The Geometry of Riemannian Curvature Radii

Abstract

We study the geometric structures associated with curvature radii of curves with values on a Riemannian manifold (M, g). We show the existence of sub-Riemannian manifolds naturally associated with the curvature radii and we investigate their properties. In the particular case of surfaces these sub-Riemannian structures are of Engel type. The main character of our construction is a pair of global vector fields $f_1,f_2$, which encodes intrinsic information on the geometry of (M, g).

1 Introduction

1.1 From Contact Elements to Curvature Radii

The space of contact elements, first introduced by S. Lie (see [3] pages 6–11, or [2] pages 78–80) as a geometric tool for studying differential equations, marks the birth of contact geometry. Let us recall that given a 2-dimensional Riemannian manifold (M, g), the space of (oriented) contact elements of M is the unit tangent bundle UM, i.e. the set of couples (x, v) where $x\in M$, $v\in T_xM$ and $|v|=1$. Any regular M-valued curve can be lifted to a UM-valued one through the maps

$$\begin{aligned} \ell _{\pm }:\gamma \mapsto \left( \gamma ,\pm \frac{\dot{\gamma }}{|\dot{\gamma }|}\right) . \end{aligned}$$

(1.1)

There exists a contact distribution $\Delta$ over UM naturally associated to the lifts Eq. 1.1. At a point $(x,v)\in UM$ such distribution is defined as

$$\begin{aligned} \Delta _{(x,v)}=\langle \{v\}\rangle \oplus T_{v}(U_xM), \end{aligned}$$

where we have denoted $U_xM=\{v\in T_xM\,:\,|v|=1\}$. This distribution characterizes the images of Eq. 1.1, in the sense that a curve $(\gamma ,v):[0,1]\rightarrow UM$ is in the image of one of such lifts if and only if it is tangent to $\Delta$ and $\gamma$ is a regular curve. In this paper we study a second order generalization of the space of contact elements, in particular, given a Riemannian manifold (M, g), we study the geometric structures associated to the lifts

$$\begin{aligned} c_{\pm }:\gamma \mapsto \left( \gamma , \pm \frac{\dot{\gamma }}{|\dot{\gamma }|}, R_g(\gamma )\right) , \end{aligned}$$

(1.2)

where $R_g(\gamma )$ is the radius of curvature of $\gamma$ (the definition is recalled below Eq. 1.4). Such second order construction has affinities with Cartan’s prolongation of contact structures (see 6.3 of [2]). Recall that, given a Riemannian manifold (M, g) (throughout the whole paper we assume $\dim (M)>1$), the geodesic curvature of a regular curve $\gamma :[0,T]\rightarrow M$ is defined as

$$\begin{aligned} \kappa _{g}(\gamma ):=\frac{|\pi _{\dot{\gamma }^\perp }\left( D_t\dot{\gamma }\right) |}{|\dot{\gamma }|^2}, \end{aligned}$$

(1.3)

where $D_t$ denotes the covariant derivative along $\gamma$, and $\pi _{\dot{\gamma }^\perp }:T_{\gamma (t)}M\rightarrow T_{\gamma (t)}M$ denotes the orthogonal projection to $\{\dot{\gamma }(t)\}^\perp$. If the geodesic curvature is never vanishing, we can define the radius of curvature of $\gamma$, computed with respect to g, as

$$\begin{aligned} R_g(\gamma ):=\frac{1}{\kappa _g(\gamma )}\frac{\pi _{\dot{\gamma }^\perp }\left( D_t\dot{\gamma }\right) }{|\pi _{\dot{\gamma }^\perp }\left( D_t\dot{\gamma }\right) |}. \end{aligned}$$

(1.4)

With the sole purpose of simplifying the exposition and the expression of certain equations, we study the following modified versions of Eq. 1.2

$$\begin{aligned} c_{\pm }:\gamma \mapsto \left( \gamma ,\pm \frac{|R_g(\gamma )|}{|\dot{\gamma }|}\dot{\gamma },R_g(\gamma )\right) . \end{aligned}$$

(1.5)

We define the manifold of curvature radii of (M, g), denoted with $\mathcal R(M,g)$, as the space of triples (x, V, R) such that $x\in M$, $R\in T_x M\setminus \{0\}$ and $V \in \{R\}^\perp$, $|V|=|R|$. The map Eq. 1.5 lifts regular M-valued curves with never vanishing geodesic curvature to $\mathcal R(M,g)$-valued ones. The first central result of this work is the following theorem.

Theorem 1.1

Let (M, g) be a smooth n-dimensional Riemannian manifold, and let $\mathcal R(M,g)$ be the corresponding manifold of curvature radii. There exists a smooth, rank-n distribution, $\mathcal D(M,g)$, over $\mathcal R(M,g)$ with the property that a smooth curve $(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ is in the image of one of the lifts Eq. 1.5 if and only if it is tangent to $\mathcal D(M,g)$ and $\gamma$ is a regular curve. A local basis for $\mathcal D(M,g)$ is given by n local vector fields $\{f_1,\dots ,f_n\}$, which can be characterized in terms of the ODEs satisfied by their integral curves:

$$\begin{aligned} f_1:{\left\{ \begin{array}{ll} \dot{x}=V,\\ D_t R=-V,\\ D_t V=R, \end{array}\right. } f_2:{\left\{ \begin{array}{ll} \dot{x}=0,\\ \dot{R}=R,\\ \dot{V}=V, \end{array}\right. } f_j:{\left\{ \begin{array}{ll} \dot{x}=0,\\ \dot{R}=b_j(x,V,R),\\ \dot{V}=0, \end{array}\right. } \end{aligned}$$

(1.6)

$j=3,\dots ,n$, where $\left\{ b_3(x,V,R),\dots ,b_n(x,V,R)\right\}$ is a norm-|R| local orthogonal basis of $\{R,V\}^{\perp }$. The distribution $\mathcal D(M,g)$ is bracket generating of step 3, it holds:

$$\begin{aligned} \begin{aligned}&\mathcal D(M,g)=\langle \{ f_1,\dots ,f_n\}\rangle ,\\&\mathcal D^2(M,g)=\mathcal D(M,g)\oplus \langle \{f_{1k}\}_{k=2}^n\rangle ,\\&\mathcal D^3(M,g)=\mathcal D^2(M,g)\oplus \langle \{f_{1k1}\}_{k=2}^n\rangle =T\mathcal R(M,g). \end{aligned} \end{aligned}$$

(1.7)

The fields $f_1,f_2$ described in Eq. 1.6 are closely related to the geometry of (M, g). Indeed, as shown in Section 4, many geometric features of the original Riemannian manifold can be recovered considering their Lie brackets. The first bracket $[f_1,f_2]$ gives us back the geodesic flow of (M, g) (see Proposition 4.1); in a way these fields factorize the geodesic flow. Moreover we can read the Riemann curvature tensor in the structure constant of the frame Eq. 1.6 (see Proposition 4.3).

1.2 Metric Structures

The knowledge of the curvature radii of curves of a Riemannian manifold (M, g), is enough to characterize the metric g up to a homothetic transformation, indeed, as shown in Section 3, two Riemannian manifolds having the same curvature radii are homothetic (see Definition 3.2 for the precise statement).

Theorem 1.2

Two Riemannian manifolds (M, g) and $(N,\eta )$ have the same curvature radii if and only if they are homothetic .

It is then natural to endow the distribution $\mathcal D(M,g)$ with a metric which is invariant under the action of the homothety group of (M, g). In Section 3 we show the existence of a family of metrics $\eta _{a,b}$ on $\mathcal D(M,g)$, parametrized by two real numbers a and b, having this invariance property. The triple $(\mathcal R(M,g), \mathcal D(M,g),\eta _{a,b})$ is a sub-Riemannian manifold ([2, 8]), which we denote with $\mathcal R_{a,b}(M,g)$. For any $(\gamma , R, V)$ in the image of the lift Eq. 1.5 the metric $\eta _{a,b}$ satisfies the following equation

$$\begin{aligned} \begin{aligned} \left| \frac{d}{dt}(\gamma ,V,R)\right| ^2_{\eta _{a,b}}=a^2\frac{|\dot{\gamma }|^2}{|R|^2}+b^2\frac{|D_tR|^2}{|R|^2}. \end{aligned} \end{aligned}$$

(1.8)

The central result regarding these metrics is stated in the following theorem.

Theorem 1.3

Let (M, g) be a Riemannian manifold, let $a,b\in \mathbf {\mathbb {R}},b>0$, and let $\mathcal R_{a,b}(M,g)$ be the corresponding sub-Riemannian manifold of curvature radii. The following map is a group isomorphism

$$\begin{aligned} \begin{aligned} Homothety(M,g)&\rightarrow Isometry(\mathcal R_{a,b}(M,g))\\ \varphi&\mapsto (\varphi _\star \oplus \varphi _{\star })_{|\mathcal R(M,g)}. \end{aligned} \end{aligned}$$

(1.9)

In particular, if $(N,\eta )$ is another Riemannian manifold, $\mathcal R_{a,b}(M,\eta )$ is isometric to $\mathcal R_{a,b}(M,g)$ if and only $(N,\eta )$ and (M, g) have the same curvature radii.

In Section 5 we show that the sub-Riemannian manifolds $\mathcal R_{a,b}(\mathbb {R}^2,g_e)$, where $g_e$ is the standard Euclidean metric, are all isometric to left-invariant sub-Riemannian structures on the group orientation preserving homothetic transformations of $(\mathbb R^2,g_e)$, and we give a characterization of their geodesics. These geometries are related to the left-invariant sub-Riemannian structure on the group of rigid motions of $\mathbb R^2$ ([4, 5, 7, 9]).

1.3 Structure of the Paper

In Section 2 we describe the sub-Riemannian manifold of curvature radii in the 2-dimensional case. In Section 3 we generalize the constructions of Section 2 to an arbitrary Riemannian manifold (M, g) and we prove Theorems 1.1, 1.2, 1.3. In Section 4 we show that taking Lie brackets of the fields $f_1,f_2$, mentioned in the abstract, we can reconstruct the geodesic flow of the original Riemannian manifold and its Riemann curvature tensor. Finally in Section 5 we study the manifold of curvature radii of $\mathbb R^2$ endowed with an homothetic invariant metric.

2 Curvature Radii on Surfaces

Let (M, g) be a 2-dimensional Riemannian manifold, which for simplicity we assume to be oriented. Let $\gamma :[0,1]\rightarrow M$ be an arc-length parametrized curve, then its curvature can be computed as

$$\begin{aligned} k_g(\gamma )=|D_t\gamma |, \end{aligned}$$

(2.1)

and, provided that $k_g(\gamma )$ is never vanishing, its radius of curvature is

$$\begin{aligned} R_g(\gamma )=\frac{1}{k_g(\gamma )^2}D_t\gamma . \end{aligned}$$

(2.2)

For every $t\in [0,1]$, the curvature radius of $\gamma$ at the point $\gamma (t)$ is a non-zero tangent vector, for this reason we define the manifold of curvature radii of (M, g) as $\mathcal R(M)=TM\setminus s_o$, where $s_o$ is the zero-section of TM. Every regular curve with non-vanishing curvature can be lifted to a $\mathcal R(M)$-valued curve through the radius of curvature map

$$\begin{aligned} \gamma \mapsto (\gamma , R_g(\gamma )). \end{aligned}$$

(2.3)

We would like to find a distribution $\mathcal D(M,g)$ over $\mathcal R(M)$ characterizing the image of such lift, in the sense that a curve $(\gamma , R):[0,1]\rightarrow \mathcal R(M)$ is in the image of Eq. 2.3 if and only if it is tangent to $\mathcal D(M,g)$ and $\gamma$ is regular. To build such a distribution at a point $(x_0,R_0)\in \mathcal R(M)$ we collect all the velocities of all radii of curvature going through this point, and we take the vector space generated by them

$$\begin{aligned} \mathcal D(M,g)_{(x_0,R_0)}:=\left\langle \left\{ \frac{d}{dt}(\gamma ,R_g(\gamma ))(0)\,:\,\gamma (0)=x_0,\,R_g(\gamma )(0)=R_0\right\} \right\rangle . \end{aligned}$$

(2.4)

Since M is oriented we have a complex strcuture

$$\begin{aligned} ^{\perp _g}:TM\setminus s_o \quad \rightarrow TM \setminus s_o\\ (x,R)\quad \mapsto (x,R^{\perp _g}) \nonumber \end{aligned}$$

(2.5)

where $R^{\perp _g}$ is the unique vector orthogonal to R, positively oriented with it, satisfying $|R|_g=|R^{\perp _g}|_g$.

