text:
\section{Introduction}
A weighted complex projective space is a quotient
\[
X = S^{2d-1}/S^1,
\]
where $S^{2d-1}\subset\mathbb{C}^d$ is the unit sphere in $\mathbb{C}^d$ and the circle acts on the sphere as
\[
e^{it}\cdot (z_1,\ldots ,z_d) = (e^{iN_1t}z_1,\ldots ,e^{iN_d t}z_d),
\]
where $N_1,\ldots N_d$ are positive integers, called the weights.
Although naturally a (singular) complex manifold, we will think of $X$ as a real Riemannian orbifold, of real dimension $2(d-1)$.

In \cite{ADFG}, Abreu, Dryden, Freitas and Godinho raise the interesting question
of whether the spectrum of the Laplace-Beltrami operator on $X$ determines the weights $N_1,\ldots , N_d$, and they show that this is indeed the case for $d=3$, if one knows the spectrum of the Laplacian
on zero and 1-forms.  Their methods center on the heat invariants and use equivariant
cohomology.

In a different direction, the wave-trace formula of \cite{DG} can be extended to compact Riemannian
orbifolds, see  \S 2 and \cite{SU}.  On the basis of this result one can expect that,
generically, the spectrum of the Laplacian on an orbifold determines the set of lengths (with multiplicities) of
closed geodesics, together with the dimension of the manifold of closed
geodesics of a given length. This is actually the case for weighted projective spaces.
 Let's call the multiset of such pairs of data, (lengths of closed geodesics with multiplicities, and the dimension of their set), the weighted length spectrum.  From the wave-trace point of view, the inverse spectral problem of the previous
paragraph is closely related to the problem:  {\em Does the weighted length
spectrum of $X$ determine the set of weights $\{N_j\}$?}

Geodesics on $X$ are projections of geodesics on $S^{2d-1}$ that are ``horizontal", meaning orthogonal to $S^1$ orbits.   Therefore all geodesics on $X$ are periodic with a common period, $2\pi$.
However, there can exist exceptional, shorter geodesics (they are ``exceptional" in the sense that the measure of their set in the space of all geodesics is zero).  They arise because of the non-trivial isotropy of some points on the sphere $S^{2d-1}$.

In this paper we show that, in many cases, the weighted length spectrum of
$X$ corresponding to the exceptional geodesics
determines the set of sums of pairs of weights, $\{N_j+N_k, j<k\}$, with multiplicities.
This in turn raises the following problem:  {\em Is a set of $d$ positive numbers, $\{N_j\}$, determined by the multiset of the sums of the pairs $\{N_j+N_k, j<k \}$?}
The answer is obviously ``no" for $d=2$ and obviously ``yes" for $d=3$.  It
has been known for some time that the answer is ``yes" if $d$ is not a power of
two.  For convenience we present a proof of this result in an appendix, since
the only reference we found, \cite{MSU}, may not be readily available to the reader.
Conversely, Elizabeth Chen and Jeffrey Lagarias have shown recently
that, if $d$ is any power of two, then there exists sets of $d$ numbers that
are not determined by the multiset of the sums of pairs.
(personal communication).

To briefly summarize our results, our calculations of the weighted length spectrum of $X$, together with the wave-trace formula for orbifolds, constitue a proof that for
$d$ not a power of two,
generically (with respect to the weights) one can ``hear" the weights of a weighted projective space.

About the organization of the paper:  After reviewing the wave trace for Riemannian orbifolds in \S 2, we compute in \S 3 the lengths of the exceptional geodesics in the case $d=2$, which we feel makes clearer the general discussion in \S\S 4 and 5.  Our main inverse spectral result is Theorem \ref{thm:7.1}.

{\sc Acknowledgments:}  We are indebted to Jeffrey Lagarias for his help with the ``sums-of-pairs" inverse problem mentioned above.

\section{The wave trace for orbifolds}
\label{sec:1}
In this section we outline a proof of the ``wave trace formula"
for orbifolds, referring to \cite{SU} for details.
We recall that an orbifold can always be represented as the
quotient of a manifold by a group, i.e.,~as a quotient $X/G$
where $X$ is a manifold, $G$ a compact connected Lie group and
$\tau: G \times X \to X$ a locally free action of $G$ on $X$.
If in addition $X$ is equipped with a $G$-invariant Riemannian
metric the Laplace operator $\Delta_X : C^\infty (X) \to C^\infty (X)$
maps $C^\infty (X)^G$ into itself and via the identification, $C^\infty
(X)^G = C^\infty (X/G)$, defines a Laplace operator, $\Delta_{X/G}$ on $X/G$.
Assuming that $X$ is compact the wave trace formula for $X$
asserts that the expression
\begin{displaymath}
  e(t) = : \mbox{trace} \exp it \sqrt{\Delta_X}
\end{displaymath}
is well-defined as a tempered distribution and that its singular
support is contained in (and modulo some slightly technical
clean intersection assumptions which we'll avoid going into at
this juncture) coincides with the period spectrum of
$X$.\footnote{A point, $T \in {\mathbb R}$ is in the \emph{period
spectrum} of $X$ if there exists a closed geodesic, $\gamma$,
on $X$ of period $T$, i.e.,~$\gamma (t +T) = \gamma (t)$ for all
$t$.}  We will prove below that the analogue of this result is
also true for $X/G$ providing one defines carefully what one means
by a ``closed geodesic'' and by its ``period''.

Let $T^\#X$ be the punctured cotangent bundle of $X$.  From
$\tau$ one gets an action, $\tau^\#$, of $G$ on $\mathbb{T}^\# X$ and from
the symbol of $\Delta_X$ a Hamiltonian vector field, $v_H$, where
$H=\sigma (\Delta_X)^\frac12$.  Moreover $\tau^{\#} $ and the
flow generated by $v_H$ commute, so one has an action,
\begin{displaymath}
  (g,t) \to \tau^\#_g \exp t v_H
\end{displaymath}
of $G \times {\mathbb R}$ on $T^\# X$.  The \emph{moment Lagrangian} for
this action (see \cite {We}) is the set of points
\begin{displaymath}
  (x,\xi,y,\eta ,t,\tau ,g, \gamma) \in T^\# (X \times X \times
  {\mathbb R} \times  G)
\end{displaymath}
satisfying
\begin{equation}
  \label{eq:1.1}
  (y,\eta) = (\tau^\#_g \exp t v_H) (x,\xi)\, , \quad
  \tau = H (x,\xi)\, , \quad \gamma = \Phi (x,\xi)
\end{equation}
where $\Phi : T^\# X \to {\mathfrak g}^*$ is the moment map associated with
the action of $G$ on $T^\# X$.  The set (\ref{eq:1.1}) is also the
microsupport of the operator, $\tau^\#_g \exp it \sqrt{\Delta_X}$,
and we'll use this fact and functorial properties
of microsupports to compute the microsupport of the distribution
\begin{equation}
  \label{eq:1.2}
  \mbox{trace} \exp it \sqrt{\Delta_{X/G}}
\end{equation}
Some observations:

\arabiclist
\begin{enumerate}

\item
Modulo the identification, $C^\infty (X)^G = C^\infty (X/G)$ the expression
(\ref{eq:1.2}) can be written as an integral
  \begin{equation}
    \label{eq:1.3}
    \int_G \mbox{trace} (\tau^\#_g \exp it \sqrt{\Delta_X})\, dg
  \end{equation}
where $dg$ is Haar measure.

\item
Let $e(x,y,t,g)$ be the Schwartz kernel of the operator,
$\tau^\#_g \exp it \sqrt{\Delta_X}$.  Then (\ref{eq:1.3}) can
also be written as a double integral
\begin{equation}
  \label{eq:1.4}
  \int e (x,x,t,g)\, dg \, dx
\end{equation}
where $dx$ is the Riemannian volume form on $X$.

\item
Let $\iota_\Delta : X \times {\mathbb R} \times G \to X \times X \times
{\mathbb R} \times G$ be the diagonal embedding and $\pi : X \times {\mathbb R}
\times G \to {\mathbb R}$ the projection, $(x,t,g) \to t$.  Then
(\ref{eq:1.4}) can be written more functorially as a
``pull-back'' followed by a ``push-forward'':
\begin{equation}
  \label{eq:1.5}
  \pi_* \iota^*_\Delta e (t) \, .
\end{equation}

\item
Functorial properties of ``pull-backs'' tell us that if $(x,\zeta
,t,\tau ,g,\gamma)$ is in the microsupport of $\iota^*_\Delta e$
there exists a point $(x,\xi,y,\eta,t,\tau g,\gamma)$ in the
microsupport of $e$, i.e., in the set (\ref{eq:1.1}), such that
\begin{equation}
  \label{eq:1.6}
  x=y \hbox{   and   } \zeta = \xi -\eta
\end{equation}
and, by (\ref{eq:1.1}),
\begin{equation}
  \label{eq:1.7}
  (x,\eta) = (\tau^\#_g \exp t v_H) (x,\xi)\, .
\end{equation}

\item
Functorial properties of ``push-forwards'' tell us that if
$(T,\tau)$ is in the microsupport of $\pi_* \iota^*_\Delta e$
there exists a point, $(x,\zeta,T,\tau ,g,\gamma)$ in the
microsupport of $\iota^*_\Delta e$, i.e., in the set
(\ref{eq:1.6})--(\ref{eq:1.7}) such that $\zeta = \gamma =0$.
Thus by (\ref{eq:1.6})--(\ref{eq:1.7}) we've proved:

\begin{theorem*}

If $T$ is in the singular support of the distribution,
(\ref{eq:1.5}) there exists an $(x,\xi,g) \in (T^\# X) \times G$
with
\begin{eqnarray}
  \label{eq:1.8}
  %
(T^\#_g \exp T v_H) (x,\xi) = (x,\xi)\\
\noalign{\mbox{and}}\notag\\
\label{eq:1.9}
\Phi (x,\xi) =0 \, .
\end{eqnarray}
Moreover, modulo appropriate clean intersection assumptions,
one can compute the leading term of the singularity of
(\ref{eq:1.5}) at $T$ in terms of the geometry of the
manifold of curves $(T^\#_g \exp t v_H) (x,\xi)$.

