text:
{\bf Problem.} The average of Amy's, Ben's, and Chris's ages is 9. Four years ago,  Chris was the same age as Amy is now. In 3 years, Ben's age will be $\frac{2}{3}$ of Amy's age at that time. How many years old is Chris now?
{\bf Level.} Level 4
{\bf Type.} Algebra
{\bf Solution.} Let Amy's, Ben's, and Chris's ages be $a$, $b$, and $c$, respectively. We have the equations \begin{align*} \tag{1}
\frac{a+b+c}{3}=9 \Rightarrow a+b+c&=27 \\ \tag{2}
c-4&=a\\ \tag{3}
b+3&=\frac{2}{3}(a+3)
\end{align*} From Equation (3), we have $b=\frac{2}{3}(a+3)-3$. We substitute Equation (2) into Equation (3) to eliminate $a$, to get $b=\frac{2}{3}(c-1)-3$. Substituting this last equation and Equation (2) into Equation (1) to eliminate $a$ and $b$, we have  \[[c-4]+[\frac{2}{3}(c-1)-3]+c=27\] Solving for $c$, we find that $c=13$. Thus, Chris's age is $\boxed{13}$.

set_name: MATH


text:
\begin{document}

\title{Fubini Theorem for pseudo-Riemannian metrics.\footnote{{\bf MSC:} 53A20,  53B21, 53C22, 53C50,  53D25, 70G45, 70H06,  70H33,  58J60}}
\date{} \author{Alexey V. Bolsinov,

Volodymyr  Kiosak,

Vladimir S. Matveev

}
\maketitle

 \begin{abstract}
 We generalize the following classical result of Fubini \cite{Fubini1}  for pseudo-Riemannian metrics: if three essentially  different  metrics on $M^{n\ge 3}$ share  the same unparametrized geodesics, and two of them (say, $g$ and $\bar g$) are strictly nonproportional (i.e., the minimal polynomial  of $g^{i\alpha } \bar g_{\alpha j}$ coincides with the characteristic polynomial) at least at  one point, then they have constant curvature.
\end{abstract}

\section{Introduction}
\subsection{Definitions and results}

Two Riemannian or pseudo-Riemannian metrics on the same manifold are said
to be \emph{geodesically equivalent}, if they have the same
geodesics considered as unparametrized curves.  Two metrics $g$ and
$\bar g$ are \emph{strictly nonproportional} at a point $x\in M$, if
the minimal polynomial  of the (1,1)-tensor $G^i_j:= g^{i k} \bar
g_{k j}$  coincides with the characteristic polynomial.  If one of
the metrics is Riemannian, strict nonproportionality  means that all
eigenvalues of $G$ have multiplicity 1.

The main result of the present paper  is the following

\begin{Th} \label{thm1}  Let $g$, $\bar g$ and $\hat g$ be three geodesically  equivalent metrics on a connected manifold $M^n $ of dimension $n\ge 3$. Suppose   there exists a  point  at which
$g$ and $\bar g$ are strictly nonproportional,  and    a point  at which
  $g$, $\bar   g$ and $\hat g$ are linearly independent. Then, the metrics $g, \bar g $ and $\hat g$  have constant curvature.
\end{Th}

If the metrics are Riemannian,  the local version of Theorem \ref{thm1} was proved by
  Fubini  in  \cite{Fubini1,Fubini2}.  The proof of Fubini is short and elegant, but unfortunately, does not work in the pseudo-Riemannian case. More precisely, Fubuni's proof is based on the following classical result:

\begin{Th}[Levi-Civita \cite{Levi-Civita}]
{\it  Assume $g$ and $\bar g$ are Riemannian geodesically equivalent
metrics that are strictly nonproportional at a point $p$.  Then, in
a neighborhood of this point there exists a coordinate system
$(x_1,...,x_n)$ such that the metrics $g,\  \bar g $ are  as
follows: \label{Levi-Civita}

\begin{eqnarray} \label{lcg}
ds^2_g &= & \sum_{i=1}^n  \left[\left|\prod_{\stackrel{j=1}{ j\ne i}}^n\left(X_i(x_i)- X_j(x_j)\right)\right| dx_i^2  \right], \\
ds^2_{\bar g} &= & \sum_{i=1}^n  \left[\frac{1}{X_i(x_i)\prod_{ \alpha = 1 }^n X_\alpha(x_\alpha)} \left|\prod_{\stackrel{j=1}{ j\ne i}}^n \left(X_i(x_i)- X_j(x_j)\right)\right| dx_i^2  \right],  \label{lcbg}
\end{eqnarray}
where  $X_i$ is a
  positive function of the
variable $x_i$ only. Moreover, every two metrics of this form are geodesically equivalent.     } \end{Th}

Fubini calculated the  riemannian curvature $R_{ij,km}$ of the metric \eqref{lcg} and observed that  the component $R_{ij,km}$ is non-zero only if $(i,j)= (k,m)$ or if $(i,j)= (m,k)$. Nowadays,  such curvature tensors are called \emph{diagonal}, see for example \cite{zaharov}.  Then he showed that, unless the curvature is constant,   the curvature tensor determines the coordinates lines $(x_1,...,x_n)$ uniquely. Thus, the metrics $g$, $\bar g$ and $\hat g$ must be simultaneously diagonalisable in a certain coordinate system and a short analysis shows that they are    linearly dependent at  every point.

Our proof, which is also valid in the pseudo-Riemannian setting, is
as follows. We study the partial differential equation \eqref{basic}
which is responsible for the fact that the metric $\bar g$ is
geodesically equivalent to $ g$. The unknown functions in this
equation are the components of a $(0,2)$-tensor $a_{ij}$ canonically
constructed by metric $\bar g$.   Then,   we find the integrability
conditions for this equation. This is a system of linear equations
on $a$ and the Hessian of $\lambda:= \tfrac{1}{2}a_{\alpha \beta }
g^{\alpha \beta}$; the coefficients  in these equations are algebraic
expressions of entries of the curvature tensor and the metric. We show that if the system has two solutions, then for certain $K\in \mathbb R$
the components of the curvature must satisfy the
condition $a_{i}^{ \alpha} Z_{\alpha jkl} +  a^{\alpha}_{j}Z_{\alpha ikl}=0$,
where $Z_{ijkl}= R_{ij,kl}- K \cdot ( g_{il}g_{jk} -g_{ik}g_{jl} )$. Then, it is an easy exercise in linear algebra to show that if in addition  the metrics $g$ and $\bar g$ are strictly nonproportional, then this  condition  on the curvature implies that the curvature is actually constant. Then, the curvature of the metrics $\bar g$ and $\hat g$ is  constant as well by the Beltrami Theorem.

\begin{Rem} We emphasize that the essential part of our proof is, in fact,  the analysis of algebraic properties of the integrability conditions  for  the  ''geodesic equivalence equation"  \eqref{basic}.  Algebraic aspects  of  the integrability conditions will  also be clarified in Section \ref{bolsinov} where we discuss an  unexpected relationship between geodesically equivalent metrics and the so-called sectional operators on semisimple Lie algebras.  As an application of this observation, we give  an alternative,  pure algebraic  proof of the (local version of the) Fubini theorem,  and we believe that  this idea  might be useful in a wider context.

\end{Rem}

\begin{Rem}  All assumptions in Theorem \ref{thm1} are important. Indeed:
\begin{itemize}
\item if the dimension $n$ is 2,  counterexamples were constructed in \cite{koenigs}, see also \cite{bryant};

\item if the metrics are not strictly nonproportional, counterexamples can be found in \S9 of  \cite{s1}, see also  \cite{shandra,s2,s3};

\item if we omit the
assumption that $g$, $\bar g$, and $\hat g$ are linearly
independent, then  we can take $\hat g= \bar g$, and  the
Levi-Civita metrics (\ref{lcg}) and (\ref{lcbg})  for generic $X_i$
give a counterexample to Theorem~\ref{thm1}.

\end{itemize}
\end{Rem}

\subsection{Motivation  }

The first  motivation, which was the reason why Fubini studied this question,  came from the study of projective vector fields of pseudo-Riemannian metrics. Recall that a vector field $v$ is \emph{projective} \weg{(\emph{affine}, respectively)}   with respect to the metric $g$ if its (local) flow takes geodesics to geodesics\weg{(preserves the Levi-Civita connection of $g$, respectively)}.   Projective and affine vector fields are  very classical objects  of investigation: both Lie \cite{Lie} and Schouten \cite{schouten}
explicitly formulated the problem of constructing all metrics admitting one or sufficiently many projective vector fields.
As a direct corollary of Theorem ~\ref{thm1} we obtain

\begin{Cor} \label{cor1} Suppose  that $g$ and  $\bar g$ are  geodesically equivalent metrics of a nonconstant curvature  on a connected manifold $M^n $ of dimension $n\ge 3$. Suppose   there exists a  point such that
$g$ and $\bar g$ are strictly nonproportional.    Then, the dimension of the space of projective vector fields minus the dimension of the space of homothety    vector fields is  at most one.
\end{Cor}

Recall that the vector field $v$ is \emph{ a homothety vector field,} if $L_v g = \const\cdot g$, where $L_v$ is the Lie derivative. We allow the case $\const =0$, so  Killing vector fields are also homothety vector fields.

\begin{Rem} In   dimension  2, Corollary \ref{cor1}  is wrong. Indeed, Darboux-superintegrable metrics  admit three projective vector fields and, as a rule,  only one homothety  vector field, which is a Killing vector field. The definition of Darboux-superintegrable metrics  and a description of their projective vector fields can be found in \cite{bryant}.
Moreover, it is possible to show that there are no other counterexamples to the 2-dimensional version  of Corollary \ref{cor1}. This  is a very nontrivial statement which follows  from  the results of  \cite{bryant, alone}, where all 2-dimensional metrics admitting projective vector fields were constructed.
\end{Rem}

Moreover, it is possible to use Theorem ~\ref{thm1} in order to describe all projective vector fields of a metric $g$  under the assumptions that there exists a metric $\bar g$
which is geodesically equivalent to $g$ and strictly nonproportional to $g$.
Indeed, in this case the system of PDE on $v$ is a system of ODE's,
which is much easier to analyse than a system of PDE's, see \cite{Aminova1, Aminova2, bryant, Fubini1,  obata, sol,CMH, archive,    alone}, where this ODE-system was obtained and   completely solved in particular cases.

One more  motivation comes from the theory of superintegrable
systems. Recall that a metric is called \emph{superintegrable}, if
the number of independent integrals of special form is greater than
the dimension of the manifold. There are different possibilities for
the special form of integrals; de-facto the most standard special
form  of the integrals is  the so-called Benenti-integrals, which
are essentially the same as geodesically equivalent strictly
nonproportional metrics, see \cite{Benenti3,benenti,era1}.  The results
of the present paper show that    Benenti-superintegrable  metrics of
nonconstant curvature cannot exist, which was a folkloric
conjecture.

  \subsection{ History  }
The theory of  projective transformations has a long and fascinating
history. The first  non-trivial examples of projective transformations
were discovered  by   Lagrange \cite{lagrange,beltrami_short}.
Geodesically equivalent metrics were studied by Beltrami, Darboux,
Levi-Civita, Painleve,  Eisenhart, Weyl and other classics.  One can
find more historical details in the surveys
\cite{Aminova2,mikes,duna} and in the introduction to the papers
\cite{archive,hyperbolic,topology,dedicata,fomenko,sbornik,integrable,quantum}.

The pseudo-Riemannian version of Fubini's  Theorem  was investigated
by  the Kazan school of geometry, in particular  by A.\,Aminova and her
collaborators. In particular, they tried to mimic the Fubini proof for
pseudo-Riemannian metrics. They used the description of geodesically
equivalent pseudo-Riemannian metrics obtained by Aminova
\cite{Aminova}, which is a generalization of Levi-Civita's Theorem
\ref{Levi-Civita}. Unfortunately, the description by Aminova is very
complicated, which makes this programm to be computationally very
hard. Besides, there are infinitely many different types of normal
forms for pseudo-Riemannian metrics (depending on the Jordan form of
the tensor $G$), and each of these types require a separate
inverstigation. According to \cite{zakirova},  they proved
Theorem~\ref{thm1} in  dimensions up to 6, though we did not find
the place where  the proof is written.

\subsubsection*{Acknowledgements}
Alexey Bolsinov thanks Russian Foundation  for Fundamental Research  for partial financial support
and Mathematisches Institut, Friedrich-Schiller-Universit\"at, Jena  for hospitality.
Vladimir Matveev   is grateful to   R.~Bryant and   M.~Eastwood for useful discussions, and MSRI for the hospitality.  Volodymyr Kiosak and Vladimir Matveev thank
Deutsche Forschungsgemeinschaft (Priority Program 1154 --- Global Differential Geometry)
  and FSU Jena for partial financial support.

 \section{Proof}
 \subsection{ Schema of the proof} \label{schema}

We first show (Section \ref{tmp}) that if geodesically equivalent
metrics are strictly non-proportional at least at one point, than it
is so at almost every point.  Then, we will show (Section
\ref{tmp1}) that if $g$, $\bar g$, and $\hat g$  are linearly
dependent at every point of a neighborhood, they are linearly
dependent at every point of the manifold.

Combining these two observation, we see that if the assumptions of
Theorem ~\ref{thm1} are fulfilled at one point, they are fulfilled
at almost every point, i.e., though the theorem is global, it is
sufficient to prove it locally. This will be done in Sections
\ref{main}, \ref{111}.

