text:
\begin{document}
\title{Compact Hermitian symmetric spaces, coadjoint orbits, and the dynamical stability of the Ricci flow}
\author{Stuart James Hall}
\address{School of Mathematics and Statistics, Herschel Building, Newcastle University, Newcastle-upon-Tyne, NE1 7RU}
\email{stuart.hall@ncl.ac.uk}
\author{Thomas Murphy}
\address{Department of Mathematics, California State University Fullerton, 800 N. State College Bld., Fullerton, CA 92831, USA.}
\email{tmurphy@fullerton.edu}
\author{James Waldron}
\address{School of Mathematics and Statistics, Herschel Building, Newcastle University, Newcastle-upon-Tyne, NE1 7RU}
\email{james.waldron@ncl.ac.uk}
\maketitle  \vspace{-10pt}

\begin{abstract}
Using a stability criterion due to Kr\"oncke, we show, providing ${n\neq 2k}$, the K\"ahler--Einstein metric on the Grassmannian $Gr_{k}(\mathbb{C}^{n})$ of complex $k$-planes in an $n$-dimensional complex vector space is dynamically unstable as a fixed point of the Ricci flow. This generalises the recent results of Kr\"oncke and Knopf--Sesum on the instability of the Fubini--Study metric on $\mathbb{CP}^{n}$ for $n>1$. The key to the proof is using the description of Grassmannians as certain coadjoint orbits of $SU(n)$. We are also able to prove that Kr\"oncke's method will not work on any of the other compact, irreducible, Hermitian symmetric spaces.
\end{abstract}
\section{Introduction} \label{sec:1}
In 2013 Kr\"oncke proved the surprising result that the Fubini--Study K\"ahler--Einstein metric on $\mathbb{CP}^{n}$, $n>1$,  is unstable as a fixed point of the Ricci flow \cite{KlKr1}. More precisely, he showed that there are certain conformal (and hence non-K\"ahler) deformations of the Fubini--Study metric from which the Ricci flow never returns. This is in stark contrast to the behaviour of the K\"ahler--Ricci flow where Tian and Zhu \cite{TZ} have shown that K\"ahler--Einstein metrics are essentially global attractors within their K\"ahler class. In \cite{KS} Knopf and Sesum give an independent verification of Kr\"oncke's result.\\
\\
The behaviour of Ricci flow on manifolds admitting K\"ahler metrics is a topic of current interest (see for example \cite{HMPJM}, \cite{HMAGAG1}, \cite{IKS}, and \cite{Max}). What Kr\"oncke's result suggests is that behaviour of the Ricci flow near the space of K\"ahler metrics is more complicated than was initially believed. If a Fano manifold $M$ with Hodge number $h^{1,1}(M)>1$  admits a K\"ahler--Einstein metric then it can be destabilised by a harmonic perturbation within the K\"ahler cone. This method can be used to show many known examples of K\"ahler--Einstein metrics are unstable. However, as the complex dimension of the Fano manifold grows, there are numerous examples of K\"ahler--Einstein manifolds with $h^{1,1}(M)=1$.  One such class of K\"ahler--Einstein manifolds are the compact, irreducible, Hermitian symmetric spaces. These manifolds were completely classified by E. Cartan into six types; there are four infinite families and two exceptional spaces. Each of these spaces admits a K\"ahler--Einstein metric unique up to automorphisms of the complex structure; this metric is the symmetric metric on each manifold. We will henceforth implicitly assume all manifolds in this paper are equipped with their symmetric space (and K\"ahler--Einstein) metrics.\\
\\
The stability criterion employed by Kr\"oncke is very simple to state (c.f. Theorem \ref{KKstabthm}); if an Einstein metric with Einstein constant $\frac{1}{2\tau}>0$  admits an eigenfunction of the Laplacian, $f$ say, with eigenvalue $-\frac{1}{\tau}$ and the integral over the manifold of $f^{3}$ does not vanish,  then the metric is dynamically unstable. It is a classical result of Matsushima \cite{Mat} that, on a Fano K\"ahler--Einstein manifold, there is a bijection between Killing fields and the eigenspace corresponding to $-\frac{1}{\tau}$.  Hence K\"ahler--Einstein manifolds with large symmetry groups are ideal candidates on which to attempt to use Kr\"oncke's result to investigate stability. The first theorem we prove says that, when the symmetric space is not a Grassmannian of complex $k$-planes in an $n$-dimensional complex vector space (which we denote $Gr_{k}(\mathbb{C}^{n})$), the integral of $f^{3}$ will necessarily vanish.

\begin{theorem}\label{ThmCST}
Let $(M,g)$ be a compact, irreducible, Hermitian symmetric space other than a Grassmannian $Gr_{k}(\mathbb{C}^{n})$ and let $g$ be the canonical K\"ahler--Einstein metric normalised to have Einstein constant $\frac{1}{2\tau}$. Then any ($-\dfrac{1}{\tau}$)-eigenfunction of the Laplacian, $f$, satisfies
\begin{equation*}
\int_{M}f^{3} \ dV_{g} = 0.
\end{equation*}
\end{theorem}
The proof of this theorem uses the Chevalley--Shepherd--Todd theorem which classifies the degrees in which the generators of certain polynomial algebras can exist when the polynomial is required to be invariant under the action of a compact simple Lie group $G$. Applied to our stability problem, this classification only permits a generator in the required degree when $G=SU(n)$ and so the only class of space where the integral can be non-zero is the complex Grassmannians.\\
\\
We are able to compute the stability integral for $Gr_{k}(\mathbb{C}^{n})$ and show that, in some sense, `generic' Grassmannians are unstable. This result generalises the $\mathbb{CP}^{n}$ calculation of Kr\"oncke and Knopf--Sesum.
\begin{theorem}\label{ThmGr}
If $k,n \in \mathbb{N}$ with $1\leq k<n$ and $n\neq 2k$, then  $Gr_{k}(\mathbb{C}^{n})$ is dynamically unstable as a fixed point of the Ricci flow.
\end{theorem}
The integral of $f^{3}$ vanishes for the spaces $Gr_{k}(\mathbb{C}^{2k})$; the proof of Theorem \ref{ThmGr} shows this directly but this can be seen without performing any calculation. The duality map ${\Psi:Gr_{k}(\mathbb{C}^{n})\rightarrow Gr_{n-k}(\mathbb{C}^{n})}$ (where a subspace is mapped to its complement) is an isometry and in the case when $n=2k$ is also an involution. One can show (see, for example, \cite{GG}) that $\Psi^{\ast}f = -f$ and thus the integral of any odd power of $f$ vanishes.\\
\\
We cannot yet say anything about the stability of the spaces $Gr_{k}(\mathbb{C}^{2k})$ apart from the case when $k=1$ as then ${Gr_{1}(\mathbb{C}^{2})\cong \mathbb{CP}^{1} \cong \mathbb{S}^{2}}$. In this case, the K\"ahler--Einstein metric is the round metric and this is known to be dynamically stable by a result of Hamilton \cite{Ham} (Chow later proved that the round metric on $\mathbb{S}^{2}$ is a global attractor for the normalised Ricci flow starting at any initial metric \cite{Chow}).\\
\\
The methods used in \cite{KS} and \cite{KlKr1} to show a destabilising eigenfunction exists on $\mathbb{CP}^{n}$ use the generalised Hopf fibration to lift the problem to finding certain $U(1)$-invariant functions on the sphere $\mathbb{S}^{2n+1}\subset \mathbb{C}^{n+1}$. This paper takes a totally different approach by viewing the Grassmannians as adjoint orbits of $SU(n)$ and using techniques coming from symplectic geometry (such as the Duistermaat--Heckman formula) to construct eigenfunctions and make calculations of the relevant integrals. \begin{comment}\\
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Finally, in the case that the manifold is $\mathbb{CP}^{n}$ with $n>1$, we can give a detailed characterisation of the unstable perturbations.  This is achieved by combining our calculation of Kr\"oncke's stability integral with the results of Koiso \cite{}.  As the Grassmannians $Gr_{k}(\mathbb{C}^{n})$ admit infinitesimal Einstein perturbations when $1< k <n-1$, it is not possible to completely characterise the unstable perturbations for these manifolds. Hence we believe, at the time of writing, the Fubini--Study metric is the only K\"ahler--Einstein metric on a Fano manifold where this information is known.
\begin{theorem}\label{ThmCPn}
For $n>1$, the space of unstable  infinitesimal perturbations of the Fubini--Study metric on $\mathbb{CP}^{n}$ can be realised as an open subset of $\mathfrak{su} (n+1)$.
\end{theorem}
\end{comment}
\section{Background}\label{sec:2}
\subsection{Stability}
Einstein metrics $g$ satisfying $\Ric(g)=\frac{1}{2\tau} g$ for $\tau \in \mathbb{R}$, evolve via homothetic scaling under the Ricci flow
\begin{equation*}
\dfrac{\partial g}{\partial t} =-2\Ric(g).
\end{equation*}
 It is therefore useful to  view Einstein metrics as fixed points of the Ricci flow up to a normalisation of the volume of the metric by homothetic scaling. A natural question is whether a given Einstein metric is stable as a fixed point in the sense that the Ricci flow starting at a small perturbation of the metric will return to the original Einstein metric.  Perelman \cite{Per1} introduced a functional $\nu(g)$ which is stationary at shrinking gradient Ricci solitons (every Einstein metric with $\tau>0$ is such a soliton) and which is otherwise strictly increasing along the Ricci flow. This allows the stability of an Einstein metric to be investigated by calculation of the second variation of the $\nu$ functional along potentially destabilising directions. If the entropy increases along a particular direction then the corresponding perturbation of the Einstein metric will never return under the flow. This process was first carried out for Einstein metrics by Cao, Hamilton, and Ilmanen \cite{CHI} and later generalised by Cao and Zhu \cite{CZ}.

\begin{theorem2}[Cao--Hamilton--Ilmanen \cite{CHI}]\label{CHIstab}
	Let $(M,g)$ be an Einstein metric with Einstein constant $\frac{1}{2\tau}>0$. Let $h\in s^{2}(TM^{\ast})$. Then
	$$
	\frac{d^{2}}{ds^{2}}\nu(g+sh) |_{s=0} = \frac{\tau}{\vol(M,g)}\int_{M}\langle N(h),h\rangle dV_{g},
	$$
	where
	\begin{equation}\label{Stabop}
	N(h) = \frac{1}{2}\Delta h+\Riem(h,\cdot)+\div^{\ast}\div (h)+\frac{1}{2}\nabla^{2}v_{h} -\frac{g}{2n\tau\vol(M,g)}\int_{M}\tr(h)dV_{g},
	\end{equation}
	and $v_{h}$ is the unique solution to
	$$ \Delta v_{h}+\frac{v_{h}}{2\tau}=\div(\div (h)). $$
\end{theorem2}
The diffeomorphism and scale invariance of $\nu(g)$ means that to check linear stability one only needs to consider perturbations  $h\in s^{2}(TM^{\ast})$ satisfying
$$\div(h) = 0 \textrm{ and } \langle h,g\rangle_{L^{2}} = \int_{M}\tr(h)dV_{g} = 0.$$
In this case, the stability operator $N$ in Equation (\ref{Stabop}) reduces to
$$N(h)  = \frac{1}{2}\Delta h +\Riem(h,\cdot) = \frac{1}{2}(\Delta_{L}+\frac{1}{\tau})h,$$
where  $\Delta_{L}$ is the Lichnerowicz Laplacian
$$\Delta_{L}h := \Delta h +2\Riem(h,\cdot)-\Ric\cdot h-h\cdot \Ric.$$
We thus have the following definitions:
\begin{definition}[Linear stability of Einstein metrics]
Let $(M,g)$ be a compact Einstein manifold satisfying $\Ric(g) =\dfrac{1}{2\tau} g$ and let $-\kappa$ be the largest eigenvalue of the Lichnerowicz Laplacian restricted to the space of divergence-free, $g$-orthogonal tensors.
\begin{enumerate}
	\item If $\kappa > \dfrac{1}{\tau}$, $g$ is called \textit{linearly stable}. \vspace{5pt}
	\item If $\kappa =\dfrac{1}{\tau}$, $g$ is called \textit{neutrally linearly stable}. \vspace{5pt}
	\item If $\kappa < \dfrac{1}{\tau}$, $g$ is called \textit{linearly unstable}.
\end{enumerate}
\end{definition}

