text:
\section{Introduction}

\subsection*{Intrinsic area of convex bodies}

A \emph{convex body} is  a non-empty compact convex subset of $\mathbb{R}^n$. In this article, we set $n>1$.
For a plane convex body $K$ (i.e. a convex body in $\mathbb{R}^2$), speaking about the ``area''  of $K$ usually means to look at its two  dimensional Lebesgue measure. Note that the area of plane convex bodies is positively homogeneous of degree $2$: for $\lambda>0$,
$\a(\lambda K)=\lambda^2 \a(K)$. For a convex body in $\mathbb{R}^3$,  the ``area'' usually refers to its surface area, i.e. the $2$-dimensional total Hausdorff measure of its boundary $\partial K$.
 Here also, the surface area is
positively homogeneous of degree two.

There are two ways to generalize the notion of ``area'' to convex bodies in $\mathbb{R}^n$ for $n>3$.
Both are coming from the
 Steiner Formula.
 Let $B^n$ be the closed unit ball centred at the origin in $\mathbb{R}^n$, and let  $\kappa_n$ be its volume. Let us set $\kappa_0=1$ and $\kappa_1=2$. If $K$ is a convex body in $\mathbb{R}^n$, then there exists non-negative real numbers $V_i(K)$, $i=0,\ldots,n$ such that, for any $\epsilon >0$,
  \begin{equation}\label{steiner}\mathrm{vol}_n(K+\epsilon B^n)=\sum_{i=0}^n \epsilon^{n-i}\kappa_{n-i}V_i(K)~.\end{equation}
Here $\mathrm{vol}_n$ is the Lebesgue measure of $\mathbb{R}^n$, and the sum is the Minkowski addition: $A+B=\{a+b|a\in A,b\in B\}$.
It appears that $V_0(K)=1$ and $V_n(K)=\mathrm{vol}_n(K)$.

The first way to generalize the notion of surface area of convex bodies in $\mathbb{R}^3$ is to consider $V_{n-1}(K)$ as the ``area'', given by the first order variation of $\operatorname{vol}_n(K+\epsilon B^n)$, seen as a function of $\epsilon$. Note that this ``area'' is homogeneous of degree $(n-1)$, and that for $n=2$, this is related to the perimeter of the convex body and not to its area.

Instead, in this paper, we will consider the \emph{intrinsic area} $V_2(K)$. Let us mention some relevant properties.

\begin{enumerate}[label={A\arabic*)}]
\item \label{homo} For any $\lambda >0$, $V_2(\lambda K)=\lambda^2V_2(K)$;
\item \label{prop 1V2} $V_2(K)\geq 0$;
\item \label{propV22} $K_1\subset K_2 \Rightarrow V_2(K_1) \leq V_2(K_2)$;
\item  \label{propV23} $V_2(K)=0$ if and only if $K$ is a point
or a segment;
\item \label{isom} for any $A\in \mathrm{SL}(n,\mathbb{R})$ and $p\in \mathbb{R}^n$,
$V_2(A(K)+\{p\})=V_2(K)$;
\item \label{mcmullen}
Let $\iota:\mathbb{R}^n\to \mathbb{R}^{n+1}$ be a linear isometric embedding.
Then $V_2(\iota(K))=V_2(K)$.
\end{enumerate}

\begin{figure}
\begin{center}
\psfrag{O}{$\ell$}
\psfrag{e}{$\epsilon$}
\psfrag{k}{$K$}
\includegraphics[width=0.4\linewidth]{segment}\caption{Let $K$ be a segment of length $\ell$ in $\mathbb{R}^n$. Then $\mathrm{vol}_n(K+\epsilon B^n)$ is the sum of the volume of two half $n$-dimensional balls of radius $\epsilon$, plus $\ell$ times the volume of a $(n-1)$-dimensional ball. By \eqref{steiner}, $V_2(K)=0$ and $V_1(K)=\ell$.}\label{fig:segments}
\end{center}
\end{figure}

These properties are well-known, however we will give proofs. Note that Property \ref{homo} above is straightforward from \eqref{steiner}, as well as \ref{isom}, by invariance of the Lebesgue measure.
It is also clear from \eqref{steiner} that points and segments have zero intrinsic area, see Figure~\ref{fig:segments}.
The  property \ref{mcmullen} explains the denomination ``intrinsic''.
   For example, if $n=2$, then $V_2(K)$ is the $\mathbb{R}^2$ Lebesgue measure of $K$. If $n=3$, then the intrinsic area is half of the surface area (see \cite{schneider}). If $K$ is a convex body in $\mathbb{R}^3$ contained in a $2$-plane $P$, then the surface area of $K$ is two times the area of $K$ in $P$: the area has to be taken into account two times, as $K$ has two support planes that coincide with $P$. This is coherent with property \ref{mcmullen}, that says that the intrinsic area of $K$ does not depend on the dimension of the ambient space.

 For proofs and comments around property~\ref{mcmullen}, see see pp.~208 and 214  in \cite{schneider}, as well as  \cite{mcmullen,mcmullen2} and Proposition 3.2 in \cite{chevet} and the references given there.

There is a geometric interpretation of the intrinsic area of a convex body $K$ that we won't use: it
 is, up to a dimensional constant, the mean value of the areas of the orthogonal projections of $K$ onto two dimensional vector planes \cite[(5.72)]{schneider}.

In the sequel we will also consider $V_1(K)$, which is related to the \emph{mean width} of the convex body $K$, see Remark \ref{remark:mean width}. We will see in Section \ref{Canonical embeddings} that this quantity is also intrinsic. Let us note that $V_1$ of a segment is its length, see Figure~\ref{fig:segments}.

For future references, let us give the following example. Writing $\mathrm{vol}_n(B^n+\epsilon B^n)=\mathrm{vol}((1+\epsilon)B^n)
=(1+\epsilon)^n\kappa_n$,
from \eqref{steiner} we have
$\kappa_n\sum_{i=0}^n\binom{n}{i} \epsilon^{n-i}=\sum_{i=0}^n \epsilon^{n-i}\kappa_{n-i}V_i(B^n)$
so, using classical equalities, we obtain,
\begin{equation}\label{V2B}V_2(B^n)=(n-1)\frac{n\kappa_n}{2\kappa_{n-2}}=(n-1)\pi~, \end{equation}
and also, if  $W_n=\int_0^{\pi/2} \cos^n$ is the Wallis' integral,
\begin{equation}\label{V1B}V_1(B^n)=\frac{n\kappa_n}{\kappa_{n-1}}= \frac{\operatorname{vol}(\mathbb{S}^{n-1})}{\kappa_{n-1}} =2\pi\frac{\kappa_{n-2}}{\kappa_{n-1}}=\frac{\pi}{W_{n-1}}=2nW_n\sim 2n\sqrt{\frac{\pi}{2n}}=\sqrt{2\pi n}~. \end{equation}
Note that in particular we have $V_1(B^1)=2$.
Let us introduce the following dimensional constants:

\begin{equation}\label{v1v2}
r_1(n)=V_1(B^n)^{-1}, \, r_2(n)=V_2(B^n)^{-1/2}~,
\end{equation}
which are such that a $n$-th dimensional ball of radius $r_1(n)$ has $V_1=1$, and a $n$-th dimensional ball of radius $r_2(n)$ has $V_2=1$.

\subsection*{Mixed-area of convex bodies}

The (intrinsic) area can be ``polarized'', in the sense that there exists a function called the (intrinsic) \emph{mixed-area} $V_2(\cdot,\cdot)$, that can be defined as
\begin{equation}\label{def mixed v2}
V_2(K_1,K_2)=\frac{1}{2}\left(V_2(K_1+K_2)-V_2(K_1)-V_2(K_2) \right)~,
\end{equation}
and satisfying the following properties:
\begin{enumerate}[label={M\arabic*)}]
\item \label{mixed 1} $V_2(K_1,K_1)=V_2(K_1)$;
\item \label{mixed 2} $V_2(K_1,K_2)=V_2(K_2,K_1)$;
\item \label{mixed 3} $V_2(K_1+K_2,K_3)=V_2(K_1,K_3)+V_2(K_2,K_3)$;
\item \label{mixed 4} for $\lambda >0$, $V_2(\lambda K_1,K_2)=\lambda  V_2(K_1,K_2)$;
 \item \label{prop mixed incl} $K_1\subset K_2 \Rightarrow V_2(K_1,K_3)\leq V_2(K_2,K_3)$;
\item \label{isotrope} $K$ is a point if and only if  for any convex body $Q$,
$V_2(K,Q)=0$;
\item \label{mixed 2'} $V_2(K_1,K_2) \geq 0$; and $V_2(K_1,K_2)=0$ if and only if $K_1$ or $K_2$ is a point, or both are segments with the same direction;
\item \label{alex fench} we have
\begin{equation} \label{eq alex fench}
\delta(K_1,K_2)=V_2(K_1,K_2)^2 -  V_2(K_1)V_2(K_2)\geq 0
\end{equation}
and if $K_1$ and $K_2$ are not points, then equality occurs if and only if $K_1$ and $K_2$ differ by a translation and a positive homothety;
\item \label{alex fench2}  we have
\begin{equation*}
V_2(K_1,K_2) \geq \sqrt{V_2(K_1)V_2(K_2)}
\end{equation*}
and if $K_1$ and $K_2$ are not points, then equality occurs if and only if $K_1$ and $K_2$ differ by a translation and a positive homothety.
\end{enumerate}

Only properties \ref{mixed 1} and \ref{mixed 2} are obvious from \eqref{def mixed v2}.
The property \ref{alex fench} is a particular case of the famous Alexandrov--Fenchel inequality. \ref{alex fench2}  is \ref{alex fench} written with the help of \ref{mixed 2'}.

For future references, let us note the following particular case. Developing using Steiner formula \eqref{steiner}
both sides of $\mathrm{vol}_n((K+B^n)+\epsilon B^n)=
\mathrm{vol}_n(K+(1+\epsilon) B^n)$ and using \eqref{V2B} and \eqref{V1B}, we have $V_2(K+B^n)=V_2(K)+V_1(B^{n-1})V_1(K)+V_2(B^n)$, and from  \eqref{def mixed v2} we finally obtain
\begin{equation}\label{V2 avec V1} V_2(K,B^n)=\frac{1}{2}V_1(B^{n-1})V_1(K)~. \end{equation}

If $K\subset \mathbb{R}^2$, then $V_1(K)$ is half the perimeter $\mathrm{per}(K)$, and $V_2(K)$ is the area $\operatorname{vol}_2(K)$, so property \ref{alex fench} is the isoperimetric inequality: indeed it reads
\begin{eqnarray*}
0 \leq V_2(K,B^2)^2 - V_2(K) V_2(B^2) &=& \frac{1}{4}V_1(B^1)^2 V_1(K)^2 - \operatorname{vol}_2(K) V_2(B^2) \\
&=& \frac{\mathrm{per}(K)^2}{4}-\pi \operatorname{vol}_2(K)~.
\end{eqnarray*}

Even if the space of convex bodies is not a  vector space, from its properties the mixed-area reminds a symmetric bilinear form, whose kernel is the space of  points, and whose isotropic cone is the space of points and segments. Moreover the Alexandrov--Fenchel inequality \eqref{eq alex fench} reminds a reversed Cauchy--Schwarz inequality. In the present paper we give the good framework to formalize those analogies.
Apart from our main results that are stated below, we will also give proofs of the properties mentioned above within this framework ---although they are classical, as intrinsic volumes are particular cases of mixed-volumes \cite{schneider}.

\subsection*{The area metric on the space of shapes}

In the sequel we denote by $\mathcal{K}^n$ the set of convex bodies in $\mathbb{R}^n$, and by $\mathcal{K}^{n*}$ the subset of convex bodies of positive intrinsic area. In other terms,  by \ref{prop 1V2} and  \ref{propV23},  $\mathcal{K}^{n*}$  is $\mathcal{K}^n$ minus points and segments.

By property \ref{alex fench2} of the mixed-area, for any $K_1,K_2 \in \mathcal{K}^{n*}$ we can set
$$\tilde{d}_1(K_1,K_2) = \operatorname{argch}\left( \frac{V_2(K_1,K_2)}{\sqrt{V_2(K_1)V_2(K_2)}}\right)~.
$$
This is clear that this quantity is invariant under  positive homotheties of $K_1$ and $K_2$.
Moreover, by \ref{isom} and \eqref{def mixed v2},
for all $p\in\mathbb{R}^n$,
$$V_2(K_1+\{p\},K_2)=V_2(K_1,K_2+\{p\})=V_2(K_1,K_2)~, $$
hence $\tilde{d}_1$ is invariant under translations of $K_1$ or $K_2$. By the case of equality in property \ref{alex fench2}, $\tilde{d}_1(K_1,K_2)=0$ if and only if $K_1$ is the image of $K_2$ by a translation and a positive homothety.

Let us define the space $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n}$ (resp. $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$) as the quotient of $\mathcal{K}^{n}$ (resp. $\mathcal{K}^{n*}$) by the action of translations and positive homotheties. For a convex body $K$, we denote by $[K]$ the \emph{oriented shape} of $K$, that is the equivalence class of $K$ for the action of translations and positive homotheties.
We can then define the \emph{area metric} on $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$: for any $[K_1],[K_2] \in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ we set
$$ d_1([K_1],[K_2])=\tilde d_1(K_1,K_2)~.$$

 Let $K_1,K_2 \in \mathcal{K}^{n*}$. Assume that $V_2(K_1)=V_2(K_2)=a>0$. Assume also that $[K_1] \neq [K_2]$, that is, $K_1$ is not the image of $K_2$ by a translation and a positive homothety. Consider the following equation:
\begin{equation}\label{equation}
V_2((1-t)K_1+t K_2)=0~.
\end{equation}
By properties of the mixed-area, the left-hand side is a polynomial in $t$, and the coefficient of $t^2$ is $2a-2V_2(K_1,K_2)$.
Since $[K_1] \neq [K_2]$, by Alexandrov-Fenchel's inequality \ref{alex fench} we have $V_2(K_1,K_2) > a$: the coefficient of $t^2$ is negative, in particular this is a second order polynomial. An easy calculation shows that its discriminant is equal to $4\delta(K_1,K_2) >0$ (see \eqref{eq alex fench}). Let $t_1<0<1<t_2$ be the two solutions of the equation \eqref{equation}, and let us define
$$\tilde{d}_2(K_1,K_2)=\frac{1}{2} \ln [0,1,t_1,t_2]~,$$
 where  $[0,1,t_1,t_2]$  is the cross-ratio (see Section~\ref{sec proj model}).
By \eqref{equation}, it is clear that $\tilde{d}_2$ is invariant by translation of $K_1$ or $K_2$. Let $[K_1], [K_2]\in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, and let $K_1, K_2$ be two
representatives having the same intrinsic area. We can then define
$$d_2([K_1],[K_2])=\tilde{d}_2(K_1,K_2)~,$$
if $[K_1]\not= [K_2]$, and zero otherwise.

Classical trigonometry will show that  $d_1=d_2$ (see Section~\ref{sec proj model}). We will denote
it as $\d$.
For futur reference, let us note that in particular we have, by \eqref{V1B} and \eqref{V2 avec V1},
\begin{equation}\label{eq distance a B}
\d([B^n],[K])=\operatorname{argch}\left(\sqrt{\frac{n-1}{\pi}}W_{n-1} \frac{V_1(K)}{\sqrt{V_2(K)}}\right)~.
\end{equation}
The main part of the present article is to prove the following properties for $\d$.

\begin{theoremO}\label{main}
$(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is a uniquely geodesic proper  metric space of infinite Hausdorff dimension and infinite diameter.
The unique shortest path between $[K_1]$ and $[K_2]$ is  $[(1-t) K_1+t K_2]$, $t\in [0,1]$. Any element of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is the endpoint of a shortest path that is not extendable beyond this point.
Moreover, $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ has curvature bounded from below and above by $-1$ in the sense of Alexandrov.

Its boundary is $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^n \setminus \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ and is homeomorphic to the real projective space of dimension $(n-1)$.
\end{theoremO}

Let us recall the following definitions and basic properties:
\begin{itemize}
\item a metric space is \emph{geodesic} if any two points are joined by a shortest path, it is \emph{uniquely geodesic} if the shortest path is unique;
\item a metric space is \emph{proper} if every bounded closed subset is compact, and a proper  metric space is locally compact and
complete;
\item a shortest path is \emph{extendable} if it is strictly contained in another shortest path;
\item the boundary of a metric space  is the set of equivalence classes of geodesic rays at bounded distance, endowed with a natural topology \cite{BH}.
\item see Definition~\ref{def cbb} for the property to have bounded curvature in the sense of Alexandrov.
\end{itemize}

Some assertions in Theorem~\ref{main} are rather straightforward. The description of the boundary follows because it is the set of segments up to translations and positive homotheties. $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is infinite dimensional because it contains finite dimensional hyperbolic convex polyhedra of arbitrary dimension constructed in \cite{BG} and \cite{FI}.

The fact that $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is  proper is non immediate, as it will follow from (the Blaschke selection theorem and) a theorem of R.A.~Vitale, see Section~\ref{haus}.

The other properties stated in Theorem~\ref{main} will be a consequence of the description of the extrinsic geometry of $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$: it is isometric to a convex subset of a hyperbolic space of infinite dimension, see the next section.

It is interesting to compare the fact that $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ has curvature bounded from above and below to the following classical result of V.~Beretovskij:   any locally compact  metric space with curvature bounded from above and below and such that shortest paths are extendable is isometric to  a finite dimensional Riemannian manifold \cite{B,BN}. Here we have a strong property of non-extendability, proved in Section~\ref{sec terminal}.

\subsection*{Convex subset of infinite dimensional hyperbolic space}

Mimicking the finite dimensional case,  given any separable Hilbert space $H$ of infinite dimension, one can define a subset which will deserve the name infinite dimensional hyperbolic space, see Section~\ref{sec hyp}.
In the present paper, we will define the infinite dimensional hyperbolic space  $\mathbb{H}_n^\infty$
by using as Hilbert space a vector subspace of  the Sobolev space $H^1(\mathbb{S}^{n-1})$, endowed with a bilinear form related to the intrinsic area, see Section~\ref{sec_laplacian}.
The support function of convex bodies will give a map from $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$
into $\mathbb{H}_n^\infty$. More precisely we have the following.

\begin{theoremO}\label{main1b}
 $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is isometric to an infinite dimensional closed convex subset with empty interior and ideal points of  $\mathbb{H}_n^\infty$.
\end{theoremO}

The construction of the infinite dimensional hyperbolic space $\mathbb{H}_n^\infty$ actually depends on $n$. Although for different $n$, all the resulting $\mathbb{H}_n^\infty$
are isometric (as separable infinite dimensional Hilbert spaces are all isometric), it is interesting to keep in mind this dependence,
as we will define canonical totally geodesic isometric embeddings of $\mathbb{H}_n^\infty$ into
$\mathbb{H}^\infty_{n+1}$, that have the following interpretation for convex bodies.
For $P$ a $k$-dimensional vector subspace of $\mathbb{R}^n$, let us denote by  $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}_P$
the subset  of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$  made of
oriented shapes of convex bodies in $\mathbb{R}^n$ contained in an affine space  directed by $P$. It will be
obvious from the proof of Theorem~\ref{main1b} that
$\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}_P$ is a convex subspace of $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$, isometric to $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{k*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^k})$. See also Section~\ref{infini}.

We don't know if some results of hyperbolic geometry can be translated into results for convex bodies.

\subsection*{Space of shapes}

In  Section~\ref{sec shapes} we investigate $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, the quotient of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ by linear isometries of the Euclidean space $\mathbb{R}^n$: $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is the space of convex bodies in $\mathbb{R}^n$ (not reduced to points or segments) up to Euclidean similarities.

From general properties
of quotient of metric spaces with curvature bounded from below by $-1$ (in short, CBB($-1$)) by a compact isometry group (see Section~\ref{sec shapes}), the metric $ \d$ induces a metric $d_{\mathscr{S}^n}$ on  $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, with the following properties.

\begin{theoremO}\label{main2}
$(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},d_{\mathscr{S}^n})$ is a CBB($-1$) proper geodesic metric space with boundary reduced to a point. It is not uniquely geodesic. It contains many totally geodesic hyperbolic surfaces.
\end{theoremO}

The fact that $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},d_{\mathscr{S}^n})$ is not uniquely geodesic implies that it is not CAT($0$), hence not CAT($-1$). However we don't know if it could be locally CAT($0$).

The case $n=3$ is of some interest, as
by famous results of Alexandrov and Pogorelov, there is a natural homeomorphism between $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*},d_{\mathscr{S}^3})$  and the space of metrics of non-negative curvature on the sphere $\S^2$ of total area one, up to isometries (endowed with the Gromov--Hausdorff topology). In turn, we are providing a CBB($-1$) metric on this space of metrics on the sphere, see Section~\ref{section metrics}. This is close from a celebrated construction of W.P.~Thurston \cite{th}
using the area form in order to give a (complex) hyperbolic structure on the space of flat metrics with prescribed conical singularities of positive curvature on the sphere $\mathbb{S}^ 2$, up to orientation-preserving isometries ---see \cite{FIgr} for the relations between Thurston's construction and the one of the present paper.

As an example of open questions related to the present work, the last section of the present paper introduce the inductive limits of
 $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ and $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, that is allowed by the intrinsic nature of the distances we defined.
They are spaces of the oriented shapes and shapes of all the convex bodies.

Let us mention the fact that a similar construction can be performed for spaces of convex set in
a Lorentzian Minkowski space (instead of convex sets of the Euclidean space as in the present paper), which are invariant under a cocompact lattice of linear isometries. For such sets, a Steiner formula holds, but the intrinsic area form leads to a positive definite form on the suitable Sobolev space. In turn, one obtains a convex subset of an infinite dimensional spherical (instead of hyperbolic) space. We refer to \cite{FF1,FIgr} for more details.

\subsection*{Notations}
To avoid confusions, let us describe here the notations we will use in the article. Everything that appears here will be defined precisely later.

$\mathcal{K}^n$ (resp. $\mathcal{K}^{n*}$) is the set of convex bodies in $\mathbb{R}^n$ (resp. convex bodies with positive intrinsic area), and $\mathcal{K}^n_S$ (resp. $\mathcal{K}^{n*}_S$) is the subset of convex bodies with Steiner point at the origin (resp. convex bodies with positive intrinsic area).

In the sequel,  a star as upperscript mean that we consider only convex bodies with positive intrinsic area (that is, we exclude points and segments). In the following table, it is obvious that all the sets without a star are in bijection, as well as all the sets with a star:

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
\textbf{convex bodies} & \textbf{up to positive} &  \textbf{with $V_2=1$} & \textbf{with $V_1=1$} \\
\textbf{in $\mathbb{R}^n$...} & \textbf{homotheties} &  & \\
\hline
\textbf{up to translations} & $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^n$ and $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ & &\\
\hline
\textbf{with Steiner point} & $\mathcal{K}^n_{SH}$ and $\mathcal{K}^{n*}_{SH}$  & $\mathcal{K}^n_{SV_2}$ and $\mathcal{K}^{n*}_{SV_2}$ & $\mathcal{K}^n_{SV_1}$ and $\mathcal{K}^{n*}_{SV_1}$ \\
\textbf{at the origin} & & & \\
\hline
\end{tabular}
\end{center}
We have $
\operatorname{Supp}(\mathcal{K}_S^{n*}) \subset \mathcal{C}_n
$ and
\begin{center}
\begin{tabular}{ c c c}
$\operatorname{Supp}(\mathcal{K}^{n*}_{SH}) \subset \mathbb{H}^\infty_n$, & $\operatorname{Supp}(\mathcal{K}^{n*}_{SV_2}) \subset \mathcal{H}^\infty_n$, & $\operatorname{Supp}(\mathcal{K}^{n*}_{SV_1}) \subset \mathbb{K}\mathbf{lein}^\infty_n$~. \\
\end{tabular}
\end{center}
The map $\operatorname{Supp}$ defines isometries
\begin{eqnarray*}
(\mathcal{K}^{n*}_{SH},d_{SH}) &\overset{\sim}\longrightarrow &(\operatorname{Supp}(\mathcal{K}^{n*}_{SH}),d_\mathbb{H})~, \\
(\mathcal{K}^{n*}_{SV_2},d_{SV_2}) &\overset{\sim}\longrightarrow &(\operatorname{Supp}(\mathcal{K}^{n*}_{SV_2}),d_\mathcal{H})~, \\
(\mathcal{K}^{n*}_{SV_1},d_{SV_1})& \overset{\sim}\longrightarrow &(\operatorname{Supp}(\mathcal{K}^{n*}_{SV_1}),d_\mathbb{K})~,
\end{eqnarray*}
and all these sets are isometric to $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$.