 \subsection{What we will use in the proof}  \label{mittel}

There are many  tensor reformulations of the condition ``metrics $g$
and $\bar g $ are geodesically equivalent". In our paper we will use
the following one, which was suggested by Sinjukov \cite{Sinjukov},
see also \cite{benenti,eastwood}: given two metrics $g$ and $\bar g$,
consider the $(0,2)-$tensor
 \begin{equation} \label{a}
 a_{ij}:=  \left|\frac{\det(\bar g)}{\det(g)}\right|^{\frac{1}{n+1}}\cdot g_{i\alpha} \bar g^{\alpha \beta }  g_{j\beta}, \end{equation}
 and the function
 \begin{equation}\label{lambda}  \lambda := \frac{1}{2} a_{\alpha \beta} g^{\alpha \beta} \end{equation}
 where and $g^{\alpha\beta}$ and $\bar g^{\alpha\beta}$ denote the tensors dual to $g_{ij}$ and $\bar g_{ij}$ respectively, i.e.,
$g^{i\alpha} g_{\alpha j}= \delta_j^i$ and $\bar g^{i\alpha} \bar
g_{\alpha j}= \delta_j^i$.
 \begin{Th}\cite{Sinjukov, benenti}
 The metrics $g$ and $\bar g$ are geodesically equivalent, if and only if
 \begin{equation} \label{basic}
 a_{ij,k}= \lambda_{,i} g_{jk} + \lambda_{,j}  g_{ik},
 \end{equation}
 where the covariant derivative is taken with respect to the
 (Levi-Civita connection of the) metric $g$.
 \end{Th}

 We will also use the following connection between geodesically equivalent metrics and integrable geodesic flows due to \cite{MT}.

Let $a_j^i$ be as in \eqref{a}, with one index lifted by $g$ (so now $a$ is a $(1,1)-$tensor,
   self-adjoint  with respect to $g$).
Consider the family $S_t$ of $(1,1)$-tensors
\begin{equation}\label{st}
 S_t\eqdef \det(a - t\ \mbox{\rm Id})\left(a-t\ \mbox{\rm Id}\right)^{-1}, \ \  \ t\in \mathbb{R}.
 \end{equation}
\begin{Rem}
Although $\left(a-t\ \Id\right)^{-1}$ is not defined for
$t$
lying
in the spectrum of $a$, the tensor  $S_t$  is well-defined
for every  $t$. Actually, the matrix of $S_t$ is the comatrix of the matrix of $a- t \ \Id$. In particular,   $S_t$
is a  polynomial
 in $t$ of degree $n-1$
with coefficients being  (1,1)-tensors.
\end{Rem}
\noindent We will identify the tangent and  cotangent bundles of $M^n$ by $g$.
This identification allows us to transfer the
 natural  Poisson structure from  $T^*M^n$ to   $TM^n$.

\begin{Th}[\cite{ MT, Topalov, dedicata}]
\label{integrability}
 If $a$ is a solution of \eqref{basic},  then  for every  $t\in \mathbb{R}$  the function
\begin{equation}\label{integral}
I_{t}:TM^n\to \mathbb{R}, \ \ I_{t}(\xi)\eqdef g(S_{t}(\xi),\xi)
\end{equation}
is an  integral for the geodesic flow
 of  $g$.
\end{Th}

Recall that a  function is an \emph{integral} of the geodesic flow, if it is constant  along the  trajectories, i.e., along the curves on the tangent bundle of the form $(\gamma(t), \dot \gamma(t))$, where $\gamma$ is a geodesic and $\dot\gamma(t)$ is its velocity vector at a point $t$.

We will also  use the following statement, whose Riemannian version was obtained in  \cite{MT}, and pseudo-Riemannian generalisation is  due to Topalov \cite{involutivity}. As in the introduction, we denote by $G$ the $(1,1)-$tensor $G^i_j:= g^{i k} \bar
g_{k j}$.

\begin{Th} [follows from  Theorem 2 and Section III B  of \cite{involutivity}] \label{4}
Suppose the degree of the minimal polynomial of $G$  is $r$ at every point of the neighbourhood $U(p)$. Then, for arbitrary mutually  different  $t_1,...,t_{r+1}\in \mathbb{R}$ the functions $I_{t_1}$,...,$I_{t_r}$ are functionally independent, and
the function $I_{t_{r+1}}$ is a linear combination of the
functions $I_{t_1}$,...,$I_{t_r}$ (the coefficients of the linear combination are constant).
\end{Th}
Recall that functions $f_1,...,f_r$ are \emph{functionally independent}, if their differentials are linearly independent almost everywhere.

In other words, the number of independent integrals among  $I_t$ is
the degree of the minimal polynomial. In particular, if the metrics
are strictly nonproportional at a point $p$ (which immediately
implies that  they are strictly non-proportional at every point
in a small neighbourhood $U(p)$), then the differentials of the
integrals are linearly independent at almost every point of $TU(p)$.

\subsection{If geodesically equivalent metrics are strictly nonproportional at one point, then they are strictly nonproportional at almost every point. }  \label{tmp}

Let geodesically equivalent metrics $g$ and $\bar g$ be  strictly
nonproportional at $p$. Consider a  geodesic $\gamma$ passing
through $p:= \gamma(0)$. Let us show that  every point $q:=
\gamma(\tau)$ of the geodesic   has a neighborhood $U(q)$ such
that at almost every point of the neighborhood the metrics are
strictly nonproportional.  Since every point can be reached by a
finite sequence of geodesics, and since the condition ``the minimal
polynomial of $G$ has degree $n$" is an open condition,  this will
prove the statement formulated in the title of this section.

As we recalled in Section \ref{mittel}, for almost every point of
$TU(p)$ the differentials of the integrals $I_{t_1},..., I_{t_n}$
are linearly independent. Take a sequence of points $(p_k, v_k)\in
TU(p)$ converging to $(\gamma(0), \dot\gamma(0))$ such that
$dI_{t_1}(p_k, v_k),\dots , dI_{t_n}(p_k, v_k)$ are linearly independent. Consider the
sequence of the geodesics $\gamma_k$ such that $\gamma_k(0)= p_k$
and $\dot\gamma_k(0)= v_k$. For sufficiently large $k$, the geodesics
exist up to time $\tau$ and the sequence
$(\gamma_k(\tau),\dot\gamma_k(\tau))$ converges to $(\gamma(\tau),
\dot\gamma(\tau))$.

Since the  integrals are preserved by the geodesic flow, the
differentials of the integrals are also preserved by the geodesic
flow. Then, at the points $(\gamma_k(\tau),\dot\gamma_k(\tau))$ the
differentials of $I_{t_1},..., I_{t_n}$  are linearly
independent. By Theorem \ref{4}, in an arbitrary small neighborhood
of $\gamma_k(\tau)$ there exists a point $q_k$ such that  the
metrics are strictly nonproportional at  $q_k$ implying the claim, \qed

\subsection{ If  $g_{i k}$, $ \bar  g_{k j}$, and $\hat  g_{k j}$
 are linearly dependent at every point of a neighborhood $U$, then they are linearly dependent
 at every point  of $M$.} \label{tmp1}

Within this section we assume that $g$, $\bar g$, $\hat g$ are geodesically equivalent metrics on a connected manifold $M$ of dimension $n\ge 3$.
We consider the tensor $a_{ij}$ and the function $\lambda$ given by (\ref{a},\ref{lambda}).
The same objects for the pair of metrics $g, \hat g$ will be denoted by the  capital letters $A$ and $\Lambda$, i.e.,
\begin{equation} \label{A}
 A_{ij}:=  \left|\frac{\det(\hat g)}{\det(g)}\right|^{\frac{1}{n+1}}\cdot g_{i\alpha} \hat g^{\alpha \beta }  g_{j\beta} , \ \ \  \  \  \Lambda := \frac{1}{2} A_{\alpha \beta} g^{\alpha \beta} \end{equation}

We will first prove the following statement (essentially due to Weyl \cite{Weyl}).

\begin{Lemma}\label{weyl} Suppose  $a$ and $A$ are solutions of \eqref{basic}. Assume
 $a = C \cdot A$, where $C$ is a function.  Then, $C$ is a constant.
\end{Lemma}

{\bf Proof. } Our proof is different from the proof of Weyl and is
based on the ideas developed in  \cite{dedicata}.  Note that in the
proof we use only the fact that the dimension is greater than one,
i.e., it works in  dimension 2 as well.

Consider the integrals $I_0$ from \eqref{integral} constructed by
$a$ (we keep the notation  $I_0$ for it) and by $\hat g$ (we denote it by
$\mathcal{I}_0$). If $a = C \cdot A$, then
 the integrals  $I_0$ and $\mathcal{I}_0$ are proportional as well, direct calculations show that
   $I_0(\xi)= \pm C^{n-1}\cdot  \mathcal{ I}_0(\xi)$. Since the  functions $I_0$ and $\mathcal{I}_0$ are constant along  every trajectory of the geodesic flow, the coefficient of proportionality of these functions  is also constant along every  trajectory of the geodesic flow implying  that  it is constant everywhere.  \qed

Now let us   assume that at every point of the neighborhood $U$  the
tensors  $g$, $\bar g ,$ and  $\hat g$ are linearly dependent. Then,
for certain functions $c , d $  the tensors  $g, a, A$ satisfy
(probably in a smaller neighbourhood $U'\subseteq U$; without loss
of generality we can think that $U' $ coincides with $U$.)
\begin{equation} \label{3}
a_{ij} = c \, g_{ij} + d \, A_{ij}.
\end{equation}

We will   show that  the functions  $c , d $  are actually
constants.

Differentiating \eqref{3}  and substituting  \eqref{basic} and its analog for the solution $A$, we obtain

\begin{equation} \label{a1}
\lambda_{,i} g_{jk} + \lambda_{,j} g_{ik} = c_{,k}\, g_{ij} + d \,
\Lambda_{,i} g_{jk} + d \, \Lambda_{,j}g_{ik} + d _{,k}\,
A_{ij},\end{equation} which is evidently equivalent to
\begin{equation} \label{a2}
\tau_{i} g_{jk} + \tau_j g_{ik} = c _{,k}g_{ij} + d
_{,k}A_{ij},\end{equation}
where $\tau_i= \lambda_{,i} - d\,
\Lambda_{,i}$. We see that for every fixed $k$ the left-hand side of (\ref{a2}) is
a symmetric matrix  of the form $\tau_i v_j + \tau_j v_i.  $   If $c
_{,k}$ is not proportional to $d _{,k}$, this will imply that
$g_{ij}$ also is of the form $\tau_i v_j + \tau_j v_i,   $ which
contradicts the non-degeneracy of $g$.  If $c _{,k}=f\cdot d _{,k}$,
then the coefficient $f$  of the proportionality should be a constant
implying $d = \const\cdot  {c }$  and $a_{ij} = c \, (g_{ij} +
\const\cdot A_{ij})$. Since the equation \eqref{basic} is linear and
since $g_{ij} $ and $A_{ij}$ are its solutions, their sum $g_{ij} +
\const\cdot  A_{ij}$ is also solution of \eqref{basic}. Then,  Lemma
\ref{weyl}   implies  that $c $ is constant. Thus, $c $ and $d $ are
constant in a neighbourhood $U$. Since the equation \eqref{basic}
is linear and of finite type, see \cite{eastwood}, linear
dependence of solutions in a neighbourhood implies linear dependence
of the solutions everywhere.   \qed

\begin{Rem} Though we used that the dimension of the manifold is at least three, the statement is true in dimension two as well provided the curvature of $g$ is not constant, see \cite{kruglikov}.

\end{Rem}

 \subsection{ Main step of the proof of Theorem \ref{thm1}} \label{main}
As we explained in Section \ref{schema}, in view of Sections \ref{tmp}, \ref{tmp1}, it is sufficient to prove Theorem locally only.

Assume the metrics $g, \bar g, $ and $\hat g$ are linearly independent at a point $p$.
Assume  the  metrics $g, \bar g $ are strictly nonproportional at $p$.
Then, for a certain neighborhood $U(p)$
 the metrics $g, \bar g, $ and $\hat g$ are linearly independent, and the metrics $g, \bar g $ are strictly nonproportional  at every  point of $U(p)$.

 Then, the solution $a$ corresponding to $\bar g$,   the solution $A$ corresponding to $\hat g$, and the metric $g$ are linearly independent at every point of $U(p)$, and the minimal polynomial of the
 (1,1)-tensor $a_j^i$ has degree $n$.

As in Section \ref{tmp}  we denote by $\Lambda $ the function \eqref{lambda} constructed by $A$.

 Let us first consider the case when $\lambda$ is constant. In this case, the equation \eqref{basic} implies that $a_{ij}$ is covariant constant. Since $a^i_{j}$ is self-adjoint with respect to $g$,
 at every point
  there exists a  basis $(b_1(p), ..., b_n(p))$
   such that the matrix of $g$ is diagonal with $\pm 1$ on the diagonal
   and the matrix of $a^i_{j}$ is Jordan-matrix.
   The basis is unique up to signs of the vectors, so locally $b_1$ can be taken to be smooth vector fields on the manifold.   Since the vectors of the basis are invariantly constructed by two  covariantly-constant object, they  are covariantly constant as well.
   Then, the metric $g$ is flat implying that $\bar g$ and $\hat g$ have constant curvature.

 In what follows we assume that $\lambda$ is not constant.