Proposition 2.1

The collection of vector spaces defined in Eq. 2.4 is a smooth Engel distribution over $\mathcal R(M)$ (for basic facts on Engel distributions see for instance [1], chapter 2). Moreover a curve $(\gamma ,R):[0,T]\rightarrow \mathcal R(M)$ is in the image of the lift Eq. 2.3 if and only if it is tangent to $\mathcal D(M,g)$ and $\gamma$ is a regular curve. A basis for $\mathcal D(M,g)$ is given by two vector fields $f^g_1,f^g_2$, which are characterized in terms of the ODEs satisfied by their integral curves as

$$\begin{aligned} f^g_1:\,{\left\{ \begin{array}{ll} \dot{x}=R^{\perp _g},\\ D_t R=-R^{\perp _g}, \end{array}\right. } f^g_2:\,{\left\{ \begin{array}{ll} \dot{x}=0,\\ \dot{R}=R, \end{array}\right. } \end{aligned}$$

(2.6)

where $D_tR$ denotes the covariant derivative of R(t) along the curve x(t).

Proof

It is a special case of Theorem 1.1, which is proved in Section 3. $\square$

What is the geometric significance of the fields $f_1^g,f_2^g$? It is quite explicit that the integral curves of $f_2^g$ are dialation of R in a fixed fiber, i.e. curves of the kind $t\mapsto (x_0,e^tR_0)$, whereas if $(\gamma ,R):[0,1]\rightarrow \mathcal R(M)$ is an integral curve of $f_1^g$, as a consequence of Proposition 2.1, R is the radius of curvature of $\gamma$, and hence $\kappa _g(\gamma )=1/|R|_g$. On the other hand according to Eq. 2.6$D_tR=-R^{\perp _g}$, therefore

$$\begin{aligned} \frac{d}{dt}|R|^2_g=-2\langle R,R^{\perp _g}\rangle _g=0, \end{aligned}$$

hence $\kappa _g(\gamma )$ is constant. Thus the integral curves of $f_1^g$ are exactly lifts of curves with constant geodesic curvature. Homothetic transformations preserve the curvature radius map. It is then natural to endow $\mathcal D(M,g)$ with a metric which is invariant under the action of the homothety group of (M, g). For every $a,b\in \mathbb R$, $b\ne 0$, we define a metric $\eta ^g_{a,b}$ on $\mathcal D(M,g)$ by declaring the fields

$$\begin{aligned} \begin{aligned}&f_1^{g,a,b}:=\frac{1}{\sqrt{a^2+b^2}}f^g_1,\\&f_2^{g,a,b}:=\frac{1}{b}f^g_2, \end{aligned} \end{aligned}$$

(2.7)

an orthonormal frame. A simple computation, which is a particular case of Eq. 3.47, shows that if $(\gamma ,R)$ is an admissible curve, then the metric $\eta _{a,b}^g$ satisfies the following equation

$$\begin{aligned} \begin{aligned} \left| \frac{d}{dt}(\gamma ,R)\right| ^2_{\eta _{a,b}^g}=a^2\frac{|\dot{\gamma }|^2}{|R|^2}+b^2\frac{|D_tR|^2}{|R|^2}. \end{aligned} \end{aligned}$$

(2.8)

The homogeneity of the right hand side shows the homothetic invariant nature of the metric $\eta ^g_{a,b}$. The triple $(\mathcal R(M), \mathcal D(M,g), \eta _{a,b}^g)$ defines a sub-Riemannian manifold, which we denote with $\mathcal R_{a,b}(M,g)$.

3 The Sub-Riemannian Manifold of Curvature Radii

We would like to extend the construction of $\mathcal R_{a,b}(M,g)$, presented for surfaces in Section 2, to an arbitrary Riemannian manifold. One of the advantages of working with surfaces is that, given a regular curve $\gamma$ with never vanishing geodesic curvature, the direction of $\dot{\gamma }$ is uniquely determined by the radius of curvature. This is no more true when $\dim (M)\ge 3$, and we need to keep track of the velocity’s direction in some way. For this reason we cannot define the manifold of curvature radii as $TM\setminus s_o$ anymore. Instead we define it as a subset of $TM\oplus TM$.

Definition 3.1

Let (M, g) be a Riemannian manifold. The manifold of curvature radii of (M, g), denoted by $\mathcal R(M,g)$, is defined by

$$\begin{aligned} \mathcal R(M,g):=\{(x,V,R)\in TM\oplus TM\,:\,\langle R,V\rangle _g=0,\,|R|_g=|V|_g>0\}. \end{aligned}$$

(3.1)

From now on, when the metric g we are referring to is clear from the context, we drop the g subscript, for instance we denote $|\cdot |=|\cdot |_g$, $\perp =\perp _g$ and so on. One can check that, if M is an n-dimensional manifold, then $\mathcal R(M,g)$ is a smooth embedded sub-manifold of $TM\oplus TM$ of dimension $3n-2$. Moreover $\mathcal R(M,g)$ has the structure of a $\mathbb S^{n-2}$ bundle over $TM\setminus s_o$, where the fiber at $(x,R)\in TM\setminus s_o$ is the sphere of radius |R| contained in the plane $R^{\perp }$,

$$\begin{aligned} \mathcal R(M,g)_{(x,R)}=\{V\in T_xM\,:\,V\perp R,\,\,|V|=|R|\,\}\simeq \mathbb S^{n-2}. \end{aligned}$$

(3.2)

As expected, if $n=2$ these spheres have dimension 0, and $\mathcal R(M,g)$ reduces to a 0-dimensional fibration over $TM{\setminus } s_o$. If M is a surface, oriented for simplicity, we have

In general $\mathcal R(M,g)$ has also the structure of a fiber bundle over M, whose fibers are

$$\begin{aligned} \mathcal R(M,g)_x=\{(V,R)\in T_xM\oplus T_xM\,:\,\langle R,V\rangle =0,\,|R|=|V|>0\,\}. \end{aligned}$$

There exist two lifts which allows us to map regular curves $\gamma :[0,1]\rightarrow M$ with never vanishing geodesic curvature, to $\mathcal R(M,g)$-valued curves

$$\begin{aligned} c_{\pm }:\gamma \mapsto \left( \gamma , \pm |R_g(\gamma )|\frac{\dot{\gamma }}{|\dot{\gamma }|}, R_g(\gamma )\right) , \end{aligned}$$

(3.3)

where $R_g$ is the map defined in Eq. 1.4. We would like to find a distribution characterizing the image of the lift Eq. 3.3, i.e. a distribution $\mathcal D(M,g)$ such that $(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ is the lift of some curve $\gamma$ if and only if it is tangent to $\mathcal D(M,g)$, and $\gamma$ is regular. To construct this distribution at $q_0:=(x_0,V_0,R_0)\in \mathcal R(M,g)$ we collect the velocities of all such curves going through this point, and we take the vector space generated by them:

$$\begin{aligned} \mathcal D(M,g)_{q_0}:=\left\langle \left\{ \frac{d}{dt}(\gamma , R,V)(0)\,:\,(\gamma ,V,R)(0)=q_0, (\gamma ,V,R)\,\text {as in}\, (3.3) \,\right\} \right\rangle . \end{aligned}$$

(3.4)

Before moving to the proof of Theorem 1.1, let us fix the notation

$$\begin{aligned} f_{i_1i_2\dots i_k}=[f_{i_1},[f_{i_2},[\dots f_{i_k}]]\dots ], \end{aligned}$$

to indicate the iterated brackets of the fields $f_1,\dots ,f_n$. Moreover we will often make use of the abbreviation

$$\begin{aligned} X\partial _x=\sum _{i=1}^nX^i\partial _{x^i}. \end{aligned}$$

(3.5)

where $(x^1,\dots ,x^n)$ are local coordinates and $X\in \mathbb {R}^n$ is the vector $X=(X^1,\dots ,X^n)$.

Proof

(Theorem 1.1) The vectorfields in Eq. 1.6 are, by definition, local sections of $T(TM\oplus TM)$. We need to show that they are actually tangent to $\mathcal R(M,g)\subset TM\oplus TM$. Let $(x_0,V_0,R_0)\in \mathcal R(M,g)$ and let (x, V, R)(t) be an integral curve of $f_i$ with initial point $(x_0,V_0,R_0)$. We have to show that $\langle V(t), R(t)\rangle = 0$ and that $|V(t)|=|R(t)|$ for each t such that the flow of $f_i$ is defined. Let us start with $f_1$, any of its integral curves satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}=V,\\ D_t R=-V,\\ D_t V=R. \end{array}\right. } \end{aligned}$$

Therefore we have

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\langle R, V\rangle =|R|^2-|V|^2,\,\,\,\frac{d}{dt}|R|^2-|V|^2=-4\langle R, V\rangle , \end{aligned} \end{aligned}$$

but $|R(0)|^2-|V(0)|^2=0=\langle R(0), V(0)\rangle$, because $(x,V,R)(0)\in \mathcal R(M,g)$, then from uniqueness of ODEs solutions we obtain $|V(t)|-|R(t)|\equiv \langle V(t), R(t)\rangle \equiv 0$. Let (x, R, V)(t) be an integral curve of $f_2$ with initial point in $\mathcal R(M,g)$, then from Eq. 1.6 we compute

$$\begin{aligned} \frac{d}{dt}\left( |R|^2-|V|^2\right) =2\left( |R|^2-|V|^2\right) ,\,\,\,\,\frac{d}{dt}\langle R,V\rangle =2\langle R,V\rangle . \end{aligned}$$

Since $(\gamma ,V,R)(0)\in \mathcal R(M,g)$, we deduce $|V(t)|-|R(t)|=\langle V(t), R(t)\rangle \equiv 0$. Let $(\gamma , R, V)(t)$ be an integral curve of $f_j$, $j=3,\dots , n$, then from Eq. 1.6 we have

$$\begin{aligned} \frac{d}{dt}\left( |R|^2-|V|^2\right) =0,\,\,\,\,\frac{d}{dt}\langle R,V\rangle =0, \end{aligned}$$

hence $f_j$ is tangent to $\mathcal R(M,g)$.

