Statistical models and
probabilistic methods
on Riemannian manifolds
Salem Said – CNRS, Université de Bordeaux

CHAPTER: A GUIDE TO THIS THESIS

This thesis reflects the major themes of my work, which I have carried out in the past four years.It does leave out some of this work, especially  on the subject of warped information metrics. I hope readers may find time for at least a glance at this “missing" part, (for example, in [1]). However, the thesis is rather self-contained, and I feel that the best way of reading it is just from beginning to end, uninterrupted.

At any rate, I would like to ask readers to begin with Chapter <ref>. Then, once this is done, they can skip to Chapters <ref> and <ref>, which I would like to ask them to read together, or go on toChapter <ref>, which is quite independent from following. The same goes for Chapter <ref>, which can be read right after Chapter <ref>, provided just a little bit of familiarity with Chapter <ref>.

Each chapter begins with a table of contents, followed by a sort of “abstract", which provides some additional details, on the table of contents, and points to some of the more interesting results. I have done my best to avoid the thesis being a copy-paste of published research papers. Chapter <ref> uncovers several new connections between Riemannian Gaussian distributions and random matrix theory, while Chapter <ref> is entirely made up of previously unpublished material. The other chapters stick more closely to my existing papers (published, or under review), although I have made a consistent effort to improve the presentation, and to include useful background and historical discussion.

I hope that readers will find it stimulating, to read an “original thesis". On the other hand, exploring new ideas exposes one to the risk of making mistakes (of various magnitude), and I also hope these are duly pointed out, and the appropriate criticism is served up, without restraint. On the whole, writing this thesis has been a humbling experience for me. I have found out, once and again, that I was unable to answer questions or to prove statements,even when they seemed very natural. Chapter <ref>, a short final chapter, contains a list of such “open problems" (they are open to me, but others may find them easy).

I should acknowledge the input of many colleagues, who have shaped the ideas layed out in the following. Chapter <ref> was born out of discussions with Marc Arnaudon, and Chapter <ref> relies heavily on joint work with Alain Durmus, Pablo Jimenez, and Eric Moulines [2][3].

For Chapter <ref>, the idea of a useful connection between Riemannian Gaussian distributions and random matrix theory was first suggested to me by my colleague Yannick Berthoumieu. During the summer of 2020, I worked on this idea with Cyrus Mostajeran and Simon Heuveline. Later on, when I was nearly finished writing this thesis, I was very excited to discover the work of Leonardo Santilli and Miguel Tierz [4], who were simultaneously developing the same idea. It is really a great satisfaction to see a whole project unfold out of an “innocent" discussion. For this, I want to thank all of the colleagues I just mentioned.

Perhaps nobody will ever write a better preface than Cervantes, whose following famous words certainly apply here are now. Idle reader : Without my swearing to it, you can believe that I would like this book, the child of my understanding, to be the most beautiful, the most brilliant, and the most discreet that anyone could imagine. But I have not been able to contravene the natural order; in it, like begets like.

CHAPTER: NOTATION AND BACKGROUND

Certainly, this thesis is intended for specialised readers, who are already familiar with the basics of Riemannian geometry. This first chapter is not a stand-alone introduction to Riemannian geometry, but merely hopes to help the readers ease into the material in subsequent chapters : by recalling some elementary notions in Riemannian geometry, I hope to find a shared language with my benevolent readers. Some original, or even unpublished, material is still included. As discussed in the following :

* <ref> – <ref> lead up to the second-order Taylor expansion of a function defined on a Riemannian manifold. Proposition <ref> of <ref> states that geodesic curves are exactly those curves which admit a Taylor expansion of any $C^2$ function, in terms of its gradient and Hessian.

* <ref> recalls real Grassmann manifolds, and the PCA objective function. It presents a calculation of the gradient and Hessian of this function, based on the symmetric space structure of real Grassmann manifolds.

* <ref> and <ref> contain some original material, from [2]. In <ref>, the original concept of regular retraction (a retraction which avoids the cut locus) is introduced. In <ref>, the usual “projection retraction" on the real Grassman manifold is shown to be a regular retraction.

* <ref> recalls the usual metric and Hessian comparison theorems of Riemannian geometry.

* <ref> is an application of <ref>, studied in [2]. It introduces the robust Riemanian barycentre of a probability measure on a Hadamard manifold, proving its existence and uniqueness.

* <ref> is concerned with Riemannian volume : integration in geodesic spherical coordinates, volume comparison, and integral formulae for Riemannian symmetric spaces. All of these will be very important, in Chapters <ref> and <ref>.

* <ref> provides the previously unpublished Propositions <ref> and <ref>, which may be used to compute geodesics in symmetric spaces.

§ THE LEVI-CIVITA CONNECTION

A smooth (0,2)-tensor field $g$, on a finite-dimensional smooth manifold $M$, is called a Riemannian metric, if the bilinear form

\begin{equation} \label{eq:metric}
\langle u,\!v\rangle_{\scriptscriptstyle x} =   g_{\scriptscriptstyle x}(u,v)  \hspace{1cm} u\,,v \in T_xM
\end{equation}

is a true scalar product, for each $x \in M$. In this case, for $u \in T^{\phantom{*}}_xM$ and $t \in T^*_xM$, the identities

(u^{\flat},v) = \langle u,\!v\rangle_{\scriptscriptstyle x} \hspace{0.3cm}\mbox{and}\hspace{0.3cm}
\langle t^{\,\scriptscriptstyle \#},\!v\rangle_{\scriptscriptstyle x}=  \left(t,v\right) \hspace{1cm} v \in T^{\phantom{*}}_xM

uniquely define $u^{\scriptscriptstyle \flat} \in T^*_xM$ and $t^{\,\#} \in T^{\phantom{*}}_xM$. By a useful abuse of notation,

\begin{equation} \label{eq:pairing}
u^{\flat} = g(u) \hspace{1cm}  t^{\,\scriptscriptstyle \#} = g^{\scriptscriptstyle -1}(t)
\end{equation}

The Levi-Civita connection of $g$ is the unique affine connection $\nabla$ which is metric, so that

\begin{equation} \label{eq:metricconnection}
\nabla g = 0
\end{equation}

and tortionless, so that the exterior derivative $d\theta$, of any $1$-form $\theta$, reads

\begin{equation} \label{eq:tortionless}
d\theta(X,Y) =  \nabla_X\theta(Y) - \nabla_Y\theta(X)
\end{equation}

for vector fields $X$ and $Y$. In effect, by (<ref>) and (<ref>), the $(1,1)$-tensor field $\nabla X$ (the covariant derivative of the vector field $X$), decomposes into self-adjoint and skew parts,

\begin{equation} \label{eq:koszul}
2\,\langle\nabla_YX,Z\rangle = \mathcal{L}_Xg(Y,Z) + dX^{\flat}(Y,Z)
\end{equation}

where $\mathcal{L}_Xg$ denotes the Lie derivative of the metric $g$ along $X$, the “the linear elasticity tensor" (the equivalence between (<ref>)–(<ref>) and (<ref>) is the content of Koszul's theorem).

Given local coordinates $(x^i\,;i=1,\ldots,n)$ on an open $U \subset M$, there is a coordinate frame $(\partial_i)$, along with a coframe $(dx^i)$ — of course,  $\partial_i$ stands for $\left.\partial\middle/\partial x^i\right.$. In terms of these coordinates, the metric $g$ takes on the form of a length element

\begin{equation} \label{eq:lengthelement}
g = g_{ij}\,dx^i\otimes dx^j \hspace{1cm} g_{ij} = \langle \partial_i\,,\partial_j\rangle
\end{equation}

and covariant derivatives may be expressed, in coordinate form,

\begin{equation} \label{eq:christoffel}
\nabla X = \left\lbrace\partial_j X^i + \Gamma^i_{jk} X^k\right\rbrace \partial_i \otimes dx^j \hspace{1cm} \nabla_{\partial j}\,\partial_k = \Gamma^i_{jk}\,\partial_i
\end{equation}

using the Christoffel symbols $(\Gamma^i_{jk})$.

§ PARALLEL TRANSPORT AND GEODESICS

A vector field $X$, along a smooth curve $c:I\rightarrow M$, defined on some interval $I \subset \mathbb{R}$, is a map $X:I \rightarrow TM$ such that
$\pi \circ X = c$ —  of course, $\pi:TM \rightarrow M$ denotes the canonical projection. The Levi-Civita connection $\nabla$ can be used to compute the covariant derivative of $X$ along $c$, itself a vector field along $c$, here denoted $\nabla_{\dot{c}\,}X$. In local coordinates,

\begin{equation} \label{eq:dotX}
\nabla_{\dot{c}\,}X(t) = \left \lbrace \frac{d}{dt}X^i(t) + (\Gamma^i_{jk}\circ c(t))\, \dot{c}^{\,j}(t)X^k(t)\right\rbrace (\partial_i\circ c)(t)
\end{equation}

and this suggests writing $\nabla_{\dot{c}\,}X = \nabla_{t\,}X$ or even $\dot{X}$, when $c$ is understood from the context. Now, $X$ is called parallel along $c$ if $\nabla_{\dot{c}\,}X = 0$. From (<ref>), this means that the components $X^i(t)$ satisfy a linear differential equation with smooth coefficients.

Thus, if $X$ is parallel along $c$, then $X$ is completely determined by its value at any instant, say $t_{\scriptscriptstyle o} \in I$. Equivalently, if $v$ is tangent to $M$ at $c(t_{\scriptscriptstyle o})$, then there exists a unique parallel vector field $X$ along $c$, with $X(t_{\scriptscriptstyle o}) = v$. It follows that, for $t \in I$, there exists a linear operator $\Pi^t_{t_{\scriptscriptstyle o}}$ which maps $T_{c(t_{\scriptscriptstyle o})}M$ onto $T_{c(t)}M$, by $\Pi^t_{t_{\scriptscriptstyle o}}(v) = X(t)$. This linear operator $\Pi^t_{t_{\scriptscriptstyle o}}$ is called parallel transport along $c$, from  $c(t_{\scriptscriptstyle o})$ to $c(t)$, and has the following properties,

\begin{equation} \label{eq:hemigroup}
\text{hemigroup property} \hspace{1cm} \Pi^t_{t_{\scriptscriptstyle o}} = \Pi^t_{t_{\scriptscriptstyle 1}} \circ \Pi^{t_{\scriptscriptstyle 1}}_{t_{\scriptscriptstyle o}}
\end{equation}

\begin{equation} \label{eq:isometry}
\phantom{abcd}\text{isometry property} \hspace{1cm} \Vert\Pi^t_{t_{\scriptscriptstyle o}}(v)\Vert_{\scriptscriptstyle c(t)} = \Vert v\Vert_{\scriptscriptstyle c(t_{\scriptscriptstyle \scriptscriptstyle o})}
\end{equation}

where $\Vert \cdot \Vert_x$ is the norm associated with the scalar product in (<ref>), for any $x \in M$. Clearly, if one knows how to compute parallel transports, then one is able to recover covariant derivatives,

\begin{equation} \label{eq:pinabla}
\nabla_{\dot{c}\,}X(t_{\scriptscriptstyle o}) = \left.\frac{d}{dt}\right|_{t=t_{\scriptscriptstyle o}}\Pi^{t_{\scriptscriptstyle o}}_{t}(X(t))
\end{equation}

A smooth curve $c:I\rightarrow M$ is called a geodesic curve, if its velocity vector field $\dot{c}$ is parallel. This means that $c$ satisfies the geodesic equation,

\begin{equation} \label{eq:geodesicequation}
\nabla_{\dot{c}\,}\dot{c} = 0 \text{ or } \ddot{c} = 0
\end{equation}

Written out in local coordinates, this is a non-linear ordinary differential equation,

\begin{equation} \label{eq:accelerationcoordinates}
\frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}c^i(t) + \Gamma^i_{jk}(c(t))\frac{d}{dt}c^{\,j}(t)\frac{d}{dt}c^k(t) = 0
\end{equation}

If its solutions $c(t)$ exists at all finite $t \in \mathbb{R}$, for any initial conditions $c(t_{\scriptscriptstyle o}) = x$ and $\dot{c}(t_{\scriptscriptstyle o})=v$, then the metric $g$ on the manifold $M$ is called geodesically complete. In this case, the Riemannian exponential map $\mathrm{Exp}:TM\rightarrow M$,  given by $\mathrm{Exp}_x(v) = c(1)$ is well-defined.

The geodesic equation (<ref>) states that the curve $c$ has zero acceleration (just like a particle in free motion). This means that geodesic curves are extremals of the energy functional

\begin{equation} \label{eq:energy}
E(c) = \int_I\,\Vert \dot{c}(t)\Vert^2\,dt
\end{equation}

and that re-parameterised geodesic curves (of the form $c\circ \varphi$ where $c$ is a geodesic, and $t^\prime = \varphi(t)$ a new parameterisation) are extremals of the length functional

\begin{equation} \label{eq:length}
L(c) = \int_I\,\Vert \dot{c}(t)\Vert\,dt
\end{equation}

This leads to the notion of Riemannian distance, which will be discussed in <ref> below.

§ TAYLOR EXPANSION OF A FUNCTION

Let $f:M \rightarrow \mathbb{R}$ be a $C^2$ function, denote $df$ its differential. The gradient of $f$ is the vector field

\begin{equation} \label{eq:gradient}
\mathrm{grad}\,f = g^{\scriptscriptstyle -1}(df)
\end{equation}

In the notation of (<ref>). The Hessian of $f$ is the $(1,1)$-tensor field

\begin{equation} \label{eq:hessian}
\mathrm{Hess}\,f = \nabla\,\mathrm{grad}\,f
\end{equation}

The following proposition says that geodesic curves are exactly the curves which admit a Taylor expansion of any $C^2$ function, in terms of its gradient and Hessian.

A smooth curve $c:I\rightarrow M$ is a geodesic curve, if and only if, for any $s,t \in I$, and any $C^2$ function $f:M \rightarrow \mathbb{R}$,

\begin{equation} \label{eq:taylor1}
(f\circ c)(t) = (f\circ c)(s) + \langle \mathrm{grad}\,f,\dot{c}\rangle_{\scriptscriptstyle c(s)}(t-s) + \frac{1}{2}\, \langle\mathrm{Hess}\,f\cdot\dot{c}\,,\dot{c}\rangle_{\scriptscriptstyle c(s)}(t-s)^{\scriptscriptstyle 2} + o(|t-s|^{\scriptscriptstyle 3})
\end{equation}

The proof of this proposition follows from the identity,

\frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}(f\circ c)(t) = \frac{d}{dt}\langle \mathrm{grad}\,f,\dot{c}\rangle_{\scriptscriptstyle c(t)} =
\langle\nabla_{\dot{c}\,} \mathrm{grad}\,f,\dot{c}\rangle_{\scriptscriptstyle c(t)} + \langle \mathrm{grad}\,f,\ddot{c}\rangle_{\scriptscriptstyle c(t)}

Indeed, the last term is identically zero (i.e., for any $C^2$ function $f$), if and only if $\ddot{c}$ is identically zero, as in the geodesic equation (<ref>).

In (<ref>), $\mathrm{Hess}\,f$ is a $(1,1)$-tensor field. This tensor field is self-adjoint, and it is common practice to identify it with a $(0,2)$-tensor field,

\begin{equation} \label{eq:hessianbis}
\mathrm{Hess}\,f = \nabla\,df = \frac{1}{2}\mathcal{L}_{\mathrm{grad}\,f}\,g
\end{equation}

where the second equality follows from (<ref>) and (<ref>). This yields a lighter notation,

\langle\mathrm{Hess}\,f\cdot u\,,v\rangle = \mathrm{Hess}\,f(u,v)

Recall the Riemannian exponential map $\mathrm{Exp}$, from <ref> (always assume geodesic completeness). Proposition <ref> can be used to write down a Taylor expansion with Lagrange remainder,

\begin{equation} \label{eq:taylor2}
f\left(\mathrm{Exp}_x(v)\right) = f(x) + \langle \mathrm{grad}\,f,v\rangle_{\scriptscriptstyle x} + \frac{1}{2}\,\mathrm{Hess}\,f_{\scriptscriptstyle c(t^*)}(\dot{c},\dot{c})
\end{equation}

where $c(t^*)$ is a point along the geodesic $c(t) = \mathrm{Exp}_x(t\,v)$, corresponding to an instant $t^* \in (0,1)$.

Remark : writing (<ref>) in local coordinates,

\begin{equation} \label{eq:hesscoordinates}
\mathrm{Hess}\,f = \left\lbrace \partial_{ ij} f - \Gamma^k_{ij\,} \partial_k f\right\rbrace dx^i\otimes dx^{\,j}
\end{equation}

The second derivatives $\partial_{ ij} f$ do not transform like a covariant tensor, but the Christoffel symbols correct for this problem, yeilding a true covariant tensor, $\mathrm{Hess}\,f$. A very nice way of saying this is that the Levi-Civita connection transforms second-order differentials, into covariant tensors.

The concepts of second-order vectors and of second-order differentials are reviewed in [5], where they are used as a starting point for stochastic analysis in manifolds.