\end{theorem*}

\end{enumerate}

It remains to show that (\ref{eq:1.8}) and (\ref{eq:1.9}) can be
interpreted as saying that $T$ is in the period spectrum of
$X/G$.  Since the action of $G \times {\mathbb R}$ on $T^\# X$ is locally
free the group of $(g,T)$'s satisfying (\ref{eq:1.8}) is a discrete
group.  Thus the projection, $G \times {\mathbb R} \to {\mathbb R}$ maps this
onto a discrete subgroup of ${\mathbb R}$ and hence onto a lattice group
\begin{equation}
  \label{eq:1.10}
  \{ n T_0 \, , \, n \in {\mathbb Z} \} \, .
\end{equation}
Moreover since $G$ is compact the projection of this group onto
$G$ is a finite subgroup of order~$m$.  Thus by (\ref{eq:1.8})
\begin{displaymath}
  (\exp mT v_H) (x,\xi) = (x, \xi)
\end{displaymath}
showing that the curve
\begin{displaymath}
(\exp t v_H) (x,\xi) \, , \, -\infty < t< \infty
\end{displaymath}
is periodic of period $mT$ and hence its projection onto $X$ is a
closed geodesic, $\gamma (t)$, of period~$mT$.

Let's next unravel the meaning of (\ref{eq:1.9}).  This asserts
that the covector, $\xi \in T^\#_x$, is conormal to the orbit of
$G$ through $x$ and hence for the projected curve, $\gamma (t)$,
that $\frac{d\gamma}{dt} (0)$ is orthogonal at $x$ to the orbit
of $G$ through $x$.  Since $\Phi^{-1}(0)$ is $G \times {\mathbb R}$
invariant, this holds as well at $\gamma (t)$, therefore the
implication of (\ref{eq:1.9}) is that at every point, $\gamma
(t)$, $-\infty <t<\infty$, the orbit of $G$ through $\gamma (t)$
is orthogonal to $\gamma (t)$.  In other words, $\gamma$ is
horizontal with respect to the quasi-fibration, $\mbox{pr} : X \to X/G$,
and therefore its projection, $\mbox{pr} \circ \gamma$, onto $X/G$ is a
geodesic on $X/G$.\footnote{Away from singular points
these projections are locally length-minimizing, and therefore worthy of the
name ``geodesics" on $X/G$.  Across singular strata they fail to be
locally lenght-minimizing, but we will continue to call them geodesics, as no
locally length minimizing curves exist in that case.  Of
  course one has to check that this definition is independent of
  the presentation of $X/G$ as the quotient of $X$ by $G$, but
  this isn't hard to do.}  Finally note that if $T$ belongs to
the set (\ref{eq:1.10}) then by (\ref{eq:1.8}) there exists a $g
\in G$ such that
\begin{displaymath}
  (\tau^\#_g \exp T v_H) (x,\xi) = (x,\xi)
\end{displaymath}
and hence
\begin{displaymath}
  \gamma (0) = \tau_g \gamma (T) \, .
\end{displaymath}
Thus for the projection of $\gamma$ onto $X/G$
\begin{displaymath}
  \mbox{pr} \circ \gamma (T) = \mbox{pr} \circ \gamma (0) \, .
\end{displaymath}
In other words, $T$ is the period of the closed geodesic, $\mbox{pr} \circ
\gamma (t)$, $0 \leq t \leq T$.  In addition, the calculation alluded to in the
previous theorem shows that in positive curvature there cannot be
singularity cancellations from geodesics of a given length.
Therefore one has:

\begin{theorem*}

If $T \in {\mathbb R}$ is in the singular support of the distribution
(\ref{eq:1.2}), there exists a closed geodesic on $X/G$ of period
  $T$.  Moreover, if appropriate clean intersection hypotheses
  are satisfied and $X$ has positive
  sectional curvature, this result is an ``if and only if'' result.

\end{theorem*}

\section{Weighted projective lines}
\label{sec:2}

These spaces are obtained as quotients of the three sphere $S^3
\subseteq {\mathbb C}^2$ by circle actions.  More explicitly, let $p$
and $q$ be positive integers with $1< p <q$ and g.c.d.$(p,q) =1$, and let
\begin{displaymath}
  \tau : S^1 \times S^3 \to S^3
\end{displaymath}
be the action $\tau (e^{i\theta}) (z_1,z_2) = (e^{ip}
z_1,e^{iq}z_2)$.  Then $X = S^3/S^1$ is an orbifold with
singularities at its north pole $[1,0]$ and south pole $[0,1]$.
Moreover we can define a Riemannian metric on the non-singular
part of $X$, as follows.  Let $v$ be the infinitesimal generator
of the action $\tau$.  Since this action is locally free $v
(p)\neq 0$ at all $p \in S^3$.  Let $\pi$ be the projection of
$S^3$ onto the quotient $X$.  Then, except at the north and
south poles, we can define the metric on $T_qX$, $q
= \pi (p)$, by identifying $T_q X$ with the horizontal part $H_p$
of $T_pS^3$, i.e.~with the space in $T_p$ orthogonal to $v_p$.  Since
the $S^1$ action is by isometrics this definition doesn't depend
on $p$.  Moreover if $\gamma : [0,2\pi] \to S^3$ is a geodesic
which is horizontal at $p = \gamma (0)$, i.e.~satisfies
\begin{displaymath}
 \Big\langle \frac{d\gamma}{dt}(0)\, , \, v_p \Big\rangle =0,
\end{displaymath}
then by parallel transport it's horizontal at $q=\gamma (t)$ so
it makes sense to talk about \emph{horizontal} geodesics in
$S^3$, and it's clear from our definition of the Riemann metric
on $X$ that their projections are geodesics on $X$.  In
particular $X$ is Zoll: all its geodesics are closed.

\subsection{Horizontal geodesics}
\label{sec:3}

Here's a more down-to-earth description of these horizontal
geodesics.  Every geodesic on $S^3$ is the intersection of $S^3$
with a two-dimensional subspace, $V$, of ${\mathbb R}^4$.  Identify ${\mathbb R}^4$
with ${\mathbb R}^2 \oplus {\mathbb R}^2$ and let $J: {\mathbb R}^2 \to {\mathbb R}^2$ be the map
$J (a,b) = (-b,a)$.  (In other words, via the identification
\begin{displaymath}
  {\mathbb R}^2 \to {\mathbb C} \qquad (x,y) \to x+ \sqrt{-1}\,y\, ,
\end{displaymath}
let $J$ be multiplication by $\sqrt{-1}$.)  Then the vector field
$v(\text{v})$ at $\text{v} = (\text{v}_1,\text{v}_2) \in {\mathbb R}^2 \oplus {\mathbb R}^2$ is just
\begin{equation}
  \label{eq:3.1}
  \tilde{J} \text{v} = (p J \text{v}_1 , q J \text{v}_2)\, .
\end{equation}
Thus if $V$ is a two-dimensional subspace of ${\mathbb R}^4$ spanned by
$\text{v}$ and $w$ with $|\text{v} | = |w| =1$ and $\text{v} \perp w$, and
$\gamma : [0,2\pi] \to S^3$ is the parametrized geodesic lying on
the circle $V \cap S^3$ with $\gamma (0) = \text{v}$ and
$\displaystyle{\frac{d\gamma}{dt}(0) =w}$,
\begin{displaymath}
  \Big\langle v (\text{v}),\frac{d\gamma}{dt}(0)\Big\rangle = 0
     \Leftrightarrow  \Big\langle \tilde{J}\text{v} ,w \Big\rangle
     =0\, .
\end{displaymath}
We conclude:

\begin{theorem}
  \label{th:3.1}
Let $\tilde{\omega} \in \Lambda^2 ({\mathbb R}^4)^*$ be the two-form
\begin{equation}
  \label{eq:3.2}
  \tilde{\omega} ({\rm v} ,w) = \langle \tilde{J}{\rm v} ,w \rangle
\end{equation}
and let $V \subseteq {\mathbb R}^4$ be a two-dimensional subspace of
${\mathbb R}^4$.  Then the geodesic $V \cap S^3$ is horizontal iff $V$ is
Lagrangian with respect to the two-form $\tilde{\omega}$.