\begin{definition}[Dynamical stability of Einstein metrics]
Let $(M^{n},g_{E})$ be a compact Einstein manifold. The metric $g_{E}$ is said to be \textit{dynamically stable} for the Ricci flow if for any $m\geq 3$ and any $C^{m}$-neighbourhood $U$ of $g_{E}$ in the space of sections $\Gamma(s^{2}(TM^{\ast}))$, there exists a $C^{m+2}$ neighbourhood of $g_{E}$, $V \subset U$, such that:
\begin{enumerate}
	\item for any $g_{0}\in V$, the volume normalised Ricci flow
	$$ \dfrac{\partial g}{\partial t} =-2\Ric(g)+\frac{2}{n}\left(\int_{M}\textrm{scal}(g)dV_{g}\right)g,$$
	with $g(0) = g_{0}$ exists for all time,
	\item the metrics $g(t)$ converge modulo diffeomorphism to an Einstein metric in $U$.
\end{enumerate}
	If the metric $g_{E}$ is not dynamically stable then it is said to be \textit{dynamically unstable}.
\end{definition}
If a metric is linearly unstable, then Sesum \cite{Sesum} showed that it is dynamically unstable. Similarly (though much more technically difficult to prove), under the assumption that all infinitesimal Einstein deformations are integrable, Sesum proved that linear stability implies dynamical stability.  In \cite{KKCVP}, Kr\"oncke built on the work of Haslhofer and M\"uller \cite{HasMul} and showed that an Einstein metric with positive Einstein constant is dynamically stable if and only if it is a local maximum of the $\nu$ functional. This characterisation does not require an infinitesimal perturbation to satisfy any integrability assumptions and forms the basis of Kr\"oncke's stability criterion in Theorem \ref{KKstabthm}.\\
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In general, it is very difficult to analyse the spectrum of the Lichnerowicz Laplacian for an arbitrary Einstein metric.  If the metric has some extra structure then more can be said.  In the case the Einstein metric is K\"ahler--Einstein then there is the following topological condition (originally stated in \cite{CHI} and proved for the more general class of K\"ahler--Ricci solitons in \cite{HMPAMS})
\begin{theorem2}[Cao--Hamilton--Ilmanen]
Let $(M,J,g)$ be a K\"ahler--Einstein metric. If the Hodge number $h^{1,1}(M)>1$ then $g$ is linearly unstable.
\end{theorem2}
This proposition can be seen as generalising the fact that any product of Einstein metrics with fixed Einstein constant $\frac{1}{2\tau}$ is unstable under the Ricci flow. The product of any two K\"ahler--Einstein metrics always has $h^{1,1}(M)>1$.\\
\\
In \cite{CH}, Cao and He made a complete study of the stability of the simply-connected, compact, irreducible, symmetric spaces. The spaces where the metric is K\"ahler--Einstein can be written in the form $M=G/H$ where $G$ is a connected compact simple Lie group and $H$ is the isotropy subgroup.  We note that the identity component of the isometry group $\textrm{Iso}_{0}=G$ and so the Lie algebra of Killing fields $\mathfrak{k} = Lie(\textrm{Iso}_{0})$  is isomorphic to the Lie algebra $\mathfrak{g}$. All the manifolds in the following theorem have $h^{1,1}(M)=1$.
\begin{theorem2}[Cao--He, c.f. Theorem 4.3 in \cite{CH}] \label{CHT}
The linear stability of the irreducible compact Hermitian symmetric spaces $M ={G}/{H}$ is as follows:
	\begin{enumerate}
		\item $M$ is linearly unstable if:
		\begin{itemize}
			\item $M$ is the space of compatible complex structures on $\mathbb{H}^{n}$,  $M = {Sp(n)}/{U(n)},$			for $n>1$.
		\end{itemize}
		\item $M$ is neutrally linearly stable if:
	\begin{itemize}
		\item $M$ is a complex Grassmannian ${Gr_{k}(\mathbb{C}^{n})}= {SU(n)}/{S(U(k)\times U(n-k))},$
		where $n>2$ and ${0<k<n}$,\vspace{3pt}
		\item $M$ is a complex hyperquadric  $Q_{n} = {SO(n + 2)}/{(SO(n) \times SO(2))},$
		where $n \geq 4$,
		\item $M$ is a space of orthogonal almost complex structures on $\mathbb{R}^{2n}$,
		$M = {SO(2n)}/{U(n)},$
		for $n>2$,
		\item $M$ is one of the exceptional spaces,
		${M= {E_{6}}/{(SO(10)\times SO(2))}},$ or \newline
		${M = {E_{7}}/{(E_{6}\times SO(2))}}.$
	\end{itemize}
\item If $M$ is the sphere $\mathbb{S}^{2} \cong Gr_{1}(\mathbb{C}^{2}) \cong \dfrac{SO(4)}{U(2)}$ then $M$ is dynamically stable and so linearly stable.
\end{enumerate}
\end{theorem2}
Missing from this list (as it is not irreducible) is the hyperquadric $${Q_{2} = \dfrac{SO(4)}{SO(2)\times SO(2)}  \cong \mathbb{CP}^{1}\times\mathbb{CP}^{1}}.$$ It is unstable as it is a product. The hyperquadric $Q_{3}\cong \dfrac{Sp(2)}{U(2)}$ has $h^{1,1}=1$ but is nevertheless linearly unstable by a result of Gasqui and Goldschmidt \cite{GG}.	\\
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What Theorem \ref{CHT} shows is that most of the Hermitian symmetric spaces are neutrally linearly stable.  In particular, the complex projective spaces $\mathbb{CP}^{n} = Gr_{1}(\mathbb{C}^{n+1})$ with $n>1$ are all neutrally linearly stable.\\
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On any Einstein manifold $(M,g)$ with Einstein constant $\frac{1}{2\tau}$, if there is an eigenfunction $f$ satisfying $\Delta f=-\frac{1}{\tau}f$ then we define the tensor
$$h_{f} := (\Delta f)g-\Hess(f)+\frac{f}{2\tau}g.$$
It can be shown (\cite{CHI}, \cite{CH}) that $h_{f}$ is divergence free, $L^{2}$-orthogonal to $g$ and satisfies
$$ \Delta_{L}h_{f} = -\frac{1}{\tau}h_{f}.$$
In 2013 Kr\"oncke proved the following stability criterion by computing the third variation of the $\nu$ functional.
\begin{theorem2}[Kr\"oncke, Theorem 1.7  in \cite{KlKr1}] \label{KKstabthm}
Let $(M,g)$ be an Einstein metric with Einstein constant $\frac{1}{2\tau}$ and let $f$ be an eigenfunction of the Laplacian with eigenvalue $-\frac{1}{\tau}$. If the integral
\begin{equation}\label{KKstab}
\int_{M}f^{3}dV_{g} \neq 0,
\end{equation}
then $g$ is dynamically unstable as a fixed point of the Ricci flow and is destabilised by the tensor $h_{f}$.
\end{theorem2}
Kr\"oncke then constructed a eigenfunction satisfying the condition (\ref{KKstab}) for the spaces $\mathbb{CP}^{n}$ with $n>1$ and proved:
\begin{cor}\label{C1}
The Fubini--Study metrics on $\mathbb{CP}^{n}$, $n>1$ are dynamically unstable as fixed points of the Ricci flow.
\end{cor}
This result was somewhat unexpected as a long-standing conjecture in the field had included $\mathbb{CP}^{2}$ on the list of stable, four-dimensional geometries for the Ricci flow. Theorem \ref{ThmGr} can be seen as a generalisation of the $\mathbb{CP}^{n}$ results of Kr\"oncke and Knopf--Sesum; however, as mentioned in the introduction, our construction of eigenfunctions and method of evaluating the integral is totally different from the methods used in \cite{KS} and \cite{KlKr1}.\\
\\
Given Theorem \ref{ThmGr}, it seems likely that the other compact irreducible Hermitian symmetric spaces are also unstable but that they represent metrics with high degrees of degeneracy as critical points of the $\nu$ functional. Further evidence for their instability might come from the behaviour of other non-Hermitian compact symmetric spaces. The first author \cite{Hall19} has investigated the stability of the canonical metric on the compact simple Lie group $G_{2}$  which is also neutrally linearly stable with the neutral directions coming from conformal perturbations corresponding to eigenfunctions of the Laplacian. In this case, Kr\"oncke's stability integral is non-zero for a certain eigenfunction and so $G_{2}$ is unstable.

