Then (16) is the equation of the level set of $D_{\circledast }^{\eta }$ passing through the point p.

Proof. Choose an orthonormal basis $F_1,\dots ,F_m$ of $\mathfrak z$ and set $J_{\alpha }=J_{F_{\alpha }}$ . Then $\bar v,J_{1}\bar v,\dots , J_{m}\bar v$ is an orthonormal basis of $J_{\mathfrak z}v \oplus \mathbb R v$ , where $\bar v=v/\|v\|$ . Thus,

$$ \begin{align*} \|v\|^2\delta (V,\mathcal F_{\circledast}^{\eta})^2 &= \left\langle V-\bar V-\frac{2s\sqrt{\bar t}}{\|v\|^2}v,v\right\rangle^2 +\sum_{\alpha=1}^m\left\langle V-\bar V-\frac{2s\sqrt{\bar t}}{\|v\|^2}v, J_{\alpha}v\right\rangle^2\\&= \left(\langle V-\bar V,v\rangle-2s\sqrt{\bar t}\right)^2 +\sum_{\alpha=1}^m\left\langle V-\bar V, J_{\alpha}v\right\rangle^2. \end{align*} $$

The proof is completed by the identity

$$\begin{align*}\sum_{\alpha=1}^m\langle V-\bar V, J_{\alpha}v\rangle^2=\sum_{\alpha=1}^m\langle [V-\bar V,v], F_{\alpha}\rangle^2=\|[V-\bar V,v]\|^2.\\[-42pt] \end{align*}$$

Proof. Consider a focal variety $\mathcal F_{x_0}$ and a geodesic curve intersecting it orthogonally at the point p. Applying the left translation $L_{p^{-1}}$ to the configuration of the focal variety and the geodesic, we see by Corollary 7.2, that it suffices to prove the theorem for the case $p=e\in \mathcal F_{x_0}$ . Condition $e\in \mathcal F_{x_0}$ is equivalent to the restrictions

(18) $$ \begin{align} t_0=-\tfrac{1}{4}\left\|V_0\right\|^2-1,\qquad Z_0=0 \end{align} $$

on the point $x_0=(V_0,Z_0,t_0)$ . The system of equations of such a focal submanifold $\mathcal F_{x_0}$ is

(19) $$ \begin{align} t=1-\tfrac{1}{4}\left\|V\right\|^2+\tfrac{1}{2}\langle V,V_0\rangle,\qquad Z=-\tfrac12[V,V_0], \end{align} $$

from which its tangent space at the identity is

(20) $$ \begin{align} T_e\mathcal F_{x_0}=\left\{(v',z',s')\in \mathfrak v\oplus \mathfrak z\oplus \mathbb R \,\,\mid \,\, s'=\tfrac{1}{2}\langle v',V_0\rangle,\,\, z'=-\tfrac{1}{2} [v',V_0]\right\}. \end{align} $$

The prolongation of a geodesic starting from e is parameterised by a map $\gamma $ of the form (8). The initial velocity of $\gamma $ is $\gamma '(0)=2(v,z,s)$ . The pregeodesic $\gamma $ intersects $\mathcal F_{x_0}$ orthogonally if and only if

$$\begin{align*}0=\left\langle 2(v,z,s),\left(v',-\tfrac{1}{2} [v',V_0],\tfrac{1}{2}\langle v',V_0\rangle\right)\right\rangle=\langle v',2v+sV_0\rangle-\langle z,[v',V_0]\rangle=\langle v',2v+sV_0+J_zV_0\rangle \end{align*}$$

for all $v'\in \mathfrak v$ . This is equivalent to the equation

(21) $$ \begin{align} v=-\tfrac{1}{2}(sV_0+J_zV_0), \end{align} $$

consequently $s^2+\|z\|^2\neq 0$ , otherwise we would have $(v,z,s)=0$ , contradicting $\|(v,z,s)\|=1$ . Using the assumption $\|v\|^2+\|z\|^2+s^2=1$ , equation (21) gives

(22) $$ \begin{align} \frac{1}{s^2+\|z\|^2}= 1+\tfrac{1}{4}\|V_0\|^2=-t_0. \end{align} $$