\end{theorem}

Note that if $\text{v} = (\text{v}_1,\text{v}_2) \in {\mathbb R}^2 \oplus {\mathbb R}^2$ and
$w=(w_1,w_2) \in {\mathbb R}^2 \oplus {\mathbb R}^2$ then
\begin{equation}
  \label{eq:3.3}
  \tilde{\omega} (\text{v} ,w) = p \omega (\text{v}_1,w_1) + q \omega (\text{v}_2,w_2),
\end{equation}
where
\begin{displaymath}
  \omega (\text{v}_i,w_i) = \langle J \text{v}_i , w_i \rangle \, .
\end{displaymath}
From this formula one easily deduces

\begin{theorem}
\label{th:3.2}
There are two kinds of Lagrangian subspaces of ${\mathbb R}^4$:

\arabiclist
\begin{enumerate}
\item
Subspaces of the form
\begin{equation}
  \label{eq:3.4}
  V_1 \oplus V_2 \subseteq {\mathbb R}^2 \oplus {\mathbb R}^2
\end{equation}
where each $V_i$ is a one-dimensional subspace of ${\mathbb R}^2$.

\item
Graphs of maps $A:{\mathbb R}^2 \to {\mathbb R}^2$ satisfying
\begin{equation}
  \label{eq:3.5}
  A^* \omega = - \frac{p}{q}\omega \, .
\end{equation}

\end{enumerate}

\end{theorem}

Note that if $V \subseteq {\mathbb R}^4$ is a Lagrangian subspace of
type~1, $V \cap S^3$ is a horizontal geodesic whose projection
onto $X$ goes through the north and south poles and if $V$ is a
Lagrangian subspace of type~2, $V \cap S^3$ is a horizontal
geodesic whose projection onto $X$ doesn't go through
\emph{either} the north or south pole.  We'll call Lagrangians of
type~1 \emph{polar} Lagrangians and those of type~2
\emph{non-polar} Lagrangians.

\subsection{The action of $S^1$ on the Grassmannian of
  two-dimensional subspaces of ${\mathbb R}^4$}
\label{sec:4}

To simplify slightly the exposition below, we'll assume that $p$
and $q$ are odd.  The linear action of $S^1$ on ${\mathbb R}^2 \oplus
{\mathbb R}^2$, defined by
\begin{equation}
  \label{eq:4.1}
  \tau (e^{it} ) = (\exp tpJ \, , \, \exp tq J),
\end{equation}
induces an action $\tau^\#$ of $S^1$ on the Grassmannian of
two-dimensional subspaces of ${\mathbb R}^4$.  If $p$ and $q$ are both
odd this action isn't effective since $\tau (e^{i\pi}) =-I$.
However we do get an effective action of $S^1 / \{ \pm 1 \}$.
Next note the following:

\begin{lemma*}
  If $V$ is a two-dimensional subspace of ${\mathbb R}^2 \oplus {\mathbb R}^2$
  then either
  \begin{equation}
    \label{eq:4.2}
    V=V_1 \oplus V_2
  \end{equation}
where $V_1$ and $V_2$ are one-dimensional subspaces of ${\mathbb R}^2$, or
\begin{equation}
  \label{eq:4.3}
  V={\rm graph}\,  A
\end{equation}
where $A: {\mathbb R}^2 \to {\mathbb R}^2$ is a map of the first summand of ${\mathbb R}^2
\oplus {\mathbb R}^2$ into the second, or
\begin{gather}
  \label{eq:4.3p}
  V= {\rm graph} \, A' \tag{\ref{eq:4.3}$^\prime$}
\end{gather}
where $A'$ is a map of the second summand of ${\mathbb R}^2$ into the first.

\end{lemma*}

It's easy to see that $S^1 /\{ \pm 1 \}$ acts freely at points
(\ref{eq:4.2}) of $G_2 ({\mathbb R}^4)$.  What about points of type
(\ref{eq:4.3}) or (\ref{eq:4.3p})?  These two cases are basically
the same so let's concentrate on (\ref{eq:4.3}).  Note that $\tau
(e^{it})$ maps the subspace (\ref{eq:4.3}) into itself iff
\begin{equation}
  \label{eq:4.4}
    (\exp tq J)A = A (\exp tp J),
\end{equation}
or, equivalently, iff
\begin{eqnarray}
  \label{eq:4.5}
  (\cos qt) A &=& (\cos pt) A\\
\noalign{\hbox{and}}\notag\\
\label{eq:4.6}
 (\sin qt) JA &=& (\sin pt) AJ\, .
\end{eqnarray}
Hence if we exclude the trivial case $A=0$ the element
$e^{it}$, $0 < t < \pi$, of $S^1$ acts trivially on the
Grassmannian at the point (\ref{eq:4.3}) iff
\begin{eqnarray}
  \label{eq:4.7}
  JA=AJ \hbox{   and   } e^{itp} = e^{itq}\\
\noalign{\hbox{or}}\notag\\
\label{eq:4.8}
  JA= -AJ \hbox{   and   }  e^{itp} = e^{-itq}\, .
\end{eqnarray}
But $e^{itp} = e^{itq}$ iff $e^{it(q-p)} =1$ and hence iff
\begin{equation}
  \label{eq:4.9}
  t=\frac{\pi k}{q-p} \hbox{   for some   } 0<k<q-p,
\end{equation}
and $ e^{itp} = e^{iqt}$ iff $e^{it (p+q)} =1$ and hence iff
\begin{equation}
  \label{eq:4.10}
  t=\frac{\pi k}{p+q} \hbox{   for some   } 0<k<q+p\, .
\end{equation}
Moreover if (\ref{eq:4.7}) or (\ref{eq:4.8}) holds and $A \neq 0$
then $A$ is invertible, and therefore graph~$A$ is also the graph of a map,
$A'$, from the second summand of ${\mathbb R}^2 \oplus {\mathbb R}^2$ to the
first.  Thus we've proved

\begin{theorem}
  \label{th:4.1}

Suppose $p$ and $q$ are odd.  Then from the action $\tau^\#$ one gets
an effective action of $S^1 / \{ \pm 1 \}$ which fixes the
subspaces ${\mathbb R}^2 \oplus \{ 0 \}$ and $\{ 0 \} \oplus {\mathbb R}^2$ and is
elsewhere locally free.  Moreover $\tau^\#$ is free except at
subspaces $V$ of type (\ref{eq:4.3}) with $AJ = \pm JA$.  If
$AJ =JA$ the isotropy group of $V$ in $S^1$ is ${\mathbb Z}_{q-p}$, and if
$AJ = -JA$ its isotropy group  is ${\mathbb Z}_{p+q}$.

\end{theorem}

If either $p$ or $q$ is even this result has to be slightly
modified.

\begin{theorem}
  \label{th:4.2}
Suppose that either $p$ or $q$ is even.  Then the action
$\tau^\#$ fixes the subspaces ${\mathbb R}^2 \oplus \{ 0 \}$ and $\{ 0 \}
\oplus {\mathbb R}^2$ and is elsewhere locally free except at subspaces
of type (\ref{eq:4.2}) and at subspaces of type (\ref{eq:4.3}),
with $AJ = \pm JA$.  If $V$ is of type (\ref{eq:4.2}) its
isotropy group is ${\mathbb Z}_2$, and if $V$ is of type (\ref{eq:4.3})
with $AJ = \pm JA$ its isotropy group is the same as in the
theorem above.

\end{theorem}

Since the action $\tau$ of $S^1$ on ${\mathbb R}^4$ preserves the
symplectic form $\tilde{\omega}$, the induced action $\tau^\#$
on the Grassmannian of two-dimensional subspaces of ${\mathbb R}^4$
preserves the set of Lagrangian subspaces.  Moreover, if $V = {\rm
graph}\,  A$ is Lagrangian then by (\ref{eq:3.5}) $\det A = -
\frac{p}{q}$.  Therefore, if $A$ is such that $AJ=JA$,
graph~$A$ is \emph{not} Lagrangian since  $\det A$ is positive.
Hence from the theorems
above we get as a corollary:

\begin{theorem}
  \label{th:4.3}

If $p$ and $q$ are odd the action of $S^1 / \{ \pm 1 \}$ on the
Lagrangian Grassmannian is free except on the one-dimensional set
\begin{equation}
  \label{eq:4.11}
  \{ {\rm graph}\,  A_\lambda \, , \quad | \lambda |^2 =
      \frac{p}{q} \},
\end{equation}
where $A_\lambda : {\mathbb C} \to {\mathbb C}$ is the map $A_\lambda z =\lambda
\overline{z}$.  If either $p$ or $q$ is even the action of $S^1$ is
free except on the set of polar subspaces (\ref{eq:4.2}) (which
are stabilized by ${\mathbb Z}_2$) and on the set (\ref{eq:4.11}).
Moreover, no matter what the parity of $p$ and $q$, the
stabilizer group of points in the set (\ref{eq:4.11}) is ${\mathbb Z}_{p+q}$.