 Theorem \ref{thm1} is an easy consequence of the following two lemmas.

\begin{Lemma}
\label{13}
Assume that $a$ and $A$ are solutions of \eqref{basic} such that $a,A,g$ are linearly independent. Then,
at  every point  there exists $K$ such that  the tensor  $Z_{ijkl}:= R_{ij,kl} -
K\cdot  \left( g_{il}g_{jk} -  g_{ik}g_{jl} \right)$ satisfy the condition
\begin{equation}  a_{i}^{ \alpha} Z_{ \alpha jkl} +  a^{\alpha}_jZ_{\alpha i kl}=0,  \label{int33b}
\end{equation}
\end{Lemma}

We see that by construction the tensor $Z_{ijkl}$ is skew-symmetric with respect to the first two indexes. Note that  in Lemma \ref{13} we did not assume that the minimal polynomial of $a$ has degree $n$, i.e., the lemma is valid in a slightly more general setting than we need it.

\begin{Rem}  Tensor $Z$ plays important role in the theory of geodesically equvialent metrics; it appears quite naturally in the investigation of special metrics such that Einstein metrics, pseudo-symmetric metrics, K\"ahler metrics,     see for example \cite{kiosak,kiosak-mikes} for details.
\end{Rem}

The next lemma shows that if in addition   the minimal polynomial of $a$ has degree $n$, then the condition \eqref{int13} implies $Z_{ijkl}= 0$.
 Then, the curvature of $g$ is constant. By Beltrami's Theorem  (see for example \cite{beltrami_short}, or the original papers \cite{Beltrami} and \cite{schur}),  the metrics $\bar g $ and $\hat g$ have constant curvature as well which proves Theorem ~\ref{thm1}.

Moreover, we will not use the indexes $k$ and $l$ in the proof, so Lemma \ref{last} is the matrix reformulation of the condition  $a_{i}^{ \alpha} Z_{ \alpha j} +  a^{\alpha}_jZ_{\alpha i }$ for arbitrary $(1,1)-$tensor $a$ and  a skew-symmetric (0,2)-tensor $Z$.

\begin{Lemma} \label{last}
Let $Z$, $a$  be $n\times n$-matrices such that $Z$ is skew-symmetric. Assume
 the minimal polynomial of $a$ has degree $n$. If $Za - a^t Z=0$, then $Z=0$.
\end{Lemma}

{The proof of Lemma \ref{last} }  is an easy exercise in linear algebra and  will be  left to
 the reader (it
 is  a kind of problem such that it is easier to prove than to understand the proof.)
 We recommend to consider the coordinates such that the matrix
 $a$ is in   Jordan form, and then to  calculate the matrix  $Za$. One immediately sees that if the matrix $Za$ is symmetric, then $Z=0$. \qed

 \subsection{Proof of Lemma \ref{13}.}  \label{111}

The proof is straightforward tensor calculations. The geometry behind the calculation could be understood with the help of \cite{eastwood}. There, the equations \eqref{basic} were written in the projectively invariant form (so that the equations for $g$ and for the geodesically equivalent metric $\bar g$ are the same). The prolongation of the equations was written as a connection on the projective tractor  bundle, and the curvature of the connection  was  calculated.  Its first part  related to (the analog of the  tensor) $a$ is a trace-free  object.  Moreover, for most objects, the trace-free  part of the covariant derivative coincides with the trace-free part of the corresponding derivative on the
tractor bundle.

We will see that in the proof we consider  the integrability conditions for the equations
\eqref{basic}, and then    "artificially" write all objects   in trace-free form.  At the end we obtain the required equation \eqref{int33b}.

Note that in the proof we  will essentially use the symmetries of the Riemannian curvature tensor, which have sense
 only if the affine connection is the Levi-Civita connection of a metric. That means, it is important for us that among the solutions of the projective-invariant analog of $a$ there is a non-degenerate solution.

In  Section \ref{bolsinov},  we give another proof of Lemma \ref{13} which is, in fact, pure algebraic and  based on some ideas from the theory of integrable Hamiltonian systems on Lie algebras.

  {\bf Proof of Lemma  \ref{13}.}
 Integrability conditions for the equation \eqref{basic} are
 (we use the standard fact that  $a_{ij,kl}- a_{ij,lk}= a_{i \alpha }R^{\alpha}_{jkl} +  a_{\alpha j}R^{\alpha}_{ikl}$  for any  $(0,2)-$tensor $a_{ij}$)

\begin{equation}  a_{i \alpha }R^{\alpha}_{jkl} +  a_{\alpha j}R^{\alpha}_{ikl} =\lambda_{, li} g_{jk}+\lambda_{, lj} g_{ik}-\lambda_{, ki} g_{jl}-\lambda_{, kj} g_{il}. \label{int1} \end{equation}

 The same is true for the other solution $A$
 \begin{equation}  A_{i \alpha }R^{\alpha}_{jkl} +  A_{\alpha j}R^{\alpha}_{ikl} =\Lambda_{, li} g_{jk}+\Lambda_{, lj} g_{ik}-\Lambda_{, ki} g_{jl}-\Lambda_{, kj} g_{il}. \label{int2} \end{equation}

Starting from this point, the proof is purely algebraic:  the statement of  Lemma  \ref{13} is an algebraic corollary of  (\ref{int1}) and (\ref{int2}). We emphasize this once again in the next section by giving another version of the proof in the Lie-algebraic language.

 Now let us  multiply the equation \eqref{int1} by $A^{l}_{s}$ and sum over $l$.
After renaming indexes,   we obtain
\begin{equation}  a_{i \alpha }R^{\alpha}_{jk\beta} A^{\beta}_l +
a_{\alpha j}R^{\alpha}_{ik\beta} A^{\beta}_l =\lambda_{, \alpha i} A^{\alpha}_l   g_{jk}+
\lambda_{, \alpha j}  A^{\alpha}_l  g_{ik}-\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}. \label{int3} \end{equation}

 Using the symmetry of the Riemann tensor we obtain   $a_{\alpha j}R^{\alpha}_{ik\beta} = a^{\alpha }_i R_{\alpha j, k \beta} A^{\beta}_l = a_i^\alpha R_{\beta k ,j \alpha } A^{\beta}_l  =a_i^\alpha A_{\beta l} R^{\beta}_{k j \alpha}$. Substituting this in  \eqref{int3}, we get

\begin{equation}  a_{i}^\alpha A_{\beta l} R^{\beta}_{ k i\alpha}  +  a_{j}^\alpha A_{\beta l} R^{\beta}_{ k j\alpha}  = \lambda_{, \alpha i} A^{\alpha}_l   g_{jk}+
\lambda_{, \alpha j}  A^{\alpha}_l  g_{ik}-\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}. \label{int4} \end{equation}

Let us now symmetrise  \eqref{int4} by $l,k$ to obtain
\begin{equation}\begin{array}{l}
a_{i}^\alpha \left(A_{\beta l} R^{\beta}_{ k j\alpha}  +   A_{\beta k} R^{\beta}_{ l j\alpha} \right)+
a_{j}^\alpha\left( A_{\beta k} R^{\beta}_{ l i\alpha}  +   A_{\beta l} R^{\beta}_{ k i\alpha}\right)  \\
= \lambda_{, \alpha i} A^{\alpha}_l   g_{jk}+
\lambda_{, \alpha j}  A^{\alpha}_l  g_{ik}-\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}+
\lambda_{, \alpha i} A^{\alpha}_k   g_{jl}+
\lambda_{, \alpha j}  A^{\alpha}_k  g_{il}-\lambda_{, li}   A_{jk} -\lambda_{, lj} A_{ik}.\end{array} \label{int5}\end{equation}

We see that the components staying in brackets are the left-hand side of the equation
\eqref{int2} with other indexes. Substituting  \eqref{int2} in the first term of the left-hand side we obtain

\begin{equation}\begin{array}{cc}
 a_{i}^\alpha \left(A_{\beta l} R^{\beta}_{ k j\alpha}  +   A_{\beta k} R^{\beta}_{ l j\alpha} \right) & =
a_{i}^\alpha \left(\Lambda_{,\alpha l} g_{jk}  +   \Lambda_{, \alpha k} g_{jl } -
\Lambda_{,j  l} g_{k\alpha }  -   \Lambda_{, j  k} g_{l \alpha}  \right) \\&  = a^\alpha_i \Lambda_{,\alpha l} g_{jk} + a_i^\alpha \Lambda_{,\alpha k}g_{jl}- \Lambda_{,jl} a_{ik} -  \Lambda_{,jk} a_{il}.\end{array}
\label{int6}
\end{equation}
Similarly, the second term of the left-hand side is
\begin{equation}\begin{array}{cc}
 a_{j}^\alpha \left(A_{\beta l} R^{\beta}_{ k i\alpha}  +   A_{\beta k} R^{\beta}_{ l i\alpha} \right) & =
a_{j}^\alpha \left(\Lambda_{,\alpha l} g_{ik}  +   \Lambda_{, \alpha k} g_{il } -
\Lambda_{,i  l} g_{k\alpha }  -   \Lambda_{, i  k} g_{l \alpha}  \right) \\&  = a^\alpha_j \Lambda_{,\alpha l} g_{uk} + a_j^\alpha \Lambda_{,\alpha k}g_{il}- \Lambda_{,il} a_{jk} -  \Lambda_{,ik} a_{jl}.\end{array}
\label{int7}
\end{equation}
Substituting (\ref{int6},\ref{int7}) in \eqref{int5}, we obtain
\begin{equation}\begin{array}{l}
a^\alpha_i \Lambda_{,\alpha l} g_{jk} + a_i^\alpha \Lambda_{,\alpha k}g_{jl}- \Lambda_{,jl} a_{ik} -  \Lambda_{,jk} a_{il}+
a^\alpha_j \Lambda_{,\alpha l} g_{ik} + a_j^\alpha \Lambda_{,\alpha k}g_{il}- \Lambda_{,il} a_{jk} -  \Lambda_{,ik} a_{jl}  \\
= \lambda_{, \alpha i} A^{\alpha}_l   g_{jk}+
\lambda_{, \alpha j}  A^{\alpha}_l  g_{ik}-\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}+
\lambda_{, \alpha i} A^{\alpha}_k   g_{jl}+
\lambda_{, \alpha j}  A^{\alpha}_k  g_{il}-\lambda_{, li}   A_{jk} -\lambda_{, lj} A_{ik}.\end{array} \label{int8}\end{equation}

 Collecting the terms by $g$, we  see that \eqref{int8}  can be written as

 \begin{equation}\begin{array}{l}
\left(a^\alpha_i \Lambda_{,\alpha l} -  \lambda_{, \alpha i} A^{\alpha}_l  \right) g_{jk}
  + \left(a_i^\alpha \Lambda_{,\alpha k}-
\lambda_{, \alpha i} A^{\alpha}_k  \right) g_{jl}
+  \left(a^\alpha_j \Lambda_{,\alpha l}  -\lambda_{, \alpha j}  A^{\alpha}_l \right) g_{ik}
+ \left(a_j^\alpha \Lambda_{,\alpha k}-
\lambda_{, \alpha j}  A^{\alpha}_k \right) g_{il} \\
=
 \Lambda_{,jl} a_{ik} +  \Lambda_{,jk} a_{il} + \Lambda_{,il} a_{jk} +  \Lambda_{,ik} a_{jl}   -\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}-\lambda_{, li}   A_{jk} -\lambda_{, lj} A_{ik}.\end{array} \label{int9}\end{equation}

We denote $\tau_{il}:= a^{\alpha}_i \Lambda_{, \alpha l} - A^{\alpha}_l \lambda_{, \alpha i}$. In this notation,  the equation \eqref{int9}  is
\begin{equation}\begin{array}{lll}
&&\tau_{il}   g_{jk}
  + \tau_{ik}  g_{jl}
+  \tau_{jl} g_{ik}
+ \tau_{jk} g_{il} \\
&=&
 \Lambda_{,jl} a_{ik} +  \Lambda_{,jk} a_{il} + \Lambda_{,il} a_{jk} +  \Lambda_{,ik} a_{jl}   -\lambda_{, ki}   A_{jl} -\lambda_{, kj} A_{il}-\lambda_{, li}   A_{jk} -\lambda_{, lj} A_{ik}.\end{array} \label{int10}\end{equation}
 Let us show that $\tau$ is symmetric.
 Multiplying with $g^{jk}$ and contracting with respect to $j,k$,  we obtain
   \begin{equation}\begin{array}{lcl}
(n+2) \tau_{il}
+ \left(\tau_{jk} g^{jk}\right)  g_{il} &
=&
 \Lambda_{,\alpha l} a_{i}^\alpha  +  \left(g^{jk}\Lambda_{,jk}\right) a_{il} +
 \Lambda_{,il} \left(a_{jk}g^{jk} \right) +  \Lambda_{,i\alpha} a^{\alpha}_l   \\ && -\lambda_{, \alpha i}   A^{\alpha}_l -\left(\lambda_{, kj} g^{kj}\right) A_{il}-\lambda_{, li}   \left(A_{jk} g^{jk}\right) -\lambda_{, l\alpha } A_{i}^\alpha .\end{array} \label{int11}\end{equation}
We see that the right-hand side is symmetric w.r.t. $i,l$. Then, so should be the left-hand-side implying
$\tau_{il}=\tau_{li}$.
We also see that the sum of the  first, fourth,  fifth and last terms of the right-hand side is $\tau_{il}+ \tau_{li} = 2 \tau_{il}$. Then,  the equation \eqref{int11} is equivalent to
\begin{equation}
 \tau_{il}
=\frac{1}{n}\left(-\left(\tau_{jk} g^{jk}\right)   g_{il} +
 \left(g^{jk}\Lambda_{,jk}\right) a_{il} +
 \Lambda_{,il} \left(a_{jk}g^{jk} \right)
  -\left(\lambda_{, kj} g^{kj}\right) A_{il}-\lambda_{, li}   \left(A_{jk} g^{jk}\right)\right) . \label{int12}\end{equation}

 Now we return to the equation \eqref{int10}. We alternate the equation with respect to $j, k$.
 \begin{equation}\begin{array}{lll}
&&   \tau_{ik}  g_{jl}
+  \tau_{jl} g_{ik} -   \tau_{ij}  g_{kl}
-   \tau_{kl} g_{ij}
  \\
&=&
 \Lambda_{,jl} a_{ik}   +  \Lambda_{,ik} a_{jl}   -\lambda_{, ki}   A_{jl}  -\lambda_{, lj} A_{ik} - \Lambda_{,kl} a_{ij}   -  \Lambda_{,ij} a_{kl}   +\lambda_{, ji}   A_{kl}  +\lambda_{, lk} A_{ij}     . \end{array} \label{int13}\end{equation}