Now we claim that a curve $(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ is in the image of one of the lifts Eq. 3.3 if and only if it is tangent to $\langle \{ f_1,\dots ,f_n\}\rangle$, and $\gamma$ is regular. Notice that, by definition of $\mathcal D(M,g)$, this would imply $\mathcal D(M,g)=\langle \{ f_1,\dots ,f_n\}\rangle$. To prove our claim we have to show that $(\gamma ,V,R)$ is in the image of one of the lifts Eq. 3.3 if an only if there exist n smooth functions, $u_1,\dots ,u_n:[0,1]\rightarrow \mathbb R$ , with $u_1\ne 0$, such that $\frac{d}{dt}(\gamma , R,V)=u_1f_1+\dots +u_nf_n$, or in other words, according to Eq. 1.6

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{\gamma }=u_1V,\\ D_t R=-u_1V+u_2R+u_3b_3+\dots +u_nb_n,\\ D_t V=u_1R+u_2V. \end{array}\right. } \end{aligned}$$

(3.6)

Assume that $\sigma :=(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ is in the image of Eq. 3.3, then, by definition $R=R_g(\gamma )$ and $\dot{\gamma }=u_1 V$, for some never vanishing smooth function $u_1:[0,1]\rightarrow \mathbb R$. Since the condition $\dot{\sigma }\in \langle \{f_1,\dots ,f_n\}\rangle$ is independent of the parametrization, without loss of generality we may assume $|\dot{\gamma }|=1$, i.e. $u_1=1/|R|$. Since $\gamma$ is arc-length parametrized, as a consequence of formulae Eqs. 1.4 and 1.3, we have

$$\begin{aligned} D_t\dot{\gamma }=\frac{1}{|R|^2}R. \end{aligned}$$

On the other hand, since $\dot{\gamma }=1/|R|V$, we have

$$\begin{aligned} D_t\dot{\gamma }=D_t\frac{V}{|R|}=\frac{d}{dt}\left( \frac{1}{|R|}\right) V+\frac{1}{|R|}D_tV, \end{aligned}$$

(3.7)

therefore

$$\begin{aligned} D_tV=\frac{1}{|R|}R-|R|\frac{d}{dt}\left( \frac{1}{|R|}\right) V=u_1R+u_2V, \end{aligned}$$

where we have denoted $u_2:=-|R|\frac{d}{dt}1/|R|=\frac{d}{dt}\log {|R|}$. It remains to compute the covariant derivative of R along $\gamma$. Notice that $\{V(t),R(t),b_3(t),\dots ,b_n(t)\}$ is a basis for $T_{\gamma (t)}M$ for every $t\in [0,1]$, therefore there exist $\lambda _1,\dots ,\lambda _n:[0,1]\rightarrow \mathbb R$ smooth functions such that

$$\begin{aligned} D_tR=\lambda _1 V+\lambda _2 R+\lambda _3 b_3+\dots \lambda _n b_n. \end{aligned}$$

To prove that Eq. 3.6 holds we just have to show that $\lambda _1=-u_1$, $\lambda _2=u_2$. We begin with $\lambda _1$:

$$\begin{aligned} \begin{aligned} \lambda _1=\langle D_t R, \frac{V}{|V|^2}\rangle =-\frac{1}{|R|^2}\langle R, D_t V\rangle =-\frac{1}{|R|^2}\langle R, u_1R\rangle =-u_1. \end{aligned} \end{aligned}$$

Concerning the second coefficient $\lambda _2$ we have

$$\begin{aligned} \begin{aligned} \lambda _2&=\langle D_t R, \frac{R}{|R|^2}\rangle =\frac{1}{2|R|^2}\frac{d}{dt}|R|^2=\frac{1}{2}\frac{d}{dt}\log {|R|^2}=\frac{d}{dt}\log {|R|}=u_2. \end{aligned} \end{aligned}$$

Conversely let $\sigma =(\gamma ,V,R)$ be a curve satisfying Eq. 3.6, with $u_1\ne 0$. We want to show that $R=R_g(\gamma )$, i.e. that R is the radius of curvature of $\gamma$. Exploiting Eq. 3.6 we compute

$$\begin{aligned} \pi _{\dot{\gamma }^{\perp }}D_t\dot{\gamma }=\pi _{\dot{\gamma }^{\perp }}(u_1D_tV+\dot{u}_1 V)=u_1^2 R, \end{aligned}$$

(3.8)

then, by definition Eq. 1.4, the radius of curvature of $\gamma$ satisfies

$$\begin{aligned} R_g(\gamma )=\frac{\pi _{\dot{\gamma }^\perp } D_t\dot{\gamma }}{\kappa _g(\gamma )|\pi _{\dot{\gamma }^\perp } D_t\dot{\gamma }|}=\frac{R}{\kappa _g(\gamma )|R|}. \end{aligned}$$

(3.9)

On the other hand combining Eq. 1.3 with 3.8, and considering that $|R|=|V|$, we deduce

$$\begin{aligned} \kappa _g(\gamma )= \frac{|\pi _{\gamma ^\perp } D_t\dot{\gamma }|}{|\dot{\gamma }|^2}=\frac{u_1^2|R|}{u_1^2|V|^2}=\frac{1}{|R|}. \end{aligned}$$

(3.10)

Substituting Eq. 3.10 into 3.9 we obtain $R_g(\gamma )=R$, as requested.

We now show that $\mathcal D(M,g)$ is an equiregular bracket generating distribution of step 3. To do this we need to be more precise about the definition of the local fields $f_3,\dots ,f_n$ in Eq. 1.6, in particular given $(x,V,R)\in \mathcal R(M,g)$ we have to make a choice of a local basis for $\{R,V\}^\perp$. Let $\mathcal U\subset M$ be an open subset and let $E_1,\dots ,E_n$ be a local orthonormal frame for g on $\mathcal U$. For every $i\ne j$ we define the following two-form $\omega _{ij}\in \Omega ^2(\mathcal U)$:

$$\begin{aligned} \omega _{ij}(R,V):=\det \begin{pmatrix} \langle R,E_i\rangle _g &{} \langle V,E_i\rangle _g \\ \langle R,E_j\rangle _g &{} \langle V,E_j\rangle _g \end{pmatrix} \end{aligned}$$

and the following open subset of $TM\oplus TM$

$$\begin{aligned} \mathcal U_{ij}:=\left\{ (x,V,R)\,:\, \omega _{ij}(R,V) \ne 0\right\} . \end{aligned}$$

(3.11)

By construction, the sets $\{V_{ij}:=\mathcal U_{ij}\cap \mathcal R(M,g)\}_{i\ne j}$, constitute an open cover for $\mathcal R(M,g)\cap T\mathcal U\oplus T\mathcal U$. Moreover the vectors

$$\begin{aligned} \{R,V,E_3(x),\dots ,\hat{E}_{i}(x),\dots ,\hat{E}_{j}(x),\dots ,E_n(x)\}, \end{aligned}$$

are linearly independent for every $(x,V,R)\in V_{ij}$. We define

$$\begin{aligned} \{e_3(x,V,R),\dots ,e_n(x,V,R)\} \end{aligned}$$

to be the norm-|R| basis of $\{R,V\}^\perp$ obtained by applying the Gram-Schmidt algorithm to the vectors

$$\begin{aligned} \{R,V,E_3(x),\dots ,\hat{E}_{i}(x),\dots ,\hat{E}_{j}(x),\dots ,E_n(x)\}, \end{aligned}$$

and, on $V_{ij}$, we set

$$\begin{aligned} f_k:=e_k\partial _R=\sum _{\mu =1}^n e_k^\mu \partial _{R^\mu },\,\,\,k=3,\dots ,n. \end{aligned}$$

Defined in this way, the local fields $e_k$ satisfy two useful properties :

$$\begin{aligned} e_k(x,\lambda R,\lambda V)=\lambda e_k(x,V,R), \end{aligned}$$

(3.12)

for every $\lambda >0$ and

$$\begin{aligned} e_k(x,\cos \theta R-\sin \theta V, \sin \theta R+\cos \theta V)=e_k(x,V,R), \end{aligned}$$

(3.13)

for every $\theta \in [0,2\pi ]$. Let $(x^\mu )$ be local coordinates on M and let $(x^\mu ,R^\mu ,V^\mu )$ be the local coordinates induced by $(x^\mu )$ on $TM\oplus TM$. Let $\Gamma ^{\mu }_{\alpha \beta }$ be the corresponding Christoffel symbols of the metric g. Given $X,Y\in \mathbb R^n$, we denote with $\Gamma (X,Y)$ the row vector

$$\begin{aligned} \Gamma (X,Y)=\sum _{\alpha ,\beta =1}^n(\Gamma _{\alpha \beta }^1X^\alpha Y^\beta ,\dots , \Gamma _{\alpha \beta }^nX^\alpha Y^\beta ), \end{aligned}$$

(3.14)

then, making use of the notation Eq. 3.5, the vector fields $f_1,f_2$ read

$$\begin{aligned} \begin{aligned} f_1&=V\partial _x-\left( V+\Gamma (V,R)\right) \partial _R+\left( R-\Gamma (V,V)\right) \partial _V,\\ f_2&=R\partial _R+V\partial _V, \end{aligned} \end{aligned}$$

(3.15)

and their commutator reads

$$\begin{aligned} \begin{aligned} \mathrm {[}f_2,f_1\mathrm {]}=&[R\partial _R+V\partial _V,V\partial _x-\left( V+\Gamma (V,R)\right) \partial _R+\left( R-\Gamma (V,V)\right) \partial _V]\\ =&-\Gamma (V,R)\partial _R+R\partial _V+V\partial _x-\left( V+\Gamma (V,R)\right) \partial _R -2\Gamma (V,V)\partial _V\\&+\left( V+\Gamma (V,R)\right) \partial _R-\left( R-\Gamma (V,V)\right) \partial _V\\ =&\,V\partial _x-\Gamma (V,R)\partial _R-\Gamma (V,V)\partial _V. \end{aligned} \end{aligned}$$

(3.16)

We define the vector field $X_{12}$ as

$$\begin{aligned} X_{12}:=f_1-f_{21}=-V\partial _R+R\partial _V. \end{aligned}$$

(3.17)

We define

$$\begin{aligned} f_{121}:=[f_{12},f_1]=[X_{12},f_1], \end{aligned}$$

(3.18)

and we notice that this vector field satisfies

$$\begin{aligned} \pi ^{M}_\star f_{121}=\pi ^{M}_\star [X_{12},f_1]=[R\partial _V,V\partial _x]=R\partial _x. \end{aligned}$$

(3.19)

Finally, to span the whole tangent space, we need the following vector fields from the third layer

$$\begin{aligned} f_{1k1}=[[f_1,f_k],f_1],\,\,\, k=3,\dots ,n. \end{aligned}$$

We start by calculating the field $f_{1k}:=[f_1,f_k]$. Let $e_k(t)=e_k(x(t),R(t),V(t))$, where (x(t), R(t), V(t)) is an integral curve of $f_1$. Then, according to Eq. 1.6, we have

$$\begin{aligned} \begin{aligned}&\langle D_te_k,R\rangle =-\langle e_k,D_tR\rangle =\langle e_k,V\rangle =0,\\&\langle D_te_k,V\rangle =- \langle e_k,D_tV\rangle =-\langle e_k,R\rangle =0,\\&F_{k}^i:=\frac{1}{|R|^2}\langle D_te_k,e_i\rangle . \end{aligned} \end{aligned}$$

Therefore

$$\begin{aligned} f_1(e_k)=\sum _{i=3}^nF_{k}^ie_i-\Gamma (V,e_k), \end{aligned}$$

(3.20)

and hence we have

$$\begin{aligned} \begin{aligned} f_{1k}=[f_1,f_k]&=[V\partial _x-\left( V+\Gamma (V,R)\right) \partial _R+\left( R-\Gamma (V,V)\right) \partial _V,e_k\partial _R]\\&=f_1(e_k)\partial _R+\Gamma (V,e_k)\partial _R-e_k\partial _V\\&=\left( \sum _{i=3}^nF_{k}^ie_i-\Gamma (V,e_k)\right) \partial _R+\Gamma (V,e_k)\partial _R-e_k\partial _V\\&=\sum _{i=3}^nF_{k}^ie_i\partial _R-e_k\partial _V=\sum _{i=3}^nF_{k}^if_i-e_k\partial _V. \end{aligned} \end{aligned}$$

(3.21)

Observe that the field $f_{1k1}$ satisfies

$$\begin{aligned} \pi _{\star }^Mf_{1k1}=\pi _{\star }^M[f_{1k},f_1] =\pi _{\star }^M\left[ \sum _{i=3}^nF_{k}^ie_i\partial _R-e_k\partial _V,V\partial _x\right] =-e_k\partial _x. \end{aligned}$$

(3.22)

Now we claim that the following $3n-2$ vector fields

$$\begin{aligned} f_1,\dots ,f_n,f_{12},f_{13},\dots ,f_{1n},f_{121},f_{131},\dots ,f_{1n1}, \end{aligned}$$

are linearly independent. Notice that, since $X_{12}=f_1+f_{12}$, we may equivalently claim that

$$\begin{aligned} f_1,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n},f_{121},f_{131},\dots ,f_{1n1}, \end{aligned}$$