\paragraph*{Acknowledgements.}
The authors want to thank Nicola Gigli, Julien Maubon, Graham Smith, Pierre-Damien Thizy and Giona Veronelli for useful conversations. This work was completed during a visit of the second author at SISSA. He wants to thank the institution for its hospitality.

\section{Infinite dimensional Minkowski space}

In this section, we recall elementary properties of the spherical Laplacian and we introduce the bilinear form $\overline{V_2}^n$ on a Sobolev space, that will be the main object of this paper.

\subsection{Laplacian on the sphere}\label{sec_laplacian}

Let us denote by $\|\cdot\|_{L^2}$ the $L^2$ norm on the round sphere $\S^{n-1}$.
Let $H^1(\mathbb{S}^{n-1})$ be the Sobolev space of $\S^{n-1}$, i.e. the space  of functions $\S^{n-1}\to \mathbb{R}$ which are in $L^2(\S^{n-1})$
 as well as their first order derivatives in the weak sense. The space $H^1(\mathbb{S}^{n-1})$ is implicitly endowed with  the norm
$$\|h\|_{H^1}=\left(\|h\|_{L^2}^2 + \| \nabla h\|_{L^2}^2\right)^{1/2}
=\left(\int_{\mathbb{S}^{n-1}}h^2+\|\nabla h\|^2\right)^{1/2} $$
where the gradient $\nabla$ is the one of the round sphere.

Let us recall basic facts about  the Laplace--Beltrami operator $\Delta$ on $\mathbb{S}^{n-1}$. A reference for the results mentioned here is \cite{chavel}.
The first eigenvalues are denoted as follows:

$$0=\lambda_0<\lambda_1=n-1<\lambda_2 < \cdots~. $$

The eigenspace associated to $\lambda_0$ is the space of constant functions.
We will denote by $H^1(\mathbb{S}^{n-1})_0$ the subspace
of $H^1(\mathbb{S}^{n-1})$ of functions $L^2$-orthogonal
to the constant functions, i.e.
$$H^1(\mathbb{S}^{n-1})_0=\{h\in H^1(\mathbb{S}^{n-1}) | \left(h,1 \right)_{L^2}=0 \}~, $$
where $\left(\cdot,\cdot\right)_{L^2}$ is the $L^2$ scalar product on $\S^{n-1}$, and $1$ is the constant function equal to $1$ on $\S^{n-1}$.
Obviously, this is also the space of functions in $H^1(\mathbb{S}^{n-1})$ which are $H^1$-orthogonal to the constant functions.

The eigenspace associated to $\lambda_1$ is the vector space
spanned by the restrictions to the sphere of the
coordinates functions of $\mathbb{R}^{n}$
(we identify the round sphere $\S^{n-1}$ with the sphere of unit vectors in the Euclidean space $\mathbb{R}^n$). We will denote by $H^1(\mathbb{S}^{n-1})_1$ the subspace
of $H^1(\mathbb{S}^{n-1})$ of functions $L^2$-orthogonal
to the eigenspace of $\lambda_1$, i.e.
\begin{eqnarray*}
H^1(\mathbb{S}^{n-1})_1 & = & \{h\in H^1(\mathbb{S}^{n-1}) | \left( h,x^i\right)_{L^2}=0~, i=1,\ldots,n \} \\
& = & \{h\in H^1(\mathbb{S}^{n-1}) | \int_{\S^{n-1}} h(x)x \,\mbox{d}\S^{n-1}(x)=0\}~.
\end{eqnarray*}
Note that this is also the space of functions $H^1$-orthogonal to the vector space spanned by the restrictions to the sphere of the
coordinates functions of $\mathbb{R}^{n}$. Indeed, for $i \in \{1,\ldots,n\}$ we have $\Delta x^i= \lambda_1 x^i$, hence $\int h x^i=0$  implies
$\int h \Delta x^i=0$, and by Stoke's theorem $ \int \langle\nabla h, \nabla x^i\rangle = 0$.
Let us denote
$$H^1(\mathbb{S}^{n-1})_{01}=H^1(\mathbb{S}^{n-1})_0 \cap H^1(\mathbb{S}^{n-1})_1~. $$
By Rayleigh theorem,
 for $h\in H^1(\mathbb{S}^{n-1})_{01}\setminus \{0\}$ we have
\begin{equation}\label{rayleigh2}\lambda_2 \leq \frac{ \| \nabla h \|_{L^2}^2}{\|h\|_{L^2}^2}~.\end{equation}

\subsection{The form $\overline{V_2}^n$}

We are interested in the following quadratic form  on $H^1(\mathbb{S}^{n-1})$, which will be related to the intrinsic area $V_2$ in Section~\ref{sec:pt area}.
For $h\in H^1(\mathbb{S}^{n-1})$,
\begin{equation}\label{eq def A}\overline{V_2}^n(h)=c_{n}\left(\|h\|_{L^2}^2 -\lambda_1^{-1} \|\nabla h\|_{L^2}^2\right)~, \end{equation}
where $c_{n}$ is a positive dimensional constant (recall that $\lambda_1=n-1$). Its precise value
 will be relevant only in Section~\ref{Canonical embeddings}.
 This quadratic form comes from  the following bilinear form $\overline{V_2}^n(\cdot,\cdot)$: for $h,k\in H^1(\mathbb{S}^{n-1})$,
$$\overline{V_2}^n(h,k)=c_{n} \left( \mathbf{(}h,k\mathbf{)}_{L^2} - \lambda_1^{-1}(\nabla h,\nabla k)_{L^2}\right)~.$$
To avoid confusions, let us emphasis that
$$\overline{V_2}^n(h,h)=\overline{V_2}^n(h)~. $$

\begin{fact}\label{noyau A}
The kernel of $\overline{V_2}^n(\cdot,\cdot)$ on $H^1(\mathbb{S}^{n-1})$ is the eigenspace of $\lambda_1$. In turn,
	$\overline{V_2}^n$ is non-degenerate on $H^1(\mathbb{S}^{n-1})_1$.
\end{fact}

\begin{proof}
Let $h \in H^1(\mathbb{S}^{n-1})$. The function $h$ belongs to the kernel of $\overline{V_2}^n(\cdot,\cdot)$ if and only if for any $k \in H^1(\mathbb{S}^{n-1})$ we have
$$
\int_{\S^{n-1}} hk = \lambda_1^{-1} \int_{\S^{n-1}}  \langle \nabla h, \nabla k \rangle~.
$$
By density of smooth functions on $\S^{n-1}$ for the $H^1-$norm and by Green's formula, this is equivalent to the following property: for any smooth function $k$ on $\S^{n-1}$ we have
$$
\int_{\S^{n-1}} hk = \lambda_1^{-1} \int_{\S^{n-1}} h \Delta k~,
$$
and this means $h = \lambda_1^{-1} \Delta h$ in the weak (hence smooth) sense.
\end{proof}

In the following, we will consider the restriction of $\overline{V_2}^n$  to  $H^1(\mathbb{S}^{n-1})_1$, on which it  is a non-degenerate bilinear form. Furthermore, we will see that the restriction of $\overline{V_2}^n$ to $H^1(\mathbb{S}^{n-1})_{01}$ is negative definite.

\begin{lemma} \label{lemme equiv distance} For $h\in H^1(\mathbb{S}^{n-1})_{01}$,
\begin{equation}\label{lem:laplacien}
c_{n}\left(\frac{\lambda_2-\lambda_1}{\lambda_1}\right)\|h\|_{L^2}^2 \leq -\overline{V_2}^n(h)\end{equation}
 and
\begin{equation}\label{lem:laplacienH1}
c_{n} \left( \frac{\lambda_2 - \lambda_1}{\lambda_1\lambda_2} \right) \|h\|^2_{H^1} \leq - \overline{V_2}^n(h) \leq c_{n}\frac{1}{\lambda_1} \|h\|^2_{H^1}~.
\end{equation}
\end{lemma}

\begin{proof}
\eqref{lem:laplacien} is immediate from
\eqref{rayleigh2}, and
the right-hand side inequality in \eqref{lem:laplacienH1} follows from
$$-\overline{V_2}^n(h)\leq c_{n}\lambda_1^{-1}\|\nabla h\|^2_{L^2}\leq c_{n} \lambda_1^{-1}\|h\|^2_{H^1}~. $$
The left-hand side inequality in \eqref{lem:laplacienH1} follows by adding the two following inequalities:
as $\lambda_2>\lambda_1=n-1 \geq 1$, \eqref{lem:laplacien} gives
$$c_{n}\frac{1}{\lambda_2}\left(\frac{\lambda_2-\lambda_1}{\lambda_1}\right)\|h\|^2_{L^2}\leq -\overline{V_2}^n(h)~, $$
and
on the other hand, using again \eqref{rayleigh2}, the equality \eqref{eq def A} gives
$$c_{n}\left(\frac{1}{\lambda_1 }-\frac{1}{\lambda_2}\right)\|\nabla h\|_{L^2}^2\leq -\overline{V_2}^n(h)~. $$

\end{proof}

\begin{proposition}\label{prop:hilbert}
$(H^1(\mathbb{S}^{n-1})_{01},-\overline{V_2}^n(\cdot,\cdot))$ is a separable Hilbert space.
\end{proposition}
\begin{proof}
By \eqref{lem:laplacien} or \eqref{lem:laplacienH1},
$-\overline{V_2}^n$ is a scalar product on $H^1(\mathbb{S}^{n-1})_{01}$.
As $H^1(\mathbb{S}^{n-1})_{01}$ is orthogonal  to a vector subspace, it is a closed subspace, hence complete and separable for the $H^1$ norm.  The result follows from \eqref{lem:laplacienH1}.

 \end{proof}

Let $\operatorname{L}$ be the line of constant functions in $H^1(\mathbb{S}^{n-1})_1$.
For $h\in H^1(\mathbb{S}^{n-1})_1$, let us introduce the number
$$ \mathbf{mean}(h)= \frac{1}{\operatorname{vol}(\mathbb{S}^{n-1})}\int_{\S^ {n-1}}h~.$$
Note that we have $\mathbf{mean}(h) = c_n^{-1} \operatorname{vol}(\S^{n-1})^{-1} \overline{V_2}^n(h,1)$.
Abusing notation, we will denote by $\mathbf{mean}$ the function from  $H^1(\mathbb{S}^{n-1})_1$ to $\operatorname{L}$ sending $h$ to the function on the sphere having constant value $\mathbf{mean}(h)$.
Obviously $H^1(\mathbb{S}^{n-1})_{01}$ is the set of elements $h$ of $H^1(\mathbb{S}^{n-1})_1$ such that $\mathbf{mean}(h)=0$, and from the definitions  it is immediate that
$H^1(\mathbb{S}^{n-1})_{01}$ and $\operatorname{L}$ are $\overline{V_2}^n$-orthogonal.

For future reference, note that for any $h \in H^1(\mathbb{S}^{n-1})_1$ we have $h-\mathbf{mean}(h)\in H^1(\mathbb{S}^{n-1})_{01}~,$
hence
  for any constant function $c$,
$$\overline{V_2}^n(h-\mathbf{mean}(h),c)=0~. $$
For any $h,k \in H^1(\mathbb{S}^{n-1})_1$ that gives, writing $h=(h-\mathbf{mean}(h))+\mathbf{mean}(h)$ and $k=(k-\mathbf{mean}(k))+\mathbf{mean}(k)$,
\begin{equation}\label{eq:ecrit supp}\overline{V_2}^n(h,k)=\overline{V_2}^n(h-\mathbf{mean}(h),k-\mathbf{mean}(k))+\mathbf{mean}(h)\mathbf{mean}(k)\overline{V_2}^n(1)~.\end{equation}

\subsection{Inequalities and signature}\label{section:ineq and sign}

Let us introduce the following cone (see Figure~\ref{fig:notations})

\begin{equation}\label{def cone}\mathcal{C}_n=\{h\in H^1(\mathbb{S}^{n-1})_1 | \overline{V_2}^n(h)> 0, \mathbf{mean}(h)>0 \}~,\end{equation}
and let
$$\overline{\mathcal{C}}_n=\{h\in H^1(\mathbb{S}^{n-1})_1 | \overline{V_2}^n(h)\geq  0, \mathbf{mean}(h)>0 \}~.$$
\begin{fact}\label{factRCSbis}
 For any $h,k\in\mathcal{C}_n$ we have
\begin{equation}\label{eq pos}\overline{V_2}^n(h,k)> 0
\end{equation}
and for any $h,k \in \overline{\mathcal{C}}_n$ we have
\begin{equation}\label{eq pos bis}\overline{V_2}^n(h,k) \geq 0~.
\end{equation}
\end{fact}
\begin{proof}
Notice that \eqref{eq pos bis} is a consequence of \eqref{eq pos}: indeed, for every $h,k \in \overline{\mathcal{C}}_n$ and $\epsilon>0$ we have $h + \epsilon,k + \epsilon \in \mathcal{C}_n$: we have
$$
\overline{V_2}^n(h+\epsilon) = \overline{V_2}^n(h) +2\epsilon \overline{V_2}^n(h,1) + \epsilon^2 \overline{V_2}^n(1)>0
$$
(remember that $\overline{V_2}^n(h,1)=c_n \operatorname{vol}(\S^{n-1}) \mathbf{mean}(h)>0$), and the same holds for $\overline{V_2}^n(k+\epsilon)$.
So if \eqref{eq pos} is true we have
$$\overline{V_2}^n(h+\epsilon,k+\epsilon) =\overline{V_2}^n(h,k) + \epsilon\overline{V_2}^n(h,1)+\epsilon\overline{V_2}^n(1,k)+\epsilon^2 \overline{V_2}^n(1)>0~,$$
and when $\epsilon$ goes to zero this gives \eqref{eq pos bis}.

Now, let us prove \eqref{eq pos}. If $h\in \mathcal{C}_n$, then $\overline{V_2}^n(h)>0$, hence by \eqref{eq:ecrit supp},
$$0<\overline{V_2}^n(h)=\overline{V_2}^n(h-\mathbf{mean}(h)) + \mathbf{mean}(h)^2 \overline{V_2}^n(1)~. $$
As $-\overline{V_2}^n$ is non-negative on $H^1(\mathbb{S}^{n-1})_{01}$ (Lemma~\ref{lemme equiv distance}), this gives
$$0\leq -\overline{V_2}^n(h-\mathbf{mean}(h))< \mathbf{mean}(h)^2 \overline{V_2}^n(1)~. $$
  Using the classic Cauchy--Schwarz inequality in $H^1(\mathbb{S}^{n-1})_{01}$ (Proposition~\ref{prop:hilbert})
and the above equation we obtain
\begin{eqnarray*}\ -\overline{V_2}^n(h-\mathbf{mean}(h),k-\mathbf{mean}(k))&\leq& \sqrt{-\overline{V_2}^n(h-\mathbf{mean}(h))}\sqrt{-\overline{V_2}^n(k-\mathbf{mean}(k))}\\ \ &<&\mathbf{mean}(h)\mathbf{mean}(k) \overline{V_2}^n(1)\end{eqnarray*}
(recall that $\mathbf{mean}(h)$ and $\mathbf{mean}(k)$ are both positive).
The result follows from \eqref{eq:ecrit supp}.
\end{proof}

The following is an immediate consequence of \eqref{eq pos}.

\begin{fact}\label{fact C cvxe}
The cone $\mathcal{C}_n$ is convex.
\end{fact}

The following fact explains why $(H^1(\mathbb{S}^{n-1})_1,\overline{V_2}^n)$ may be considered as an infinite dimensional Minkowski space (in the Lorentzian sense).

\begin{fact}\label{signature}
The restriction of $\overline{V_2}^n(\cdot,\cdot)$ to any
vector subspace of finite dimension $p$ of $H^1(\mathbb{S}^{n-1})_1$, containing an element of $\mathcal{C}_n$, has Lorentzian signature $(+,-,\ldots,-)$, with 1 positive direction and $p-1$ negative directions.
\end{fact}
\begin{proof}
By definition, $\overline{V_2}^n$ is positive when evaluated on any vector of $\mathcal{C}_n$, and the intersection of the vector space with  $H^1(\mathbb{S}^{n-1})_{01}$ has dimension $(p-1)$, on which $\overline{V_2}^n$ is negative definite from Proposition~\ref{prop:hilbert}.
\end{proof}

\begin{fact}[Reversed Cauchy--Schwarz inequality]\label{factRCS}
 For any $h,k\in \overline{\mathcal{C}}_n$, we have
 \begin{equation}\label{RCS}\overline{V_2}^n(h,k)^2 \geq \overline{V_2}^n(h)\overline{V_2}^n(k)~, \end{equation}
or if one prefers, by Fact \ref{factRCSbis},
 \begin{equation} \overline{V_2}^n(h,k) \geq \overline{V_2}^n(h)^{1/2}\overline{V_2}^n(k)^{1/2}~. \end{equation}
Equality holds if and only if $h=\lambda k$, $\lambda>0$.
\end{fact}
\begin{proof}
The equality is obvious if $h,k\in \overline{\mathcal{C}}_n$ are colinear.

Let $h,k\in \mathcal{C}_n$ be non-colinear. Then they span a plane, which, by Fact~\ref{signature}, contains a positive and a negative vector for $\overline{V_2}^n$.
So  there exists a $t$ such that $\overline{V_2}^n(h+tk)$ is positive, and a $t$ such that  $\overline{V_2}^n(h+tk)$ is negative, hence the
second order polynomial $t \mapsto \overline{V_2}^n(h+tk)$ has a positive discriminant. This gives \eqref{RCS} with a strict inequality.

Now let $h,k \in \overline{\mathcal{C}}_n$ be non-colinear. If $h,k \in \mathcal{C}_n$ then we know that \eqref{RCS} holds with a strict inequality, so for example assume that $\overline{V_2}^n(k)=0$. Then $\overline{V_2}^n(h+tk)=\overline{V_2}^n(h)+2t\overline{V_2}^n(h,k)$, and $h$ and $k$ span a plane, hence there exists some $t$ such that $\overline{V_2}^n(h+tk)<0$. This shows that $\overline{V_2}^n(h,k) \neq 0$, hence $\overline{V_2}^n(h,k)^2>0=\overline{V_2}^n(h)\overline{V_2}^n(k)$.
\end{proof}

 \subsection{Intrinsic nature of $\overline{V_1}^n$ and $\overline{V_2}^n$}\label{Canonical embeddings}

We settle that, in the definition of $\overline{V_2}^n$,
\begin{equation}\label{def cn}c_{n}=\frac{(n-1)W_{n-1}}{\kappa_{n-1}}\stackrel{\eqref{V1B}}{=}\frac{n-1}{2\kappa_{n-2}}\end{equation}
(recall that $W_n$ is the Wallis' integral). For future reference, let us note that in particular we have $c_2=1/2$, and that the dimensional constant $r_2(n)$ introduced in \eqref{v1v2} satisfies
$$r_2(n)=\overline{V_2}^n(1)^{-1/2}~.$$
This definition of $c_n$ allows to prove that, for natural embeddings of $H^1(\mathbb{S}^{n-1})$ into $H^1(\mathbb{S}^{n})$ (defined below), the corresponding value of $\overline{V_2}^n$ will not change.

In $\mathbb{R}^{n+1}$, choose a point $N \in \S^n$, and choose a direct orthonormal base $\mathcal{B}$ of the hyperplane $N^\bot$, such that $(\mathcal{B},N)$ is a direct orthonormal basis of $\mathbb{R}^{n+1}$. Let $R_{\mathcal{B},N} \in SO(n+1)$ be the unique rotation sending the canonical basis of $\mathbb{R}^{n+1}$ on $(\mathcal{B},N)$. It is clear that $R_{\mathcal{B},N}$, restricted to $\mathbb{S}^{n-1} \equiv \mathbb{S}^{n-1}\times \{0\} \subset \mathbb{R}^{n+1}$ is an isometric embedding of $\mathbb{S}^{n-1}$ into $\mathbb{S}^n$ (given by the intersection $N^\bot \cap \mathbb{S}^{n}$). Moreover, every such embedding is obtained in this way.

This gives the following coordinates on $\S^n\setminus\{N,-N\}$:
$$\Phi_{\mathcal{B},N}: \mathbb{S}^{n-1}\times ]-\frac{\pi}{2},\frac{\pi}{2}[\to \S^n,\, (x,t)\mapsto R_{\mathcal{B},N}(\cos(t)x,\sin(t))~. $$

\begin{remark}\label{remarkjacobien}{\rm
Assume that $R_{\mathcal{B},N}$ is the identity, and let $\Phi = \Phi_{\mathcal{B},N}$. Then an orthonormal basis of the tangent space of $\mathbb{S}^n$ at the point $\Phi(x,t)$ is
$$\left( \frac{1}{\cos(t)} \nabla _{\partial \theta_1}\Phi (x,t), \ldots, \frac{1}{\cos(t)} \nabla_{\partial \theta_{n-1}} \Phi (x,t), \nabla_{\partial t} \Phi (x,t) \right)~,$$
where $\partial \theta_1,\ldots, \partial \theta_{n-1}$ is an orthonormal basis of the tangent space of $\mathbb{S}^{n-1}$ at the point $x$. In particular, the Jacobian of $\Phi$ at the point $(x,t)$ is $\cos^{n-1}(t)$.
}\end{remark}

For $h\in H^1(\mathbb{S}^{n-1})$ let us define
$E_{\mathcal{B},N}(h):\S^n \to \mathbb{R}$ by $0$ on $\{N,-N\}$, and otherwise
$$E_{\mathcal{B},N}(h)\circ \Phi_{\mathcal{B},N}(x,t)=\cos(t) h(x)~. $$
The geometric meaning of the  functions $E_{\mathcal{B},N}(h)$ will be clarified
in Section~\ref{sec supp} and Fact~\ref{factexplicationEh}.

\begin{proposition}
We have defined an injective linear map $E_{\mathcal{B},N}$ from $H^1(\S^{n-1})$ to 	$H^1(\S^n)$. Moreover, $E_{\mathcal{B},N}(H^1(\S^{n-1})_1)\subset H^1(\S^n)_1$ and  $E_{\mathcal{B},N}(H^1(\S^{n-1})_{01})\subset H^1(\S^n)_{01}$.
\end{proposition}

The first part of the proposition is obvious, and the two inclusions are direct consequences of the following lemma.
\begin{lemma}\label{lem EB}
For $h \in H^1(\mathbb{S}^{n-1})$ we have
\begin{equation}
\int_{\S^n} E_{\mathcal{B},N}(h)=2W_n \int_{\S^{n-1}}h~
\end{equation}
and
$$ \int_{\S^n} E_{\mathcal{B},N}(h)(v) v  = R_{\mathcal{B},N}\left(2W_{n+1}\int_{\S^{n-1}}h(u) u , 0 \right) \in \mathbb{R}^{n+1}~.$$
\end{lemma}
\begin{proof}
The Jacobian of $\Phi_N$ at the point $(x,t)$ is $\cos^{n-1}(t)$, hence a change of variable gives the first identity.