Let us now rename $i\leftrightarrow k$ in \eqref{int13}. We obtain
\begin{equation}\begin{array}{lll}
 &&  \tau_{ik}  g_{jl}
+  \tau_{jl} g_{ik} -   \tau_{kj}  g_{il}
-   \tau_{il} g_{kj}
  \\
&=&
 \Lambda_{,jl} a_{ik}   +  \Lambda_{,ik} a_{jl}   -\lambda_{, ki}   A_{jl}  -\lambda_{, lj} A_{ik} - \Lambda_{,il} a_{kj}   -  \Lambda_{,kj} a_{il}   +\lambda_{, jk}   A_{il}  +\lambda_{, li} A_{kj}     . \end{array} \label{int14}\end{equation}

 Adding  \eqref{int10} with \eqref{int14}  and dividing by $2$ for cosmetic
 reasons, we obtain

 $$
\tau_{ik}  g_{jl}
+   \tau_{jl} g_{ik}
=
 \Lambda_{,jl} a_{ik}  +  \Lambda_{,ik} a_{jl}
  -\lambda_{, ki}   A_{jl}      -\lambda_{, lj} A_{ik}    . $$

  Substituting the expression for $\tau$
  from  \eqref{int12}, we obtain

  $$\begin{array}{ccl}
&&\frac{1}{n}\left(-\left(\tau_{\beta\alpha } g^{\beta\alpha }\right)   g_{ik} +
 \left(g^{\alpha \beta }\Lambda_{,\alpha \beta }\right) a_{ik} +
 \Lambda_{,ik} \left(a_{\alpha \beta  }g^{\alpha \beta } \right)
  -\left(\lambda_{, \alpha \beta  } g^{\alpha \beta }\right) A_{ik}-\lambda_{, ki}   \left(A_{\alpha \beta} g^{\alpha \beta }\right)\right)  g_{jl}\\
&+&  \frac{1}{n}\left(-\left(\tau_{\alpha \beta } g^{\alpha \beta }\right)   g_{jl} +
 \left(g^{\alpha \beta }\Lambda_{,\alpha \beta}\right) a_{jl} +
 \Lambda_{,jl } \left(a_{\alpha \beta }g^{\alpha \beta } \right)
  -\left(\lambda_{, \alpha \beta } g^{\alpha \beta }\right) A_{jl}-\lambda_{, lj}   \left(A_{\alpha \beta } g^{\alpha \beta }\right)\right) g_{ik} \\
&=&
 \Lambda_{,jl} a_{ik}  +  \Lambda_{,ik} a_{jl}
  -\lambda_{, ki}   A_{jl}      -\lambda_{, lj} A_{ik}    . \end{array} $$

  Denoting $\left(\tau_{\beta\alpha } g^{\beta\alpha }\right)$ by $\tau$,
  $\left(\lambda_{,\beta\alpha } g^{\beta\alpha }\right)$ by $\mu$,
   $\left(\Lambda_{,\beta\alpha } g^{\beta\alpha }\right)$ by $\mathcal{M}$,   and using that
   $\left(a_{\alpha \beta  }g^{\alpha \beta } \right)=2 \lambda $, and  $\left(A_{\alpha \beta  }g^{\alpha \beta } \right)=2 \Lambda $,   one obtains

  $$\begin{array}{ll}
&\frac{1}{n}\left(-\tau   g_{ik} +
 \mathcal{M} a_{ik} +
 2\Lambda_{,ik}\lambda
  -\mu A_{ik}-2 \lambda_{, ki}  \Lambda \right)  g_{jl}\\
+& \frac{1}{n}\left(-\tau   g_{jl} +
 \mathcal{M}  a_{jl} +
2 \Lambda_{,jl } \lambda
  -\mu A_{jl}-2 \lambda_{, lj}   \Lambda \right) g_{ik} \\
=&
 \Lambda_{,jl} a_{ik}  +  \Lambda_{,ik} a_{jl}
  -\lambda_{, ki}   A_{jl}      -\lambda_{, lj} A_{ik}    . \end{array}
 $$

  Combining the terms, we obtain
\weg{   $$\begin{array}{ll}
&\frac{1}{n}\left(-\tau   g_{ik} +
 2 \Lambda_{,ik}\lambda
 -2 \lambda_{, ki}  \Lambda  \right)  g_{jl}
+ \frac{1}{n}\left(-\tau   g_{jl} +
 2 \Lambda_{,jl } \lambda
 -2 \lambda_{, lj}   \Lambda \right) g_{ik} \\
=& \left(-\frac{1}{n} \mathcal{M}    g_{jl} +
 \Lambda_{,jl}\right) a_{ik}  +
\left( \Lambda_{,ik}- \frac{1}{n}\mathcal{M}  g_{ik}  \right)   a_{jl}
  -\left(\lambda_{, ki}     -\frac{1}{n}\mu g_{ik}\right) A_{jl}    -\left( \lambda_{, lj}    -\frac{1}{n} \mu g_{jl}\right)A_{ik}  . \end{array} $$}

   \begin{equation}\begin{array}{ll}
&\left(-\frac{\tau}{2}   g_{ik} +
 2 \Lambda_{,ik}\lambda  \right)  \frac{g_{jl}}{n}  -\left(\frac{\tau}{2}   g_{ik}
 +\lambda_{, ki}  \Lambda  \right)  \frac{g_{jl}}{n}
+ \left(-\frac{\tau}{2}   g_{jl}
 -2 \lambda_{, lj}   \Lambda \right)\frac{g_{ik}}{n}  +
 \left(-\frac{\tau}{2}   g_{jl} +2  \Lambda_{,jl } \lambda  \right)\frac{g_{ik}}{n}   \\
=& \left(-\frac{1}{n} \mathcal{M}    g_{jl} +
 \Lambda_{,jl}\right) a_{ik}  +
\left( \Lambda_{,ik}- \frac{1}{n}\mathcal{M}  g_{ik}  \right)   a_{jl}
  -\left(\lambda_{, ki}     -\frac{1}{n}\mu g_{ik}\right) A_{jl}    -\left( \lambda_{, lj}    -\frac{1}{n} \mu g_{jl}\right)A_{ik}  . \end{array} \label{int18}\end{equation}

Now,  let us  calculate $\tau$: we multiply \eqref{int12} by $g^{il}$ and sum over  $i,l$. After dividing by $2$,   we obtain
  $$
 \tau
=\frac{2}{n}\left(
  \mathcal{M} \lambda
 - \mu   \Lambda \right) .$$

   Substituting in \eqref{int18}, we obtain

  $$\begin{array}{ll}
&\left(-\frac{\mathcal{M}}{n}     g_{ik} +
 \Lambda_{,ik} \right)  \frac{2\lambda}{n}g_{jl}  -\left(            -         \frac{\mu}{n}    g_{ik}
 +\lambda_{, ki}   \right)  \frac{2\Lambda}{n}g_{jl}
+ \left(\frac{\mu}{n}    g_{jl}
 -\lambda_{, lj}   \right)\frac{2\Lambda }{n}g_{ik}  +
 \left(-\frac{\mathcal{M}}{n}   g_{jl} + \Lambda_{,jl } \right)\frac{2\lambda }{n} g_{ik}  \\
=& \left(-\frac{1}{n} \mathcal{M}    g_{jl} +
 \Lambda_{,jl}\right) a_{ik}  +
\left( \Lambda_{,ik}- \frac{1}{n}\mathcal{M}  g_{ik}  \right)   a_{jl}
  -\left(\lambda_{, ki}     -\frac{1}{n}\mu g_{ik}\right) A_{jl}    -\left( \lambda_{, lj}    -\frac{1}{n} \mu g_{jl}\right)A_{ik}  , \end{array} $$
 which is equivalent to

  \begin{equation}\begin{array}{cl}
&\left( \Lambda_{,ik}-\frac{\mathcal{M}}{n}     g_{ik}  \right)
\left(a_{jl} -\frac{2\lambda }{n}g_{jl}\right) +\left( \Lambda_{,jl }-\frac{\mathcal{M}}{n}   g_{jl}  \right)\left(a_{ik}-\frac{2 \lambda  }{n} g_{ik} \right)\\
 = & \left(           \lambda_{, ki}  -         \frac{\mu}{n}    g_{ik}    \right)
 \left(  A_{jl} -\frac{2\Lambda}{n}g_{jl}\right)+
 \left(\lambda_{, lj} -\frac{\mu}{n}    g_{jl}   \right) \left(A_{ik} -\frac{2 \Lambda  }{n}g_{ik} \right)
 \end{array} \label{int20}\end{equation}

 Denoting $B_{ik}:= \left(\Lambda_{,ik}-\frac{\mathcal{M}}{n}     g_{ik}
  \right) $ ,     $ b_{ik}:=  \left(\lambda_{, ki}
 - \frac{\mu}{n}    g_{ik} \right) $,
 $d_{jl}:=   \left(a_{jl} -\frac{2\lambda }{n}g_{jl}\right) $,
 $D_{jl}:= \left(A_{ik} - \frac{2\Lambda  }{n}g_{ik}\right)$, we see that \eqref{int20} is equivalent to
 \begin{equation}
B_{ik} d_{jl} + B_{jl} d_{ik} = b_{ik} D_{jl} + b_{jl} D_{ik}.
 \label{int21}\end{equation}

Now,  it is easy to see that the condition $B_{\alpha} d_{\beta} + B_{\beta} d_{\alpha} = b_{\alpha} D_{\beta} + b_{\beta} D_{\alpha}$ for dimensions $\ge 3$ implies
that $B$ is proportional to $b$ and $D$ to $d$, or that  $B$ is proportional to $D$ and $b$ to $d$.
We see that the condition \eqref{int21} is essentially the same as this  condition, the role of $\alpha $
and  $\beta$ play the multi-indexes $ik$ and $jl$. Since $g,$ $  a,$ and $A$ are linearly independent,
 $D$ can not be  proportional to $d$. Thus,   $b$ is proportional to $d$ which imply that
 $\lambda_{,ij}$
 is a linear combination of  $a_{ij}$  and $g_{ij} $. The coefficients of the linear combination are not important for us, let
 $$ \lambda_{,ij} = \rho \cdot g_{ij} +  K\cdot a_{ij}$$

 Substituting this equation in \eqref{int1}, we obtain

\begin{equation}  a_{i \alpha} Z^{\alpha}_{jkl} +  a_{\alpha j}Z^{\alpha}_{ikl}=0,  \label{int32}
\end{equation}
 Where $Z^{i}_{jkl}= R^i_{jkl} - K\cdot  ( \delta^i_lg_{jk} - \delta^i_kg_{jl}  )$.
Clearly, the equation \eqref{int32} is equivalent to  the equation \eqref{int33b}.

We see that by construction the tensor $Z_{ijkl}$ is skew-symmetric with respect to the first two indexes
(actually, the tensor $Z$ has the same symmetries as the curvature tensor, i.e., for example,  skew-symmetric with respect to the last  two indexes   as well,   but we will need only the first  two indexes).  Thus,
Lemma~\ref{13} and, therefore,  Theorem~\ref{thm1}  are proved, \qed

\section{Fubini theorem and  sectional operators  \\ on semisimple Lie algebras}

In this section we discuss an unexpected and remarkable relationship between geodesically equivalent  metrics and some special operators on semisimple Lie algebras which appeared in the theory of integrable systems.

We start with a brief  overview on (one special type of)  integrable Euler equations on semisimple Lie algebras (see \cite{bogo, fomtrofbook, marsden, MF1, pere,  reyman}  for details).

Let  $\goth g$ be a semisimple Lie algebra, $R: \goth g \to \goth g$  an operator  symmetric with respect to the Killing form $\langle~,~\rangle$ on $\goth g$. The differential equation
\begin{equation}
\dot x = [R(x),x], \quad x\in\goth g,
\label{Euler}
\end{equation}
is Hamiltonian on $\goth g$ with respect to the standard Lie-Poisson structure  and called the {\it Euler equation} related to the Hamiltonian function $H(x)=\frac{1}{2}\langle R(x), x\rangle$.

A classical, interesting  and extremely difficult problem is to find those operators
$R: \goth g \to \goth g$ for which the system (\ref{Euler}) is completely integrable.

One of such operators was discovered by S.\,Manakov in \cite{Manakov} and his idea then led to an elegant general construction developed by A.\,Mischenko and A.\,Fomenko \cite{MF1} and  called {\it argument shift method}. This construction in brief can be presented as follows.

Assume that $R : \goth g \to \goth g$ satisfies the following identity
\begin{equation}
[R(x),a]=[x,b],   \quad x\in \goth g,
\label{sectional}
\end{equation}
for certain $a,b\in \goth g$,  $a\ne 0$.   Then the following statement holds

\begin{Th}{\rm  \cite{MF1}} \label{MFth}
Let $R: \goth g \to \goth g$ be symmetric and satisfy {\rm (\ref{sectional})}. Then

1) the system {\rm (\ref{Euler})}
admits the following Lax representation with a parameter:
$$
 \frac{d}{dt}(x + \lambda a) = [ R(x)+\lambda b, x+\lambda a];
$$

2) the functions  $f(x+\lambda a)$, where $f:\goth g\to \R$ is an invariant of the adjoint representation,  are first integrals of {\rm (\ref{Euler})}  for any $\lambda\in \R$ and, moreover, these integrals commute;

3) if $a\in \goth g$ is  regular, then {\rm (\ref{Euler})} is completely integrable.
\end{Th}

This construction has a very important particular case. If the Lie algebra $\goth g$ admits  a  $\Z_2$-grading,  i.e., a decomposition $\goth g = \goth h  + \goth v$  (direct sum of  subspaces) such that $[\goth h,\goth h]\subset \goth h$, $[\goth h,\goth v]\subset \goth v$, $[\goth v,\goth v]\subset \goth h$,
then  we may consider  $R:\goth h \to \goth h$  satisfying   (\ref{sectional}) with $a,b\in \goth v$,  and
Theorem   \ref{MFth} still holds if we replace $\goth g$ by $\goth h$.