Combination of equation (9) for the case $\|v\|\neq 1$ , and equations (21), (22),(18) provides

$$ \begin{align*} \gamma(\infty) =\left(\frac{(s^2V_0+sJ_zV_0)-(sJ_zV_0-\|z\|^2V_0)}{{s}^2+\|z\|^2},0,-\frac{1}{{s}^2+\|z\|^2} \right)=(V_0,Z_0,t_0)=x_0. \end{align*} $$

Consider now a focal variety $\mathcal F_{\circledast }^{\eta }$ and a geodesic curve meeting it orthogonally. Referring to Proposition 8.3, we may assume that they meet at the identity element e. Let us write $\eta $ as the left translate $L_{(\bar V,\bar Z,\bar t)}\circ \gamma $ of a pregeodesic $\gamma $ of the type (8) with initial velocity $\gamma '(0)=2(v,0,s)$ . Then $e\in \mathcal F_{\circledast }^{\eta }$ holds if and only if $[\bar V,v]=0$ and $\langle \bar V,v\rangle =-2s\sqrt {\bar t}$ . In this case, the tangent space $T_e\mathcal F_{\circledast }^{\eta }$ is the space $(J_{\mathfrak z}v\oplus \mathbb R v)^{\perp }$ . Then any geodesic curve intersecting $\mathcal F_{\circledast }^{\eta }$ orthogonally at e starts with initial velocity in $J_{\mathfrak z}v\oplus \mathbb R v\subseteq \mathfrak v$ . The prolongation of any such geodesic goes through $\circledast $ by Theorem 5.1.

Proof. The points of $\mathrm {im}\, \eta $ in the boundary hyperplane of our model are $\eta (\pm 1)$ . As $\eta $ is a birational equivalence, $\eta $ preserves cross-ratio. The four-tuple $\big (\theta ,1/\theta ,1,-1\big )$ is a harmonic range in $\mathbb R\mathbf P^1$ , therefore, if we write $x_0$ in the form $\eta (\theta )$ , then the unique point $p\in \mathrm {im}\, \eta $ for which the cross ratio of the points $\big (x_0,p,\eta (1),\eta (-1)\big )$ with respect to $\mathrm {im}\, \eta $ is equal to $-1$ is $p=\eta (1/\theta )$ . Since $\big (x_0,p,\eta (1),\eta (-1)\big )$ is a harmonic range, the points $\eta (1)$ and $\eta (-1)$ separate the points $x_0$ and p in $\mathrm {im}\, \eta $ , which implies that $p\in \mathfrak n\times \mathbb R_+$ .

We show that the focal variety $\mathcal F_{x_0}$ passes through p, and that $\eta $ is orthogonal to $\mathcal F_{x_0}$ at p. The statement does not depend on the choice of the parameterisation of the geodesic and its prolongation. Thus, we may assume without loss of generality, that $\eta (0)=p$ and $\eta (\infty )=x_0$ . The statement is also invariant under left translations by Corollary 7.2; therefore we may also assume that $p=e$ , and $\eta =\gamma $ , where $\gamma $ is the pregeodesic given by equation (8). Since $x_0\neq \circledast $ , equation (9) yields $\|v\|\neq 1$ and

$$ \begin{align*} x_0=(V_0,Z_0,t_0)=\gamma(\infty)= \left(-\dfrac{2s}{{s}^2+\|z\|^2}v+\dfrac{2}{{s}^2+\|z\|^2}J_{z}v,0,-\dfrac{1}{{s}^2+\|z\|^2}\right). \end{align*} $$

A simple computation shows that the components of $x_0$ satisfy the equations in (18); therefore $\mathcal F_{x_0}$ is passing through e. The orthogonality condition (21) is also fulfilled, hence $\gamma $ intersects $\mathcal F_{x_0}$ orthogonally.

If $t_0$ is chosen so that $\tanh (t_0/2)={1}/{\theta }$ , then $\hat \eta (t_0)=p$ and the distance of $\hat \eta (t)$ from $\mathcal F_{x_0}$ is equal to $|t-t_0|$ , which implies that p is the only intersection point of $\hat \eta $ and the focal variety $\mathcal F_{x_0}$ .