For the second one, with the same change of variable we obtain
$$\int_{\S^n} E_{\mathcal{B},N}(h)(v) v  = \int_{\mathbb{S}^{n-1}\times ]-\frac{\pi}{2},\frac{\pi}{2}[} \cos^n (t) h(x) R_{\mathcal{B},N}(\cos(t)x, \sin(t))~,$$
and this gives the result since $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n(t) \sin(t)\mbox{d} t=0$.
\end{proof}

Let us define
$$\overline{V_1}^n(h)=\frac{1}{\kappa_{n-1}}\int_{\S^{n-1}}h~.$$
Note that $\overline{V_1}^n(h)=\frac{n\kappa_n}{\kappa_{n-1}}\mathbf{mean}(h)$. As we will show later, this quantity is related to $V_1(K)$ for a convex body $K$. As $2W_n=\frac{\kappa_n}{\kappa_{n-1}}$,  the first formula in Lemma~\ref{lem EB}
says that
\begin{equation}\label{Vbar1}
\overline{V_1}^{n+1}( E_{\mathcal{B},N}(h))=\overline{V_1}^n(h)~.
\end{equation}

\begin{proposition}\label{prop A intrins}
Let $h\in H^1(\S^{n-1})$. The quantity $\overline{V_2}^n(h)$ is intrinsic in the following sense:
$$\overline{V_2}^{n+1}(E_{\mathcal{B},N}(h))=\overline{V_2}^n(h)~.$$
\end{proposition}

\begin{proof}
Since the integral of $h^2$ and $\|\nabla h\|^2$ are invariant by rotation, we may assume that $R_{\mathcal{B},N}$ is the identity (that is, the embedding of $\S^{n-1}$ in $\S^n$ is given by $\S^{n-1} \subset \mathbb{R}^n \times \{0\}$). For simplicity, let $E = E_{\mathcal{B},N}$ and $\Phi = \Phi_{\mathcal{B},N}$.

We have the relation
$$
\|\nabla E (h)(\Phi (x,t))\|^2 = \|\nabla h(x)\|^2 + \sin^2(t) h(x)^2~.$$
Indeed, using the notations of Remark \ref{remarkjacobien}, the relation $E(h)\circ \Phi (x,t)=\cos(t) h(x)$ and the chain rule formula we get
$$
\nabla E(h)(\Phi(x,t)) \cdot \left(\frac{1}{\cos(t)}\nabla_{\partial \theta_i} \Phi (x,t) \right) = \frac{1}{\cos(t)} \nabla_{\partial \theta_i} (E(h) \circ \Phi)(x,t)= \nabla_{\partial \theta_i} h(x)
$$
and
$$
\nabla E(h)(\Phi(x,t)) \cdot \nabla_{\partial t} \Phi (x,t) = \nabla_{\partial t} (E(h) \circ \Phi)(x,t) = -\sin(t)h(x)~.
$$
Hence by the change of variables given by $\Phi$ we obtain
\begin{eqnarray*}
\int_{\S^n} \|\nabla E(h)\|^2 & = & \int_{\S^{n-1}\times ]-\frac{\pi}{2}, \frac{\pi}{2}[} \cos^{n-1}(t)\left( \|\nabla h \|^2 + \sin^2(t) h^2 \right) \\
& =& 2 W_{n-1} \int_{\S^{n-1}} \|\nabla h\|^2 + 2(W_{n-1}-W_{n+1}) \int_{\S^{n-1}} h^2~,
\end{eqnarray*}
and by the same change of variables we get
$$ \int_{\S^n} E(h)^2  = \int_{\S^{n-1}\times ]-\frac{\pi}{2}, \frac{\pi}{2}[} \cos^{n-1}(t) \cos^2(t) h^2 = 2W_{n+1} \int_{\S^{n-1}} h^2~. $$
This gives
\begin{eqnarray*}
\overline{V_2}^{n+1}(E(h)) &=& c_{n+1} \left( \int_{\S^n} E(h)^2 - \frac{1}{n} \int_{\S^n} \|\nabla E(h)\|^2 \right) \\
& = & 2 c_{n+1} \left( \left(W_{n+1} - \frac{W_{n-1} - W_{n+1}}{n} \right) \int_{\S^{n-1}} h^2 - \frac{W_{n-1}}{n} \int_{\S^{n-1}} \|\nabla h\|^2 \right)~.
\end{eqnarray*}
Using the relation $ W_{n+1}=\frac{n}{n+1}W_{n-1}$ we get
$$ W_{n+1} - \frac{W_{n-1} - W_{n+1}}{n} = \frac{n+1}{n} W_{n+1} - \frac{1}{n}W_{n-1} =(n-1)\frac{W_{n-1}}{n}~,$$
and noting that $\frac{2 c_{n+1} W_{n-1} (n-1)}{n} = c_{n}$ for any $n>1$ (see equation \eqref{V1B}),
 we finally obtain
$$
\overline{V_2}^{n+1}(E(h)) = \frac{2c_{n+1} W_{n-1} (n-1)}{n} \left( \int_{\S^{n-1}} h^2 - \frac{1}{n-1} \int_{\S^{n-1}} \|\nabla h\|^2 \right) = \overline{V_2}^n(h)~.
$$
\end{proof}

\section{Infinite dimensional hyperbolic space}\label{sec hyp}

In this section, we use the bilinear form $\overline{V_2}^n$ defined in the preceding section, in order to define an infinite dimensional hyperbolic space. The content of this section is almost elementary. Actually, apart from some convergence properties, the constructions are formally the same as in the finite dimensional case. However we are provinding details for the sake of completeness.

We introduce three models of the infinite dimensional hyperbolic space:
\begin{itemize}
\item a hyperboloid model  $\mathcal{H}_n^\infty$ in $(H^1(\mathbb{S}^{n-1})_1,\overline{V_2}^n)$, which is in natural bijection with $H^1(\mathbb{S}^{n-1})_{01}$, that will allow to check some completeness properties;
\item a projective model $\mathbb{H}_n^\infty$, which allows to get rid of normalizations, and on which we can define a Hilbert distance;
\item a Klein ball model $\mathbb{K}\mathbf{lein}^\infty_n$, an affine chart of the projective model, in which the geodesics are very easy to describe (these are convex combinations).
\end{itemize}

\subsection{Hyperboloid model}

Let us introduce \begin{equation*}\label{hinf}
\mathcal{H}_n^\infty=\{h \in \mathcal{C}_n | \overline{V_2}^n(h)=1 \}~,
\end{equation*}
where $\mathcal{C}_n$ is the cone defined in \eqref{def cone}.

\begin{figure}
\begin{center}
\psfrag{C}{$\mathcal{C}_n$}
\psfrag{K}{$\mathbb{K}\mathbf{lein}^\infty_n$}
\psfrag{H}{$\mathcal{H}_n^\infty$}
\psfrag{H+}{$\mathbf{mean}>0$}
\psfrag{H01}{$H^1(\mathbb{S}^{n-1})_{01}$}
\psfrag{L}{$\operatorname{L}$}
\psfrag{1}{$r_1(n)$}
\psfrag{c}{$r_2(n)$}
\psfrag{P}{$P_n$}
\psfrag{V1}{$V_1=1$}
\psfrag{V}{$V_1=0$}
\includegraphics[width=0.5\linewidth]{Hyperbola}\caption{Notations for subspaces of $H^1(\mathbb{S}^{n-1})_1$.}\label{fig:notations}
\end{center}
\end{figure}

Let $\mathbf{p}_{\mathcal{H}} : \mathcal{H}_n^\infty \rightarrow H^1(\mathbb{S}^{n-1})_{01}$
be the map $\mathbf{p}_{\mathcal{H}}(h)=h-\mathbf{mean}(h)$.
 It is a bijection between $\mathcal{H}_n^\infty$ and $H^1(\mathbb{S}^{n-1})_{01}$, and its inverse is the map $\mathbf{p}_{\mathcal{H}}^{-1} : H^1(\mathbb{S}^{n-1})_{01} \rightarrow \mathcal{H}_n^\infty$ defined by $\mathbf{p}_{\mathcal{H}}^{-1} (h) = h + \mathbf{c}(h)$, where $\mathbf{c}(h)$ is the constant function on $\S^{n-1}$ defined as
$$\mathbf{c}(h)=r_2(n)\sqrt{1-\overline{V_2}^n(h)}~: $$
indeed, since $h \in H^1(\mathbb{S}^{n-1})_{01}$, we have $\overline{V_2}^n(h+\mathbf{c}(h))=\overline{V_2}^n(h) + \mathbf{c}(h)^2 \overline{V_2}^n(1)$, hence the equation $\overline{V_2}^n(h+\mathbf{c}(h))=1$ gives the formula above (recall that $r_2(n)=\overline{V_2}^n(1)^{-1/2}$).
Note that $\mathbf{c}(h)$ is well defined on $H^1(\mathbb{S}^{n-1})_{01}$, on which $\overline{V_2}^n\leq 0$.

As the Hilbert structure on $H^1(\mathbb{S}^{n-1})_{01}$ is given by $\overline{V_2}^n$, the map $\overline{V_2}^n$ is smooth, so  $\mathbf{c}:H^1(\mathbb{S}^{n-1})_{01}\to \mathbb{R}$ is also smooth.
It follows that as a graph, $\mathcal{H}_n^\infty$ is an infinite dimensional smooth manifold.
We implicitly endow it with the restriction of $-\overline{V_2}^n( \cdot,\cdot)$ on its tangent spaces.
The intersection of $\mathcal{H}_n^\infty$ with any vector $p$-plane of $H^1(\mathbb{S}^{n-1})_1$ containing a vector of $\mathcal{C}_n$ is a hyperboloid model of the hyperbolic space of dimension $(p-1)$ by Fact~\ref{signature}. In turn,
$\mathcal{H}_n^\infty$  is a Riemannian manifold of constant sectional curvature $-1$. However, it will deserve the name \emph{infinite dimensional hyperbolic space} once we will have checked its completeness, that is the content of the remainder of this section, see Corollary~\ref{cor complete}.

At a metric level,
Fact~\ref{factRCS} implies that for any $h,k\in \mathcal{H}_n^\infty$ we have
$\overline{V_2}^n(h,k)\geq 1$, so
 we can define the following:
 for $h,k\in \mathcal{H}_n^\infty$ we set
\begin{equation}\label{eq: dist hyperboloide}d_{\mathcal{H}}(h,k)=\operatorname{argch} \overline{V_2}^n\left(h,k \right)~. \end{equation}
Using again an intersection with a finite dimensional vector space, it is well known that $d_{\mathcal{H}}$ is indeed a distance, actually the one given by the Riemannian structure.
In the same way we obtain the following.
\begin{fact}\label{fact geod}
  $(\mathcal{H}_n^{\infty},d_{\mathcal{H}})$  is a uniquely  geodesic metric space, and the shortest path between $h$ and $k$ is the intersection of $\mathcal{H}_n^\infty$ with the vector plane spanned by $h$ and $k$.
 \end{fact}

 We will denote by  $d_{01}$ the distance induced by $\sqrt{-\overline{V_2}^n}$
on $H^1(\mathbb{S}^{n-1})_{01}$: for $h,k \in~H^1(\mathbb{S}^{n-1})_{01}$ we set
 $$d_{01}(h,k)=\sqrt{-\overline{V_2}^n(h-k)}~.$$

\begin{lemma}\label{proj0}
  The map  $\mathbf{p}_{\mathcal{H}} : \mathcal{H}_n^\infty \rightarrow H^1(\mathbb{S}^{n-1})_{01}$
is expanding:
for every $h,k \in \mathcal{H}_n^\infty$ we have
\begin{equation} \label{inegproj1}
d_{\mathcal{H}}(h,k) \leq d_{01}(\mathbf{p}_{\mathcal{H}} (h),\mathbf{p}_{\mathcal{H}} (k))~.
\end{equation}
\end{lemma}
\begin{proof}
For $h,k\in \mathcal{H}_n^\infty$, the inequality $\overline{V_2}^n(h,k) \geq 1+d_\mathcal{H}(h,k)^2/2$ follows directly from $\cosh(u) \geq 1+u^2/2$ and \eqref{eq: dist hyperboloide}, and with $\overline{V_2}^n(h-k)=2-2\overline{V_2}^n(h,k)$ we obtain
\[
d_\mathcal{H}(h,k) \leq \sqrt{-\overline{V_2}^n(h-k)}~.
\]
Moreover by \eqref{eq:ecrit supp}, for every $f \in H^1(\mathbb{S}^{n-1})_{1}$, $$-\overline{V_2}^n(f-\mathbf{mean}(f)) = -\overline{V_2}^n(f) + \mathbf{mean}(f)^2 \overline{V_2}^n(1) \geq -\overline{V_2}^n(f)~,$$ hence
$$\sqrt{-\overline{V_2}^n(\mathbf{p}_{\mathcal{H}}(h)-\mathbf{p}_{\mathcal{H}}(k))}=
\sqrt{-\overline{V_2}^n(h-k-\mathbf{mean}(h-k)))} \geq \sqrt{-\overline{V_2}^n(h-k)}$$
 and this gives the result.
\end{proof}

\begin{lemma}\label{lem c bilip}
The map  $\mathbf{c}$  is $(r_2(n)\tanh(t))$-Lipschitz on the ball  of radius $\sinh(t)$ centered
at $0$ in $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$.
\end{lemma}

\begin{proof}
For $u,v \geq 0$ we have the formula $\sqrt{1+u}-\sqrt{1+v} = \frac{\sqrt{u} + \sqrt{v}}{\sqrt{1+u} + \sqrt{1+v}} (\sqrt{u}-\sqrt{v})$, and this gives
\begin{eqnarray*}
|\mathbf{c}(h)-\mathbf{c}(k)| &=& r_2(n)\left|\sqrt{1-\overline{V_2}^n(h)}-\sqrt{1-\overline{V_2}^n(k)}\right| \\
& = & r_2(n)\frac{\sqrt{-\overline{V_2}^n(h)}+\sqrt{-\overline{V_2}^n(k)}}{\sqrt{1-\overline{V_2}^n(h)}+\sqrt{1-\overline{V_2}^n(k)}} \left|\sqrt{-\overline{V_2}^n(h)}-\sqrt{-\overline{V_2}^n(k)}\right|~.
\end{eqnarray*}
Moreover, $\left|\sqrt{-\overline{V_2}^n(h)}-\sqrt{-\overline{V_2}^n(k)}\right| = |d_{01}(0,h)-d_{01}(0,k)| \leq d_{01}(h,k)$, and there exists $s,s'<t$ such that $\sqrt{-\overline{V_2}^n(h)}=\sinh(s)$ and
$\sqrt{-\overline{V_2}^n(k)}=\sinh(s')$, hence $\frac{\sqrt{-\overline{V_2}^n(h)}+\sqrt{-\overline{V_2}^n(k)}}{\sqrt{1-\overline{V_2}^n(h)}+\sqrt{1-\overline{V_2}^n(k)}}=\tanh(\frac{s+s'}{2})\leq \tanh(t)$.

\end{proof}

 \begin{lemma}\label{proj}
 Let $O$ be the ball centered at $0$ and of radius $\sinh(t)$ in
 $H^1(\mathbb{S}^{n-1})_{01}$.
For every $h,k \in \mathbf{p}_{\mathcal{H}}^{-1}(O)\subset \mathcal{H}_n^\infty$,
 we have
\begin{equation} \label{inegproj}
(1-\tanh(t)) d_{01}(\mathbf{p}_{\mathcal{H}}(h),\mathbf{p}_{\mathcal{H}}(k)) \leq d_{\mathcal{H}}(h,k)~.
\end{equation}
\end{lemma}
\begin{proof}
Let $\gamma:[0,d_{\mathcal{H}}(h,k)]\to \mathcal{H}_n^\infty$ be
 the arc-length parametrized geodesic between $h$ and $k$ in $\mathcal{H}_n^\infty$.
We have
 $d_{\mathcal{H}}(h,k)=\int \sqrt{-\overline{V_2}^n((\mathbf{p}_{\mathcal{H}}^{-1}\circ\bar\gamma)')}$, with $\bar \gamma = \mathbf{p}_\mathcal{H}(\gamma)$. Note that
 $(\mathbf{p}_{\mathcal{H}}^{-1}\circ\bar\gamma)'=
 (\bar\gamma+c \circ \bar\gamma)'=\bar\gamma'+(c\circ \bar\gamma)'$. This gives, using the inequality $\sqrt{a-b} \geq \sqrt{a} - \sqrt{b}$ valid for $a \geq b \geq 0$,
$$\sqrt{-\overline{V_2}^n((\mathbf{p}_{\mathcal{H}}^{-1}\circ\bar\gamma)')}
=\sqrt{-\overline{V_2}^n(\bar\gamma')-(c\circ \bar\gamma)'^2r_2(n)^{-2}}
\geq \sqrt{-\overline{V_2}^n(\bar\gamma')} -
|(c\circ \bar\gamma)'|r_2(n)^{-1}$$
(recall that $\overline{V_2}^n(1) = r_2(n)^{-2}$).
By Lemma~\ref{lem c bilip}, $| (c \circ  \bar\gamma)'| \leq r_2(n)\tanh(t) \sqrt{-\overline{V_2}^n(\bar\gamma')}$. Finally we obtain
\[
d_{\mathcal{H}}(h,k) \geq (1-\tanh(t)) \int \sqrt{-\overline{V_2}^n(\bar\gamma')}~,
\]
and since $\int \sqrt{-\overline{V_2}^n(\bar\gamma')}$ is the length of a curve in
$H^1(\mathbb{S}^{n-1})_{01}$ between $\mathbf{p}_{\mathcal{H}}(h)$ and $\mathbf{p}_{\mathcal{H}}(k)$,  it is $\geq d_{01}(\mathbf{p}_{\mathcal{H}}(h),\mathbf{p}_{\mathcal{H}}(k))$ and this ends the proof.
\end{proof}

Lemma \ref{proj0} and \ref{proj} give the following.

\begin{corollary}\label{cor bilip} The map $\mathbf{p}_{\mathcal{H}}:(\mathcal{H}_n^\infty,d_{\mathcal{H}})\to (H^1(\mathbb{S}^{n-1})_{01},d_{01})$ is locally bi-Lipschitz.
\end{corollary}

Hence from Proposition~\ref{prop:hilbert} one obtains the following.

\begin{corollary}\label{cor complete}
$(\mathcal{H}_n^\infty,d_{\mathcal{H}})$ is complete.
\end{corollary}

We also obtain the following.
\begin{corollary} \label{cor topo hyperboloid}
On $\mathcal{H}_n^\infty$, $d_\mathcal{H}$ and $\|\cdot \|_{H^1}$ induce the same topology.
\end{corollary}
\begin{proof}
Let $(h_i)$ be a sequence in $\mathcal{H}_n^\infty$ converging to $h \in \mathcal{H}_n^\infty$ for $\|\cdot\|_{H^1}$. Then $(h_i)$ (resp. $(\nabla h_i)$) converges to $h$ (resp. $\nabla h$) in $L^2$, hence $\overline{V_2}^n(h_i-h) \rightarrow 0$. But $\overline{V_2}^n(h_i-h) = 2 - 2 \overline{V_2}^n(h_i,h)$, hence $\overline{V_2}^n(h_i,h) \rightarrow 1$ and $d_{\mathcal{H}}(h_i,h) = \operatorname{argch}(\overline{V_2}^n(h_i,h)) \rightarrow 0$.

On the other hand, let $(h_i)$ be a sequence in $\mathcal{H}_n^\infty$ converging to $h \in \mathcal{H}_n^\infty$ for $d_\mathcal{H}$. Then by Corollary \ref{cor bilip} we have $\mathbf{p}_{\mathcal{H}}(h_i) = h_i - \mathbf{mean}(h_i) \rightarrow \mathbf{p}_{\mathcal{H}}(h)=h-\mathbf{mean}(h)$ in $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$, hence  $\nabla \mathbf{p}_{\mathcal{H}}(h_i) \rightarrow \nabla \mathbf{p}_{\mathcal{H}}(h)$ in $L^2$ (see Lemma \ref{lemme equiv distance}), so $\nabla h_i \rightarrow \nabla h$ in $L^2$. Moreover, $\overline{V_2}^n(h-h_i) = c_{n}(\|h-h_i\|_{L^2}^2 - \lambda_1^{-1} \|\nabla h-\nabla h_i\|_{L^2}^2) \rightarrow 0$, hence $h_i \rightarrow h$ in $L^2$, and this gives $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$.
\end{proof}

\begin{remark}{\rm
From Proposition~\ref{prop A intrins}, the maps $E_{\mathcal{B},N}$ induce totally geodesic isometric
immersions of $\mathcal{H}_n^\infty$ into $\mathcal{H}_{n+1}^\infty$.
}\end{remark}

\begin{remark}{\rm
As in the finite dimensional case, the isometry group of $(\mathcal{H}_n^\infty,d_{\mathcal{H}})$ is given by the linear maps preserving $\overline{V_2}^n$, up to a quotient by $\pm\mathrm{Id}$. We refer to \cite{BIM} for this fact and others about isometries of the infinite dimensional hyperbolic space.
}\end{remark}

\subsection{Projective model}\label{sec proj model}

Let $h,k \in \mathcal{C}_n$, with $h \neq k$, be such that $\overline{V_2}^n(h)=\overline{V_2}^n(k)=a>0$. Consider the equation
$$\overline{V_2}^n((1-t)h+t k)=0$$
(see also equation \eqref{equation} in the introduction).
The left-hand side gives the polynomial
$$t^2\overline{V_2}^n(h-k)+2t\overline{V_2}^n(h,k-h)+\overline{V_2}^n(h)~.$$
Since $\overline{V_2}^n(h)=\overline{V_2}^n(k)=a$, we have
$$\overline{V_2}^n(h-k) =  2\left(a - \overline{V_2}^n(h,k)\right)~,$$
and by Fact \ref{factRCS} this is negative (remember that $h \neq k$): hence this is a second order polynomial. A direct computation shows that the discriminant is $4\delta(h,k)$, where we set
$$
\delta(h,k)=\overline{V_2}^n(h,k)^2 - a^2>0~.$$
The two real solutions are $t_1<0<1<t_2$ with
$$t_1=\frac{\overline{V_2}^n(h,h-k)-\sqrt{\delta(h,k)}}{\overline{V_2}^n(h-k)} \mbox{ and } t_2=\frac{\overline{V_2}^n(h,h-k)+\sqrt{\delta(h,k)}}{\overline{V_2}^n(h-k)}~.$$

Let us denote by $\mathbb{H}_n^\infty$ the projective quotient of
$\mathcal{C}_n$, and, for $[h],[k]\in \mathbb{H}_n^\infty$, let us define
\begin{equation*
d_{\mathbb{H}}([h],[k]):=\frac{1}{2} \ln [0,1,t_1,t_2]~, \end{equation*}
where $h,k$ are representative of
$[h],[k]$ respectively, such that $\overline{V_2}^n(h)=\overline{V_2}^n(k)$, and
$[0,1,t_1,t_2]=\frac{t_1}{t_2}\frac{1-t_2}{1-t_1}$ is the cross-ratio.