\subsection{Geometry of coadjoint orbits}
In this section $G$ is a compact, semisimple Lie group. Hence the Killing form $\langle\cdot,\cdot \rangle$ is non-degenerate and so the adjoint representation of $G$ is orthogonal. We can also use the Killing form (or any $\mathrm{Ad}_{G}$ -invariant inner product) to identify $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$. The coadjoint action of $G$ on $\mathfrak{g}^{\ast}$ is defined by
$$\mathrm{Ad}_{g}^{\ast}(\xi)(X) := \xi(\mathrm{Ad}_{g^{-1}}(X))$$
for $g\in G$,  $\xi \in \mathfrak{g}^{\ast}$ and $X \in \mathfrak{g}$.  If ${\xi(\cdot) = \langle\cdot,X\rangle}$ then ${\mathrm{Ad}^{\ast}_{g}(\xi)(\cdot) = \langle \cdot, \mathrm{Ad}_{g}(X)\rangle}$ and we have a straightforward identification of coadjoint and adjoint orbits via the Killing form. \\
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For $\xi \in \mathfrak{g}$ we consider the orbit $\mathcal{O}_{\xi}$ of $\xi$ under the adjoint action of $G$. Denote by $H$ stabiliser of $\xi$ and let $\mathfrak{h}\subset \mathfrak{g}$ be its Lie algebra. Then $\mathfrak{g}  =\mathfrak{h} \oplus \mathfrak{m}$ where $\mathfrak{m}$ can be identified with the tangent space to $\mathcal{O}_{\xi}$ at $\xi$. The subalgebra $\mathfrak{h}$ is the kernel of the map ${\mathrm{ad}({\xi}):\mathfrak{g} \rightarrow \mathfrak{g}}$ and thus $\mathfrak{m}$ is the image.\\
\\
We let $T$ be a maximal torus of $G$ and take $\mathfrak{t} = Lie(T)$ to be its Lie algebra. The Weyl group ${W=N_{G}(T)/T}$ where $N_{G}(T)$ is the normaliser of $T$ in $G$. A classical theorem (see for example Bott \cite{Bott}) yields:
\begin{enumerate}
	\item $\mathcal{O}_{\xi}\cap \mathfrak{t} \neq \emptyset$,
	\item $\mathcal{O}_{\xi}\cap \mathfrak{t}$ is a $W$-orbit.
\end{enumerate}
This means, without loss of generality, we can take the element representing the orbit $\xi \in \mathfrak{t}$.\\
\\
The orbits have the structure of a complex manifold. Decomposing the complexified Lie algebra $\mathfrak{g} \otimes \mathbb{C}$ we get
$$ \mathfrak{g} \otimes \mathbb{C} =  \mathfrak{t}_{\mathbb{C}} \oplus\left( \bigoplus_{\alpha: \ \langle \alpha, \xi\rangle=0} R_{\alpha} \right)\oplus A\oplus \bar{A}$$
where $\alpha \in \mathfrak{t}_{\mathbb{C}}$ are the roots of $G$, $R_{\alpha}$ is the root space of $\alpha$, and $A$ is the span of the root spaces satisfying
$$[\xi, r_{\alpha}] =i\langle \alpha, \xi\rangle r_{\alpha},  \textrm{ with } \langle \alpha, \xi\rangle >0 \textrm{ for all } r_{\alpha}\in R_{\alpha}.$$
One can identify $\mathfrak{m}_{\mathbb{C}}\cong A\oplus \bar{A}$ and show that  $A$ and $\mathfrak{t}_{\mathbb{C}} \oplus A$ are Lie subalgebras of $ \mathfrak{g} \otimes \mathbb{C}$. By defining $\mathfrak{m}^{(1,0)}=A$ we get a $G$-invariant complex structure on $\mathcal{O}_{\xi}$.\\
\\
The Kirillov--Kostant--Souriau symplectic form is defined as
$$\omega_{\xi}(x,y) = -\langle\xi,[x,y]\rangle,$$
for $x,y \in \mathfrak{m}$. This is extended over  $\mathcal{O}_{\xi}$ using the adjoint action. This form is compatible with the complex structure and gives the orbit the structure of a K\"ahler manifold. In the case when the center of $H$ has dimension 1 (which will be the cases that we wish to consider in this article), the induced metric is K\"ahler--Einstein (c.f.\cite{Bes} Proposition 8.85).
\subsection{Properties of the eigenfunctions}
We will now show how to construct eigenfunctions for the Laplacian of the K\"ahler--Einstein metric on $\mathcal{O}_{\xi}$. We begin by defining functions ${f_{\eta} \in C^{\infty}(\mathcal{O}_{\xi})}$ by
\begin{equation} \label{ef_def}
f_{\eta}(Z): =  \langle Z,\eta \rangle,
\end{equation}
where $Z\in \mathcal{O}_{\xi}$ and $\langle \cdot, \cdot \rangle$ is the inner product on $\mathfrak{g}$ coming from the Killing form. These functions satisfy some important properties.
\begin{lemma} \label{f_mv_lem}
Suppose that $\eta \in \mathfrak{g}$ and $\tilde{\eta} \in \mathfrak{g}$ are in the same $G$-orbit. Then the functions $f_{\eta},f_{\tilde{\eta}} \in C^{\infty}(\mathcal{O}_{\xi})$ defined by Equation (\ref{ef_def}) satisfy
$$
\int_{\mathcal{O}_{\xi}}f_{\eta}^{k} \ \omega^{n}  = \int_{\mathcal{O}_{\xi}}f_{\tilde{\eta}}^{k} \ \omega^{n},
$$
where $k\in \mathbb{N}$, $n$ is the compex dimension of $\mathcal{O}_{\xi}$, and $\omega$ is the Kirillov--Kostant--Souriau symplectic form. Furthermore, in the case that $k=1$ we have
$$\int_{\mathcal{O}_{\xi}} f_{\eta} \ \omega^{n} = 0,$$
for all $\eta \in \mathfrak{g}$.
\end{lemma}
\begin{proof}
The conditions of the lemma mean there is a $g \in G$ such that $\tilde{\eta} = \mathrm{Ad}_{g}(\eta)$. Hence by the Ad-invariance of the inner product we have
$$ f_{\eta}(Z) = \langle \eta,Z\rangle  = \langle \mathrm{Ad}_{g^{-1}}(\tilde{\eta}),Z\rangle  = \langle \tilde{\eta}, \mathrm{Ad}_{g}(Z)\rangle = f_{\tilde{\eta}}(\mathrm{Ad}_{g}(Z)).$$
In other words, ${f_{\eta} = \mathrm{Ad}_{g}^{\ast}(f_{\tilde{\eta}})}$. As ${\mathrm{Ad}_{g}:\mathcal{O}_{\xi}\rightarrow\mathcal{O}_{\xi}}$ is an orientation preserving isometry, we have
$$
\int_{\mathcal{O}_{\xi}}f_{\eta}^{k} \ \omega^{n}=  \int_{\mathcal{O}_{\xi}}\mathrm{Ad}_{g}^{\ast}(f_{\tilde{\eta}})^{k} \ \omega^{n} = \int_{\mathcal{O}_{\xi}}\mathrm{Ad}^{\ast}_{g}(f_{\tilde{\eta}})^{k} \ (\mathrm{Ad}_{g}^{\ast}\omega^{n}) =  \int_{\mathcal{O}_{\xi}}f_{\tilde{\eta}}^{k} \ \omega^{n}.
$$
To prove the second part of the lemma, we note that the function ${F: G\rightarrow \mathbb{R}}$ given by
$$F(g) = \int_{\mathcal{O}_{\xi}}f_{\mathrm{Ad}_{g}(\eta)} \omega^{n},$$
is constant. Taking the derivative at the identity yields
$$\int_{\mathcal{O}_{\xi}}f_{[\eta,\zeta]}\omega^{n} = 0,$$
for all $\eta, \zeta \in \mathfrak{g}$. This means that the map
$$\eta \rightarrow \int_{\mathcal{O}_{\xi}}f_{\eta} \ \omega^{n}$$ is a Lie algebra homomorphism. The fact that $G$ is simple means that this map must be the trivial homomorphism and so the result follows.
\end{proof}
 Next we recall  a theorem of Matsushima \cite{Mat} which says that for any Fano K\"ahler--Einstein manifold $(M,g,J)$ there is an isomorphism between the $(-1/\tau)$-eigenspace, $E_{(-1/\tau)}$, and the Lie algebra of Killing vector fields $\mathfrak{k}$  given by
$$\phi \rightarrow -J\nabla\phi,$$
where $J$ is the complex structure.\\
\\
All the connected, compact, irreducible Riemannian symmetric spaces  can be constructed in the form $M=\textrm{Iso}_{0}/\textrm{Iso}_{p}$ where $\textrm{Iso}_{0}$ is the connected component of the identity of the isometry group of $(M,g)$ and $\textrm{Iso}_{p}$ is the isotropy group of isometries fixing a point. For the spaces $G/H$ in Theorem \ref{CHT} we have
$$\mathfrak{g}\cong \textrm{Lie}(\textrm{Iso})\cong \mathfrak{k}.$$
This map can be realised by the assignment
$$\eta\rightarrow \frac{d}{dt} \bigg|_{t=0} \mathrm{Ad}_{\exp(t\eta)}(Z) = [\eta,Z],$$
for $\eta \in \mathfrak{g}$ and $Z \in \mathcal{O}_{\xi}$.
\begin{lemma}\label{eflem}
If the K\"ahler--Einstein metric on $\mathcal{O}_{\xi}$ has Einstein constant $\frac{1}{2\tau}$, then the functions $f_{\eta}$ defined in Equation (\ref{ef_def}), satisfy
$$\Delta f_{\eta} = -\frac{1}{\tau}f_{\eta}.$$
\end{lemma}
\begin{proof}
	Let $X \in \mathfrak{m}$ and consider
	$$F(t)=f_{\eta}(\mathrm{Ad}_{\exp(tX)}\xi) = \langle  \eta, \mathrm{Ad}_{\exp(tX)}\xi \rangle = \langle \mathrm{Ad}_{\exp(-tX)}\eta, \xi \rangle. $$
	Taking derivatives we see
	$$\dfrac{dF}{dt}\Bigr\vert_{t=0} =-\langle \xi, [X,\eta]\rangle = -\omega(X,\eta) = g(X,J\eta).$$
	Hence $\nabla f_{\eta} =J\eta$ (here we identify $\eta$ with the Killing field it generates on $\mathcal{O}_{\xi}$).  As $\eta$ is a holomorphic Killing field we invoke Matsushima's theorem  which says the map
	$$ \phi \rightarrow \nabla \phi,$$
	is an isomorphism between the eigenspace $E_{-1/\tau}$ and $J\mathfrak{k}$ where $\mathfrak{k}$ is the space of Killing fields. Hence, as $f_{\eta}$ has mean value zero by Lemma \ref{f_mv_lem}, we see $f_{\eta}$ is an eigenfunction of the Laplacian with eigenfunction $-\frac{1}{\tau}$.
\end{proof}
An element $X\in \mathfrak{g}$ is said to be $\textit{regular}$ if its centraliser in $\mathfrak{g}$ is of smallest possible dimension. The regular elements of $\mathfrak{t}$ we will denote by $\mathfrak{t}_{\textrm{reg}}$. The following lemma will be vital in our analysis.
\begin{lemma}[Bott \cite{Bott}]\label{Bottlem}
Let $D \in \mathfrak{t}_{\mathrm{reg}}$ be regular. Then the function $f_{D}$ defined by Equation (\ref{ef_def}) has the following properties.
\begin{enumerate}
 \item  The critical points of $f_{D}$ are non-degenerate and of even index.
 \item  The critical points are the orbit of $\xi \in \mathfrak{t}$ under the action of Weyl group $W$.
\end{enumerate}
\end{lemma}
We remark that $f_{D}$ is a Hamiltonian function for the  action on $\mathcal{O}_{\xi}$ generated by $D$. If $\Lambda \subset \mathfrak{t}$ is the weight lattice, then choosing $D\in \Lambda\otimes_{\mathbb{Z}}\mathbb{Q}$ will generate an $\mathbb{S}^{1}$-action. We denote the set of elements in $\mathfrak{t}$ that generate closed orbits by $\mathfrak{t}_{c}$.  The set $\mathfrak{t}_{reg}\cap \mathfrak{t}_{c}$ is dense in $\mathfrak{t}$ (with respect to the Euclidean topology). It turns out that the eigenfunctions $f_{D}$ generated by  $D \in \mathfrak{t}_{reg}\cap \mathfrak{t}_{c}$  are the only ones that one needs to check the stability condition (\ref{KKstab}) on.
\begin{proposition}\label{Proptorstab}
Let $G/H=\mathcal{O}_{\xi}$ be one of the symmetric spaces in Theorem \ref{CHT} and let $T$ be a maximal torus in $G$ with Lie algebra $\mathfrak{t}$. Suppose that there exists $f \in E_{-(1/\tau)}$ such that
$$\int_{\mathcal{O}_{\xi}}f^{3}\omega^{n} \neq0,$$
then there exists $D\in \mathfrak{t}_{reg}\cap \mathfrak{t}_{c}$ such that
$$\int_{\mathcal{O}_{\xi}}f_{D}^{3} \ \omega^{n} \neq0.$$
\end{proposition}
\begin{proof}
As the map $\varrho:\mathfrak{g}\rightarrow E_{-1/\tau}$ given by $\varrho(X) = f_{X}$ is an isomorphism, there must exist $\eta\in \mathfrak{g}$ such that $f=f_{\eta}$. As remarked in Section 2.2, given a fixed maximal torus $T$ with Lie algebra $\mathfrak{t}$, any element of $\mathfrak{g}$ is in the adjoint orbit of some element in $\mathfrak{t}$. If $D = \mathrm{Ad}_{g}(\eta)\in \mathfrak{t}$ for some $g \in G$ then, by Lemma \ref{f_mv_lem}
$$ \int_{\mathcal{O}_{\xi}}f_{D}^{3} \ \omega^{n} \neq 0.$$
Finally we see we can take $D\in \mathfrak{t}_{reg}\cap \mathfrak{t}_{c}$ as this set is dense in $\mathfrak{t}$.
\end{proof}
\subsection{Computing integrals}
We will need to be able to compute integrals of powers of $f$ over the orbit.  This is achieved by the famous Duistermaat--Heckman formula \cite{DH} (see \cite{MS} for the form we are using). On a symplectic manifold $(M^{2n},\omega)$ with a Hamiltonian circle action that has an associated Hamiltonian function $\varphi$ with non-degenerate critical points, the  Duistermaat--Heckman formula  is
$$\int_{M}e^{-t\varphi(p)}\omega^{n}(p) = \frac{n!}{t^{n}}\sum_{q \textrm{ critical }}\frac{e^{-t\varphi(q)}}{\varpi(q)},$$
where the $\varpi(q)$ is the product of the weights of the circle action that is induced on the tangent space at each fixed point $q$.\\
\\
The holomorphic tangent space at a critical point $q$ is identified with the span of the root spaces $R_{\alpha}$ satisfying ${\langle q, \alpha\rangle>0}$, we will denote this set $P(q)$. The derivative of ${r_{\alpha} \rightarrow ad_{\exp(tD)}r_{\alpha}}$ at $t=0$  is $\langle \alpha,D \rangle r_{\alpha}$. Hence the weight at each fixed point $q \in \mathfrak{t}$ is given by
$$\varpi(q) = \prod_{\alpha \in P(q)}\langle \alpha, D\rangle.$$
\begin{proposition}
Let $D\in \mathfrak{t}_{reg}\cap\mathfrak{t}_{c}$ and let ${f_{D}(Z) = \langle Z, D\rangle}.$ Then
\begin{equation}\label{DHGroup}
\int_{\mathcal{O}_{\xi}}e^{-tf(p)}\omega^{n}(p) = \frac{n!}{t^{n}}\sum_{w\in W /\textrm{stab}(\xi)}\frac{e^{-tf(w\cdot \xi)}}{\prod_{\alpha \in P(w\cdot \xi)}\langle \alpha, D\rangle},
\end{equation}
where $n$ is the complex dimension of the orbit $\mathcal{O}_{\xi}$.
\end{proposition}
\begin{proof}
We simply apply the Duistermaat--Heckman formula to the function $f_{D}$ which is the Hamiltonian for the circle action generated by $D\in \mathfrak{t}_{reg}\cap\mathfrak{t}_{c}$. As the element $D$ is regular, Lemma \ref{Bottlem}, says the critical points are non-degenerate precisely the orbit of $\xi$ under the action of the Weyl group $W$. The value of the weights follows from the previous discussion.
\end{proof}
Formulae similar to (\ref{DHGroup}) occur in the theory of the orbit method developed by Kirillov \cite{Kir}. In this theory integrals over certain coadjoint orbits correspond to the characters of the representation corresponding to the orbit. Similar expressions also appear elsewhere in the literature, for example in the papers of Berline and Vergne \cite{BV}, Paradan \cite{Par}, and Rossman \cite{Ros}.
\section{The proof of Theorem \ref{ThmCST}} \label{sec:3}

The proof of Theorem \ref{ThmCST} is based on the classification of certain invariant polynomial algebras.  If we let $G$ be one of the compact simple Lie groups appearing as the larger group in Theorem \ref{CHT}, then we denote by $\mathbb{R}[\mathfrak{g}]^{G}$ the graded algebra of $G$-invariant polynomials functions on $\mathfrak{g}$. If $T$ is a maximal torus of $G$ with Lie algebra $\mathfrak{t}$ and Weyl group $W$, then the Chevalley restriction theorem (see for example Section 23 in \cite{Hum}) yields an isomorphism ${r:\mathbb{R}[\mathfrak{g}]^{G}\rightarrow \mathbb{R}[\mathfrak{t}]^{W}}$ where $r(P)$ is simply restriction of a polynomial $P\in \mathbb{R}[\mathfrak{g}]^{G}$ to $\mathfrak{t}$. By the Chevalley--Shephard--Todd theorem and the Shephard--Todd classification of complex reflection groups, see \cite{Coxe} and \cite{ST}, the algebra ${\mathbb{R}[\mathfrak{t}]^{W}}$  is a polynomial algebra, with generators of well defined degrees. We list them in the following table with the groups in brackets having Lie algebra with the corresponding root system:
\begin{center}
\begin{tabular}{ |c|c| }
 \hline
 Root system  & Degree of generators  in ${\mathbb{R}[\mathfrak{t}]^{W}}$  \\
 \hline
 $A_{n} \ (SU(n+1))$ &  $2,3,\dots, n+1$  \\
 $B_{n} \ (SO(2n+1))$ &  $2,4, \dots, 2n$\\
 $C_{n} \ (Sp(n))$ & $2,4, \dots, 2n$\\
 $D_{n} \ (SO(2n))$ & $2,4,\dots, 2(n-1); n$ \\
 $E_{6}$ & $2,5,6,8,9,12$\\
 $E_{7}$ & $2,6,8,10,12,14,18$\\
 \hline
\end{tabular}
\end{center}
\vspace{10pt}
The following lemma shows that the stability integral (\ref{KKstab}) can be thought of as an element of ${\mathbb{R}[\mathfrak{g}]^{G}}$.
\begin{lemma}\label{homflem}
Let $I^{k}:\mathfrak{g}\rightarrow\mathbb{R}$ be defined by
$$I^{k}(\eta) = \int_{\mathcal{O_{\xi}}}(f_{\eta})^{k}\omega^{n}.$$
Then $I^{k}$ is a (possibly trivial) $\mathrm{Ad}_{G}$-invariant, homogenous, degree$-k$ polynomial.
\end{lemma}
\begin{proof}
The $\mathrm{Ad}_{G}$-invariance was demonstrated in Lemma \ref{f_mv_lem}. The fact that the function is a homogenous, degree-$k$ polynomial is straightforward if one picks a basis of $\mathfrak{g}$ and then calculates in coordinates.
\end{proof}
We now give the proof of Theorem \ref{ThmCST}.
\begin{proof}
We note by Proposition \ref{Proptorstab}, we might as well assume that a destabilising eigenfunction is of the form $f_{D}$ for $D \in \mathfrak{t}_{reg}\cap\mathfrak{t}_{c}$. If the symmetric space is not a Grassmanian then it is of the form $G/H$ with $G$ being one of the groups $B_{n},D_{n},E_{6}$ or $E_{7}$. Lemma  \ref{homflem} shows that $I^{3}$ is a degree $3$, homogenous $G$-invariant polynomial and so by the Chevalley restriction theorem yields a degree 3 element of  ${\mathbb{R}[\mathfrak{t}]^{W}}$.  However, the above table shows this function must vanish unless $G=D_{3} = SO(6) =A_{3}$. One can show that $SO(6)/U(3) \cong \mathbb{CP}^{3}$ and  $Q_{4}  \cong Gr(2,4)$  and so the stability of these spaces follows from the type $A_{n}$ consideration.
\end{proof}
The fact that the integral \ref{KKstab} vanishes on the spaces $Gr_{k}(\mathbb{C}^{2k})$ is a consequence of the calculation in section \ref{sec:5}, specifically Equation (\ref{Efinal}).\\
\\
We also note that this method of proof also shows that the functions $f_{\eta}$ defined by Equation (\ref{ef_def}) have mean value $0$ (which was demonstrated explicitly in Lemma \ref{f_mv_lem}). This follows as there are no non-zero homogeneous, degree $1$ polynomials that are invariant under the Weyl groups of the compact connected Lie groups we are considering in this article.
\section{Combinatorial properties of certain determinants} \label{sec:4}
In order to prove Theorem \ref{ThmGr} we first need to collect some results on the power series of certain matrices that will appear after the manipulation of the righthand side of Equation (\ref{DHGroup}).  For $m_{1},m_{2},\dots,m_{n} \in \mathbb{R}$ and ${0<k \leq [n/2]}$ we consider the matrix valued function ${\mathcal{M}:\mathbb{R}\rightarrow \mathrm{Mat}^{n\times n}_{\mathbb{R}}}$ given by
\begin{equation}\label{DHMatrix}
\mathcal{M}(t) = \left(
\begin{array}{cccc}
e^{-m_{1}t} & e^{-m_{2}t} & \dots  & e^{-m_{n}t} \\
e^{-m_{1}t}m_{1} & e^{-m_{2}t}m_{2} & \dots  & e^{-m_{n}t}m_{n}\\
\vdots &\vdots &\vdots &\vdots\\
e^{-m_{1}t}m_{1}^{k-1} & e^{-m_{2}t}m_{2}^{k-1} & \dots  & e^{-m_{n}t}m_{n}^{k-1}\\
1& 1 & \dots  & 1 \\
m_{1} & m_{2} & \dots  & m_{n} \\
\vdots &\vdots &\vdots &\vdots\\
m_{1}^{n-k-1} & m_{2} ^{n-k-1}& \dots  & m_{n}^{n-k-1}
\end{array}
\right).
\end{equation}
 We will be interested in computing the derivatives of $\det(\mathcal{M}(t))$ when $t=0$; it is therefore useful to think of $\det(\mathcal{M}(t))$ as the sum of various products of $k$ functions. To compute the $p^{th}$ derivative we can use the formula
$$\dfrac{d^{p}}{dt^{p}}\det(\mathcal{M}(t)) = \sum_{d_{1}+d_{2}+\dots+d_{k}=p}\left(\dfrac{p!}{ d_{1}! d_{2}!\dots d_{k}!}\right)\det(\mathcal{M}(R_{1}^{(d_{1})}, R_{2}^{(d_{2})},\dots,R_{k}^{(d_{k})})), $$
where $d_{i}\in \mathbb{N}\cup\{0\}$ and $\mathcal{M}(R_{1}^{(d_{1})}, R_{2}^{(d_{2})},\dots,R_{k}^{(d_{k})})$ is the matrix formed by taking $d_{i}$ derivatives of the $i^{th}$ row.\\
\\
It is clear that, evaluating at $t=0$, this formula is going to require the calculation of determinants of matrices of the form
$$
A =\left(
\begin{array}{cccc}