\end{theorem}

\subsection{Geodesics on the weighted projective line}
\label{sec:5}

By (\ref{eq:4.1}), $\tau^\# (e^{it})$ maps the subspace $V=
{\rm graph}\, A_\lambda$  to the subspace $V' = {\rm graph}\, A_{\lambda'}$, where
 $\lambda' = e^{it (p+q)}\lambda$.  Thus in particular $S^1$ acts
 transitively on the set of horizontal geodesics
 \begin{equation}
   \label{eq:5.1}
   \left\{ {\rm graph}\, A_\lambda \cap S^3 \, , \quad
     | \lambda |^2 = \frac{p}{q}\right\}.
 \end{equation}
Therefore these horizontal geodesics project to a single geodesic in
$X$.  We will call this the \emph{exceptional} geodesic.  Note
that this geodesic is non-polar, i.e.,~lies entirely in the
non-singular part of $X$.  Another key observation:  Suppose that
for some $0 < \theta < 2 \pi$, $\tau (e^{i\theta})$ maps a
Lagrangian subspace $V$ onto itself.  Then for every point
$\text{v}$ on the geodesic $V \cap S^3$, $\text{v}$ and its image,
$\tau (e^{i\theta})\text{v}$, project onto the same point in $X$.  Now
let $w \in V \cap S^3$ be a vector in $V$ perpendicular to $\text{v}$
and let
\begin{displaymath}
  \gamma (t) = \cos t \text{v} + \sin t w \,  \quad
    0 \leq t \leq 2 \pi
\end{displaymath}
be the geodesic parametrization of the circle $V \cap S^3$, with
$\gamma (0) = \text{v}$, $\frac{d}{dt} \gamma (0) =w$ and $\gamma
(t_0) = \tau (e^{i\theta} \text{v} )$.  Since the action of $S^1$ on
the space of horizontal geodesics is locally free, $\theta =
\frac{2\pi}{k}$ for some $k \in {\mathbb Z}_+$ and $\gamma $ is the union
of $k$ geodesic segments
\begin{displaymath}
  \gamma_m (t) = \tau (e^{\frac{i 2\pi m}{k}}) \gamma (t) \, ,
     \quad 0 < t < t_0 \, .
\end{displaymath}
Thus $t_0 = \theta$ and each of these segments projects onto a
closed geodesic in $X$ of length $\frac{2 \pi}{k}$.  Hence from
Theorem~\ref{th:4.3} we conclude:

\begin{theorem}
  \label{th:5.1}

If $p$ and $q$ are odd the geodesics on the Zoll orbifold $X$
are all of length $\pi$ except for the exceptional geodesic
which is of length $\frac{2\pi }{ p+q}$.  If either $p$ or $q$ is even
the geodesics on $X$ are all of length $2\pi$ except for the
polar geodesics (which are of length $\pi$) and the exceptional
geodesic (which is again of length $\frac{2\pi }{ p+q}$).

\end{theorem}

Thus, coming back to the results of \S1 we've proved

\begin{theorem}
  \label{th:5.2}

From the spectrum of the Laplace operator on $X$ one can ``hear''
whether $p$ or $q$ is odd and can also hear their sum $p+q$.

\end{theorem}

\section{Circle actions on ${\mathbb C}^d$}
\label{sec:6}

As in the previous sections we'll identify ${\mathbb C}^d$ with ${\mathbb R}^{2d}$, i.e.,~with the
sum
\begin{equation}
  \label{eq:6.1}
  {\mathbb R}^2 \oplus \cdots \oplus {\mathbb R}^2
\end{equation}
of $d$ copies of ${\mathbb R}^2$, and we'll view the standard representation of $S^1$
on ${\mathbb C}^d$ with weights $N_i$, $i=1,\ldots ,d$  as an
${\mathbb R}$-linear representation:
\begin{equation}
  \label{eq:6.2}
  \tau (e^{it}) = (\exp N_1 tJ , \ldots , \exp N_dtJ)\, .
\end{equation}
From (\ref{eq:6.2}) we get an action, $\tau^\#$, of $S^1$ on the
Grassmannian of two-dimensional subspaces of ${\mathbb R}^{2d}$, and we'll
try to ascertain below what the stabilizer groups are for this
action.  Our approach to this problem will be an inductive one:
Assuming we know what these isotropy groups are for ${\mathbb R}^{2d-2}$
we'll try to determine what they are for ${\mathbb R}^{2d}$.  Henceforth
we'll assume that the $N_i$'s \emph{are greater than one and are
  pair-wise relatively prime}.

We first note that each of the summands of (\ref{eq:6.1}) is a
fixed point for the action $\tau^\#$, and that on the complement
of these $d$ fixed points this action is locally free.    If all the
$N_i$'s are odd then $\tau (e^{i\pi})=-I$ and therefore $\tau^\#
(e^{i\pi})$  is the identity map, and so, as in \S4, one gets from
$\tau^\#$ an effective action of $S^1 /\{ \pm 1\}$.  If, one the
other hand, $N_1$ is even, the action $\tau^\#$ itself is an
effective action; however, if $V$ is a two-dimensional subspace
of the sum
\begin{equation}
  \label{eq:6.3}
  \{ 0 \} \oplus {\mathbb R}^2 \oplus \cdots \oplus {\mathbb R}^2
\end{equation}
its stabilizer contains ${\mathbb Z}_2$.  Also, no matter what the parity
of the $N_i$'s
 is, if $V$ is a subspace which is spanned by $\text{v}_1$ and
 $\text{v}_2$ where $\text{v}_1$ lies in (\ref{eq:6.3}) and $\text{v}_2$ lies in
 the first summand of (\ref{eq:6.1}) then its stabilizer is
 \emph{exactly} ${\mathbb Z}_2$.

\noindent{\bf Definition.} We will call a two-dimensional
subspace, $V$, of ${\mathbb R}^{2d}$ \emph{exceptional} if its stabilizer
is not the identity and is not ${\mathbb Z}_2$.

We'll now outline an inductive method for determining these
exceptional points in $\mbox{Gr}_2 ({\mathbb R}^{2d})$ and their stabilizers.
Let
\begin{equation}
  \label{eq:6.4}
  {\mathbb R}^2 \oplus \{ 0 \} \oplus \cdots \oplus \{ 0 \}
\end{equation}
be the first summand of (\ref{eq:6.1}).

\begin{lemma}
  \label{lem:6.1}
If $V$ is exceptional then its projection onto (\ref{eq:6.4}) is
either zero or two-dimensional.

\end{lemma}

\begin{proof}
If the projection is one-dimensional
then $V$ is spanned by vectors $\text{v}_1$ and $\text{v}_2$ where
$\text{v}_1$ lies in (\ref{eq:6.4}) and $\text{v}_2$ lies in
(\ref{eq:6.3}).  Therefore, as we noted above, the stabilizer of
$V$ is ${\mathbb Z}_2$ and $V$ is not exceptional.

\end{proof}

The same argument shows

\begin{lemma}
  \label{lem:6.2}

If $V$ is exceptional then its projection onto (\ref{eq:6.3}) is
either zero or two-dimensional.
\end{lemma}

From these results we conclude

\begin{lemma}
  \label{lem:6.3}

If $V$ is exceptional then either it has to be the space
(\ref{eq:6.4}), or be contained in the space (\ref{eq:6.3}), or be
the graph of a map $A$ of (\ref{eq:6.4}) into (\ref{eq:6.3}).
Moreover this map has to be injective.

\end{lemma}

If $V$ is the space (\ref{eq:6.4}) then it is a fixed point for
the action $\tau^\#$, and if $V$ is contained in (\ref{eq:6.3})
then it is an exceptional subspace of ${\mathbb R}^{2d-2}$ and hence
determined by our induction procedure.  Let's focus therefore on
the case where $V = {\rm graph} \, A$, $A$ an injective map of
(\ref{eq:6.4}) into (\ref{eq:6.3}).  We can write this map as a
direct sum
\begin{displaymath}
  A= (A_2,\ldots ,A_d),
\end{displaymath}
where $A_k$ is a map of ${\mathbb R}^2$ into the $k$\st{th} summand of
(\ref{eq:6.3}).  The following result is a consequence of (\ref{eq:4.7}) and
(\ref{eq:4.8}):

\begin{lemma}
  \label{lem:6.4}
Suppose $V$ is stabilized by $\tau^\# (e^{it})$
with $0<t<2\pi$ and $t \neq \pi$.  Then, for each $k\in\{\, 2,\ldots ,d\,\}$,
either

\begin{eqnarray}
 \rm{(i)}  & A_k &=0\,\, \mbox{or}\notag\\
\label{eq:6.5}
 \rm{(ii)} & JA_k &= A_kJ\,\, \mbox{and}\,\, e^{itN_1}=e^{itN_k}
      \,\, \mbox{or}\\
 \rm{(iii)} & JA_k &= -A_kJ \,\, \mbox{and}\,\,
      e^{itN_1}=e^{-itN_k}\, .\notag
\end{eqnarray}

\end{lemma}

\noindent{\emph{Some notation:}} Let $S_1 (A)$ be the set of $k$'s
for which (ii) holds and let $S_2(A)$ be the set of $k$'s for
which (iii) holds.  Also, if $k \in S_1 (A)$ let $\sigma_k =
|N_k-N_1|$ and if $k \in S_2 (A)$ let $\sigma_k (A) = N_1 + N_k$.

Putting together the lemmas above we get the main result of this
section.

\begin{theorem}
\label{th:6.5}
Let $  A= (A_2,\ldots ,A_d)$ be a above, and
let $S(A) = S_1 (A) \cup S_2 (A)$.  Then the graph of $A$ is exceptional
if the components of $A$ satisfy (\ref{eq:6.5}) and the greatest common
divisor of the set of numbers
\begin{equation}
  \label{eq:6.6}
  \{ \sigma_k (A) \, , \quad k \in S (A) \}
\end{equation}
is greater than $2$.

\end{theorem}

Thus one can determine the isotropy groups of the exceptional
points of $\mbox{Gr}_2 ({\mathbb R}^{2d})$ by the following procedure:

\arabiclist
\begin{enumerate}
\item
Fix two disjoint subsets $S_1$ and $S_2$ of the set $\{ 2 ,\ldots
,d \}$.