(3.23)

are linearly independent. Consider a vanishing linear combination, with coefficients in $\mathcal C^\infty (\mathcal R(M,g))$, of Eq. 3.23:

$$\begin{aligned} \sum _{k=1}^na_kf_k+a_{12}X_{12}+\sum _{k=3}^na_{1k}f_{1k}+a_{121}f_{121}+\sum _{k=3}^n a_{1k1}f_{1k1}=0. \end{aligned}$$

(3.24)

If we apply $\pi ^{M}_{\star }$ to both sides of Eq. 3.24, since according to Eqs. 1.6, 3.17, 3.21, the fields $f_2,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n}$ have x-component equal to zero, we are left with

$$\begin{aligned} \pi _\star ^M\left( a_1f_1+a_{121}f_{121}+\sum _{k=3}^n a_{1k1}f_{1k1}\right) =0, \end{aligned}$$

which thanks to Eqs. 3.15, 3.19 and 3.22 translates to

$$\begin{aligned} a_1V+a_{121}R-\sum _{k=3}^n a_{1k1}e_k=0, \end{aligned}$$

(3.25)

therefore $a_1=a_{121}=a_{131}=\dots =a_{1n1}=0$ and Eq. 3.24 becomes

$$\begin{aligned} \sum _{k=2}^na_kf_k+a_{12}X_{12}+\sum _{k=3}^na_{1k}f_{1k}=0. \end{aligned}$$

(3.26)

If we compute the V-component of both sides of Eq. 3.26, according to Eqs. 3.15, 3.17, 3.21 we get

$$\begin{aligned} a_2V+a_{12}R-\sum _{k=3}^na_{1k}e_k=0, \end{aligned}$$

thus $a_2=a_{12}=a_{123}=\dots =a_{12n}=0$, and Eq. 3.26 reads

$$\begin{aligned} \sum _{k=3}^na_kf_k=0, \end{aligned}$$

(3.27)

but $f_3,\dots ,f_n$ are linearly independent by definition, therefore also $a_3=\dots =a_n=0$. We have just proved that $\mathcal D^3(M,g)=T\mathcal R(M,g)$. It remains to show that $\text {rank}\,\mathcal D^2(M,g)< \text {rank}\,T\mathcal R(M,g)=3n-2$. Notice that

$$\begin{aligned} \mathcal D^2(M,g)=\mathcal D(M,g)+\langle \{[f_1,f_j]\}_{j=2}^n \rangle +\langle \{[f_i,f_j]\}_{i,j=2}^n \rangle . \end{aligned}$$

We claim that the distribution $\langle \{ f_2,\dots ,f_n\}\rangle$ is integrable. If that is the case then $\langle \{[f_i,f_j]\}_{i,j=2}^n \rangle$ $\subset$ $\mathcal D(M,g)$ and as a consequence

$$\begin{aligned} \mathcal D^2(M,g)=\mathcal D(M,g)+\langle \{[f_1,f_j]\}_{j=2}^n \rangle , \end{aligned}$$

(3.28)

therefore, since $f_1,\dots ,f_n,f_{12},\dots ,f_{1n}$ are linearly independent, we deduce

$$\begin{aligned} \text {rank}\,\mathcal D^2(M,g)=\text {rank}\, \mathcal D(M,g)+\text {rank} \mathcal \langle \{[f_1,f_j]\}_{j=2}^n \rangle =2n-1<3n-2. \end{aligned}$$

(3.29)

To prove our claim observe that, for $j=2,\dots ,n$, $[f_2,f_j]=0$, indeed $e^{tf_2}(x,V,R)=(x,e^tR,e^tV)$, hence by construction $e_j(e^{tf_2}(x,V,R))=e^te_j(x,V,R)$, or in other words $f_2(e_j)=e_j$, but then

$$\begin{aligned}{}[f_2,f_j]=f_2(e_j)\partial _R-e_j\partial _R=0. \end{aligned}$$

Now consider $f_j,f_i$ with $i,j\ge 3$, to lighten the notation in the following we have denoted $e_{ij}(t)=e_j(e^{tf_i}(x_0,V_0,R_0))$, $V(t)=V(e^{tf_i}(x_0,V_0,R_0))$ and $R(t)=R(e^{tf_i}(x_0,V_0,R_0))$. Let us compute

$$\begin{aligned} \begin{aligned}&\langle \dot{e}_{ij}, R\rangle =-\langle e_{ij},\dot{R}\rangle =-\langle e_{ij},e_{ii}\rangle =0,\\&\langle \dot{e}_{ij}, V\rangle =-\langle e_{ij},\dot{V}\rangle =0, \end{aligned} \end{aligned}$$

and let us denote $a_{ijk}=\langle \dot{e}_{ij}, e_k\rangle /|R|^2$, then we have

$$\begin{aligned} f_j(e_i)=\dot{e}_{ij}=\sum _{k=3}^na_{ijk}e_k, \end{aligned}$$

meaning that

$$\begin{aligned}{}[f_i,f_j]=(f_i(e_j)-f_j(e_i))\partial _R =\sum _{k=3}^n(a_{jik}-a_{ijk})e_k\partial _R=\sum _{k=3}^n(a_{jik}-a_{ijk})f_k, \end{aligned}$$

and this concludes the proof of the claim. Equation 3.28, together with the fact that the fields $\{f_j\}_{j=1}^n$,$\{f_{1k}\}_{k=2}^n$, $\{f_{1k1}\}_{k=3}^n$ are a basis of $T\mathcal R(M,g)$, gives us the following local characterization of $\mathcal D(M,g)$’s flag:

$$\begin{aligned} \begin{aligned}&\mathcal D(M,g)=\langle f_1,\dots ,f_n\rangle ,\\&\mathcal \mathcal \mathcal D^2(M,g)=\mathcal D(M,g)\oplus \langle \{f_{1k}\}_{k=2}^n\rangle ,\\&\mathcal D^3(M,g)=\mathcal D^2(M,g)\oplus \langle \{f_{1k1}\}_{k=2}^n\rangle =T\mathcal R(M,g). \end{aligned} \end{aligned}$$

(3.30)

In particular we have

$$\begin{aligned} \text {Growth}\,\mathcal D(M,g)=(n,2n-1,3n-2). \end{aligned}$$

(3.31)

$\square$

We obtain the following corollary.

Corollary 3.1

The following inclusions hold:

$$\begin{aligned} \begin{aligned}&f_{jk}\in \mathcal D(M,g),\,\,\,j,k=2\dots ,n, \\&f_{j1k}\in \mathcal D^2(M,g),\,\,\,j,k=2,\dots ,n. \end{aligned} \end{aligned}$$

(3.32)

Proof

We have already proved the first inclusion of Eq. 3.32 since we have shown that $\langle \{f_2,\dots ,f_n\}\rangle$ is integrable. To prove the second inclusion observe that the kernel of $\pi ^M_\star$ has dimension $(3n-2)-n=2n-2$. On the other hand Eqs. 1.6, 3.17, 3.21, imply that the following $2n-2$ linearly independent local smooth sections of $\mathcal D^2(M,g)$

$$\begin{aligned} f_2,f_3,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n}, \end{aligned}$$

(3.33)

are contained in $\ker \pi _\star ^M$, and hence they constitute a basis for it:

$$\begin{aligned} \ker \pi _\star ^M=\langle \left\{ f_2,f_3,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n}\right\} \rangle . \end{aligned}$$

The distribution $\ker \pi _\star ^M$ is integrable: if X, Y are sections of $\ker \pi _\star ^M$, then they are $\pi ^M$-related to the zero-section of TM, and therefore also their bracket is so. Thanks to the integrability of $\ker \pi _\star ^M$, recalling that $X_{12}=f_1+f_{12}$, we deduce that $f_{j1k}\in \mathcal D^2(M,g)$, $j,k=2,\dots ,n$. $\square$

If (M, g) is a Riemannian surface, which for simplicity we assume to be orientable, then the fields $f_1,f_2$ defined in Eq. 1.6 can be related with the fields $f_1^g$, $f_2^g$ defined in Eq. 2.6. Indeed, in this case $\mathcal R(M,g)$ has two connected components:

The projection $p:\mathcal R(M,g)\rightarrow TM\setminus s_o$ restricted to the first of such components is a diffeomorphism satisfying

$$\begin{aligned} p_\star f_1=f^g_1,\,\,\,p_\star f_2=f^g_2. \end{aligned}$$

This fact, combined with Theorem 1.1, constitutes a proof of Proposition 2.1. Focusing back on the general case, we would like to define a metric on $\mathcal D(M,g)$ that preserves the symmetries of the curvature radii lift

$$\begin{aligned} \begin{aligned} R_g: \{\gamma \in \mathcal C^2([0,1],M)\,:\,\gamma \,\, \text {regular},\,\kappa _g(\gamma )\ne 0\}&\rightarrow \mathcal C^{0}\left( [0,1],TM \right) \\\gamma&\mapsto R_g(\gamma ), \end{aligned} \end{aligned}$$

(3.34)

hence before defining such a metric, we need to have a better understanding of how much information is encoded in the mapping Eq. 3.34.

Definition 3.2

Let (M, g) and $(N,\eta )$ be Riemannian manifolds, we say that they have the same curvature radii if and only if there exists a diffeomorphism $\varphi :(M,g)\rightarrow (N,\eta )$ such that

$$\begin{aligned} R_\eta \circ \varphi =\varphi _\star \circ R_g. \end{aligned}$$

(3.35)

Remark 3.1

Equality Eq. 3.35 is meant as an equality of maps, therefore the diffeomorphism $\varphi$ of Definition 3.2 maps the domain of $R_g$ to the one of $R_\eta$. This implies that $\gamma :[0,1]\rightarrow M$ is the reparametrization of a geodesic of (M, g) if and only if $\varphi \circ \gamma$ is the reparametrization of a geodesic of $(N,\eta )$. In particular

$$\begin{aligned} \kappa _g(\gamma )\equiv 0\iff \kappa _\eta (\varphi \circ \gamma )\equiv 0. \end{aligned}$$

(3.36)

Definition 3.2 gives a precise meaning to the statement of Theorem 1.2, which we now prove.

Proof

(Theorem 1.2) It is sufficient to prove the theorem for two Riemannian metrics $\eta ,g$ on the same manifold M. Assume firs that (M, g) and $(M,\eta )$ are homothetic manifolds, then $\eta =\lambda g$ for some $\lambda >0$. The two metrics have the same Levi-Civita connection $\nabla$, moreover for every $X,Y\in TM$, $X\perp _{g}Y$ if and only if $X\perp _{\eta }Y$, therefore we simply denote $\perp =\perp _{g}=\perp _{\eta }$. Let $\gamma :[0,T]\rightarrow M$ be a regular curve, then the curvature computed with respect to $\eta$ can be easily related to the curvature computed with respect to g:

$$\begin{aligned} \kappa _\eta (\gamma )=\frac{|\pi _{\dot{\gamma }^\perp }\left( D_t\dot{\gamma }\right) |_{\lambda \,g}}{|\dot{\gamma }(t)|_{\lambda \,g}^2}=\frac{1}{\sqrt{\lambda }}\frac{|\pi _{\dot{\gamma }^\perp }\left( D_t\dot{\gamma }\right) |_{g}}{|\dot{\gamma }(t)|_{g}^2}=\frac{1}{\sqrt{\lambda }}\kappa _g(\gamma ). \end{aligned}$$

Therefore $k_\eta (\gamma )$ is never vanishing if and only if also $k_g(\gamma )$ is so, and in that case using Eq. 1.4 we deduce $R_\eta (\gamma )=R_g(\gamma )$.