§ EXAMPLE : THE PCA OBJECTIVE FUNCTION

The problem of principal component analysis consists in maximising the objective function

\begin{equation} \label{eq:pcaf}
f(x) = \mathrm{tr}\left(x\Delta\right) \hspace{1cm} x \in \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)
\end{equation}

where $\Delta$ is a symmetric positive-definite matrix, of size $(p+q)\times(p+q)$. The maximisation is over $x$ in the real Grassmann manifold $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$, identified with a space of orthogonal projectors

\begin{equation} \label{eq:grassconst}
\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)= \left \lbrace x \in \mathbb{R}^{\scriptscriptstyle (p+q)\times(p+q)}\,:x^{\dagger} - x = 0 \hspace{0.1cm},\hspace{0.1cm} x^2 - x = 0 \hspace{0.1cm},\hspace{0.1cm} \mathrm{tr}(x) = p\right\rbrace
\end{equation}

where $^\dagger$ denotes the transpose. Remarkably, it is possible to show that $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ is a submanifold of $\mathrm{S}(p+q)$, the affine space of symmetric matrices of size $(p+q)\times(p+q)$, with tangent spaces (the proof of this statement may be found in [6]),

\begin{equation} \label{eq:tangentgrassconst}
T_x \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)= \left\lbrace v \in \mathrm{S}(p+q)\,: xv + vx = v \right\rbrace
\end{equation}

It then follows that $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ is of dimension $pq$. Clearly, $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ admits of a Riemannian metric, which is the restriction of the trace scalar product of $\mathrm{S}(p+q)$,

\begin{equation} \label{eq:grasssp}
\langle u,\!v\rangle_{\scriptscriptstyle x}  = \mathrm{tr}(uv) \hspace{1cm} u\,,v \in T_x\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)
\end{equation}

By (<ref>) and (<ref>), it follows from (<ref>) that the gradient of $f(x)$ is given by

\begin{equation} \label{eq:pcapx}
\mathrm{grad}\,f(x) = \mathrm{P}_x(\Delta)
\end{equation}

where $\mathrm{P}_x:\mathrm{S}(p+q)\rightarrow \mathrm{S}(p+q)$ is the orthogonal projection onto $T_x \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$. Now, let $x = o$, the projector onto the span of the first $p$ vectors in the canonical basis of $\mathbb{R}^{\scriptscriptstyle p+q}$. One readily checks from (<ref>) that

\begin{equation} \label{eq:grassto}
T_o \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)= \left\lbrace \tilde{\omega} =
\left(\begin{array}{cc} 0_{\scriptscriptstyle p\times p} & \omega^\dagger \\[0.15cm] \omega & 0_{\scriptscriptstyle q\times q}\end{array}\right)\,;\, \omega \in \mathbb{R}^{\scriptscriptstyle q\times p}
\right\rbrace
\end{equation}

Therefore, $\mathrm{P}_o(\Delta)$ is just $\Delta$ with its main diagonal blocks of size $p \times p$ and $q\times q$ set to zero.Then, note that the orthogonal group $O(p+q)$ acts transitively on $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$, by $g\cdot x = gxg^\dagger$for $g \in O(p+q)$ and $x \in \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$, and that this action preserves the  Riemannian metric (<ref>). Therefore, one has the following alternative to (<ref>)[By an abuse of notation, $g\cdot a= g\,a\,g^\dagger$, for any matrix $a$ of size $(p+q)\times(p+q)$.],

\begin{equation} \label{eq:tangentgrassgroup}
T_x \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)= \left \lbrace v = g\cdot \tilde{\omega}\,;\, g\cdot o = x\hspace{0.1cm},\hspace{0.1cm} \tilde{\omega} \in T_o \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q) \right\rbrace
\end{equation}

where $g\cdot o = x$ simply means the first $p$ columns of $g$ span the image space of $x$. Since the action of $O(p+q)$ preserves the  Riemannian metric (<ref>), it easily follows

\begin{equation} \label{eq:grasspx}
\mathrm{P}_x(\Delta) = g \cdot \mathrm{P}_o(g^\dagger\cdot \Delta) \text{ for any $g$ such that $g\cdot o = x$}
\end{equation}

which can be used to evaluate the gradient of $f(x)$, in (<ref>). For the Hessian of $f(x)$, note that, according to Propositon <ref>,

\begin{equation} \label{eq:pcahess}
\mathrm{Hess}\,f_x(v,v) = \left.\frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}\right|_{t=0} f\left(\mathrm{Exp}_x(tv)\right)
\end{equation}

Here, the Riemannian exponential can be transformed into a matrix exponential (see Proposition <ref>, in <ref>). For $g \in O(p+q)$, note that $g \cdot o = o$ if and only if $g \in O(p) \times O(q) \subset O(p+q)$. Denote $\mathfrak{g}$ and $\mathfrak{k}$ the Lie algebras of $O(p+q)$ and $O(p) \times O(q) \subset O(p+q)$. Let $\mathfrak{p}$ denote the orthogonal complement of $\mathfrak{k}$ (with respect to the bilinear form $Q(\xi\hspace{0.02cm},\eta) = \mathrm{tr}(\xi\eta)$, for $\xi\,,\eta \in \mathfrak{o}(p+q)$). Then,

\begin{equation} \label{eq:grassp}
\mathfrak{p} = \left\lbrace \hat{\omega} =
\left(\begin{array}{cc} 0_{\scriptscriptstyle p\times p} & -\omega^\dagger \\[0.15cm] \omega & 0_{\scriptscriptstyle q\times q}\end{array}\right)\,;\, \omega \in \mathbb{R}^{\scriptscriptstyle q\times p}
\right\rbrace
\end{equation}

From (<ref>) and (<ref>), it is clear there exists a canonical isomorphism $\pi_o: T_o \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q) \rightarrow \mathfrak{p}$ (just add a minus sign in front of $\omega^\dagger$ in (<ref>)). In terms of this isomorphism,

\begin{equation} \label{eq:grasslift}
\mathrm{Exp}_x(tv) = \exp(t\, \hat{\omega}_{\scriptscriptstyle v})\cdot x \hspace{1cm} \hat{\omega}_{\scriptscriptstyle v} = g\cdot \pi_o(g^\dagger \cdot v)
\end{equation}

Replacing (<ref>) into (<ref>), the second derivative is easily computed,

\begin{equation} \label{eq:pcahessbis}
\mathrm{Hess}\,f_x(v,v) = \mathrm{tr}\!\left( \Delta\,\hat{\omega}^2_{\scriptscriptstyle v}\,x \right) +
\mathrm{tr}\!\left( \Delta\,x\,\hat{\omega}^2_{\scriptscriptstyle v}  \right) -2\,\mathrm{tr}\!\left(\Delta\,\hat{\omega}_{\scriptscriptstyle v}\,x\, \hat{\omega}_{\scriptscriptstyle v} \right)
\end{equation}

Remark : a nice property of the linear map $v \mapsto \hat{\omega}_{\scriptscriptstyle v}$ is that $\mathrm{tr}(\hat{\omega}^2_{\scriptscriptstyle v}) = \langle v,v\rangle_{\scriptscriptstyle x\,}$.

§ REGULAR RETRACTIONS

A retraction is a map $\mathrm{Ret}:TM \rightarrow M$, taking $v \in T_xM$ to $\mathrm{Ret}_x(v)$, and which verifies [7][8],

\begin{equation} \label{eq:retdefinition}
\mathrm{Ret}_x(0_x) = x \hspace{1cm} d\,\mathrm{Ret}_x(0_x) = \mathrm{Id}_x
\end{equation}

where $0_x \in T_xM$ is the zero vector in $T_xM$, and $\mathrm{Id}_x$ is the identity map of $T_xM$. While the Riemannian exponential $\mathrm{Exp}:TM \rightarrow M$ is itself a retraction[Recall that it is always assumed $M$ is geodesically complete.], other retractions are often used as computationally cheap (or numerically stable) substitutes for the Riemannian exponential.

From (<ref>), for any retraction $\mathrm{Ret}$, $\mathrm{Ret}_x$ agrees with $\mathrm{Exp}_x$ up to first-order derivatives. Further, $\mathrm{Ret}$ will be called geodesic, if $\mathrm{Ret}_x$ agrees with $\mathrm{Exp}_x$ up to second-order derivatives. This means the curve $c(t) = \mathrm{Ret}_x(tv)$ has zero initial acceleration : $\ddot{c}(0) = 0_{x\,}$, for any $v \in T_xM$ (in the notation of (<ref>).

To compare a retraction $\mathrm{Ret}$ with the exponential $\mathrm{Exp}$, it is useful to introduce the maps

\begin{equation} \label{eq:PHI}
\Phi_x : T_xM \rightarrow T_xM \hspace{1cm} \Phi_x(v) = \left(\mathrm{Exp}^{-1}_x \circ \mathrm{Ret}_x\right)(v)
\end{equation}

These maps are well-defined if $\mathrm{Ret}$ is regular. That is, if $\mathrm{Ret}_x(v) \notin \mathrm{Cut}(x)$ for any $v \in T_xM$ ($\mathrm{Cut}(x)$ denotes the cut locus of $x$, whose definition is recalled in <ref>). In addition, they satisfy the following propositions.

Let $\mathrm{Ret}:TM\rightarrow M$ be a regular retraction. Then, $\Phi_x : T_xM \rightarrow T_xM$ verify

(a) $\Phi_x(0_x) = 0_x$ and $\Phi^\prime_x(0_x) = \mathrm{Id}_x$ (the prime denotes the Fréchet derivative).

(b) $\Phi^{\prime\prime}_x(0_x)(v,v) = \ddot{c}(0)$, where the curve $c(t)$ is given by $c(t) = \mathrm{Ret}_x(tv)$.

Let $\mathrm{Ret}:TM\rightarrow M$ be a regular retraction and $f:M \rightarrow \mathbb{R}$ be a $C^2$ function.

\begin{equation} \label{eq:taylor2ret}
f\left(\mathrm{Ret}_x(v)\right) = f(x) + \langle \mathrm{grad}\,f,\Phi_x(v)\rangle_{\scriptscriptstyle x} + \frac{1}{2}\,\mathrm{Hess}\,f_{\scriptscriptstyle \gamma(t^*)}(\dot{\gamma},\dot{\gamma})
\end{equation}

where $\gamma(t^*)$ is a point along the geodesic $\gamma(t) = \mathrm{Exp}_x(t\Phi_x(v))$, corresponding to some $t^* \in (0,1)$.

As an application of Proposition <ref>, consider the following examples.

Example 1 : let $M = S^n \subset \mathbb{R}^{n+1}$, the unit sphere of dimension $n$, with its usual (round) Riemannian metric. The retraction $\mathrm{Ret}_x(v) = \left(x+v)\middle/\Vert x + v \Vert\right.$ ($\Vert \cdot \Vert$ is the Euclidean norm) is regular, and the maps $\Phi_x$ are given by

\begin{equation} \label{eq:phisphere}
\Phi_x(v) = \arctan(\Vert v \Vert)\,\frac{v}{\Vert v \Vert}
\end{equation}

Example 2 : let $M = U(d)$, the Lie group of $d \times d$ unitary matrices, with its bi-invariant metric $\langle u,\!v\rangle_{\scriptscriptstyle x} = -(1/2)\mathrm{tr}(uv)$. The retraction $\mathrm{Ret}_x(v) = \mathrm{Pol}(x+v)$ ($\mathrm{Pol}$ denotes the left polar factor) is regular, and the maps $\Phi_x$ are given by

\begin{equation} \label{eq:phiun}
\Phi_x(v) = x\left(u \exp(i\arctan(\theta))\, u^\dagger\right)
\end{equation}

where $^\dagger$ denotes the conjugate-tranpose, and $\omega = x^\dagger v$ has spectral decomposition $\omega = u(i\theta)u^\dagger$, where $u$ is unitary and $\theta$ is real and diagonal — as one may expect, $\arctan(\theta) = \mathrm{diag}(\arctan(\theta_{ii}))$.

Now, (b) of Proposition <ref> implies the retractions in question are geodesic, since the Taylor expansion at zero of the arctangent only contains odd powers. Both of these retractions are based on orthogonal projection onto the manifold $M$, which is embedded in a Euclidean space.

Proof of Proposition <ref> : note that (a) is immediate, by (<ref>), and the fact that $\mathrm{Exp}$ is a retraction. To prove (b), note that

\begin{equation} \label{eq:PHInorm}
\Phi_x(v) = \tau^i(\mathrm{Ret}_x(v))\,\partial_i(x)
\end{equation}

where $(\tau^i\,;i=1,\ldots,n)$ are normal coordinates with origin at $x$, and where $\partial_i = \left.\partial\middle/\partial \tau^i\right.$. Since $\Phi_x$ is smooth (precisely, $C^2$),

\Phi^{\prime\prime}_x(0_x)(v,v) = \left. \frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}\right|_{t=0}\! \Phi_x(tv)

Thus, if $c(t) = \mathrm{Ret}_x(tv)$ and $c^i(t) = (\tau^i \circ c)(t)$, then

\Phi^{\prime\prime}_x(0_x)(v,v) = \frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}c^i(0)\,\partial_i(x) =
\left\lbrace\frac{d^{\scriptscriptstyle\hspace{0.03cm}2}}{dt^{\scriptscriptstyle 2}}c^i(0) + \Gamma^i_{jk}(c(0))\frac{d}{dt}c^{\,j}(0)\frac{d}{dt}c^k(0)\right\rbrace\partial_i(x)

where the second equality holds since $\Gamma^i_{jk}(c(0)) = \Gamma^i_{jk}(x) = 0$, by the definition of normal coordinates. Comparing to (<ref>) and (<ref>), it is clear $\Phi^{\prime\prime}_x(0_x)(v,v) = \ddot{c}(0)$.

Proof of Proposition <ref> : this is a direct application of (<ref>), using $\mathrm{Ret}_x(v)  =\mathrm{Exp}_x(\Phi_x(v))$.

Remark : the claims in Examples 1 and 2 above will not be proved in detail. Example 1 is quite elementary, and only requires one to recall that $\mathrm{Cut}(x) = \lbrace - x\rbrace$. For Example 2, the cut locus on a point $x$ in $U(d)$ is described in [9], and (<ref>) follows by a straightforward matrix calculation.

§ EXAMPLE : A RETRACTION FOR $\MATHRM{GR}_{\SCRIPTSCRIPTSTYLE \MATHBB{R}}(P\,,Q)$

Let $\mathrm{St}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ denote the Stiefel manifold, whose elements are the $d \times p$ matrices $b$ with
$b^\dagger b = \mathrm{I}_p$($\mathrm{I}_p$ is the $p \times p$ identity matrix, and $d = p+q$). Note that $T_{b\,}\mathrm{St}_{\scriptscriptstyle \mathbb{R}}(p\,,q) = \lbrace w\,: w^\dagger b + b^\dagger w = 0\rbrace$. For $w \in T_{b\,}\mathrm{St}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$, let $[b] = bb^\dagger$ and $[w] = wb^\dagger + bw^\dagger$. If $v \in T_x \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$, one says that $(b,w)$ is representative of $(x,v)$, whenever $x = [b]$ and $v = [w]$.

Recall that $x$ and $v$ may always be expressed $x = g\cdot o$ and $v = g\cdot \tilde{\omega}$, using (<ref>). If $x = [b]$, then $g$ may be chosen
$g = (b,b^{\scriptscriptstyle \perp})$ (the columns of $b^{\scriptscriptstyle \perp}$ span the orthogonal complement of the image space of $x$). Then, a direct calculation shows $v = [w_{\scriptscriptstyle v}]$, where $w_{\scriptscriptstyle v} = b^{\scriptscriptstyle \perp}\omega$. Now, define

\begin{equation} \label{eq:retgrassman1}
\mathrm{Ret}_x(v) = \mathrm{Proj}\left( b + w_{\scriptscriptstyle v}\right) \text{ for some $b$ such that } x = [b]
\end{equation}

where $\mathrm{Proj}(h)$ denotes the orthogonal projector onto the span of the columns of $h$, for $h \in \mathbb{R}^{\scriptscriptstyle d\times p}$. This is well-defined, since it does not depend on the choice of $b$ and $b^{\scriptscriptstyle \perp}$, and is indeed a retraction, since it verifies (<ref>).

For a nicer expression of (<ref>), identify each $x \in \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ with its image space $\mathrm{Im}(x)$.In other word, consider $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ as the space of all $p$-dimensional subspaces of $\mathbb{R}^d$. Then,

\begin{equation} \label{eq:retgrassman}
\mathrm{Ret}_x(v) = \mathrm{Span}\left( b + w_{\scriptscriptstyle v}\right)
\end{equation}

where $\mathrm{Span}(h)$ denotes the span of the column space of $h \in \mathbb{R}^{\scriptscriptstyle d\times p}$.
[Whenever $a = (\alpha\,,0_{\scriptscriptstyle p\times q})^\dagger$, where $\alpha$ is $p\times p$ and diagonal, let $\arctan(a) = (\arctan(\alpha)\,,0_{\scriptscriptstyle p\times q})^\dagger$ where $\arctan(\alpha) = \mathrm{diag}(\arctan(\alpha_{ii}))$. For the proof on the following page, define $\cos(a)$ and $\sin(a)$ in the same way.]  The retraction $\mathrm{Ret}$ in (<ref>) is regular, and the corresponding maps $\Phi_x$ (defined as in (<ref>) are given by

\begin{equation} \label{eq:phigrass}
\Phi_x(v) = \left[ b^{\scriptscriptstyle \perp}(r\arctan(a) s^\dagger) \right]
\end{equation}

for $x = [b]$ and $v = [b^{\scriptscriptstyle \perp}\omega]$, where $\omega$ has s.v.d. $\omega = ras^\dagger$ with $r \in O(q)$ and $s \in O(p)$.

As for Examples 1 and 2 in <ref>, (b) of Proposition <ref> now implies $\mathrm{Ret}$ is a  geodesic retraction.

Proof of Proposition <ref> : here, $x \in \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ is identified with its image space, $\mathrm{Im}(x)$. Without loss of generality, it is assumed $p \leq q$.

With $\Phi_x$ given by (<ref>), the aim will be to show that, for $x \in \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$ and $v \in T_x \mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q)$,

\begin{equation} \label{eq:phigrassproof1}
\mathrm{Exp}_x(\Phi_x(v)) = \mathrm{Ret}_x(v)
\end{equation}

In [9], the cut locus of $x$ is obtained under the form

\begin{equation} \label{eq:cutgrass}
\mathrm{Cut}(x) = \left\lbrace\mathrm{Exp}_x\left([b^{\scriptscriptstyle \perp}\omega]\right)\,;\,\omega = ras^\dagger\,,\,\Vert a\Vert_{\scriptscriptstyle \infty} = \frac{\pi}{2}\right\rbrace
\end{equation}

where $\omega= ras^\dagger$ is the s.v.d. of $\omega \in \mathbb{R}^{\scriptscriptstyle\,q\times p}$, with $r \in O(q)$ and $s \in O(p)$, and $\Vert a\Vert_{\scriptscriptstyle \infty} = \max_{ij}|a_{ij}|$. Since $\Vert \arctan(a)\Vert_{\scriptscriptstyle \infty} < \pi/2$, it follows from(<ref>) and (<ref>) that $\mathrm{Ret}_x(v) \notin \mathrm{Cut}(x)$, so $\mathrm{Ret}$ is a regular retraction. Thus, to prove the proposition, one only has to prove (<ref>).

Starting with the left-hand side of (<ref>), let $\varphi = r \arctan(a) s^\dagger$, so $\Phi_x(v) = [ b^{\scriptscriptstyle \perp}\varphi]$. By the discussion before (<ref>), it follows that

\begin{equation} \label{eq:proofgrass1}
\Phi_x(v) = g \cdot \tilde{\varphi} \hspace{1cm} \text{(where $g = (b,b^{\scriptscriptstyle \perp})$)}
\end{equation}

However, then, by (<ref>),

\begin{equation} \label{eq:proofgrass2}
\mathrm{Exp}_x(\Phi_x(v)) =   \exp(g\cdot \hat{\varphi})\cdot x = \left(g\exp(\hat{\varphi})\right)\cdot o
\end{equation}

where the second equality follows from $g^\dagger x = o$, using $g\cdot \hat{\varphi} = g\,\hat{\varphi}\,g^\dagger$. Using the s.v.d. of $\varphi$ ($\varphi = r\arctan(a) s^\dagger$), a straightforward matrix multiplication yields

\begin{equation} \label{eq:proofgrass3}
\hat{\varphi} = k\cdot\hat{q} \hspace{0.5cm} \text{where } k = \left(\begin{array}{cc} s & 0 \\[0.1cm] 0 & r \end{array}\right) \;,\;
q = \arctan(a)
\end{equation}

Thus, from (<ref>) and (<ref>), using the fact that $k \in O(p) \times O(q)$, so $k\cdot o = o$ (or $k^\dagger \cdot o = o$),

\mathrm{Exp}_x(\Phi_x(v)) = \left(g\,k \exp(\hat{q})\right)\cdot o

That is, by the group action property,

\begin{equation} \label{eq:proofgrass33}
\mathrm{Exp}_x(\Phi_x(v)) = g\,k\cdot\left(\exp(\hat{q})\cdot o\right)
\end{equation}

Now, let $b_o = (\mathrm{I}_p\,,0_{\scriptscriptstyle p\times q})^\dagger$, so $o = \mathrm{Span}(b_o)$ and

\begin{equation} \label{eq:proofgrass4}
\exp(\hat{q})\cdot o = \mathrm{Span}\left( \exp(\hat{q})\,b_o\right)
\end{equation}

Then, let $a = (\alpha\,,0_{\scriptscriptstyle p\times q})^{\dagger\,}$, where $\alpha$ is $p \times p$ and diagonal. It will be shown below that

\begin{equation} \label{eq:proofgrass5}
\exp(\hat{q})\,b_o \,=\, \left(\begin{array}{c} \cos(\arctan(\alpha))\\[0.1cm] \sin(\arctan(a))\end{array}\right) =
\left(\begin{array}{c} \mathrm{I}_p\\[0.1cm] a\end{array}\right)\left(\mathrm{I}_p + \alpha \right)^{-\frac{1}{2}}
\end{equation}

where the second equality follows from the identities

\cos(\arctan(\alpha_{ii})) = (1+ \alpha^2_{ii})^{-\frac{1}{2}} \text{ and } \sin(\arctan(\alpha_{ii})) = \alpha_{ii}(1+ \alpha^2_{ii})^{-\frac{1}{2}}

By (<ref>) and (<ref>), after ignoring the invertible matrix $\left(\mathrm{I}_p + \alpha \right)^{-\frac{1}{2}}$,

\exp(\hat{q})\cdot o = \mathrm{Span}\left(\begin{array}{c} \mathrm{I}_p\\[0.1cm] a\end{array}\right) = \mathrm{Span}\left(b_o + b^{\scriptscriptstyle\perp}_oa\right)

Replacing this into (<ref>), it follows that

\begin{equation} \label{eq:proofgrass6}
\mathrm{Exp}_x(\Phi_x(v)) = g\,k\cdot\mathrm{Span}\left(b_o + b^{\scriptscriptstyle\perp}_oa\right) =
g\cdot \mathrm{Span}\left( k(b_o + b^{\scriptscriptstyle\perp}_oa)\right)
\end{equation}

and, by carrying out the matrix products, one may perform the simplification,

\mathrm{Span}\left( k(b_o + b^{\scriptscriptstyle\perp}_oa)\right) =
\mathrm{Span}\left(b_o + b^{\scriptscriptstyle\perp}_ora\right) =
\mathrm{Span}\left(b_o + b^{\scriptscriptstyle\perp}_o\omega\right)

to obtain from (<ref>),

\mathrm{Exp}_x(\Phi_x(v)) = g\cdot\mathrm{Span}\left(b_o + b^{\scriptscriptstyle\perp}_o\omega\right)

which immediately yields (<ref>), since $g\,b_o = b$ and  $g\,b^{\scriptscriptstyle \perp}_o = b^{\scriptscriptstyle \perp}$.