m_{1}^{e_{1}} & m_{2}^{e_{1}} & \dots & m_{n}^{e_{1}}\\
m_{1}^{e_{2}} & m_{2}^{e_{2}} & \dots & m_{n}^{e_{2}}\\
\vdots &\vdots &\vdots &\vdots\\
m_{1}^{e_{k}} & m_{2}^{e_{k}} & \dots & m_{n}^{e_{k}}\\
m_{1}^{n-k-1} & m_{2} ^{n-k-1}& \dots  & m_{n}^{n-k-1}\\
m_{1}^{n-k-2} & m_{2}^{n-k-2} & \dots  & m_{n}^{n-k-2} \\
\vdots &\vdots &\vdots &\vdots\\
m_{1} & m_{2} & \dots & m_{n}\\
1 & 1 & \dots & 1
\end{array}
\right),
$$
for exponents ${e_{1},e_{2},\dots,e_{k}\in \mathbb{N}}$. In fact, a matrix of the form $A$ that gives a non-zero determinant can be written (after possibly reordering rows) in the form
\begin{equation}\label{lambdamat}
A_{ij} = m_{j}^{\lambda_{i}+n-i},
\end{equation}
for some vector $\lambda = (\lambda_{1}, \lambda_{2},\dots, \lambda_{n}) \in\mathbb{Z}_{\geq 0}^{n}$ with $\lambda_{i}\geq \lambda_{i+1}$. Such determinants are all multiples of the Vandermonde determinant $V$ given by
$$ V = |m_{j}^{n-i}| = \prod_{1\leq i<j\leq n}(m_{i}-m_{j}),$$
which is the determinant of the $\lambda=(0,0,\dots,0)$ case of Equation (\ref{lambdamat}). This leads to the definition of the Schur polynomials.
\begin{definition}[Schur polynomial]
	Given $\lambda = (\lambda_{1}, \lambda_{2},\dots, \lambda_{n}) \in\mathbb{Z}_{\geq 0}^{n}$ with $\lambda_{i}\geq \lambda_{i+1}$ the Schur polynomial $S_{\lambda}$ is given by
	$$S_{\lambda}(m_{1},m_{2},\dots,m_{n}) = \frac{|m_{j}^{\lambda_{i}+n-i}|}{V}.$$
\end{definition}
The Schur polynomial $S_{\lambda}$ is a homogeneous, $\textrm{Sym}_{n}$-invariant multinomial of degree $\sum_{i=1}^{i=n}\lambda_{i}$.  It is straightforward to write a short list of these in degrees 0 to 3:
\begin{align*}
S_{(0,0,\dots,0)} &= 1,\\
S_{(1,0,\dots,0)} &= m_{1}+m_{2}+\dots+m_{n},\\
S_{(2,0,\dots,0)} &= \sum_{i=1}^{i=n}m_{i}^{2}+\sum_{1\leq i<j\leq n}m_{i}m_{j}, \\
S_{(1,1,0,\dots,0)} &= \sum_{1\leq i<j\leq n}m_{i}m_{j}, \\
S_{(3,0,0,\dots,0)} &=\sum_{i=1}^{i=n}m_{i}^{3}+\sum_{1\leq i\neq j\leq n}m_{i}m_{j}^{2}+\sum_{1\leq i<j<l\leq n}m_{i}m_{j}m_{l}, \\
S_{(2,1,0,\dots,0)} &= \sum_{1\leq i\neq j\leq n}m_{i}m_{j}^{2}+2\sum_{1\leq i<j<l\leq n}m_{i}m_{j}m_{l},\\
S_{(1,1,1,0,\dots,0)} &= \sum_{1\leq i<j<l\leq n}m_{i}m_{j}m_{l}.
\end{align*}

In order to keep track of signs we introduce the following function $\sigma:\mathbb{Z}\rightarrow \{0,1\}$ defined by
$$
\sigma(N) = \left\{\begin{array}{c}0 \ \mathrm{ if } \ N \equiv \ 0,1 \ (\mathrm{mod } \ 4),\\
1  \ \mathrm{ if } \ N\equiv \ 2,3 \ (\mathrm{mod } \ 4).
\end{array} \right.
$$
\begin{lemma}\label{AL1}
Let $k,n,$ and $\mathcal{M}(t)$ be as in Equation (\ref{DHMatrix}) and let $V$ be the Vandemonde determinant.  Then the power series expansion about $0$ of $\det(\mathcal{M}(t))$ begins
	$${\det(\mathcal{M}(t))} =  \dfrac{\varepsilon_{k,n}V}{(k(n-k))!}c_{0}t^{k(n-k)}+\dots,$$
	where
	$$
	\varepsilon_{k,n} =(-1)^{k(n-k)+\sigma(k)+\sigma(n-k)},
	$$
	and
	\begin{equation}\label{deggrass}
	c_{0} = (k(n-k))!\prod_{i=1}^{i=k}\dfrac{(i-1)!}{(n-k+i-1)!}.
	\end{equation}
\end{lemma}

\begin{proof}
	In order to get a non-zero determinant when $t=0$, we need the first $k$ rows to yield the powers ${m_{j}^{n-k}, m_{j}^{n-k+1},\dots,m_{j}^{n-1}}.$ Given that there are existing powers ${m_{i}^{0},m_{i}^{1},\dots m_{i}^{k-1}}$ we require
	$$\sum_{j=n-k}^{j=n-1}j - \sum_{j=0}^{j=k-1} = \frac{k(2n-k-1)}{2} - \frac{k(k-1)}{2} = k(n-k),$$
	derivatives. The $(k(n-k))^{th}$ derivative of $\det(M(t))$ evaluated at $t=0$ is then some multiple of the Vandemonde determinant $V$.\\
	\\
	For $1\leq i \leq k$ let ${l_{i} =  (n-k)+k-i = n-i}$. For each $\sigma \in \textrm{Sym}_{k}$, let the number of derivatives of the $\sigma(i)^{th}$ line be given by
	$$(l_{i}-\sigma(i)+1).$$
	(Hence the power of $m$ contributed by the $\sigma(i)^{th}$ row is $l_{i}$).
	Using the multinomial version of the Leibniz rule and weighting the resulting matrix by the sign of  the permutation $\sigma$ (needed to reorder the rows so the powers of $m_{i}$ run from $1$ to $n-1$) we obtain the $(k(n-k))^{th}$ derivative of $\det(\mathcal{M}(t))$ at $t=0$ is given by
	$$\varepsilon_{k,n}V\left(\sum_{\sigma \in \textrm{Sym}_{k}} \textrm{sgn}(\sigma)\dfrac{[k(n-k)]!}{(l_{1}-\sigma(1)+1)!(l_{2}-\sigma(2)+1)!\dots(l_{k}-\sigma(k)+1)!}\right).$$
	The result now follows from the discussion in \cite{FH} where the quantity inside the brackets is shown to compute the number of standard Young tableau of row structure
	$${(n-k,n-k,\dots,n-k)\in \mathbb{N}^{k}}.$$
	The formula for $c_{0}$ can be computed by the famous Hook Length formula (see Section 4.1 in \cite{FH}).  This gives the result.
\end{proof}
The formula for $c_{0}$ (up to factors of $\pi$) recovers the volume of $Gr_{k}(\mathbb{C}^{n})$ as first computed by Schubert \cite{GrHa}. We will see that $c_{0}$ is essentially the first term in the expansion of the Duistermaat--Heckman integral (\ref{DHGroup}) which indeed should be the volume of the manifold computed with respect to the symplectic form. \\
\\
In order to compute the stability integral (\ref{KKstab}), we will require the coefficient of $t^{k(n-k)+3}$ in the power series expansion about $0$ of $\det(\mathcal{M}(t))$.  After factoring out $V$, this coefficient will be a combination of the Schur polynomials $S_{(3,0,\dots,0)}$, $S_{(2,1,0\dots,0)}$, and $S_{(1,1,1,0\dots,0)}$. The coefficient of each polynomial can be computed in terms constant $c_{0}$ given by Equation (\ref{deggrass}).
\begin{lemma}\label{AL2}
	Let $\dfrac{c_{3}}{(k(n-k)+3)!}$ be the coefficient of $t^{k(n-k)+3}$ in the power series expansion about $0$ of $\det(\mathcal{M}(t))$ and let $\varepsilon_{k,n}$ be as in Lemma \ref{AL1}. Then
	$$c_{3} = \varepsilon_{k,n}V(c_{(3,0,0)}S_{(3,0,\dots,0)}+c_{(2,1,0)}S_{(2,1,0\dots,0)}+c_{(1,1,1)}S_{(1,1,1,0\dots,0)}),$$
	where
	\begin{eqnarray}
	c_{(3,0,0)} & = \dfrac{[k(n-k)+1][k(n-k)+2][k(n-k)+3]}{6}\dfrac{k(k+1)(k+2)}{n(n+1)(n+2)}c_{0},\\
	c_{(2,1,0)} & = \dfrac{[k(n-k)+1][k(n-k)+2][k(n-k)+3]}{3}\dfrac{(k-1)k(k+1)}{(n-1)n(n+1)}c_{0},\\
	c_{(1,1,1)} & = \dfrac{[k(n-k)+1][k(n-k)+2][k(n-k)+3]}{6}\dfrac{(k-2)(k-1)k}{(n-2)(n-1)n}c_{0}.
	\end{eqnarray}

\end{lemma}
\begin{proof}
	Let ${\chi_{1} = (3,0,\dots,0)},$ ${\chi_{2} = (2,1,0,\dots,0)},$ and ${\chi_{3} = (1,1,1,0,\dots,0) \in \mathbb{Z}^{k}}$. For ${i=1,2}$ or $3$, let $\upsilon = (n-k,n-k,\dots,n-k)+\chi_{i}$. Furthermore, for ${j=1,2,\dots,k}$ let
	$$l_{j} = (\upsilon_{j}+k-j).$$
	Let row $\sigma(j)$ have $(l_{j}-\sigma(j)+1)$ derivatives so, as in the proof of Lemma \ref{deggrass}, the power of the $m$s in row $\sigma(j)$ is $l_{j}$. For a fixed element $\sigma \in \textrm{Sym}_{k}$, computing the derivative of $\det(\mathcal{M}(t))$ at $0$ and after dividing through by $V$, the distribution of the powers shows we get (abusing notation slightly) the Schur polynomial $S_{\chi_{i}}$.\\
	\\
	Using the multinomial version of the Leibniz rule where we sum only over the distribution of derivatives that yield powers of $m$ running  ${1,2,\dots,n+2}$ and weighting the resulting matrix by the sign of the permutation $\sigma$ we obtain that this part of the $(k(n-k)+3)^{rd}$ derivative of $\det(\mathcal{M}(t))/V$ at $t=0$ (and hence the coefficient of $S_{\chi_{i}}$) is given by
	$$\varepsilon_{k,n}V\left(\sum_{\sigma \in \textrm{Sym}_{k}} \textrm{sgn}(\sigma)\dfrac{[k(n-k)+3]!}{(l_{1}-\sigma(1)+1)!(l_{2}-\sigma(2)+1)!\dots(l_{k}-\sigma(k)+1)!}\right).$$
	Again, the discussion in \cite{FH} shows that the quantity inside the brackets computes the number of standard Young tableau of row structure $(n-k,n-k,\dots,n-k)+\chi_{i}$. The result follows from the Hook Length formula.
\end{proof}