\begin{fact}  For $h,k\in \mathcal{C}_n$ we have
\begin{equation}\label{dh et dh}d_{\mathbb{H}}([h],[k])=d_{\mathcal{H}}(\overline{V_2}^n(h)^{-1/2}h,\overline{V_2}^n(k)^{-1/2}k) = \operatorname{argch}\left( \frac{\overline{V_2}^n(h,k)}{\sqrt{\overline{V_2}^n(h) \overline{V_2}^n(k)}} \right)~. \end{equation}
 \end{fact}
 \begin{proof}
Let $a = \overline{V_2}^n(h)=\overline{V_2}^n(k)$. A direct computation yields to
$t_1t_2=\frac{\overline{V_2}^n(h)}{\overline{V_2}^n(h-k)}$, and this gives
$$
[0,1,t_1,t_2]=\frac{t_1-t_1 t_2}{t_2 - t_1 t_2}=\frac{\overline{V_2}^n(h,h-k)-\sqrt{\delta(h,k)}-\overline{V_2}^n(h)}{\overline{V_2}^n(h,h-k)+\sqrt{\delta(h,k)}-\overline{V_2}^n(h)}=\frac{\overline{V_2}^n(h,k)+\sqrt{\delta(h,k)}}{\overline{V_2}^n(h,k)-\sqrt{\delta(h,k)}}~, $$
hence
\begin{eqnarray*}
 d_{\mathbb{H}}([h],[k]) & = & \frac{1}{2} \ln \left( \frac{\overline{V_2}^n(h,k)+\sqrt{\delta(h,k)}}{\overline{V_2}^n(h,k)-\sqrt{\delta(h,k)}} \right) = \frac{1}{2} \ln \left( \frac{\left(\overline{V_2}^n(h,k)+\sqrt{\delta(h,k)}\right)^2}{\overline{V_2}^n(h,k)^2-\delta(h,k)} \right) \\
 & = &\frac{1}{2} \ln \left( \frac{\left(\overline{V_2}^n(h,k)+\sqrt{\delta(h,k)}\right)^2}{a^2} \right) = \ln \left( \frac{\overline{V_2}^n(h,k)}{a}+\frac{\sqrt{\delta(h,k)}}{a} \right)~.
 \end{eqnarray*}
And the formula $\operatorname{argch}(u)=\ln(u+\sqrt{u^2-1})$ valid for $u \geq 1$ gives
$$\operatorname{argch}\left( \frac{\overline{V_2}^n(h,k)}{a} \right) = \ln\left( \frac{\overline{V_2}^n(h,k)}{a} + \sqrt{\frac{\overline{V_2}^n(h,k)^2}{a^2}-1}\right) = \ln\left( \frac{\overline{V_2}^n(h,k)}{a} + \frac{\sqrt{\delta(h,k)}}{a}\right)
$$
and this ends the proof.
 \end{proof}

Hence the map
$$\operatorname{Nor}_{\overline{V_2}^n}: \mathbb{H}_n^\infty  \to \mathcal{H}_n^\infty~,$$
defined by choosing the unique representative of an equivalence class in $\mathbb{H}_n^\infty$ which is in $\mathcal{H}_n^\infty$ (that is, which satisfies $\overline{V_2}^n=1$), is an isometry between
$(\mathbb{H}_n^\infty,d_{\mathbb{H}})$ and $(\mathcal{H}_n^\infty, d_{\mathcal{H}})$.
\subsection{Klein ball model}\label{sec klein}

It is useful to consider an affine model
of $(\mathbb{H}_n^\infty,d_\mathbb{H})$.
Let us consider the intersection $\mathbb{K}\mathbf{lein}^\infty_n$ of $\mathcal{C}_n$
with the affine hyperplane
$$P_n=\{h \in H^1(\mathbb{S}^{n-1})_1 ~|~ \overline{V_1}^n(h)=1\}$$
(see Figure~\ref{fig:notations}), that is
$$\mathbb{K}\mathbf{lein}^\infty_n=\{h \in \mathcal{C}_n ~|~ \overline{V_1}^n(h)=1\}~.$$

\begin{remark}{\rm

From \eqref{Vbar1},   the maps $E_{\mathcal{B},N}$ induce
injective maps of $\mathbb{K}\mathbf{lein}^\infty_n$ into $\mathbb{K}\mathbf{lein}^\infty_{n+1}$, which will appear to be totally geodesic isometric immersions.
}\end{remark}

Recall the dimensional constant $r_1(n)$ from \eqref{v1v2}. One can check that
$$r_1(n)=\overline{V_1}^n(1)^{-1}~,$$
that is $\overline{V_1}^n(r_1(n))=1$, hence $r_1(n) \in \mathbb{K}\mathbf{lein}^\infty_n$ (we also denote by $r_1(n)$ the constant function equal to $r_1(n)$
on $\mathbb{S}^{n-1}$).
From the following trivial fact, $\mathbb{K}\mathbf{lein}^\infty_n$ is an open ball in the affine space
$P_n$,
 in particular it is invariant under convex combinations.
\begin{fact}\label{fact K=B+1}
We have $\mathbb{K}\mathbf{lein}^\infty_n=\{h+r_1(n)| h\in  H^1(\mathbb{S}^{n-1})_{01}, -\overline{V_2}^n(h)<  \overline{V_2}^n(r_1(n))  \}~.$
\end{fact}

More formally, we have a map $$\operatorname{Nor}_{\overline{V_1}^n}: \mathbb{H}_n^\infty  \to \mathbb{K}\mathbf{lein}^\infty_n~,$$
defined by choosing the unique representative of an equivalence class in $\mathbb{H}_n^\infty$ which is in $\mathbb{K}\mathbf{lein}^\infty_n$ (that is, which satisfies $\overline{V_1}^n=1$).
The map $\operatorname{Nor}_{\overline{V_1}^n}$ is an isometry when
$ \mathbb{K}\mathbf{lein}^\infty_n$ is endowed with the metric
\begin{equation*}\label{def:dK}
d_{\mathbb{K}}(h,k)=d_{\mathbb{H}}([h],[k])~.\end{equation*}
Note that by equation \eqref{dh et dh} we have, for any $h,k \in \mathbb{K}\mathbf{lein}^\infty_n$,
\begin{equation}\label{eq:dk dh}
d_{\mathbb{K}}(h,k) =  d_{\mathcal{H}}(\overline{V_2}^n(h)^{-1/2}h,\overline{V_2}^n(k)^{-1/2}k)=\operatorname{argch}\left( \frac{\overline{V_2}^n(h,k)}{\sqrt{\overline{V_2}^n(h) \overline{V_2}^n(k)}} \right)~.
\end{equation}
One of the main interest of $\mathbb{K}\mathbf{lein}^\infty_n$  is the following result, which is immediate from Fact~\ref{fact geod}.

\begin{lemma}\label{lem shortest klein}
Let $h,k\in \mathbb{K}\mathbf{lein}^\infty_n$. Then the convex combination of $h$ and $k$ is the unique shortest path between $h$ and $k$ for the metric $d_{\mathbb{K}}$.
\end{lemma}

We need the following elementary fact.
\begin{fact} \label{fact continuité mean}
The function $(\mathcal{H}_n^\infty,d_{\mathcal{H}}) \rightarrow \mathbb{R}$, $h\mapsto \int_{\mathbb{S}^{n-1}}h$ is continuous.
\end{fact}
\begin{proof}
This function is linear on $H^1(\mathbb{S}^{n-1})_1$, and by H\"older inequality, $|\int_{\mathbb{S}^{n-1}}h| \leq \sqrt{\operatorname{vol}(\mathbb{S}^{n-1})} \|h\|_{L^2} \leq \sqrt{\operatorname{vol}(\mathbb{S}^{n-1})} \|h\|_{H^1}$, hence $\int_{\mathbb{S}^{n-1}} : (\mathcal{H}_n^\infty,\|\cdot\|_{H^1}) \rightarrow \mathbb{R}$ is continuous. Then Corollary \ref{cor topo hyperboloid} proves the claim.
\end{proof}

\begin{lemma}\label{lem conv klein}
Let $(h_i)_i$ converge to $h$ in $(\mathbb{K}\mathbf{lein}^\infty_n,d_{\mathbb{K}})$. Then $\overline{V_2}^n(h-h_i)\to 0$.
\end{lemma}

\begin{proof}
By \eqref{eq:dk dh}, $$\overline{V_2}^n(h_i)^{-1/2} h_i \rightarrow \overline{V_2}^n(h)^{-1/2} h$$ in $(\mathcal{H}_n^\infty,d_{\mathcal{H}})$. As $\int_{\mathbb{S}^{n-1}}h_i=\int_{\mathbb{S}^{n-1}}h$ on $\mathbb{K}\mathbf{lein}^\infty_n$, by Fact \ref{fact continuité mean} we obtain $$\overline{V_2}^n(h_i)^{-1/2} \rightarrow \overline{V_2}^n(h)^{-1/2}~.$$ Since $(\mathcal{H}_n^\infty,d_{\mathcal{H}})$ is locally bi-Lipschitz to $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$, we have $$\mathbf{p}_{\mathcal{H}}(\overline{V_2}^n(h_i)^{-1/2} h_i) \rightarrow \mathbf{p}_{\mathcal{H}}(\overline{V_2}^n(h)^{-1/2} h)$$ in $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$, that is $\overline{V_2}^n(h_i)^{-1/2} (h_i-\mathbf{mean} (h_i)) \rightarrow \overline{V_2}^n(h)^{-1/2} (h-\mathbf{mean} (h))$ in $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$. Since $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$ is a normed vector space and
$\mathbf{mean}(h_i)=\mathbf{mean}(h)$, we obtain  $h_i \rightarrow h$ in $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$, that is $\sqrt{-\overline{V_2}^n(h-h_i)} \rightarrow 0$.
\end{proof}

We use this Lemma, as well as the equivalence of topologies on $\mathcal{H}_n^\infty$ (as stated in Corollary~\ref{cor topo hyperboloid}) to prove the following. We denote by $d_{H^1}$ the distance given by
$\|\cdot\|_{H^1}$ on $H^1(\mathbb{S}^{n-1})_1$.
\begin{proposition} \label{prop equiv topo Klein}
On $\mathbb{K}\mathbf{lein}^\infty_n$, $d_\mathbb{K}$ and $d_{H^1}$ define the same topology.
\end{proposition}

\begin{proof}
If $d_\mathbb{K}(h_i,h) \rightarrow 0$, then by Lemma \ref{lem conv klein} we have $\overline{V_2}^n(h-h_i) \rightarrow 0$, hence since $h-h_i \in H^1(\mathbb{S}^{n-1})_{01}$ we have by Lemma \ref{lemme equiv distance} $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$.
Now suppose that $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$. By \eqref{eq:dk dh} and Corollary \ref{cor topo hyperboloid}, to prove $d_\mathbb{K}(h_i,h) \rightarrow 0$ this is sufficient to prove that $\frac{1}{\sqrt{\overline{V_2}^n(h_i)}} h_i \rightarrow \frac{1}{\sqrt{\overline{V_2}^n(h)}} h$ for $\|\cdot\|_{H^1}$. Since $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$, we have $\overline{V_2}^n(h_i) \rightarrow \overline{V_2}^n(h) >0$, and then the right-hand side of the following inequality goes to zero:
\[
\|\frac{1}{\sqrt{\overline{V_2}^n(h_i)}} h_i - \frac{1}{\sqrt{\overline{V_2}^n(h)}} h\|_{H^1} \leq \frac{1}{\sqrt{\overline{V_2}^n(h_i)}} \|h_i-h\|_{H^1} + \left| \frac{1}{\sqrt{\overline{V_2}^n(h_i)}} - \frac{1}{\sqrt{\overline{V_2}^n(h)}} \right| \|h\|_{H^1}.
\]
\end{proof}

Concerning the intersection of the isotropic cone of $\overline{V_2}^n$ with $P_n$, we have the following property.

\begin{proposition}\label{lem bord infini}
Let $h_i\in \mathbb{K}\mathbf{lein}^\infty_n$ and $k \in \mathbb{K}\mathbf{lein}^\infty_n$. Then
$$\overline{V_2}^n(h_i)\to 0 \iff d_\mathbb{K}(h_i,k)\to +\infty~.$$
\end{proposition}
\begin{proof}
First, let us remark that $\mathbf{mean}(h_i)=\mathbf{mean}(k) = r_1(n)$, hence equation \eqref{eq:ecrit supp} gives
\begin{equation}\label{eq id above}\overline{V_2}^n(h_i,k) = \overline{V_2}^n(h_i-r_1(n),k-r_1(n)) + \overline{V_2}^n(r_1(n))\end{equation}
and
\begin{equation}\label{eq id above2}\overline{V_2}^n(h_i) = \overline{V_2}^n(h_i-r_1(n)) + \overline{V_2}^n(r_1(n)) \mbox{ and } \overline{V_2}^n(k) = \overline{V_2}^n(k-r_1(n)) + \overline{V_2}^n(r_1(n))~.\end{equation}

Assume that $d_\mathbb{K}(h_i,k) = \operatorname{argch}\left( \frac{\overline{V_2}^n(h_i,k)}{\sqrt{\overline{V_2}^n(h_i)\overline{V_2}^n(k)}}\right) \to +\infty$. Since \eqref{eq id above} implies $\overline{V_2}^n(h_i,k) \leq \overline{V_2}^n(r_1(n))$, we have $\frac{\overline{V_2}^n(h_i,k)}{\sqrt{\overline{V_2}^n(h_i)\overline{V_2}^n(k)}} \leq \frac{\overline{V_2}^n(r_1(n))}{\sqrt{\overline{V_2}^n(h_i) \overline{V_2}^n(k)}}$, hence $\overline{V_2}^n(h_i) \rightarrow 0$.

Now assume that $\overline{V_2}^n(h_i) \rightarrow 0$. To prove that $d_\mathbb{K}(h_i,k)\to +\infty$ it is sufficient to show that for $i$ large enough, $\overline{V_2}^n(h_i,k)$ is bounded from below by a positive constant. By \eqref{eq id above} and Cauchy--Schwarz inequality in $H^1(\mathbb{S}^{n-1})_{01}$ we have
$$\overline{V_2}^n(h_i,k) \geq -\sqrt{\overline{V_2}^n(h_i-r_1(n))\overline{V_2}^n(k-r_1(n))}+\overline{V_2}^n(r_1(n))~.$$
By equation \eqref{eq id above2} we have $\overline{V_2}^n(h_i-r_1(n)) = \overline{V_2}^n(h_i)-\overline{V_2}^n(r_1(n)) \rightarrow -\overline{V_2}^n(r_1(n))$, hence we obtain
$$-\sqrt{\overline{V_2}^n(h_i-r_1(n))\overline{V_2}^n(k-r_1(n))}+\overline{V_2}^n(r_1(n)) \rightarrow -\sqrt{-\overline{V_2}^n(r_1(n))\overline{V_2}^n(k-r_1(n))}+\overline{V_2}^n(r_1(n))~.$$
By equation \eqref{eq id above2} we have $-\overline{V_2}^n(k-r_1(n))<\overline{V_2}^n(r_1(n))$, hence this limit is positive, and this proves that for $i$ large enough, $\overline{V_2}^n(h_i,k)$ is bounded from below by a positive constant.
\end{proof}

\section{Spaces of convex bodies}

In this section, we are considering spaces of convex bodies in $\mathbb{R}^n$. Convex bodies are determined by their support functions, that are functions living in the spaces introduced in the preceding sections. The intrinsic area $V_2$ can be written with the help of the bilinear form $\overline{V_2}^n$.

\subsection{Support functions}\label{sec supp}

The \emph{support function} $\operatorname{Supp}(K)$ of a convex body $K$ in $\mathbb{R}^n$ gives, at the point $x\in \S^{n-1}$, the distance from the origin of $\mathbb{R}^n$ to the support hyperplane of
$K$ with outward normal $x$. More precisely,
$\operatorname{Supp}(K):\S^{n-1}\to \mathbb{R}$ is defined as
$$\operatorname{Supp}(K)(x)=\max_{p\in K} \langle x,p\rangle~, $$
where $\langle\cdot,\cdot\rangle$ is the usual  scalar product of $\mathbb{R}^n$.

The support functions are characterized among functions on $\mathbb{S}^{n-1}$ by the following fact, see \cite{schneider}.
\begin{fact}\label{fact:ext supp}
A function $h:\mathbb{S}^{n-1}\to \mathbb{R}$ is the support function of a convex body in $\mathbb{R}^n$ if and only if its one homogeneous extension $\tilde{h}:\mathbb{R}^n\to \mathbb{R}$, $\tilde{h}(x)=\|x\|h(x/\|x\|)$,  is a convex function.
\end{fact}

Let us recall some elementary facts about the support functions. For the next one, see \cite[(2.2.3)]{groemer}.
Recall that we denote by $\mathcal{K}^n$ the space of convex bodies in $\mathbb{R}^n$.

\begin{lemma}\label{lem lips}
Let $K\in \mathcal{K}^n$. If $\operatorname{Supp}(K)\leq R$, then $\operatorname{Supp}(K)$ is $R$-Lipschitz.
\end{lemma}
In particular, $\operatorname{Supp}(K)$ belongs to $H^1(\mathbb{S}^{n-1})$, so in turn the support function defines a map
$$\operatorname{Supp}:\mathcal{K}^n\toH^1(\mathbb{S}^{n-1})~. $$

\begin{fact}\label{fact:pte supp}
The map $\operatorname{Supp}$  satisfies  the following properties:
\begin{itemize}
\item $\operatorname{Supp}(K_1+K_2)=\operatorname{Supp}(K_1)+\operatorname{Supp}(K_2)$;
\item $\operatorname{Supp}(\lambda K)=\lambda \operatorname{Supp}(K)$, $\lambda>0$;
\item $\operatorname{Supp}$ is a bijection onto its image;
\item if $K_1\subset K_2$, then $\operatorname{Supp}(K_1)\leq \operatorname{Supp}(K_2)$;
\item $\operatorname{Supp} (\mathcal{K}^n)$ is a convex cone in $H^1(\mathbb{S}^{n-1})$.
\end{itemize}
\end{fact}

The last point follows because  the convex combination of convex bodies is a convex body.

\begin{remark}{\rm
Let us warn the reader that if $\operatorname{Supp}(\lambda K)=\lambda \operatorname{Supp}(K)$, $\lambda>0$, we don't have $\operatorname{Supp}(- K)=- \operatorname{Supp}(K)$ in general, where $-K=\{-x | x\in K \}$. Indeed, both $\operatorname{Supp}(- K)$ and $\operatorname{Supp}(K)$ are positive if the origin of $\mathbb{R}^n$ is in the interior of $K$.
Actually, $\operatorname{Supp}(-K)(v)=\operatorname{Supp}(K)(-v)$, and $- \operatorname{Supp}(K)$ is like the support function of $K$ but with the  support planes defined by their  \emph{inward} unit normals.
}\end{remark}

The two next lemmas show that support functions of convex bodies have strong convergence properties.
For this lemma, see \cite[Theorem 1.8.15]{schneider}.
\begin{lemma}\label{lem:point impl unif}
Let $K_i,K\in \mathcal{K}^n$. If $(\operatorname{Supp}(K_i))_i$ converges pointwise to $\operatorname{Supp}(K)$, then
the convergence is uniform.
\end{lemma}
For the next result, see \cite[Lemma~2.3.4]{groemer} or \cite{heil}. \begin{lemma}\label{lem conv gradient}
Let $K_i,K\in \mathcal{K}^n$ such that $(\operatorname{Supp}(K_i))_i$ converges to
$\operatorname{Supp}(K)$. Then, almost everywhere, $(\nabla\operatorname{Supp}(K_i))_i$ converges pointwise to
$\nabla\operatorname{Supp}(K)$.
\end{lemma}

In the following, $C^0(\S^{n-1})$ is the set of continuous functions on $\S^{n-1}$, and
for $h\in C^0(\S^{n-1})$ we set $\|h\|_\infty=\sup \{|h(x)|, x\in \S^{n-1}\}$. Let $d_\infty$ be the distance induced by $\|\cdot\|_\infty$ on $\operatorname{Supp}(\mathcal{K}^{n})$.
Abusing notation, we also denote by
$\operatorname{Supp}$ the map from $\mathcal{K}^n$ to
$C^0(\S^{n-1})$ (instead of Sobolev space) which associates to a convex body
its support function.

\begin{remark}\label{rem dis hausdorff}{\rm
The pull-back of  $d_\infty$ onto $\mathcal{K}^{n}$ is the classical Hausdorff distance, \cite[Lemma~1.8.14]{schneider}. Recall that the Hausdorff distance between $K_1$ and $K_2$ in $\mathbb{R}^n$ is the min of the non-negative $\lambda$ such that $K_1\subset K_2+\lambda B^n$ and
 $K_2\subset K_1+\lambda B^n$.
}\end{remark}

Lemma~\ref{lem conv gradient} gives the following
\begin{corollary} \label{cor conv H1}
In $\operatorname{Supp}(\mathcal{K}^n)$, if $d_\infty(h_i,h)\to 0$ then $d_{H^1}(h_i,h)\to 0$.
\end{corollary}
\begin{proof}
This is obvious that $h_i \rightarrow h$ in $L^2$. Moreover, let $R>0$ be such that $h_i \leq R$ for every $i$. Then $(\nabla h_i)_i$ almost everywhere converges pointwise to $\nabla h$, hence the convergence holds in $L^2$ via Lebesgue's dominated convergence: these functions are uniformly bounded by $R$ by Lemma \ref{lem lips}. Hence $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$.
\end{proof}

By definition, $\overline{V_2}^n(\cdot,\cdot)$ is continuous in each entry on $H^1(\mathbb{S}^{n-1})$, hence Corollary \ref{cor conv H1} gives the following.

\begin{corollary}\label{prop cont aire}
 $\overline{V_2}^n(\cdot,\cdot)$  is continuous in each entry on $(\operatorname{Supp}(\mathcal{K}^{n}),d_\infty)$.
\end{corollary}

Let us end this section giving a geometric interpretation of the  functions $E_{\mathcal{B},N}(h)$ from Section~\ref{Canonical embeddings}, see also Figure~\ref{fig:extension}.

\begin{fact}\label{factexplicationEh}
Let $K \in \mathcal{K}^n$ and $\iota : \mathbb{R}^n \rightarrow \mathbb{R}^{n+1}$ be a linear isometric embedding.
If $\mathcal{B}$ be is the image by $\iota$ of the canonical basis of $\mathbb{R}^n$, and $N$ is such that $(\mathcal{B},N)$ is a direct orthonormal basis of $\mathbb{R}^{n+1}$, then
$$\operatorname{Supp}(\iota(K))= E_{\mathcal{B},N}(\operatorname{Supp}(K))~. $$
\end{fact}

\begin{figure}
\begin{center}
\psfrag{N}{$N$}
\psfrag{x}{$x$}
\psfrag{xt}{$(x,t)$}
\psfrag{h}{$h(x)$}
\psfrag{K}{$K$}
\psfrag{ht}{$\cos(t)h(x)$}
\includegraphics[width=0.7\linewidth]{extension}\caption{The extension of a support function to a higher dimensional space, see Fact~\ref{factexplicationEh}. Here $h(x)=\operatorname{Supp}(K)(x)$.}\label{fig:extension}
\end{center}
\end{figure}

\begin{proof}
Denote by $h' \in H^1(\mathbb{S}^n)$ the support function of $\iota(K)$, and let $R_{\mathcal{B},N} \in SO(n+1)$ be the unique rotation sending the canonical basis of $\mathbb{R}^{n+1}$ on $(\mathcal{B},N)$: we have $R_{\mathcal{B},N}(x,0)=\iota(x)$ for $x \in \mathbb{R}^n$, so in particular $R_{\mathcal{B},N}(p,0)=\iota(c)$ for any $p \in K$. So for any $(x,t) \in \mathbb{S}^{n-1}\times ]-\frac{\pi}{2}, \frac{\pi}{2}[$ we have
\begin{eqnarray*}
h'(\Phi_{\mathcal{B},N}(x,t))& =& \max_{p \in K} \langle\Phi_{\mathcal{B},N}(x,t) , \iota(p) \rangle \\
& = & \max_{p \in K} \langle R_{\mathcal{B},N}(\cos(t) x,\sin(t)) , R_{\mathcal{B},N} (p,0) \rangle \\
& = & \max_{p \in K} \langle \cos(t) x,p \rangle = \cos(t) h(x)~.
\end{eqnarray*}

\end{proof}

\subsection{Properties of the intrinsic area}\label{sec:pt area}

A $C^2_+(n)$ convex body $K$ is a convex body in $\mathbb{R}^n$ with $C^2$ boundary and such that its Gauss map is a diffeomorphism from the boundary of $K$ onto $\mathbb{S}^{n-1}$.
We will repeatedly use the following approximation result, see  e.g. \cite[Lemma~2.3.3]{groemer}.
\begin{fact}\label{fact:approxC2}
$\operatorname{Supp}(C^2_+(n))$ is dense in  $(\operatorname{Supp}(\mathcal{K}^{n}),d_\infty)$.
\end{fact}

 If $K\in C^2_+(n)$, then the eigenvalues  of the second differential of
 $\widetilde{\operatorname{Supp}(K)}$  (the one-homogeneous extension of $\operatorname{Supp}(K)$) at $x\in \mathbb{S}^{n-1}$ are $0$ (with eigenvector $x$) and positive numbers $r_1,\ldots,r_{n-1}$, which are the principal radii of curvature of (the boundary of) $K$. We can consider the $r_i$ as functions on $\mathbb{S}^{n-1}$ \cite[Corollary 2.5.2]{schneider}.
A change of variable provided by the Gauss map gives  \cite[p.~58, 119]{schneider}
$$\mathrm{vol}_n(K)=\frac{1}{n}\int_{\mathbb{S}^{n-1}}\operatorname{Supp}(K) r_1\cdots r_{n-1}~. $$
Now for $\epsilon>0$, as on the one hand the Gauss map of $B^n$ is the identity, and on the other hand if $\lambda$ is an eigenvalue of a matrix $A$, then $\lambda+\epsilon$ is an eigenvalue of
$A+\epsilon \operatorname{Id}$, we obtain
\begin{equation}\label{eq K+e}\mathrm{vol}_n(K+\epsilon B^n)=\frac{1}{n}\int_{\mathbb{S}^{n-1}}(\operatorname{Supp}(K)+\epsilon) (r_1+\epsilon)\cdots (r_{n-1}+\epsilon)~,\end{equation}
and Steiner formula \eqref{steiner} follows for $C^2_+(n)$ convex bodies.
Note that we can make a polynomial interpolation for the intrinsic volumes: given $n+1$ different positive numbers $\epsilon$, \eqref{steiner} can be considered as a solvable linear system of
$d+1$ equations with $V_i(K)$ as unknowns ---the matrix of the system is a Vandermonde matrix. In turn, the $V_i(K)$ are linear combination of volumes, hence the maps $V_i:\mathcal{K}^n\to \mathbb{R}$ are continuous for the Hausdorff topology as the volume is continuous for the Hausdorff topology on the space of convex bodies \cite{schneider}.