\item
For each $k \in S_1$ let $\sigma_k = |N_k - N_1 |$.

\item
For each $k \in S_2$ let $\sigma_k = N_k + N_1 $.

\item
Let $\sigma (S_1,S_2)$ be the greatest common divisor of the set
of numbers
\begin{displaymath}
\{\,  \sigma_k \, , \quad k \in S_1 \cup S_2\,\}\, .
\end{displaymath}

\end{enumerate}

Then if $N=\sigma (S_1,S_2)$ is greater than $2$, ${\mathbb Z}_N$ is the
isotropy group of an exceptional point, $V= {\rm graph}\, A$, of
$\mbox{Gr}_2 ({\mathbb R}^{2d})$.

This observation gives us, by induction, the result below.

\begin{theorem}
  \label{th:6.6}

  For $1 \leq r \leq d-1$ let $S_1$ and $S_2$ be disjoint subsets
  of the set $\{ r+1 , \ldots, d \}$, and let $\sigma_k =
  |N_k-N_r|$ if $k$ is in $S_1$ and $\sigma_k= N_k + N_r$ if $k$ is in
  $S_2$.  Then, if the greatest common divisor $N=\sigma
  (S_1,S_2)$ of the set of numbers $\{\,\sigma_k$, $k \in S_1 \cup
  S_2\,\}$ is greater than $2$, ${\mathbb Z}_N$ is the isotropy group of an
  exceptional point of $\mbox{Gr}_2 ({\mathbb R}^d)$.

\end{theorem}

\section{Geodesics on weighted $(d-1)$-dimensional projective
spaces}
\label{sec:7}

As mentioned in the introduction, this space is the quotient of $S^{2d-1}$ by the
circle action (\ref{eq:6.2}), i.e.,~it is the space
\begin{equation}
  \label{eq:7.1}
  X= S^{2d-1} /S^1 \, .
\end{equation}
(The title of this section is a bit of a misnomer since $X$, as a
real $C^\infty$ Riemannian orbifold, is ($2d-2$)-dimensional.) There are ${d}\choose {2}$ totally-geodesic embedded weighted
projective lines $X_{i,j} \ 1\leq i<j\leq d$
in $X$, corresponding to the direct sum of two complex
coordinate lines in ${\mathbb C}^d$.  As we showed in \S 3, each weighted
projective line has a unique exceptional geodesic $\gamma_{i,j}$,
and it has length
\[
\ell_{i,j} = \frac{2\pi}{N_i+N_j}
\]
(assuming, as we always do in this paper, that the weights are pairwise
relatively prime).  In this section we
investigate conditions on the weights which guarantee that the lengths
$\ell_{i,j},\ 1\leq i<j\leq d$, are the {\em shortest} in the length spectrum of $X$.
As a corollary of the discussion of \S 2 in such cases one can ``hear" the
multiset of the $\ell_{i,j}$, and therefore the multiset of pair-wise sums $\{ N_i+N_j,
1\leq i<j\leq d\}$.  This analysis requires that we have a description of the lengths of
the exceptional geodesics on $X$ (other than the $\gamma_{i,j}$).

Let
$\tilde{\omega} \in \Lambda^2 ({\mathbb R}^{2d})^*$ be the symplectic form
\begin{displaymath}
  \tilde{\omega} (\text{v} ,\text{w}) = \sum^d_{i=1} N_i \langle
     J\text{v}_i,\text{w}_i \rangle
\end{displaymath}
where $\text{v}_i$ and $\text{w}_i$ are the projections of $\text{v}$ and $\text{w}$
onto the $i$\st{th} summand of (\ref{eq:6.1}).  The geodesics
on the sphere $S^{2d-1}$ are just the intersections $V \cap
S^{2d-1}$, where $V$ is a two-dimensional subspace of ${\mathbb R}^{2d}$,
and one can show by exactly the same argument as in \S3 that $V
\cap S^{2d-1}$ is \emph{horizontal} if and only if $\iota^*_V
\tilde{\omega}=0$, i.e.,~if and only if $V$ is an isotropic
two-dimensional subspace of ${\mathbb R}^{2d}$.  Given such a geodesic, the
projection $S^{2d-1} \to X$ maps $V \cap S^{2d-1}$ onto a closed
geodesic of $X$ of period $\frac{2\pi}{m}$, where ${\mathbb Z}_m$ is the
stabilizer of $V$ with respect to the action $\tau^\#$ of $S^1$
on $\mbox{Gr}_2 ({\mathbb R}^{2d})$.
In particular, suppose $V= {\rm graph} (A)$ as in the previous section.
Then $V$ is isotropic if and only if
\begin{equation}
  \label{eq:7.2}
  N_1 + \sum^d_{m=2} N_m \det A_m =0,
\end{equation}
and if this is the case there exists $m$ such that $\det A_m$ is negative.
Therefore, if
$V$ is exceptional and isotropic, (\ref{eq:6.5})(iii) must hold for some $m$,
or, equivalently, $S_2(A)\not=\emptyset$.
It follows that the stabilizer
group of $V$ has to be contained in ${\mathbb Z}_{N_1 + N_m}$ for
some~$m$.

We summarize:
\begin{proposition}\label{k}
The lengths of the exceptional geodesics on $X$ are of the form $\ell_k = {2\pi}/k$,
where $k$ is the greatest common divisor of a set of numbers of the form
\[
\mathcal{F}_{S_1,\, S_2} = \{\, \sigma_m\;;\; m\in S_1\cup S_2 \},
\]
and where $S_1,\, S_2$ are sets such that
\[
S_1\cup S_2\subset \{r+1,\ldots d\},\ S_1\cap S_2= \emptyset,\
S_2\not=\emptyset\,
\]
and such that g.c.d.$(\mathcal{F}_{S_1,\, S_2}) > 2$.  Here
\[
\sigma_m =
\bigl\{
\begin{array}{c}
|N_m-N_r|\quad  \text{if}\ m\in S_1\\
N_m+N_r\quad  \text{if}\ m\in S_2.
\end{array}
\bigr.
\]
\end{proposition}

Note that if one takes $S_1=\emptyset$ and $S_2 = \{m\}$ (a singleton), then
$k=N_r+N_m$ gives rise to the length of $\gamma_{r,m}$.

\bigskip
In low dimensions we have the following results:
\begin{theorem}
  \label{thm:7.1}
  Suppose all the $N_i$'s are pair-wise relatively prime.  Then:\\
  $\mathrm{(1)}$ If $d=3$, one can ``hear" the weights
                 $N_1, N_2$ and $N_3$. \\
  $\mathrm{(2)}$ If $d=4$, one can ``hear" all the pairwise
                 sums $\{\,N_i+N_j,\; i<j\,\}$, which will determine at most
                 two different choices of $N_i$'s.
\end{theorem}

\subsection{The case $d=3$}
\label{sec:7.1}

The exceptional
geodesics other than the $\gamma_{i,j}$, if they exist, have to be associated with
isotropics of the form $V = {\rm graph} \, A$, where $A =
(A_2,A_3)$ and each of the $A_i$'s is bijective.  If we order
the $N_i$'s so that $N_1 < N_2 < N_3$, these geodesics will have lengths
of the form $2\pi /k$ where the possibilities for $k$ are
\begin{eqnarray}
  \label{eq:7.3}
  k &=& \text{g.c.d} (N_1 + N_3 , N_2 - N_1)\\
\noalign{\hbox{or}}\notag \\
  \label{eq:7.4}
  k &=& \text{g.c.d} (N_1 + N_2 , N_3 - N_1)\\
\noalign{\hbox{or}}\notag \\
  \label{eq:7.5}
  k &=& \text{g.c.d} (N_1 + N_2 , N_1 + N_3)\, .
\end{eqnarray}
(These possibilities correspond to the set $S_2$ of Theorem \ref{th:6.5}
being equal to $\{3\}$, $\{2\}$ and $\{2, 3\}$, respectively.)
If $k$ is given by (\ref{eq:7.3}) then $k < N_2<N_1+N_2$, and the
geodesic corresponding to $k$ is strictly longer  than the
geodesics $\gamma_1$, $\gamma_2$ and $\gamma_3$.  If $k$ is given
by (\ref{eq:7.4}) then $k<N_1+N_3$ and therefore the geodesic
is strictly longer than $\gamma_2$ and $\gamma_3$.
The geodesic will also be longer than $\gamma_1$ unless $N_1 +
N_2$ divides $N_3 -N_1$, or, equivalently (since $N_2 + N_3 = N_3 -
N_1 + N_1 + N_2$), unless $N_1 + N_2 $ divides $N_2 + N_3$.
Similarly, if $k$ is given by (\ref{eq:7.5}) the geodesic will be
strictly longer unless
$N_1 + N_2$ divides $N_1 + N_3$.  Thus if neither of these two
worst case scenarios occurs, $\gamma_1$, $\gamma_2$ and $\gamma_3$
are the three shortest geodesics on $X$. Moreover, they lie on the
non-singular part of $X$ and are isolated and non-degenerate.
Hence by \cite{DG}, $\gamma_i$ makes a contribution of the form
\begin{equation}
  \label{eq:7.6}
  c_i (t-T_i)^{-1}_+ + \cdots
\end{equation}
to the wave trace with $c_i \neq 0$ where $T_i$ is the period of
$\gamma_i$ and hence one can ``hear'' the $T_i$'s, i.e.,~one can
hear $N_1 + N_2$, $N_1 + N_3$ and $N_2 + N_3$ and hence one can
hear $N_1,N_2$ and $N_3$.