\section{Proof of Theorem \ref{ThmGr}} \label{sec:5}
The Lie algebra of $SU(n)$, $\mathfrak{su}(n)$, is identified with trace-free $n\times n$  skew-Hermitian matrices. The rank of $SU(n)$ is $n-1$ with a maximal torus $T$ being given by diagonal matrices. Hence the Lie algebra of $T$, $\mathfrak{t}$, can be identified with
$$ \textrm{Diag}(\sqrt{-1}\mu_{1},\sqrt{-1}\mu_{2},\dots,\sqrt{-1}\mu_{n}) \textrm{ with } \sum_{i}\mu_{i}=0.$$
The roots can identified with $e_{j}-e_{l}$ for $j\neq l$ where $e_{j}$ is the diagonal matrix with the entry $\sqrt{-1}$ in the $j^{th}$ coefficient and we are using the inner product $\langle X,Y\rangle = \tr(X^{\ast}Y)$ to identify $\mathfrak{su}(n)$ and $\mathfrak{su}^{\ast}(n)$. The Weyl group $W\cong \textrm{Sym}_{n}$ acts on $T$ by permuting the elements of the diagonal and $W$ acts on $\mathfrak{t}$ permuting the $\sqrt{-1}\mu_{i}$.\\
\\
The adjoint orbits we consider can be represented by an element $\xi \in \mathfrak{t}$. We let
 $$\xi = \textrm{Diag}(\underbrace{\sqrt{-1}\mu_{1},\dots,\sqrt{-1}\mu_{1}}_{k\text{ entries}},\underbrace{\sqrt{-1}\mu_{2}, \sqrt{-1}\mu_{2},\dots,\sqrt{-1}\mu_{2}}_{n-k\text{ entries}})$$
where $\mu_{1}>0$ and ${k\mu_{1}+(n-k)\mu_{2}=0}$. In fact, it will be useful to fix
$$\mu_{1} = \frac{n-k}{n} \textrm{ and } \mu_{2} = -\frac{k}{n},$$
so that $\mu_{1}-\mu_{2}=1.$ We get an identification of $\mathcal{O}_{\xi}$ with $Gr_{k}(\mathbb{C}^{n})$ by considering the $k$-plane generated by the span of the $\sqrt{-1}\mu_{1}$ eigenspace at each point in the orbit.\\
\\
The vector $D\in \mathfrak{t}$ given by  $D=\textrm{Diag}(2\pi \sqrt{-1}m_{1}, 2\pi \sqrt{-1}m_{2},\dots,2\pi \sqrt{-1}m_{n})$, where  ${m_{j}\in \mathbb{Z}}$, $m_{j}\neq m_{l}$ for $j\neq l$, and $\sum_{j}m_{j}=0$, generates a circle action on $\mathcal{O}_{\xi}$.  Recall the function $f_{D}:\mathcal{O}_{\xi}\rightarrow\mathbb{R}$ defined by Equation (\ref{ef_def}). By Lemma \ref{eflem}, $f_{D}$ is an eigenfunction of the Laplacian and  by Lemma \ref{Bottlem}, the fixed points of the circle action (or equivalently the critical points of $f_{D}$) are the orbit of $\xi$ under the Weyl group $\textrm{Sym}_{n}$ which are the vectors consisting of to the $^{n}C_{k}$ possible placements of the $\mu_{1}$s. If we index a fixed point of the circle action by the $k$-element set ${J\subset \{1,2,\dots,n\}}$ corresponding to this placement and denoting such a fixed point $q_{J}$, then the value of the function $f_{D}$ at this point is
$$f_{D}(q_{J}) = 2\pi\left(\mu_{1}\sum_{j\in J}m_{j}+\mu_{2}\sum_{j\in J^{c}}m_{j}\right).$$
The set of roots $\alpha \in \Lambda_{R}$ such that $\langle \alpha,q_{J}\rangle>0$ are $e_{j_{1}}-e_{j_{2}}$ where $j_{1}\in J$ and $j_{2}\in J^{c}$.  Hence the weight of the induced action on the holomorphic tangent space at $q_{J}$ is
$$\varpi(q_{J}) = \prod_{j\in J, \ l\in J^{c}}(m_{j}-m_{l}).$$
The Duistermaat--Heckman formula yields
$$\int_{\mathcal{O}_{\xi}}e^{-tf_{D}(p)}\omega^{n(n-k)}(p) =  \frac{[k(n-k)]!}{t^{n(n-k)}}\sum_{J\subset \{1,2,\dots,n\} \ : \ |J|=k}\frac{e^{-\left(\mu_{1}\sum_{j\in J}m_{j}+\mu_{2}\sum_{j\in J^{c}}m_{j}\right)2\pi t}}{\prod_{j\in J, \ l\in J^{c}}(m_{j}-m_{l})}.$$
We manipulate this expression by pulling out the Vandemonde factor
$$V=\prod_{1\leq j<l \leq n}(m_{j}-m_{l}),$$ yielding
$$\frac{[k(n-k)]!}{Vt^{k(n-k)}}\sum_{J\subset \{1,2,\dots,n\}: \ |J|=k}\varepsilon_{J}e^{-\left(\mu_{1}\sum_{j\in J}m_{j}+\mu_{2}\sum_{j\in J^{c}}m_{j}\right)2\pi t}\prod_{j<l\in J}(m_{j}-m_{l})\prod_{j<l\in J^{c}}(m_{j}-m_{l}),$$
where $\varepsilon_{J}$ is the sign of the permutation sending $1,\dots,k$ to the sequence ${j_{1}<j_{2}<\dots<j_{k} \in J}$ and $k+1,k+2,\dots n$ to the elements of $J^{c}$.\\
\\
The sum is the (Laplace) expansion of the determinant of the following matrix
$$ M(t) = \left(
\begin{array}{cccc}
e^{-\mu_{1}m_{1}t} & e^{-\mu_{1}m_{2}t} & \dots  & e^{-\mu_{1}m_{n}t} \\
e^{-\mu_{1}m_{1}t}m_{1} & e^{-\mu_{1}m_{2}t}m_{2} & \dots  & e^{-\mu_{1}m_{n}t}m_{n}\\
\vdots &\vdots &\vdots &\vdots\\
e^{-\mu_{1}m_{1}t}m_{1}^{k-1} & e^{-\mu_{1}m_{2}t}m_{2}^{k-1} & \dots  & e^{-\mu_{1}m_{n}t}m_{n}^{k-1}\\
e^{-\mu_{2}m_{1}t} & e^{-\mu_{2}m_{2}t} & \dots  & e^{-\mu_{2}m_{n}t} \\
e^{-\mu_{2}m_{1}t}m_{1} & e^{-\mu_{2}m_{2}t}m_{2} & \dots  & e^{-\mu_{2}m_{n}t}m_{n} \\
\vdots &\vdots &\vdots &\vdots\\
e^{-\mu_{2}m_{1}t}m_{1}^{n-k-1} & e^{-\mu_{2}m_{2}t}m_{2} ^{n-k-1}& \dots  & e^{-\mu_{2}m_{n}t}m_{n}^{n-k-1}
\end{array}
\right).$$
Hence, we have
\begin{equation}\label{DHeqDetM}
\int_{\mathcal{O}_{\xi}}e^{-tf_{D}(p)}\omega^{k(n-k)}(p) =\frac{[k(n-k)]!}{Vt^{k(n-k)}}\det(M(2\pi t)) = \varepsilon_{k,n}\frac{[k(n-k)]!}{Vt^{k(n-k)}}\det(\mathcal{M}(2\pi t)),
\end{equation}
where $\mathcal{M}(t)$ is the matrix (\ref{DHMatrix}) and $\varepsilon_{k,n}$ is the signed quantity from Lemma \ref{AL1}. The final equality in Equation \ref{DHeqDetM} follows as  $\sum_{j}m_{j}=0$ and $\mu_{1}-\mu_{2}=1$. \\
\\
We remark again, in order to be able to apply Kr\"oncke's test we need $f_{D}$ to be a genuine eigenfunction of the Laplacian. This means we must normalise $f_{D}$ so that it has mean value zero or, equivalently, so that the first derivative of
$$\int_{\mathcal{O}_{\xi}}e^{-tf_{D}(p)}\omega^{k(n-k)}(p)$$
vanishes when $t=0$. The $t^{k(n-k)+1}$ coefficient of the power series expansion of $\mathcal{M}(t)$ is a multiple of $S_{(1,0,\dots,0)} = \sum_{j}m_{j}$ which vanishes and hence $f_{D}$ is an eigenfunction. As mentioned after the proof of Theorem \ref{ThmCST} in Section \ref{sec:3}, the function $f_{D}$ does not really need normalising as there are no homogenous, degree 1, $\mathrm{Sym}_{n}$-invariant polynomials by considering the $A_{n}$ part of the Chevalley--Shephard--Todd theorem.\\
\\
To prove the theorem we need to show that the third derivative of the integral expansion does not vanish. By Lemma \ref{AL2}, the third derivative of the expansion at $0$ will be a combination of the Schur polynomials $S_{(3,0,\dots,0)}$, $S_{(2,1,0,\dots,0)}$, and $S_{(1,1,1,0,\dots,0)}$. This yields a symmetric, homogeneous degree 3 multinomial in the variables ${m_{1},m_{2},\dots,m_{n}}$ which we denote $\mathcal{P}_{3}(m_{1},\dots m_{n})$.  The fact that the $c_{(3,0,0)}$ coefficient in Lemma \ref{AL2} is non-zero means that the coefficient  of the $m_{i}^{3}$ in $\mathcal{P}_{3}$ (which is $c_{(3,0,0)}$) is non-zero. Hence the stability integral (\ref{KKstab}) will not be identically zero if we can show that $\mathcal{P}_{3}$ is not a multiple of $S_{(1,0,\dots,0)}$.\\
\\
Using the notation of Lemma \ref{AL2}, suppose that
$$c_{3} = S_{(1,0,\dots,0)}(\alpha S_{(2,0,\dots,0)}+\beta S_{(1,1,\dots,0)}),$$
for some $\alpha,\beta \in \mathbb{R}$. Then by equating coefficients of $m_{i}^{3}, m_{i}^{2}m_{j},$ and $m_{i}m_{j}m_{l}$ we obtain
\begin{eqnarray*}
c_{(3,0,0)} = &  \alpha,\\
c_{(3,0,0)}+c_{(2,1,0)} = &  2\alpha+\beta,\\
c_{(3,0,0)}+2c_{(2,1,0)}+c_{(1,1,1)} = & 3(\alpha + \beta).
\end{eqnarray*}
This system is consistent if
$$c_{(3,0,0)}-c_{(2,1,0)}+c_{(1,1,1)}=0.$$
Evaluating using Lemma \ref{AL2} this yields
\begin{equation}\label{Efinal}
k(n-k)(2k-n) = 0.
\end{equation}
Hence we see that the third derivative vanishes identically if and only if $n=2k$. The if part is easy to see in fact as, in the case $2k=n$, $\mu_{1}=-\mu_{2}$ and so the determinant of the matrix $M(t)$ is clearly an even function of $t$ if $k$ is even and an odd function if $k$ is odd.  This finishes the proof of Theorem \ref{ThmGr}.

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\maketitle

\begin{abstract}
\paragraph{Abstract.}

We look at the number of permutations $\beta$ of $[N]$ with $m$ cycles such that $(1\ 2\ \ldots\ N) \beta^{-1}$ is a long cycle. These numbers appear as coefficients of linear monomials in Kerov's and Stanley's character polynomials. D. Zagier, using algebraic methods, found an unexpected connection with Stirling numbers of size $N+1$. We present the first combinatorial proof of his result, introducing a new bijection between partitioned maps and thorn trees. Moreover, we obtain a finer result, which takes the type of the permutations into account.

\paragraph{R\'esum\'e.}

Nous \'etudions le nombre de permutations $\beta$ de $[N]$ avec $m$ cycles telles que $(1\ 2\ \ldots\ N) \beta^{-1}$ a un seul cycle. Ces nombres apparaissent en tant que coefficients des mon\^omes lin\'eaires des polyn\^omes de Kerov et de Stanley. \`A l'aide de m\'ethodes alg\'ebriques, D. Zagier a trouv\'e une connexion inattendue avec les nombres de Stirling de taille N+1. Nous pr\'esentons ici la premi\`ere preuve combinatoire de son r\'esultat, en introduisant une nouvelle bijection entre des cartes partitionn\'ees et des arbres \'epineux. De plus, nous obtenons un r\'esultat plus fin, prenant en compte le type des permutations.
\end{abstract}

\section{Introduction}
The question of the number of factorizations of the long cycle $(1\ 2\ \ldots\ N)$ into two permutations with given number of cycles has already been studied via algebraic or combinatorial\footnote{It can be reformulated in terms of unicellular bicolored maps with given number of vertices, see paragraph \ref{subsect:partitioned_maps}.} methods (see \cite{
Adrianov1998serie_carte_bicolore, SchaefferVassilieva:factorizations_long_cycle}). In these papers, the authors obtain nice generating series for these numbers. Note that the combinatorial approach has been refined to state a result on the number of factorizations of the long cycle $(1\ 2\ \ldots\ N)$ in two permutations with given type (see \cite{MoralesVassilieva:factorizations_long_cycle}).\\
Unfortunately, in all these results, extracting one coefficient of the generating series gives complicated formulas. The case where one of the two factors has to be also a long cycle is particularly interesting. Indeed, the number $B'(N,m)$ of permutations $\beta$ of $[N]$ with $m$ cycles, such that $(1\ 2\ \ldots\ N) \beta^{-1}$ is a long cycle, is known to be the coefficient of some linear monomial in Kerov's and Stanley's character polynomials (see \cite[Theorem 6.1]{Biane2003Kerov} and \cite{Stanley2003rectangles,Feray:stanley_formula}). These polynomials express the character value of the irreducible representation of the symmetric group indexed by a Young diagram $\lambda$ on a cycle of fixed length in terms of some coordinates of $\lambda$.\\
A very simple formula for these numbers was found by D. Zagier \cite[Application 3]{ZagierCycles}
(see also \cite[Corollary 3.3]{Stanley2009product_cycles}):
\begin{Th}[Zagier, 1995]\label{th:stanley}
Let $m \leq N$ be two positive integers such that $m \equiv N [2]$. Then
\begin{equation}\label{eq:stanley}
 \frac{N(N+1)}{2} B'(N,m) = s(N+1,m),
\end{equation}
where $s(N+1,m)$ is the unsigned Stirling number of the first kind.
\end{Th}
It is well-known that $s(N+1,m)$ is the number of permutations of $[N+1]$ with $m$ cycles. As former proofs of this result are purely algebraic, R. Stanley \cite{Stanley2009product_cycles} asked for a combinatorial proof of Theorem \ref{th:stanley}. This paper presents the first bijective approach proving this formula.