Proof of (<ref>) : write $q = (\kappa\,,0_{\scriptscriptstyle p\times q})^{\dagger\,}$, where $\kappa$ is $p \times p$ and diagonal. It is enough to show

\begin{equation} \label{eq:finalgrassproof1}
\exp(\hat{q}) = \left(\begin{array}{cc}\cos(\kappa) & -\sin(q)^\dagger \\[0.1cm] \sin(q) & \cos(\kappa)_{\scriptscriptstyle q\times q}\end{array}\right)
\end{equation}

wehre $\cos(\kappa)_{\scriptscriptstyle q\times q}$ is the $q \times q$ matrix,

\cos(\kappa)_{\scriptscriptstyle q\times q} = \left(\begin{array}{cc} \cos(\kappa) & \\[0.1cm] & \mathrm{I}_{q-p}\!\end{array}\right)

This follows by writing

\hat{q} = \sum^p_{i=1}\,\kappa_{ii}\,\hat{f}_{i} \hspace{1cm} f_{i} = (\delta_{i}\,,0_{\scriptscriptstyle p\times q})^\dagger

where $\delta_i$ is $p\times p$, diagonal, with its only non-zero element on the $i$-th line, and equal to $1$. Indeed, the matrices $\hat{f}_{i}$ commute with one another, so that

\begin{equation} \label{eq:finalgrassproof3}
\exp(\hat{q}) = \prod^p_{i=1}\,\exp(\kappa_{ii}\,\hat{f}_{i})
\end{equation}

and one readily checks $\hat{f}^{\scriptscriptstyle\hspace{0.03cm}2}_i = -e_{i\,}$, where $e_i$ is  $d\times d$, diagonal, with its only non-zero elements on the $i$-th and $(p+i)$-th lines, and equal to $1$. Therefore,

\begin{equation} \label{eq:finalgrassproof2}
\exp(t\,\hat{f}_i) = \mathrm{I}_d + (\cos(t)-1)\,e_i + \sin(t)\,\hat{f}_i
\end{equation}

Then, (<ref>) obtains after replacing (<ref>) into (<ref>), and using

\begin{array}{ccccc}
e_i\,e_j = 0 && e_i\,\hat{f}_j = 0 & &\\[0.2cm]
\hat{f}_i\,e_j = 0 && \hat{f}_i\,\hat{f}_j = 0 & & \text{for $i \neq j$}
\end{array}

which may be shown by performing the matrix products.

Remark : the above proof has a flavor of the structure theory of Riemannian symmetric spaces. In fact, $\mathrm{Gr}_{\scriptscriptstyle \mathbb{R}}(p\,,q) = \left. O(p+q) \middle/ O(p) \times O(q)\right.$ is a Riemannian symmetric space. The associated Cartan decomposition is

\begin{equation} \label{eq:grasscartan}
\mathfrak{o}(p+q) = \mathfrak{k} \,+\,\mathfrak{p}
\end{equation}

where $\mathfrak{k}$ is the Lie algebra of $K = O(p)\times O(q)$, and where $\mathfrak{p}$ was given in (<ref>). Then,

\begin{equation} \label{eq:grassaa}
\mathfrak{a} = \left\lbrace \hat{a}\,;\, a = (\alpha\,,0_{\scriptscriptstyle p\times q})^{\dagger\,}\,,\, \alpha \text{ is $p\times p$ diagonal}\right\rbrace
\end{equation}

is a maximal Abelian subspace of $\mathfrak{p}$. From [10] (Lemma 6.3, Chapter V), it follows that any $\hat{\omega} \in \mathfrak{p}$ is of the form $\hat{\omega} = \mathrm{Ad}(k)\,\hat{a}$ where $\mathrm{Ad}$ denotes the adjoint representation, $k \in K$ and $\hat{a} \in \mathfrak{a}$.In the present context, this reads $\hat{\omega} = k\cdot \hat{a}$, which is indeed realised if $\omega$ has s.v.d. $\omega = ras^\dagger$, and $k$ is the same as in (<ref>).

§ THE SQUARED DISTANCE FUNCTION

§.§ Cut locus

A Riemannian manifold $M$ becomes a metric space, when equipped with the distance function

\begin{equation} \label{eq:distance}
d(x,y) = \inf \left\lbrace  L(c)\,;\, c \in C^1([0,1]\,,M):c(0) = x \,,\,c(1) = y\,\right\rbrace
\end{equation}

known as the Riemannian distance.  Here, $L(c)$ is the length functional (<ref>). When $M$ is geodesically complete, the infimum in (<ref>) is always achieved by some curve $c^*$, which is then said to be length-minimising. In addition, any length-minimising curve is a geodesic.

This is not to say that all geodesics are length-minimising. A geodesic curve $c$, with $c(0) = x$, may reach a point $c(t) = y$, such that $L(\left.c\right|_{\scriptscriptstyle [0,t]}) \geq d(x,y)$. Roughly, this happens when $t$ is so large that $c$ becomes too long.

For $v \in T_xM$ with $\Vert v \Vert_x = 1$ ($\Vert\cdot \Vert_x$ is the norm given by the scalar product $\langle\cdot,\cdot\rangle_x$), define

\begin{equation} \label{eq:tv}
\mathrm{t}(v) = \sup\left\lbrace t\geq 0 : L(\left.c_{\scriptscriptstyle v}\right|_{\scriptscriptstyle [0,t]}) = d(x,c_{\scriptscriptstyle v}(t))\right\rbrace
\end{equation}

where $c_{\scriptscriptstyle v}$ denotes the geodesic curve with $\dot{c}_{\scriptscriptstyle v}(0) = v$. The following sets

\begin{equation} \label{eq:tangentcut}
\mathrm{TC}(x) = \left\lbrace t\,v\,;\, t = \mathrm{t}(v)\,,\,\Vert v\Vert_x = 1 \right\rbrace \hspace{1cm}
\mathrm{TD}(x) = \left\lbrace t\,v\,;\, t < \mathrm{t}(v)\,,\,\Vert v\Vert_x = 1 \right\rbrace
\end{equation}

are known as the tangent cut locus and tangent injectivity domain of $x$. The cut locus and injectivity domain of $x$ are the sets $\mathrm{Cut}(x) = \mathrm{Exp}\left( \mathrm{TC}(x)\right)$ and $\mathrm{D}(x) = \mathrm{Exp}\left( \mathrm{TD}(x)\right)$.

Since any two points $x$ and $y$ in $M$ are connected by a length-minimising geodesic $c^*$,

\begin{equation} \label{eq:cutdecomp}
M = \mathrm{D}(x) \,\cup\,\mathrm{Cut}(x)
\end{equation}

It is interesting to note that $\mathrm{Cut}(x)$ is a closed and negligible set.

§.§ Normal coordinates

The exponential map $\mathrm{Exp}_x$ is a diffeomorphism of $\mathrm{TD}(x)$ onto $\mathrm{D}(x)$ ($\mathrm{TD}(x)$ is the largest subset of $T_xM$ with this property).  Pick some orthonormal basis $(u_i)$ of $T_xM$, and define, for $y \in \mathrm{D}(x)$,

\begin{equation} \label{eq:normalcoordinates}
\tau^i(y) = \left\langle \mathrm{Exp}^{-1}_x(y)\hspace{0.02cm}, u_i\hspace{0.02cm}\right\rangle_x \hspace{1cm} i = 1\,,\ldots,\,n
\end{equation}

Then, $\tau^i:\mathrm{D}(x) \rightarrow \mathbb{R}$ are well-defined local coordinates, known as normal coordinates. These coordinates satisfy

\begin{equation} \label{eq:normal0}
\tau^i(x) = 0 \hspace{0.4cm} g_{ij}(x) = \delta_{ij} \hspace{0.5cm} \Gamma^i_{jk}(x) = 0
\end{equation}

in the notation of (<ref>) and (<ref>). Even more, (<ref>) is equivalent to the property that geodesics through $x$ appear as straight lines through $0 \in \mathbb{R}^n$, in the normal coordinate map $\tau:\mathrm{D}(x) \rightarrow \mathbb{R}^n$.

Now, the coordinate vector fields $\partial_i = \left.\partial\middle/\partial \tau^i\right.$ are given by

\begin{equation} \label{eq:dexpnorm}
\partial_i(y) \,=\,\mathrm{d\hspace{0.02cm}Exp}_x(v)(u_i) \hspace{1cm} \text{where } v = \tau^i(y)\hspace{0.02cm}u_i
\end{equation}

where $\mathrm{d\hspace{0.02cm}Exp}_x$ is the derivative of $\mathrm{Exp}_{x}:T_xM\rightarrow M$. This may be computed using Jacobi fields,

\begin{equation} \label{eq:dexpjacobi}
\mathrm{d\hspace{0.02cm}Exp}_x(tv)(tu) = J(t)
\end{equation}

where $J$ is a vector field (Jacobi field) along the geodesic $c(t) = \mathrm{Exp}_x(tv)$, which solves the Jacobi equation

\begin{equation} \label{eq:jacobiequation}
\nabla^2_{t\,}{J} - R(\dot{c},J)\dot{c} = 0
\end{equation}

where $J(0) = 0$, $\nabla_{t\,}J(0) = u$, and where $R$ denotes the Riemann curvature tensor. Of course, there do exist other means of computing the derivative $\mathrm{d\hspace{0.02cm}Exp}_x$ (e.g. when $\mathrm{Exp}_x$ coincides with a matrix exponential).

§.§ Distance function

For $x \in M$, consider the distance function $r_x(y)  = d(x,y)$. For $y \in \mathrm{D}(x)$, it is possible to show

\begin{equation} \label{eq:localdistance}
r_x(y) = \left( \sum^n_{i=1} \tau^i(y)^{\hspace{0.02cm}2}\right)^{\!\!\frac{1}{2}} \hspace{0.5cm}
%  f_x(y) = \frac{1}{2}\,\sum^n_{i=1} \tau^i(u)^{\hspace{0.02cm}2}
\end{equation}

in terms of the normal coordinates $\tau^i$. From (<ref>), the distance function $r_x$ is smooth on
$\mathrm{U}_x = \mathrm{D}(x) - \lbrace x \rbrace$.

When $y \in \mathrm{U}_{x}$ is of the form $y = c_{\scriptscriptstyle v}(t)$, where $c_{\scriptscriptstyle v}$ is a geodesic with $\dot{c}_{\scriptscriptstyle v}(0) = v$ and $\Vert v \Vert_x = 1$, define $\partial_{\hspace{0.02cm}r}(y) = \dot{c}_{\scriptscriptstyle v}(t)$.  By the first variation of arc length formula (Theorem II.4.1 in [11]),

\begin{equation} \label{eq:gradr}
\mathrm{grad}\,r_x(y) = \partial_{\hspace{0.02cm}r}(y) \hspace{1cm} \text{for } y \in \mathrm{U}_x
\end{equation}

Introduce geodesic spherical coordinates $(r,\theta^{\scriptscriptstyle\,\alpha})$ on $\mathrm{U}_{x\,}$. If $y = c_{\scriptscriptstyle v}(t)$ these are given by $r = t$ and $(\theta^{\scriptscriptstyle\,\alpha}) = \theta(v)$, where $\theta$ identifies the unit sphere in $T_xM$ with the Euclidean unit sphere $S^{n-1}$.In these coordinates, the metric is given by

\begin{equation} \label{eq:lengthspherical}
g = dr \otimes dr \,+\, g^{\hspace{0.03cm} r}_{\scriptscriptstyle \alpha\beta}\;d\theta^{\scriptscriptstyle\,\alpha}\!\otimes\!d\theta^{\scriptscriptstyle\,\beta}
\end{equation}

reflecting the fact that $\partial_{\hspace{0.02cm}r}$ is orthogonal to constant $r_x$ surfaces, here parameterised by $(\theta^{\scriptscriptstyle\,\alpha})$.
The coordinate vector fields $\partial_{\hspace{0.02cm}\scriptscriptstyle \alpha}$ are given by (<ref>) : $\partial_{\hspace{0.02cm}\alpha}(y) = J(r)$ for $y = c_{\scriptscriptstyle v}(r)$, where $J(0) = 0$ and $\nabla_{t\,}J(0) = u_{\hspace{0.02cm}\scriptscriptstyle \alpha}$ (where $u_{\hspace{0.02cm}\scriptscriptstyle \alpha} = \left.\partial\middle/\partial \theta^{\scriptscriptstyle\,\alpha}\right.$ are coordinate vector fields on the unit sphere in $T_xM$). In particular, if $A:T_xM \rightarrow T_yM$ solves the operator Jacobi equation (along the geodesic $c_{\scriptscriptstyle v}$)

\begin{equation} \label{eq:jacobioperator}
\nabla^2_{t\,}{A} - R_{\dot{c}_{\scriptscriptstyle v}}\hspace{0.02cm}A = 0 \hspace{1cm} A(0) = 0 \,,\, \nabla_{t\,}A(0) = \mathrm{Id}_x
\end{equation}

where $R_{\hspace{0.02cm}\dot{c}_{\scriptscriptstyle v}}(\cdot) = R(\dot{c}_{\scriptscriptstyle v\hspace{0.02cm}},\cdot)\hspace{0.02cm}\dot{c}_{\scriptscriptstyle v\,}$, then $\partial_{\hspace{0.02cm}\alpha}(y) = A(r)\hspace{0.02cm} u_{\hspace{0.02cm}\scriptscriptstyle \alpha\,}$. Thus, if $\mathcal{A}(y):T_xM\rightarrow T_xM$ is given by $\mathcal{A}(y) = \Pi^{\scriptscriptstyle 0}_{r} \circ A(r)$, then $g^{\hspace{0.03cm} r}(y) = (\mathcal{A}(y))^*(h)$, the pullback under $\mathcal{A}(y)$ of the metric $h$ of the unit sphere in $T_xM$. It should be noted $\mathcal{A}(y)$ maps tangent spaces of this unit sphere to themselves.

The Hessian of $r_x$ follows from (<ref>) and (<ref>), which yield (after using the fact that the vector fields $\partial_{\hspace{0.02cm}r}$ and $\partial_{\hspace{0.02cm}\scriptscriptstyle \alpha}$ commute)

\mathrm{Hess}\,r_x\cdot \partial_{\hspace{0.02cm}r\hspace{0.02cm}} = 0 \hspace{0.3cm}\text{and}\hspace{0.3cm}\mathrm{Hess}\,r_x\cdot \partial_{\hspace{0.02cm}\alpha} =\nabla_{\partial_{\hspace{0.02cm}r}}\hspace{0.02cm}\partial_{\hspace{0.02cm}\alpha}

Then, using the expression of the $\partial_{\hspace{0.02cm}\alpha}$ as Jacobi fields,

\begin{equation} \label{eq:hessr}
\mathrm{Hess}\,r_x(y) \,=\, \left.\nabla_{t\,}A(t)A^{-1}(t)\right|_{t=r}
\end{equation}

Taking the covariant derivative $\nabla_{t}$ of this formula yields the Ricatti equation

\begin{equation} \label{eq:ricatti}
\nabla_{\partial_{\hspace{0.02cm}r}}  \mathrm{Hess}\,r_x \,=\, R_{\scriptscriptstyle \partial_{\hspace{0.02cm}r}} - \left(\mathrm{Hess}\,r_x\right)^{\hspace{0.02cm}2}
\end{equation}

The Jacobi equation (<ref>) and the Ricatti equation (<ref>) lead up to the comparison theorems[The inequalities (<ref>) and (<ref>) are in the sense of the usual Loewner order for self-adjoint operators.].

Assume the sectional curvatures of $M$ lie within the interval $[\kappa_{\min}\hspace{0.02cm},\kappa_{\max}\hspace{0.03cm}]\hspace{0.02cm}$. Then,

\begin{equation} \label{eq:metcomp}
\mathrm{sn}^2_{\kappa_{\max}}(r)\,h\,\leq\, g^{\hspace{0.03cm} r}(y) \,\leq\,    \mathrm{sn}^2_{\kappa_{\min}}(r)\,h   %g^{\hspace{0.03cm} r}(y)
\end{equation}

\begin{equation} \label{eq:hescomp}
\mathrm{ct}_{\kappa_{\max}}(r)\,g^{\hspace{0.03cm} r}(y)\,\leq\, \mathrm{Hess}\,r_x(y)\,\leq\,
\mathrm{ct}_{\kappa_{\min}}(r)\,g^{\hspace{0.03cm} r}(y)
\end{equation}

for $y \in \mathrm{U}_x\,$. Here, $\mathrm{sn}^{\prime\prime}_\kappa(r) + \kappa\,\mathrm{sn}_{\kappa}(r) = 0$ with $\mathrm{sn}_{\kappa}(0) = 0$ and $\mathrm{sn}^\prime_{\kappa}(0) = 1$, and $\mathrm{ct}_{\kappa} = \left. \mathrm{sn}^\prime_{\kappa}\middle/\mathrm{sn}_{\kappa}\right.$.

Remark : in addition to its singularity at $x$, the distance function $r_x$ is singular on $\mathrm{Cut}(x)$.If $y\in\mathrm{Cut}(x)$, then either $y$ is a first conjugate point ($A(r)$ is singular, for the first time after $x$),or there exist two distinct length-minimising geodesics connecting $x$ to $y$. In the first case, $\mathrm{Hess}\,r_x(y)$ has an eigenvalue equal to $-\infty$. In the second case, $\mathrm{grad}\,r_x$ is discontinuous at $y$.The distributional Hessian of $r_x$ was studied in [12].

Remark : the reader may have noted, or recalled, that $y \in \mathrm{Cut}(x)$ if and only if $x \in \mathrm{Cut}(y)$.