Conversely, assume that (M, g) and $(M,\eta )$ have the same curvature radii, i.e. Eq. 3.35 holds and

$$\begin{aligned} R_g(\gamma )=R_\eta (\gamma ) \end{aligned}$$

(3.37)

for any regular curve $\gamma$ with never vanishing geodesic curvature. Observe that for any $X,Y\in T_xM$, $X\perp _g Y$ if and only if there exists $\gamma :[0,1]\rightarrow M$ such that $\gamma (0)=x$, $\dot{\gamma }(0)=X$ and $R_g(\gamma )(0)=Y$, but then, since $R_g=R_\eta$, we obtain $\perp _g=\perp _\eta$. According to Remark 3.1 we have

$$\begin{aligned} D^g_t\dot{\gamma }\propto \dot{\gamma }\,\,\iff \,\, D^\eta _t\dot{\gamma }\propto \dot{\gamma }. \end{aligned}$$

Hence for any curve $\gamma :[0,1]\rightarrow M$ the following proportionality relationship holds

$$\begin{aligned} D_t^g\dot{\gamma }-D_t^\eta \dot{\gamma }\propto \dot{\gamma }, \end{aligned}$$

(3.38)

and in particular

$$\begin{aligned} \pi _{\dot{\gamma }^\perp }D_t^g\dot{\gamma }=\pi _{\dot{\gamma }^\perp }D_t^\eta \dot{\gamma }. \end{aligned}$$

(3.39)

In light of Eq. 3.39, condition Eq. 3.37 reduces to

$$\begin{aligned} \kappa _g(\gamma )|\pi _{\dot{\gamma }^\perp }D_t^g\dot{\gamma }|_g=\kappa _\eta (\gamma )|\pi _{\dot{\gamma }^\perp }D_t^\eta \dot{\gamma }|_\eta , \end{aligned}$$

which, by definition of geodesic curvature, in turn is equivalent to

$$\begin{aligned} \frac{|\pi _{\dot{\gamma }^\perp }D_t^g\dot{\gamma }|_g}{|\dot{\gamma }|_g}=\frac{|\pi _{\dot{\gamma }^\perp }D_t^\eta \dot{\gamma }|_\eta }{|\dot{\gamma }|_\eta }. \end{aligned}$$

(3.40)

For any $X,Y\in T_xM$, X is perpendicular to Y if and only if there exists $\gamma :[0,1]\rightarrow M$ satisfying

$$\begin{aligned} \dot{\gamma }(0)=X,\,\,\,\pi _{\dot{\gamma }^\perp }D_t^g\dot{\gamma }(0)=Y. \end{aligned}$$

Hence Eq. 3.40 implies that

$$\begin{aligned} \frac{|Y|_g}{|X|_g}=\frac{|Y|_\eta }{|X|_\eta },\,\,\,\forall \,X\perp Y, \end{aligned}$$

(3.41)

which in turn implies that the two metrics are conformally related: there exists a smooth function $f:M\rightarrow \mathbb R$ such that

$$\begin{aligned} \eta =e^{2f}g. \end{aligned}$$

(3.42)

Since the metrics are conformal, their Levi-Civita connection difference tensor can be easily computed

$$\begin{aligned} T(X,Y):=\nabla ^\eta _XY-\nabla ^g_XY=X(f)Y+Y(f)X-\langle X,Y\rangle _g\,\text {grad}_g f. \end{aligned}$$

On the other hand thanks to Eq. 3.38 we know that

$$\begin{aligned} T(X,X)\propto X, \end{aligned}$$

and hence for any $Y\perp X$ we have

$$\begin{aligned} 0=\langle T(X,X),Y\rangle _g=-| X|^2_g\langle \text {grad}_g f,Y\rangle _g=-|X|^2_gdf(Y). \end{aligned}$$

Since X, Y are arbitrary orthogonal vectors, this implies that f is constant. $\square$

The curvature radius lift $R_g$ of a Riemannian manifold (M, g) is a complete homothety invariant of the metric g. It is then natural to define on $\mathcal D(M,g)$ a metric which is invariant under homothetic transformations. For every $a,b\in \mathbb R$, $b\ne 0$, we define a metric $\eta _{a,b}$ on $\mathcal D(M,g)$ by declaring the fields

$$\begin{aligned} \begin{aligned}&f_1^{a,b}:=\frac{1}{\sqrt{a^2+b^2}}f_1,\\&f_j^{a,b}:=\frac{1}{b}f_j,\,\,j=2,\dots ,n,\\ \end{aligned} \end{aligned}$$

(3.43)

a local orthonormal frame. For any $(\gamma , R, V)$ in the image of the lift Eq. 3.3 the metric $\eta _{a,b}$ satisfies the following equation

$$\begin{aligned} \begin{aligned} \left| \frac{d}{dt}(\gamma ,V,R)\right| ^2_{\eta _{a,b}}=a^2\frac{|\dot{\gamma }|^2}{|R|^2}+b^2\frac{|D_tR|^2}{|R|^2}. \end{aligned} \end{aligned}$$

(3.44)

To prove Eq. 3.44 recall that according to Theorem 1.1 there exists $u_1,\dots ,u_n:[0,1]\rightarrow \mathbb R$ such that

$$\begin{aligned} \frac{d}{dt}(\gamma ,V,R)=u_1f^{a,b}_1+u_2f^{a,b}_2+\dots u_nf^{a,b}_n, \end{aligned}$$

(3.45)

or equivalently according to Eq. 3.6,

$$\begin{aligned} \begin{aligned} \dot{\gamma }&=\frac{u_1}{\sqrt{a^2+b^2}}V,\\ D_t R&=-\frac{u_1}{\sqrt{a^2+b^2}}V+\frac{u_2}{b}R+\frac{u_3}{b}e_3+\dots +\frac{u_n}{b}e_n,\\ D_t V&=\frac{u_1}{\sqrt{a^2+b^2}}R+\frac{u_2}{b}V. \end{aligned} \end{aligned}$$

(3.46)

Hence

$$\begin{aligned} \begin{aligned} a^2\frac{|\dot{\gamma }|^2}{|R|^2}+b^2\frac{|D_tR|^2}{|R|^2}=&a^2\frac{u_1^2}{|R|^2(a^2+b^2)}|R|^2+b^2\frac{u_1^2}{|R|^2(a^2+b^2)}|R|^2\\&+\frac{b^2}{|R|^2}\left( \frac{|R|^2}{b^2}u_2^2+\dots \frac{|R|^2}{b^2}u_n^2\right) \\ =&u_1^2+u_2^2+\dots +u_n^2=\left| \frac{d}{dt}(\gamma ,V,R)\right| ^2_{\eta _{a,b}}. \end{aligned} \end{aligned}$$

(3.47)

The sub-Riemannian manifold $(\mathcal R(M,g),\mathcal D(M,g), \eta _{a,b})$ is called the sub-Riemannian manifold of curvature radii of (M, g) and it is denoted with $\mathcal R_{a,b} (M,g)$. We now prove the main result regarding these metrics, Theorem 1.3.

Proof

(Theorem 1.3) We begin by making sure that the map Eq. 1.9 is well defined. Let $\varphi :(M,g)\rightarrow (M,g)$ be a homothety of Riemannian manifolds. By construction $\varphi _\star \oplus \varphi _\star :TM\oplus TM\rightarrow TM\oplus TM$ maps $\mathcal R(M,g)$ to $\mathcal R(M,g)$, this is a simple consequence of the fact that $\varphi _\star$ preserves orthogonality and the ratios between norms of vectors. To prove that $\Phi :=\varphi _\star \oplus \varphi _\star$ is a sub-Riemannian isometry we show that $\Phi _\star f_i=f_i$, $i=1,2$, and that the fields $\{\Phi _\star f_3,\dots , \Phi _\star f_n\}$ are related to $\{f_3,\dots ,f_n\}$ by an orthogonal transformation. Let $(\gamma ,V,R):[0,T]\rightarrow \mathcal R(M,g)$ be an integral curve of $f_1$ then we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{\gamma }=V,\\ D_t R=-V,\\ D_t V=R, \end{array}\right. } \end{aligned}$$

while the image under $\Phi$ of such curve, $\Phi (\gamma , R, V)=(\varphi \circ \gamma ,\varphi _\star R, \varphi _\star V)=:(\bar{\gamma },\bar{R} ,\bar{V})$, satisfies

$$\begin{aligned} \begin{aligned}&\dot{\bar{\gamma }}=\frac{d}{dt}\varphi (\gamma )=\varphi _\star \dot{\gamma }=\varphi _\star V=\bar{V},\\&D_t\bar{R}=\nabla _{\bar{V}}\bar{R}=\varphi _\star \nabla _V R=-\varphi _\star V=-\bar{V},\\&D_t\bar{V}=\nabla _{\bar{V}}\bar{V}=\varphi _\star \nabla _V V=\varphi _\star R=\bar{R}. \end{aligned} \end{aligned}$$

Summarazing we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{\bar{\gamma }}=\bar{V},\\ D_t\bar{R}=-\bar{V},\\ D_t \bar{V}=\bar{R}, \end{array}\right. } \end{aligned}$$

hence $\Phi _\star f_1=f_1$. The case of $f_2$ is analogous. Since $\varphi$ is a homothety it sends local orthogonal basis of TM to local orthogonal basis, and if $X,Y\in T_xM$ have the same norm, then so do $\varphi _\star X$, $\varphi _\star Y$. Given $(x,V,R)\in \mathcal R(M,g)$, let $\{e_3(x,V,R),\dots ,e_n(x,V,R)\}$ be the local orthogonal basis of $\{R,V\}^\perp$ constructed in the proof of Theorem 1.1, then both $\{e_j(\varphi (x),\varphi _\star V,\varphi _\star R) \}_{j=3}^n$ and $\{\varphi _\star e_j(x,V,R)\}_{j=3}^n$ are othogonal basis of $\{\varphi _\star R,\varphi _\star V\}^{\perp }$, therefore they are related by an orthogonal transformation: there exists $O\in O(n-2)$ such that

$$\begin{aligned} \varphi _\star e_j(x,V,R)=O^i_je_i(\varphi (x),\varphi _\star V,\varphi _\star R). \end{aligned}$$

(3.48)

Now let $(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ be an integral curve of $f_j$, for $j\ge 3$, we have

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\varphi \circ \gamma =\varphi _\star \dot{\gamma }=0,\\&\frac{d}{dt}\varphi _\star R=\varphi _\star \frac{d}{dt}R=\varphi _\star e_j(x,V,R)=O^i_je_i(\varphi (x),\varphi _\star V,\varphi _\star R),\\&\frac{d}{dt}\varphi _\star V=\varphi _\star \frac{d}{dt}V=0. \end{aligned} \end{aligned}$$

Hence the fields $\{\Phi _\star f_3,\dots , \Phi _\star f_n\}$ are related to $\{f_3,\dots ,f_n\}$ by an orthogonal transformation.

The map Eq. 1.9 is an injective homomorphism by definition. We need to prove that it is also surjective. Let $\Phi :\mathcal R_{a,b}(M,g)\rightarrow \mathcal R_{a,b}(M,g)$, given by

$$\begin{aligned} \begin{aligned} (x,V,R)\mapsto \Phi (x,V,R)=(\Phi ^x(x,V,R),\Phi ^R(x,V,R),\Phi ^V(x,V,R)), \end{aligned} \end{aligned}$$

(3.49)

be an isometry of sub-Riemannian manifolds, we have to show that there exists $\varphi :(M,g)\rightarrow (N,\eta )$ homothety of Riemannian manifolds such that

$$\begin{aligned} \Phi =(\varphi _\star \oplus \varphi _\star )|_{\mathcal R(M,g)}. \end{aligned}$$

Assume first $n>2$ and consider the projection to M

$$\begin{aligned} \begin{aligned} \pi ^M:\mathcal R(M,g)&\rightarrow M\\ (x,V,R)&\mapsto x. \end{aligned} \end{aligned}$$

Notice that the kernel of $\pi ^M_\star$ has dimension $(3n-2)-n=2n-2$. On the other hand, as shown in the proof of Theorem 1.1, the Eqs. 1.6, 3.17, 3.21, imply that the following $2n-2$ linearly independent smooth local sections of $\mathcal D^2(M,g)$

$$\begin{aligned} f_2,f_3,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n}, \end{aligned}$$

(3.50)

are contained in $\ker \pi _\star ^M$, and hence they constitute a basis for it:

$$\begin{aligned} \ker \pi _\star ^M=\langle \left\{ f_2,f_3,\dots ,f_n,X_{12},f_{13},\dots ,f_{1n}\right\} \rangle . \end{aligned}$$