Proof. By Proposition 8.3, we may assume that the two focal varieties intersect the geodesic at e. Then Proposition 8.2 (ii) gives that $F_{\circledast }^{\eta _1}=F_{\circledast }^{\eta _2}$ if and only if their tangent spaces at e are equal, that is $(J_{\mathfrak z}v_1\oplus \mathbb R v_1)^{\perp }=(J_{\mathfrak z}v_2\oplus \mathbb R v_2)^{\perp }$ , where $v_1$ and $v_2$ are non-zero elements of $\mathfrak v$ , related to $\eta _1$ and $\eta _2$ as in Proposition 8.2 (ii). The existence of a geodesic meeting both $F_{\circledast }^{\eta _1}$ and $F_{\circledast }^{\eta _2}$ orthogonally at e is equivalent to the existence of a non-zero vector $w\in (J_{\mathfrak z}v_1\oplus \mathbb R v_1)\cap (J_{\mathfrak z}v_2\oplus \mathbb R v_2)$ serving for the initial velocity of such a geodesic curve. Thus, the condition that the focal varieties $F_{\circledast }^{\eta _1}$ and $F_{\circledast }^{\eta _2}$ coincide if there is a geodesic which meets both of them orthogonally at a common point is equivalent to condition (iii) of Proposition 4.8, therefore, it is equivalent to the space being symmetric.

Proof. For the special choice of $x_0$ , the focal variety $\mathcal F_{x_0}$ and its tangent space at the identity are defined by equations (19) and (20), respectively. In particular, $(v,z,s)\in T_e\mathcal F_{x_0}$ is non-zero if and only if $v\neq 0$ . Consider the reparameterisation $\gamma \colon (-1,1)\to S$ of the unit speed geodesic curve $\hat \gamma $ with initial velocity $\hat \gamma '(0)=(v,z,s)\in T_e\mathcal F(v_0)$ given by equation (8). The point $\gamma (\theta )$ belongs to $\mathcal F_{x_0}$ for a given $\theta \in (-1,1)$ if and only if

(23) $$ \begin{align} \frac{1-\theta^2}{\chi(\theta)}=1-\frac{1}{4}\left\|\frac{2\theta(1-s\theta)}{\chi(\theta)}v+\frac{2\theta^2}{\chi(\theta)}J_zv\right\|^2+\frac{1}{2}\left\langle \frac{2\theta(1-s\theta)}{\chi(\theta)}v+\frac{2\theta^2}{\chi(\theta)}J_zv,V_0\right\rangle \end{align} $$

and

(24) $$ \begin{align} \frac{2\theta}{\chi(\theta)}z=-\frac12\left[\frac{2\theta(1-s\theta)}{\chi(\theta)}v+\frac{2\theta^2}{\chi(\theta)}J_zv,V_0\right]. \end{align} $$

Equation (23) is equivalent to the equation

$$\begin{align*}1-\theta^2=\chi(\theta)-\frac{\theta^2(1-s\theta)^2}{\chi(\theta)}\|v\|^2-\frac{\theta^4}{\chi(\theta)}\|z\|^2\|v\|^2+\theta(1-s\theta)\langle v,V_0\rangle+\theta^2\langle [v,V_0],z\rangle. \end{align*}$$

As the right hand side of this equation can be simplified as

$$ \begin{align*} \chi(\theta)&-\theta^2\frac{(1-s\theta)^2+\|z\|^2\theta^2}{\chi(\theta)}\|v\|^2+2s\theta(1-s\theta)+\theta^2\langle -2z,z\rangle\\ &=(1-s\theta)^2+\|z\|^2\theta^2-\theta^2\|v\|^2+2s\theta(1-s\theta)-2\theta^2\|z\|^2 =1-\theta^2(\|v\|^2+s^2+\|z\|^2)=1-\theta^2, \end{align*} $$

equation (23) is always fulfilled. Equation (24) holds if $\theta =0$ . If $\theta \neq 0$ , then substituting $s=\tfrac {1}{2}\langle v,V_0\rangle $ and $z=-\tfrac {1}{2} [v,V_0]$ , it can be brought to the equivalent form

$$ \begin{align*} -\langle v,V_0\rangle[v,V_0]=[J_{[v,V_0]}v,V_0]. \end{align*} $$