§.§ Squared distance

For $x \in M$, consider the squared distance function $f_x(y)  = d^{\hspace{0.03cm}2}(x,y)/2$. For $y \in \mathrm{D}(x)$,

\begin{equation} \label{eq:localfx}
f_x(y) = \frac{1}{2}\,\sum^n_{i=1} \tau^i(y)^{\hspace{0.02cm}2}
\end{equation}

in terms of the normal coordinates $\tau^i$. It follows that $f_x$ is smooth on $\mathrm{D}(x)$. Of course, $f_x = r^{\hspace{0.03cm}2}_x/2$. Therefore, applying the chain rule to (<ref>),

\begin{equation} \label{eq:gradfx}
\mathrm{grad}\,f_x(y) \,=\, -\hspace{0.02cm}\mathrm{Exp}^{-1}_y(x) \hspace{0.5cm} \text{for } y \in \mathrm{D}(x)
\end{equation}

and, by another application of the chain rule,

\begin{equation} \label{eq:hessfx}
\mathrm{Hess}\, f_x(y) = dr_x \otimes dr_x + r_x\,\mathrm{Hess}\,r_x
\end{equation}

Just like $r_{x\,}$, $f_x$ is singular on $\mathrm{Cut}(x)$. If $y \in \mathrm{Cut}(x)$ is a first conjugate point, then $ \mathrm{Hess}\, f_x(y)$ has an eigenvalue equal to $-\infty$.

The convexity of the function $f_x$ will play a significant rôle, in the following, especially when $M$ is a Hadamard manifold : a simply connected, geodesically complete Riemannian manifold of non-positive sectional curvature. When $M$ is a Hadamard manifold, the following properties hold :  any $x\hspace{0.02cm},y \in M$ are connected by a unique geodesic $c$ ; for all $x \in M$, $\mathrm{Cut}(x)$ is empty, and $f_x$ is smooth and $1/2$-strongly convex ; all geodesic balls are convex (see the remarks below,for the notions of convex set and function).

Assume $M$ is a Hadamard manifold. In addition, assume that the sectional curvature of $M$ is bounded below by $\kappa_{\min} = -c^{\hspace{0.02cm}\scriptscriptstyle 2}$. Theorem <ref> may be applied to (<ref>), after setting
$\kappa_{\max} = 0$. This yields

\begin{equation} \label{eq:hesfxcomp}
g(y) \,\leq\,\mathrm{Hess}\,f_x(y)\,\leq\, c\hspace{0.03cm} r_x(y)\coth(c\hspace{0.03cm} r_x(y))\,g(y)
\end{equation}

for $y \in M$. In addition to showing that $f_x$ is $1/2$-strongly convex, this shows that $\mathrm{Hess}\,f_x$ has, at most, linear growth

\begin{equation} \label{eq:hesfxcompbis}
\mathrm{Hess}\,f_x(y)\,\leq\, (1+c\hspace{0.03cm}r_x(y))\,g(y)
\end{equation}

since $x\coth(x) \leq 1 + x$ for $x \geq 0$.

Remark : a subset $A \subset M$ is called convex (that is, strongly convex, in the terminology of [11]) if any $x\hspace{0.02cm},y \in A$ are connected by a unique length-minimising geodesic $c$,  and $c$ lies entirely in $A$.A function $f:A \rightarrow \mathbb{R}$ is then called (strictly) convex if $f \circ c:\mathbb{R} \rightarrow \mathbb{R}$ is (strictly) convex, for any geodesic $c$ which lies in $A$. It is called $\alpha$-strongly convex (for some $\alpha > 0$) if $f \circ c:\mathbb{R} \rightarrow \mathbb{R}$ is $\alpha$-strongly convex, for any geodesic $c$ which lies in $A$,

\begin{equation} \label{eq:strongconv}
(f\circ c)(p\hspace{0.02cm}s + q\hspace{0.02cm}t) \leq p\hspace{0.02cm}(f\circ c)(s) + q\hspace{0.02cm}(f\circ c)(t) - \alpha\hspace{0.02cm}p\hspace{0.02cm}q\,d^{\hspace{0.03cm}2}(c(s),c(t))
\end{equation}

whenever $p\hspace{0.02cm},q \geq 0$ and $p+q = 1$. For example, if $M$ is a sphere and $A$ is the open northern hemisphere, then $A$ is  convex. Then,  $f_x:A \rightarrow \mathbb{R}$, where $x$ denotes the north pole, is strictly convex, but not strongly convex.

Remark : for $x \in M$, let $\mathrm{inj}(x) = d(x\hspace{0.02cm},\mathrm{Cut}(x))$ denote the injectivity radius at $x$. Then, let $\mathrm{inj}(M) = \inf_{x\in M}\mathrm{inj}(x)$, the injectivity radius of $M$. Assume all the sectional curvatures of $M$ are less than $\kappa_{\max} = c^{\hspace{0.02cm}\scriptscriptstyle 2}$. If $B(x,R)$ is a geodesic ball with radius $R \leq (1/2)\hspace{0.02cm}\min\lbrace \mathrm{inj}(M)\hspace{0.02cm},\pi\hspace{0.03cm}c^{\scriptscriptstyle -1}\rbrace$, then $B(x,R)$ is convex. Here, if $\kappa_{\max} = 0$, then $c^{\scriptscriptstyle -1}$ is understood to be $+\infty$. However, there do exist manifolds $M$ with negative sectional curvature, and with $\mathrm{inj}(M) = 0$ (e.g. the quotient of the Poincaré upper half-plane, by a discrete group of translations).

§ EXAMPLE : ROBUST RIEMANNIAN BARYCENTRE

Let $M$ be a Hadamard manifold, with sectional curvatures bounded below by $\kappa_{\min} = -c^{\hspace{0.02cm}\scriptscriptstyle 2}$. Recall that $f_x$ is $1/2$-strongly convex, and $\mathrm{Hess}\,f_x$ has, at most, linear growth (as in (<ref>)). On the other hand, consider the function

\begin{equation} \label{eq:hdist}
V_x(y) =  \delta^{\hspace{0.02cm} \scriptscriptstyle 2}\left[\mathstrut 1+ \left(d(x,y)\middle/\delta\right)^{\scriptscriptstyle 2\,}\right]^{\scriptscriptstyle\frac{1}{\mathstrut 2}}\, -\,\delta^{\hspace{0.02cm} \scriptscriptstyle 2}
\end{equation}

where $\delta > 0$ is a cutoff parameter. Note that $V_x(y) \geq 0$, and $V_x(y) = 0$ if and only if $x = y$. Moreover, $V_x \sim f_{x\,}$, when $\left.d(x\hspace{0.02cm},y)\middle/\delta\right.$ is small, and $V_x \sim \delta\hspace{0.02cm}r_{x\,}$, when $\left.d(x\hspace{0.02cm},y)\middle/\delta\right.$ is large.

Let $M$ be a Hadamard manifold, with sectional curvatures bounded below by $\kappa_{\min} = -c^{\hspace{0.02cm}\scriptscriptstyle 2}$. If $V_x:M\rightarrow \mathbb{R}$ is defined as in (<ref>), then
$V_x$ is smooth, strictly (but not strongly) convex, and $\mathrm{Hess}\,V_x$ is bounded by $1+\delta\hspace{0.02cm}c$.

Let $\pi$ be a probability measure on $M$, and consider the problem of minimising

\begin{equation} \label{eq:huberpi}
V_{\pi}(y) = \int_M\,V_x(y)\,\pi(dx)
\end{equation}

A global minimiser of $V_\pi$ will be called a robust Riemannian barycentre of $\pi$. Here, the adjective “robust"  comes from the field of robust statistics [13].

Let $\pi$ be a probability distribution on a Hadamard manifold $M$. If $\pi$ has finite first-order moments, then the function $V_{\pi}$ is a proper, strictly convex function, with a unique global minimum $x^* \in \Theta$. Therefore, $\pi$ has a unique robust Riemannian barycentre $x^*$.

Recall that $\pi$ has finite first-order moments, if and only if there exists $y_o \in M$ with

\begin{equation} \label{eq:firstorder}
\int_M\,r_{x}(y_o)\,\pi(dx) \,<\,\infty
\end{equation}

and recall that $V_{\pi}$ is said to be proper if it takes on finite values.

Proof of Proposition <ref> : by applying the chain rule to (<ref>), and using (<ref>),

\begin{equation} \label{eq:hgrad}
\mathrm{grad}\,V_x(y) = -\,\frac{\mathrm{Exp}^{-1}_y(x)}{\mathstrut \left[\mathstrut 1+ \left(d(x,y)\middle/\delta\right)^{\scriptscriptstyle 2\,}\right]^{\scriptscriptstyle\frac{1}{\mathstrut 2}}}
\end{equation}

Then, by applying (<ref>),

\begin{equation} \label{eq:hhess}
\mathrm{Hess}\,V_x(y) = -\,
\frac{{\small \mathrm{Exp}^{-1}_y(x)\otimes \mathrm{Exp}^{-1}_y(x)}}{\mathstrut\delta^{\hspace{0.02cm} \scriptscriptstyle 2} \left[\mathstrut 1+ \left(d(x,y)\middle/\delta\right)^{\scriptscriptstyle 2\,}\right]^{\scriptscriptstyle \frac{3}{\mathstrut 2}}}\,-\, \frac{\nabla\,\mathrm{Exp}^{-1}_y(x)}{\mathstrut \left[\mathstrut 1+ \left(d(x,y)\middle/\delta\right)^{\scriptscriptstyle 2\,}\right]^{\scriptscriptstyle\frac{1}{\mathstrut 2}}}
\end{equation}

To conclude, it is enough to note the inequalities,

0\,\leq\,\mathrm{Exp}^{-1}_y(x)\otimes \mathrm{Exp}^{-1}_y(x)\leq d^{\hspace{0.03cm}\scriptscriptstyle 2}(x,y)\hspace{0.03cm}g(y) %\hspace{0.4cm}\text{and}\hspace{0.4cm}
%1\,\leq\,-\,\nabla\,\mathrm{Exp}^{-1}_y(x)\,\leq\, (1+\kappa\hspace{0.02cm} r_x(y))

which follows since $\mathrm{Exp}^{-1}_y(x)\otimes \mathrm{Exp}^{-1}_y(x)$ is a rank-one operator in $T_yM$, and

g(y)\,\leq\,-\,\nabla\,\mathrm{Exp}^{-1}_y(x)\,\leq\, (1+c\hspace{0.03cm}r_x(y))\hspace{0.03cm}g(y)

which is the same as (<ref>), and follows from (<ref>) and (<ref>). Replacing these into (<ref>), a direct calculation shows

\begin{equation} \label{eq:hdistproof}
0\,<\,\mathrm{Hess}\,V_x(y)\,\leq\, (1+\delta\hspace{0.02cm}c)\hspace{0.03cm}g(y)
\end{equation}

which completes the proof. Proof of Proposition <ref> : using the sub-additivity of the square root, (<ref>) and (<ref>) imply that for any $y \in M$,

V_{\pi}(y) \,\leq\, \int_M\,r_{x}(y)\,\pi(dx)

But, by the triangle inequality, and (<ref>),

\int_M\,r_{x}(y)\,\pi(dx) \leq d(y\hspace{0.02cm},y_o) + \int_M\,r_{x}(y_o)\,\pi(dx) \,<\infty

Therefore, $V_\pi$ is proper. That $V_\pi$ is also strictly convex is an immediate result of Proposition <ref> :each function $V_x$ is strictly convex, and $V_\pi(y)$ is the expectation of $V_x(y)$ with respect to a random $x$ with distribution $\pi$. Now, to show that $V_\pi$ has a unique global minimum, it is enough to show that $V_\pi(y)$ goes to infinity as $y$ goes to infinity. Note that $\varphi(x) = (1+x^{\scriptscriptstyle 2})^{\scriptscriptstyle \frac{1}{2}}$ is convex. This implies (using the elementary fact that the graph of a convex function remains above any of its tangents),

V_x(y) \geq (\sqrt{2} - 1)\hspace{0.04cm}\delta^{\hspace{0.02cm}\scriptscriptstyle 2}\,+\, \frac{\delta}{\sqrt{2}}\hspace{0.04cm}r_x(y)

Taking the expectation with respect to $\pi$,

V_\pi(y) \geq (\sqrt{2} - 1)\hspace{0.04cm}\delta^{\hspace{0.02cm}\scriptscriptstyle 2}\,+\, \frac{\delta}{\sqrt{2}}\hspace{0.04cm}\int_M\,r_x(y)\,\pi(dx)

To see that $V_\pi(y)$ goes to infinity as $y$ goes to infinity, it is now enough to note, using the triangle inequality,

\int_M\,r_x(y)\,\pi(dx)\,\geq\,d(y\hspace{0.02cm},y_o)\,-\int_M\,r_{x}(y_o)\,\pi(dx)

where $d(y\hspace{0.02cm},y_o)$ goes to infinity as $y$ goes to infinity.

Remark : the above Proposition <ref> only requires $M$ to be a Hadamard manifold, without the additional condition that it have sectional curvatures bounded below. Indeed, Proposition <ref> only relies on the fact that $V_x$ is strictly convex, and not on the fact that the Hessian of $V_x$ is bounded above by $1+\delta\hspace{0.02cm}c$.

Remark : if a function $V:M\rightarrow \mathbb{R}$, on a Riemannian manifold $M$, has bounded Hessian, then it has Lipschitz-gradient.
That is, if there exists $\ell \geq 0$ such that $\left|\mathrm{Hess}\,V(x)(u,u)\right|\leq \ell\hspace{0.02cm}g(u,u)$ for all $x \in M$ and $v \in T_xM$, then

\begin{equation} \label{eq:lipschitzgrad}
\left\Vert \Pi^{\scriptscriptstyle 0}_{\scriptscriptstyle 1}\left(\mathrm{grad}\,V_{c(1)}\right) - \mathrm{grad}\,V_{c(0)}\hspace{0.03cm}\right\Vert_{c(0)}\,\leq\,\ell\hspace{0.02cm}L(c)
\end{equation}

for any smooth curve $c:[0,1]\rightarrow M$, where $L(c)$ is the length of $c$. This is due to the following.

Let $X$ be a vector field on a Riemannian manifold $M$. If the operator norm of the covariant derivative $\nabla X$ is bounded by $\ell \geq 0$, then

\begin{equation} \label{eq:lipschitzfield}
\left\Vert \Pi^{\scriptscriptstyle 0}_{\scriptscriptstyle 1}\left(X_{c(1)}\right) - X_{c(0)}\hspace{0.03cm}\right\Vert_{c(0)}\,\leq\,\ell\hspace{0.02cm}L(c)
\end{equation}

for any smooth curve $c:[0,1]\rightarrow M$.

Sketch of proof : let $u_i$ be a parallel orthonormal base along $c$ ($u_i$ are vector fields along $c$, with $u_i(t)$ an orthonormal basis of $T_{c(t)}M$, for each $t$). Let $X^i(t) = \langle X\hspace{0.02cm},u_i\rangle_{c(t)\,}$ and note

\left\Vert \Pi^{\scriptscriptstyle 0}_{\scriptscriptstyle 1}\left(X_{c(1)}\right) - X_{c(0)}\hspace{0.03cm}\right\Vert^2_{c(0)} =
\sum^n_{i=1}\left(X^i(1) - X^i(0)\right)^{\!2} = \sum^n_{i=1}\left(\int^1_0\,\langle \nabla_{\dot{c}\,}X\hspace{0.02cm},u_{i\hspace{0.02cm}}\rangle_{c(t)\,}dt\right)^{\!\!2}

the proof then follows by using Jensen's inequality, since $\Vert \nabla_{\dot{c}\,}X\Vert_{c(t)} \leq \ell\hspace{0.03cm}\Vert \dot{c}\Vert_{c(t)\,}$.

§ RIEMANNIAN VOLUME AND INTEGRAL FORMULAE

§.§ Elementary volume comparison

If a Riemannian manifold $M$ is orientable, then $M$ admits a volume form, called the Riemannian volume form, to be denoted $\mathrm{vol}$, in the following. In terms of local coordinates $(x^i\,;i=1,\ldots,n)$

\begin{equation} \label{eq:volumeform}
\mathrm{vol} = \det(g)^{\frac{1}{2}}\,dx^1 \wedge \ldots \wedge dx^n
\end{equation}

where $\det(g)$ is the determinant of the metric, which is equal the determinant of the matrix $(g_{ij})$, defined in  (<ref>). Then, the integral of a continuous, compactly-supported function $f:M\rightarrow \mathbb{R}$, with respect to $\mathrm{vol}$, is the integral of the $n$-form $f\hspace{0.02cm}\mathrm{vol}$ over $M$. This is denoted $\int_M\, f(x)\,\mathrm{vol}(dx)$. There exists a unique measure $|\mathrm{vol}|$ on the Borel $\sigma$-algebra of $M$, such that [14] (Chapter 8), for continuous, compactly-supported $f$,

\int_M\, f(x)\,\mathrm{vol}(dx) \,=\, \int_M\,f(x)\,|\mathrm{vol}|(dx)

where the integral on the left is a Riemann integral, and the integral on the right is a Lebesgue integral. It is quite useful to study these integrals using geodesic spherical coordinates (which were introduced in <ref>). Let $(r,\theta^{\scriptscriptstyle\,\alpha})$ be geodesic spherical coordinates, with origin at $x \in M$. Recall that these are defined on $\mathrm{U}_{x} = \mathrm{D}(x) - \lbrace x \rbrace$, where $\mathrm{D}(x)$ is the injectivity domain of $x$. Since $M$ can be decomposed as in (<ref>), and $\mathrm{Cut}(x)$ is negligible,

\begin{equation} \label{eq:integralux}
\int_M\,f(y)\,\mathrm{vol}(dy) \,=\,\int_{\,\mathrm{U}_{x}}f(y)\,\mathrm{vol}(dy)
\end{equation}

Using (<ref>) and (<ref>), $\mathrm{vol}(dy) = \det(\mathcal{A}(y))\,dr\wedge \omega_{n-1}(d\theta)$, where $\omega_{n-1}$ is the area measure on the unit sphere in $T_xM$ (as of now, this is identified with the Euclidean unit sphere $S^{n-1}$). Using (<ref>) and $\mathrm{D}(x) = \mathrm{Exp}\left( \mathrm{TD}(x)\right)$, (<ref>) yields

\begin{equation} \label{eq:integralspherical}
\int_M\,f(y)\,\mathrm{vol}(dy) \,=\, \int^{\mathrm{t}(\theta)}_0\!\!\!\int_{S^{n-1}}f(r,\theta)\,\det(\mathcal{A}(r,\theta))\,dr\hspace{0.03cm} \omega_{n-1}(d\theta)
\end{equation}

where $\mathrm{t}$ was defined in (<ref>). This formula expresses integrals, with respect to the Riemannian volume form, using geodesic spherical coordinates.

Recall the Laplacian $\Delta\hspace{0.03cm} r_x = \mathrm{div}\hspace{0.03cm}\partial_{\hspace{0.02cm}r\,}$. By definition of the divergence, $\mathcal{L}_{\partial_{\hspace{0.02cm}r}}\hspace{0.02cm}\mathrm{vol} = (\mathrm{div}\hspace{0.02cm}\partial_{\hspace{0.02cm}r})\hspace{0.02cm}\mathrm{vol}$. Writing this in geodesic spherical coordinates,

\begin{equation} \label{eq:laplacelogdet}
\Delta\hspace{0.03cm} r_x(r,\theta) \,=\, \partial_{\hspace{0.02cm}r}\hspace{0.02cm}\log\det(\mathcal{A}(r,\theta))
\end{equation}

Accordingly, the comparison theorems <ref> can be used to obtain the volume comparison theorem.