We claim that $\ker \pi ^M_\star \subset \mathcal D^2(M,g)$ is the unique integrable sub-bundle of $\mathcal D^2$ having maximal rank $2n-2$. We have already noticed that $\ker \pi _\star ^M$ is integrable in the proof of Theorem 1.1. Let $\mathcal D'\subset \mathcal D^2(M,g)$ be an integrable distribution and let X, Y be smooth sections of $\mathcal D'$. Then there exists some smooth functions $\alpha _1,\dots ,\alpha _{2n-1}$, $\beta _1,\dots ,\beta _{2n-1}$ such that

$$\begin{aligned} \begin{aligned} X&=\alpha _1f_1+\alpha _2f_2+\dots +\alpha _{n+1}X_{12}+\alpha _{n+2}f_{13}+\dots +\alpha _{2n-1}f_{1n},\\ Y&=\beta _1f_1+\beta _2f_2+\dots +\beta _{n+1}X_{12}+\beta _{n+2}f_{13}+\dots +\beta _{2n-1}f_{1n}. \end{aligned} \end{aligned}$$

On the other hand, since $\mathcal D'\subset \mathcal D^2(M,g)$ is integrable, it holds

$$\begin{aligned}{}[X,Y]\equiv 0\mod \mathcal D^2(M,g), \end{aligned}$$

(3.51)

which translates to

$$\begin{aligned} \begin{aligned} (\alpha _1\beta _{n+1}-\beta _1\alpha _{n+1})[f_1,X_{12}]+\sum _{k=3}^n(\alpha _1\beta _{n+k-1}-\beta _1\alpha _{n+k-1})[f_1,f_{1k}]\equiv 0 \end{aligned} \end{aligned}$$

$\text {mod}\,\mathcal D^2(M,g)$. Since $[f_1,X_{12}]$, $[f_1,f_{1k}]$, $k\!=\!3,\dots ,n$, are linearly independent mod$\,\mathcal D^2(M,$g), we have

$$\begin{aligned} \alpha _1\beta _{n+k-1}-\beta _1\alpha _{n+k-1}=0,\,\, k=2,\dots ,n. \end{aligned}$$

(3.52)

If $\alpha _1\equiv \beta _1\equiv 0$ then $X,Y\in \ker \pi ^M_\star$. Otherwise if, for instance, $\alpha _1\ne 0$, then $\beta _{n+k-1}=\frac{\beta _1}{\alpha _1}\alpha _{n+k-1}$ for $k=2,\dots ,n$ and setting

$$\begin{aligned} Z=\alpha _{n+1}X_{12}+\alpha _{n+2}f_{13}+\dots +\alpha _{2n-1}f_{1n}, \end{aligned}$$

we deduce that

$$\begin{aligned} X=\sum _{i=1}^n\alpha _if_i+Z,\,\,Y=\sum _{i=1}^n\beta _if_i+\frac{\beta _1}{\alpha _1}Z. \end{aligned}$$

(3.53)

As a consequence

$$\begin{aligned} \mathcal D'\subset \langle \{f_1,f_2,\dots ,f_n,Z\}\rangle , \end{aligned}$$

therefore $\textrm{rank}\mathcal D'\le n+1$. Observe that it is not possible that $\text {rank}\,\mathcal D'=n+1$, because otherwise $\mathcal D'$ would contain $\mathcal D(M,g)$, contradicting the hypothesis of integrability. We deduce that

$$\begin{aligned} \textrm{rank}\mathcal D'\le n<2n-2,\,\,\,\forall \,\,n>2, \end{aligned}$$

proving our claim. Since $\Phi$ is an isometry, it preserves every layer of $\mathcal D(M,g)$’s flag, in the sense that

$$\begin{aligned} \Phi _\star \mathcal D^i(M,g)=\mathcal D^i(M,g),\,\, i=1,2,3, \end{aligned}$$

therefore since $\ker \pi ^M_\star \subset \mathcal D^2(M,g)$ is the unique integrable sub-bundle of $\mathcal D^2(M,g)$ having maximal rank $2n-2$, we deduce that

$$\begin{aligned} \Phi _\star \ker \pi ^M_\star =\ker \pi ^M_\star , \end{aligned}$$

(3.54)

or equivalently

$$\begin{aligned} \ker (\pi ^M\circ \Phi )_\star =\ker \pi ^M_\star . \end{aligned}$$

(3.55)

Both $\ker \pi ^M_\star$ and $\mathcal D(M,g)$ are invariant under $\Phi _\star$, therefore also their intersection is so, hence

$$\begin{aligned} \Phi _\star \langle \{f_2,\dots ,f_n\}\rangle =\Phi _\star (\ker \pi ^M_\star \cap \mathcal D(M,g))=\ker \pi ^M_\star \cap \mathcal D(M,g)=\langle \{f_2,\dots ,f_n\}\rangle , \end{aligned}$$

but then, since $\Phi _\star$ is a sub-Riemannian isometry preserving $\langle \{f_2,\dots ,f_n\}\rangle$, it must preserve also its orthonormal complement, namely $\Phi _\star f_1=\pm f_1$. To show that $\Phi _\star f_2=\pm f_2$, consider the following linear map

$$\begin{aligned} \begin{aligned} L:\ker \pi ^M_{\star }\cap \mathcal D(M,g)&\rightarrow TM\\ X&\mapsto \pi ^M_\star [f_1,X],\\ \end{aligned} \end{aligned}$$

and notice that, according to Eqs. 3.16 and 3.21, $\ker L=\langle \{f_3,\dots , f_n\}\rangle$. On the other hand, according to Eq. 3.55

$$\begin{aligned} \pi ^M_\star [f_1,X]=0\iff \pi ^M_\star \circ \Phi _\star [f_1,X]=0, \end{aligned}$$

which, taking into account the fact that $\Phi _\star f_1=\pm f_1$, translates to

$$\begin{aligned} \pi ^M_\star [f_1,X]=0\iff \pi ^N_\star [f_1,\Phi _\star X]=0, \end{aligned}$$

or more simply to $\ker L=\Phi _\star \ker L$, thus $\Phi _\star \langle \{ f_3,\dots , f_n\}\rangle =\langle \{ f_3,\dots , f_n\}\rangle$. So far we have deduced that $\Phi _\star \langle \{ f_1,f_3,\dots , f_n\}\rangle =\langle \{f_1,f_3,\dots , f_n\}\rangle$, taking the orthonormal complement of this last equation with respect to the sub-Riemannian metric, we deduce that $\Phi _\star f_2=\pm f_2$. A direct consequence of Eq. 3.55 is that

$$\begin{aligned} \Phi ^x(x,V,R)=\Phi ^x(x,R',V'),\,\,\,\forall \,\,(x,V,R),(x,R',V')\in \mathcal R(M,g), \end{aligned}$$

(3.56)

therefore the following map is well defined diffeomorphism

$$\begin{aligned} \begin{aligned} \varphi :M&\rightarrow M\\ x&\mapsto \Phi ^x(x,V,R)=\pi ^M\circ \Phi (x,V,R). \end{aligned} \end{aligned}$$

Let $(\gamma ,V,R)$ be an integral curve of $f_1$, then, since $\Phi _\star f_1= \pm f_1$, it holds

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{\Phi }^x=\pm \Phi ^V,\\ D_t\Phi ^R=\mp \Phi ^V,\\ D_t\Phi ^V=\pm \Phi ^R, \end{array}\right. } \end{aligned}$$

(3.57)

but we know that $\Phi ^x=\varphi$, hence the first equation of Eq. 3.57 reads

$$\begin{aligned} \varphi _\star V=\varphi _\star \dot{\gamma }= \dot{\Phi }^x=\pm \Phi ^V, \end{aligned}$$

consequently $\Phi ^V=\pm \varphi _\star V$. Now consider the vector field $f_{21}=[f_1,f_1]$, recall that, according to Eq. 3.16, its integral curves satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}= V,\\ D_tR=0,\\ D_tV=0, \end{array}\right. } \end{aligned}$$

(3.58)

on the other hand since $\Phi _\star f_1= \pm f_1$, $\Phi _\star f_2=\pm f_2$, it also holds $\Phi _\star f_{12}=\pm f_{12}$ thus

$$\begin{aligned} \nabla _VV=0\iff \nabla _{\varphi _\star V}\varphi _\star V=0 \end{aligned}$$

or equivalently

$$\begin{aligned} \nabla =\varphi ^\star \nabla . \end{aligned}$$

(3.59)

Exploiting Eq. 3.59 together with the third equation in Eq. 3.57 we deduce

$$\begin{aligned} \pm \Phi ^R=D_t\Phi ^V=\nabla _{\varphi _\star V}\varphi _\star V=\varphi _\star \nabla _VV=\varphi _\star D_tV=\varphi _\star R, \end{aligned}$$

from which we conclude

$$\begin{aligned} \Phi (x,V,R)=(\varphi (x),\pm \varphi _\star R,\pm \varphi _\star V)=\pm \varphi _\star \oplus \varphi _\star (x,V,R). \end{aligned}$$

(3.60)

Substituting Eq. 3.60 in the second equation of Eq. 3.57 we find that we have to discard the minus sign and we are left with

$$\begin{aligned} \Phi (x,V,R)=(\varphi (x),\varphi _\star V,\varphi _\star R). \end{aligned}$$

(3.61)

Let $(\gamma ,V,R):[0,1]\rightarrow \mathcal R(M,g)$ be a curve tangent to $\mathcal D(M,g)$ with $\gamma$ regular, then according to Theorem 1.1, $R=R_g(\gamma )$. On the other hand since $\Phi$ preserves $\mathcal D(M,g)$, also $(\varphi \circ \gamma ,\varphi _\star R,\varphi _\star V)$ is tangent to $\mathcal D(M,g)$ and hence, according to Theorem 1.1 we have

$$\begin{aligned} R_g(\varphi \circ \gamma )=\varphi _\star R_g(\gamma ). \end{aligned}$$

On the other hand, since $\varphi :(M,\varphi ^\star g)\rightarrow (M,g)$ is an isometry, it holds

$$\begin{aligned} R_g(\varphi \circ \gamma )=\varphi _\star R_{\varphi ^\star g}(\gamma ), \end{aligned}$$

therefore $R_g=R_{\varphi ^\star g}$ and, according to Theorem 1.2, $\varphi :(M,g)\rightarrow (M,g)$ is a homothety.

Assume now that $n=\dim {M}=2$. Consider the following linear map

$$\begin{aligned} \begin{aligned} P:\mathcal D(M,g)&\rightarrow \mathcal D^3(M,g)/\mathcal D^2(M,g)\\ X&\mapsto \left[ X,[f_1,f_2]\right] \mod \mathcal D^2(M,g). \end{aligned} \end{aligned}$$

(3.62)

Observe that the kernel of Eq. 3.62, is invariant under pushforwards of sub-Riemannian isometries, in the sense that

$$\begin{aligned} \Phi _\star \ker P=\ker P, \end{aligned}$$

therefore

$$\begin{aligned} \Phi _\star \langle \{f_2\}\rangle =\langle \{f_2\}\rangle , \end{aligned}$$

and since $\Phi$ is an isometry we deduce that

$$\begin{aligned} \Phi _\star f_1=\pm f_1,\,\,\,\Phi _\star f_2=\pm f_2. \end{aligned}$$

(3.63)

We claim that $\Phi _\star f_2=f_2$. Assume by contradiction that $\Phi _\star f_2=-f_2$ and consider the following couple of vector fields

$$\begin{aligned} X=f_1+f_{12},\,\,\,Y=f_2. \end{aligned}$$

Since $[f_2,f_{12}]=f_{12}$, we have

$$\begin{aligned}{}[X,Y]=[f_1,f_2]-f_{12}=f_{12}-f_{12}=0. \end{aligned}$$

On the other hand since $\Phi _\star f_1=\pm f_1$, $\Phi _\star f_2=-f_2$ we find

$$\begin{aligned} 0=\Phi _\star [X,Y]=[\pm f_1\mp f_{12},-f_2]=\mp 2f_{12}\ne 0, \end{aligned}$$

which is a contradiction. Combining $\Phi _\star f_2=f_2$ and $\Phi _\star f_1=\pm f_1$ we deduce that

$$\begin{aligned} \Phi _\star X_{12}=\pm X_{12}, \end{aligned}$$

therefore, since

$$\begin{aligned} \ker \pi ^M_\star =\langle \{f_2,X_{12}\}\rangle , \end{aligned}$$

we obtain

$$\begin{aligned} \ker \pi ^M_\star =\ker (\pi ^M\circ \Phi )_\star . \end{aligned}$$

The remaining part of the proof is identical to the one of the case $n>2$. In particular we can conclude by repeating verbatim what is written between Eq. 3.56 and the discussion of the 2-dimensional case. $\square$

4 The Fields $f_1,f_2$

Given a Riemannian manifold (M, g), the vector fields $f_1,f_2$ defined in (1.6) are global sections of $T(TM\oplus TM)$, which restrict to vector fields on $\mathcal R(M,g)$. As a consequence of Theorem 1.1 we know that they are metric invariants of (M, g). Actually they are even invariant under homothetic transformation. In the current section we show that some classical metric invariants can be recovered from the iterated Lie brackets of $f_1,f_2$. Indeed, as the next proposition shows, the field $[f_1,f_2]$ already gives us a complete description of the geodesics of (M, g); in this sense the fields $f_1,f_2$ give us a factorization of the geodesic flow.