The most important  example for applications (in particular,  in the theory of integrable tops)  is $\goth  g = sl(n,\R)$,  $\goth h = so(n,\R)$, with  $a$ and $b$ symmetric matrices. This  is  the  situation  that was studied in the pioneering work by S.~Manakov \cite{Manakov}  leading  to integrability of the Euler equations of $n$-dimensional rigid body dynamics.

From the algebraic point of view,  the above construction still makes sense if we replace $so(n)$ by $so(p,q)$ and  assume $a,b$ to be  symmetric operators with respect to the corresponding indefinite form $g$.
Moreover, if  we complexify our considerations we do not even notice any difference.
However, to indicate the presence (but not influence) of the bilinear form $g$, we shall denote the space of  $g$-symmetric operators by $Sym(g)$, and the Lie algebra of $g$-skew-symmetric operators by $so(g)$.

\begin{Def}
\label{defsect}
{\rm  We shall say that $R: so(g) \to so(g)$ is a {\it sectional operator} associated with $a,b\in Sym(g)$,  if  $R$ is symmetric with respect to the Killing form and the following identity holds:
\begin{equation}
[R(x), a] = [x,b],    \quad \mbox{for all } x\in  so(g).
\label{sectionall}
\end{equation}
}\end{Def}

We follow the terminology introduced by Fomenko and Trofimov  in \cite{troffom, fomtrofbook} where they studied various generalizations of such operators. Strictly speaking, the above definition is just a particular case of  a more general construction.  The term``sectional"  was motivated by the following reason.   The  identities
(\ref{sectional}) and (\ref{sectionall}) suggest  that one may represent $R$ as $\mbox{ad}_a^{-1}\mbox{ad}_b$,  but in general we cannot do so because $\mbox{ad}_a$, as a rule,  is not invertible. That is why  the operator $R$ splits into parts each of which acts independently on its own subspace (section).

A surprising relationship between sectional operators and geodesically equivalent metrics is explained by the following observation. Notice, first of all, that due to its algebraic symmetries   (skew-symmetry with respect to $i,j$ and $k,l$  and symmetry with respect to permutation of pairs $(ij)$ and $(kl)$),   the Riemann curvature tensor $R_{ij,kl}$  can be naturally considered as
a symmetric operator  $R: so(g) \to so(g)$   (strictly speaking we need to raise indices $i$ and $k$  by means of $g$ to get the tensor of the form $R^{i,k}_{j,l}$). Having this interpretation of $R$ in mind, we immediately obtain

\begin{Th}
Let  $g$ and $\bar g$ be geodesically equivalent  nonproportional metrics, then
the Riemann curvature tensor $R^{i,k}_{j,l}$ of the metric $g$ is a sectional operator in the sense of Definition {\rm \ref{defsect}}.
More precisely,
$$
[R(x), a]=[x,b]
$$
where $a$ is  the $g$-symmetric operator associated with the form $a_{ij}$ defined by (3),
and  $b$ is the $g$-symmetric operator associated with the form $2\lambda_{,ij}$ (Hessian  of  $\mathrm{tr}\, a$).
\end{Th}

The proof of this statement is just the observation that (\ref{sectionall}) is a translation of the compatibility condition (\ref{int1}) into Lie-algebraic language.

Before discussing the proof of  the Fubini theorem in this "new" language,  we make some remarks which could also be useful.

Notice, first of all, that in our new notation the condition "curvature is constant"   (at a point) simply means that  $R: so(g) \to so(g)$ is a scalar operator, i.e. $R(x)=K\cdot x$.

Furthermore, it is a very simple Lie-algebraic fact that   (\ref{sectionall})   implies that $a$ and $b$ commute.  Indeed,
$\langle [b,a], x \rangle = \langle a, [x,b] \rangle =\langle a, [R(x),a] \rangle =
\langle [a,a], R(x) \rangle =0 $ for any $x\in so(g)$,  so $[a,b]=0$.   In the theory of projectively equivalent metrics this means that  the operator $a$ commutes with the Hessian of its trace $2\lambda =\tr\, a$. This fact is, of course,  well known, but  the above proof seems to be the simplest one.  Moreover, if instead of $a$ we substitute any element  $\xi$  from its centralizer $C(a)$,  we obviously get the same conclusion
$[b,\xi]=0$, i.e., $b$  lies in the center of the centralizer of $a$.  This means, in fact, that  $b$ is a polynomial of $a$.

Finally, if $a$ is regular in the Lie-algebraic sense, i.e. its minimal polynomial coincides with the characteristic one, then  the operator ${\rm ad}_a: so(g) \to Sym(g)$  has trivial kernel  so that  the sectional operator $R$  (i.e., the curvature tensor!) can be reconstructed from $a$ and $b$.   Namely, $R(x)={\rm ad}_a^{-1}{\rm ad}_b (x)$, a well-known formula in the theory of integrable systems on Lie algebras.  If we take into account the fact that  $b=P(a)= \lambda_{n-1} a^{n-1} + \lambda_{n-2} a^{n-2} + \dots + \lambda_1 a + \lambda_0$  (polynomial of $a$), then this formula can be rewritten as
\begin{equation}
R(x) = \frac{d}{dt}\left. P(a+tx) \right|_{t=0} .
\label{rx}
\end{equation}
Indeed,  $[P(a+tx), a+tx]=0$ implies
$$
0=\frac{d}{dt} [P(a+tx), a+tx]|_{t=0}=[\frac{d}{dt} P(a+tx) |_{t=0},a]+[P(a), x],
$$
i.e., $[\frac{d}{dt} P(a+tx)|_{t=0} ,a]=[x,b]$. Since $a$ is regular, we have (\ref{rx}).

This shows, in particular,  that the algebraic structure of the curvature tensor can be understood in terms of the operator $a$ only.

Using this language we now give another proof of  the tensor part of  the Fubini theorem  (Lemmas 2 and 3).

Assume that we have three geodesically equivalent metrics  $g$, $\bar g$, and $\hat g$.
Then the Riemann curvature tensor $R$ of the metric $g$ satisfies at the same time two identities :
\begin{equation}
[ R(x), a]=[x, b]\qquad \mbox{and}  \qquad [ R(x), A]=[x, B],
\label{2sectional}
\end{equation}
where $a^k_j = g^{ki}a_{ij}$, $A^k_j = g^{ki}A_{ij}$, $b^k_j = 2g^{ki}\lambda_{, ij}$, $B^k_j = 2g^{ki}\Lambda_{,ij}$ (cf. (\ref{int1}) and (\ref{int2})).

From now on,  we may forget about the geometrical meaning of $a,b, A, B$ and start thinking of them as just certain $g$-symmetric operators.  In addition, without loss of generality we may assume all these operators to be trace free (as, of course, it should be in the semisimple Lie algebra $sl(n,\R)$ which stands behind this construction). Moreover,  we are allowed to complexify all the objects  so that instead of  $so(g)$ and $Sym(g)$  we may simply consider the spaces of symmetric and skew-symmetric complex matrices.

The reformulation of the (algebraic part of) Fubini theorem are the following  analogs of Lemmas 2 and 3 respectively.

\begin{Lemma}
Let $R: so(g) \to so(g)$  be symmetric and satisfy {\rm (\ref{2sectional})}.
If $a$ and $A$  are not proportional,  then
$b$ is proportional to $a$  and, therefore,  $[R(x) - K\cdot x, a]=0$ for some $K\in \R$.
\label{lemma4}
\end{Lemma}

\begin{Lemma}
If  $a$ is regular, i.e., its minimal polynomial coincides with the characteristic one, then
the identity $[R(x) - K\cdot x, a]=0$   implies  $R = K\cdot {\rm{id}}$  (i.e., the curvature is constant).
\label{lemma5}
\end{Lemma}

{\bf Proof of Lemma \ref{lemma4}.}
Let $y$ and $z$ be arbitrary $g$-symmetric matrices, then $[A, y], [a, z] \in so(g)$
and we have:
$$
[ R([A, y]), a]=[[A,y], b], \quad  [ R([ a , z]), A]=[[a, z], B].
$$
Since $R$ is symmetric with respect to the Killing form $\langle \ , \ \rangle$  we
have
$$
\langle [[A, y], b] , z\rangle= \langle [ R([A, y]), a], z\rangle =
\langle  R([A, y]), [ a , z] \rangle = \langle  [A, y], R([ a , z]) \rangle =
$$
$$
 \langle  y, [ R([ a , z]), A] \rangle =
\langle  y , [[a, z], B] \rangle = \langle [[B,y], a], z \rangle
$$
Since $z$ is an arbitrary symmetric matrix, we conclude that
\begin{equation}
[[A, y], b] =[[B,y], a].
\label{mainident}
\end{equation}

This relation is an analog of (\ref{int8}).
Similarly,  $ [[a, y], B] =[[b,y], A]$.
Using the Jacobi identity, it is not hard to see that
$$
[b, A]=[a, B]
$$

Rewriting (\ref{mainident}) as
$$
y (Ba - A b) + (aB - b A) y = Bya+ ayB - byA - Ayb
$$
and noticing that $[b, A]=[a, B] $ implies $Ba - Ab
=aB - b A$,  we get
$$
y  T + T  y = Bya+ ayB - byA - Ayb
$$
where $T$ denotes $aB - b A$  (this is an analog of $\tau$ from (\ref{int10})).

This formula can be considered as a relation between two  linear operators acting on the space of symmetric matrices  (the argument of both operators is $y\in Sym(g)$).
To get some consequences from this identity,  we take a kind of its trace. Recall that we consider $A, a, B, b, y, T$ as usual  symmetric (complex) matrices.

Instead of $y$ we substitute the symmetric matrix of the form  $e_i  v^\top + v e_i^\top$,  where $e_i$ and $v$ are vector-columns ($e_1,\dots, e_n$  is the standard (orthonormal) basis), then apply the result to  $e_i$ and take the sum over $i$. Here is the result:
$$
(e_i  v^\top + v e_i^\top) T e_i + T ( e_i  v^\top + v e_i^\top) e_i=
B ( e_i  v^\top + v e_i^\top)a e_i + ...
$$
$$
e_i  (Tv, e_i) +  v (Te_i, e_i)+ Te_i (v,e_i) + Tv (e_i, e_i) =
B e_i (a v, e_i) + B v (a e_i, e_i) + ...
$$

Using obvious facts from Linear Algebra such as
$$
\sum_i(Te_i, e_i) = \tr \,T,  \quad  \sum_i(e_i,e_i)=n, \quad
\sum_i e_i (v, e_i)=v,
$$
we get
$$
Tv +  \tr \, T\cdot v  + Tv + n \cdot Tv  =
B a v  +  \tr \, a\cdot B v  + ...
$$
Taking into account that $a, A, b, B$ are all trace free we have
$$
((n+2) T +   \tr\, T \cdot {\rm Id} ) v  =
(B a   + aB -   A b - bA) v
$$
Since $v$ is arbitrary and $T=B a -   A b = aB  - bA$, we finally get
$$
nT + \tr \, T \cdot {\rm Id }= 0
$$

But this simply means that $T=0$. Hence we come to the identity of the form
\begin{equation}
Bya+ ayB = byA + Ayb.
\label{mainident2}
\end{equation}

It remains to use the following simple statement:
if  $a,  b,  A, B$  are symmetric, $a\ne 0$ and (\ref{mainident2}) holds for any symmetric $y$,
then either    $b = K\cdot  a$, or  $A = K\cdot a$ for some constant $K\in \R$.

By our assumption, $a$ and $A$ are not proportional, so we conclude that
$b= K\cdot a$ and therefore the identity $[ R(x), a]=[x, b]$ becomes $[ R(x)- K\cdot x, a]=0$, as needed.

Notice that  (\ref{mainident2}) and  the rest of the proof almost literally repeat (\ref{int21})
and the end of the proof of Lemma 2, \qed

{\bf Proof of Lemma \ref{lemma5}.}
Let $[ R(x)- K\cdot x, a]=0$ and $a$ be regular.  It is a well known algebraic fact that the centralizer of a regular matrix $a$ is generated by the powers of $a$. In particular, the centralizer of $a$ consists of $g$-symmetric matrices.  On the other hand, $R(x)- K\cdot x$ is skew-symmetric. Thus, $R(x)- K\cdot x$ has to be zero for any $x$, i.e., $R = K\cdot {\rm{id}}$, as was to be proved, \qed

\label{bolsinov}

config: arxiv


TITLE: A set containing more than half elements of a group
QUESTION [4 upvotes]: I wish to prove the exercise which states that for a set $A$ containing more than half elements of a group $G$, every element of $G$ is a product of two elements of $A$.
My attempt:
By Lagrange Theorem, subgroup generated by $A$ must coincide with $G$, so every element in $G$ is a product of elements from $A$. How can I prove that the product is of exactly two elements?

REPLY [5 votes]: Let $g\in G$. Then $A$ and $gA^{-1}$ are two sets larger than half of $G$, hence intersect. But if $a_1=ga_2^{-1}$ then $g=a_1a_2$


TITLE: compute line integral where line is boundary of $(x-5)^2 + (y-2)^2 \le 1$
QUESTION [2 upvotes]: Line integral of $$\frac{-y^3}{(x^2+y^2)^2} dx + \frac{xy^2}{(x^2+y^2)^2} dy$$
Using greene's theorem, this integral is equal to zero. Am i right? $P(x,y)$ and $Q(x,y)$ are not continuous at $(0,0)$. Since the circle does not contain $(0,0)$, green's theorem applies?