Let's next examine the first worst case scenario:  $N_1 + N_2$ divides
$N_1 + N_3$.  Then $V= {\rm graph}\, A$ is isotropic and stabilized
by ${\mathbb Z}_{N_1 + N_2}$ if and only if $A= (A_{\lambda_2},
A_{\lambda_3}) $ with
\begin{equation}
  \label{eq:7.7}
  N_1 - N_2 |\lambda_2 |^2 - N_3 |\lambda_3 |^2 = 0
\end{equation}
(see (\ref{eq:4.11})).  Thus the set of geodesics stabilized by
${\mathbb Z}_{N_1 + N_2} $ consists of the isolated geodesic $\gamma_1$
on the projective line $X_1$, and completely disjoint from it the
2-parameter family of geodesics lying on the quotient by $S^1$ of
the three sphere (\ref{eq:7.7}).  In this case the wave trace
contains contributions at $T_1 = 2\pi / (N_1 + N_2)$ from both
$\gamma_1$ and from the geodesics (\ref{eq:7.7}).  However, since
the geodesics (\ref{eq:7.7}) also lie on the non-singular part of
$X$ the clean intersection techniques of \cite{DG} show that their
contribution to the wave trace is a singularity of the form
\begin{equation}
  \label{eq:7.8}
  c (t-T_1)^{-3} + \cdots
\end{equation}
which dominates the singularity (\ref{eq:7.6}), so again one can
hear $N_1 + N_2$.

A similar situation occurs if $N_1 + N_2$ divides $N_2 + N_3$. In
this case one gets again a 2-parameter family of geodesics on $X$
corresponding to $A$'s of the form $A= (A_{\lambda_2},
A_{\lambda_3})$ where
\begin{equation}
  \label{eq:7.9}
  N_2 | \lambda_2 |^2 = N_1 + N_3 |\lambda_3 |^2\, .
\end{equation}
The projection of this set onto the space of geodesics on $X$
isn't closed, but its closure  consists of the geodesics in this
set plus the geodesic $\gamma_3$ (which is stabilized by ${\mathbb Z}_{N_2
+ N_3}$ and hence by ${\mathbb Z}_{N_1 + N_2}$).  Thus the clean
intersection techniques of \cite{DG} show that the contribution of
this family to the wave trace is of the form (\ref{eq:7.8}), which
again dominates the contribution (\ref{eq:7.6}) coming from
$\gamma_1$.  Thus, in both these two worst cases scenarios one can
hear $N_1 + N_2$  and therefore all three $N_i$.

\subsection{The case $d=4$}
\label{sec:7.2}

Without loss of generality, we suppose $N_1 < N_2 < N_3 <N_4$.
There are now ${4\choose2}=6$ embedded totally
geodesic weighted projective lines, each one with a unique exceptional geodesic
$\gamma_{i,j}$,
of length $\frac{2\pi}{N_i+N_j}$ (these are the ``desirable" lengths that we wish
to ``hear").   Note that $\gamma_{4,3}$ is the shortest geodesic, $\gamma_{4,2}$
the second shortest, $\gamma_{2,1}$ the longest and $\gamma_{3,1}$ the
second longest of these geodesics, and in general we can't order the lengths of
$\gamma_{4,1}$ with $\gamma_{3,2}$ (in fact they could be equal, e.g.:
$17+3 = 13+7$).

The lengths of undesirable
exceptional geodesics are of the form
$L_k = \frac{2\pi}{k}$, $k$ a positive integer.
We want to bound the possible values of $L_k$ from below, to determine which
desirable lengths are spectrally determined.

According to our inductive procedure for finding the possible $k$'s, we should first consider possible two-dimensional planes contained in
\[
\{ 0\}\oplus {\mathbb R}^2\oplus {\mathbb R}^2\oplus {\mathbb R}^2.
\]
These correspond to the exceptional
geodesics of a $d=3$ weighted projective space, with
weights $N_2,\, N_3,\, N_4$.  As we have seen, the possibilities
for $k$ are the greatest common divisor of the following sets:
\begin{equation}\label{uno}
\{N_4+N_2,\,N_3-N_2\},\quad
\{N_3+N_2,\,N_4-N_2\},\quad
\{N_3+N_2,\,N_4+N_2\}.
\end{equation}
The remaining exceptional geodesics are graphs of maps
$A = (A_2,\,A_3,\,A_4)$.  The possible values of $k$ are the greatest common
divisors of the following seven sets:
\begin{equation}\label{dos}
\begin{array}{cc}
\{N_2+N_1,\,N_3+N_1\}, &\{N_2+N_1,\,N_3+N_1,\, N_4-N_1\}, \\
\{N_2+N_1,\,N_4+N_1\}, &\{N_2+N_1,\,N_4+N_1,\, N_3-N_1\}, \\
\{N_3+N_1,\,N_4+N_1\}, &\{N_3+N_1,\,N_4+N_1,\, N_2-N_1\}
\end{array}
\end{equation}
and
\begin{equation}\label{tres}
\{N_2+N_1,\,N_3+N_1,\, N_4+N_1\}.
\end{equation}

Using only that the g.c.d. of any set of numbers is
less than or equal to any element in the set, one can check that in all
cases $k\leq N_3+N_2$, so we can always hear the lengths of the
two shortest $\gamma_{i,j}$ and determine $N_4+N_3$ and $N_4+N_2$.
In addition, we claim that we can always hear $N_4+N_1$, because the
g.c.d.'s of all the sets listed above are all less than $N_4+N_1$.
The only set for which this is not immediately obvious is $\{N_3 + N_2,\, N_4+N_2\}$;
however, note that
\[
\text{g.c.d}\{N_3 + N_2,\, N_4+N_2\} =
\text{g.c.d}\{N_3 + N_2,\, N_4-N_3\} < N_4 < N_4+N_1.
\]

One can also check that the g.c.d of the sets (\ref{dos})  satisfies
$k\leq N_3+N_1$, and
the g.c.d of the sets (\ref{tres}) $k\leq N_1+N_2$.
Therefore the lengths of the corresponding geodesics are greater than
the length of $\gamma_{3,2}$. If the g.c.d.'s of the sets (\ref{uno}) is less
than $N_3+N_2$, then all undesirable exceptional geodesics are longer
than $\gamma_{3,2}$ as well.   However, by our analysis of the ``worst case scenario"
when one of the g.c.d.'s of the sets (\ref{uno}) equals $N_3+N_2$, we can
conclude that we can always hear  $N_3+N_2$.

In conclusion, the multiset $\{ N_2+N_3, N_1+N_4, N_2 +N_4,
N_3+N_4\}$ is always spectrally determined.
In this set we can identify the individual sums $ N_3 +N_4$ and
$N_2+N_4$ as the largest and second largest elements,
respectively.  Moreover, although we are not able to distinguish between the
two remaining elements in the set which one is
$N_2+N_3$ and which one is $N_1+N_4$, we can form their sum,
\[
N_1+N_2+N_3+N_4,
\]
and from this determine $N_1+N_2$ and $N_1+N_3$,
(by subtracting the known quantities $ N_3 +N_4$ and
$N_2+N_4$).  This shows that we can always hear the set
of all six sums of pairs $\{N_i+N_j, i<j\}$.

To reconstruct the possible weights giving rise to a given multiset of
sums of pairs, proceed as follows:  $N_1+N_2$ and $N_1+N_3$
are the smallest and second smallest elements of the set,
respectively.  However, $N_2+N_3$ may be the third or fourth
element in the set if the set has six elements (the set may have only
five elements, exactly when $N_2+N_3 = N_4+N_1$, in which case this sum
is the middle element and the reconstruction of the weights is unique).
Each one of those possibilities for $N_2+N_3$ gives a unique
solution for $N_1,\, N_2,\, N_3$, and the remaining elements
in the set uniquely determine the fourth weight.

It is clear that picking $N_2+N_3$ to be the third or the fourth element
from a set of six distinct pairwise sums gives different solutions $\{N_j\}$.
However, the two solutions ``usually" do not both consist of four pair-wise
relatively prime numbers, although this can happen.  For example,
\begin{equation*}
   \{25, 29, 41, 61\} \mbox{\ and\ } \{17, 37, 49, 53\}
\end{equation*}
have the same set of pairwise sums.

\subsection{Remarks for $d$ general}

In general, if the ${d}\choose{2}$ shortest geodesics of $X$ are the
$\gamma_{i,j}$ then one can ``hear" the weights if $d$ is not a power
of two.  A way to ensure that the $\gamma_{i,j}$ are the
${d}\choose{2}$ shortest geodesics is the following:

\begin{theorem}
 If $N_1<N_2<\ldots <N_d$,
 $N_d \le 2 N_1$ and the list $N_i+N_j\,,\,1\leq i< j\leq d,$ contains no
repeated elements, then the  $\gamma_{i,j}$ are the ${d}\choose{2}$ shortest geodesics,
(and therefore, by the result in the appendix, the set of weights is spectrally
determined if $d$ is not a power of two).
\end{theorem}
\begin{proof}
Let $\frac{2\pi}{k}$ be an undesirable length of an exceptional geodesic.
How short can the exceptional geodesic be or, equivalently, how large can $k$ be?
Using Proposition \ref{k}, it is not hard to see that
\[
k\leq \text{g.c.d}\{N_i+N_j,\, N_l+N_m\}
\]
for some $i<j$, $l<m$ and $(i,j)\not= (l,m)$.  Thus it suffices to show that such g.c.d.'s are
less than $N_1+N_2$.