With the help of Lemma 3.3, this condition can be transformed into the equivalent condition

$$\begin{align*}\langle v,V_0\rangle[v,V_0]=[v,J_{[v,V_0]}V_0], \end{align*}$$

which gives by Corollary 3.5 the condition

(25) $$ \begin{align} -\langle v,V_0\rangle[v,V_0]=\|V_0\|^2 [v,P_{V_0}(v)], \end{align} $$

where $P_{V_0}$ is the orthogonal projection onto $J_{\mathfrak z}V_0$ . This computation shows that the focal variety $\mathcal F_{x_0}$ contains all geodesic curves starting from e with initial velocity belonging to $T_e\mathcal F_{x_0}$ if and only if equation (25) holds for any $v\in \mathfrak v$ . If $V_0=0$ , then $V_0$ satisfies both equation (25) and the $J^2$ -condition, thus, it is enough to consider the case $V_0\neq 0$ .

Decompose v as $v=v_1+v_2+v_3$ , where $v_1=P_{V_0}(v)=J_zV_0\in J_{\mathfrak z}V_0$ , $v_2=\lambda V_0\in \mathbb R V_0$ , and $v_3\in (J_{\mathfrak z}V_0\oplus \mathbb R V_0)^{\perp }\cap \mathfrak v=\ker _{\mathfrak v}(\operatorname {\mathrm {ad}} V_0)\cap V_0^{\perp }$ . Plugging this decomposition into equation (25), we obtain

$$ \begin{align*} -\lambda \|V_0\|^2 [J_zV_0,V_0]=\|V_0\|^2 [\lambda V_0+v_3,J_zV_0]. \end{align*} $$

Since $v_3\perp V_0$ , we have $[v_3,J_zV_0]=[J_zv_3,V_0]$ by Lemma 3.3, thus, the above equation reduces to

$$ \begin{align*} 0=\|V_0\|^2 [J_zv_3,V_0]. \end{align*} $$

Observe that if v is running over $\mathfrak v$ , then z can be an arbitrary element of $\mathfrak z$ , and $v_3$ can be an arbitrary element of $\ker _{\mathfrak v} (\operatorname {\mathrm {ad}} V_0)\cap V_0^{\perp }$ independently, so condition (25) holds for every v if and only if $J_{\mathfrak z}(\ker _{\mathfrak v} (\operatorname {\mathrm {ad}} V_0)\cap V_0^{\perp })\subseteq \ker _{\mathfrak v} (\operatorname {\mathrm {ad}} V_0)$ , however, by Corollary 4.4, this latter condition is fulfilled if and only if $V_0$ satisfies the $J^2$ -condition.

Proof. Apply to $\mathcal F_{x_0}$ the left translation by $\Upsilon (\bar V)^{-1}$ . This moves the point $\Upsilon (\bar V)$ to e and the focal variety $\mathcal F_{x_0}$ to $\mathcal F_{\Upsilon (\bar V)^{-1}x_0}$ by Corollary 7.2. Setting $(\bar V,\bar Z,\bar t)=\Upsilon (\bar V)$ , we have

$$\begin{align*}\Upsilon(\bar V)^{-1}x_0=\left(-\frac{\bar V}{\sqrt{\bar t}},-\frac{\bar Z}{\bar t},\frac{1}{\bar t}\right)(V_0,Z_0,t_0)= \left(\frac{V_0-\bar V}{\sqrt{\bar t}},\frac{Z_0-\bar Z-\frac{1}{2}[\bar V,V_0]}{\bar t},\frac{t_0}{\bar t}\right)=\left(\frac{V_0-\bar V}{\sqrt{\bar t}},0,\frac{t_0}{\bar t}\right). \end{align*}$$

Since left translations are isometries of the Damek–Ricci space, $\mathcal F_{x_0}$ is the exponential image of $T_{\Upsilon (\bar V)}\mathcal F_{x_0}$ if and only if $\mathcal F_{\Upsilon (\bar V)^{-1}x_0}$ is the exponential image of $T_e\mathcal F_{\Upsilon (\bar V)^{-1}x_0}$ , and by Proposition 10.1, this is equivalent to the condition that the vector $\frac {V_0-\bar V}{\sqrt {\bar t}}$ or simply the vector $\bar V-V_0$ satisfies the $J^2$ -condition.