We even prove a finer result, which takes the type\footnote{The type of a permutation is the sequence of the lengths of its cycles, sorted in increasing order.} of the permutations into consideration and not only their number of cycles. To state it, we need to introduce a few notations. First, we refine the numbers $s(N+1,m)$ and $B'(N,m)$: if $\lambda \vdash n$ (\textit{i.e.} $\lambda$ is a partition of $n$), let $A(\lambda)$ (resp. $B(\lambda)$) be the number of permutations $\beta \in S_n$ of type $\lambda$ (resp. with the additional condition that $(1\ 2\ \ldots\ N) \beta^{-1}$ is a long cycle). Then, as Stanley's result deals with permutations of $[N]$ and $[N+1]$, we need operators on partitions which modify their size, but not their length. If $\mu$ (resp. $\lambda$) has at least one part $i+1$ (resp. $i$), let $\mu^{\downarrow (i+1)}$ (resp. $\lambda^{\uparrow (i)}$) be the partition obtained from $\mu$ (resp. $\lambda$) by erasing a part $i+1$ (resp. $i$) and adding a part $i$ (resp. $i+1$). For instance, using exponential notations (see \cite[chapter 1, section 1]{Macdo}), $(1^2 3^1 4^2)^{\downarrow (4)}=1^2 3^2 4^1$ and $(2^2 3^2 4)^{\uparrow (2)}=2^1 3^3 4^1$.\\
Now we can state our main theorem, which implies immediately Theorem \ref{th:stanley}:
\begin{Th}[Main result]\label{th:main}
 For each partition $\mu \vdash N+1$ of length $p$ with $p \equiv N [2]$, one has:
\begin{equation}
\frac{N+1}{2} \sum_{\lambda = \mu^{\downarrow (i+1)}, i>0}  i\ m_i(\lambda)\ B(\pnv\lambda) = A(\mu) = \frac{(N+1)!}{z_\mu},
\end{equation}
where $m_i(\lambda)$ is the number of parts $i$ in $\lambda$ and $z_\mu=\prod_i i^{m_i(\mu)} m_i(\mu)!$.
\end{Th}
To be comprehensive on the subject, we mention that G. Boccara found an integral formula for $B(\lambda)$ (see \cite{Boccara1980produit2cycles}), but there does not seem to be any direct link with our result.\\
As in paper \cite{MoralesVassilieva:factorizations_long_cycle}, the first step (section \ref{sect:reformulation}) of our proof of Theorem \ref{th:main} consists in a change of basis in the ring of symmetric functions in order to show the equivalence with the following statement:
\begin{Th}\label{th:reformulation}
 Let $\lambda$ be a partition of $N$ of length $p$. Choose randomly (with uniform probability) a set-partition $\pi$ of $\{1,\ldots,N\}$ of type $\lambda$ and then (again with uniform probability) a permutation $\beta$ in $S_\pi$ (that means that each cycle of $\beta$ is contained in a block of $\pi$). Then the probability for $(1\ 2\ \ldots\ n) \beta^{-1}$ to be a long cycle is exactly $1/(N-p+1)$.
\end{Th}
Once again, such a simple formula is surprising.  We give a bijective proof in sections \ref{sect:def_psi}, \ref{sect:inj_psi} and \ref{sect:im_psi}.

\begin{Rem}\label{rem:syst_inv}
Theorem \ref{th:main}, written for all $\mu \vdash N+1$, gives the collection of numbers $B(\pnv\lambda)$ as solution of a sparse triangular system. Indeed, if we endow the set of partitions of $N$ with the lexicographic order, Theorem \ref{th:main}, written for $\mu=\lambda^{\uparrow (\lambda_1)}$, gives $B(\pnv\lambda)$ in terms of the quantities $A(\mu)$ and $B(\pnv\nu)$ with $\nu > \lambda$.
\end{Rem}

\section{Link between Theorems \ref{th:main} and \ref{th:reformulation}}\label{sect:reformulation}

\subsection{Black-partitioned maps}\label{subsect:partitioned_maps}
By definition, a map is a graph drawn on a two-dimensional oriented surface (up to deformation), i.e. a graph with a cyclic order on the incident edges to each vertex.\\

As usual, a couple of permutations $(\alpha,\beta)$ in $S_N$ can be represented as a bicolored map with $N$ edges labeled with integers from $1$ to $N$. In this identification, $\alpha(i)$ (resp. $\beta(i)$) is the edge following $i$ when turning around its white (resp. black) extremity. White (resp. black) vertices correspond to cycles of $\alpha$ (resp. $\beta$). The condition $\alpha \cdot \beta = (1\ 2\ \ldots\ N)$ (which we will assume from now on) means that the map is unicellular (i.e. if we remove the edges of the maps from the oriented surface, the resulting surface is homeomorphic to an open disc) and that the positions of the labels are determined by the choice of the edge labeled by $1$ (which can be seen as a \emph{root}). In this case, the couple of permutations is entirely determined by $\beta$.\\

Therefore, if $\lambda \vdash N$, the quantity $A(\lambda)$ (resp. $B(\lambda)$) is the number of rooted unicellular maps (resp. \emph{star} maps, that means that we make the additional assumption that the map has only one white vertex) with black vertices' degree distribution $\lambda$.\\
As in the papers \cite{SchaefferVassilieva:factorizations_long_cycle} and \cite{MoralesVassilieva:factorizations_long_cycle}, our combinatorial construction deals with maps with additional information:

\begin{Def}
 A black-partitioned (rooted unicellular) map is a rooted unicellular map with a set partition $\pi$ of its black vertices. We call degree of a block of $\, \pi$ the sum of the degrees of the vertices in $\pi$. The \emph{type} of a black-partitioned map is its blocks' degree distribution.
\end{Def}

In terms of permutations, a black-partitioned map consists in a couple $(\alpha,\beta)$ in $S_N$ with the condition $\alpha \, \beta=(1\ 2\ \ldots\ N)$ and a set partition $\pi$ of $\{1,\ldots,N\}$ coarser than the set partition in orbits under the action of $\beta$ (in other words, if $i$ and $j$ lie in the same cycle of $\beta$, they must be in the same part of $\pi$).

\begin{example}
\label{exspm}
 Let $\beta = (1)(2 5)(3 7)(4)(6)$, $\alpha = (1 2 3 4 5 6 7) \beta^{-1} = (1 2 6 7 4 5 3)$,  and $\pi$ be the partition $\left\{ \{1, 3, 6, 7\};\{2, 5\};\{4\}\right\}$.
Associating the triangle, circle and square shape to the blocks, $(\beta, \pi)$ is the black-partitioned star map pictured on figure \ref{spm}.
\begin{figure}[h]
\label{spm}
\begin{center}
\includegraphics[width=50mm]{SemiPartMap}
\end{center}
\caption{The black-partitioned map defined in example \ref{exspm}}
\end{figure}
\end{example}

We denote by $C(\lambda)$ (resp. $D(\lambda)$) the number of black-partitioned maps (resp. black-partitioned star maps) of partition type $\lambda$. Equivalently, $C(\lambda)$ (resp. $D(\lambda)$) is the number of couples $(\beta,\pi)$ as above such that $\pi$ is a partition of type $\lambda$ (resp. and $(1\ 2\ \ldots\ N) \beta^{-1}$ is a long cycle). Quantities $A$ and $C$ (resp. $B$ and $D$) are linked by the following equations (whose proofs are identical to the one of \cite[Proposition 1]{MoralesVassilieva:factorizations_long_cycle}, see also \cite[Chapter 1, equation (6.9)]{Macdo})
\begin{align}
\label{eq:C2A} \sum_{\mu \vdash N+1} C(\pnuv\mu) \Aut(\mu) m_\mu & = \sum_{\nu \vdash N+1} A(\nu) p_\nu;\\
\label{eq:D2B} \sum_{\lambda \vdash N} D(\pnv\lambda) \Aut(\lambda) m_\lambda & = \sum_{\pi \vdash N} B(\pnv\pi) p_\pi,
\end{align}
where $m_\bullet$ and $p_\bullet$ denote the monomial and power sum basis of the ring of symmetric functions.

\subsection{Permuted star thorn trees and Morales'-Vassilieva's bijection}\label{subsect:thorn_trees}

The main tool of this article is to encode black-partitioned maps into star thorn trees, which have a very simple combinatorial structure. Note that they are a particular case of the notion of thorn trees, introduced by A. Morales and the second author in \cite{MoralesVassilieva:factorizations_long_cycle}.

\begin{Def}[star thorn tree]

An \emph{(ordered rooted bicolored) star thorn tree} of size $N$ is a tree with a white root vertex, $p$ black vertices and $N - p$ thorns connected to the white vertex (the order in which they are connected matters) and $N - p$ thorns connected to the black vertices. A thorn is an edge connected to only one vertex.

We call \emph{type} of a star thorn tree its black vertices' degree distribution (taking the thorns into account).

If $\mu$ is an integer partition, we denote by $\BT(\mu)$ the number of star thorn trees of type $\mu$.
\end{Def}

An example is given on Figure \ref{fig:ex_arbre_permute} (for the moment, please do not pay attention to the labels). The interest of this object lies in the following theorem, which corresponds to the case $\lambda=(N)$ of \cite[Theorem 2]{MoralesVassilieva:factorizations_long_cycle} (note that the proof is entirely bijective).

\begin{Th}\label{th:bij_MV}
Let $\mu \vdash N$ be a partition of length $p$. One has:
\begin{equation}\label{eq:bij_MV}
C(\mu) = (N-p)! \cdot \BT(\mu).
\end{equation}
\end{Th}

The right-hand side of \eqref{eq:bij_MV} is the number of couples $(\tau,\sigma)$ where:
\begin{itemize}
 \item $\tau$ is a star thorn tree of type $\mu$.
 \item $\sigma$ is a permutation of $[N - p]$, which happens to be exactly the number of thorns with a white (resp. black) extremity in $\tau$. So $\sigma$ may be seen as a bijection between the thorns with a white extremity and thorns with a black extremity.
\end{itemize}
We call such a couple a permuted (star) thorn tree. By definition, the type of $(\tau,\sigma)$ is the type of $\tau$. Examples of graphical representations are given on Figure \ref{fig:ex_arbre_permute}: we put symbols on edges and thorns with the following rule. Two thorns have the same label if they are associated by $\sigma$ and except from that rule, all labels are different (the chosen symbols and their order do not matter, we call that a symbolic labeling).\\

\begin{figure}[ht]
$$\includegraphics[width=30mm]{taubarsymbols} \hspace{2cm}
\includegraphics[width=3cm]{taubarpart5-0}$$
\caption{Example of two permuted star thorn trees $\ex{1}$ of type $1^1 2^1$ and $\ex{2}$ of type $2^2 3^2$}
\label{fig:ex_arbre_permute}
\end{figure}

It is easy to transform a permuted thorn tree $(\tau,\sigma)$ where $\tau$ has type $\lambda \vdash N$ to a permuted thorn tree $(\tau',\sigma')$ where $\tau'$ has type $\mu=\lambda^{\uparrow (i)}$. We just add a thorn anywhere on the white vertex ($N+1$ possible places) and a thorn anywhere on a black vertex of degree $i$ (there are $i$ possible places on each of the $m_i(\lambda)$ black vertices of degree $i$). Then we choose $\sigma'$ to be the extension of $\sigma$ associating the two new thorns. This procedure is invertible if we remember which thorn of black extremity is the new one (it must be on a black vertex of degree $i+1$, so there are $i \cdot m_{i+1}(\mu)$ choices). This leads immediately to the following relation:
\begin{equation}\label{eq:rec_BT}
\BT(\pnuv\mu) \cdot (N+1-p)! \cdot i \cdot m_{i+1}(\mu) = (N+1) \cdot i \cdot m_i(\lambda) \cdot \BT(\pnv\lambda) \cdot (N-p)!.
\end{equation}

\subsection{Reduction of the main theorem}

\begin{Prop}\label{prop:reformulation}
 For any partition $\lambda \vdash N$ of length $p$, one has:
$$D(\pnv\lambda) = \frac{1}{N-p+1} (N-p)! \BT(\pnv\lambda).$$
\end{Prop}

Sections \ref{sect:def_psi}, \ref{sect:inj_psi} and \ref{sect:im_psi} are devoted to the proof. It is easy to see, with the definition of subsection \ref{subsect:partitioned_maps} and the bijection of subsection \ref{subsect:thorn_trees}, that this proposition is a reformulation of Theorem \ref{th:reformulation}.