From \eqref{eq K+e} we obtain in particular
$$V_1(K)=\frac{1}{n\kappa_{n-1}}\int_{\mathbb{S}^{n-1}} \operatorname{Supp}(K) + r_1+\cdots + r_{n-1}~.$$
But $r_1+\cdots + r_{n-1}$ is the trace of the second differential of $\widetilde{\operatorname{Supp}(K)}$, and since for $1$-homogeneous functions $f$ on $\mathbb{R}^n$ we have, as functions on $\S^{n-1}$,
$$\Delta_e f = (n-1)f + \Delta f~,$$
where $\Delta_e$ is the Laplacian of $\mathbb{R}^n$ and $\Delta$ is the Laplacian of $\mathbb{S}^{n-1}$,
we have \begin{equation}\label{somme ri}r_1+\cdots+r_{n-1}=(n-1)\operatorname{Supp}(K) + \Delta \operatorname{Supp}(K)~.\end{equation}
As by Green formula $\int_{\mathbb{S}^{n-1}}\Delta \operatorname{Supp}(K)=0$ we obtain
$$V_1(K)=\frac{1}{\kappa_{n-1}}\int_{\S^{n-1}} \operatorname{Supp}(K)~.$$
Note that this gives
$$V_1(K)=
\overline{V_1}^n(\operatorname{Supp}(K))~.$$
Clearly, by Fact~\ref{fact:approxC2}, the formula above holds for any convex body.
\begin{remark}\label{remark:mean width}{\rm
By Fact~\ref{factexplicationEh} and \eqref{Vbar1}, we obtain that if
$\iota:\mathbb{R}^n\to \mathbb{R}^{n+1}$ is a linear isometric embedding, then
 $$V_1(\iota(K))=V_1(K)~.$$
 Since $\overline{V_1}^n(\operatorname{Supp}(K))=\frac{n\kappa_n}{\kappa_{n-1}}\mathbf{mean}(\operatorname{Supp}(K))$, the quantity $V_1(K)$ is intimately related to the \emph{mean width} of $K$, which is equal to $2\mathbf{mean}(\operatorname{Supp}(K))$.
}\end{remark}

We will use a Minkowski type formula, see Proposition~4.3.1 as well as the formula in the bottom of p.~45 and Remark~4.3.2 in \cite{OS83}:
$$\int_{\mathbb{S}^{n-1}} \sum_{i\not= j}r_ir_j = \frac{n-2}{2}\int_{\mathbb{S}^{n-1}}\operatorname{Supp}(K)(r_1+\cdots+r_{n-1})~.$$
On the other hand, we also obtain from \eqref{eq K+e} that
$$V_2(K)=\frac{1}{n\kappa_{n-2}}\int_{\mathbb{S}^{n-1}}\operatorname{Supp}(K)(r_1+\cdots+r_{n-1}) + \sum_{i\not=j}r_ir_j $$
so
\begin{equation}\label{eq V2po}V_2(K)=\frac{1}{2\kappa_{n-2}}\int_{\mathbb{S}^{n-1}}\operatorname{Supp}(K)(r_1+\cdots+r_{n-1}) \end{equation}
$$\stackrel{\eqref{def cn},\eqref{somme ri}}{=} c_{n}\int_{\S^{n-1}}\operatorname{Supp}(K)(\operatorname{Supp}(K)+ (n-1)^{-1} \Delta \operatorname{Supp}(K))   $$
and using Green formula,
$$V_2(K) =  \overline{V_2}^n(\operatorname{Supp}(K))~.$$
By density and continuity on both sides of the equation above, the equality above  holds actually for any convex body, that gives the bridge with the preceding sections. Seemly this formula first appeared in  \cite[Theorem~4.2a]{heil}, see also \cite[p.~298]{schneider} or \cite[Proposition~2.4.2]{groemer}.

Let us check the properties of the intrinsic area stated in the introduction.
Most of them need to be checked for the  $C^2_+(n)$ case,  and then use again an approximation argument.
For any $p\in \mathbb{R}^n$, $V_2(K+\{p\})=V_2(K)$, so up to a translation we can consider that the origin is in the relative interior of $K$, that gives $\operatorname{Supp}(K)\geq 0$.
 As in the $C^2_+(n)$ case, $r_i>0$, \ref{prop 1V2} is immediate from \eqref{eq V2po}.
 \ref{mcmullen}  follows immediately from Proposition~\ref{prop A intrins} and Fact~\ref{factexplicationEh}.

 \ref{propV22} and \ref{propV23} will be obtained  as a byproduct of properties of the mixed-area: \ref{propV22} is a consequence of \ref{prop mixed incl}, and \ref{propV23} is a consequence of \ref{mixed 2'}.

From \eqref{def mixed v2} and Green formula, one has
$$V_2(K_1,K_2)=\overline{V_2}^n(\operatorname{Supp}(K_1),\operatorname{Supp}(K_2))~, $$
and this gives properties \ref{mixed 3} and \ref{mixed 4} of the mixed-area. Moreover in the $C^2_+(n)$ case,
$$V_2(K_1,K_2)=c_{n}\int_{\mathbb{S}^{n-1}}\operatorname{Supp}(K_1)(r_1(K_2)+\cdots+r_{n-1}(K_2))~, $$
and this gives property \ref{prop mixed incl}.
  If $K$ is a point, it is immediate from \eqref{def mixed v2} that $V_2(K,Q)=0$ for any $Q$, that gives an implication of \ref{isotrope}. The first part of \ref{mixed 2'} follows from that and from
 \ref{prop mixed incl}, by taking a point for $K_1$.
 The second part of \ref{mixed 2'} follows from Fact~\ref{factRCSbis} and the following observation: if $Q$ is a segment,  for any other segment $K$ with a different direction, $Q+K$ is a parallelogram with non-empty interior, and \eqref{def mixed v2} gives that $V_2(K,Q)>0$. The remaining implication in \ref{isotrope} follows from \ref{mixed 2'}.

The \emph{Steiner point} of a convex body is the following point of $\mathbb{R}^n$:
$$\mathbf{stein}(K)=\frac{1}{\kappa_n}\int_{\mathbb{S}^{n-1}} \operatorname{Supp}(K)(x)x\mbox{d}\mathbb{S}^{n-1}~. $$
We will denote by $\mathcal{K}^n_S$ the space of convex bodies in $\mathbb{R}^n$ with the origin as Steiner point, and by $\mathcal{K}^{n*}_S$ the subset of convex bodies with positive intrinsic area. For clarity, let us note the following obvious fact.
\begin{fact} For $K \in \mathcal{K}^n$ we have
$$\mathbf{stein}(K)=0 \Longleftrightarrow \operatorname{Supp}(K)\in H^1(\mathbb{S}^{n-1})_1~,$$
so in particular
$$
\operatorname{Supp}(\mathcal{K}_S^n) \subset H^1(\mathbb{S}^{n-1})_1~.
$$
Moreover, if $K \in \mathcal{K}_S^n$ is not a point then $\operatorname{Supp}(K) \in \overline{\mathcal{C}}_n$
(see Section \ref{section:ineq and sign}),
and if $K \in \mathcal{K}_S^n$ is not a point or a segment we have $\operatorname{Supp}(K) \in \mathcal{C}_n$, that is
$$
\operatorname{Supp}(\mathcal{K}_S^{n*}) \subset \mathcal{C}_n~.
$$
\end{fact}

Let $K_1, K_2$ be two convex bodies.  \ref{alex fench}  does not change if a translation is performed on $K_1$ or $K_2$, so we can consider that the Steiner point of both $K_1$ and $K_2$ is the origin. If $K_1$ or $K_2$ are points then \ref{alex fench} is obvious. Otherwise, $\operatorname{Supp}(K_1)$ and $\operatorname{Supp}(K_2)$ belong to $\overline{\mathcal{C}}_n$, so Fact~\ref{factRCS} (reversed Cauchy--Schwarz inequality)  gives
 \ref{alex fench} (Alexandrov--Fenchel inequality). Note that this result
corresponds to
classical geometric inequalities in $\mathbb{R}^3$ (see
\cite[p.387]{schneider}, \cite{mmarx}). In $\mathbb{R}^2$,
it is the famous Minkowski inequality, and as we have seen in the introduction, if one of the two convex bodies is a disc, this gives the
isoperimetric inequality.

\begin{remark}{\rm Note that \ref{isotrope} is related to Fact~\ref{noyau A}. Indeed, support functions of points are exactly the restriction of linear forms to the sphere: if $p\in \mathbb{R}^n$, then
$\operatorname{Supp}(\{p\})=\langle p,\cdot \rangle$. And these functions are exactly  the kernel of $\overline{V_2}^n$ on $H^1(\mathbb{S}^{n-1})$.}\end{remark}

\subsection{Comparison of topologies}\label{haus}

Recall that $d_\infty$ is the distance induced by  $\|h\|_\infty=\sup \{|h(x)|, x\in \S^{n-1}\}$.
One of the principal features of  $d_\infty$ is the classical Blaschke selection theorem, which says that from each bounded sequence of convex bodies in $\mathbb{R}^n$ one can extract a converging subsequence for the Hausdorff distance.
Actually it is a direct consequence of
 Lemma~\ref{lem lips} and the Arzela--Ascoli Theorem. Let us state it in the following way.

\begin{theorem}[Blaschke selection theorem]\label{thm:proper}
$(\operatorname{Supp}(\mathcal{K}^{n}),d_\infty)$ is a proper metric space.
\end{theorem}

Let $\mathcal{K}_{SV_1}^{n}$ be the subset of $\mathcal{K}_{S}^{n}$ (convex bodies with Steiner point at the origin) of convex bodies $K$ such that
$V_1(K)=1$. Elements of  $\mathcal{K}_{SV_1}^{n}$ are sometimes called
 \emph{normalized convex bodies}, see  p.~164 and after in \cite{schneider}.
Note that $\mathcal{K}_{SV_1}^{n}$ is stable under convex combinations of convex bodies.

\begin{fact} \label{fact convexes dans B(1)}
If $K \in \mathcal{K}^n_{SV_1}$, then $K$ is included in the closed unit ball in $\mathbb{R}^n$.
\end{fact}
\begin{proof}
If this is not true, then there exists $x \in K$ with $\|x\|>1$. The Steiner point of $K$ is $0$ and belongs to $K$, hence $[0,x] \subset K$, which gives
$V_1(K)\geq V_1([0,x])=\|x\|>1$: this is a contradiction.
\end{proof}
Then Blaschke selection theorem (Theorem \ref{thm:proper}) implies the following
\begin{corollary} \label{cor compact}
$(\operatorname{Supp}(\mathcal{K}_{SV_1}^{n}),d_\infty)$ is compact.
\end{corollary}
\begin{proof}
This is sufficient to show that $\operatorname{Supp}(\mathcal{K}^n_{SV_1})$ is closed and bounded in $(\operatorname{Supp}(\mathcal{K}^n),d_\infty)$.
The boundedness is a direct consequence of Fact \ref{fact convexes dans B(1)}: for every $h \in \operatorname{Supp}(\mathcal{K}^n_{SV_1})$ we have $\|h\|_\infty \leq 1$.
And closeness is clear, since the maps $h \mapsto \int_{\S^{n-1}} h(x)x \mbox{d}\S^{n-1}(x)$ and $h \mapsto \int_{\S^{n-1}} h(x) \mbox{d}\S^{n-1}(x)$ are continuous on $(\operatorname{Supp}(\mathcal{K}^n),d_\infty)$.
\end{proof}

Let $\mathcal{K}_{SV_1}^{n*}$ the subset of $\mathcal{K}_{SV_1}^{n}$ of convex bodies of positive intrinsic area: it is clear that $\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$ is a convex subset of $\mathbb{K}\mathbf{lein}^\infty_n$.
The goal is  to prove that the analog of Theorem~\ref{thm:proper} holds on  $(\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*}),d_{\mathbb{K}})$. As a tool, we will use the distance $d_{L^2}$ induced by the $L^2$ norm on $\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$, as well as the following theorem, see  \cite{vitale} and \cite[Proposition~2.3.1]{groemer}.

\begin{theorem}[{Vitale}]\label{vitale}
The  distances $d_{\infty}$ and $d_{L^2}$ induce the same topology on $\operatorname{Supp}(\mathcal{K}^n) \subset C^0(\mathbb{S}^{n-1})$.
\end{theorem}
The result is weaker than saying that the two norms are equivalent on the space of convex bodies, that is not true, see \cite{vitale} for details.

\begin{corollary} \label{cor equi L2 H1}
The  distances $d_\infty$, $d_{L^2}$ and $d_{H^1}$ induce the same topology on $\operatorname{Supp}(\mathcal{K}^n)$.
\end{corollary}
\begin{proof}
We prove that $d_{L^2}$ and $d_{H^1}$ induce the same topology.
If $h_i \rightarrow h$ for $\|\cdot\|_{H^1}$ then obviously $h_i \rightarrow h$ for $\|\cdot\|_{L^2}$. And if $h_i \rightarrow h$ for $\|\cdot\|_{L^2}$ then by Theorem \ref{vitale} we have $h_i \rightarrow h$ for $d_\infty$, and Corollary \ref{cor conv H1} proves the claim.
\end{proof}
A direct consequence of Proposition \ref{prop equiv topo Klein} and Corollary \ref{cor equi L2 H1} is the following corollary, which relate the distances $d_\infty$ and $d_\mathbb{K}$.
\begin{corollary}\label{same cor}
On  $\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$, $d_\infty$ and $d_{\mathbb{K}}$ (as well as $d_{L^2}$ and $d_{H^1}$) induce the same topology.
\end{corollary}

\begin{proposition}\label{prop proper}
$(\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*}), d_{\mathbb{K}})$ is a proper metric space.
\end{proposition}
\begin{proof}

Let $A$ be a closed bounded subset of $(\operatorname{Supp}(\mathcal{K}^{n*}_{SV_1}),d_\mathbb{K})$. We want to show that $A$ is compact for $d_\mathbb{K}$; by Corollary \ref{same cor}, it suffices to show that it is compact for $d_\infty$. And by Corollary \ref{cor compact}, it suffices to show that $A$ is closed in $(\operatorname{Supp}(\mathcal{K}_{SV_1}^{n}),d_\infty)$.

So assume $(h_i)_i$ is a sequence of elements of $A$ converging to  $h\in\operatorname{Supp}(\mathcal{K}^{n}_{SV_1})$ for $d_\infty$; we want to show that $h \in A$. If $h \in \operatorname{Supp}(\mathcal{K}^{n*}_{SV_1})$, then this is true, because Corollary \ref{same cor} implies that $A$ is a closed subset of $(\operatorname{Supp}(\mathcal{K}^{n*}_{SV_1}),d_\infty)$. Otherwise, $h\in\operatorname{Supp}(\mathcal{K}^{n}_{SV_1})\setminus  \operatorname{Supp}(\mathcal{K}^{n*}_{SV_1})$, hence $\overline{V_2}^n(h)=0$ and by Corollary~\ref{prop cont aire} we have $\overline{V_2}^n(h_i)\to 0$.
Then by Proposition~\ref{lem bord infini}, the distance in $(\mathbb{K}\mathbf{lein}^\infty_n,d_\mathbb{K})$ between $h_i$ and any given point $k \in \mathbb{K}\mathbf{lein}^\infty_n$ goes to infinity, and that contradicts the fact that $A$ is a bounded subset of $(\operatorname{Supp}(\mathcal{K}^{n*}_{SV_1}),d_\mathbb{K})$.
\end{proof}

\begin{remark}{\rm
	Let us give an illustration of the fact that, even if $d_\infty$ and $d_\mathbb{K}$ induce the same topology, their behavior is quite different. First, recall that segments are shortest paths for the Hausdorff metric. Indeed,
 for any $K,L\in \mathcal{K}^n$ and $t\in (0,1)$, $d_\infty(\operatorname{Supp}(K), (1-t)\operatorname{Supp}(K)+t \operatorname{Supp}(L))=\|\operatorname{Supp}(K)- (1-t)\operatorname{Supp}(K)-t \operatorname{Supp}(L)\|_\infty=t \|\operatorname{Supp}(K)-\operatorname{Supp}(L)\|_\infty=t d_\infty(K,L)$.
		But conversely to  $d_\mathbb{K}$, in general, segments are not the only shortest path between two given convex bodies, see note 11 of Section~1.8 in \cite{schneider}.

}\end{remark}

\subsection{Terminal points of segments}\label{sec terminal}

Let $K_1,K_2 \in \mathcal{K}_{SV_1}^{n}$. The \emph{segment} between
$K_1$ and $K_2$ is $\{(1-t) K_1+ t K_2, t \in [0,1]\}$.
We say that  $K_1\in \mathcal{K}_{SV_1}^{n}$ is a \emph{terminal point} of the segment
if for any $t<0$, $(1-t)\operatorname{Supp}(K_1)+t\operatorname{Supp}(K_2) \notin \operatorname{Supp}( \mathcal{K}_{SV_1}^{n})$.
An \emph{extreme point} $K$ of  $\mathcal{K}_{SV_1}^{n}$ is
such that there does not exist $K_1,K_2\in  \mathcal{K}_{SV_1}^{n}$, $K_1\not=K_2$, and $t\in(0,1)$ such that $\operatorname{Supp}(K)=(1-t)\operatorname{Supp}(K_1)+t\operatorname{Supp}(K_2)$.
In the plane, extreme points of $ \mathcal{K}_{SV_1}^{2}$ are
segments and triangles \cite[Theorem 3.2.14]{schneider}.
For $n \geq 3$, extreme points of  $ \mathcal{K}_{SV_1}^{n}$ are dense for the Hausdorff metric \cite[3.2.18]{schneider}.

Clearly, an extreme point is a terminal point for all the segments ending at this point. But there are much more terminal points. For example, one can find convex bodies with a non smooth point on the boundary (i.e. a point of the convex body contained in more than one support plane) which are terminal points for the segment
starting at the unit ball ---this idea is illustrated in Figure~\ref{fig:extreme}.

\begin{figure}
\begin{center}
\psfrag{K}{$K$}
\psfrag{e}{$\epsilon$}
\psfrag{K+}{$K+\epsilon B^2$}
\psfrag{K-}{$K-\epsilon B^2$}
\includegraphics{terminal2}\caption{
If a plane convex body $K$ has a non-smooth point, then for any $\epsilon >0$, $\operatorname{Supp}(K)+\epsilon \operatorname{Supp}(B^2)$ is the support function of a convex body, while $\operatorname{Supp}(K)-\epsilon \operatorname{Supp}(B^2)$ is not.}\label{fig:extreme}
\end{center}
\end{figure}

In this section, we will use a different argument to prove
 that \emph{any} convex body is the terminal point of some segment (Proposition \ref{propositionpointsterminaux}). The simplest case illustrating our argument is depicted in Figure~\ref{fig:terminaux}.

\begin{figure}
\begin{center}
\includegraphics[width=1\linewidth]{segment-terminaux}\caption{The disc and the segment $[-1,1]$ are both terminal points of the segment joining them.}\label{fig:terminaux}
\end{center}
\end{figure}

By Fact~\ref{fact:ext supp}, if a function $h\in \mathbb{K}\mathbf{lein}^\infty_n$ belongs to $\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$, then
 the Laplacian of $\tilde{h}$ is non-negative, in the weak sense ($\tilde{h}$ is the one-homogenous extention of $h$). Let $C^\infty_c (\mathbb{R}^n)$ be the set of smooth functions with compact support in $\mathbb{R}^n$: this means that for every non-negative function $\varphi \in C^\infty_c (\mathbb{R}^n)$ we have
\[
\int_{\mathbb{R}^n} \tilde{h}(x) \Delta_e \varphi(x)\mbox{d}x \geq 0~.
\]

For $1 \leq p < n$ we will denote by $B_{p,n}$ the $p$-dimensional ball with radius $r_1(p)$ in $\mathbb{R}^n$, which is the set of points $x\in \mathbb{R}^n$  with $x_1^2 +\cdots+ x_p^2 \leq r_1(p)^2$ and $x_{p+1}= \cdots=x_n=0$. We have $V_1(B_{p,n})=1$, hence $B_{p,n} \in \mathcal{K}_{SV_1}^{n}$ (note that $B_{p,n} \in \mathcal{K}_{SV_1}^{n*}$ if and only if $p \geq 2$).
 Let $b_{p,n} = \operatorname{Supp}(B_{p,n}) \in \operatorname{Supp}(\mathcal{K}_{SV_1}^{n})$ and
let $\widetilde{b_{p,n}}(x)=r_1(p) \sqrt{x_1^2+\cdots+x_p^2}$ be the 1-homogeneous extension of $b_{p,n}$ (if $p=1$, then $\widetilde{b_p}(x)=r_1(1)|x_1|=\frac{|x_1|}{2}$).

\begin{example}[The $n-$dimensional ball is the terminal point of a segment]\label{exemplepointserminaux}{\rm
Assume $n \geq 3$, and let $p \in \mathbb{N}$ be such that $2 \leq p<n$. Let $K$ be the $n-$dimensional ball with radius $r_1(n)$ in $\mathbb{R}^n$: we have $V_1(K)=1$, hence $K \in \mathcal{K}_{SV_1}^{n}$, and let $k = \operatorname{Supp}(K) \in \operatorname{Supp}(\mathcal{K}_{SV_1}^{n})$ be its support function.
 Then $k$ is the terminal point of the segment starting at $b_{p,n}$.