Proof. Left translations of the Damek–Ricci space are given by affine transformations (7) in the half-space model, the linear part of which leaves invariant every linear subspace containing the subspace $\mathfrak z\oplus \mathfrak a$ , in particular all subspaces of the form $(J_{\mathfrak z}v\oplus \mathbb Rv)^{\perp }$ . As the direction space of the affine half-space representing the focal variety $\mathcal F_{\circledast }^{\eta }$ in the half-space model is $(J_{\mathfrak z}v\oplus \mathbb Rv)^{\perp }$ , any left translation which maps an arbitrary point of $\mathcal F_{\circledast }^{\eta }$ to e maps the focal variety $\mathcal F_{\circledast }^{\eta }$ onto the focal variety

(27) $$ \begin{align} \mathcal F(v)=\{(V,Z,t)\mid [ V,v]=0\text{ and } \langle V,v \rangle=0\}. \end{align} $$

This implies that the focal variety $\mathcal F_{\circledast }^{\eta }$ is totally geodesic if and only if any unit speed geodesic curve $\hat \gamma \colon \mathbb R\to S$ starting from $\hat \gamma (0)=e$ with initial velocity $\hat \gamma '(0)\in T_e\mathcal F(v)$ stays in $\mathcal F(v)$ . The tangent space of $\mathcal F(v)$ at e is

$$\begin{align*}T_e\mathcal F(v)=\{(v',{z'},s')\in \mathfrak v\oplus\mathfrak z\oplus\mathbb R \mid [v',v]=0 \text{ and } \langle v',v \rangle=0\}=(J_{\mathfrak z}v\oplus \mathbb R v)^{\perp}. \end{align*}$$

The unit speed geodesic curve $\hat \gamma $ starting with initial velocity $\hat \gamma '(0)=(v',{z'},s')\in T_e\mathcal F(v)$ can be reparameterised by the pregeodesic $\gamma \colon (-1,1)\to S$ given by equation (8) substituting $(v',z',s')$ for $(v,z,s)$ . The point $\gamma (\theta )$ belongs to $\mathcal F(v)$ if and only if

$$\begin{align*}\left[\frac{2\theta(1-s'\theta)}{\chi(\theta)}v'+\frac{2\theta^2}{\chi(\theta)}J_{z'}v',v\right]=0\quad\text{ and }\quad\left\langle \frac{2\theta(1-s'\theta)}{\chi(\theta)}v'+\frac{2\theta^2}{\chi(\theta)}J_{z'}v',v \right\rangle=0,\end{align*}$$

thus, $\hat \gamma $ stays in $\mathcal F(v)$ if and only if

$$\begin{align*}\left[J_{z'}v',v\right]=0\quad\text{ and }\quad\left\langle J_{z'}v',v \right\rangle=0,\end{align*}$$

which is equivalent to $J_{z'}v'\in \ker _{\mathfrak v}(\operatorname {\mathrm {ad}} v)\cap v^{\perp }$ . We conclude that $\mathcal F_{\circledast }^{\eta }$ is totally geodesic if and only if $J_{z'}v'\in \ker _{\mathfrak v}(\operatorname {\mathrm {ad}} v)\cap v^{\perp }$ for all ${z'}\in \mathfrak z$ and for all $v'\in \ker _{\mathfrak v}(\operatorname {\mathrm {ad}} v)\cap v^{\perp }$ , meaning that $\ker _{\mathfrak v}(\operatorname {\mathrm {ad}} v)\cap v^{\perp }$ is a $\mathrm {Cl}(\mathfrak z,q)$ -submodule of $\mathfrak v$ . This completes the proof by Corollary 4.4.

Proof. If the space is symmetric, then it is a hyperbolic space $\mathbb K\mathbf H^{k}$ over an algebra $\mathbb K\in \{\mathbb R,\mathbb C,\mathbb H,\mathbb O\}$ , ( $k=2$ if $\mathbb K=\mathbb O)$ , and its tangent spaces are $\mathbb K$ -modules. In that case, each focal variety $\mathcal F_{x_0}$ is totally geodesic, and its tangent spaces are $\mathbb K$ -submodules of corank $1$ . Thus, the focal varieties are congruent to a totally geodesic $\mathbb K\mathbf H^{k-1}$ . Totally geodesic submanifolds of $\mathbb K\mathbf H^{k}$ that are the singular orbits of a cohomogeneity one isometric action were classified by J. Berndt and M. Brück [Reference Berndt and Brück1], and the submanifolds $\mathbb K\mathbf H^{k-1}$ are contained in their list; therefore the tubes about these focal surfaces are homogeneous.