Assume the sectional curvatures of $M$ lie within the interval $[\kappa_{\min}\hspace{0.02cm},\kappa_{\max}\hspace{0.03cm}]\hspace{0.02cm}$. Then,

\begin{equation} \label{eq:volcomp}
\mathrm{sn}^{n-1}_{\kappa_{\max}}(r)\,\leq\, \det(\mathcal{A}(r,\theta)) \,\leq\,    \mathrm{sn}^{n-1}_{\kappa_{\min}}(r)  %g^{\hspace{0.03cm} r}(y)
\end{equation}

\begin{equation} \label{eq:laplacecomp}
(n-1)\hspace{0.02cm}\mathrm{ct}_{\kappa_{\max}}(r)\,\leq\, \partial_{\hspace{0.02cm}r}\hspace{0.02cm}\log\det(\mathcal{A}(r,\theta))\,\leq\,
\end{equation}

This volume comparison theorem is quite elementary, as stronger and deeper comparison results do exist[For example, Gromov's volume comparison theorem can be used to give a short proof of the famous “sphere theorem", Theorem III.4.6 in [11]. ]. Moreover, in this theorem, the lower bound on sectional curvature may be replaced by a lower bound on Ricci curvature, without any change to the conclusion.

Remark : roughly, (<ref>) states that “more curvature means less volume". If $f:M\rightarrow \mathbb{R}$ is a non-negative function of distance to $x$, so $f(y) = f(r)$ in terms of the coordinates $(r,\theta^{\scriptscriptstyle\,\alpha})$, then

\begin{equation} \label{eq:integralcomp}
\omega_{n-1}\,\int^{\scriptscriptstyle R}_{\scriptscriptstyle 0}\,f(r)\,\mathrm{sn}^{n-1}_{\kappa_{\max}}(r)\hspace{0.02cm}dr\leq\,\int_{\scriptscriptstyle B(x,R)}f(y)\,\mathrm{vol}(dy)\,\leq\,\omega_{n-1}\,
\int^{\scriptscriptstyle R}_{\scriptscriptstyle 0}\,f(r)\,\mathrm{sn}^{n-1}_{\kappa_{\min}}(r)\hspace{0.02cm}dr
\end{equation}

for any $R \leq \min\lbrace \mathrm{inj}(x)\hspace{0.03cm},\pi\hspace{0.03cm}c^{\scriptscriptstyle -1}\rbrace$. Here, $\mathrm{inj}(x)$ is the injectivity radius at $x$, $c = |\kappa_{\max}|^{\scriptscriptstyle 1/2\hspace{0.03cm}}$, and $\omega_{n-1}$ denotes the area of $S^{\hspace{0.02cm}n-1}$. In addition, if $\kappa_{\max} \leq 0$, then $c^{\scriptscriptstyle -1}$ is understood to be $+\infty$.

In general, it may be impossible to apply the integral formula (<ref>), since $\mathrm{t}(\theta)$ may be unknown. Here are two examples where $\mathrm{t}(\theta)$ is known, and quite tractable (in fact, constant).

Example 1 : if $M$ is a Hadamard manifold, then for any choice of the origin $x$, and any $\theta \in S^{n-1}$, one has $\mathrm{t}(\theta) = \infty$, and (<ref>) becomes

\begin{equation} \label{eq:integralsphericalhadamard}
\int_M\,f(y)\,\mathrm{vol}(dy) \,=\, \int^{\infty}_0\!\!\!\int_{S^{n-1}}f(r,\theta)\,\det(\mathcal{A}(r,\theta))\,dr\hspace{0.03cm} \omega_{n-1}(d\theta)
\end{equation}

Example 2 : compact rank-one symmetric space are the following manifolds : spheres, real projective spaces, complex projective spaces,  quaternion projective spaces, or the Cayley plane. These are manifolds all of whose geodesics are closed (i.e. periodic) and isometric to one another
(see [15], for a detailed account). Therefore, $\mathrm{t}(\theta)$ does not depend on $x$ nor on $\theta$, but is always equal to $l/2$, where $l$ is the length of a simple geodesic loop. Scaling the metric so the maximum sectional curvature is equal to $1$, it can be shown $l = \pi$ for real projective spaces, and $l = 2\pi$in all other cases. Moreover (<ref>) takes on the form (this may be found by looking up the solution of the Jacobi equation in [15], Page 82),

\begin{equation} \label{eq:integralsphericalcross}
\int_M\,f(y)\,\mathrm{vol}(dy) \,=\, \int^{\frac{l}{\mathstrut 2}}_0\!\!\!\int_{S^{n-1}}f(r,\theta)\left(\sin(r)\right)^{k-1}\left(2\sin(r/2)\right)^{n-k}\,dr\hspace{0.03cm} \omega_{n-1}(d\theta)
\end{equation}

where $k= n$ for spheres and real projective spaces, and $k = 2$ or $4$ for complex or quaternion projective spaces, respectively. For the Cayley plane, $n = 16$ and $k = 8$.

§.§ Riemannian symmetric spaces

A Riemannian symmetric space is a Riemannian manifold $M$, such that, for each $x \in M$, there exists an isometry $s_x : M \rightarrow M$, with $s_x(x) = x$ and $d\hspace{0.02cm}s_x(x) = -\mathrm{Id}_x\,$. This isometry $s_x$ is called the geodesic symmetry at $x$.

Let $G$ denote the identity component of the isometry goup of $M$, and $K = K_o$ be the stabiliser in $G$ of some point $o \in M$[According to the Myers-Steenrod theorem, $G$ is a connected Lie group, and $K$ a compact subgroup of $G$.]. Then, $M = G/K$ is a Riemannian homogeneous space. The mapping $\theta : G \rightarrow G$, where $\theta(g) = s_o\circ g \circ s_o$ is an involutive isomorphism of $G$.

Let $\mathfrak{g}$ denote the Lie algebra of $G$, and consider the Cartan decomposition, $\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$, where $\mathfrak{k}$ is the $+1$ eigenspace of $d\hspace{0.02cm}\theta$ and $\mathfrak{p}$ is the $-1$ eigenspace of $d\hspace{0.02cm}\theta$. One clearly has the commutation relations,

\begin{equation} \label{eq:sscommute}
[\mathfrak{k},\mathfrak{k}] \subset \mathfrak{k}\hspace{0.2cm};\hspace{0.2cm}
[\mathfrak{k},\mathfrak{p}] \subset \mathfrak{p}\hspace{0.2cm};\hspace{0.2cm}
[\mathfrak{p},\mathfrak{p}] \subset \mathfrak{k}
\end{equation}

In addition, it turns out that $\mathfrak{k}$ is the Lie algebra of $K$, and that $\mathfrak{p}$ may be identified with $T_oM$.

The Riemannian metric of $M$ may always be expressed in terms of an $\mathrm{Ad}(K)$-invariant scalar product $Q$ on $\mathfrak{g}$. If $x \in M$ is given by $x = g\cdot o$ for some $g \in G$ (where $g\cdot o = g(o)$), then

\begin{equation} \label{eq:ssmetric}
\langle u,\!v\rangle_{\scriptscriptstyle x} = Q(g^{\scriptscriptstyle -1}\cdot u\hspace{0.02cm},g^{\scriptscriptstyle -1}\cdot v)
\end{equation}

where the vectors $g^{\scriptscriptstyle -1}\cdot u$ and $g^{\scriptscriptstyle -1}\cdot v$, which belong to $T_oM$, are identified with elements of $\mathfrak{p}$. Here, by an abuse of notation, $d\hspace{0.02cm}g^{\scriptscriptstyle -1}\cdot u$ is denoted $g^{\scriptscriptstyle -1}\cdot u$.

Let $\exp:\mathfrak{g} \rightarrow G$ denote the Lie group exponential. If $v \in T_oM$, then the Riemannian exponential $\mathrm{Exp}_o(v)$ is given by

\begin{equation} \label{eq:ssexp1}
\mathrm{Exp}_o(v) = \exp(v)\cdot o
\end{equation}

Moreover, if $\Pi^t_{\scriptscriptstyle 0}$ denotes parallel transport along the geodesic $c(t) = \mathrm{Exp}_o(tv)$, then

\begin{equation} \label{eq:ssparallel}
\Pi^t_{\scriptscriptstyle 0}(u) = \exp(tv)\cdot u
\end{equation}

for any $u \in T_o M$ (note that the identification $T_oM \simeq \mathfrak{p}$ is always made, implicitly). Using (<ref>), one can derive the following expression for the Riemann curvature tensor at $o$,

\begin{equation} \label{eq:sscurvature}
R_o(v,u)w = -[[v\hspace{0.03cm},u]\hspace{0.02cm},w] \hspace{1cm} v,u,w \in T_oM
\end{equation}

A fundamental property of symmetric spaces is that the curvature tensor is parallel :$\hspace{0.03cm}\nabla\,R = 0$. This is often used to solve the Jacobi equation (<ref>), and then express the derivative of the Riemannian exponential, using in (<ref>),

\begin{equation} \label{eq:dexpss}
\mathrm{d\hspace{0.02cm}Exp}_x(v)(u) \,=\, \exp(v)\cdot \mathrm{sh}(R_v)(u)
\end{equation}

where $\mathrm{sh}(R_v) = \sum^{\infty}_{n=0} (R_v)^n/(2n+1)!$ for the self-adjoint curvature operator $R_v(u) = [v\hspace{0.03cm},[v\hspace{0.02cm},u]]$. Since $\exp(v)$ is an isometry, the following expression of the Riemannian volume is immediate

\begin{equation} \label{eq:ssvol}
\mathrm{Exp}^*_o(\mathrm{vol}) = \left|\det(\mathrm{sh}(R_v))\right|\hspace{0.02cm}dv
\end{equation}

where $dv$ denotes the volume form on $T_oM$, associated with the restriction of $Q$ to $\mathfrak{p}$.

Expression (<ref>) yields applicable integral formulae, when $\mathfrak{g}$ is a reductive Lie algebra ($\mathfrak{g} = \mathfrak{z} + \mathfrak{g}_{ss}$ : $\mathfrak{z}$ the centre of $\mathfrak{g}$ and $\mathfrak{g}_{ss}$ semisimple). If $\mathfrak{a}$ is a maximal Abelian subspace of $\mathfrak{p}$[Recall that the dimension of $\mathfrak{a}$ is known as the rank of $M$. In fact, $\mathrm{Exp}_o(\mathfrak{a})$ is a totally flat submanifold of $M$,of maximal dimension, and the only such submanifold, up to isometry.], any $v \in \mathfrak{p}$ is of the form $v = \mathrm{Ad}(k)\,a$ for some $k \in K$ and $a \in\mathfrak{a}$ (see [10], Lemma 6.3, Chapter V).Moreover, using the fact that $\mathrm{Ad}(k)$ is an isomorphism of $\mathfrak{g}$,

\begin{equation} \label{eq:raeigen}
\mathrm{Ad}(k^{\scriptscriptstyle -1})\circ R_v \circ \mathrm{Ad}(k) = R_a = \sum_{\lambda \in \Delta_+} (\lambda(a))^2\;\Pi_{\lambda}
\end{equation}

where each $\lambda \in \Delta_+$ is a linear form $\lambda : \mathfrak{a} \rightarrow \mathbb{R}$, and $\Pi_{\lambda}$ is the orthogonal projectors onto the corresponding eigenspace of $R_{a\,}$. Here, $\Delta_+$ is the set of positive roots of $\mathfrak{g}$ with respect to $\mathfrak{a}$ [10] (see Lemma 2.9, Chapter VII).

It is possible to use the diagonalisation (<ref>), in order to evaluate the determinant (<ref>). To obtain a regular parameterisation, let $S = K/K_{\mathfrak{a\,}}$, where $K_{\mathfrak{a}}$ is the centraliser of $\mathfrak{a}$ in $K$. Then, let $\varphi : S \times \mathfrak{a} \rightarrow M$ be given by $\varphi(s\hspace{0.02cm},a) = \mathrm{Exp}_o(\beta(s\hspace{0.02cm},a))$ where $\beta(s,a) = \mathrm{Ad}(s)\,a$. Now, by (<ref>) and (<ref>),

\varphi^*(\mathrm{vol}) = \prod_{\lambda \in \Delta_+} \left| \frac{\sinh\hspace{0.02cm}\lambda(a)}{\lambda(a)}\right|^{m_\lambda}\hspace{0.03cm}\beta^*(dv)

where $m_\lambda$ is the multiplicity of $\lambda$ (the rank of $\Pi_\lambda$). On the other hand, one may show that

\begin{equation} \label{eq:beta*}
\beta^*(dv) = \prod_{\lambda \in \Delta_+} |\lambda(a)|^{ m_\lambda}\hspace{0.05cm}da\,\omega(ds)
\end{equation}

where $da$ is the volume form on $\mathfrak{a}$, and $\omega$ is the invariant volume induced onto $S$ from $K$.

Finally, the Riemannian volume, in terms of the parameterisation $\varphi$, takes on the form

\begin{equation} \label{eq:ssvolka}
\varphi^*(\mathrm{vol}) = \prod_{\lambda \in \Delta_+} \left| \sinh\hspace{0.02cm}\lambda(a)\right|^{ m_\lambda}\hspace{0.03cm}da\,\omega(ds)
\end{equation}

Using (<ref>), it will be possible to write down integral formulae for Riemannian symmetric spaces, either non-compact or compact.

§.§.§ The non-compact case

This is the case were $\mathfrak{g}$ admits an $\mathrm{Ad}(G)$-invariant, non-degenerate, symmetric bilinear form $B$, such that $Q(u\hspace{0.02cm},z) = - B(u,d\hspace{0.02cm}\theta(z))$ is an $\mathrm{Ad}(K)$-invariant scalar product on $\mathfrak{g}$. In this case, $B$ is
negative-definite on $\mathfrak{k}$ and positive-definite on $\mathfrak{p}$. Moreover, $\mathrm{ad}(z) = [z,\cdot]$ is skew-symmmetric or symmetric (with respect to $Q$), according to whether $z \in \mathfrak{k}$ or $z \in \mathfrak{p}$.

If $u_{\scriptscriptstyle 1\hspace{0.02cm}},u_{\scriptscriptstyle 2} \in \mathfrak{p}$ are orthonormal, the sectional curvature of $\mathrm{Span}(u_{\scriptscriptstyle 1\hspace{0.02cm}},u_{\scriptscriptstyle 2})$ is found from (<ref>), $\kappa(u_{\scriptscriptstyle 1\hspace{0.02cm}},u_{\scriptscriptstyle 2}) = - \Vert [u_{\scriptscriptstyle 1\hspace{0.02cm}},u_{\scriptscriptstyle 2}] \Vert^2_o \leq 0$. Therefore, $M$ has  non-positive sectional curvature.

In fact, $M$ is a Hadamard manifold. It is geodesically complete by (<ref>). It is moreover simply connected, because $\mathrm{Exp}_o:\mathfrak{p} \rightarrow M$ is a diffeomorphism [10] (Theorem 1.1, Chapter VI). Thus, (<ref>) yields a first integral formula,

\begin{equation} \label{eq:ssvolnc}
\int_M\,f(x)\,\mathrm{vol}(dx) = \int_{\mathfrak{p}}\, f(\mathrm{Exp}_o(v))\hspace{0.02cm}\left|\det(\mathrm{sh}(R_v))\right|\hspace{0.02cm}dv
\end{equation}

To obtain an integral formula from (<ref>), one should first note that $\beta : S \times \mathfrak{a} \rightarrow \mathfrak{p}$ is not regular, nor one-to-one. Recall the following :

$\bullet$ the hyperplanes $\lambda(a) = 0$, where $\lambda \in \Delta_+\,$, divide $\mathfrak{a}$ into finitely many connected components, which are open and convex sets, known as Weyl chambers. From (<ref>), $\beta$ is regular on each Weyl chamber.

$\bullet$ let $K^\prime_{\mathfrak{a}}$ denote the normaliser of $\mathfrak{a}$ in $K$. Then, $W = \left.K^\prime_{\mathfrak{a}}\middle/K_{\mathfrak{a}}\right.$ is a finite group of automorphisms of $\mathfrak{a}$, called the Weyl group, which acts freely transitively on the set of Weyl chambers [10] (Theorem 2.12, Chapter VII).

Then, for each Weyl chamber $C$, $\beta$ is regular and one-to-one, from $S\times C$ onto its image in $\mathfrak{p}$. Moreover, if $\mathfrak{a}_r$ is the union of the Weyl chambers ($a \in \mathfrak{a}_r$ if and only if $\lambda(a) \neq 0$ for any $\lambda \in \Delta_+$), then $\beta$ is regular and $|W|$-to-one from $S \times \mathfrak{a}_r$ onto its image in $\mathfrak{p}$.

To obtain the desired integral formula, it only remains to note that $\varphi$ is a diffeomorphism from $S\times C$ onto its image in $M$. However, this image is the set $M_r$ of regular values of $\varphi$. By Sard's lemma, its complement is negligible [16].

Let $M = G/K$ be a Riemannian symmetric space, which belongs to the “non-compact case", just described. Then, for any bounded continuous function $f:M\rightarrow \mathbb{R}$,

\begin{equation} \label{eq:ssvolncka1}
\int_M\,f(x)\,\mathrm{vol}(dx) \,=\,
\int_{C_+}\int_{S}f(\varphi(s,a))\hspace{0.03cm}\prod_{\lambda \in \Delta_+}\left( \sinh\hspace{0.02cm}\lambda(a)\right)^{ m_\lambda}\hspace{0.03cm}da\,\omega(ds)
\end{equation}

\begin{equation} \label{eq:ssvolncka2}
\phantom{\int_M\,f(x)\,\mathrm{vol}(dx)} \,=\,
\frac{1}{|W|}\,\int_{\mathfrak{a}}\int_{S}\,f(\varphi(s,a))\hspace{0.03cm}\prod_{\lambda \in \Delta_+}\left| \sinh\hspace{0.02cm}\lambda(a)\right|^{ m_\lambda}\hspace{0.03cm}da\,\omega(ds)
\end{equation}

Here, $C_+$ is the Weyl chamber $C_+ = \lbrace a \in \mathfrak{a}\,: \lambda \in \Delta_+ \Rightarrow \lambda(a) > 0\rbrace$.

Example 1 : consider $M = \mathrm{H}(N)$ the space of $N \times N$ Hermitian positive-definite matrices. Here, $G = \mathrm{GL}(N,\mathbb{C})$ and $K = U(N)$, the groups of $N \times N$, complex, invertible and unitary matrices. Moreover, $B(u,\!z) = \mathrm{Re}(\mathrm{tr}(uz))$ and $d\hspace{0.02cm}\theta(z) = -z^\dagger$. Thus, $\mathfrak{p}$ is the space of $N \times N$ Hermitian matrices, and one may choose $\mathfrak{a}$ the space of $N \times N$ real diagonal matrices. The positive roots are the linear maps $\lambda(a) = a_{ii} - a_{jj}$ where $i < j$, and each one has its multiplicity equal to $2$. Thus, $C_+$ is the cone of real diagonal matrices $a$ with $a_{\scriptscriptstyle 11} > \ldots > a_{\scriptscriptstyle NN} > 0$. The Weyl group $W$ is the group of permutation matrices in $U(N)$ (so $|W| = N!$). Finally, $S= U(N)/T_{\scriptscriptstyle N} \equiv S_{\scriptscriptstyle N\,}$, where $T_{\scriptscriptstyle N}$ is the torus of diagonal unitary matrices. By (<ref>),

\begin{equation} \label{eq:ssvolncka2hn}
\int_{\mathrm{H}(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{1}{N!}\,\int_{\mathfrak{a}}\int_{S_{\scriptscriptstyle N}}\,f\left(s\hspace{0.02cm}\exp(2a)\hspace{0.02cm}s^\dagger\right)\hspace{0.03cm}\prod_{i < j} \sinh^2(a_{ii} - a_{jj})\hspace{0.03cm}da\,\omega(ds)
\end{equation}

where $da = da_{\scriptscriptstyle 11}\ldots da_{\scriptscriptstyle NN\,}$. Example 2 : pursuing the previous example, assume $f$ is a class function : $f(k\cdot x) = f(x)$ for $k \in K$ and $x \in \mathrm{H}(N)$. That is, $f(x)$ depends only on the eigenvalues $x_i = e^{r_i}$ of $x$. By (<ref>),

\begin{equation} \label{eq:integralhn}
\int_{\mathrm{H}(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{\omega(S_{\scriptscriptstyle N})}{2^{\scriptscriptstyle N}N!}\, \int_{\mathbb{R}^{\scriptscriptstyle N}}\,f\left(\exp(r)\right)\hspace{0.03cm}\prod_{i < j}\sinh^2((r_i - r_j)/{2})\hspace{0.03cm}dr
\end{equation}

or, by introducing the eigenvalues $x_i$ as integration variables,

\begin{equation} \label{eq:integralhnvmonde}
\int_{\mathrm{H}(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{\omega(S_{\scriptscriptstyle N})}{2^{\scriptscriptstyle N^2}N!} \,\int_{\mathbb{R}^{\scriptscriptstyle N}_+}\,f\left(x_{\scriptscriptstyle 1\,},\ldots,x_{\scriptscriptstyle N}\right)\hspace{0.03cm}|V(x)|^2\hspace{0.03cm}\prod^N_{i=1} x^{\scriptscriptstyle -N}_i\hspace{0.02cm}dx_i
\end{equation}

where $V(x) = \prod_{i<j} (x_j - x_i)$ is the Vandermonde determinant. Integrals of this form are well-known in random matrix theory [17].