Indeed, the 1-homogeneous extension of the support function of $K$ is $$\tilde{k}(x)=r_1(n) \sqrt{x_1^2 +\cdots+x_n^2}~,$$
and $\Delta_e \tilde{k}(x)=\frac{r_1(n)(n-1)}{\sqrt{x_1^2 + \cdots + x_n^2}}$. Also  $\Delta_e \widetilde{b_{p,n}}(x)=\frac{r_1(p)(p-1)}{\sqrt{x_1^2 + \cdots + x_p^2}}$.
  Now let $\epsilon>0$ and consider the point $x_\epsilon \in \mathbb{R}^n$ such that $x_1=\cdots=x_p=\epsilon$ and $x_{p+1}=\cdots= x_n=1$. For $t<0$ we have
\[
\Delta_e((1-t)\tilde{k} + t\widetilde{b_{p,n}})(x_\epsilon) = \frac{(1-t)r_1(n) (n-1)}{\sqrt{p \epsilon^2 + (n-p)}} + \frac{t r_1(p) (p-1)}{\sqrt{p}\epsilon}~,
\]
and this quantity goes to $-\infty$ when $\epsilon$ goes to zero. This shows that for $t<0$, $(1-t)k + t b_{p,n} \notin \operatorname{Supp}(\mathcal{K}_{SV_1}^{n})$.
}\end{example}

By using the same arguments we obtain the following.

\begin{proposition} \label{propositionpointsterminaux}
Let $p \in \mathbb{N}$ such that $1 \leq p < n$. Then any $K \in \mathcal{K}_{SV_1}^{n*}$ is the terminal point of a segment in $\mathcal{K}_{SV_1}^{n*}$, which starts at some embedded $p-$dimensional ball in $\mathbb{R}^n$.
\end{proposition}

In fact the proof shows that there are infinitely many such segments.

\begin{remark}{\rm
If $p=1$, this ball is in fact a segment and lies on the boundary of $\mathbb{K}\mathbf{lein}^\infty_n$.
}\end{remark}

To prove this Proposition we need the following theorem due to Alexandrov (see \cite{hessian}):
\begin{theorem}
A convex function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ is \emph{twice differentiable} at almost every $\bar{x} \in \mathbb{R}^n$, which means that for almost every $\bar{x} \in \mathbb{R}^n$ there exists a quadratic polynomial $Q_{\bar{x}}$ and a function $R_{\bar{x}}$ such that
\[
f(x) = Q_{\bar{x}}(x) + R_{\bar{x}}(x) \mbox{ and } \lim_{u \rightarrow 0} \frac{R_{\bar{x}}(\bar{x}+u)}{\|u\|^2}=0~.
\]
\end{theorem}

\begin{proof}[Proof of Proposition \ref{propositionpointsterminaux}]
Let $k = \operatorname{Supp}(K) \in \operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$, and let $\tilde{k}$ be its 1-homogeneous extension. Let $\bar{x} \in \mathbb{R}^n$ be a point at which $\tilde{k}$ is twice differentiable, and let $Q_{\bar{x}}$ and $R_{\bar{x}}$ be as in Alexandrov's theorem. Since $n >p$, the vector space $\{x_1=\cdots=x_p=0\}$ has positive dimension, hence up to a rotation of $K$ we may assume that the first components of $\bar{x}$ are $\bar{x}_1=\cdots=\bar{x}_p=0$.

Let $\varphi \in C^\infty_c (\mathbb{R}^n)$ be a non-negative function, with support in the unit ball in $\mathbb{R}^n$, positive in a neighborhood of $0$, and with $\int_{\mathbb{R}^n} \varphi=1$. For $\epsilon>0$, let $\varphi_\epsilon \in C^\infty_c (\mathbb{R}^n)$ be the function $\varphi_\epsilon(x)=\frac{1}{\epsilon^n} \varphi(\frac{x-\bar{x}}{\epsilon})$: this function is non-negative, has support in $B(\epsilon,\bar{x})$ (the ball centered at $\bar{x}$ and with radius $\epsilon$), and $\int_{\mathbb{R}^n}\varphi_\epsilon=1$.

Let $t <0$. We want to show that $(1-t)k + t b_{p,n} \notin \operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$. We argue by contradiction: assume that $(1-t)k + t b_{p,n} \in \operatorname{Supp}(\mathcal{K}_{SV_1}^{n*})$. Then $(1-t)\tilde{k} + t \widetilde{b_{p,n}}$ is a convex function on $\mathbb{R}^n$, hence its Laplacian is non-negative in the weak sense, so in particular we have
\begin{equation} \label{eqdemopointterminal}
\int_{\mathbb{R}^n}((1-t)\tilde{k} + t \widetilde{b_{p,n}}) \Delta_e \varphi_\epsilon 	 \geq 0~.
\end{equation}
 We will first show that we always have
\begin{equation} \label{eqdemopointterminalbis}
\int_{\mathbb{R}^n}\tilde{k} \Delta_e \varphi_\epsilon \underset{\epsilon \rightarrow 0} \longrightarrow + \infty ~.
\end{equation}
Since $t$ is negative, with equation \eqref{eqdemopointterminal} it is sufficient to show that
\begin{equation} \label{eqdemopointterminalter}
\int_{\mathbb{R}^n}\widetilde{b_{p,n}} \Delta_e \varphi_\epsilon \underset{\epsilon \rightarrow 0} \longrightarrow + \infty ~.
\end{equation}
Now we need to argue depending whether $p=1$ or $p \geq 2$.
\begin{itemize}
\item If $p \geq 2$ we have $\Delta_e \widetilde{b_{p,n}} (x) = \frac{r_1(p)(p-1)}{\sqrt{x_1^2 + \cdots+x_p^2}}$, and since $\bar{x}_1=\cdots=\bar{x}_p=0$ we have $\sqrt{x_1^2+\cdots+x_p^2} \leq \|x-\bar{x}\|$, hence $\Delta_e \widetilde{b_{p,n}} (x) \geq \frac{r_1(p)(p-1)}{\epsilon}$ for every $x \in B(\epsilon,\bar{x})$, so we have (by using Green's formula)
\[
\int_{\mathbb{R}^n} \widetilde{b_{p,n}} \Delta_e \varphi_\epsilon =  \int_{B(\epsilon,\bar{x})} \varphi_\epsilon \Delta_e \widetilde{b_{p,n}} \geq \frac{r_1(p)(p-1)}{\epsilon} \int_{B(\epsilon,\bar{x})}  \varphi_\epsilon=\frac{r_1(p)(p-1)}{\epsilon}~,
\]
and this gives \eqref{eqdemopointterminalter}.

\item If $p=1$, then we have
\begin{eqnarray*}
\int_{\mathbb{R}^n} \widetilde{b_{p,n}}(x) \Delta_e \varphi_\epsilon(x) \mbox{d}x &=& \frac{1}{2} \int_{\mathbb{R}^n} |x_1| \Delta_e \varphi_\epsilon(x) \mbox{d}x\\
 &=&  \int_{\mathbb{R}^{n-1}} \varphi_\epsilon (0,x_2,\ldots,x_n) \mbox{d}x_2 \ldots \mbox{d}x_n \\
&=& \frac{1}{\epsilon^n} \int_{\mathbb{R}^{n-1}} \varphi \left(0,\frac{x_2-\bar{x}_2}{\epsilon},\ldots,\frac{x_n-\bar{x}_n}{\epsilon}\right) \mbox{d}x_2 \ldots \mbox{d}x_n \\
&=& \frac{1}{\epsilon} \int_{\mathbb{R}^{n-1}} \varphi (0,y_2,\ldots ,y_n) \mbox{d}y_2 \ldots \mbox{d}y_n ~:
\end{eqnarray*}
the second equality is a classical computation, the third is true because $\bar{x}_1=0$, and for the last one we use the change of variable $y_i = \frac{x_i - \bar{x}_i}{\epsilon}$. Since $\varphi$ is positive in a neighborhood of zero we have $\int_{\mathbb{R}^{n-1}} \varphi (0,y_2,\ldots,y_n) \mbox{d}y_2 \ldots \mbox{d}y_n >0$, and this gives (\ref{eqdemopointterminalter}).
\end{itemize}

Moreover, since $\widetilde{k}=Q_{\bar{x}} + R_{\bar{x}}$ we have
\[
\int_{\mathbb{R}^n}\widetilde{k} \Delta_e \varphi_\epsilon = \int_{\mathbb{R}^n}Q_{\bar{x}} \Delta_e \varphi_\epsilon + \int_{\mathbb{R}^n} R_{\bar{x}}\Delta_e \varphi_\epsilon~.
\]
The function $Q_{\bar{x}}$ is a quadratic polynomial, hence its Laplacian is equal to a constant $C \in \mathbb{R}$, which gives $\int_{\mathbb{R}^n}Q_{\bar{x}} \Delta_e \varphi_\epsilon = \int_{\mathbb{R}^n} C \varphi_\epsilon =C$. And since $\Delta_e \varphi_\epsilon(x) =\frac{1}{\epsilon^{n+2}} \Delta_e \varphi(\frac{x-\bar{x}}{\epsilon})$, with the change of variable $y=\frac{x-\bar{x}}{\epsilon}$ we have
\begin{eqnarray*}
\int_{\mathbb{R}^n} R_{\bar{x}}(x) \Delta_e \varphi_\epsilon(x) \mbox{d}x &=& \frac{1}{\epsilon^{n+2}} \int_{B(\epsilon,\bar{x})} R_{\bar{x}}(x) \Delta_e \varphi \left(\frac{x-\bar{x}}{\epsilon} \right)\mbox{d}x \\
& =& \frac{1}{\epsilon^2} \int_{B(1,0)} R_{\bar{x}}(\bar{x}+\epsilon y) \Delta_e \varphi (y) \mbox{d}y ~.
\end{eqnarray*}
Since $\frac{R_{\bar{x}}(\bar{x}+u)}{\|u\|^2} \underset{u \rightarrow 0} \longrightarrow 0$, there exists $M>0$ such that $| R_{\bar{x}}(\bar{x}+u)| \leq M \|u\|^2$ for $\|u\|$ small enough, hence for $\epsilon$ small enough we have, for every $y \in B(1,0)$, $|R_{\bar{x}}(\bar{x}+ \epsilon y)| \leq M \epsilon^2 \|y\|^2$, hence we obtain
\[
|\int_{\mathbb{R}^n} R_{\bar{x}}(x) \Delta_e \varphi_\epsilon(x) \mbox{d}x | \leq M \int_{B(1,0)} \|y\|^2 |\Delta_e \varphi(y)| \mbox{d}y~.
\]
The integral $\int_{\mathbb{R}^n} R_{\bar{x}} \Delta_e \varphi_\epsilon$ does not go to $+\infty$ when $\epsilon$ goes to zero, and by (\ref{eqdemopointterminalbis}) this is a contradiction.
\end{proof}

\subsection{Proof of Theorems~\ref{main} and \ref{main1b}}\label{sec oriented shapes}

Remember that the notations used in this article are described at the end of the introduction.
Let $\mathcal{K}^{n*}_{SH}$ be the set of convex bodies in $\mathbb{R}^n$, which are not points nor segments, with Steiner point equal to 0, and modulo positive homotheties. Let $d_{SH}$ be the distance on $\mathcal{K}^{n*}_{SH}$ such that the projection map
$$
(\mathcal{K}^{n*}_{SH},d_{SH}) \rightarrow(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)
$$
is an isometry. We obviously have the inclusion
$\operatorname{Supp}(\mathcal{K}^{n*}_{SH}) \subset \mathbb{H}^\infty_n~,$
and by construction of the metric we have the following:
$$
(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d) \mbox{ is isometric to } (\operatorname{Supp}(\mathcal{K}^{n*}_{SH}),d_\mathbb{H}) \mbox{ and } (\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*}), d_{\mathbb{K}})~.
$$

Moreover, $\operatorname{Supp}(\mathcal{K}^{n*}_S)$ is a convex cone in $\mathcal{C}_n$. Since $\mathcal{K}^{n*}_{SH}$ is the quotient of $\mathcal{K}^{n*}_S$ by the action of positive homotheties, and $\mathbb{H}_n^\infty$ is the projective quotient of
$\mathcal{C}_n$, we obtain that $\operatorname{Supp}(\mathcal{K}^{n*}_{SH})$ is a convex subset of $\mathbb{H}_n^\infty$.

This gives part of Theorem \ref{main1b}, as well as part of Theorem \ref{main}: $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is isometric to a convex subset of $\mathbb{H}^\infty_n$. It is a uniquely geodesic metric space, the unique shortest path between $[K_1]$ and $[K_2]$ being $[(1-t) K_1+t K_2]$, $t\in [0,1]$, and it has curvature bounded from below and above by $-1$ in the sense of Alexandrov (see Section~\ref{eq bounded curv} for the definition of this property). It is a proper metric space by Proposition \ref{prop proper}, and Proposition~\ref{propositionpointsterminaux} gives the non-extendability property for shortest paths.

Since $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ is proper, it is complete, hence $(\operatorname{Supp}(\mathcal{K}^{n*}_{SH}),d_\mathbb{H})$ is also complete, so $\operatorname{Supp}(\mathcal{K}^{n*}_{SH}) \subset \mathbb{H}_n^\infty$ is a closed subspace. Now, let us prove that $\operatorname{Supp}(\mathcal{K}^{n*}_{SH})$ has empty interior. If this is not true, then there exists a ball $B$ in $(\mathbb{H}^\infty_n,d_\mathbb{H})$ such that $B \subset \operatorname{Supp}(\mathcal{K}^{n*}_{SH})$; we can even assume that $\bar B$ (the closure of $B$) satisfies $\bar B \subset \operatorname{Supp}(\mathcal{K}^{n*}_{SH})$. Since $(\operatorname{Supp}(\mathcal{K}^{n*}_{SH}),d_\mathbb{H})$ is proper, closed balls are compact, hence $\bar B$ is compact. Hence there exists a non-empty relatively compact open set in $(\mathbb{K}\mathbf{lein}^\infty_n,d_{\mathbb{K}})$. Hence by Corollary~\ref{cor bilip} this is also true for the infinite-dimensional Banach space $(H^1(\mathbb{S}^{n-1})_{01},d_{01})$, and that is impossible: a closed ball would be compact.

Moreover $\operatorname{Supp}(\mathcal{K}^{n*}_{SH})$ contains an entire geodesic of $\mathbb{H}_n^\infty$: it is sufficient to do it for $n=2$. In the plane, consider the following segments: $K_1=[-1,1]\times\{0\}$ and $K_2=\{0\} \times [-1,1]$. For $0\leq t \leq 1$, the convex combination $(1-t)K_1+t K_2$ is the rectangle $[-(1-t),1-t] \times [-t,t]$, whose Steiner point is 0. This gives an entire geodesic of $\mathbb{H}_2^\infty$ contained in $\operatorname{Supp}(\mathcal{K}^{2*}_{SH})$.

The convex hyperbolic polyhedra  constructed
in \cite{BG} parametrize the shapes of convex polygons with fixed angles; by construction, they isometrically
embed into $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2})$.
As the dimension of the hyperbolic polyhedra is $(s-3)$ if the polygons have $s$ edges, and by the isometric immersions of  $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2})$ into any  $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ induced by any linear isometric immersion of
$\mathbb{R}^2$ into $\mathbb{R}^n$,  we arrive at the following fact, which in particular shows that $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$ has infinite Hausdorff dimension:
\begin{fact}
For any $p \in \mathbb{N}$, there is an open ball of the finite dimensional hyperbolic space $\mathbb{H}^p$ that isometrically embeds into  $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$.
\end{fact}

Let us prove the assertion about the boundary $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ of $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$. It is the space of segments, up to translations and homotheties: indeed, for example by looking at the isometric model $(\operatorname{Supp}(\mathcal{K}_{SV_1}^{n*}), d_{\mathbb{K}})$, we see that the convex bodies $K$ on the boundary are the one for which $V_2(K)=0$ (see Proposition \ref{lem bord infini}) and $V_1(K)=1$, and these are exactly segments. Hence $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is in bijection with $P^{n-1}(\mathbb{R})$, the real projective space of dimension $n-1$ (that is, the space of lines in $\mathbb{R}^n$).

We can endow $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ with the visibility metric from $[B^n]$: the distance between $a,b \in \partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, denoted by $<_B(a,b)$,   is the angle (with value in $[0,\pi]$) between the two lines $c_a$ and $c_b$ from $[B^n]$ and with endpoints $a$ and $b$ respectively. But clearly
the element of $O(n)$ sending the line $a$ to the line $b$  is also a $\d$-isometry sending $c_a$ to $c_b$. In turn, $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ endowed with the visibility metric is isometric to $P^{n-1}(\mathbb{R})$ endowed with its round metric.
From \cite[Proposition II.9.2]{BH}, $<_B:\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*} \times \partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*} \to \mathbb{R}$ is continuous for the classical topology on $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. Hence for this topology, $\partial \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is homeomorphic to $P^{n-1}(\mathbb{R})$.

\begin{remark}{\rm
The smallest vector space containing $\operatorname{Supp}(\mathcal{K}^{n})$  as a convex cone
is the vector space spanned by the cone:
$$\mathrm{Sonic}^n=\{ h-k | h,k\in  \operatorname{Supp}(\mathcal{K}^{n})\}~, $$
the space of \emph{$n$-dimensional hedgehogs}.
See \cite[9.6]{schneider}, \cite{schnei-h} and the references therein for more informations.
Let us say that
the name was coined in \cite{LLR1}, although
they previously appeared in the literature  under different names, see \cite{oliker}.
If $h\in \mathrm{Sonic}^n$, there is a way to associate a geometric object in $\mathbb{R}^n$, see  \cite{schnei-h,MM}. See  figures~\ref{fig:extreme}, \ref{fig:terminaux} and \ref{fig:quadrangl} for  illustration.

A description of $\mathrm{Sonic}^2$ in $C^0(\mathbb{S}^1)$ is contained in \cite{MM}.
But $\mathrm{Sonic}^n$ is not complete for any reasonable norm on it
---it contains
 $C^2(\mathbb{S}^{n-1})$, so it is dense in both $H^1(\mathbb{S}^{n-1})$ and $C^0(\S^{n-1})$ endowed with their classical norms.

 Partial results were achieved in this setting (mostly in the regular case) in \cite{mmarx0,mmarx,mmarx2}.
}\end{remark}

\begin{figure}
\begin{center}
\psfrag{P}{$(\overline{V_1}^n)^{-1}(0)$}
\psfrag{a}{$(\overline{V_2}^n)^{-1}(0)$}
\psfrag{K}{$\operatorname{Supp}(\mathcal{K}_S^{n*})$}
\psfrag{dK}{$\partial \mathcal{K}$}
\psfrag{+}{$\scriptsize{+}$}
\includegraphics[width=0.7\linewidth]{quadrangle}\caption{The minus sign outside of the polygons indicates edges with negative algebraic length, while the minus sign inside a polygon indicates a negative $\overline{V_2}^n$. The dotted line is the extension of the segment between the segment and the triangle.}\label{fig:quadrangl}
\end{center}
\end{figure}

\begin{remark}[Plane convex bodies]{\rm
As a particular example, in the plane $\mathbb{R}^2$ we have
$$d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K],[B^2])= \operatorname{argch} \left( \frac{\mathrm{per}(K)}{2\sqrt{\pi \operatorname{vol}_2(K)}} \right)~,$$ that is coherent with the isoperimetric inequality (we denote by $\mathrm{per}(K)$ is the perimeter of $K$). In the plane, all the preceding results can be proven directly from Wirtinger's inequality, as the intrinsic area is given by
$$V_2(K)= \frac{1}{2} \int_0^{2\pi} \operatorname{Supp}(K)^2- (\operatorname{Supp}(K)')^2~,$$
where a support function defined on the unit circle
is considered as a $2\pi$-periodic function
on $\mathbb{R}$. See the last chapter of \cite{debin} for more details.
}\end{remark}

\section{The space of shapes $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$}\label{sec shapes}

In this section, we will investigate $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, the quotient of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ by the action of linear isometries $O(n)$ of the Euclidean space $\mathbb{R}^n$: $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is the space of convex bodies in $\mathbb{R}^n$ (not reduced to points or segments) up to Euclidean similarities. We will also study its quotient metric $d_{\mathscr{S}^n}$.

In  particular we obtain that, as a quotient of a CBB($-1$) space,  $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},d_{\mathscr{S}^n})$ is CBB($-1$). Although this is a well-known fact
(see Section~10.2.2 in \cite{BBI}),
for a matter of presentation we give a complete argument.

\subsection{Metric spaces with bounded curvature}\label{eq bounded curv}

Consider the hyperbolic plane $\mathbb{H}^2$, with its hyperbolic metric $d_h$. Let $ABC$ be a triangle in $\mathbb{H}^2$, and let us define $a=d_h(B,C)$, $b=d_h(A,C)$ and $c=d_h(A,B)$. Let $\operatorname{mid}(a,b,c)$ be the distance between $C$ and the midpoint of the shortest path $AB$. For futur references, let us note that the function $\operatorname{mid}$ is increasing in $a$ and $b$: using two times the hyperbolic cosine law, one gets
$$\cosh(\operatorname{mid}(a,b,c))=\frac{\cosh a + \cosh b}{2\cosh (c/2)}~. $$

\begin{definition}\label{def cbb} We say that a metric space $(X,d)$ is a CAT($-1$) (resp. CBB($-1$)) space if $(X,d)$ is geodesic and, for any triangle with length sides $a,b,c$ and vertices $A,B,C$ of $(X,d)$, if $m$  is the middle point of the shortest path $AB$, then
$$d(C,m) \leq \operatorname{mid}(a,b,c) \, (\mbox{resp. } d(C,m) \geq \operatorname{mid}(a,b,c))~.$$
\end{definition}

Note that the condition about the triangles in Definition~\ref{def cbb} above is global, and that we don't require completeness of $(X,d)$.

\subsection{Properties of a quotient metric}

Let $(X,d)$ be a metric space, and let $O$ be a subgroup of its isometry group. Assume that $O$ is endowed with a topology for which it is compact, and such that the maps $A \in O \mapsto A(x) \in X$ are continuous for every $x \in X$.
Let $\bar X$ be the quotient space $\bar X=X/O$. For $x \in X$, we denote by $\bar x$ the equivalent class. Let $\bar d:\bar X \times \bar X \rightarrow \mathbb{R}$ be the following function:
\[
\bar d(\bar x, \bar y) := \inf_{A,B \in O} d(A(x),B(y))~.
\]
Since $O$ is a group of isometries we have $d(A(x),B(y))=d(x,A^{-1}B(y))$, so $\bar d(\bar x, \bar y) = \inf_{A \in O} d(x,A(y))$. Moreover, by compactness of $O$ and continuity of the map $A \in O \mapsto A(y) \in X$ we have
\[
\bar d(\bar x, \bar y) = \min_{A \in O} d(x,A(y))~.
\]

\begin{lemma} \label{propquotientmetricisametric}
The function $\bar d$ is a metric on $\bar X$.
\end{lemma}

\begin{proof}
The symmetry property and the fact that $\bar d$ separates points are obvious. Now let $\bar x, \bar y, \bar z \in \bar X$, and let $A,B\in O$ such that $\bar d(\bar x,\bar y)=d(x,A(y))$ and $\bar d (\bar y,\bar z)=d(y,B(z))$. Then
\begin{eqnarray*}
\bar d(\bar x,\bar z) &\leq&  d(x,A B (z)) \leq d(x,A(y)) + d(A(y),AB(z)) \\
 & = & \bar d(\bar x,\bar y) + d(y,B(z)) = \bar d(\bar x,\bar y) + \bar d(\bar y, \bar z)~.
 \end{eqnarray*}
\end{proof}

Let us denote by $p:X\to \bar X$ the projection map. It may exist an element $x$ of $X$ such that for a continuous path $A_t\in O$ we have $A_t(x)=x$ (for example this will be the case with $X=\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ and $x=[B^n]$). In this case, $p$ is not locally injective, so in general $p$ is not a covering map. Nevertheless, the following holds.

\begin{lemma}\label{lem quotient topo}
The topology given by $\bar d$ on $\bar X$ is the quotient topology.
\end{lemma}
\begin{proof}
We need to show that a set $\bar \Omega \subset \bar X$ is open (for the topology given by $\bar d$) if and only if $p^{-1}(\bar \Omega) \subset X$ is an open set.