Proposition 4.1

Let (M, g) be a Riemannian manifold, let $x\in M$ and let $\exp _x^{(M,g)}$ be the corresponding exponential map at x. Let $f_{21}=[f_2,f_1]$ and $f_{121}=[f_1,f_{21}]$, then, for every $(x,V,R)\in TM\oplus TM$, the following formulae hold

$$\begin{aligned} \begin{aligned}&\pi ^M\circ e^{tf_{21}}(x,V,R)=\exp _x^{(M,g)}(tV),\\&\pi ^M\circ e^{tf_{121}}(x,V,R)=\exp _x^{(M,g)}(tR), \end{aligned} \end{aligned}$$

(4.1)

for every $t\in \mathbb R$ such that the flow of the fields is defined.

Proof

According to Eq. 3.16, the vector field $f_{21}$ reads

$$\begin{aligned} f_{21}=[f_2,f_1]=V\partial _x-\Gamma (V,R)R\partial _R-\Gamma (V,V)\partial _V, \end{aligned}$$

where the symbols $\Gamma$ are the ones defined in Eq. 3.14. Consequently, any integral curve of $f_{21}$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}=V,\\ D_t R=0,\\ D_tV=0. \end{array}\right. } \end{aligned}$$

(4.2)

Therefore the curve $x(t)=\pi ^M\circ e^{t[f_2,f_1]}(x_0,V_0,R_0)$ is the unique geodesic with initial point $x_0$ and initial velocity $V_0$.

Recall the vector field $X_{12}$ defined in Eq. 3.17 and observe that

$$\begin{aligned} f_{121}=[f_1,f_{21}]=[f_1,f_1-X_{12}]=-[f_1,X_{12}], \end{aligned}$$

therefore

$$\begin{aligned} \begin{aligned} f_{121}=&[V\partial _x-(V+\Gamma (R,V))\partial _R+(R-\Gamma (V,V))\partial _V, V\partial _R-R\partial _V]\\ =&[V\partial _x-\Gamma (R,V)\partial _R-\Gamma (V,V)\partial _V, V\partial _R-R\partial _V]\\ =&\Gamma (R,V)\partial _V-\Gamma (V,V)\partial _R+\Gamma (V,V)\partial _R+R\partial _x\\&-\Gamma (R,R)\partial _R-2\Gamma (R,V)\partial _V\\ =&R\partial _x-\Gamma (R,R)\partial _R-\Gamma (R,V)\partial _V. \end{aligned} \end{aligned}$$

Hence any integral curve of $f_{121}$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}=R,\\ D_t R=0,\\ D_tV=0, \end{array}\right. } \end{aligned}$$

(4.3)

concluding the proof. $\square$

If we consider another layer of the Lie algebra generated by $f_1,f_2$, the components of Riemann curvature tensor appear.

Proposition 4.2

Let (M, g) be a Riemannian manifold, the integral curves of the vector field $f_{1121}:=[f_1,f_{121}]$ satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{x}=-V,\\ D_t R=-\mathcal R^{\nabla }(V,R)R,\\ D_t V=-\mathcal R^{\nabla }(V,R)V, \end{array}\right. } \end{aligned}$$

(4.4)

where $\mathcal {R}^{\nabla }:\mathfrak { X}(M)\times \mathfrak {X}(M)\times \mathfrak {X}(M)\rightarrow \mathfrak {X}(M)$ is the Riemann curvature tensor of (M, g).

Proof

Given a vector field X we denote with $X^x$ its x-component, with $X^R$ its R-component and with $X^V$ its V-component, meaning that $X=X^x\partial _x+X^R\partial _R+X^V\partial _V$. We compute Eq. 4.4 component by component:

$$\begin{aligned} \begin{aligned} f_{1121}^x&=f_1(f_{121}^x)-f_{121}(f_1^x)=f_1(R)-f_{121}(V)\\&=-V-\Gamma (R,V)+\Gamma (R,V)=-V. \end{aligned} \end{aligned}$$

(4.5)

Concerning the R-component we compute

$$\begin{aligned} \begin{aligned} f_1(f_{121}^R)&=(V\partial _x-(V+\Gamma (R,V))\partial _R+(R-\Gamma (V,V))\partial _V)(-\Gamma (R,R))\\&=-V\partial _x\Gamma (R,R)+2\Gamma (V,R)+2\Gamma (\Gamma (R,V),R), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} f_{121}(f_1^R)&=(R\partial _x-\Gamma (R,R)\partial _R-\Gamma (R,V)\partial _V)(-V-\Gamma (R,V))\\&=-R\partial _x\Gamma (R,V)+\Gamma (\Gamma (R,R),V)+\Gamma (R,V)+\Gamma (R,\Gamma (R,V)), \end{aligned} \end{aligned}$$

therefore

$$\begin{aligned} \begin{aligned} f_{1121}^R=&f_1(f_{121}^R)-f_{121}(f_1^R)\\ =&\Gamma (R,V)-V\partial _x\Gamma (R,R)+R\partial _x\Gamma (R,V)\\&-\Gamma (V,\Gamma (R,R))+\Gamma (R,\Gamma (R,V))\\ =&\Gamma (R,V)-\mathcal R^{\nabla }(V,R)R. \end{aligned} \end{aligned}$$

(4.6)

The quantity $f_1(f_{121}^V)$ can be computed as

$$\begin{aligned} \begin{aligned} f_1(f_{121}^V)=&(V\partial _x-(V+\Gamma (R,V))\partial _R+(R-\Gamma (V,V))\partial _V)(-\Gamma (R,V))\\ =&-V\partial _x\Gamma (R,V)+\Gamma (V,V)+\Gamma (\Gamma (R,V),V)-\Gamma (R,R)+\Gamma (R,\Gamma (V,V), \end{aligned} \end{aligned}$$

whereas $f_{121}(f_1^V)$ satisfies

$$\begin{aligned} \begin{aligned} f_{121}(f_1^V)&=(R\partial _x-\Gamma (R,R)\partial _R-\Gamma (R,V)\partial _V)(R-\Gamma (V,V))\\&=-R\partial _x\Gamma (V,V)-\Gamma (R,R)+2\Gamma (\Gamma (R,V),V), \end{aligned} \end{aligned}$$

hence

$$\begin{aligned} \begin{aligned} f_{1121}^V=&f_1(f_{121}^V)-f_{121}(f_1^V )\\ =&\Gamma (V,V)-V\partial _x\Gamma (R,V)+R\partial _x\Gamma (V,V)\\&-\Gamma (V,\Gamma (R,V))+\Gamma (R,\Gamma (V,V))\\ =&\Gamma (V,V)-\mathcal R^{\nabla }(V,R)V. \end{aligned} \end{aligned}$$

(4.7)

Equations 4.5, 4.6 and 4.7 together imply that any integral curve of $f_{1121}$ satisfies the ODEs system Eq. 4.4.$\square$

As stated in Theorem 1.1, the fields

$$\begin{aligned} \{f_1,\dots ,f_n,f_{12},\dots ,f_{1n},f_{121},\dots ,f_{1n1}\} \end{aligned}$$

(4.8)

constitute a local basis for $T\mathcal R(M,g)$, therefore there exists some real valued smooth functions $\{c_1,\dots ,c_{3n-2}\}$ on $\mathcal R(M,g)$ such that

$$\begin{aligned} f_{1121}=\sum _{i=1}^nc_if_i+\sum _{i=2}^nc_{n+i-1}f_{1i}+\sum _{i=2}^nc_{2n+i-2}f_{1i1}. \end{aligned}$$

(4.9)

Proposition 4.3

Let (M, g) be a Riemannian manifold, then the function $c_1:\mathcal R(M,g)\rightarrow \mathbb R$ defined in Eq. 4.9 is a homothetic invariant of (M, g), which can be expressed in terms of sectional curvatures as

$$\begin{aligned} c_1(x,V,R)=|R|^2\sec (R,V). \end{aligned}$$

(4.10)

In particular if (M, g) is a Riemannian surface, then

$$\begin{aligned} c_1(x,V,R)=|R|^2K(x), \end{aligned}$$

(4.11)

where K is the Gaussian curvature of (M, g).

Proof

From Eqs. 4.4 and 4.2 it follows that

$$\begin{aligned} f_{1121}+f_{21}=-\mathcal R^\nabla (V,R)R\partial _R-\mathcal R^\nabla (V,R)V\partial _V, \end{aligned}$$

(4.12)

therefore

$$\begin{aligned} \begin{aligned} f_{1121}+f_{21}=&-\frac{1}{|R|^2}\langle \mathcal R^\nabla (V,R)R,V\rangle V\partial _R-\frac{1}{|R|^2}\langle \mathcal R^\nabla (V,R)V,R\rangle R\partial _V\\&-\frac{1}{|R|^2}\sum _{i=3}^n\left( \langle \mathcal R^\nabla (V,R)R,e_j\rangle e_j\partial _R +\langle \mathcal R^\nabla (V,R)V,e_j\rangle e_j\partial _V\right) ,\\ =&|R|^2\sec (R,V)(-V\partial _R+R\partial _V)-X, \end{aligned} \end{aligned}$$

where $\{e_3,\dots ,e_n\}$ is the basis of $\{R,V\}^\perp$ described in the proof of Theorem 1.1 and we have denoted

$$\begin{aligned} X=\frac{1}{|R|^2}\sum _{i=3}^n\left( \langle \mathcal R^\nabla (V,R)R,e_j\rangle e_j\partial _R +\langle \mathcal R^\nabla (V,R)V,e_j\rangle e_j\partial _V\right) . \end{aligned}$$

Observe that

$$\begin{aligned} -V\partial _R+R\partial _V=f_1-f_{21}, \end{aligned}$$

hence

$$\begin{aligned} f_{1121}+f_{21}=|R|^2\sec (R,V)(f_1-f_{21})-X. \end{aligned}$$

(4.13)

The unique expression of the vector field X in terms of the basis Eq. 4.8 is a linear combination which does not involve $f_1$, hence we deduce

$$\begin{aligned} c_1(x,V,R)=|R|^2\sec (R,V). \end{aligned}$$

(4.14)

$\square$

The fields $f_1,f_2$ allow us to characterize the homotheties of Riemannian surfaces with one synthetic equation. Let M be a smooth manifold and let $X\in \mathfrak X(M)$ be a vector field, which we assume to be complete for simplicity. The family of maps $\{e^{tX}_\star \}_{t\in \mathbb R}$ constitutes a one-parameter group of diffeomorphisms of TM; we denote its infinitesimal generator with $\overset{\rightarrow }{X}$.