Theorem 10.2 shows that if the ambient space is not symmetric, then each focal variety $\mathcal F_{x_0}$ has at least one point p with the property that $\mathcal F_{x_0}$ is the exponential image of the tangent space $T_p\mathcal F_{x_0}$ , and it also has points which do not have this property. This means that the focal variety cannot be homogeneous.

Proof. The focal varieties are special cases of the construction of [Reference Díaz-Ramos and Domínguez-Vázquez17] with the choice $\mathfrak w=\ker _{\mathfrak v}(\operatorname {\mathrm {ad}} v)\cap v^{\perp }$ of the subspace $\mathfrak w\leq \mathfrak v$ . As it is proved in [Reference Díaz-Ramos and Domínguez-Vázquez17], the tubes about $\mathcal F_{\circledast }^{\eta }$ have constant principal curvatures if and only if the Kähler angles of the non-zero vectors $u\in \mathfrak w^{\perp }=J_{\mathfrak z}v\oplus \mathbb R v$ are constant. Recall that the Kähler angles of $u\in \mathfrak w^{\perp }$ are defined to be the principal angles between the subspaces $J_{\mathfrak z} u$ and $\mathfrak w^{\perp }$ . In our case, $v\in \mathfrak w^{\perp }=J_{\mathfrak z}v\oplus \mathbb R v$ and $J_{\mathfrak z}v\subseteq \mathfrak w^{\perp }$ , thus, all the Kähler angles of the vector v are equal to $0$ . For this reason, the Kähler angles of the non-zero vectors of $\mathfrak w^{\perp }$ are constant if and only if they are all equal to $0$ for any $u\in \mathfrak w^{\perp }$ . But this happens if and only if $J_{\mathfrak z}u\subseteq J_{\mathfrak z}v\oplus \mathbb R v$ for any $u\in J_{\mathfrak z}v\oplus \mathbb R v$ and the latter condition is equivalent to the $J^2$ -condition for v.

Proof. If the tubes about $\mathcal F_{\circledast }^{\eta }$ are homogeneous, then they have constant principal curvatures and v satisfies the $J^2$ -condition by Theorem 11.2.

Conversely, assume that $v\neq 0$ satisfies the $J^2$ -condition. The existence of such a vector implies $m=\dim \mathfrak z\in \{0,1,3,7\}$ by Proposition 4.5.

If $m\in \{0,1\}$ , then the ambient space is isometric with $\mathbb K\mathbf H^k$ for some k and $\mathbb K\in \{\mathbb R,\mathbb C\}$ , hence it is symmetric. By Theorem 10.3, the focal varieties $\mathcal F_{\circledast }^{\eta }$ are totally geodesic in a symmetric space. The tangent spaces of the ambient space are linear spaces over $\mathbb K$ , and the tangent space of $\mathcal F_{\circledast }^{\eta }$ at any point is a $\mathbb K$ -linear subspace of codimension $1$ . Thus, the tubes about $\mathcal F_{\circledast }^{\eta }$ are homogeneous by the same reason as in the proof of the first part of Theorem 11.1.

Assume that $m\in \{3,7\}$ . Consider two arbitrary points $p_1$ , $p_2$ on the tube $\mathcal T$ of radius r about $\mathcal F_{\circledast }^{\eta }$ , and denote by $\tilde p_1$ and $\tilde p_2$ the points of the focal variety $\mathcal F_{\circledast }^{\eta }$ lying nearest to them. The left translations by $\tilde p_1^{-1}$ and $\tilde {p}_2^{-1}$ map the tube $\mathcal T$ onto the tube of radius r about the focal variety $\mathcal F(v)$ defined by equation (27) and map the points $\tilde {p}_1$ and $\tilde p_2$ to e. We can write the points $\tilde p_1^{-1}p_1$ and $\tilde p_2^{-1}p_2$ as $\mathrm {exp}_e(\xi _1)$ and $\mathrm {exp}_e(\xi _2)$ respectively with some uniquely defined vectors $\xi _1,\xi _2\in (T_e\mathcal F(v))^{\perp }$ of length r.