§.§.§ The compact case

In this case, $\mathfrak{g}$ admits an $\mathrm{Ad}(G)$-invariant scalar product $Q$. Therefore, $\mathrm{ad}(z)$ is skew-symmmetric, with respect to $Q$, for each $z \in \mathfrak{g}$. Using (<ref>), it follows that $M$ is compact, with non-negative sectional curvature.

In fact, the compact case may be obtained from the previous non-compact case by duality. Denote $\mathfrak{g}_{\scriptscriptstyle\, \mathbb{C}}$ the complexification of $\mathfrak{g}$, and let $\mathfrak{g}^* = \mathfrak{k} + \mathfrak{p}_*$ where $\mathfrak{p}_* = i\hspace{0.02cm}\mathfrak{p}$. Then, $\mathfrak{g}^*$ is a compact real form of $\mathfrak{g}_{\scriptscriptstyle\, \mathbb{C}}$ (that is, $\mathfrak{g}^*$ is a compact Lie algebra, and its complexification is equal to $\mathfrak{g}_{\scriptscriptstyle\, \mathbb{C}}$). Denote $G^*$ the connected Lie group with Lie algebra $\mathfrak{g}^*$.

If $M = G/K$ is a Riemannian symmetric space which belongs to the non-compact case, then $M^* = G^*\!/K$ is a Riemannian symmetric space which belongs to the compact case. Formally, to pass from the non-compact case to the compact case, all one has to do is replace $a$ by $i\hspace{0.02cm}a$. Applying this recipe to (<ref>), one obtains

\begin{equation} \label{eq:ssvolkac}
\varphi^*(\mathrm{vol}) = \prod_{\lambda \in \Delta_+} \left| \sin\hspace{0.02cm}\lambda(a)\right|^{ m_\lambda}\hspace{0.03cm}da\,\omega(ds)
\end{equation}

where $da$ is the volume form on $\mathfrak{a}_* = i\hspace{0.02cm}\mathfrak{a}$, and $\omega$ is the invariant volume induced onto $S$ from $K$.
Note that the image under $\mathrm{Exp}_o$ of $\mathfrak{a}_*$ is the torus $T_* = \mathfrak{a}_*/\mathfrak{a}_{\scriptscriptstyle K\,}$,  where $\mathfrak{a}_{\scriptscriptstyle K}$ is the lattice given by $\mathfrak{a}_{\scriptscriptstyle K} = \lbrace a \in \mathfrak{a}_*: \mathrm{Exp}_o(a) = o \rbrace$. Recall the following :

$\bullet$ $\varphi(s,a)$ only depends on $t = \mathrm{Exp}_o(a)$. Thus, $\varphi$ may be considered as a map from $S \times T_*$ to $M$.

$\bullet$ if $a \in \mathfrak{a}_{\scriptscriptstyle K}$ then $\exp(2a) = e$ (the identity element in $G^*$). Thus, $\lambda(a) \in i\hspace{0.02cm}\pi\,\mathbb{Z}$ for all $\lambda \in \Delta_+$ [10] (Page 383).  Therefore, there exists a function $D:T \rightarrow \mathbb{R}$, such that

D(t) = \prod_{\lambda \in \Delta_+} \left| \sin\hspace{0.02cm}\lambda(a)\right|^{ m_\lambda} \hspace{0.5cm}
\text{whenever $t = \mathrm{Exp}_o(a)$}

Now, $T_*$ is a totally flat submanifold of $M$. Therefore, $\mathrm{Exp}^*(dt) = da$, where $dt$ denotes the invariant volume induced onto $T_*$ from $M$. With a slight abuse of notation, (<ref>) now reads,

\begin{equation} \label{eq:ssvolkac1}
\varphi^*(\mathrm{vol}) = D(t)\hspace{0.03cm}dt\,\omega(ds)
\end{equation}

Denote $(T_*)_r$ the set of $t \in T_*$ such that $D(t) \neq 0$. By the same arguments as in the non-compact case, $\varphi$ is a regular $|W|$-to-one map from $S \times (T_*)_r$ onto $M_r\,$, the set of regular values of $\varphi$.

Let $M = G^*\!/K$ be a Riemannian symmetric space, which belongs to the “compact case", just described. Then, for any bounded continuous function $f:M\rightarrow \mathbb{R}$,

\begin{equation} \label{eq:ssvolkac2}
\int_M\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{1}{|W|}\,\int_{T_*}\int_{S}\,f(\varphi(t,a))\hspace{0.03cm}D(t)\hspace{0.03cm}dt\,\omega(ds)
\end{equation}

Example 1 : the dual of $\mathrm{H}(N)$ is the unitary group $U(N)$. Here, $G^* = U(N)\times U(N)$ and $K \simeq U(N)$, is the diagonal group $K = \lbrace (x\hspace{0.02cm},x)\,;x \in U(N)\rbrace$. The Riemannian metric is given by the trace scalar product $Q(u,\!z) = -\mathrm{tr}(uz)$. Moreover, $T_* = T_{\scriptscriptstyle N}$ and $S= S_{\scriptscriptstyle N}$ (this is $U(N)/T_{\scriptscriptstyle N}$). The positive roots are $\lambda(ia) = a_{ii} - a_{jj}$ where $i < j$ and where $a$ is $N \times N$, real and diagonal[Please do no confuse the imaginary number $i$ with the subscript $i$.]. By writing the integral over $T_{\scriptscriptstyle N}$ as a multiple integral, (<ref>) reads,

\begin{equation} \label{eq:ssvolkacun}
\int_{U(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{1}{N!}\,\int_{[0\hspace{0.02cm},2\pi]^N}\int_{S_{\scriptscriptstyle N}}\,f\left(s\hspace{0.02cm}\exp(2ia)\hspace{0.02cm}s^\dagger\right)\hspace{0.03cm}\prod_{i < j} \sin^2(a_{ii} - a_{jj})\hspace{0.03cm}\omega(ds)\,da
\end{equation}

where $da = da_{\scriptscriptstyle 11}\ldots da_{\scriptscriptstyle NN\,}$.

Example 2 : assume $f$ is a class function.  That is, $f(x)$ depends only on eigenvalues $e^{i\theta_i}$ of $x$. Integrating out $s$, from (<ref>), it follows,

\begin{equation} \label{eq:integralun}
\int_{U(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{\omega(S_{\scriptscriptstyle N})}{2^{\scriptscriptstyle N}N!}\, \int_{[0\hspace{0.02cm},2\pi]^N}\,f\left(\exp(i\theta)\right)\hspace{0.03cm}\prod_{i < j}\sin^2((\theta_i - \theta_j)/{2})\hspace{0.03cm}d\theta
\end{equation}

or, after an elementary manipulation,

\begin{equation} \label{eq:integralunvmonde}
\int_{U(N)}\,f(x)\,\mathrm{vol}(dx) \,=\,
\frac{\omega(S_{\scriptscriptstyle N})}{2^{\scriptscriptstyle N^2}N!} \,\int_{[0\hspace{0.02cm},2\pi]^N}\,f\left(\theta_{\scriptscriptstyle 1\,},\ldots,\theta_{\scriptscriptstyle N}\right)\hspace{0.03cm}|V(e^{i\theta})|^2\hspace{0.03cm}d\theta_{\scriptscriptstyle 1}\ldots\theta_{\scriptscriptstyle N}
\end{equation}

where $V(e^{i\theta}) = \prod_{i<j} (e^{i\theta_j} - e^{i\theta_i})$ is the Vandermonde determinant. Integrals of this form are well-known in the random matrix theory of compact groups [18].

§ GEODESICS IN SYMMETRIC SPACES

Let $M = G/K$ be a Riemannian symmetric space, and assume $G$ has reductive Lie algebra $\mathfrak{g}$.Let the metric of $M$ be given by an $\mathrm{Ad}(K)$-invariant scalar product $Q$ on $\mathfrak{g}$, according to (<ref>).
Recall the Cartan decomposition, $\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$. Assume $\mathfrak{k}$ and $\mathfrak{p}$ are orthogonal, with respect to $Q$, and extend $Q$ to a left-invariant Riemannian metric $(\cdot,\cdot)$ on $G$.

Then, consider the natural projection $\pi:G \rightarrow M$,  which is given by $\pi(g) = g\cdot o$ for $g \in G$. Denote by $V_g$ the kernel of $d\pi(g)$, and by $H_g$ its orthogonal complement with respect to $(\cdot,\cdot)_{\scriptscriptstyle g\,}$.  Since $(\cdot,\cdot)$ is left-invariant, and $\mathfrak{k}$ and $\mathfrak{p}$ are orthogonal,

\begin{equation} \label{eq:kptovh}
V_g = dL_g(\mathfrak{k}) \hspace{1cm}   H_g = dL_g(\mathfrak{p})
\end{equation}

where $L_g$ denotes left translation by $g$ (below, $R_g$ will denote right translation).

For $v \in T_xM$, there exists a unique $v^{\scriptscriptstyle H}(g) \in H_g$ such that $d\pi\left( v^{\scriptscriptstyle H}(g)\right) = v$. By an abuse of notation, denote $v^{\scriptscriptstyle H}(e) = dL_{\scriptscriptstyle g^{-1}}\left( v^{\scriptscriptstyle H}(g)\right)$. Recall that for any $u\hspace{0.03cm},v \in T_xM$ (see [19], Chapter X),

\begin{equation} \label{eq:vhtometric}
\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} \,=\, \left(u^{\scriptscriptstyle H}(g)\hspace{0.02cm},v^{\scriptscriptstyle H}(g)\right)_{\scriptscriptstyle g} \,=\, Q\left(u^{\scriptscriptstyle H}(e)\hspace{0.02cm},v^{\scriptscriptstyle H}(e)\right)
\end{equation}

For Propositions <ref> and <ref>, consider the “infinitesimal action",

\begin{equation} \label{eq:infaction}
\xi\cdot x = \left.\frac{d}{dt}\right|_{t=0} \exp(t\hspace{0.02cm}\xi)\cdot x \hspace{1cm} \xi \in \mathfrak{g}\text{ and } x \in M
\end{equation}

In addition, let $\mathrm{B}$ be the bilinear form on $\mathfrak{g}$, given by $\mathrm{B} = B$, if $M$ belongs to the non-compact case, and by $\mathrm{B} = Q$, if $M$ belongs to the compact case (these two cases were described in <ref>).

Let $M = G/K$ be a symmetric space, of the “non-compact case", or of the “compact case".  For $g \in G$, $x = \pi(g)$ and $v \in T_xM$, let $\omega_{\scriptscriptstyle v} = \mathrm{Ad}(g)\hspace{0.02cm}v^{\scriptscriptstyle H}(e)$. Then, $\omega_{\scriptscriptstyle v}\cdot x = v$ and $\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} = \mathrm{B}(\xi\hspace{0.02cm},\omega_{\scriptscriptstyle v})$ whenever $u = \xi \cdot x$.

In the notation of the previous proposition,

\begin{equation} \label{eq:geodesiclift}
\mathrm{Exp}_x(v) = \exp(\omega_{\scriptscriptstyle v})\cdot x
\end{equation}

for $x \in M$ and $v \in T_xM$.

Propositions <ref> and <ref> offer a straightforward computational route to the Riemannian exponential map $\mathrm{Exp}$. To compute $\mathrm{Exp}_x(v)$, one begins by “lifting" $v$ from $T_xM$ to $\mathfrak{g}$, under the form of $\omega_{\scriptscriptstyle v\,}$. Then, it is enough to compute the action of $\exp(\omega_{\scriptscriptstyle v})$, which is just a matrix exponential, in practice.

Example 1 : consider an example of the non-compact case, $M = \mathrm{H}(N)$, the space of $N \times N$ Hermitian positive-definite matrices. Here, $G = \mathrm{GL}(N,\mathbb{C})$ and $\pi(g) = gg^\dagger$ for $g \in G$. Then,

d\pi(g)\cdot h = h\hspace{0.02cm}g^\dagger + g\hspace{0.02cm}h^\dagger

for $h \in T_gG$. For $x = \pi(g)$ and $v \in T_xM$, it follows that $v^{\scriptscriptstyle H}(g) = \left.v\hspace{0.03cm}\theta(g)\middle/2\right.$, where $\theta(g) = (g^\dagger)^{-1}$. By definition, $\omega_{\scriptscriptstyle v} = dR_{\scriptscriptstyle g^{-1}}(v^{\scriptscriptstyle H}(g))$. Since $gg^\dagger = x$, this gives $\omega_{\scriptscriptstyle v} = (v/2)\hspace{0.03cm}x^{-1}$. Therefore, using the fact that $g\cdot x = g\hspace{0.02cm}x\hspace{0.02cm}g^\dagger$, it follows

\mathrm{Exp}_x(v) = \exp\left(v\hspace{0.03cm}x^{-1}\!\middle/2\right) \,x \,\exp\left(x^{-1}\hspace{0.03cm}v\middle/2\right)

Accordingly, by an elementary property of the matrix exponential[Matrix functions (powers, logarithms, etc.) of Hermitian arguments should be understood as Hermitian matrix functions, obtained using the spectral decomposition — see [20].],

\begin{equation} \label{eq:pennec}
\mathrm{Exp}_x(v) = x^{\frac{1}{2}}\exp\left(x^{-\frac{1}{2}}\hspace{0.04cm}v\hspace{0.04cm}x^{-\frac{1}{2}}\right)x^{\frac{1}{2}}
\end{equation}

which is the formula made popular by [21].

Example 2 : let $M=G/K$ be a Riemannian symmetric space of the compact case. That is, the scalar product $Q$ on $\mathfrak{g}$ is $\mathrm{Ad}(G)$-invariant. Write $\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$ the Cartan decomposition of $\mathfrak{g}$. For $x \in M$, denote $K_x$ the stabiliser of $x$ in $G$. If $x = \pi(g)$, this has Lie algebra $\mathfrak{k}_x =\mathrm{Ad}(g)(\mathfrak{k})$ (that is, the image under $\mathrm{Ad}(g)$ of $\mathfrak{k}$).  For $v \in T_xM$, by Proposition <ref>, its “lift" $\omega_{\scriptscriptstyle v}$ should verify (note that, for the present example, $\mathrm{B} = Q$)

\omega_{\scriptscriptstyle v}\cdot x = v \hspace{0.3cm}\text{and}\hspace{0.3cm} Q(\xi\hspace{0.02cm},\omega_{\scriptscriptstyle v}) = 0 \text{ for } \xi \in
\mathfrak{k}_x

where the second identity is because $\xi \cdot x = 0$ for $\xi \in \mathfrak{k}_x\,$. Because $Q$ is $\mathrm{Ad}(G)$-invariant, this second identity is equivalent to
$Q(\kappa\hspace{0.02cm},\mathrm{Ad}(g^{\scriptscriptstyle -1})(\omega_{\scriptscriptstyle v})) = 0$ for $\kappa \in \mathfrak{k}$. That is, $\omega_{\scriptscriptstyle v} = \mathrm{Ad}(g)(\omega_{\scriptscriptstyle v}(o))$ for some $\omega_{\scriptscriptstyle v}(o) \in \mathfrak{p}$. This $\omega_{\scriptscriptstyle v}(o)$ is determined from $\omega_{\scriptscriptstyle v}\cdot x = v$, which yields $\omega_{\scriptscriptstyle v}(o) \cdot o = g^{\scriptscriptstyle -1}\cdot v$. However, the map $\omega \mapsto \omega \cdot o$ is an isomorphism from $\mathfrak{p}$ onto $T_oM$. Denoting its inverse by $\pi_o : T_oM \rightarrow \mathfrak{p}$, it follows that $\omega_{\scriptscriptstyle v}(o) = \pi_o(g^{\scriptscriptstyle -1}\cdot v)$. Finally,

\begin{equation} \label{eq:compactlift}
\omega_{\scriptscriptstyle v} = \mathrm{Ad}(g)\left( \pi_o(g^{\scriptscriptstyle -1}\cdot v)\right)
\end{equation}

A special case of this formula was used in (<ref>) of <ref>.

Proof of Proposition <ref> : to begin, one must prove

\begin{equation} \label{eq:proofgeolemma1}
\omega_{\scriptscriptstyle v}\cdot x = v
\end{equation}

From the definition of $\omega_{\scriptscriptstyle v}$ and $v^{\scriptscriptstyle H}(e)$, it is clear $\omega_{\scriptscriptstyle v} = dR_{g^{\scriptscriptstyle -1}}\hspace{0.02cm}(v^{\scriptscriptstyle H}(g))$. Replacing this into (<ref>), the left-hand side of (<ref>) becomes,

\omega_{\scriptscriptstyle v}\cdot x = \left.\frac{d}{dt}\right|_{t=0}\exp(t\,dR_{g^{\scriptscriptstyle -1}}\hspace{0.03cm}v^{\scriptscriptstyle H}(g))\cdot x = \left.\frac{d}{dt}\right|_{t=0} \left(\gamma(t)\hspace{0.04cm}g^{\scriptscriptstyle-1}\right)\cdot x

where $\gamma$ is any curve in $G$, through $g$ with $\dot{\gamma}(0) = v^{\scriptscriptstyle H}(g)$. Therefore,

\omega_{\scriptscriptstyle v}\cdot x = \left.\frac{d}{dt}\right|_{t=0} \gamma(t)\cdot o = d\pi\left( v^{\scriptscriptstyle H}(g)\right) = v

from the definition of $v^{\scriptscriptstyle H}(g)$. This proves (<ref>). It remains to show,

\begin{equation} \label{eq:proofgeolemma2}
\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} = Q(\xi\hspace{0.02cm},\omega_{\scriptscriptstyle v}) \hspace{1cm}
\text{for } u = \xi \cdot x
\end{equation}

The proof is separated into two cases.

non-compact case : in this case, $Q(\xi\hspace{0.02cm},\omega) = - B(\xi,d\hspace{0.02cm}\theta(\omega))$, where $B$ is an $\mathrm{Ad}(G)$-invariant, non-degenerate, symmetric bilinear form. To prove (<ref>), note that

d\pi(g)\left( dR_{\scriptscriptstyle g}(\xi)\right) = \left.\frac{d}{dt}\right|_{t=0} \left(\exp(t\xi)\,g\right)\cdot o = \left.\frac{d}{dt}\right|_{t=0} \exp(t\xi) \cdot x = u

Therefore, $dR_{\scriptscriptstyle g}(\xi) = u^{\scriptscriptstyle H}(g) + w$ where $w \in V_g$. From (<ref>), using left-invariance of $(\cdot,\cdot)$,

\begin{equation} \label{eq:proofgeolemma21}
\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} \,=\, \left(dR_{\scriptscriptstyle g}(\xi)\hspace{0.02cm},v^{\scriptscriptstyle H}(g)\right)_{\scriptscriptstyle g} \,=\, Q\left(\mathrm{Ad}(g^{\scriptscriptstyle -1})(\xi)\hspace{0.02cm},v^{\scriptscriptstyle H}(e)\right)
\end{equation}

Thus, using the definition of $Q$, and the fact that $v^{\scriptscriptstyle H}(e) \in \mathfrak{p}$,

\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} = - B(\mathrm{Ad}(g^{\scriptscriptstyle -1})(\xi),d\hspace{0.02cm}\theta(v^{\scriptscriptstyle H}(e))) =
B(\mathrm{Ad}(g^{\scriptscriptstyle -1})(\xi),v^{\scriptscriptstyle H}(e))

Finally, since $B$ is $\mathrm{Ad}(G)$-invariant,

\langle u\hspace{0.02cm},v\rangle_{\scriptscriptstyle x} = B(\mathrm{Ad}(g^{\scriptscriptstyle -1})(\xi),v^{\scriptscriptstyle H}(e)) =
B(\xi,\mathrm{Ad}(g)(v^{\scriptscriptstyle H}(e)))

which is the same as (<ref>), by the definition of  $\omega_{\scriptscriptstyle v\,}$. Indeed, in the present case, $\mathrm{B} = B$.

compact case : this follows from (<ref>), using the fact that $Q$ is $\mathrm{Ad}(G)$-invariant. Indeed, in the present case, $\mathrm{B} = Q$.

