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\title{Metric Geometry}
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\section{Properties of the CRT}

Let $\mathbf{A}_{n}
$ be the set of all rooted oriented trees with $n$ edges. If $\tau\in \mathbf{A}_{n}$  and $k \in \{0, ... , 2n-1\}$ we define $\op{Reroot}(\tau,k)$ the tree $\tau$ re-rooted at the $k$-th corner in the counter clockwise contour of $\tau$. See figure below.

\vspace{-0.5cm}\begin{figure}[h]
\begin{center}
\includegraphics[height=5cm]{reroot}
\end{center}
\end{figure}
\vspace{-1.4cm}\begin{exo} Let $k \in \{0, ... , 2n-1\}$.

\begin{enumerate}
\item Show that $\op{Reroot}(\cdot,k): \mathbf{A}_{n} \to \mathbf{A}_{n}$ is a bijection. Deduce that if $\tau_{n}$ is uniform over $\mathbf{A}_{n}$ then $\op{Reroot}(\tau_{n},k)$ is also uniform over $\mathbf{A}_{n}$.
\end{enumerate}

 Let $\tau\in \mathbf{A}_{n}$ and denote its contour function by $(C_{n}(t), t \in [0,2n])$. We also write for $x \in [0,1]$, $\mathbf{C}_{n}(x)= C_{n}(2nx)$. For every $r\geq$ let $\overline{r}= r- \lfloor r \rfloor$ be the fractional part of $r$. If $g : [0,1] \longrightarrow [0,+\infty[$ such that $g(0)=g(1)=0$ and $s,t \in [0,1]$ we recall the notation 
 $$m_{g}(s,t)= \inf \big\{ g(u), u \in [s\wedge t, s \vee t]\big\}.$$
\begin{enumerate}
\setcounter{enumi}{1}
\item Show that the contour function $(\mathfrak{C}_{n}(x),x \in [0,2n])$ of the rooted oriented tree $\op{Reroot}(\tau,k)$ is given for every $t\in[0,1]$ by
$$ \mathfrak{C}_{n}(t)= \mathbf{C}_{n}\left(\frac{k}{2n}\right)+ \mathbf{C}_{n}\left(\overline{\frac{t+k}{2n}}\right) - 2 m_{\mathbf{C}_{n}}\left(\frac{k}{2n}, \overline{\frac{k+t}{2n}}\right).$$
\item \vspace{-0.3cm} Deduce from the previous considerations that if $(\mathbf{e}(t))_{t\in[0,1]}$ is a normalized Brownian excursion and $x \in [0,1]$ then the process $(\mathfrak{e}(t))_{t\in [0,1]}$ defined by
$$ \mathfrak{e}(t)= \mathbf{e}(x)+\mathbf{e}(\overline{x+ t})-2 m_{\mathbf{e}}(x, \overline{x+t}),$$
has the same distribution as $(\mathbf{e}(t))_{t \in [0,1]}$. What does it imply for the Brownian Continuum Random Tree ?

\end{enumerate}
\end{exo}

Let $\mathcal{T}$ be a real tree. For $x \in \mathcal{T}$, the \emph{multiplicity} of $x$ is the number of connected components of $\mathcal{T} \backslash\{x \}$. A \emph{leaf} is a point of multiplicity one. In the following $(\mathbf{e}(t))_{t \in [0,1]}$ denotes a normalized Brownian excursion and $\mathcal{T}_{\mathbf{e}}$ its associated $\R$-tree, in particular we denote $p_{\mathbf{e}} : [0,1] \longrightarrow \mathcal{T}_{\mathbf{e}}$ the canonical projection.

\begin{exo}
\begin{enumerate}
\item Show that almost surely $p_{\mathbf{e}}(0)$ is a leaf of $\mathcal{T}_{\mathbf{e}}$. Deduce that for every $x \in [0,1]$, $p_{\mathbf{e}}(x)$ is almost surely a leaf of $\mathcal{T}_{\mathbf{e}}$.
\item  Show that a.s.\,the local minima of a Brownian motion are pairwise distinct.
\item Deduce that almost surely $\mathcal{T}_{\mathbf{e}}$ has only countable points of multiplicity $3$ but no point of multiplicity strictly larger than $3$.
\end{enumerate}
\end{exo}

\clearpage
\begin{exo} Who are these charming gentlemen ?
\begin{figure}[h]
\begin{center}
\includegraphics[height=5.5cm]{brown}
\includegraphics[height=5.5cm]{wiener}
\end{center}
\end{figure}
\end{exo}

\nocite{LG06}
\nocite{MM06}
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