\begin{Lem}\label{lem:reduction}
 Proposition \ref{prop:reformulation} implies Theorem \ref{th:main}.
\end{Lem}

\begin{proof}
We fix a partition $\mu \vdash N+1$ of length $p<N+1$ and sum equation \eqref{eq:rec_BT} on $\lambda=\mu^{\downarrow (i+1)}$:
$$\BT(\pnuv\mu) \cdot (N+1-p)! \cdot (N+1-p) = (N+1) \sum_{\lambda = \mu^{\downarrow (i+1)}, i>0} i \cdot m_i(\lambda) \cdot \BT(\pnv\lambda) \cdot (N-p)!.$$
Using Morales'-Vassilieva's bijection and Proposition \ref{prop:reformulation}, this equality becomes:
\begin{align*}
C(\pnuv\mu) \cdot (N+1-p) & = (N+1) \sum_{\lambda = \mu^{\downarrow (i+1)}, i>0} i \cdot m_i(\lambda) \cdot D(\pnv\lambda) \cdot (N+1-p).\\
\text{Hence, }\sum_{\mu \vdash N+1 \atop \mu \neq 1^{(N+1)}} C(\pnuv\mu) \Aut(\mu) m_\mu & = (N+1) \sum_{\mu \vdash N+1 \atop \mu \neq 1^{(N+1)}} \sum_{i>0 \atop \lambda = \mu^{\downarrow (i+1)}} i \cdot m_i(\lambda) \Aut(\mu) D(\pnv\lambda) m_\mu ;\\
& = (N+1) \sum_{\lambda \vdash N} \Aut(\lambda) D(\pnv\lambda) \left(\sum_{i>0 \atop \mu = \lambda^{\uparrow (i)}} i \cdot m_{i+1}(\mu) m_\mu \right).
\end{align*}
The last equality has been obtained by changing the order of summation and using the trivial fact that, if $\mu = \lambda^{\uparrow (i)}$, one has $\Aut(\mu) \cdot m_i(\lambda)= \Aut(\lambda) \cdot m_{i+1}(\mu)$. Now, observing that the expression in the bracket can be written $\Delta (m_\lambda)$, where $\Delta$ is the differential operator $\sum_i x_i^2 \di$, one has:
$$\sum_{\mu \vdash N+1} C(\pnuv\mu) \Aut(\mu) m_\mu - (N+1)! m_{1^{N+1}} = (N+1) \cdot \Delta \left( \sum_{\lambda \vdash N} \Aut(\lambda) D(\pnv\lambda) m_\lambda \right).$$
Let us rewrite this equality in the power sum basis. The expansion of the two summations in this basis are given by equations \eqref{eq:C2A} and \eqref{eq:D2B}. Furthermore, one has:
$\Delta(p_\pi) = \sum_i i \cdot m_i(\pi) p_{\pi^{\uparrow (i)}}$.
Indeed, the one-part case ($\Delta(p_k)=k \cdot p_{k+1}$) is trivial and the general case follows because $\Delta$ is a derivation. We also need the power-sum expansion of $(N+1)! m_{1^{N+1}}$, which is (see \cite[Chapter I, equation (2.14')]{Macdo}):
$$(N+1)! m_{1^{N+1}} = (N+1)! e_{N+1} = (N+1)! \sum_{\nu \vdash N+1} \frac{(-1)^{N+1-\ell(\nu)}}{z_\nu} p_\nu = \sum_{\nu \vdash N+1} A(\nu) (-1)^{N+1-\ell(\nu)} p_\nu,$$
where $e_{N+1}$ is the $N+1$-th elementary function. Putting everything together, we have:
$$\sum_{\nu \vdash N+1} A(\nu) p_\nu + \sum_{\nu \vdash N+1} A(\nu) (-1)^{N-\ell(\nu)} p_\nu = (N+1) \sum_\pi B(\pnv\pi) \sum_{\rho = \pi^{\uparrow (i)},i>0} i \cdot m_i(\pi) p_\rho.$$
If we identify the coefficients of $p_\mu$ in both sides, we obtain exactly Theorem \ref{th:main}
\end{proof}

\begin{Rem}
 Using Remark \ref{rem:syst_inv}, the converse statement of Lemma \ref{lem:reduction} can be proved the same way.
\end{Rem}

\section{Mapping black-partitioned star maps to permuted thorn trees}\label{sect:def_psi}

The following sections provide a combinatorial proof of Proposition \ref{prop:reformulation}. We proceed in a three step fashion. First we define a mapping $\Psi$ from the set of black-partitioned star maps of type $\lambda$ (counted by $D(\lambda)$) to a set of permuted star thorn trees of the same type and show it is injective. As a final step, we compute the cardinality of the image set of $\Psi$ and show it is exactly $\left (1/(N-p+1)\right)(N-p)! \BT(\pnv\lambda)$.\\

Although there are some related ideas, $\Psi$ is not the restriction of the bijection of paper \cite{MoralesVassilieva:factorizations_long_cycle}.

\subsection{Labeled thorn tree}
Let $(\beta,\pi)$ be a black-partitioned star map. We first construct a labeled star thorn tree $\overline{\tau}$:
\begin{enumerate}[(i)]

\item \label{item:root_place} Let $(\alpha_k)_{(1\leq k \leq N)}$ be integers such that $\alpha_1 = 1$ and the long cycle $\alpha=(1\ 2\ \ldots\ N) \beta^{-1}$ is equal to $(\alpha_1 \alpha_2 \alpha_3 \dots \alpha_N)$. The root of $\overline{\tau}$ is a white vertex with $N$ descending edges labeled from right to left with $\alpha_1,  \alpha_2,  \alpha_3,  \dots, \alpha_N$ ($\alpha_1$ is the rightmost descending edge and $\alpha_N$ the leftmost).

\item \label{item:edge_root} Let $m_i$ be the maximum element of block $\pi_i$. For $k=1\ldots N$, if $\alpha_k = \beta(m_i)$ for some $i$, we draw a black vertex at the other end of the descending edge labeled with $\alpha_k$. Otherwise the descending edge is a thorn.
\begin{Rem}\label{rem:leftmost_edge}
As $\alpha_N=\alpha^{-1}(1) = \beta(N)$ the leftmost descending edge isn't a thorn and is labeled with $\beta(N)$ ($N$ is necessarily the maximum element of the block containing it).
\end{Rem}

\item \label{item:black_thorns} For $i = 1\dots p$, let ${(\beta^{u}_{1} \ldots \beta^{u}_{l_u})}_{1\leq u \leq c}$ be the $c$ cycles included in block $\pi_i$ such that $\beta^{u}_{l_u}$ is the maximum element of cycle $u$. (We have $\Sigma_{u} l_u = \mid\pi_i\mid$). We also order these cycles according to their maximum, \textit{i.e.} we assume that $ \beta^{c}_{l_c} < \beta^{c-1}_{l_{c-1}} < \ldots < \beta^{1}_{l_1} = m_i$. As a direct consequence, $\beta^{1}_1 = \beta(m_i)$.\\
We connect $\mid\pi_i\mid-1$ thorns to the black vertex linked to the root by the edge $\beta(m_i)$. Moving around the vertex counter-clockwise and starting right after edge $\beta(m_i)$, we label the thorn with $\beta^{1}_{2}, \beta^{1}_{3}, \ldots, \beta^{1}_{l_1}, \beta^{2}_{1}, \dots, \beta^{2}_{l_2}, \beta^{3}_{l_3},\ldots \beta^{c}_{l_c}$. Then $\overline{\tau}$ is the resulting thorn tree.

\begin{Rem}\label{rem:betai}
Moving around a black vertex clockwise starting with the thorn right after the edge (in clockwise order), a new cycle of $\beta$ begins whenever we meet a left-to-right maximum of the labels.
\end{Rem}
\vspace{-8mm}
\begin{Rem}\label{rem:taubar2smp}
As the long cycle $\alpha$ and the repartition of the cycles of $\beta$ in the various blocks appear explicitly in $\overline{\tau}$, one can recover the black-partitioned star map from it.
\end{Rem}
\end{enumerate}

The idea behind this construction is to add a root to the map $(\alpha,\beta)$, select one edge per block, cut all other edges into two thorns and merge the vertices corresponding to the same black block together. Step (\ref{item:root_place}) tells us where to place the root, step (\ref{item:edge_root}) which edges we select and step (\ref{item:black_thorns}) how to merge vertices (in maps unlike in graphs, one has to do some choices to merge vertices).

\begin{example}
\label{exspm2lstt}
Let us take the black-partitioned star map of example \ref{exspm}. Following construction rules (\ref{item:root_place}) and (\ref{item:edge_root}), one has $m_\triangle = 7$, $m_\bigcirc = 5$, $m_\Box = 4$ and the descending edges indexed by $\beta(m_\triangle) = 3$, $\beta(m_\bigcirc) = 2$ and $\beta(m_\Box) = 4$ connect a black vertex to the white root. Other descending edges from the root are thorns. Using (\ref{item:black_thorns}), we add labeled thorns to the black vertices to get the labeled thorn tree depicted on Figure \ref{taubar}. Focusing on the one connected to the root through the edge $3$, we have $(\beta^1_1 \beta^1_{l_1}) (\beta^2_{l_2}) (\beta^3_{l_3}) = (3 7)(6)(1)$. Reading the labels clockwise around this vertex, we get $1,6,7,3$. The three cycles can be simply recovered looking at the left-to-right maxima $1$, $6$ and $7$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=140mm]{TauBar}
\end{center}
\caption{Labeled thorn tree associated to the black-partitioned star map of Figure \ref{spm}} \label{taubar}
\end{figure}
\end{example}

\subsection{Permuted thorn tree}

Using $\overline{\tau}$, we call $\tau$ the star thorn tree obtained by removing labels and $\sigma$ the permutation that associates to a white thorn in $\tau$ the black thorn with the same label in $\overline{\tau}$.

Finally, we define:
$
\Psi(\beta,\pi) = (\tau,\sigma).
$
\begin{example}
Following up with example \ref{spm}, we get the permuted thorn tree $\ex{3}$ drawn on Figure \ref{tausigma}. Graphically we use the same convention as in section \ref{sect:reformulation} to represent $\sigma$.
 \begin{figure}[h]
 \begin{center}
 \includegraphics[width=100mm]{tausigma}
 \end{center}
 \caption{Permuted thorn tree $\ex{3}$ associated to the black-partitioned star map of Figure \ref{spm}} \label{tausigma}
 \end{figure}

\end{example}

\section{Injectivity and reverse mapping}\label{sect:inj_psi}
Assume $(\tau,\sigma) = \Psi(\beta,\pi)$ for some black partitioned star map $(\beta,\pi)$. We show that $(\beta,\pi)$ is actually uniquely determined by $(\tau,\sigma)$.\\
As a first step, we recover the labeled thorn tree $\overline{\tau}$. Let us draw the permuted thorn tree $(\tau,\sigma)$ as explained in subsection \ref{subsect:thorn_trees}. We show by induction that there is at most one possible integer value for each symbolic label.

\begin{enumerate}[(i)]

\item By construction, the label $\alpha_1$ of the right-most edge or thorn descending from the root is necessarily $1$.

\item \label{item:etape_determiner_bi} Assume that for $i \in [N-1]$, we have identified the symbols of values $1,2,\ldots$, and $i$. We look at the edge or thorn with label $i$ connected to a black vertex $b$. In this step, we determine which symbol corresponds to $\beta(i)$.\\
Recall that, when we move around $b$ clockwise finishing with the edge (in this step, we will always turn in this sense), a new cycle begins whenever we meet a left-to-right maximum. So, to find $\beta(i)$, one has to know whether $i$ is a left-to-right maximum or not.\\
If all values of labels of thorns before $i$ haven't already been retrieved, then $i$ is not a left-to-right maximum. Indeed, the remaining label values are $i+1$, \ldots, $N$ and at least one thorn's label on the left of $i$ lies in this interval. Following our construction, necessarily $\beta(i)$ corresponds to the symbolic label of the thorn right at the left of $i$.\\
If all the thorns' label values on the left of $i$ have already been retrieved (or there are no thorns at all), then $i$ is a left-to-right maximum. According to the construction of $\overline{\tau}$, $\beta(i)$ corresponds necessarily to the symbolic label of the thorn preceding the next left-to-right maximum
. But one can determine which thorn (or edge) corresponds to the next left-to-right maximum: it is the first thorn (or edge) $e$ without a label value retrieved so far (again moving around the black vertex from left to right). Indeed, all the value retrieved so far are less than $i$ and those not retrieved greater than $i$. Therefore $\beta(i)$ is the thorn right at the left of $e$. If all the values of the labels of the thorns connected to $b$ have already been retrieved then $i$ is the maximum element of the corresponding block and $\beta(i)$ corresponds to the symbolic label of the edge connecting this black vertex to the root.

\item \label{item:etape_determiner_ipu} Consider the element (thorn of edge) of white extremity with the symbolic label corresponding to $\beta(i)$. The next element (turning around the root in counter-clockwise order) has necessarily label $\alpha(\beta(i))=i+1$.\\
As a result, the knowledge of the thorn or edge with label $i$ uniquely determines the edge or thorn with label $i+1$.
\end{enumerate}

Applying the previous procedure up to $i=N-1$ we see that $\overline{\tau}$ is uniquely determined by $(\tau,\sigma)$ and so is $(\beta, \pi)$ (see Remark \ref{rem:taubar2smp}).

\begin{example}
Take as an example the permuted thorn tree $\ex{1}$ drawn on the left-hand side of Figure \ref{fig:ex_arbre_permute}, the procedure goes as described on figure \ref{recons}. First, we identify $\alpha_1 = 1$. Then, as there is a non (value) labeled thorn ($\alpha_2$) on the left of the thorn connected to a black vertex with label value $1$, necessarily $1$ is not a left-to-right maximum and $\alpha_2$ is the label of the thorn right on the left of $1$, that is $\alpha_2$. Then as $\alpha_3$ follows $\alpha_2 = \beta(1)$ around the white root, we have $\alpha_3 = \alpha(\beta(1))=2$.