First, let us observe that the map $p$ is continuous: indeed, if $x_n$ in $X$ converges to $x$ for $d$, then we have
$0\leq \bar d(\bar x_n,\bar x) \leq d(x_n,x)$, hence $\bar x_n$ converge to $\bar x$. This shows that if $\bar \Omega \subset \bar X$ is open, then $p^{-1}(\bar \Omega) \subset X$ is an open set.

On the other hand, assume that $\bar \Omega \subset \bar X$ is a set such that $p^{-1}(\bar \Omega)$ is an open set of $X$. We want to show that $\bar \Omega$ is open.
Let $\bar x \in \bar \Omega$, and let $x \in X$ be such that $p(x)=\bar x$. We define the distance to $X-p^{-1}(\bar \Omega)$ as the function
$$y \in X \mapsto d(y,X - p^{-1}(\bar \Omega)) := \inf_{z \in X-p^{-1}(\bar \Omega)} d(y,z)$$
(note that if $X-p^{-1}(\bar \Omega)=\emptyset$, then $\bar X = p(X) = p(p^{-1}(\bar \Omega)) \subset \bar \Omega$, so $\bar \Omega = \bar X$ is an open set).

This function is continuous (indeed it is $1-$Lipschitz), hence the function
\[
A \ni O \mapsto d(A(x),X-p^{-1}(\bar \Omega))
\]
is also continuous. Since $O$ is compact, this function attains its minimum. And this minimum can not be zero: indeed we have $p(A(x))=p(x)=\bar x \in \bar \Omega$, so $A(x) \notin X - p^{-1}(\bar \Omega)$, which gives $d(A(x),X-p^{-1}(\bar \Omega)) >0$ (note that $X-p^{-1}(\bar \Omega)$ is a closed set). So there exists $r>0$ such that
\begin{equation} \label{eqdistancetopoquotient}
d(A(x),y) \geq r \mbox{ for every } A \in O  \mbox{ and every } y \in X-p^{-1}(\bar \Omega) ~.
\end{equation}
This shows that $B_{\bar X}(\bar x,r)$ (which is the ball in $(\bar X, \bar d)$ with center $\bar x$ and radius $r$) is included in $\bar \Omega$. Indeed, let $\bar y \in B_{\bar X}(\bar x,r)$, and let $y \in X$ be such that $p(y)=\bar y$: there exists $A \in O$ such that $\bar d(\bar x,\bar y) = d(A(x),y) <r$, and equation (\ref{eqdistancetopoquotient}) shows that $y \in p^{-1}(\bar \Omega)$, that is $\bar y = p(y) \in \bar \Omega$.

\end{proof}

The aim of the rest of this section is to prove the following.
\begin{proposition} \label{propquotientmetric}
The following properties hold:
\begin{enumerate}
\item If $d$ is proper, then $\bar d$ is proper.
\item If $d$ is a geodesic metric, then $\bar d$ is a geodesic metric.
\item If $(X,d)$ is CBB($-1$), then $(\bar X,\bar d)$ is CBB($-1$).
\end{enumerate}
\end{proposition}

\begin{proof}[Proof of property 1 in Proposition \ref{propquotientmetric}]
Suppose that $(X,d)$ is a proper metric space, and let $(\bar x_i)_{i\in \mathbb{N}}$ be a bounded sequence in $(\bar X, \bar d)$.  There are  $A_i \in O$ such that $(A_i(x_i))_{i\in \mathbb{N}}$
is a bounded sequence in $(X,d)$. Since $(X,d)$ is proper, up to extract a subsequence, there exists $y \in X$ such that $d(A_i(x_i),y)\to 0$.
As $\bar d(\bar x_i, \bar y) \leq d(A_i(x_i),y)$, we have $\bar d(\bar x_i, \bar y) \to 0$.
\end{proof}

Property 2 in Proposition \ref{propquotientmetric} is a direct consequence of the
\begin{lemma}\label{lem geod shape}
Let $x,y \in X$, and let $A \in O$ be such that $\bar d(\bar x,\bar y)=d(x,A(y))$. Suppose that $\gamma$ is a shortest path between $x$ and $A(y)$. Then the projection $\bar \gamma = p \circ \gamma$ is a shortest path between $\bar x$ and $\bar y$. Moreover the projection is an isometry from $\gamma$ to $\bar \gamma$.
\end{lemma}

\begin{proof}
Let us suppose that $\gamma:[0,1]\to X$ is  affinely parametrized. Then for any $0\leq s\leq t \leq 1$,
\[
 \bar d(\bar \gamma(s), \bar \gamma(t)) \leq d(\gamma(s),\gamma(t))= (t-s) d(x,A(y)) =(t-s) \bar d(\bar x, \bar y)~.
\]
Using three times this inequality we obtain
\begin{eqnarray*}
\bar d(\bar x, \bar y) & \leq & \bar d(\bar \gamma(0), \bar \gamma(s)) + \bar d(\bar \gamma(s), \bar \gamma(t)) + \bar d(\bar \gamma(t), \bar \gamma(1)) \\
 & \leq & (s+(t-s)+(1-t)) \bar d(\bar x, \bar y) = \bar d(\bar x, \bar y)~.
\end{eqnarray*}
All these inequalities are equalities, so in particular $\bar d(\bar \gamma(s), \bar \gamma(t)) = (t-s)\bar d(\bar x,\bar y)$.
\end{proof}

In general, this is not true that every shortest path in the quotient space is obtained as the projection of a shortest path.
To prove property 3 in Proposition \ref{propquotientmetric} we need the following Lemma:

\begin{lemma}\label{lemmeversionutile}
Assume that $d$ is geodesic (hence $\bar d$ is also geodesic). If $\bar w \in \bar X$ is the midpoint of a shortest path between $\bar x \in \bar X$ and $\bar y \in \bar X$, then there exists $A,B \in O$ such that:
\begin{itemize}
\item $\bar d (\bar x,\bar y) = d(x,A(y))$ and
\item $B(w)$ is the midpoint of a shortest path joining $x$ and $A(y)$ in $X$.
\end{itemize}
\end{lemma}

\begin{proof}
Let $B \in O$ such that $\bar d(\bar x, \bar w) = d(x,B(w))$, and $A \in O$ such that $\bar d(\bar y, \bar w) = d(y, A^{-1}B(w))$. We have
\begin{eqnarray*}
\bar d(\bar x, \bar y) & \leq & d(x,A(y))  \leq d(x,B(w)) + d(B(w),A(y)) \\
& = & \bar d( \bar x, \bar w)+d(A^{-1}B(w),y) = \bar d(\bar x, \bar w) + \bar d(\bar y, \bar w) = \bar d(\bar x, \bar y)~.
\end{eqnarray*}
Hence all these inequalities are equalities and this ends the proof.
\end{proof}

\begin{proof}[Proof of property 3 in Proposition \ref{propquotientmetric}]
Suppose that $(X,d)$ is a CBB($-1$) space. Then in particular $(X,d)$ and $(\bar X, \bar d)$ are geodesic metric spaces.

Let $T$ be a geodesic triangle in $\bar X$ with vertices $\bar x \bar y \bar z$, and let $\bar w$ be the midpoint of the geodesic between $\bar x$ and $\bar y$. In order to show that $(\bar X, \bar d)$ is CBB($-1$) we need to prove the following inequality:
\begin{equation}\label{eq cbb}
\bar d(\bar z, \bar w) \geq \operatorname{mid}(\bar d(\bar y, \bar z), \bar d(\bar x, \bar z), \bar d(\bar x, \bar y))~.
\end{equation}

By Lemma~\ref{lemmeversionutile}, there exists $A,B \in O$ such that $\bar d (\bar x,\bar y) = d(x,A(y))$,
and $B(w)$ is the midpoint of a shortest path $L$ joining $x$ and $A(y)$ in $X$. Let $C \in O$ be such that $\bar d(\bar z, \bar w) = d(C(z), B(w))$.

Now consider the geodesic triangle $xA(y)C(z)$ in $X$ (the shortest path between $x$ and $A(y)$ is $L$). Since $(X,d)$ is CBB($-1$), and $B(w)$ is the midpoint of the geodesic between $x$ and $A(y)$, we have
\begin{equation}\label{ineq: CBB}
\bar d(\bar z,\bar w) = d(C(z),B(w)) \geq \operatorname{mid}(a,b,c)~,
\end{equation}
where
\begin{equation}\label{ineq CBB eq}
\begin{array}{c c c cc}
    a & = & d(A(y),C(z)) & \geq & \bar d (\bar y,\bar z)~, \\
    b & = & d(x,C(z)) & \geq & \bar d (\bar x,\bar z)~, \\
    c & = & d(x,A(y)) &  =&  \bar d (\bar x,\bar y)~.           \\
\end{array}
\end{equation}
This proves (\ref{eq cbb}) since the function $\operatorname{mid}$ is increasing in $a$ and $b$.
\end{proof}

\subsection{The space of shapes}

Let $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ be the quotient of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ by linear isometries of the Euclidean space $\mathbb{R}^n$: the action of $O(n)$ on
$\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is defined by $\Phi[K]:=[\Phi K]$. For $K\in\mathcal{K}^{n*}$,  we will denote by $\llbracket K\rrbracket$
the set of convex bodies differing from $K$ by positive homotheties and Euclidean isometries.

Since $V_2$ is $O(n)-$invariant, we have $\d(\Phi[K_1],\Phi[K_2])=\d([K_1],[K_2])$, so $O(n)$ acts by isometries on $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$.
\begin{fact}
Let $K \in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. The map $ O(n) \ni \Phi  \mapsto \Phi[K] \in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is continuous.
\end{fact}
\begin{proof}
We have
$\operatorname{Supp}(\Phi K)(x)=\operatorname{Supp}(K)(\Phi^ {-1}(x))$. In particular, if $\Phi_n\to\Phi$ in $O(n)$,
then for any $x\in \mathbb{S}^{n-1}$, $\operatorname{Supp}(\Phi_n K)(x)$ converges to $\operatorname{Supp}(\Phi K)(x)$. The result follows from Lemma~\ref{lem:point impl unif} and Corollary~\ref{same cor}.
\end{proof}

This fact shows that we can apply the results of the previous section, for the action of the (compact) group $O(n)$ on $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d)$. Let us introduce
\begin{equation}\label{def bd}d_{\mathscr{S}^n}(\llbracket K_1\rrbracket,\llbracket K_2\rrbracket)=\inf_{\Phi,\Phi'\in O(n)} \d (\Phi[K_1],\Phi'[K_2])~.\end{equation}
We have
$d_{\mathscr{S}^n}(\llbracket K_1\rrbracket,\llbracket K_2\rrbracket)=\min_{\Phi \in O(n)} \d([K_1], \Phi[K_2])~.$ The previous section gives the following.
\begin{proposition} \label{prop:shapeCBBproper}
$(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},d_{\mathscr{S}^n})$ is a CBB($-1$) proper geodesic metric space.
\end{proposition}

\subsection{Non-uniqueness of shortest paths in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$}\label{sectionnonuniqueness}

The aim of this section is to prove that shortest paths are not unique in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. Obviously, since $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*}$ isometrically embedds in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ for $n \geq 2$, this is sufficient to prove this property for $n=2$. Hence in this section we consider convex bodies in $\mathbb{R}^2$.

Let $K$ be the intersection of the half-space $[0,\infty) \times \mathbb{R}$ with the ellipse with center 0, width $2\sqrt{2}$ and height $\frac{2}{\sqrt{2}}$. The support function of $K$ is a function on $\mathbb{S}^1$, and with the parametrization $x=(\cos s,\sin s) \in \mathbb{S}^1$, for $s\in[0,2\pi]$, we will actually define the
support function $k$ of $K$ on $[0,2\pi]$. Namely,
$$
k(s) = \sqrt{2 \cos^2 s + \frac{1}{2} \sin^2 s} \mbox{ for }  s \in [-\frac{\pi}{2}, \frac{\pi}{2}], \mbox{ and } k(s) = \frac{1}{\sqrt{2}} |\sin s| \mbox{ for } s \in [\frac{\pi}{2}, \frac{3\pi}{2}]~.
$$
Let $(\beta,0)$ be the Steiner point of $K$, and let $\alpha = V_1(K) = \frac{1}{2} \int_0^{2\pi} k \simeq 2.4$. Then the convex body $K_1 = \alpha^{-1}K + (-\alpha^{-1}\beta,0)$ has Steiner point 0, and $V_1(K_1)=1$: hence $K_1 \in\mathcal{K}_{SV_1}^{2*}$. Its support function $k_1 \in \operatorname{Supp}(\mathcal{K}_{SV_1}^{2*})$ is given by
$$ k_1(s) = \alpha^{-1} \left(\sqrt{2 \cos^2 s + \frac{1}{2} \sin^2 s}-\beta \cos s \right) \mbox{ for } s \in [\frac{-\pi}{2},\frac{\pi}{2}]$$
and
$$ k_1(s) = \alpha^{-1}\left( \frac{1}{\sqrt{2}}|\sin s| - \beta \cos s \right) \mbox{ for } s \in [\frac{\pi}{2},\frac{3\pi}{2}]~.$$

Let $K_2$ be the rectangle $[-\frac{2}{5},\frac{2}{5}]\times [-\frac{1}{10},\frac{1}{10}]$. Obviously, 0 is the Steiner point of $K_2$. Its support function is defined for any $s \in [0,2\pi]$ by
$$ k_2(s) = \frac{2}{5} |\cos s| + \frac{1}{10} |\sin s|~,$$
and since $K_2 = [-\frac{2}{5},\frac{2}{5}] \times \{0\} +  \{0\} \times [-\frac{1}{10},\frac{1}{10}]$, we have $V_1(K_2) = \operatorname{length}([-\frac{2}{5},\frac{2}{5}]) + \operatorname{length}([-\frac{1}{10},\frac{1}{10}])=1$. Hence $K_2 \in \mathcal{K}_{SV_1}^{2*}$ and $k_2 \in \operatorname{Supp} (\mathcal{K}_{SV_1}^{2*})$.

Let $\llbracket K_1 \rrbracket$ and $\llbracket K_2 \rrbracket$  be the corresponding equivalent classes in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*}$. Since $K_2$ is invariant by the symmetry with respect to the horizontal line, the distance between $\llbracket K_1 \rrbracket$ and $\llbracket K_2 \rrbracket$ is given by
$$
d_{\mathscr{S}^2} (\llbracket K_1 \rrbracket, \llbracket K_2 \rrbracket) = \min_{\theta \in \mathbb{R}} d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], R_\theta [K_2])~,
$$
where we denote by $R_\theta$ the rotation of angle $\theta$ in $\mathbb{R}^2$. We will prove the following:
\begin{proposition} \label{prop non unicite}
The minimum is obtained for $\theta=0$ and $\theta=\frac{\pi}{2}$, that is we have
$$d_{\mathscr{S}^2} (\llbracket K_1 \rrbracket, \llbracket K_2 \rrbracket) = d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], [K_2]) =d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], R_{\frac{\pi}{2}}[K_2]) ~.$$
\end{proposition}

This Proposition is sufficient to prove the non-uniqueness of shortest paths in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*}$. Indeed, Lemma~\ref{lem geod shape} shows that the projections of the shortest paths in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*}$ between $[K_1]$ and $[K_2]$, and between $[K_1]$ and $R_{\frac{\pi}{2}}[K_2]$, are again shortest paths in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{2*}$. But these two shortest paths are different: the first shortest path contains the point $\llbracket \frac{1}{2} K_1 + \frac{1}{2}K_2 \rrbracket$, and this point is not on the second shortest path $t \mapsto \llbracket (1-t)K_1 + t R_{\frac{\pi}{2}} (K_2) \rrbracket$:
$\frac{1}{2}K_1 + \frac{1}{2}K_2$ is not the image by a rotation of $(1-t)K_1 + t R_{\frac{\pi}{2}}(K_2)$, which is equivalent to say that $\frac{1}{2 }\alpha^{-1} K + \frac{1}{2}K_2$ is not the image by a rotation and a translation of
$(1-t)\alpha^{-1} K + t R_{\frac{\pi}{2}}(K_2)$. See Figure \ref{fig:ellipse}.

\begin{figure}
\begin{center}
\includegraphics[scale=0.6]{ellipse}\caption{The convex body
$\frac{1}{2 }\alpha^{-1} K + \frac{1}{2}K_2$ (middle of the upper line) is not the image by a rotation and a translation of
$(1-t)\alpha^{-1} K + t R_{\frac{\pi}{2}}(K_2)$ (represented on the bottom line for $t=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1$).}\label{fig:ellipse}
\end{center}
\end{figure}

Since $R_\pi [K_2] = [K_2]$, to compute the minimum this is sufficient to consider $\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]$. Moreover, let $T$ be the symmetry with respect to the $x$ axis: we have $T[K_1]=[K_1]$, hence we have
$$d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], R_\theta [K_2])=d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} (T[K_1], R_\theta [K_2])=d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], T \circ R_\theta [K_2]) = d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], R_{-\theta} [K_2])~.$$
This shows that in fact we need only to consider $\theta \in [0,\frac{\pi}{2}]$.

Let $k_2^\theta$ be the support function of $R_\theta[K_2]$, that is $k_2^\theta(s) = k_2(s-\theta)$. We have
$$
\cosh(d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^2} ([K_1], R_\theta [K_2])) = \frac{V_2(k_1,k_2^\theta)}{\sqrt{V_2(k_1)V_2(k_2^\theta)}} = \frac{f(\theta)}{2\sqrt{V_2(k_1)V_2(k_2)}}~,
$$
where we denote by $f(\theta)$ the function defined by
$$
f(\theta) = \int_0^{2\pi} (k_1(s)k_2(s-\theta) - k_1'(s)k_2'(s-\theta)) \mbox{d}s~.
$$
Proposition \ref{prop non unicite} is a direct consequence of the following Lemma:
\begin{lemma}
On $[0,\frac{\pi}{2}]$, $f$ attains its minimum at the points $\theta=0$ and $\theta=\frac{\pi}{2}$.
\end{lemma}

\begin{proof}
Fix $\theta \in (0,\frac{\pi}{2})$, and consider the function $s \mapsto k_1(s) k_2'(s-\theta)$. This function is piecewise $\mathcal{C}^1$, but is not continuous: the function $k_2'(s-\theta)$ has jumps, with height $\frac{1}{5}$ at the points $s= \theta$ and $s= \pi+\theta$,  and with height $\frac{4}{5}$ at the points $s= \frac{\pi}{2} + \theta$ and $s= \frac{3\pi}{2}+\theta$. Hence we have

$$\int_0^{2\pi} (k_1(s)k_2'(s-\theta))' \mbox{d}s = -\frac{1}{5} k_1(\theta) - \frac{1}{5} k_1(\pi+\theta) - \frac{4}{5}k_1(\frac{\pi}{2} + \theta) - \frac{4}{5} k_1(\frac{3\pi}{2} + \theta)
$$
$$ = -\frac{1}{5\alpha} \sqrt{2 \cos^2 \theta + \frac{1}{2}\sin^2 \theta} -\frac{4}{5\alpha} \sqrt{2 \sin^2 \theta + \frac{1}{2}\cos^2 \theta} - \frac{1}{5 \sqrt{2}\alpha} \sin \theta  - \frac{4}{5\sqrt{2}\alpha} \cos\theta~.
$$
The equality $(k_1 k'_2)'=k_1' k_2' +k_1 k_2''$ gives $-k_1' k_2' = k_1 k_2'' - (k_1k_2')'$, so
$$- \int_0^{2\pi} k_1'(s)k_2'(s-\theta) \mbox{d}s = \int_0^{2\pi}  (k_1(s)k_2''(s-\theta) -(k_1(s)k_2'(s-\theta))')\mbox{d}s~,$$
and since $k_2(s-\theta)+ k_2''(s-\theta)=0$ for almost every $s \in [0,2\pi]$ we finally obtain
\begin{eqnarray*}
f(\theta) &=& \int_0^{2\pi} (k_1(s)k_2(s-\theta) - k_1'(s)k_2'(s-\theta)) \mbox{d}s \\
&=& \int_0^{2\pi} (k_1(s)(k_2(s-\theta)+k_2''(s-\theta)) - (k_1(s)k_2'(s-\theta))')\mbox{d}s \\
&=& \frac{1}{5\alpha} \sqrt{2 \cos^2 \theta + \frac{1}{2}\sin^2 \theta} +\frac{4}{5\alpha} \sqrt{2 \sin^2 \theta + \frac{1}{2}\cos^2 \theta} + \frac{1}{5 \sqrt{2}\alpha} \sin \theta  + \frac{4}{5\sqrt{2}\alpha} \cos\theta~.
\end{eqnarray*}
We easily check that $f(0)=f(\frac{\pi}{2})=\frac{\sqrt{2}}{\alpha}$ (the parameters of the ellipse and the segment have been chosen so that this property holds). And a direct computation shows that $f'(0)=\frac{1}{5\sqrt{2}\alpha}>0$ and $f'(\frac{\pi}{2}) = -\frac{4}{5\sqrt{2}\alpha} <0$. Moreover, let $g : [0,1] \rightarrow [0,\infty)$ be defined by
$$
g(u)= \frac{1}{5\alpha} \sqrt{\frac{3}{2} u + \frac{1}{2}} +\frac{4}{5\alpha} \sqrt{2 - \frac{3}{2} u } + \frac{1}{5 \sqrt{2}\alpha} \sqrt{1-u}  + \frac{4}{5\sqrt{2}\alpha} \sqrt{u}~:
$$
with the identity $\cos^2+\sin^2=1$ we easily check that $g(\cos^2 \theta)=f(\theta)$ for any $\theta \in [0,\frac{\pi}{2}]$. Hence $f'(\theta)=-2 g'(\cos^2 \theta)\sin \theta \cos \theta$. But $g$ is strictly concave, hence $g'$ has at most one zero on $[0,1]$, hence $f'$ has also at most one zero on $(0,\frac{\pi}{2})$. And this ends the proof: if the minimum of $f$ on $[0,\frac{\pi}{2}]$ was attained at a point $\theta \notin \{0,\frac{\pi}{2}\}$, since $f'(0)>0$ and $f'(\frac{\pi}{2})<0$, $f'$ would have at least 3 zeros on $(0,\frac{\pi}{2})$, and that is impossible.
\end{proof}

\subsection{Embedding of hyperbolic planes}\label{sectionembeddinghyp}

Trivially, for any $\Phi\in O(n)$ we have $\Phi [B^n]=[B^n]$. Apart from the fact that the action of $O(n)$ on $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is not proper, this says that for any $[K]\in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$,
\begin{equation}\label{eq:metrique boule}d_{\mathscr{S}^n}(\llbracket K \rrbracket, \llbracket B^n \rrbracket)=\d([K],[B^n])~.\end{equation}
From this we  first deduce the following fact.

\begin{fact}[Uniqueness of shortest paths starting from $B^n$]  Let $\llbracket K \rrbracket \in \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. Then there is a \emph{unique} shortest path from $\llbracket  B^n \rrbracket$ to $\llbracket K \rrbracket$, which is the projection of the shortest path in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ between $[B^n]$ and $[K]$.
\end{fact}
\begin{proof}
Let $\bar \delta : [0,d_{\mathscr{S}^n}(\llbracket B^n \rrbracket, \llbracket K \rrbracket)] \rightarrow \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ be an arc-length parametrized shortest path between $\llbracket B^n \rrbracket$ and $\llbracket K \rrbracket$, and let $[\delta(t)] \in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ be such that $\bar \delta(t) = \llbracket \delta(t) \rrbracket$. Let $t \mapsto [\gamma(t)]$ be the (unique) arc-length parametrized shortest path in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ between $[B^n]$ and $[K]$: we want to show that $\llbracket \delta(t) \rrbracket = \llbracket \gamma(t) \rrbracket$.