Thus, to prove that $\mathcal T$ is homogeneous, it is enough to show that there is an isometry I of the space such that $I(\mathcal F(v))=\mathcal F(v)$ , $I(e)=e$ , and $T_eI(\xi _1)=\xi _2$ . In the case $\|\xi _1\|=\|\xi _2\|=0$ , we can choose the identity map for I, so assume $\|\xi _1\|=\|\xi _2\|\neq 0$ .

Let $\mathrm {Spin}(m)\subset \mathrm {Cl}(\mathfrak z,q)$ be the spin group and $\rho \colon \mathrm {Spin}(m)\to \mathrm {SO}(\mathfrak z,q)$ its canonical representation. Recall that $\mathfrak z$ is embedded into $\mathrm {Cl}(\mathfrak z,q)$ as a subspace, and for $\sigma \in \mathrm {Spin}(m)$ , we have $(\rho (\sigma ))(z)=\sigma z \sigma ^{-1}$ . As the vector v satisfies the $J^2$ -condition, it generates an irreducible Clifford module $J_{\mathfrak z} v\oplus \mathbb Rv=(T_e\mathcal F(v))^{\perp }$ , which contains both $\xi _1$ and $\xi _2$ . It is known that for $m\in \{3,7\}$ , the group $\mathrm {Spin}(m)$ acts transitively on the unit sphere of any irreducible $\mathrm {Cl}(\mathfrak z,q)$ -module. For $m=3$ , this follows from the fact that the group $\mathrm {Spin}(3)$ is isomorphic to the group of unit quaternions, see [Reference Lawson and Michelsohn29, Ch. I., Thm. 8.1], and this group acts on itself transitively both by left and right translations. The case $m=7$ is a theorem of A. Borel, see [Reference Lawson and Michelsohn29, Ch. I., Thm. 8.2]. As a consequence, there is an element $\sigma \in \mathrm {Spin}(m)$ , such that $(J(\sigma ))(\xi _1)=\xi _2$ . The orthogonal transformation $\iota \colon \mathfrak v\oplus \mathfrak z\oplus \mathfrak a\to \mathfrak v\oplus \mathfrak z\oplus \mathfrak a$ ,

$$ \begin{align*}\iota((v,z,a))=((J(\sigma))(v), (\rho(\sigma))(z), a) \end{align*} $$

is an automorphism of the Lie algebra $\mathfrak s=\mathfrak v\oplus \mathfrak z\oplus \mathfrak a$ . This is an easy consequence of the identity

$$\begin{align*}J_{\iota(z)}\iota(v)=\big(J(\sigma)\circ J(z)\circ J(\sigma)^{-1}\circ J(\sigma)\big)(v)= \big(J(\sigma)\circ J(z)\big)(v)=\iota(J_zv) \quad \forall v\in \mathfrak v, \, z\in \mathfrak z. \end{align*}$$

Thus, $\iota $ is the derivative of an isometric automorphism I of the group S at e. It is clear from the construction of I that $I(e)=e$ and $T_eI(\xi _1)=\iota (\xi _1)=\xi _2$ . This implies that

(28) $$ \begin{align} \iota(J_{\mathfrak z}\xi_1\oplus \mathbb R\xi_1)=J_{\iota(\mathfrak z)}\iota(\xi_1)\oplus \mathbb R\iota(\xi_1)=J_{\mathfrak z}\xi_2\oplus \mathbb R\xi_2. \end{align} $$

Since $\xi _1$ and $\xi _2$ are in the $(m+1)$ -dimensional Clifford module $J_{\mathfrak z}v\oplus \mathbb R v$ , the $(m+1)$ -dimensional linear spaces $J_{\mathfrak z}\xi _1\oplus \mathbb R\xi _1$ and $J_{\mathfrak z}\xi _2\oplus \mathbb R\xi _2$ are also contained in $J_{\mathfrak z}v\oplus \mathbb R v$ , which implies

(29) $$ \begin{align} J_{\mathfrak z}\xi_1\oplus \mathbb R\xi_1=J_{\mathfrak z}v\oplus \mathbb R v= J_{\mathfrak z}\xi_2\oplus \mathbb R\xi_2. \end{align} $$