REPLY [1 votes]: Using Green's Theorem we get:
$$\int_C\frac{-y^3}{(x^2+y^2)^2}dx+\frac{xy^2}{(x^2+y^2)^2}dy=\int\int_D\left(\frac{8xy^3}{(x^2+y^2)^3}-\frac{2x^2y}{(x^2+y^2)^2}\right)dA$$
Then switch to polor coordinates, since $D$ is a unit circle, (just shifted) we get:
$$=\int_0^{2\pi}\int_0^1\left(8\cos\theta \sin^2\theta\cdot r-2\cdot r\cos^2\theta\sin\theta\right)drd\theta=0$$
These are fairly simply integrals to integrate. For the first simply let $u=\sin\theta$ and the second let $u=\cos\theta$, after we integrate with respect to $r$, of course. So yes, you're right.


TITLE: Limit of jumping times is infinity
QUESTION [1 upvotes]: I encountered the following problem:
Problem: Assume that $(N_t)_{t\geq 0}$ be a counting process and $(T_j)_{j\geq 1}$ be the corresponding jumping times. If $$\mathbb E[N_t]<+\infty$$ for all $t\geq 0$ then $$\lim_{j\rightarrow +\infty} T_j=+\infty.$$
Clue: What I know is that
$$\{N_t\geq n\}=\{T_n\leq t\}$$
But I don't know how to apply this relation to above problem.
Thank you in advance for any hints!

REPLY [1 votes]: Since $N_t \geq 0$, you have
$$\mathbb{E}\left(N_t\right) = \sum_{n\geq 0} P\left(N_t \geq n\right),$$
whence the sum
$$\sum_n P\left(T_n\leq t\right) < \infty$$
by the identity you mention. By the Borel-Cantelli lemma, the probability that infinitely many events of the form $\left\{T_n\leq t\right\}$ occur is zero, i.e., you can have only finitely many $T_n$ that are less than $t$. This holds for any $t$, so it must be the case that $T_n\rightarrow\infty$ a.s.


TITLE: Column sum of character table
QUESTION [3 upvotes]: Consider character table of a finite group, whose rows are indexed by irreducible $\mathbb{C}$-characters and columns by conjugacy classes.
Then Huppert says in his book on character theory:
(1) For any row, the sum of entries is always non-negative integer.
(2) The sum of column entries is an integer, and may be negative.
(3) The Mathieu group $M_{11}$ has a conjugacy class such that its column sum is negative.
Q. My question is related to (3); isn't there any smaller order group for which column sum is negative?

REPLY [1 votes]: A routine computer search shows that the smallest example has order $96$. It is the group $\mathtt{SmallGroup}(96,3)$ in the small groups database, and there is an conjugacy class of elements of order $2$ for which the sum of character values is $-2$.
This group has centre of order $2$, with a normal Sylow $2$-subgroup $P$ with structure $2.(4 \times 4)$ with an element of order $3$ acting on $P$ and centralizing only the group centre. So the group has structure $2.(4 \times 4).3$.


TITLE: Uniform convergence- Point-wise convergence. Doubt regarding the difference
QUESTION [0 upvotes]: I know the definition of "pointwise convergence" and "uniform convergence", nevertheless I have some difficulties understanding the difference between those two concepts.
My book defines Uniform convergence as follows:

Let be $f_n$ a sequence of functions on $A\subseteq \mathbb{R}$. Then
  $(f_n)$ converges uniformly on $A$ to a limit function $f$ defined on
  $A$ if, for every $\epsilon>0$ there exists $N\in\mathbb{N}$ such that
  $|f_n(x)-f(x)|<\epsilon$ whenever $n\geq N$ and $x\in A$.

Now I do understand that in the case of uniform convergence we can choose $N$ irrespectively of the point $x$ while in the case of Pointwise convergence this is not the case (something like the difference between continuity and uniform continuity).
If the sequence of functions converge pointwise to $f$ it means that $\forall \epsilon>0, x\in A$ there exists $N$ such that $|f_n(x)-f(x)|<\epsilon$ whenever $n\geq N$, right? In this case $N$ depends both on $\epsilon$ and $x$.
My question is, given $\epsilon$ couldn't we pick an $N^*= \sup_{x\in A}{N(x,\epsilon)}$ (maybe this notation is a little bit messy).
Then we know that for this $\epsilon$ $, |f_n(x)-f(x)|<\epsilon$ whenever $n\geq N^* \forall x\in A$.
I know that in some cases this cannot be done, but I cannot see whether this is always false.
Would you mind to spot my error here? Thanks in advance.

REPLY [2 votes]: The value $N^*$ might not exist in $\mathbb{N}$ (i.e. it's $\infty$). You write $\max$ but really you should only write $\max$ when the underlying set (in this case, $A$) is finite. If it is not you must write $\sup$ and the $\sup$ is not always attained. If the $\sup$ is attained, i.e. $N^*(\epsilon)$ is finite, for all $\epsilon$ then there is no issue with choosing $N^*$ and you have uniform convergence.


TITLE: double sequence
QUESTION [0 upvotes]: Simple question, rather linguistic one.
Having two sequence
\begin{align*}
R_1&=\{a,b,c,d,e,\ldots\}\\
R_2&=\{a,a,b,b,c,c,d,d,e,e,\ldots\}
\end{align*}
I want to say (mathematically corect):  $R_2$ is doubled(?) sequence of $R_1$.
Any better, more informative, ideas?
Of course I will also give this sequences in my work.

REPLY [0 votes]: Let's call the first sequence $(a_n)_n$. If I were you, I would introduce the "doubled sequence" of $(a_n)_n$ as the sequence $(b_n)_n$ defined by
$$
b_{2n} = b_{2n-1} := a_n.
$$
After you've defined what it means to be a "doubled sequence" then there's no harm in using the term even if it's not a commonly used term.


TITLE: Higher point semi-classical Virasoro conformal blocks
QUESTION [1 upvotes]: I am looking for references of higher point semi-classical Virasoro Conformal blocks. I know of one paper where two heavy and arbitrary light operators(https://arxiv.org/abs/1601.06794) have been considered, and another paper with 5-point correlator(https://arxiv.org/abs/1512.07627) is considered. But I am particularly trying to find references/attempts with all equal dimensional operator light correlators. Are there any?

REPLY [1 votes]: In the light limit, there is no reason for the identity exchange to dominate. If all operators are light, including exchanged operators, an $N$-point block is a simple integral over $N-3$ variables. The case $N=4$ is given by formula (2.12) of https://arxiv.org/abs/1109.6764 , whose generalization to $N\geq 5$ is straightforward. It should also be possible to generalize the hypergeometric representation (2.13), I am not sure if this has been worked out explicitly. For $N=5$ this seems to be formula (2.10) of your reference https://arxiv.org/abs/1512.07627 .


TITLE: Limit points of a real function.
QUESTION [0 upvotes]: Given a real number $a$ and an no empty compact set $K$, get a function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that the set of the limit point of $f$ in $a$ is $K$.

Can anyone give a hint to this problem? I think I should do a partitions on the real numbers, but I don't know how.
Thanks a lot!

REPLY [1 votes]: Let $b\in K$.
Let
$$g(x)=\begin{cases}\frac1{x-a}\sin\frac1{x-a}&\text{if }x\ne a\\0&\text{if }x=a \end{cases}$$
and
$$ f(x)=\begin{cases}g(x)&\text{if }g(x)\in K\\b&\text{otherwise}\end{cases}$$


TITLE: Dimension of $ U'\subset V^*$
QUESTION [0 upvotes]: Let $V$ be a finite dimensional real vector space and let $U$ be a subspace of $V$.
Let $U':= \lbrace  \xi \in V^* , \xi(v) = 0 , \forall v \in U\rbrace$.
How can we prove that dim $ U' = $ dim $ V - $ dim $U$ ?

REPLY [1 votes]: I'll try to give you an idea and you fill in the details. Let $W$ be a subspace of $V$ such that $U \bigoplus W = V$. Let $\{e_1, \cdots, e_n\}$ be the basis of $U$ and $\{ e_{n+1}, \cdots, e_{n+m}\}$ the basis of $W$ such that $\{ e_1, \cdots, e_n, e_{n+1}, \cdots, e_{n+m}\}$ is the basis of $V$. Let $\{ f_1, \cdots,f_n, f_{n+1}, \cdots, f_{n+m} \}$ be the basis of V*, which is such that $f_i(e_j) = \delta_{ij}$. If $\xi \in U', $ $\xi(u) = 0, \,\,\ \forall u \in U$ (I imagine you mean that in the definition of $U'$). Then if $v \in V$, $v = u+w \in U \bigoplus W $. Thus
$$\xi(v) = \xi(u+w) = \xi(u)+\xi(w) = \xi(w) = \xi \left( \sum_{i=n+1}^{n+m}w_i e_i \right)= \sum_{j = 1}^{n+m}\alpha_j f_j \left( \sum_{i=n+1}^{n+m}w_i e_i \right) = \sum_{j = n+1}^{n+m}\alpha_j f_j \left( \sum_{i=n+1}^{n+m}w_i e_i \right)  =  \sum_{j = n+1}^{n+m}\alpha_j f_j \left(\sum_{i=1}^{n}u_i e_i+ \sum_{i=n+1}^{n+m}w_i e_i \right) = \sum_{j = n+1}^{n+m}\alpha_j f_j (u+v). $$
that is, the functionals of $U'$ only depend linearly on $\{f_{n+1}, \cdots, f_{n+m}\}.$ Then $\dim U' = m = (n+m) - n = \dim V - \dim U.$
If you already know, a more direct way to solve this is to note that $U'$ is the orthogonal complement of $U$.


TITLE: Which course should I take to be well-prepared for PHD in OR or Financial Math?
QUESTION [3 upvotes]: I'm currently a 2nd-year MS student in Mathematics at a mediocre state school. I'm planning to apply for the PHD programs in Operation Research/Financial Mathematics at schools like Columbia, Cornell, CMU, Georgia Tech, etc. I got straight As in Fourier Analysis (written by Elias Stein) and Linear Programming (written by I. Griva and A. Sofer) classes last year. For the incoming Fall semester, I currently consider taking 1 course out of the following, but not sure which one would benefit me the most for getting into one of the programs above:

Nonlinear Programming - Linear and Nonlinear Optimization by A. Sofer and I. Griva
Stochastic Processes - Introduction to Probability Models by Sheldon Cross
Complex Analysis - Banach spaces of Analytic Functions by Keneth Hoffman
Probability Theory - Probability by Alan Karr

I'm thinking of taking either $1$ or $3$, but I also saw that Stochastic Processes and Nonlinear Programming are core courses for the ORIE/ORFE programs at Columbia, Cornell, CMU, Georgia Tech, etc. Thus I am not entirely sure which one I should take to be well-prepared for such programs. Can anyone please give me some advice on which one of these $4$ courses should I absolutely need to take?

REPLY [0 votes]: financial maths tends to be applied stochastic processes so I'd regard that as most important. Probability is related.
Regression is used quite a lot in financial maths so linear programming is important too. Complex analysis is also useful for Fourier transform methods when doing pricing.
Non-linear programming I see as less important.


TITLE: Given $f(x)=x^n+px + q, \ n\in{\mathbb{N}}$
QUESTION [2 upvotes]: Problem: Given $f(x)=x^n+px+q, \ n\in{\mathbb{N}}, \ x\in\mathbb{R},$ determine (as a function of $n$) the maximum number of distinctive real roots that $f$ can have for $p,q\in\mathbb{R}.$
I need help to understand this problem. Below follows a solution from my professor, but I don't really understand.
Solution:
$n=1:$ One real root for $p\neq-1$ and no real root if $p=-1$.
$n\geq2:$ In this case, $f'(x)$ must have a root between the two roots for $f(x)$. The derivative $f'(x)=nx^{n-1}+p$ has one root for even $n$ and max two roots for odd $n\Rightarrow f$ has max two roots for even $n$ and max three for odd $n$.

Can someone explain to me what is going on here? I understand the first $n=1$ case but for the second case I have the following questions:

If $n=2$, then I understand why $f'(x)$ must have one root between the two roots of $f(x)$. This is because in order for $f(x)$ to have two roots, the derivative must change sign so that the function changes direction and intercepts the $x$-axis again. But she never specifies if it's for $n>2$ or $n=2.$
Why is this statement true: "the derivative has one root for even $n$ and max two roots for odd $n$"?

Any help to understand this is welcome. If there are simpler methods to prove this, only using differentiation, please let me know.

REPLY [4 votes]: Maybe if I add in some words to the case $n \ge 2$ then it will become clearer.
If $f$ has at least two real roots then the derivative has a root in between. This is true regardless of $n$.
Now, note that $f'(x)=nx^{n-1}+p$, so that $f'(x)=0 \iff x^{n-1}=-p/n$. So how many roots can $f'$ have? Well, if $n-1$ is odd then the function $x \to x^{n-1}$ is a bijection, so this equation always has a unique solution. In that case $f'(x)$ has exactly one root, so $f$ may have at most two roots (it could in principle have 0,1, or 2).
Similarly if $n-1$ is even, then $x^{n-1}$ is always positive: it could be possible that $x^{n-1}=-p/n$ has 0,1 or 2 solutions. In any case, it has at most 2 roots. Then $f$ must have at most 3 roots, since if it had 4 then we could find 3 roots of the derivative by looking in between the 3 consecutive pairs of roots.


TITLE: Let $X_1\sim Laplace(0,\sqrt{1/2})$ and $X_2 \sim Laplace(1/2,\sqrt{1/2})$. Are $X_1$ and $X_2$ independent?
QUESTION [0 upvotes]: Let $X_1\sim Laplace(0,\sqrt{1/2})$ and $X_2 \sim Laplace(1/2,\sqrt{1/2})$. Are $X_1$ and $X_2$ independent?
I understand that in case of independence, the joint pdf is the product of the marginal pdfs. I know how to find marginals from joint pdf but I'm not sure how to find $f_{X_1,X_2}$. Also, can I use the expectation to prove independence?