For any $t \in [0,d_{\mathscr{S}^n}(\llbracket B^n \rrbracket, \llbracket K \rrbracket)]$, let $\Phi_t \in O(n)$ be such that $$d_{\mathscr{S}^n} (\llbracket K \rrbracket ,\llbracket \delta(t) \rrbracket)=\d([K], \Phi_t[\delta(t)])~.$$
Since $t \mapsto \llbracket \delta(t) \rrbracket$ is a geodesic in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ we have
\begin{eqnarray*}
 \d([B^n],\Phi_t[\delta(t)]) + \d(\Phi_t[\delta(t)],[K]) &=& d_{\mathscr{S}^n} (\llbracket B^n \rrbracket ,\llbracket \delta(t) \rrbracket) + d_{\mathscr{S}^n} (\llbracket \delta(t) \rrbracket ,\llbracket K \rrbracket) \\
& = & d_{\mathscr{S}^n} (\llbracket B^n \rrbracket ,\llbracket K \rrbracket) = \d ([B^n],[K])~.
\end{eqnarray*}
Hence $\Phi_t[\delta(t)]$ is on the shortest path between $[B^n]$ and $[K]$ in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. Moreover we have $\d ([B^n], \Phi_t[\delta(t)]) = d_{\mathscr{S}^n}(\llbracket B^n \rrbracket, \llbracket \delta(t) \rrbracket)=t$ (the geodesic $t \mapsto \llbracket \delta(t) \rrbracket$ is arc-length parametrized), so $\Phi_t[\delta(t)] = [\gamma(t)]$ (remember that the geodesic
$t \mapsto [\gamma(t)]$ is also arc-length parametrized). Finally this gives
$\llbracket \delta(t) \rrbracket= \llbracket \gamma(t) \rrbracket$.
\end{proof}

In turn, we can construct totally geodesic hyperbolic surfaces in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$.

\begin{proposition}
Let $\llbracket P \rrbracket, \llbracket Q \rrbracket \in \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ be such that
$\llbracket P \rrbracket, \llbracket Q \rrbracket$ and $\llbracket B^n \rrbracket$ are three different points. Let $A \in O(n)$ be such that $d_{\mathscr{S}^n} (\llbracket P \rrbracket, \llbracket Q \rrbracket)=\d([P],A[Q])$. Then the projection $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*} \rightarrow \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, when restricted to the (plain) geodesic triangle with vertices $[B^n], [P]$ and $A[Q]$, is an isometry onto its image.
\end{proposition}

\begin{proof}
Without loss of generality, we may assume that $A$ is the identity (that is, $d_{\mathscr{S}^n} (\llbracket P \rrbracket, \llbracket Q \rrbracket)=\d([P],[Q])$). Let $[K_1]$ and $[K_2]$ be in the geodesic triangle with vertices $[B^n], [P]$ and $[Q]$: since geodesics in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ are convex combinations, we can write
$$
[K_1] = [\alpha_1 B^n + \beta_1 P + \gamma_1 Q] \mbox{ and } [K_2] = [\alpha_2 B^n + \beta_2 P + \gamma_2 Q]~,
$$
where the $\alpha_i,\beta_i,\gamma_i$ are non-negative real numbers, with $\alpha_1 + \beta_1 + \gamma_1 = \alpha_2 + \beta_2 + \gamma_2 =1$. We want to prove that $d_{\mathscr{S}^n}(\llbracket K_1 \rrbracket, \llbracket K_2 \rrbracket)=\d([K_1],[K_2])$, which means that for any $\Phi \in O(n)$ we have $\d([K_1],[K_2]) \leq \d([K_1],\Phi[K_2])$. Since $V_2$ is $O(n)-$invariant, we only need to show that
\begin{equation} \label{eq isometries triangles}
V_2(K_1,K_2) \leq V_2(K_1,\Phi(K_2))
\end{equation}
($K_1$ and $K_2$ denote two convex bodies in the equivalent classes $[K_1]$ and $[K_2]$). We have
$$
V_2(K_1,K_2) =
\alpha_1 \alpha_2 V_2(B^n) + \alpha_1 \beta_2 V_2(B^n,P) + \alpha_1 \gamma_2 V_2(B^n,Q)$$
$$
+\beta_1 \alpha_2 V_2(P,B^n) + \beta_1 \beta_2 V_2(P) + \beta_1 \gamma_2 V_2(P,Q)$$
$$
+\gamma_1 \alpha_2 V_2(Q,B^n) + \gamma_1 \beta_2 V_2(Q,P) + \gamma_1 \gamma_2 V_2(Q)~.
$$
Moreover $\Phi(K_2) = \alpha_2 B^n + \beta_2 \Phi(P) + \gamma_2 \Phi(Q)$, hence
$$
V_2(K_1,\Phi(K_2)) =
\alpha_1 \alpha_2 V_2(B^n) + \alpha_1 \beta_2 V_2(B^n,\Phi(P)) + \alpha_1 \gamma_2 V_2(B^n,\Phi(Q))$$
$$
+\beta_1 \alpha_2 V_2(P,B^n) + \beta_1 \beta_2 V_2(P,\Phi(P)) + \beta_1 \gamma_2 V_2(P,\Phi(Q))$$
$$
+\gamma_1 \alpha_2 V_2(Q,B^n) + \gamma_1 \beta_2 V_2(Q,\Phi(P)) + \gamma_1 \gamma_2 V_2(Q,\Phi(Q))~.
$$
And we obviously have $V_2(B^n,P)=V_2(B^n,\Phi(P))$ and  $V_2(B^n,Q)=V_2(B^n,\Phi(Q))$. Moreover, the Alexandrov--Fenchel inequality \eqref{eq alex fench} gives $V_2(P)=\sqrt{V_2(P) V_2(\Phi(P))} \leq V_2(P,\Phi(P))$, and $V_2(Q)=\sqrt{V_2(Q) V_2(\Phi(Q))} \leq V_2(Q,\Phi(Q))$. And $d_{\mathscr{S}^n} (\llbracket P \rrbracket, \llbracket Q \rrbracket)=\d([P],[Q])$ gives $V_2(P,Q) \leq V_2(P,\Phi(Q))$ and $V_2(Q,P) \leq V_2(Q,\Phi(P))$. Since all the real numbers $\alpha_i,\beta_i,\gamma_i$ are non-negative, this gives inequality \eqref{eq isometries triangles}.

\end{proof}

\subsection{Proof of Theorem~\ref{main2}}

Proposition \ref{prop:shapeCBBproper} and sections \ref{sectionnonuniqueness} and \ref{sectionembeddinghyp} give part of Theorem \ref{main2}. It remains to prove the assertion about the boundary of $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$. It obviously contains only one point: indeed, the boundary of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is the set of segments up to translations and homotheties, so the boundary of $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$ is the set of segments, up to translations, homotheties and isometries of $\mathbb{R}^n$, and there is only one equivalence class.

\subsection{Metrics of non-negative curvature on the sphere $\S^2$}
\label{section metrics}

Let $\mathcal{M}_{\geq 0}(\S^2)^1$ be the set of metrics of non-negative curvature on the sphere in the sense of Alexandrov, up to isometries, and with unit area.
Let $K$ be a convex body in $\mathbb{R}^3$. Then the induced inner distance on the boundary of $K$ (i.e. the infimum of the length of curves on the boundary between the two points) is isometric to a metric of non-negative curvature on $\mathbb{S}^2$ \cite{alex2}, \cite[Theorem~10.2.6]{BBI}. This gives a well defined map
$$\mathcal{I}:\mathcal{K}^{3*}_{SV_2} \to\mathcal{M}_{\geq 0}(\S^2)^1~,$$
where we denote by $\mathcal{K}^{3*}_{SV_2}$ the space of convex bodies in $\mathbb{R}^3$, not reduced to points or segments, with Steiner point at the origin and intrinsic area $V_2$ equals to one. Remember that the total surface area is two times the intrinsic area, so that for $K\in \mathcal{K}^{3*}_{SV_2}$, $\mathcal{I}(K)$ is a homothety of factor $2^{-1/2}$ of the induced inner distance on the boundary of $K$.
We have the following classical result, see \cite{alex}, as well as
\cite{BI} for an alternative proof.

\begin{theorem}[{Alexandrov}]\label{alex}
For any metric $m$ of non-negative curvature on the sphere, there exists a convex body $K$ in $\mathbb{R}^3$ such that the induced inner distance on the boundary of  $K$ is isometric to $m$.
\end{theorem}

Of course, the induced inner distance does not change if an isometry of the ambient Euclidean space is performed on $K$. In particular, a translation can be performed on $K$, such that its Steiner point becomes the origin.
Hence Theorem~\ref{alex} implies the following.

\begin{corollary}\label{alex cor}
The map $\mathcal{I}$ is surjective.
\end{corollary}

 Later Pogorelov proved that convex bodies in $\mathbb{R}^3$ are uniquely determined, up to global isometries, by the induced inner distance on their boundary. See \cite{pog}, as well as \cite{volkov} for a stronger result.

 \begin{theorem}[Pogorelov]\label{thmpogo}
Let $K_1$ and $K_2$ in $\mathcal{K}^{3*}_{SV_2}$ such that
$\mathcal{I}(K_1)=\mathcal{I}(K_2)$. Then $K_1$ and $K_2$ differ by an element of $O(3)$.
 \end{theorem}

Note that  $\operatorname{Supp}(\mathcal{K}^{3*}_{SV_2})\subset \mathcal{H}_3^\infty$.
By Corollary~\ref{cor topo hyperboloid} and Corollary~\ref{cor equi L2 H1}, one obtains the following.

\begin{lemma}
On $\operatorname{Supp}(\mathcal{K}^{3*}_{SV_2})$, $d_{\mathcal{H}}$ and  $d_\infty$ induce the same topology.
\end{lemma}

Let $d_{SV_2}$ be the distance on $\mathcal{K}^{3*}_{SV_2}$ defined as the pullback of $d_\mathcal{H}$ by the map $\operatorname{Supp} : \mathcal{K}^{3*}_{SV_2} \rightarrow \operatorname{Supp}(\mathcal{K}^{3*}_{SV_2})$. By construction of the metrics, this is also the pullback of the distance $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^3}$ by the projection map $\mathcal{K}^{3*}_{SV_2} \rightarrow \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*}$.
By Remark~\ref{rem dis hausdorff}, the lemma above can be rephrased as follows.

\begin{lemma}\label{lem top meme aire 1}
On $\mathcal{K}^{3*}_{SV_2}$, $d_{SV_2}$ and the Hausdorff distance
induce the same topology.
\end{lemma}

Also, the space $\mathcal{M}_{\geq 0}(\S^2)^1$ is endowed with a natural topology, which is the one of uniform convergence of metric spaces \cite[7.1.5]{BBI} (which is actually, in this very particular case, the Gromov--Hausdorff topology, \cite[Exercise 7.5.14]{BBI}).

\begin{lemma}
The map $\mathcal{I}$ is continuous for the topology given by $d_{SV_2}$ on $\mathcal{K}^{3*}_{SV_2}$, and the topology of the uniform convergence on $\mathcal{M}_{\geq 0}(\S^2)^1$.
\end{lemma}
\begin{proof}
By \cite[Lemma~10.2.7]{BBI}, $\mathcal{I}$ is continuous when $\mathcal{K}^{3*}_{SV_2}$ is endowed with the Hausdorff topology. Hence Lemma \ref{lem top meme aire 1} proves the claim.
\end{proof}

Let $p:\mathcal{K}^{3*}_{SV_2}\to \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*}$ be the map which associates to any element of $\mathcal{K}^{3*}_{SV_2}$ its class in $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*}$. We have $p = p_2 \circ p_1$, where $p_1$ and $p_2$ are the following projections:
$$
(\mathcal{K}^{3*}_{SV_2},d_{SV_2}) \overset{p_1}{\longrightarrow} (\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^3}) \overset{p_2}{\longrightarrow} (\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*},d_{\mathscr{S}^3})~.
$$
The projection $p$ is continuous: indeed, $p_1$ is continuous because $d_{SV_2}$ is the pullback of $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^3}$ by the map $p_1$, and $p_2$ is continuous since the topology on $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*}$ is the quotient topology (Lemma~\ref{lem quotient topo}).

Let us define the map $\bar{\mathcal{I}}$ by the following commutative diagram
\[
  \begin{tikzcd}
    \mathcal{K}^{3*}_{SV_2} \arrow{d}{p} \arrow{r}{\mathcal{I}}  & \mathcal{M}_{\geq 0}(\S^2)^1  \\
   \mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*} \arrow[swap]{ur}{\bar{\mathcal{I}}} &
  \end{tikzcd}
\]
By Corollary~\ref{alex cor} and Theorem~\ref{thmpogo}, $\bar{\mathcal{I}}$ is well-defined and
bijective.

\begin{lemma}
The map $\bar{\mathcal{I}}$ is a homeomorphism.
\end{lemma}
\begin{proof}
As $\mathcal{I}$ is continuous and as the topology on $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{3*}$ is the quotient topology (Lemma~\ref{lem quotient topo}), $\bar{\mathcal{I}}$ is continuous.
Let us check that $\bar{\mathcal{I}}^{-1}$ is continuous. This is sufficient to show that for every sequence $(m_j)$ of $\mathcal{M}_{\geq 0}(\S^2)^1$ converging to $m \in \mathcal{M}_{\geq 0}(\S^2)^1$, there exists a subsequence $(m_{j_k})$ such that $(\bar{\mathcal{I}}^{-1}(m_{j_k}))$ converges to $\bar{\mathcal{I}}^{-1}(m)$.

Let $(m_j)$ be a sequence of $\mathcal{M}_{\geq 0}(\S^2)^1$ converging to $m\in \mathcal{M}_{\geq 0}(\S^2)^1$ for the uniform topology, and let $\llbracket K_j\rrbracket=\bar{\mathcal{I}}^{-1}(m_j)$.
Let us choose representatives $K_j$ of $\llbracket K_j \rrbracket$ in $\mathcal{K}^{3*}_{SV_2}$. In particular, $\mathcal{I}(K_j)=m_j$,
and the convergence of the $m_j$
 implies that the diameters of the $\partial K_j$ are uniformly bounded \cite[7.3.14]{BBI}.
In turn,
the Euclidean distances between points on $\partial K_j\subset \mathbb{R}^3$
 are uniformly bounded. As the Steiner points of the $K_j$ are at the origin, the $K_j$ are contained in a ball.
 By the Blaschke selection theorem (Theorem~\ref{thm:proper}), there is a converging subsequence $K_{j_k}$ for the Hausdorff topology, hence for $d_{SV_2}$ by Lemma~\ref{lem top meme aire 1}; we set $K = \lim K_{j_k} \in \mathcal{K}^{3*}_{SV_2}$.
Since $p$ is continuous, $\llbracket K_{j_k} \rrbracket$ converges to $\llbracket K\rrbracket$. As $\bar{\mathcal{I}}$ is continuous, $m_{j_k} = \bar{\mathcal{I}}( \llbracket K_{j_k} \rrbracket)$ converges to $\bar{\mathcal{I}}(\llbracket K \rrbracket)$, so
$m=\bar{\mathcal{I}}(\llbracket K \rrbracket)$. Finally this gives that $\bar{\mathcal{I}}^{-1}(m_{j_k}) = \llbracket K_{j_k} \rrbracket$ converges to $\bar{\mathcal{I}}^{-1}(m) = \llbracket K \rrbracket$.
\end{proof}

More informations about the topology of $\mathcal{M}_{\geq 0}(\S^2)^1$ may  be found in \cite{Bel}.
For $n\geq 3$, the induced inner distance on the boundary of a convex body in $\mathbb{R}^n$ is (isometric to) a
metric of non-negative curvature on $\mathbb{S}^{n-1}$ in the sense of Alexandrov. But not every such metric on $\mathbb{S}^{n-1}$
arises in this way
(\cite{matv}, \cite[1.9]{alex}).

\section{The space of all the (oriented) shapes}\label{infini}

This section is an opening to the study of spaces of convex bodies, considered without making distinction between dimensions.

For $p\geq 0$ let us denote by $\iota_{n,p}$ the  canonical isometric embedding of $\mathbb{R}^n$ into $\mathbb{R}^{n+p}$ which is given by $\mathbb{R}^n \simeq \mathbb{R}^n \times \{0\}^{p} \subset \mathbb{R}^{n+p}$.

\begin{fact}
The map
$$\iota_{n,p}:(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},\d) \to (\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{(n+p)*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^{n+p}})$$
defined by $\iota_{n,p}([K])=[\iota_{n,p}(K)]$ is an isometry.
\end{fact}
\begin{proof}
Let $K_1$ and $K_2$ be convex bodies in $\mathbb{R}^n$ of unit intrinsic area. By property~\ref{mcmullen} of the intrinsic area, $\iota_{n,p}(K_1)$ and $\iota_{n,p}(K_2)$ have also unit intrinsic area. So it suffices to check that
$V_2(K_1,K_2)=V_2(\iota_{n,p}(K_1),\iota_{n,p}(K_2))$, that is clear from \ref{def mixed v2} and \ref{mcmullen}.
 \end{proof}

Let $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ be the union over $n$ of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*}$, quotiented by the following equivalence relation:  $[K_1]$ is equivalent to $[K_2]$ if and only if there exist $i,j \leq p$ such that  $K_1\subset\mathbb{R}^i$, $K_2\subset \mathbb{R}^j$ and $[\iota_{i,p-i}(K_1)]=[\iota_{j,p-j}(K_2)]$.
We will denote by $[K]_\infty$ an element of $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$.
For two representatives of  $[K_1]_\infty, [K_2]_\infty \in \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ in $\mathbb{R}^n$, let us define
$$d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([K_1]_\infty,[K_2]_\infty)=\d([K_1],[K_2])~.$$
It is easy to see that $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}$ is well-defined and that it is actually a distance on $ \mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$.

The isometric embeddings $\iota_{n,p}$ induce isometric maps
from $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{n*},d_{\mathscr{S}^n})$ to $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{(n+p)*},d_{\mathscr{S}^{n+p}})$, so in the same way we can define the set $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ and the metric space
$(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{S}^\infty})$.

It follows from Theorems \ref{main} and \ref{main2} that $ (\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}, d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty})$ and $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{S}^\infty})$
are geodesic metric spaces. But it may happen that a sequence of convex bodies with non-empty interior in $\mathbb{R}^p$ converges to a convex body in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^\infty$ when $p$ goes to infinity, see below. This suggests that there may exist other shortest paths than the ones we know, and in consequence we don't know if those metric spaces have bounded curvature.

\begin{fact}
Let $K \in \mathcal{K}^{n*}$.
Let $(\epsilon_p)_p$ be a sequence of real numbers such that $\sqrt{p} \epsilon_p \rightarrow 0$. Then the sequence $([\iota_{n,p}(K) + \epsilon_p B^{n+p}]_\infty)_p$ converges in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ to $[K]_\infty$.
\end{fact}

\begin{proof}
We have $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([K]_\infty,[\iota_{n,p}(K) + \epsilon_p B^{n+p}]_\infty) = d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^{n+p}}([\iota_{n,p}(K)],[\iota_{n,p}(K) + \epsilon_p B^{n+p}])$, so
$$
\cosh(d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([K]_\infty,[\iota_{n,p}(K)+\epsilon_p B^{n+p}]_\infty)) = \frac{V_2(\iota_{n,p}(K),\iota_{n,p}(K) + \epsilon_p B^{n+p})}{\sqrt{V_2(\iota_{n,p}(K))V_2(\iota_{n,p}(K) + \epsilon_k B^{n+p})}}~.
$$
We have $V_2(\iota_{n,p}(K)) = V_2(K)$, and equations \eqref{V2B} and \eqref{V2 avec V1} give
$$
V_2(\iota_{n,p}(K),\iota_{n,p}(K)+\epsilon_p B^{n+p})= V_2(\iota_{n,p}(K)) + \epsilon_p V_2(\iota_{n,p}(K),B^{n+p}) = V_2(K) + \frac{\epsilon_p}{2} V_1(B^{n+p-1})V_1(K)
$$
and
\begin{eqnarray*}
V_2(\iota_{n,p}(K)+\epsilon_p B^{n+p}) &=& V_2(\iota_{n,p}(K)) + 2\epsilon_p V_2(\iota_{n,p}(K),B^{n+p}) + \epsilon_p^2 V_2(B^{n+p}) \\
& =&  V_2(K) + \epsilon_p V_1(B^{n+p-1}) V_1(K) + \epsilon_p^2 (n+p-1)\pi~.
\end{eqnarray*}
Equation \eqref{V1B} gives $V_1(B^{n+p-1}) \sim \sqrt{2\pi (n+p-1)}$, so $V_2(\iota_{n,p}(K),\iota_{n,p}(K) + \epsilon_p B^{n+p}) \rightarrow V_2(K)$ and $V_2(\iota_{n,p}(K)+\epsilon_p B^{n+p}) \rightarrow V_2(K)$, and this gives $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([K]_\infty,[\iota_{n,p}(K) + \epsilon_p B^{n+p}]_\infty) \rightarrow 0$.
\end{proof}

We show in the next proposition that the spaces $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty})$ and $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{S}^\infty})$ are not complete, hence not proper.

\begin{proposition} \label{propnoncomplete}
The space $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty})$ (resp. $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{S}^\infty})$) is not complete: the sequence of balls $([B^n]_\infty)_n$ (resp. $(\llbracket B^n \rrbracket_\infty)_n$) is a diverging Cauchy sequence.
\end{proposition}

\begin{proof}
Let $p \leq n$. Since the balls are invariant under isometries we have
$$d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([B^p]_\infty,[B^n]_\infty) = d_{\mathscr{S}^\infty}(\llbracket B^p \rrbracket_\infty, \llbracket B^n \rrbracket_\infty)=\d([\iota_{p,n-p}(B^p)],[B^n])~,$$
and by equations \eqref{V2B}, \eqref{V1B} and \eqref{eq distance a B} we have
$$
\cosh(\d([\iota_{p,n-p}(B^p)],[B^n]))=\left(\frac{\sqrt{(n-1)}W_{n-1}}{\sqrt{(p-1)}W_{p-1}}\right)~.
$$
Since $W_n \sim \sqrt{\frac{\pi}{2n}}$, the sequence of balls (either in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ or $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$) are Cauchy sequences.

Moreover, suppose that $([B^n]_\infty)_n$ or $(\llbracket B^n \rrbracket_\infty)_n$ converges to $[K]_\infty$ or $\llbracket K \rrbracket_\infty$ in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$ or $\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$. Since we have $d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty}([K]_\infty,[B^n]_\infty) = d_{\mathscr{S}^\infty}(\llbracket K \rrbracket_\infty, \llbracket B^n \rrbracket_\infty)$, in any of these two cases we have that $([B^n]_\infty)_n$ converges to $[K]_\infty$ in $\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*}$. And that is impossible.

Indeed, assume that $K$ is a convex body of some Euclidean space $\mathbb{R}^p$. As $B^n$ is invariant under the action of $O(n)$, then for any $\Phi\in O(p)$, by considering $O(p)$ as a subgroup of $O(n)$, we have that $([B^n]_\infty)_n$ converges to $\Phi [K]_\infty$. So $K$ must be a ball, but from the computations above, $([B^n]_\infty)_n$ cannot converge to the class of a ball.
\end{proof}

It would be interesting to  explore more intensively the metric spaces $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty})$ and  $(\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{S}^\infty})$, that is out of the scope of the present paper.
The description of the completion of those spaces would certainly involve the study of
``infinite dimensional convex bodies'', as in \cite{hoermander,chevet,vitale-steiner,vitale-survey}. Note that in the case of $(\mathscr{O}\hspace{-0.06 cm}\mathscr{S}\hspace{-0.06 cm}\emph{hape}^{\infty*},d_{\mathscr{O}\hspace{-0.06 cm}\mathscr{S}^\infty})$, it would be relevant to describe the space ``$\H_\infty$'' (that is, the inductive limit of the $\H_n$), and its completion.

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