First, the condition  $N_d \le 2 N_1$ implies that
$N_l+N_m < 2(N_i+N_j)$.  Therefore
\[
\text{g.c.d}\{N_i+N_j,\, N_l+N_m\}\not= N_i+N_j
\]
(since $a=\text{g.c.d}\{a,b\}\,\wedge\, b<2a\Rightarrow b=a$, and we assumed
that the list $N_i+N_j\,,\,1\leq i< j\leq d,$ contains no
repeated elements).  It follows that $k\leq (N_i+N_j)/2 < N_1+N_2$.
\end{proof}

\begin{appendix}
\section{}
We present in this appendix a proof of the following:
\begin{theorem}
If $d$ is not a power of two, a set of $d$ real numbers
$\{N_1,\ldots ,N_d\}$ is determined by the multiset of sums of pairs
$\{N_j+N_k\ :\ j < k\}$.
\end{theorem}
\begin{proof}
The proof is based on the identity
\begin{equation*}
\sum_{1\le i < j \le d}(N_i+N_j)^k = (d-2^k)\sum_{i=1}^d N_i^k
+ \frac 12 \sum_{m=1}^{k-1}{k \choose m}(\sum_{i=1}^d N_i^{k-m})(\sum_{i=1}^d N_i^m).
\end{equation*}

This shows that if $d$ is not a power of two then one can find $\sum N_i^k$
from $\sum_{1\leq i<j\leq d}(N_i+N_j)^k$ and the sums $\sum N_i^m$ with $m<k$.
Proceeding inductively (starting from $\sum_{1\leq i<j\leq d}(N_i+N_j) = (d-1)\sum N_i$),
we can conclude that the elementary symmetric functions on the sums of pairs
determine the elementary symmetric functions on the $N_j$.
\end{proof}

Finally, we remark that one can generalize the previous discussion of the $d-2$ case
to obtain an algorithm for finding all possible sets of $N_i$'s, given the multiset of
their pair-wise sums.  This algorithm shows that, if
$d$ is a power of $2$ and $d>2$, then the multiset $\{N_i+N_j\}$
will determine at most $d-2$ different choices of $N_i$'s.  However,
it seems very hard to determine what the sharp bound of possible choices of $N_i$'s is, for a given $d$.

\end{appendix}

redpajama_set_name: RedPajamaArXiv


Some excellent advice in here, and some straight talk about how it works.
Hey that's me! Looking forward to our spooky stories night.
As an incentive for our members, we are pleased to announce two discounts at local merchants for Friends members only in October.
On Thursday evenings–October 5, 12, and 26–Antonia's in downtown Hillsborough will offer a 10 percent discount to all diners who present a Friends membership card. And throughout the month, the Hillsborough Wine Company will offer a 15 percent discount on all items except sale items to customers with a Friends membership card. If you have joined or renewed your membership for 2017 but have not received your membership card, contact Susan Vreeland (susanvreeland@gmail.com). Please come out and support these great local businesses all month long.
Now that the shock of Sunday night's mass shooting is wearing off, the inevitable finger-pointing has begun. As usual, the blame is being laid squarely at the feet of the mentally ill.


text:
Morgan – My Story
Ernst Ooi Chen Wei – My Story
Geoffrey – My Story
My name is Morgan. This is my story.
In May 2013 I was a senior at Sundown High School in Texas.
We were taking pictures for the school yearbook's "Senior Who's Who?" page. One of the two superlatives that I earned that year was "Most Political."
For the picture I decided to hold a political magazine to make it look like me and the other student in the picture were having a political discussion. I chose to use the now famous copy of TIME magazine that had a photograph of two women kissing on the cover.
The student photographer refused to take the photo because of the magazine I had chosen and the other student refused to be in the picture with me. The photographer and I got into a rather heated argument about the matter and when he still refused, I went to my principal, who told me that gay rights was not a topic that he wanted discussed at the school. He then told me that I either had to use a different magazine or not use a magazine at all.
I decided that it was not in my best interest to choose either option. I chose to stand up for my right to use the copy of TIME magazine that I had initially chosen.
After school was out for the day I couldn't get any of this off of my mind, and I realized that the issue was no longer about the magazine and photo but about the discrimination my school was holding against the LGBT community and the infringement of my basic constitutional rights.
The next morning when I got to school I spent over an hour decorating my locker in protest of the discrimination I was facing. However, not even an hour after I had finished, the principal tore it down and when I protested he claimed that it was "distracting." He called my mother, hoping to get me in trouble, but instead she took the day off from work and came to the school after a heated phone call with the principal.
We quickly argued our case, giving reasons why the treatment I was receiving wasn't fair and calling him out for discriminating against me and the LGBT community. He still wouldn't budge so we threatened to call a gay rights organization so that everybody could learn about the discrimination being practiced at my high school. As the threat of media attention loomed over his school he decided that he would allow me to use the magazine in the photograph for the yearbook after all.
I then posted my story to Gay Marriage USA's Facebook page. Within the first hour it already had about 1,000 likes. To date it has been shared more than 3,000 times.
My story would go on to become viral, garnering nearly 10,000 likes and thousands of comments. I had tons of support on Facebook and the Internet but at school it was a different matter. Being an ally here in Texas is just as difficult as being LGBT here in Texas. It is not a stereotype to say that Texas is full of homophobic rednecks and bible thumpers, because it's very true – especially in the "Bible Belt," which is where I went to school.
I was shunned. My story went up on a Friday and by the following Monday my best friend stopped talking to me and nobody at school wanted anything to do with me. I could tell who really cared about what had happened and what had not happened.
Students in the halls whispered and made it a point to make rude and jeering remarks, all of which my principal ignored, of course.
Later I saw an article written by Kristen Henderson, one of the women in the picture on the cover of TIME, in the Huffington Post where she said "Once I got over the jolt of seeing my wife and me kissing on the cover of Time (not because we're kissing but because it's the cover of Time), I read the story about a really brave girl doing something that I would not have had the courage to do at her age."
"It never occurred to us that our choice to pose for the cover of TIME would result in an 18-year-old high school senior from Sundown, Texas, becoming a hero of a movement.
What Morgan did by standing up for my family will save lives. And what her mother did alongside her will help other mothers – mothers whose kids are struggling with their own sexuality, mothers who need to teach their own children tolerance, love and acceptance – do the right thing."
I've always known that I am an ally but I never felt that I had proven myself to be one until the day I stood up to my principal and, by extension, to my whole town. That day everything changed for me and I knew that my story was one that had to be told because I'm sure there is more than one person in the world that's like me. Somebody that knows what matters in this fight and wants to do something but doesn't have much of a platform. This event may seem small, but in all honesty it sent a shockwave through our staunchly Christian town that has a population of only 1,300 people.
I have a whole archive of messages filled with hate towards me because of my involvement with the LGBT community or because of the story I posted to the Internet. I also have an archive of messages of love and support that warm m my heart every time I think of them.
Both the negative and positive stories are vital in learning what is most important in spreading tolerance. I have changed the hearts of a few already and have garnered respect from areas where I wasn't sure I'd receive it.
It's not always easy to change people's hearts and minds but it is possible. Some people are homophobic because that's just how it is where they live. If you make them knowledgeable of issues that stem from homophobia and issues the LGBT community is facing today I find that a lot of people will change their minds. I think I really have what it takes to teach people how to accept people who are LGBT.
One of the next steps I took to achieve LGBT equality was to continue to take advantage of the resources my university has for me. This semester I plan to take part in Ally Training. This is a free course offered by my university on how to handle LGBT issues in school, the workplace, and politics as an Ally. I am looking forward to finding out new and exciting ways I can help with the LGBT equality movement.
In the meantime I'll just keep sharing my story, because somewhere out there are others like me who don't know where to begin. It's not just a matter of gay rights but it's a matter of human rights. Or in the words of Chelsea Cannell from the movie "Bridegroom" – "It's not a gay thing. It's not a straight thing. It's a human thing."
Morgan hopes that sharing her story can help LGBT youth. A portion of your tax-deductible donation will go to The Trevor Project. CLICK HERE to donate.
Learn how to share your unique story here.
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The Personal Stories Project is a not-for-profit 501(c)(3) corporation, EIN #46-4157080. All or part of your contributions may be tax deductible.
Copyright © 2019 The Personal Stories Project. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


How would you like something different like this at your Wedding or Event, with Ribbons and Balloons it's Stunning…!
Decorating with ribbon is a relatively easy way to add lots of pizzazz to your wedding ceremony and reception. In this post I will share some ideas you may not have seen before. I won't be talking about using ribbon on bouquets, aisle markers or on little boxes for your wedding favors … you probably know about that already!