REPLY [1 votes]: People in the comments section are trying to explain that we really don't have enough information. Maybe this convinces you...
I construct $X_1$ and $X_2$, with the same distributions you have in the question, which are dependent in one case and independent in another. Let $Y\sim \text{Laplace}(0,\sqrt{1/2})$. I give you two situations.
Dependent.
Define $X_1 = Y$ and $X_2=Y+\frac 1 2$. Now we have that $X_1$ and $X_2$ are dependent, you see why?
Independent.
Let $Y_1$ and $Y_2$ be independent versions of $Y$. Now define $X_1=Y_1$ and $X_2=Y_2+\frac 1 2$. In this case they are independent, you see why?
So, we really can't tell anything about dependence if you just tell us the marginal distributions.


TITLE: $16!$ without calculator
QUESTION [9 upvotes]: Scrolling through old discrete mathematics exams I have came across this "choose correct answer" question:

$16!$ is:

a). $20 \; 922 \; 789 \; 888 \; 000$
b). $18 \; 122 \; 471 \; 235 \; 500$
c). $17 \; 223 \; 258 \; 843 \; 600$

Would you show me how your thinking process of solving this problem would look like?
The ultimate goal is to find the correct answer; how you get to it does not matter, except that you have to invest only a reasonable amount of time, and calculators or other devices are not allowed.

REPLY [3 votes]: \begin{align}16!&=1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11\times12\times13\times14\times15\times16
     \\&=(2\times5\times10)\times3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16
     \\&=100\times(3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16)\end{align}
So the final answer will have $2$ zeros at the end. If you want the first digit after the zeros, you can multiply the remaining numbers modulo $10$.
\begin{align}&3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16 \pmod {10}
\\&= 3\times4\times6\times7\times8\times9\times1\times2\times3\times4\times5\times6 \pmod {10}
\\&= 2\times6\times7\times8\times9\times2\times3\times4\times5\times6 \pmod {10}
\\&=  2\times7\times8\times9\times2\times3\times4\times5\times6 \pmod {10}
\\&= 4\times8\times9\times2\times3\times4\times5\times6 \pmod {10}
\\&= 2\times9\times2\times3\times4\times5\times6 \pmod {10}
\\&= 8\times2\times3\times4\times5\times6 \pmod {10}
\\&= 6 \times3\times4\times5\times6 \pmod {10}
\\&= 8\times4\times5\times6 \pmod {10}
\\&= 2\times5\times6 \pmod {10}
\\&= 60 \pmod {10}
\\&= 0 \pmod {10}
\end{align}
So the last 3 digits of $16!$ are $000$ and the answer must be a).


TITLE: Counter-examples to $x \not \in K \implies \Bbb Q(x) \cap K = \Bbb Q$
QUESTION [1 upvotes]: Let $K/\Bbb Q$ be a finite extension and let $x$ be an algebraic number such that $x \not \in K$.

Is it possible that $\Bbb Q(x) \cap K \neq \Bbb Q$, if the degree of $x$ over $\Bbb Q$ equals $[K:\Bbb Q]$ ?
Is it possible that $ \Bbb Q(x) \cap K \neq \Bbb Q$, if $K=\Bbb Q(a_1,\dots,a_n)$ and $x,a_1,\dots,a_n$ are conjugated over $\Bbb Q$ (i.e. have the same minimal polynomial)?

I know that if the degrees $\text{deg}_{\Bbb Q}(x)$ and $[K:\Bbb Q]$ are coprime, then $x \not \in K \implies \Bbb Q(x) \cap K = \Bbb Q$ holds. A simple counter-example to this implication is $K=\Bbb Q(\sqrt 2),x=\sqrt[4]{2}$ since the intersection is $K$. But if the degrees are the same, I did not find any counter-example.
As for $2$., we have in particular that $\text{deg}_{\Bbb Q}(x)$ divides $[K:\Bbb Q]$. However, I wasn't sure how to find a possible counter-example.
Thank you very much!

REPLY [1 votes]: Let $x=\sqrt[4]{2}$ and $K=\mathbb Q(a_1)$ where $a_1=i\sqrt[4]{2}$.


TITLE: $A\in M_{3}(\mathbb{R})$ is a matrix such that $\mbox{ Tr }(A-A^t)^{2n}=0$
QUESTION [0 upvotes]: If $A\in M_{3}(\mathbb{R})$ is a matrix such that $\mbox{ Tr }(A-A^t)^{2n}=0$ for some integer $n>0$. Show that $A=A^t$
I have tried like this: Let $X=A-A^t$. Then $X$ is a skew symmetric matrix with all its eigenvalues are zero. This follows from the trace condition. From this how I can proceed?

REPLY [0 votes]: Let $X=A-A^t$. Note that $X$ is a skew symmetric matrix. Now $X$ being a $3\times3$ skew symmetric real matrix, its eigenvalues are $0,\pm i c$ for some $c\geq0$. Then $$\mbox{ Tr }X^{2n}=0\implies2(-1)^{n}c^{2n}=0\implies c=0$$ Hence all the eigenvalues of $X$ are zero. Now consider $$\mbox{ Tr }(XX^t)=\mbox{ Tr }(X(-X))=-\mbox{ Tr }X^2=0.$$ Consequently we have $X=0\implies A=A^t.$


TITLE: $p$-adic exponential function
QUESTION [0 upvotes]: Simple question about $p$-adic numbers $\mathbb{Q}_p$. How do I evaluate $$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ where $x\in {\mathbb Q}_p$. I'm a bit confused, because as far as I understand it, each term is added in the same way as in $\mathbb R$, so why should the result be different?
As a start let's evaluate $x=1$ in $p=3$.

REPLY [2 votes]: The reals (and complexes) are truly different from the various $p$-adic fields $\Bbb Q_p$. One of the glaring differences is the domain of convergence of exponential and logarithm.
For $\Bbb R$ and $\Bbb C$, the exponential series is convergent for no matter what $z$ you plug into it, while the logarithmic series is convergent only for $|z|<1$.
For any $p$-adic field (including the complete extensions of $\Bbb Q_p$), the logarithmic series is still convergent for $z$ with $|z|<1$, but for convergence of the exponential series, you need the very restrictive condition $|z|<\varepsilon^{\frac1{p-1}}$, where $\varepsilon=|p|$ must be a positive real number $<1$, and $\varepsilon=1/p$ is the conventional choice. The above restriction means that you can calculate the $3$-adic value of $\exp(3)$, as @RobertIsrael has done in his answer, but you can not calculate the $2$-adic exponential of $2$.
Now that I’m on my hobby-horse, I must ride it a bit further. The real logarithm extends from its domain of convergence to a homomorphism from the group $\Bbb R^{>0}$ onto $\Bbb R$, multiplicative group to additive group. The complex logarithm has no such extension: there is no group containing $1+\{z\in\Bbb C:\,|z|<1\}$ on which the logarithm can be defined as a homomorphism.
The $p$-adic logarithm is much more wonderful (and useful) than the complex, for if $k$ is a complete $p$-adic field, the set $1+\{z\in k:\,|z|<1\}$ is a multiplicative group, and the logarithmic series defines a homomorphism from it to the additive group $k$. If you have a $p$-adic number $z$, you could try to define $\exp(z)$ as a number $\zeta$ such that $\log(\zeta)=z$, and this equation is always solvable, for $\zeta$ perhaps in a finite field extension of $\Bbb Q_p$. Only problem is that there are infinitely many such $\zeta$, and in most cases, there is little reason to choose one of these over the others. In my opinion, this is the deep reason why you can’t define the exponential of most $p$-adic numbers.


TITLE: Deriving Cartesian Coordinates from Cylindrical Coordinates
QUESTION [1 upvotes]: The ans given was:

$x = r \cos (\alpha)$
$y = h$
$z = -r \sin(\alpha)$

Could somebody explain to me how to arrive at the formula?
I'm probably confused with the axes because usually, the $Z$ axis was given as the depth or "height" for cylindrical coordinates. Even when the axes rotates, shouldn't the formula be

$x = r \cos(\alpha)$
$y = r \sin(\alpha)$
$z = z$

Sorry if this qns seems elementary. I'm really confused with 3D coordinates

REPLY [1 votes]: The important facts of cylindrical coordinates are:

You need an axis to determine height,
The other two coordinates will be given as distance from the origin and rotation around the other planes.

Therefore, height ("depth") doesn't have to be $z$. Here is how I would go about it:
The picture suggests you use the $y$ axis as height, therefore $y=h$. We need to a point in the $xz$ plane in polar coordinates. From the picture you can see that the vector has a positive $x$ component and a negative $z$ component, hence we perform $x = r \cos (\alpha)$ and $z = - r \sin (\alpha)$.


TITLE: Proving the uniform convergence of the Fourier series of $f(x) = x^2$
QUESTION [0 upvotes]: I calculted the Fourier series for $f(x)= x^2$ and I get that it is:
$$\frac{ \pi^{2}}{3} + 4 \sum_{m = 1}^{\infty} \frac{(-1)^m}{m^2} \cos (mx).$$
But the rest of the question asks me to show directly (without using the theorem below) that the series converges uniformly.

Let $f$ be a continuous, piecewise differentiable function in ${\rm PC}(2\pi)$, and suppose that $f’\in{\rm PC}(2\pi)$. Then the Fourier series of $f$ converges uniformly on $\Bbb R$.

Could anyone help me in doing so?
My attempt: I think that the first term is a constant so it will not make a problem for uniform convergence, and the second series is less than or equal $\sum_{m=1}^{\infty} 1/m^{2}$ which is a convergent series by $p$-test as $p=2>1$ . . . but why would this lead to uniform convergence?

REPLY [1 votes]: $|\frac{(-1)^m}{m^2} \cos (mx)| \le \frac{1}{m^2} $ for all $m \in \mathbb N$ and all $x \in \mathbb R.$
This shows that the series $\sum_{m = 1}^{\infty} \frac{(-1)^m}{m^2} \cos (mx)$ converges uniformly on $ \mathbb R.$


TITLE: Lagrangian Mechanics: semi-holonomic constraints
QUESTION [3 upvotes]: By switching to a different set of coordinates, can you make problem with semi holonomic constraints into a problem with holonomic constraints? If so, then when can you do this? I wold like to know if this is possible for all semi-holonomic problems, some semi-holonomic problems or no semi-holonomic problems.
My intuition is that it probably works for some specific cases, but not in general. However, I don't know the reason why. Would be very nice with some sort of proof.
(Constraints on the form: $f=f(q_i,\dot q_i,t)$.)

REPLY [5 votes]: A non-holonomic$^1$ constraint is by definition a constraint that is not holonomic,
e.g. on the form $f(q,\dot{q},t)=0$ or an inequality.

A semi-holonomic constraint
$$ \omega~\equiv~\sum_{j=1}^na_j(q,t)~\mathrm{d}q^j+a_0(q,t)\mathrm{d}t~=~0 $$
is equivalent to a holonomic constraint iff there exist an integrating factor $\lambda(q,t)\neq 0$ and a one-form $\eta$ such that
$$ \lambda\omega+ f\eta~\equiv~\mathrm{d}f ,$$
cf. my related Phys.SE answer here.

--
$^1$ If you are using the 3rd edition of Goldstein, be aware of erratum.


TITLE: IS $f(z) = x^3 + i(1-y)^3$ analytic and where is it differentiable?
QUESTION [1 upvotes]: Where is $f(z) = x^3 + i(1-y)^3$ analytic and where is it differentiable?
I have taken Cauchy-Riemann equations as follows:
$$u(x,y) = x^3$$
$$v(x,y) =(1-y)^3$$
$$\frac{\partial u}{\partial x}=3x^2$$
$$\frac{\partial u}{\partial y}=0$$
$$\frac{\partial v}{\partial x}=0$$
$$\frac{\partial v}{\partial y}=-3(1-y)^2$$
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \implies 0=0$$
$$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}\implies -3(1-y)^2=3x^2\iff (1-y)^2 =x^2$$
Surely this isn't open on any neighborhood?

How do I determine differentiability?

REPLY [1 votes]: The function is differentiable where the CREs hold (and the derivatives are continuous), that is,
$$x^2=-(y-1)^2\ .$$
Since a (real) square cannot be negative, this is the single point $x=0$, $y=1$.  As you state, the region of differentiability does not contain an open neighbourhood of any of its points: so the function is analytic nowhere.


TITLE: Openness and Closedness in Metric Spaces
QUESTION [2 upvotes]: Let $X$ be a metric space. Furthermore, let $E$ be an open subset of $X$.  Then, the complement of $E$, or all members of $X$ that are not in $E$, is closed, or contains all of its limit points.  I understand this to be true locally around $E$.
However, why is this true when taking into account $X$ entirely.  For instance, could there not exist a limit point of $X$ which is not a limit point of $E$?  What if there is a point not in $E$ "distant" from $E$ which is a limit point of $X$ but not in $X$?  Then, $E$ would still be open, but its complement would not be closed.

REPLY [0 votes]: The complement of any open subset of $X$ is a closed subset of $X$. This is often taken to be the definition of a closed subset. Any limit point of $X$ is a point of $X$ (since $X$ is the essentially the entire universe you are working in), so if a limit point is not in $E$ then it must be in its complement.