Proof of Proposition <ref> : for $\xi \in \mathfrak{g}$, introduce the corresponding vector fields $X_\xi$ on $M$, given by $
X_\xi(x) = \xi\cdot x$. Since this is a Killing vector field [19], if $c:\mathbb{R} \rightarrow M$ is a geodesic curve in $M$, then $\ell(\xi) = \langle X_\xi\hspace{0.03cm},\dot{c}\rangle_{\scriptscriptstyle c(t)}$ is a constant, (a law of conservation, really due to Noether's theorem!). Now, in the notation of Proposition <ref>, let $\omega(t) = \omega_{\scriptscriptstyle \dot{c}(t)\,}$. By Proposition <ref>,

\mathrm{B}(\omega(t),\xi) = \ell(\xi)

Since this is a constant, and since $\mathrm{B}$ is non-degenerate, it follows that $\omega(t) = \omega$ is a constant. Proposition <ref> also implies that $c$ satisfies the ordinary differential equation

\dot{c} = \omega \cdot c

But this differential equation is also satisfied  by $c(t) = \exp(t\hspace{0.03cm}\omega)\cdot\hspace{0.02cm} c(0)$, as one may see from (<ref>).By uniqueness of the solution, for given initial conditions,

c(t) = \exp(t\hspace{0.03cm}\omega_{\scriptscriptstyle \dot{c}(0)})\cdot c(0)

This immediately implies (<ref>), by setting $t = 1$, $c(0) = x$ and $\dot{c}(0) = v$.

CHAPTER: THE BARYCENTRE PROBLEM

State-of-the art results establish the existence and uniqueness of the Riemannian barycentre of a probability distribution which is supported inside a compact convex geodesic ball. What happens for a probability distribution which is not supported, but concentrated, inside a convex geodesic ball ?This question raises new difficulties that cannot be resolved by using the tools applicable to distributions which have compact convex support. The present chapter develops new tools, able to deal with these difficulties (at least in part), following the approach in [22].

* <ref> and <ref> review some of the major contributions to the study of Riemannian barycentres, due to Fréchet, Emery, Kendall, Afsari, and Arnaudon.

* <ref> introduces the main problem treated in the following, which is to study the existence and uniqueness of Riemannian barycentres of Gibbs distributions on compact Riemannian symmetric spaces.

* <ref> – <ref> lead up to the following conclusion : let $\pi_{\scriptscriptstyle T} \propto \exp(-U/T)$ be a Gibbs distribution on a simply connected compact Riemannian symmetric space $M$, such that the potential function $U$ has a unique global minimum at $x^* \in M$. If $M$ is simply connected, then for each $\delta < r_{\scriptscriptstyle cx}/2$ (where $r_{\scriptscriptstyle cx}$ is the convexity radius of $M$), there exists a critical temperature $T_{\scriptscriptstyle \delta}$ such that $T < T_{\scriptscriptstyle \delta}$ implies that $\pi_{\scriptscriptstyle T}$ has a unique Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ and this $\hat{x}_{\scriptscriptstyle T}$ belongs to the geodesic ball $B(x^*\!,\delta)$. The assumption that $M$ is simply connected cannot be removed (see Lemma <ref> and the following remark).

* <ref> provides expressions which can be used to analytically compute the critical temperature $T_{\scriptscriptstyle \delta\hspace{0.03cm}}$.

* <ref> introduces additional background material, on the geometry of compact symmetric spaces, which is required for the proofs of the results in <ref> – <ref>.

* <ref> details the proofs of these results, concerning the concentration, differentiability, convexity,existence and uniqueness of Riemannian barycentres of Gibbs distributions on compact symmetric spaces.

§ FRÉCHET'S FRUITFUL IDEA

In 1948, Maurice Fréchet proposed a generalisation of the concept of mean value, from Euclidean spaces to general metric spaces [23].
Today, this generalisation is known as the Fréchet mean. Precisely, a Fréchet mean, of a probability distribution $\pi$ on a metric space $M$, is any global minimum of the so-called variance function

\begin{equation} \label{eq:frechet}
\mathcal{E}_{\pi}(y) \,=\,
\frac{1}{2}\hspace{0.03cm}
\int_M\,d^{\hspace{0.03cm}\scriptscriptstyle 2}(y\hspace{0.02cm},x)\hspace{0.03cm}\pi(dx)
%ONE HALF
%mathcal{E} separately
\end{equation}

where $d(x\hspace{0.02cm},y)$ denotes the distance between $x$ and $y$ in $M$. In the following, the focus will be on the case where $M$ is a Riemannian manifold. Then, a Fréchet mean of $\pi$ will be called a Riemannian barycentre, or just a barycentre, of $\pi$.

If $\mathcal{E}_{\pi}(y)$ takes on finite values (in fact, if it is finite for just one $y = y_o$), then $\pi$ has at least one Fréchet mean. In particular, if $M$ is a Euclidean space, then this Fréchet mean is always unique, and equal to the mean value (expectation) of $\pi$. In general, the Fréchet mean of a probability distribution $\pi$ is not unique, and one may think of the Fréchet mean of $\pi$ as the set $F(\pi)$, of all global minima of its variance function $\mathcal{E}_{\pi\hspace{0.03cm}}$.

Example 1 : if $M = S^1$, the unit circle, and $\pi$ is the uniform distribution (i.e. Haar measure), on $S^1$, then $F(\pi) = S^1$. Any point on the circle is a barycentre of the uniform distribution.

If $x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N} \in M$, then an empirical Fréchet mean of $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$ is any Fréchet mean of the empirical distribution $(\delta_{x_{\scriptscriptstyle 1}}+\ldots+\delta_{x_{\scriptscriptstyle N}})/N$ ($\delta_x$ denotes the Dirac distribution concentrated at $x$). In other words, an empirical Fréchet mean of $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$ is any global minimum of the empirical variance function

\begin{equation} \label{eq:empiricalfrechet}
\mathcal{E}_{\scriptscriptstyle N}(y) \,=\,\frac{1}{2N}\sum^N_{n=1}d^{\hspace{0.03cm}2}(y\hspace{0.02cm},x_n)
%ONE HALF
%mathcal{E}_N separately
\end{equation}

When $M$ is a Riemannian manifold, the term “empirical Fréchet mean" will be replaced by the term “empirical barycentre".

Example 2 : if $M = S^1$, and $x_{\scriptscriptstyle 1\hspace{0.02cm}},x_{\scriptscriptstyle 2}$ are two opposite points on $S^1$, then the empirical barycentre of $(x_{\scriptscriptstyle 1\hspace{0.02cm}},x_{\scriptscriptstyle 2})$ is a two-point set. For example, if $x_{\scriptscriptstyle 1} = 1$ and $x_{\scriptscriptstyle 2} = -1$, then the empirical barycentre is the set $\lbrace i,-i\rbrace$ ($i$ being the square root of $-1$).

Two important problems, in relation to the concept of Fréchet mean, are establishing the uniqueness of the Fréchet mean of some probability distribution, and effectively computing this Fréchet mean, (or the set of Fréchet means, in case uniqueness does not hold).

Another type of problem is related to the large-sample theory of the Fréchet mean, and was treated in [24][25].  Assume $(x_n\,;n\geq 1)$ are independent samples from the distribution $\pi$. If $F_{\scriptscriptstyle N}$ is the set of empirical Fréchet means of $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$, then one is interested in using $F_{\scriptscriptstyle N}$ to somehow approximate $F(\pi)$.

In [24], it was shown that, if the metric space $M$ is such that any closed and bounded subset of $M$ is compact[That is, for any $x \in M$, the function $y \mapsto d(x\hspace{0.02cm},y)$ is a proper function, meaning it has compact sublevel sets.], then for any $\epsilon > 0$, the set $F_{\scriptscriptstyle N}$ almost-surely belongs to the $\epsilon$-neighborhood of the set $F(\pi)$, when $N$ is sufficiently large.

Moreover, if $\pi$ has a unique Fréchet mean, say $F(\pi) = \lbrace\hat{x}_{\pi}\rbrace$, then any sequence of empirical Fréchet means, $\bar{x}_{\scriptscriptstyle N} \in F_{\scriptscriptstyle N}$ converges almost-surely to $\hat{x}_{\pi}$ (an extension of this last result, from independent to Markovian samples, is obtained in <ref>).

In [25], a central limit theorem was added to this last convergence result. Specifically, if $M$ is a Riemannian manifold, the distribution of $N^{\scriptscriptstyle \frac{1}{2}}\hspace{0.03cm}\mathrm{Exp}_{\hat{x}_\pi}(\bar{x}_{\scriptscriptstyle N})$ converges to a multivariate normal distribution (in the tangent space at $\hat{x}_\pi$). This “central limit theorem" requires several technical conditions, in order to hold true, and should therefore only be applied after due verification.

§ EXISTENCE AND UNIQUENESS

The problem of the existence and uniqueness of Riemannian barycentres has generated a rich literature, with ramifications in stochastic analysis on manifolds, Riemannian geometry, and probability theory. The present section attempts a quick, non-exhaustive summary of some famous results from this literature.

§.§ Emery and Kendall

The works of Emery and Kendall [26], later expanded upon by Afsari [27], are related to the existence and uniqueness of the Riemannian barycentre of a probability distribution $\pi$, supported inside some geodesic ball $B(x^*\!,\delta)$, in a Riemannian manifold $M$.

Emery and Kendall, among others, considered the so-called Karcher mean of $\pi$. This is a local minimum of the variance function $\mathcal{E}_\pi$ in (<ref>). In [26], $\pi$ is assumed to have compact support, inside a so-called regular geodesic ball $B(x^*\!,\delta)$. Here, “regular geodesic ball" means

$\bullet$ $\delta < \frac{\pi}{2}c^{\scriptscriptstyle -1}$, where all sectional curvatures of $M$ are less than $\kappa_{\max} = c^{\hspace{0.02cm}\scriptscriptstyle 2}$.

$\bullet$ the cut locus of $x^*$ does not intersect $B(x^*\!,\delta)$ (that is $\delta < \mathrm{inj}(x^*)$).

These two conditions guarantee that the closed ball $\bar{B}(x^*\!,\delta)$ is weakly convex, and that it has convex geometry.

Weakly convex means for any $x\hspace{0.02cm},y \in  \bar{B}(x^*\!,\delta)$ there exists a unique geodesic $\gamma:[0,1]\rightarrow M$, such that $\gamma(0) = x$, $\gamma(1) = y$ and $\gamma(t) \in \bar{B}(x^*\!,\delta)$ for all $t \in [0,1]$ (this is equivalent to the terminology of [11][This geodesic $\gamma$ is the unique length-minimising curve, among all curves which connect $x$ to $y$ and lie in $\bar{B}(x^*\!,\delta)$. See the proof of Theorem IX.6.2, Page 405 in [11].]).

Convex geometry means there exists a positive, bounded, continuous, and convex function $\Psi$, defined on $\bar{B}(x^*\!,\delta) \times \bar{B}(x^*\!,\delta)$, such that $\Psi(x\hspace{0.02cm},y) = 0$ if and only if $x = y$.

When $\pi$ is supported inside $B(x^*\!,\delta)$, the function $\mathcal{E}_\pi$ takes on finite values, and therefore has a global minimum $\hat{x}_\pi\hspace{0.03cm}$. However, it is not immediately clear this $\hat{x}_\pi$ should lie within $B(x^*\!,\delta)$. In [26], the existence of a local minimum, i.e. Karcher mean, within $B(x^*\!,\delta)$ is guaranteed, subject to interpreting the distance in (<ref>) as geodesic distance within $B(x^*\!,\delta)$.

If $\hat{x}_\pi$ is a local minimum of $\mathcal{E}_\pi$ in $B(x^*\!,\delta)$, then the convex geometry property of the closed ball $\bar{B}(x^*\!,\delta)$ guarantees this local minimum is unique. This follows by using a general form of Jensen's inequality, due to Emery. Specifically, if $\hat{x}_{\scriptscriptstyle 1}$ and $\hat{x}_{\scriptscriptstyle 2}$ are Karcher means in $B(x^*\!,\delta)$, then $(\hat{x}_{\scriptscriptstyle 1\hspace{0.02cm}},\hat{x}_{\scriptscriptstyle 2})$ is a Karcher mean of the image distribution $\delta^*\pi$ of $\pi$, under the map $\delta(x) = (x,x)$. Then, applying Jensen's inequality to the convex function $\Psi$, it follows

\Psi(\hat{x}_{\scriptscriptstyle 1\hspace{0.02cm}},\hat{x}_{\scriptscriptstyle 2}) \leq \int_{\scriptscriptstyle \bar{B}(x^*\!,\delta)}\Psi(x,x)\hspace{0.03cm}\pi(dx)

so $\Psi(\hat{x}_{\scriptscriptstyle 1\hspace{0.02cm}},\hat{x}_{\scriptscriptstyle 2}) = 0$, and therefore $\hat{x}_{\scriptscriptstyle 1} = \hat{x}_{\scriptscriptstyle 2\hspace{0.03cm}}$.

Remark : it was conjectured by Emery that any weakly convex geodesic ball should also have convex geometry. A counterexample to this conjecture was provided by Kendall, in the form of his “propeller" [28].

§.§ Afsari's contribution

Afsari's seminal work on Riemannian barycentres was published ten years ago [27]. It provided the following statement : if $\pi$ is supported inside a geodesic ball $B(x^*\!,\delta)$, then $\pi$ has a unique Riemannian barycentre $\hat{x}_{\pi}$ and $\hat{x}_{\pi} \in B(x^*\!,\delta)$, as soon as

\begin{equation} \label{eq:afsari1}
\delta < \frac{1}{2}\hspace{0.02cm}\min\left\lbrace \pi c^{\scriptscriptstyle -1},\mathrm{inj}(M)\right\rbrace
\end{equation}

Here, $c$ is such that all sectional curvatures of $M$ are less than $\kappa_{\max} = c^{\hspace{0.02cm}\scriptscriptstyle 2}$ (if $M$ has negative sectional curvatures, $c^{\scriptscriptstyle -1}$ is understood to be $+\infty$), and $\mathrm{inj}(M)$ is the injectivity radius of $M$.

Condition (<ref>) ensures the geodesic ball $B(x^*\!,\delta)$ is convex, in the sense of <ref> (strongly convex, in the terminology of [11]), rather than just weakly convex as in <ref>. This stronger condition is required, because the Riemannian barycentre (Fréchet mean) is considered, rather than just the Karcher mean. In fact, Afsari extended his results beyond Riemannian barycentres to $L^{\scriptscriptstyle p}$ Riemannian barycentres, which are obtained by replacing the squared distance in (<ref>) with a distance elevated to the power $p$, where $p\geq 1$.

In [27], the following approach is used, for the proof of existence and uniqueness. First, it is shown that any global minimum of $\mathcal{E}_\pi$ must lie inside $B(x^*\!,\delta)$. This is done using the Alexandrov-Toponogov comparison theorem, under its form stated in [11] (Page 420). Then, the Poincaré-Hopf theorem is employed, in order to prove uniqueness of local minima, inside the geodesic ball $B(x^*\!,\delta)$.

Specifically, $\mathcal{E}_\pi$ is differentiable at any point $y$ which belongs to the closed ball $\bar{B}(x^*\!,\delta)$, and

\mathrm{grad}\,\mathcal{E}_\pi(y) \,=\, -\int_{M}\,\mathrm{Exp}^{-1}_y(x)\hspace{0.03cm}\pi(dx) \hspace{0.5cm} \text{for } y \in \bar{B}(x^*\!,\delta)

Then, it is shown that, if $y \in B(x^*\!,\delta)$ and $\mathrm{grad}\,\mathcal{E}_\pi(y) = 0$, then $\mathrm{Hess}\,\mathcal{E}_\pi(y)$ is positive-definite. In other words, the singular point $y$ of the gradient vector field $\mathrm{grad}\,\mathcal{E}_\pi$ has its index equal to $1$.Since this vector field is outward pointing on the boundary of $\bar{B}(x^*\!,\delta)$, the Poincaré-Hopf theorem implies the sum of the indices of all its singular points in $B(x^*\!,\delta)$ is equal to the Euler-Poincaré characteristic of $\bar{B}(x^*\!,\delta)$, which is equal to $1$ (since $\bar{B}(x^*\!,\delta)$ is homeomorphic to a closed ball in $\mathbb{R}^n$).

Remark : the argument just summarised not only shows that $\mathcal{E}_\pi$ has a unique local minimum in $B(x^*\!,\delta)$, but that it has a unique stationary point in $B(x^*\!,\delta)$. Moreover, the advantage of this argument, over the “convex geometry" uniqueness argument, (summarised in <ref>),is that it can be used to show the uniqueness of $L^{\scriptscriptstyle p}$ Riemannian barycentres, for general $p > 1$.

§.§ Hadamard manifolds

Existence and uniqueness of Riemannian barycentres hold under quite general conditions, when the underlying Riemannian manifold $M$ is a Hadamard manifold (recall definition from <ref>). Mostly, these existence and uniqueness properties are just special cases of the properties of Fréchet means in metric spaces of non-positive curvature, which were developed by Sturm [29].

Let $\pi$ be a probability distribution on a Hadamard manifold $M$. As already mentioned in <ref>, if the variance function $\mathcal{E}_\pi$ in (<ref>) takes on finite values, then $\pi$ has at least one Riemannian barycentre, say $\hat{x}_{\pi\hspace{0.03cm}}$. For this, it is enough that
$\mathcal{E}_\pi(y_o) < \infty$, for just one $y_o \in M$. In other words, it is enough that $\pi$ should have a finite second-order moment

\begin{equation} \label{eq:secondordermoment}
\int_M\,d^{\hspace{0.03cm} 2}(y_o\hspace{0.03cm},x)\,\pi(dx) \,<\,\infty
\end{equation}

Indeed, if (<ref>) is verified, then a straightforward application of the triangle inequality implies that $\mathcal{E}_\pi(y) < \infty$ for all $y \in M$.