Equations (28) and (29) give $\iota (J_{\mathfrak z}v\oplus \mathbb R v)=J_{\mathfrak z}v\oplus \mathbb R v$ , and since $\iota $ is orthogonal,

$$\begin{align*}\iota(T_e\mathcal F(v))=\iota \left((J_{\mathfrak z}v\oplus \mathbb R v)^{\perp}\right)=(J_{\mathfrak z}v\oplus \mathbb R v)^{\perp}=T_e\mathcal F(v). \end{align*}$$

By Theorem 10.3, the focal variety $\mathcal F(v)$ is a totally geodesic submanifold, hence it is the Riemannian exponential image of $T_e\mathcal F(v)$ . Since the derivative $\iota $ of the isometry I at e maps $T_e\mathcal F(v)$ into itself, $I(\mathcal F(v))=\mathcal F(v)$ .

Proof. The case of spheres and horospheres is well known. For a complete harmonic manifold $(M,g)$ , there is a smooth function $\omega \colon \mathbb R\to \mathbb R$ , called the volume density function, defined by the identity $\omega (\|\xi \|)=\sqrt {\det (G(\xi ))}$ , where $\xi \in T_pM$ is an arbitrary tangent vector of M, and $G(\xi )$ is the matrix of the pull-back form $(T_{\xi }\mathrm { exp}_p)^*(g_{\mathrm {exp}{}_p(\xi )})$ with respect to a $g_p$ -orthonormal basis of $T_{\xi }(T_pM)\cong T_pM$ . It is known [Reference Damek and Ricci15] that the volume density function $\omega $ of the Damek–Ricci space is

$$\begin{align*}\omega(r)=\cosh^m(r/2)\left(\frac{\sinh(r/2)}{r/2}\right)^{m+n}.\end{align*}$$

The function h for a sphere of radius r can be expressed with the help of the volume density function

$$\begin{align*}h=-\partial_r(\ln r^{m+n}\omega)\end{align*}$$

(see [Reference Szabó41]), from which we obtain the formula for the mean curvature of spheres, and taking the limit $r\to \infty $ gives the formula for the horospheres.

Consider now a regular level set $\Sigma =D_{x_0}^{-1}(c)$ of a function $D_{x_0}$ with $x_0=(V_0,Z_0,t_0)$ , $t_0<0$ . Then, according to the proof of Theorem 6.1, the functions a and b certifying that $D_{x_0}$ is isoparametric are $a(x)=(m+\frac {n}{2}+1)x-2(m+1)t_0$ and $b(x)=x^2-4t_0x$ . The minimal value of $D_{x_0}$ is $0$ and by Proposition 2.5, the hypersurface $\Sigma $ is a tube of radius

$$\begin{align*}r=\int_0^c\frac{\,\mathrm{d} x}{\sqrt{x^2-4t_0x}}=2\ln(\sqrt{c-4t_0}+\sqrt c)-2\ln(\sqrt{-4t_0}) \end{align*}$$

about $\mathcal F_{x_0}$ . Expressing c as a function of r from this equation, we get

$$\begin{align*}c=-4t_0\sinh^2(r/2). \end{align*}$$

By Proposition 2.2, the trace of the shape operator of $\Sigma $ is

$$ \begin{align*} h&=\frac{-2a(c)+b'(c)}{2\sqrt{b(c)}}=-\frac{(m+n/2)c-2mt_0}{\sqrt{c^2-4t_0c}}=-\frac {m+n}2\sqrt{\frac{c}{c-4t_0}}-\frac{m}2\sqrt{\frac{c-4t_0}{c}} \\&=-\frac {m+n}2\tanh(r/2)-\frac{m}2\coth(r/2), \end{align*} $$

as claimed.

By equation (17), the functions a and b corresponding to the isoparametric functions $D_{\circledast }^{\eta }$ are $a(x)=(m+\frac {n}{2}+1)x+2(m+1)$ and $b(x)=x^2+4x$ . Hence, substituting $t_0=-1$ into the above formulae obtained for the level sets of $D_{x_0}$ , we get the corresponding formulae for $D_{\circledast }^{\eta }$ , which completes the proof.