Proposition 4.4

Let (M, g) be a Riemannian surface and let $f_1$ be the local vector vield over $TM\setminus s_o$ defined in Eq. 2.6. A vector field $X\in \mathfrak X(M)$ is the infinitesimal generator of a one-parameter group of homotheties if and only if

$$\begin{aligned}{}[\overset{\rightarrow }{X},f_1]=0. \end{aligned}$$

(4.15)

Proof

A vector field $X\in \mathfrak X(M)$ satisfies Eq. 4.15 if and only if

$$\begin{aligned} e^{t\overset{\rightarrow }{X}}_\star f_1=f_1 \end{aligned}$$

(4.16)

for every $t\in \mathbb R$. On the other hand, any vector field $X\in \mathfrak X(M)$ satisfies

$$\begin{aligned} e^{t\overset{\rightarrow }{X}}_\star f_2=f_2, \end{aligned}$$

(4.17)

indeed

$$\begin{aligned} \begin{aligned} e^{s\overset{\rightarrow }{X}}\circ e^{tf_2}(x,R)&=e^{s\overset{\rightarrow }{X}}(x,e^t R)=(e^{sX}x,e^{sX}_\star e^tR)\\&=(e^{sX}x,e^te^{sX}_\star R)=e^{tf_2}\circ e^{s\overset{\rightarrow }{X}}(x,R). \end{aligned} \end{aligned}$$

Therefore X satisfies Eq. 4.15 if and only if $e^{t\overset{\rightarrow }{X}}$ is an isometry of $\mathcal R_{a,b}(M,g)$ for each $t\in \mathbb R$, hence, by Theorem 1.3, X satisfies Eq. 4.15 if and only if $e^{tX}$ is a one-parameter group of homotheties of (M, g). $\square$

The results obtained so far can be used to produce a flatness theorem for Riemannian surfaces having a 4-dimensional Lie algebra of homothetic vector fields.

Theorem 4.1

Let (M, g) be a 2-dimensional Riemannian manifold and let $L\subset \mathfrak X(M)$ be the corresponding Lie algebra of homothetic vector fields. Then $\dim (L)\le 4$, and (M, g) is flat if equality is achieved.

Proof

Let $\mathcal L\subset \mathfrak X(\mathcal R(M))$ be the Lie algebra of isometric vector fields for the manifold of curvature radii $\mathcal R(M,g)$. By Theorem 1.3 this Lie algebra is isomorphic to the one of homothetic vector fields of (M, g). Let $X\in \mathcal L$ be a vector field vanishing at some point $(x, R)\in \mathcal R(M)$. We claim that X is identically zero. Indeed, since X is isometric, by Proposition 4.4, it satisfies $[X,f_1]=[X,f_2]=0$. Since $\mathcal D=span\{f_1,f_2\}$ is bracket generating, for any $(x',R')\in \mathcal R$ there exist some real numbers $s_1,\dots ,s_k$ such that

$$\begin{aligned} (x',R')=e^{s_1 f_{i_1}}\circ \dots \circ e^{s_k f_{i_k}}(x,R),\,\,\,\,i_1,\dots ,i_k\in \{1,2\}, \end{aligned}$$

(4.18)

consequently

$$\begin{aligned} X_{(x',R')}=e^{s_1 f_{i_1}}_\star \circ \dots \circ e^{s_k f_{i_k}}_\star X_{(x,R)}=0. \end{aligned}$$

Let $X_1,\dots ,X_5\in \mathcal L$ and $(x,R)\in \mathcal R(M)$. The manifold of curvature radii is 4-dimensional, hence there exists a linear combination of $X_1,\dots ,X_5$ vanishing at (x, R), and thus vanishing everywhere. We deduce that $\dim \mathcal L\le 4$. Assume now that $\dim \mathcal L=4$ and let $X_1,\dots , X_4$ be a basis for $\mathcal L$. By the above argument $X_1,\dots ,X_4$ are linearly independent at every point of $\mathcal R(M)$. Thus they can be used to produce local coordinates in the neighbourhood of every point, by means of the map

$$\begin{aligned} (s_1,\dots ,s_4)\mapsto e^{s_1X_1}\circ \dots \circ e^{s_4X_4}(x,R). \end{aligned}$$

This implies that the structure coefficients of the frame $f_1,f_2,f_{12},f_{121}$ are constants (Theorem 1.1 implies that the latter is a basis for $T\mathcal R(M)$). In particular there exist real constants $c_1,c_2,c_{12},c_{121}$ such that

$$\begin{aligned} f_{1121}=[f_1,f_{121}]=c_1f_1+c_2f_2+c_{12}f_{12}+c_{121}f_{121}. \end{aligned}$$

Now we exploit Proposition 4.3, telling us that $c_1(x,R)=|R|^2 K(x)$, where K(x) is the Gaussian curvature of (M, g). Such a function is constant on a fixed fiber of the manifold of curvature radii $\mathcal R(M)=TM\setminus s_0$ if and only if it is identically zero. It follows that $K=0$. $\square$

5 Similarity Transformations of the Plane

In this section we show that the sub-Riemannian manifolds $\mathcal R_{a,b}(\mathbb R^2,g_e)$, where $g_e$ is the standard Euclidean metric, are isomorphic to left invariant sub-Riemannian structures on the group of orientation preserving homotheties of $\mathbb R^2$, which we denote with G. Moreover we give a a characterization of the sub-Riemannian geodesics of $\mathcal R_{0,1}(\mathbb R^2,g_e)$ in terms of the euclidean curvature of their projections to a plane. Such characterization in terms of curvature is similar to the one of Euler elasticae ([10]), which are projection of normal extremal trajectories of the nilpotent Engel group ([6]). All the results of this section follow from straightforward computations, therefore the proofs are omitted. Let $(y_1,y_2,R_1,R_2)$ be global coordinates for $T\mathbb R^2\setminus \{0\}=\mathbb R^2\times \mathbb R^2\setminus \{0\}$. It is convenient to define a new set of coordinates $(\theta , r, x_1,x_2)$ as

$$\begin{aligned} (y_1,y_2,R_1,R_2)=:( x_1+r\text {cos}\theta , x_2+r\text {sin}\theta , -r\text {cos}\theta , -r\text {sin}\theta ). \end{aligned}$$

(5.1)

In coordinates (y, R) a point in $\mathcal R_{a,b}(\mathbb R^2,g_e)$ is interpreted as the curvature radius of some curve going through y. In the new coordinates $(\theta , r, x_1,x_2)$ we interpret a point in $\mathcal R_{a,b}(\mathbb R^2,g_e)$ as the osculating circle of such curve, having radius $(r\cos \theta ,r\sin \theta )$ and center $(x_1,x_2)$. Observe that the data of a homothetic transformation, which in the case of $(\mathbb R^2, g_e)$ consists in a composition of rotations, dialations and translation, can be encoded into an osculating circle, i.e. a point of $\mathcal R_{a,b}(\mathbb R^2,g_e)$: given a circle $(\theta , r, x_1,x_2)$ we can dilate by r, rotate by $\theta$ and translate by $(x_1,x_2)$. We have a diffeomorphism

$$\begin{aligned} \begin{aligned} F:\mathcal R(\mathbb R^2)&\rightarrow G\\ (\theta ,r,x_1,x_2)&\mapsto Q:=\begin{pmatrix} r\text {cos}\theta &{} -r\text {sin}\theta &{} x_1\\ r\text {sin}\theta &{} r\text {cos}\theta &{} x_2\\ 0 &{} 0&{} 1 \end{pmatrix}, \end{aligned} \end{aligned}$$

(5.2)

which allows us to push forward the sub-Riemannian structure of $\mathcal R_{a,b}(\mathbb R^2,g_e)$, to G. The resulting sub-Riemannian structure is left invariant.

Proposition 5.1

The frame $(f_1,f_2)$ can be written in coordinates $(\theta , r, x_1,x_2)$ as

$$\begin{aligned} \begin{aligned} f_1=-\frac{\partial }{\partial \theta },\,\,\,\, f_2=r\frac{\partial }{\partial r} -r\cos \theta \frac{\partial }{\partial x_1} -r\,sin\theta \frac{\partial }{\partial x_2}. \end{aligned} \end{aligned}$$

(5.3)

Both of these vector fields are pushforwarded to left invariant vector fields by the map Eq. 5.2:

$$\begin{aligned} \begin{aligned}&(F_\star f_1)(Q)=-Q\begin{pmatrix} 0 &{} -1 &{} 0\\ 1 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{pmatrix},\,\,\,\,&(F_\star f_2)(Q)=Q \begin{pmatrix} 1 &{} 0 &{} -1\\ 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 0 \end{pmatrix}. \end{aligned} \end{aligned}$$

(5.4)

Remark 5.1

The geometry of G is reminiscent of the left invariant sub-Riemannian structure on the group of rigid motions of the plane, related to ‘bicycling mathematics’, which has been studied in [4, 5, 7, 9]. The group of rigid motions of $\mathbb R^2$ can be described as

$$\begin{aligned} SE_2=\left\{ \begin{pmatrix} \cos \theta &{} -\sin \theta &{} x_1\\ \sin \theta &{} \cos \theta &{} x_2\\ 0 &{} 0&{} 1 \end{pmatrix}\,:\, \theta \in [0,2\pi ],\,(x_1,x_2)\in \mathbb R^2\, \right\} . \end{aligned}$$

In coordinates $(\theta , x_1,x_2)$ we can define a left-invaraint sub-Riemann structure on $SE_2$ by declaring the fields

$$\begin{aligned} X_1=\cos \theta \frac{\partial }{\partial x_1}+\sin \theta \frac{\partial }{\partial x_2},\,\,\,X_2=\frac{\partial }{\partial \theta }, \end{aligned}$$

(5.5)

an orthonormal generating family. There exists a submersion

$$\begin{aligned} \begin{aligned} P:G&\rightarrow SE_2\\ (\theta ,r,x_1,x_2)&\mapsto (\theta ,x_1,x_2), \end{aligned} \end{aligned}$$

(5.6)

satisfying

$$\begin{aligned} P_\star \frac{1}{r}f_2=-X_1,\,\,\, P_\star f_1=-X_2. \end{aligned}$$

(5.7)

Let $h_1,h_2:T^\star \mathcal R(\mathbb R^2)\rightarrow \mathbb R$ be the Hamiltonian functions associated with $f_1,f_2$ and let $p_\theta ,p_r,p_{x_1},p_{x_2}$ be the canonical momenta associated with the coordinates $\theta ,r,x_1,x_2$, then

$$\begin{aligned} \begin{aligned}&h_1=-p_\theta ,\\&h_2=rp_r-r\cos \theta p_{x_1}-r\sin \theta p_{x_2}. \end{aligned} \end{aligned}$$

The sub-Riemannian Hamiltonian of $\mathcal R_{0,1}(\mathbb R^2,g_e)$ reads

$$\begin{aligned} 2H=h_1^2+h_2^2=p_\theta ^2+(rp_r-r\cos \theta p_{x_1}-r\sin \theta p_{x_2})^2. \end{aligned}$$

(5.8)

If we make the change of coordinates $\rho =\log r$ the Hamiltonian becomes

$$\begin{aligned} 2H=p_\theta ^2+(p_\rho -e^\rho \cos \theta p_{x_1}-e^\rho \sin \theta p_{x_2})^2. \end{aligned}$$

(5.9)

The next proposition characterizes normal extremal trajectories in terms of the euclidean curvature of their projection to the $(\rho ,\theta )$-plane.

Proposition 5.2

The following quantities

$$\begin{aligned} \epsilon =\sqrt{p_{x_1}^2+p_{x_2}^2},\,\,\, \alpha =\arg {(p_{x_1}+ip_{x_2})}, \end{aligned}$$

are first integrals of the sub-Riemannian Hamiltonian Eq. 5.9.

A curve $(\rho ,\theta ):[0,1]\rightarrow \mathbb R^2$ is the projection of a normal extremal trajectory $(\rho ,\theta ,x_1,x_2):[0,1]\rightarrow \mathbb R^4$, if and only if its Euclidean curvature $\kappa$ satisfies

$$\begin{aligned} \kappa (\rho ,\theta )=\epsilon e^\rho \sin (\theta -\alpha ). \end{aligned}$$

There are no strictly abnormal extremals.