When $M$ is a Hadamard manifold, existence of a Riemannian barycentre automatically implies its uniqueness. This can be shown using the “convex geometry" uniqueness argument, discussed in <ref>. Indeed, if $M$ is a Hadamard manifold, then $\Psi:M\times M \rightarrow \mathbb{R}$, where $\Psi(x\hspace{0.02cm},y) = d(x\hspace{0.02cm},y)$ is convex, and $\Psi(x\hspace{0.02cm},y) = 0$ if and only if $x = y$. Alternatively, uniqueness of the Riemannian barycentre follows from the strong convexity of the variance function $\mathcal{E}_{\pi\hspace{0.03cm}}$. Recall from <ref> that $f_x(y) = d^{\hspace{0.03cm} 2}(x\hspace{0.02cm},y)/2$ is a $1/2$-strongly convex function, for each $x \in M$. Then, (<ref>) says that $\mathcal{E}_\pi$ is an expectation of $1/2$-strongly convex functions, and is therefore $1/2$-strongly convex. In turn, this implies that $\mathcal{E}_\pi$ has a unique global minimum, $\hat{x}_\pi \in M$.

When $M$ is a Hadamard manifold, it should also be noted that $\mathcal{E}_\pi$ is smooth throughout $M$, and that its gradient is given by

\begin{equation} \label{eq:gradepsilonhadamard}
\mathrm{grad}\,\mathcal{E}_\pi(y) = -\int_M\,\mathrm{Exp}^{-1}_y(x)\hspace{0.03cm}\pi(dx)
\end{equation}

as can be found by applying (<ref>) under the integral in (<ref>). Strong convexity of $\mathcal{E}_\pi$ implies its global minimum $\hat{x}_\pi$ is also its unique stationary point in $M$ (i.e. the unique point where $\mathrm{grad}\,\mathcal{E}_\pi$ is equal to zero).

§.§ Generic uniqueness

The empirical barycentre of the points $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$, in any complete Riemannian manifold  $M$,is generically unique. This means that this empirical barycentre is unique, for almost all $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$ in the product Riemannian manifold $M^{\scriptscriptstyle N} = M \times \ldots \times M$, equipped with its Riemannian volume measure. This interesting result was obtained by Arnaudon and Miclo [30]. In particular, it implies that when $(x_{\scriptscriptstyle 1},\ldots,x_{\scriptscriptstyle N})$ are independent samples, from a distribution $\pi$, which has a probability density with respect to the Riemannian volume of $M$, then their empirical barycentre $\bar{x}_{\scriptscriptstyle N}$ is almost-surely unique.

§ GIBBS DISTRIBUTIONS : AN OPEN PROBLEM

Throughout the following, $M$ will be a compact, orientable Riemannian manifold, with positive sectional curvatures, all less than $\kappa_{\max} = c^{\hspace{0.02cm}\scriptscriptstyle 2}$. Afsari's statement, recalled in <ref>, says that if  $\pi$ is a probability distribution on $M$, supported inside a convex geodesic ball $B(x^*\!,\delta)$, then $\pi$ has a unique Riemannian barycentre $\hat{x}_\pi\hspace{0.03cm}$, as soon as

\begin{equation} \label{eq:afsaribis}
\delta < \frac{1}{2}\hspace{0.02cm}\min\left\lbrace \pi c^{\scriptscriptstyle -1},\mathrm{inj}(M)\right\rbrace
\end{equation}

where $\mathrm{inj}(M)$ denotes the injectivty radius of $M$.

Inequality (<ref>) is optimal. Indeed, it is easy to think of examples which show that, if it is replaced by an equality, then $\hat{x}_\pi$ will immediately fail to be unique. On the other hand, this inequality does not tell us what happens in the important case where $\pi = \pi_{\scriptscriptstyle T}$ is a Gibbs distribution,

\begin{equation} \label{eq:gibbs}
\pi_{\scriptscriptstyle T}(dx) = \left(Z(T)\right)^{\scriptscriptstyle -1}\exp\left[-\frac{U(x)}{T}\right]\hspace{0.03cm}\mathrm{vol}(dx) \\[0.12cm]
\end{equation}

for some temperature $T$, and potential function $U:M\rightarrow \mathbb{R}$, where $Z(T)$ is a normalising constant ($\mathrm{vol}$ denotes the Riemannian volume form).

The present chapter will introduce several results, which deal with this case. These are concerned with the concentration, differentiability, convexity, and uniqueness properties, of the Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ of the Gibbs distribution $\pi_{\scriptscriptstyle T}$.

The starting assumption for these results is that the potential function $U$ has a unique global minimum at $x^* \in M$. Under this assumption, while $\pi_{\scriptscriptstyle T}$ is not supported inside any convex geodesic ball $B(x^*\!,\delta)$, it is still concentrated on any such ball, provided the temperature $T$ is sufficiently small. Then, the aim is to know exactly how small $T$ should be made, in order to ensure the required properties of $\hat{x}_{\scriptscriptstyle T\hspace{0.02cm}}$. This aim can be fully achieved, under the further assumption that $M$ is a simply connected compact Riemannian symmetric space.

Given these two assumptions, the following conclusion will be obtained : for each $\delta < \frac{1}{2}r_{\scriptscriptstyle cx}$ ($r_{\scriptscriptstyle cx}$ denotes the convexity radius of $M$), there exists a critical temperature $T_{\scriptscriptstyle \delta}$ such that $T < T_{\scriptscriptstyle \delta}$ implies $\pi_{\scriptscriptstyle T}$ has a unique Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ and this $\hat{x}_{\scriptscriptstyle T}$ belongs to the geodesic ball $B(x^*\!,\delta)$. Moreover, if $U$ is invariant by geodesic symmetry about $x^*$, then $\hat{x}_{\scriptscriptstyle T} = x^*$.

Remark : if $M$ is a Riemannian manifold, the convexity radius $r_{\scriptscriptstyle cx}(x)$ of $x \in M$ is the supremum of $R > 0$ such that the geodesic ball $B(x\hspace{0.02cm},R)$ is convex (this is strictly positive, for any $x \in M$). The convexity radius $r_{\scriptscriptstyle cx}(M)$ of $M$ is the infimum of $r_{\scriptscriptstyle cx}(x)$, over all $x \in M$ (if $M$ is compact, this is strictly positive). Here, $r_{\scriptscriptstyle cx}(M)$ is just denoted $r_{\scriptscriptstyle cx\hspace{0.03cm}}$.

§ CONCENTRATION OF BARYCENTRES

Denote the variance function of the Gibbs distribution $\pi_{\scriptscriptstyle T}$ in (<ref>) by $\mathcal{E}_{\scriptscriptstyle T\hspace{0.03cm}}$. According to (<ref>),

\begin{equation} \label{eq:ETT}
\mathcal{E}_{\scriptscriptstyle T}(y) =
\frac{1}{2}\hspace{0.03cm}
\int_M\,d^{\hspace{0.03cm}\scriptscriptstyle 2}(y\hspace{0.02cm},x)\hspace{0.03cm}\pi_{\scriptscriptstyle T}(dx)
\end{equation}

Throughout the following, it will be assumed that the potential function $U$, which appears in (<ref>), has a unique global minimum at $x^* \in M$. While $U$ is not required to be smooth, it is required to be well-behaved near $x^*$, in the sense that there exist $\mu_{\min}\hspace{0.02cm},\mu_{\max} > 0$ and $\rho > 0$ such that

\begin{equation} \label{eq:wellbehaved}
\mu_{\min}\hspace{0.03cm}d^{\hspace{0.03cm}\scriptscriptstyle 2}(x\hspace{0.02cm},x^*) \,\leq\, 2(U(x) - U(x^*))\,\leq
\mu_{\max}\hspace{0.03cm}d^{\hspace{0.03cm}\scriptscriptstyle 2}(x\hspace{0.02cm},x^*)
\end{equation}

whenever $d(x\hspace{0.02cm},x^*) \leq \rho$. This is always verified if $U$ is twice differentiable at $x^*$, and the spectrum of $\mathrm{Hess}\,U(x^*)$ is contained in the open interval $(\mu_{\min}\hspace{0.02cm},\mu_{\max})$.

The following Proposition <ref> establishes the concentration property of the Riemannian barycentres of $\pi_{\scriptscriptstyle T}$ as the temperature $T$ is made small. In this proposition, $W$ denotes the Kantorovich ($L^{\scriptscriptstyle 1}$-Wasserstein) distance, and $\delta_{x^*}$ the Dirac distribution concentrated at $x^*$.

Let $M$ be a compact, orientable Riemannian manifold, with positive sectional curvatures, and dimension equal to $n$.

(i) Let $\eta > 0$. For any Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ of $\pi_{\scriptscriptstyle T}$

\begin{equation} \label{eq:concentration1}
W(\pi_{\scriptscriptstyle T}\hspace{0.02cm},\delta_{x^*}) < \frac{\eta^2}{4\hspace{0.02cm}\mathrm{diam}\,M} \;\Longrightarrow\; d(\hat{x}_{\scriptscriptstyle T}\hspace{0.02cm},x^*) < \eta
\end{equation}

where $\mathrm{diam}\,M$ is the diameter of $M$.

(ii) There exists a temperature $T_{\scriptscriptstyle W}$ such that $T \leq T_{\scriptscriptstyle W}$ implies

\begin{equation} \label{eq:concentration2}
W(\pi_{\scriptscriptstyle T}\hspace{0.02cm},\delta_{x^*}) \leq (8\pi)^{\!\frac{1}{2}}\hspace{0.02cm}B^{-1}_n\left(\frac{\pi}{2}\right)^{\!n-1}
\left(\frac{\mu_{\max}}{\mu_{\min}}\right)^{\!\!\frac{n}{2}}\left(\frac{T}{\mu_{\min}}\right)^{\!\!\frac{1}{2}}
\end{equation}

where $B_n = B(1/2\hspace{0.02cm},n/2)$ in terms of the Euler Beta function.

Proposition <ref> shows exactly how small $T$ should be made, in order to ensure that all the Riemannian barycentres $\hat{x}_{\scriptscriptstyle T}$ concentrate within an open ball $B(x^*\!,\eta)$. Roughly, (i) states that, if $\pi_{\scriptscriptstyle T}$ is close to $\delta_{x^*}\hspace{0.03cm}$, then all $\hat{x}_{\scriptscriptstyle T}$ will be close to $x^*$. On the other hand, (ii) bounds the distance between $\pi_{\scriptscriptstyle T}$ and $\delta_{x^*}\hspace{0.03cm}$, as a function of $T$. The temperature $T_{\scriptscriptstyle W}$ mentioned in (ii) will be expressed explicitly  in <ref>, below.

Here, two things should be noted, concerning (<ref>). First, this inequality is both optimal and explicit. It is optimal because the dependence on $T^{\frac{1}{2}}$ in its right-hand side cannot be improved. Indeed, the multi-dimensional Laplace approximation (for example, see [31]), shows the left-hand side is equivalent to $\mathrm{L}\cdot T^{\frac{1}{2}}$ when $T \rightarrow 0$. While this constant $\mathrm{L}$ is not tractable, the constants appearing in (<ref>) depend explicitly on the manifold $M$ and the function $U$. In fact, (<ref>) does not follow from the multi-dimensional Laplace approximation, but rather from the volume comparison theorems, in <ref>.

Second, in spite of these nice properties, (<ref>) does not escape the curse of dimensionality. Indeed, for fixed $T$, its right-hand side increases exponentially with the dimension $n$ of $M$ (note that $B_n$ decreases like $n^{\scriptscriptstyle -\frac{1}{2}}$). In fact, the temperature $T_{\scriptscriptstyle W}$ also depends on $n$, but it is typically much less affected by it, and decreases slower than $n^{\scriptscriptstyle -1}$ as $n$ increases.

§ DIFFERENTIABILITY OF THE VARIANCE FUNCTION

Assume that $M$ is a simply connected compact Riemannian symmetric space. Under this assumption, it turns out that the variance function $\mathcal{E}_{\scriptscriptstyle T}(y)$ is $C^2$ throughout $M$, for any value $T > 0$ of the temperature $T$. This surprising result is contained in Proposition <ref>.

To state  Proposition <ref>, consider for $x \in M$ the function $f_x(y) = d^{\hspace{0.03cm}2}(x,y)/2$. Recall from <ref> that this function is $C^2$ on the open set $\mathrm{D}(x) = M - \mathrm{Cut}(x)$. When $y \in \mathrm{D}(x)$, denote $G_y(x)$ and $H_y(x)$ the gradient and Hessian of $f_x(y)$.

With this notation, for any $x \in M$, the gradient $G_y(x)$ belongs to $T_yM$, and the Hessian $H_y(x)$ defines a symmetric bilinear form on $T_yM$. However (recall the remarks in <ref>), both $G_y(x)$ and $H_y(x)$ are singular on $\mathrm{Cut}(x)$, where $H_y(x)$ will even blow up, as it has an eigenvalue equal to $-\infty$.

Let $M$ be a simply connected compact Riemannian symmetric space.

(i) The following integrals converge for any temperature $T > 0$

\begin{equation} \label{eq:GH}
G_y = \int_{\mathrm{D}(y)}G_y(x)\hspace{0.03cm}\pi_{\scriptscriptstyle T}(dx) \hspace{0.2cm};\hspace{0.2cm}
H_y = \int_{\mathrm{D}(y)}H_y(x)\hspace{0.03cm}\pi_{\scriptscriptstyle T}(dx)
\end{equation}

and both depend continuously on $y$.

(ii)  The gradient and Hessian of the variance function $\mathcal{E}_{\scriptscriptstyle T}(y)$ are given by

\begin{equation} \label{eq:derivatives}
\mathrm{grad}\,\mathcal{E}_{\scriptscriptstyle T}(y) = G_y \hspace{0.2cm};\hspace{0.2cm}
\mathrm{Hess}\,\mathcal{E}_{\scriptscriptstyle T}(y) = H_y
\end{equation}

so that $\mathcal{E}_{\scriptscriptstyle T}(y)$ is $C^2$ throughout $M$.

The proof of Proposition <ref> relies on the following lemma.

Assume $M$ is a simply connected compact Riemannian symmetric space. Let $\gamma : I \rightarrow M$ be a geodesic defined on a compact interval $I$. Denote by $\mathrm{Cut}(\gamma)$ the union of all cut loci $\mathrm{Cut}(\gamma(t))$ for $t \in I$. Then, the Hausdorff dimension of $\mathrm{Cut}(\gamma)$ is strictly less than the dimension of $M$. In particular, $\mathrm{Cut}(\gamma)$ is a set with Riemannian volume equal to zero.

Remark : the assumption that $M$ is simply connected cannot be removed. For example, the conclusion of Lemma <ref> does not hold if $M$ is a real projective space.

The proof of Lemma <ref> uses the structure of Riemannian symmetric spaces, as well as some results from dimension theory, found in [32]. The notion of Hausdorff dimension is needed, because $\mathrm{Cut}(\gamma)$ may fail to be a manifold.

Lemma <ref> is crucial to Proposition <ref>, because it leads to the following expression,

\mathcal{E}_{\scriptscriptstyle T}(\gamma(t)) = \int_M\,f_x(\gamma(t))\hspace{0.03cm}\pi_{\scriptscriptstyle T}(dx) =
\int_{\mathrm{D}(\gamma)}f_x(\gamma(t))\hspace{0.03cm}\pi_{\scriptscriptstyle T}(dx) \hspace{0.5cm} \text{for all $t \in I$}

where $\mathrm{D}(\gamma) = M - \mathrm{Cut}(\gamma)$, and the second inequality follows since $\mathrm{Cut}(\gamma)$ has Riemannian volume equal to zero. Then, recalling that $x \in \mathrm{Cut}(\gamma(t))$ if and only if $\gamma(t) \in \mathrm{Cut}(x)$, it becomes possible to differentiate $f_x(\gamma(t))$ under the integral. This leads to the proof of (ii).

§ UNIQUENESS OF THE BARYCENTRE

The following Proposition <ref> establishes the uniqueness of $\hat{x}_{\scriptscriptstyle T}$ as the temperature $T$ is made small. As in the previous Proposition <ref>, $M$ is a simply connected compact Riemannian symmetric space. The convexity radius of $M$ is denoted $r_{\scriptscriptstyle cx\hspace{0.03cm}}$. This is given by $r_{\scriptscriptstyle cx} = \frac{\pi}{2}\hspace{0.03cm}c^{\scriptscriptstyle -1}$ (see <ref>, below).

Recall the definition (<ref>) of the Gibbs distribution $\pi_{\scriptscriptstyle T\hspace{0.03cm}}$, where the potential function $U$ has a unique global minimum at $x^* \in M$. Let $s_{x^*}$ denote the geodesic symmetry at $x^*$ (recall definition from <ref>). The potential function $U$ is said to be invariant by geodesic symmetry about $x^*$, if $U \circ s_{x^*} = U$.

Let $M$ be a simply connected compact Riemannian symmetric space, with convexity radius $r_{\scriptscriptstyle cx\hspace{0.03cm}}$. For $\delta < \frac{1}{2}r_{\scriptscriptstyle cx\hspace{0.03cm}}$, there exists a critical temperature $T_{\scriptscriptstyle \delta}$ such that

(i) When $T < T_{\scriptscriptstyle \delta\hspace{0.03cm}}$, the Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ of $\pi_{\scriptscriptstyle T}$ is unique and $\hat{x}_{\scriptscriptstyle T} \in B(x^*\!,\delta)$.

(ii) If, in addition, $U$ is invariant by geodesic symmetry about $x^*$, then $\hat{x}_{\scriptscriptstyle T} = x^*$.

Proposition <ref> shows exactly how small $T$ should be made, in order to ensure that the Riemannian barycentre $\hat{x}_{\scriptscriptstyle T}$ is unique. In turn, this uniqueness of $\hat{x}_{\scriptscriptstyle T}$ follows from the convexity of the variance function $\mathcal{E}_{\scriptscriptstyle T}(y)$, obtained in the following Lemma <ref>.

To state this lemma, consider the function $f(T)$ of the temperature $T$

\begin{equation} \label{eq:fT}
f(T) = \left(\frac{2}{\pi}\right)\left(\frac{\pi}{8}\right)^{\!n-1}\left(\frac{\mu_{\max}}{T}\right)^{\!\!\frac{n}{2}}\exp\left(-\frac{U_\delta}{T}\right)
\end{equation}

for any given $\delta$, where $U_\delta = \inf\lbrace U(x) - U(x^*)\,; x \notin B(x^*\!,\delta)\rbrace$. Note that $f(T)$ decreases to zero as $T$ is made arbitrarily small.

Under the same assumptions as Proposition <ref>, let $\delta < \frac{1}{2}r_{\scriptscriptstyle cx\hspace{0.03cm}}$.

(i) For all $y \in B(x^*\!,\delta)$,

\begin{equation} \label{eq:ethlower}
\mathrm{Hess}\,\mathcal{E}_{\scriptscriptstyle T}(y) \,\geq\, \mathrm{Ct}(2\delta)\hspace{0.03cm}[1 - \mathrm{vol}(M)f(T)] - \pi A_{\scriptscriptstyle M}\hspace{0.03cm}f(T)
\end{equation}

where $\mathrm{Ct}(2\delta) = 2c\delta\hspace{0.02cm}\cot(2c\delta)$ and $A_{\scriptscriptstyle M} > 0$ is a constant which depends only on the symmetric space $M$.

(ii) There exists a critical temperature $T_{\scriptscriptstyle \delta}$ such that $T < T_{\scriptscriptstyle \delta}$ implies the variance function $\mathcal{E}_{\scriptscriptstyle T}(y)$ is strongly convex on $B(x^*\!,\delta)$.

The inequality in (<ref>) should be understood as saying all the eigenvalues of $\mathrm{Hess}\,\mathcal{E}_{\scriptscriptstyle T}(y)$ are greater than the right-hand side (of course, this is an abuse of notation). The critical temperature $T_{\scriptscriptstyle \delta}$ will be expressed in the following section.

§ FINDING $T_{\SCRIPTSCRIPTSTYLE W}$ AND $T_{\SCRIPTSCRIPTSTYLE \DELTA}$