\noindent We apply the procedure up to the full retrieval of the edges' and thorns' labels. We find $\alpha_2 = 3$, $\alpha_4 = 4$, $\alpha_5 = 5$. Finally, we have $\alpha = (1 3 2 4 5)$, $\beta = (213)(4)(5)$, $\pi = \{\{1, 2, 3\}; \{4, 5\} \}$ as shown on figure \ref{endrecons}.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=30mm]{taubarsymbols1_1}\hspace{15mm}

\includegraphics[width=30mm]{taubarsymbols5}\hspace{15mm}
\includegraphics[width=30mm]{taubarsymbols7}
\end{center}
\begin{center}
\includegraphics[width=30mm]{taubarsymbols8}\
\hspace{15mm}
\includegraphics[width=30mm]{reconsmap}
\caption{Reconstruction of the map}\label{recons}\label{endrecons}
\end{center}
\end{figure}
\end{example}

\section{Characterization and size of the image set $\Im(\mm)$}\label{sect:im_psi}

\subsection{A necessary and sufficient condition to belong to $\Im(\mm)$}

\paragraph{Why $\mm$ is not surjective?}

Let us fix a permuted star thorn tree $(\tau,\sigma)$. We can try to apply to it the procedure of section \ref{sect:inj_psi} and we distinguish two cases:
\begin{itemize}
\item it can happen, for some $i<N$, when one wants to give the label $i+1$ to the edge following $\beta(i)$ (step (\ref{item:etape_determiner_ipu})), that this edge has already a label $j$. If so, the procedure fails and $(\tau,\sigma)$ is not in $\Im(\mm)$.
\item if this never happens, the procedure ends with a labeled thorn tree $\overline{\tau}$. In this case, one can find the unique black-partitioned star map $M$ corresponding to $\overline{\tau}$ and by construction $\mm(M)=(\tau,\sigma)$.
\end{itemize}
For instance, if we take as $(\tau,\sigma)$ the couple $\ex{2}$ on the right of Figure \ref{fig:ex_arbre_permute}, the procedure gives successively : $\alpha_1=1,\ \alpha_9=2,\ \alpha_{10}=3,\ \alpha_6=4,\ \alpha_5=5$ and then we should choose $\alpha_1=6$, but this is impossible because we already have $\alpha_1=1$.
\begin{Lem}
 If the procedure fails, the label $j$ of the edge that should get a second label $i+1$ is always $1$.
\end{Lem}
\begin{proof}
 If $j >1$, this means that $\beta(j-1)=\beta(i)$. Let us distinguish two cases.\\
If $j-1$ is a left-to-right maximum, the label $i$ must be at the right of $\beta(j-1)$ and not a left-to-right maximum. But this is impossible because all thorns at the left of $\beta(j-1)$ (including $\beta(j-1)$) have labels smaller than $j$.\\
If  $j-1$ is not a left-to-right maximum, the label $j-1$ must be at the right of $\beta(j-1)=\beta(i)$ and $i$ is a left-to-right maximum. Then $\beta(i)$ is before the next left-to-right maximum. So the edge at the right of $\beta(i)$ has a label greater than $i$ and can not be $j-1$.
\end{proof}

\paragraph{An auxiliary oriented graph}
Remark \ref{rem:leftmost_edge} gives a necessary condition for $(\tau,\sigma)$ to be in $\Im(\mm)$: its left-most edge leaving the root must be a real edge $e_0$, and not a thorn. From now on, we call this property $(P1)$: note that, among all permuted thorn tree of a given type $\lambda \vdash N$ of length $p$, exactly $p$ over $N$ have this property. When $(P1)$ is satisfied, we denote $\pi_0$ the black extremity of $e_0$. The lemma above shows that the procedure fails if and only if $e_0$ is chosen as $\beta(i)$ for some $i<N$. But this can not happen at any time. Indeed, the following lemma is a direct consequence from step (\ref{item:etape_determiner_bi}) of the inverting procedure:
\begin{Lem}\label{lem:sommet_complete}
A real edge (\textit{i.e.} which is not a thorn) $e$ can be chosen as $\beta(i)$ only if the edge and all thorns leaving the corresponding black vertex have labels smaller or equal to $i$. If this happen, we say that the black vertex is completed at step $i$.
\end{Lem}

\begin{Corol}
 Let $e$ be a real edge of black extremity $\pi \neq \pi_0$. Let us denote $e'$ the element (edge or thorn) right at the left of $e$ on the white vertex. Let $\pi'$ be the black extremity of the element $e''$ associated to $e'$ (\textit{i.e.} $e'$ itself if it is an edge, its image by $\sigma$ else). Then $\pi'$ can not be completed before $\pi$.
\end{Corol}
\begin{proof}
If $\pi'$ is completed at step $i$, by Lemma \ref{lem:sommet_complete}, element $e''$ has a label $j \leq i$. As $e'$ has the same label, this implies that $e$ has label $\beta(j-1)$ or in other words, that $\pi$ is completed at step $j-1 < i$.
\end{proof}

When applied for every black vertex $\pi \neq \pi_0$, this corollary gives some partial information on the order in which the black vertices can be completed. We will summarize this in an oriented graph $G(\tau,\sigma)$: its vertices are the black vertices of $\tau$ and its edges are $\pi \to \pi'$, where $\pi$ and $\pi'$ are in the situation of the corollary above.
This graph has one edge leaving each of its vertex, except for $\pi_0$. As examples, the graphs corresponding to $\ex{2}$ and to $\ex{3}$ (see Figures \ref{fig:ex_arbre_permute} and \ref{tausigma}) are drawn on Figure \ref{fig:ex_graphe_ordre_completion}.

\begin{figure}[t]
$$\includegraphics[width=2cm]{grapheaux} \qquad \qquad \includegraphics[width=1.5cm]{grapheaux2}$$
\caption{Two examples of auxiliary graphs.}
\label{fig:ex_graphe_ordre_completion}
\end{figure}

\paragraph{The graph $G(\tau,\sigma)$ gives all the information we need!}
Can we decide, using only $G(\tau,\sigma)$, whether $(\tau,\sigma)$ belongs to $\Im(\mm)$ or not? There are two cases, in which the answer is obviously yes:
\begin{enumerate}
 \item Let us suppose that $G(\tau,\sigma)$ is an oriented tree of root $\pi_0$ (all edges are oriented towards the root). In this case, we say that $(\tau,\sigma)$ has property $(P2)$. Then, the vertex $\pi_0$ can be completed only when all other vertices have been completed, \textit{i.e.} when all edges and thorns have already a label. That means that $e_0$ can be chosen as $\beta(i)$ only for $i=N$. Therefore, in this case, the procedure always succeeds and $(\tau,\sigma)$ belongs to $\Im(\mm)$. This is the case of $\ex{3}$.
 \item Let us suppose that $G(\tau,\sigma)$ contains an oriented cycle (eventually a loop). Then all the vertices of this cycle can never be completed. Therefore, the procedure always fails in this case and $(\tau,\sigma)$ does not belong to $\Im(\mm)$. This is the case of $\ex{2}$.
\end{enumerate}
In fact, we are always in one of these two cases (the proof of the following lemma is left to the reader):
\begin{Lem}
 Let $G$ be an oriented graph whose vertices have out-degree $1$, except for one vertex $v_0$ which has out-degree $0$. Then $G$ is either an oriented tree of root $v_0$ or contains an oriented cycle.
\end{Lem}

Finally, one has the following result:
\begin{Prop}
 The mapping $\mm$ defines a bijection:
\begin{equation}\label{eq:main_bijection}
\left\{\begin{array}{c} \text{black-partitioned star maps} \\
          \text{of type }\lambda
         \end{array}\right\}
\simeq
\left\{\begin{array}{c} \text{permuted star thorn trees of type }\lambda\\
        \text{with properties }(P1) \text{ and }(P2)
       \end{array}\right\}.
\end{equation}
\end{Prop}

\subsection{Proportion of permuted thorn trees $(\tau,\sigma)$ in $\Im(\mm)$}

To finish the proof of Proposition \ref{prop:reformulation}, one just has to compute the size of the right-hand side of \eqref{eq:main_bijection}:

\begin{Prop}\label{prop:proportion_P2}
 Let $\lambda$ be a partition of $N$ of length $p$. Denote by $P(\lambda)$ (resp. $P'(\lambda)$) the proportion of couples $(\tau,\sigma)$ with properties $(P1)$ and $(P2)$ among all the permuted thorn trees of type $\lambda$ (resp. among permuted thorn trees of type $\lambda$ with property $(P1)$) of type $\lambda$. Then, one has:
$$P'(\lambda)=\frac{N}{p (N - p+1)} \text{ and, hence, }P(\lambda)=\frac{1}{N-p+1}.$$
\end{Prop}
\begin{proof}[(by induction on $p$)]

The case $p=1$ is easy: as $G(\tau,\sigma)$ has only one vertex and no edges, it is always a tree. Therefore, for any $N \geq 1$, one has $P'((N))=1$.\\
Suppose that the result is true for any $\lambda$ of length $p-1$ and fix a partition $\mu \vdash N$ of length $p>1$. Consider the permuted thorn trees $(\tau,\sigma)$ of type $\mu$, verifying $(P1)$, with a marked black vertex $\overline{\pi} \neq \pi_0$: as there are always $p-1$ choices for the marked vertex, the proportion of these objects verifying $(P2)$ is still $P'(\mu)$.\\
Let us now split this set, depending on the degrees (in $\tau$) of the marked vertex and of the end of the edge leaving $\overline{\pi}$ in the graph $G(\tau,\sigma)$. The proportion of marked star thorn trees of type $\mu$ (with property $(P1)$) whose marked vertex has degree $k_0$ is $m_{k_0}(\mu)/p$. We denote $k=\deg_\tau(\overline{\pi})$ and $\mu' = \mu \backslash k$ (\textit{i.e.} the partition obtained from $\mu$ by deleting one part $k$).
\begin{itemize}
 \item In $k-1$ cases over $N-1$, this second extremity is also $\overline{\pi}$. So $G(\tau,\sigma)$ contains a loop and $(\tau,\sigma)$ does not fulfill $(P2)$.
 \item For every $j$, in $j\cdot m_j(\mu')$ cases over $N-1$, this second extremity is a vertex $\pi' \neq \overline{\pi}$ of degree $j$ (in $\tau$). But one has an easy bijection $\varphi$:
\begin{align*}
\hspace{-1cm}\left\{\begin{array}{c} (\tau,\sigma) \text{ of type $\mu$ verifying $(P1)$} \\ \text{with a marked black vertex $\overline{\pi} \neq \pi_0$} \\ \text{of size $k$ such that }\overline{\pi} \to_{G(\tau,\sigma)} \pi' \\ \text{with $\pi' \neq \overline{\pi}$ of size $j$} \end{array}\right\}
&\simeq
\left\{\begin{array}{c} (\tau',\sigma') \text{ of type $\mu^{\downarrow (j,k)} := \mu \backslash (j,k) \cup (j+k-1)$} \\ \text{verifying $(P1)$ with the edge or one of the first }j-1 \\ \text{thorns of a black vertex of size $j+k-1$ marked} \\ \text{(always $j \cdot m_{j+k-1}(\mu^{\downarrow (j,k)})$ choices)}
\end{array}\right\} \\
\begin{array}{c}\includegraphics[width=4cm]{taubarpart5-1}\end{array} &\leftrightarrows \begin{array}{c}\includegraphics[width=4cm]{{taubarpart5-2}} \end{array}.
\end{align*}

From left-to-right: erase the marked black vertex $\overline{\pi}$ with its edge $e_{\overline{\pi}}$ and move its thorns to the black vertex $\pi'$ (at the right of its own thorns). Choose as marked the element (edge of thorn) with a black extremity with the same symbolic label as the element right at the left of $e_{\overline{\pi}}$.\\
From right to left: look at the white thorn corresponding to the marked thorn $\overline{e}$ (if the marked element is an edge, just take the edge itself). Then add a new edge with a black vertex just at the right of this thorn (or edge). Finally, move the $k-1$ right-most thorns of the black extremity of $\overline{e}$ to this new black vertex. The marked black vertex is the new one.\\
This bijection keeps property (P2). Indeed, if $\varphi(\tau,\sigma,\overline{\pi}) = (\tau',\sigma',\overline{e})$, the graph $G(\tau',\sigma')$ is obtained from $G(\tau,\sigma)$ by contracting the edge of origin $\overline{\pi}$. Therefore, the proportion of couples having property $(P2)$ on the left-hand side is the same as on the right-hand side. But, as $\mu^{\downarrow (j,k)}$ has length $p-1$ and size $N-1$, by induction hypothesis, this proportion is:
$$\frac{N-1}{(p-1)\big((N-1)-(p-1)+1\big)}.$$
\end{itemize}
We can now put the different cases together to compute $P'(\mu)$:
\begin{align*}
P'(\mu) & = \sum_k \frac{m_k(\mu)}{p} \left(
\sum_j \frac{j \cdot m_j(\mu')}{N-1} \cdot \frac{N-1}{(p-1)\big(N-p+1\big)} \right)

=\frac{N}{p\big(N-p+1\big)}.
\end{align*}
The last equality is obtained by a straight-forward computation and ends the proof of Proposition \ref{prop:proportion_P2} and, therefore, of Proposition \ref{prop:reformulation}.
\end{proof}

\bibliographystyle{alpha}
\bibliography{fpsac}

\end{document}

config: arxiv


TITLE: Prove that $\liminf [X_n = 2]) \subseteq ([\lim S_n] = \infty)$
QUESTION [0 upvotes]: Original problem:
Let $X_1, X_2, ...$ be random variables in $(\Omega, \mathscr{F}, \mathbb{P})$
where $X_1 = 0$ and for $n > 1$
$$P(X_n=-n^2)=\frac{1}{n^2}$$
$$P(X_n=-n^3)=\frac{1}{n^3}$$
$$P(X_n=2)=1-\dfrac{1}{n^2}-\dfrac{1}{n^3}$$
Let $S_n=\sum_{i=1}^{n} X_i$. Prove that $P([\lim S_n] = \infty) = 1$

REPLY [0 votes]: I used BCL1 to deduce that
$$P(\liminf [X_n = 2]) = 1$$
Let $\omega \in \Omega$. If $\omega \in \liminf [X_n = 2]$, then $\exists m \ge 1$ s.t.
$$2 = X_m(\omega) = X_{m+1}(\omega) = ...$$
$$\to S_{m+k}(\omega) = S_m(\omega) + 2k \ \forall k \ge 0$$
$$\to \lim _{k \to \infty} S_{m+k}(\omega) = \infty$$
$$\to \lim _{k \to \infty} S_{k}(\omega) = \infty$$
$$\to \lim _{n \to \infty} S_{n}(\omega) = \infty$$
The result follows by monotonicity of probability. QED


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TITLE: Least upper bound property and greatest lower bound property
QUESTION [2 upvotes]: (a)
$$S = (-∞, 1) \cup [2, 3) \cup (3, 10]$$
(b)
$$S = (-∞, 1) \cup [2, 3) \cup [9, 10]$$

Can someone explain to me, simply, why the first question, $S$ wouldn't have LUP?  And why the second one does?

REPLY [0 votes]: In the first one, $[2,3)$ has no least upper bound in $S$, whereas in the second one, the least upper bound of $[2,3)$ in $S$ is $9$.