The latest leotards from Milano are now available from Dancemania. Milano's Spring 2013 range of gymnastics leotards will be available until 30th September 2013, so longer as stocks last.
To add extra kudos to these leotards they have been modelled by three of GB's Artistic Gymnastics Team: Imogen Cairns, Hannah Whelan and Rebecca Tunney – all veterans of the London 2012 Olympics.
Milano Annie Bodice in Indian Ocean Sparkle, Amazon and Perfect Pearl with Scattered Cobalt Diamante.
Colour: Indian Ocean Sparkle, Amazon and Perfect Pearl .
Milano Calla Bodice in Lime Sparkle and Platinum with detailing in Indian Ocean Sparkle and Aqua and Christie Diamante.
Colour: Lime Sparkle and Platinum with detailing in Indian Ocean Sparkle and Aqua.
Milano Chiara Bodice in Damson Delight Sparkle and Onyx with Everest Drop Diamante.
Colour: Damson Delight Sparkle and Onyx.
Milano Jupiter Bodice in Powder Blue and Ebony with Platinum swirls, Powder Blue Stars and Clear Nadine Diamante.
Colour: Powder Blue and Ebony with Platinum swirls, Powder Blue Stars.
Milano Jupiter Leotard with Long Sleeves in Navy Glitter and Kingfisher with Platinum swirls, Navy Glitter Stars and Clear Nadine Diamante.
Colour: Navy Glitter and Kingfisher with Platinum swirls, Navy Glitter Stars.
Milano Loire Bodice in Kingfisher and Lime Sparkle with Platinum detailing and Loire Diamante.
Colour: Kingfisher and Lime Sparkle with Platinum detailing.
Milano Perry Bodice in Sugar Plum with a dramatic shock of Ice fabric and glistening Perry Diamante.
Colour: Sugar Plum with a dramatic shock of Ice fabric.
Milano Spektro Bodice in Pineapple Crush and Ebony with Graphite details and Makalu Drop Diamante.
Colour: Pineapple Crush and Ebony with Graphite details.
Milano Waverly Bodice in Damson Delight Sparkle, Party Pink Sparkle, Marshmallow with Ebony trim and Waverly Diamante.
Colour: Damson Delight Sparkle, Party Pink Sparkle, Marshmallow with Ebony trim.
Last updated by Dancemania at February 20, 2013 .


text:
Adopt-a-Beach team leader connects the ocean to the Great Lakes
Adopt-a-Beach Team of the Week
A weekly celebration of our awesome Adopt-a-Beach volunteers!
Long-time Adopt-a-Beach team leader Stephanie Kreuger may have been absent from the cleanup she coordinated June 9th, but she was far from absent from the cause. While her new organization — the Buffalo NY Ocean Coalition — cleaned up Woodlawn State Park Beach, Kreuger was in Washington, D.C. to have a presence in the fight to protect the world's oceans.
Volunteers clean up Woodlawn State Park Beach
Maximizing her influence, Krueger attended the March for the Ocean, an event with more than 150 partners in Washington, D.C., and with many sister marches in other countries. According to the event's website, the March for the Ocean is about "the survival of our blue planet. But it's also about the fact that it's not too late to turn the tide – restore and protect what we love…The ocean is rising and so are we."
Back in Buffalo, 13 members and friends of her new chapter of the Inland Ocean Coalition tackled the Woodlawn Beach cleanup. Demonstrating the connectivity of our waterways via Adopt-a-Beach seemed only natural for the group. Each chapter addresses issues unique to its locale and leads programs to help relay the land-to-sea concept.
The event was held the weekend after World Oceans Day on Friday, June 8th. "Although June has been declared World Oceans Month, so we should celebrate all month," Krueger quipped.
Even as sea levels rise, the Great Lakes lower in response to climate change. Two independent ecosystems, they are nonetheless united in human impact. For Krueger, it's important to pair advocacy for our oceans with mindfulness of our freshwater.
"The coalition focuses on how we affect our oceans even though we don't live on a coast, and now we can show how the two are connected and how plastic from Lake Erie can end up in the ocean," she said.
Krueger continues to advance opportunities for the Coalition to partner with the Alliance for the Great Lakes and Adopt-a-Beach, to seize the opportunity to network and increase volunteer engagement by advertising via multiple platforms. "And it's fun just joining forces," she said.
Volunteers removed 270 lbs. of litter from Woodlawn Beach State Park
With cleanups dating back to 2011, Krueger is a tireless supporter of the Adopt-a-Beach program. She says she admires the Alliance's efforts to collect litter data. "I like being a team leader with the Alliance because…I enjoy trying to raise awareness about the amount of plastic pollution in our Great Lakes and waterways."
Krueger's crew also ventured out for an Earth Day cleanup earlier this year. Between the two events, they picked up about 270 pounds of trash, 2,774 pieces of which are plastic items.
Keep Plastic Out of the Great Lakes
Plastic pollution is a big problem on Great Lakes beaches — just ask our Adopt-a-Beach volunteers. But you can help stop the flow of plastic into the Great Lakes. Sign the Plastic-Free Great Lakes pledge today!


The Babystyle prestige is where tradition seamlessly meets modern elegance. The Prestige comes with everything you need for your little one from birth to approximately 3 years old. In the Babystyle Prestige your little one will travel in style - whatever the weather, whatever the adventure, from winter to summer the matching footmuff and parasol will keep them protected. With so many chassis and colour options you're sure to find one that suits you.
Suitable from birth. Available on a fully sprung Classic chassis in grey or chrome and a swivel wheel Active chassis in grey and white.Compatible with Prestige matching car seat, and Maxi-Cosi Pebble/Pebble Plus and CabrioFix car seats. Height adjustable handle and adjustable pushchair footrest. Parasol. All models have a ventilated hood and ventilated pram body base. Removable and washable pram body liner.Suitable for occasional overnight sleeping.Rocking pram body with stabilising feet. Large shopping basket. Choice of wheel designs for the Classic chassis. Optional "zippy" rain cover for car seat and pram body. Adjustable multi position pram body sit up back rest. Multi position reclining seat unit, lie flat. Front and rear facing seat unit.
Chrome white wall, chrome rim, pneumatic tyre with white stripe.
Chrome solid,chrome rim, with solid EVA composite tyre.
14″ Chrome white air wheel, chrome rim with pneumatic tyre.(Surcharge).
Chassis options: Active Grey, Active White, Classic Grey, Gold and Classic Chrome.
Marlin (Black top/ White base). Thunder Cloud (White top/ Black base). Ballerina (Pink top/ white base). Shark (Black top/ Black base). Dolphin (Light grey top/ Black base). Blizzard (White top/ White base). Blaze Grey (Mid grey top/ Black base). Horizon (Black top/ City Scape on white base). Carbon Gold (Black top/ Metalled looking grey base). Frost (Light grey top/ Light grey base). Flint (Mid grey top/ Mid grey base).
Classic chassis includes: Chrome air whitewall or chrome solid wheels.
Optional Classic chassis wheel upgrade: £20 surcharge per set. Options: 14″ Chrome white air wheel or 14″ Chrome black air wheel.


text:
Caitlyn Jenner says 'hello world' on cover of Vanity Fair
Caitlyn Jenner has been struggling with her gender identity for decades. Now the former Olympic champion shares her true identity with the world.
By Cyd Zeigler@CydZeigler Jun 1, 2015, 9:55am PDT
Share All sharing options for: Caitlyn Jenner says 'hello world' on cover of Vanity Fair
Caitlyn Jenner has made a splash on the cover of Vanity Fair, in a photo by Annie Leibowitz that will have everyone talking (watch the behind-the-scenes video below). Jenner is the former Olympic decathlete who made a name for herself in the mid-Seventies, and who has finally revealed the name she had been hiding for so long.
Jenner came out publicly in an interview with Diane Sawyer last month, yet had elected to continue using her birth name and male pronouns. Now she takes the plunge publicly - on the first day of Pride Month - to let the world know her true identity. How powerful was her need to share her identity with the world?
"If I was lying on my deathbed and I had kept this secret and never ever did anything about it, I would be lying there saying, 'You just blew your entire life,'" she tells Vanity Fair.
Congratulations, Caitlyn Jenner, on an incredible cover photo, and thank you so much for being true!
Note to readers: It is important to now consistently use female pronouns for Caitlyn and to not use her former name. Her name is Caitlyn Jenner. Thank you.


text:
Red Cross admits $6-million fraud during Ebola crisis
By AFP Nov 4, 2017 in World
The Red Cross has uncovered several cases of fraud by officials during efforts to combat the Ebola outbreak that struck West Africa in 2014-2016, estimating losses of $6-million (5.2 million euros).
In a statement, the Geneva-based International Federation of Red Cross and Red Crescent Societies (IFRC) said it was "outraged", and said it would "ensure any staff involved are held to account."
The Ebola virus killed more than 11,300 people and infected an estimated 29,000, mainly in Guinea, Sierra Leone and Liberia.
The IFRC said the operations set up to contain the outbreak "were among the most complex in recent humanitarian history," with several groups moving quickly to deploy staff and mobilise volunteers, and spending tens of millions of dollars on medical equipment and other supplies.
In Sierra Leone, it said it had found evidence of "likely collusion" between former IFRC employees and a bank, leading to a potential loss of $2.1-million.
Overbilling and fake invoices by a provider of customs clearance services in Guinea cost the organisation $1.2 million, and two other investigations in the country are under way.
The IFRC also said it had previously found that in Liberia, inflated prices for relief items and payroll costs had cost it $2.7 million.
The organisation said it was "committed to holding all those involved in any form of fraud to account, and to reclaiming all misappropriated, diverted, or otherwise illegally taken funds."
It also said that since 2014 it has bolstered efforts to root out corruption in its operations, including cash spending limits in "high-risk settings" and deploying trained auditors when sending out teams.
More about Ebola, Liberia, sleone, Guinea, redcross
Ebola Liberia sleone Guinea redcross Fraud