TITLE: Logical correlation from Oedipus myth
QUESTION [1 upvotes]: My girlfriend likes the myths and she found an MIT article about Oedipus myth which is looks interesting for her. She showed me, but for me it is looks like no correlation between the logical formulas. Maybe because is it normalized.
If you have time guys, please try to describe me. Which are the correlations between the 2 parts of the formula.
The article is here article. The formulas are the following on the second page:
    [∃x: outcome x] [∀y: action y] (If Oedipus were to perform y, then x would come about)
    [∀x: actual outcome x] [∀y: possible action y] (If Oedipus were to perform y,
then x would come about)
    [∀x: actual action x] [∀y: possible choice y] (If Oedipus were to make y,
    then Oedipus would perform x)

I can't understand how [∃x: outcome x] connected with [∀y: action y].
Thanks for your help.

REPLY [1 votes]: It's not clear exactly what your question is.  Where Holton writes

[∃x: outcome x] [∀y: action y] (If Oedipus were to perform y, then x would come about)

he is expressing something like
$$ \exists x.\left(\mathit{outcome}(x) \land \forall y.( \mathit{action}(y) \to (\mathit{perform}(O,y) > \mathit{happen}(x)) \right)$$
Since Holton's "if … then …" is a counterfactual, I used the connective $>$ instead of the material conditional, $\to$.
By request, here are the other sentences:

[∀x: actual outcome x] [∀y: possible action y] (If Oedipus were to perform y,
  then x would come about)

\begin{multline}
\forall{x}.( \mathit{actualOutcome}(x) \to \\
\forall y.(\mathit{possibleAction}(y) \to \\
(\mathit{perform}(O,y) > \mathit{comeAbout}(x))))
\end{multline}

[∀x: actual action x] [∀y: possible choice y] (If Oedipus were to make y,
      then Oedipus would perform x)

\begin{multline}
\forall{x}.( \mathit{actualAction}(x) \to \\
\forall y.(\mathit{possibleChoice}(y) \to \\
(\mathit{makeChoice}(O,y) > \mathit{perform}(O,x))))
\end{multline}


TITLE: Question on random bits
QUESTION [0 upvotes]: Sample a uniformly random bit. Let $X$ be the corresponding random variable.
Here is a randomized procedure to create two more output bits, denoted by random variables $Y_1$ and $Y_2$. To generate each output bit $Y$:

First toss $t$ fair coins. If all of these coins are heads, let $Y = X$.
Else, output a uniformly random $Y$.

I am trying to compute:
\begin{equation}
\Pr[Y_1=x_1, Y_2=x_2],
\end{equation}
for each of the bits $x_1 \in \{0, 1\}$ and $x_2 \in \{0, 1\}$.

Here is my attempt:
\begin{equation}
\Pr[Y_1=x_1, Y_2=x_2] = \Pr[Y_1=x_1]\Pr[Y_2=x_2~|~Y_1=x_1].
\end{equation}
Now, for any $x_1 \in \{0, 1\}$,
\begin{align}
\Pr[Y_1 = x_1] &= \Pr[X = x_1] \Pr[Y_1=x_1|X=x_1] + \Pr[X = \bar{x_1}] \Pr[Y_1=x_1|X=\bar{x_1}] \\
&= \frac{1}{2}\left(\frac{1}{2^{p}} + \left(1 - \frac{1}{2^{p}}\right)\frac{1}{2} \right) + \frac{1}{2} \left(1 - \frac{1}{2^{p}}\right)\frac{1}{2} \\
&= \frac{1}{2}.
\end{align}
This is because $Y_1$ is filled with $X$ with probability $\frac{1}{2^{p}}$ and is a random bit with probability $1 -\frac{1}{2^{p}}$.
However, I cannot find the probability
\begin{equation}
\Pr[Y_2=x_2~|~Y_1=x_1].
\end{equation}

REPLY [1 votes]: You can use the following equality:
$$p(A) = \sum\limits_n p(A|B_n)p(B_n)$$
For you, $A=Y_2=x_2|Y_1=x_1$. I am using the notation $x^*$ for bit inversion
$$P[Y_2=x_2|Y_1=x_1] $$
$$= P[Y_2=x_2|Y_1=x_1, X=x_1]P[X=x_1] + P[Y_2=x_2|Y_1=x_1, X=x_1^*]P[X=x_1^*]$$
You know that $P[X=x_1]=P[X=x_1^*]=0.5$. You can calculate the rest in the same way that you calculated the other part.


TITLE: Collapse of quantum state upon measurement - why not some other unit scalar multiple
QUESTION [1 upvotes]: I have learnt that upon measurement, the state of the system changes to the eigenstate corresponding to the observed eigenvalue. Specifically, we obtain the eigenstate that is the normalized projection of the original state onto the eigenspace. However, if you multiply a state by a scalar of absolute value 1, it will still represent the same state but I thought the scalars can affect future calculations (e.g. when interference is brought into the picture).
So my question is: why not a unit scalar multiple of the projected state instead when we performed a measurement?

REPLY [1 votes]: There is a fascinating discussion in the final sections of Feynman's lectures on physics (available online for free) that actually shows that in some cases the complex phase can play a significant physical role, in superconductors. However, this does not invalidate the general wisdom that a complex phase does not matter, however, as you will see if you decide to read it: In Feynman's explanation of superconductors the wave function takes on a more classical meaning, producing weird results :) This is not even a completely correct description of the phenomenon (this would require quantum field theory), but it was very enlightening to me and I just thought it was cool and should be mentioned.
Now, more to the point: Generally, the reason why we say that a complex phase $e^{i\theta}$ cannot be physical is because, when we consider a general expectation value of an operator $\hat{O}$ in the state $|\psi\rangle$, we have
$$\langle \psi|\hat{O}|\psi\rangle=\langle \psi|e^{-i\theta}\hat{O}e^{i\theta}|\psi\rangle$$
So that any overall phase cancels immediately. It is important here to always keep in mind what is observable and what is not. On a side note, it is important to realize that relative phases  such as the one occurring in the state $|\phi\rangle =\frac{1}{\sqrt{2}}(|0\rangle +e^{i\theta}|1\rangle$ do produce important physical effects (see, for instance, this wikipedia article). It can be quickly understood why this should be the case by remembering how interference works.


TITLE: Probability of winning a ticket with a red dot
QUESTION [2 upvotes]: Question:
You have obtained some interesting information about the local lottery. There was a malfunction at the printer that accidentally marked a bunch of tickets with a red dot. This malfunction disproportionately affected winning lottery tickets. In total $40\%$ of winning tickets were marked with a red dot, while only $20\%$ of losing tickets were marked with a red dot. You have a probability of $\frac{3}{39}$ of winning the lottery.
You have found a ticket  marked with a red dot. What is the probability that this is a winning ticket?
What i have done is the following:
A= probability of a winning ticket.
B= probability of having a red dot.
P(A|B)= P(A intersection B) / P(B)
P(B)= P(red dot| winning ticket)$\times$ P(winning ticket) + P(red dot| losing ticket)$\times$ P(losing ticket)
$0.4 \times \frac{3}{39} + 0.20 \times \frac{12}{13}= 0.215$
P(A intersection B)= P(A|B)$\times$P(B)= $0.4\times 0.215=0.086$
So P(A|B)= $\frac{0.086}{0.215}=0.4$
I'm not sure if I'm doing this right. Maybe someone can give feedback.

REPLY [0 votes]: Apply Bayes' rule:
$$
\begin{aligned}
P(\text{winning}|\text{red})
&=\frac{P(\text{red}|\text{winning})P(\text{winning})}{P(\text{red}|\text{winning})P(\text{winning})+P(\text{red}|\text{not winning})P(\text{not winning})}\\ &=
\frac{\frac{4}{10} \cdot \frac{3}{39} }{\frac{4}{10} \cdot \frac{3}{39} +\frac{2}{10} \cdot \frac{36}{39} }\\
&=\frac{4 \cdot 3 }{4 \cdot 3 +2\cdot 36 }\\
&= \frac{1}{7}
\end{aligned}$$


TITLE: Two sphere - a spacelike hypersurface
QUESTION [3 upvotes]: Is the surface of a 2-sphere spacelike? What are the corresponding tangent vectors to the surface of a 2-sphere?
This question arises from the point of a trapped surface. In Schwarzschild spacetime the surfaces inside the event horizon are trapped and they are all 2-spheres. Does this mean that trapped surfaces are always 2-spheres? Are there trapped surfaces which are not 2-spheres but other spacelike hypersurfaces?

REPLY [2 votes]: The Schwarzschild metric in radial coordinates $(t, r, \theta, \phi)$ is given by
$ds^2 = -(1 - 2M/r) dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\theta^2 +r^2 \sin^2 \theta d\phi^2$
where $M$ is the black hole mass.
A hypersurface is defined by a constant function $f$, in case of a 2-sphere $r = constant$. A vector normal to a hypersurface is described by $\xi^\mu = g^{\mu \nu} \nabla_\nu f$, where $\nabla_\mu$ is the covariant derivative (reducing to the partial derivative when applied to a scalar) and $g^{\mu \nu}$ is the inverse metric tensor. As for a 2-sphere, we have
$\xi_\mu = (0, 1, 0, 0)$ dual vector
$\xi^\mu = (0, (1 - 2M/r), 0, 0)$ vector
The norm squared is $\xi^\mu \xi_\mu = (1 - 2M/r)$, which is positive (space-like) when $r > 2M$, that is outside the event horizon, zero (null) when $r = 2M$, that is on the event horizon and negative (time-like) when $r < 2M$, that is inside the event horizon.
As a hypersurface is time-like if the nornal is space-like, null if the normal is null and space-like if the normal is time-like, we have
A 2-sphere is time-like when $r > 2M$, outside the event horizon
A 2-sphere is null when $r = 2M$, on the event horizon
A 2-sphere is space-like when $r < 2M$, inside the event horizon
The tangent vectors to the 2-sphere are
$\zeta^t = (1, 0, 0, 0)$
$\zeta^\theta = (0, 0, 1, 0)$
$\zeta^\phi = (0, 0, 0, 1)$
Trapped surfaces not necessarily are 2-spheres, that depends on the metric. For instance trapped surfaces in Kerr metric (rotating black hole) do not exibit a radial symmetry.


TITLE: Does mass of an object has different values in different directions
QUESTION [0 upvotes]: I was studying " Force" (book Kleppner and kolenkow)  . I m unable to understand  a para which is saying  "if mass were different in different directions , acceleration would not be parallel to force and force and acceleration could not be related by a simple vector eq. Although the concept of mass having different values in different directions might sound absurd , but it  is not impossible "...   ??

REPLY [1 votes]: Probably. In general Cayley-Klein spaces, the mass and the principle moments of inertia of a 3D rigid body are all parts of a 6-dimensional structure. In flat Euclidean space the mass is isotropic (the same in all directions). But in hyperbolic and elliptic spaces the translational mass behaves a bit like the rotational mass in that it depends on the distribution of mass in the rigid body and can vary in different directions.
Charles Gunn goes into the gory mathematical details in his thesis here, but to make the idea slightly more plausible, we can consider the problem of a compact 2D rigid body moving on the surface of a giant frictionless sphere. Intuitively, and taking a close-up view, we can see that the body has two 'translational' degrees of freedom sliding across the surface of the sphere, and one 'rotational' degree of freedom spinning around its own centre. However, on a sphere, the 'translational'/'rotational' distinction breaks down, since each 'translation' is actually a rotation about a distant point 90 degrees away around the sphere. The 'translational mass' is actually the moment of inertia of the rigid body defined about this distant point.
In the limit as the sphere gets larger, these two points recede to infinity, and we find that the 'translational mass' is simply the moment of inertia of the body measured about a point at infinity. Since all parts of the body are equally distant from infinity in all directions, the distribution of mass within the rigid body has no effect and all the translational inertia terms are equal. The rotational inertia term is about the centre of the body, and for this one the distribution of mass does matter.
In 3D we get the same sort of effect, but now there are 3 rotational terms constituting the moment of inertia and 3 translational terms about axes that have been pushed away to infinity.
(See Remarks 191-193 and section 9.9.3 on pages 115-117 of Gunn's thesis.)
Since space where we happen to be in the universe is pretty flat, looking like an infinite Euclidean volume in all directions, any effects from the curvature are expected to be tiny - on the order of one over the radius of the universe. But in the presence of gravity, the distinction between 'translation' and 'rotation around some very distant point' becomes a bit more approximate, and the translational mass of extended bodies may turn out to be something a bit more complex than just a scalar number.


TITLE: If $|f(x)| \leq |x|^\alpha$ then $f$ is differentiable at $0$
QUESTION [0 upvotes]: In an exercise I'm asked to prove the following:

Let $\alpha > 1$. Prove that if there exists a function $f: \Bbb R \to \Bbb R$ such that:
$$\forall x \in \Bbb R, \ |f(x)| \leq |x|^\alpha$$
then $f$ is differentiable at $x = 0$

I think that we are supposed to use squeeze theorem or something like that to prove that the limit exists but the absolutes values on the condition are a little bit confusing to me to work with. How can I do this?

REPLY [4 votes]: Clearly, $f(0)=0$. And then, since$$\left|\frac{f(x)-f(0)}x\right|=\left|\frac{f(x)}x\right|\leqslant|x|^{\alpha-1}.$$So, since $\alpha-1>0$, $\lim_{x\to0}\frac{f(x)-f(0)}x=0$.


TITLE: Can you use Oseledet's theorem to numerically approximate the Lyapunov spectra?
QUESTION [5 upvotes]: Let's say you have a set of first order differential equations with known Jacobian $J$. Let $x_0, x_1, ..., x_n$ be sampled points on the trajectory near the attractor.
Let $T_n = J(x_{n-1})J(x_{n-2})...J(x_0)$.
Oseledet says that
$$\frac{1}{n} \log(\mbox{the singular values of } T_n)$$
converge to the Lyapunov spectra as $n \to \infty$.
However, can this be used in practice to numerically approximate the Lyapunov spectra (say just to the 2nd decimal place)? Is it known how big $n$ must be to do so?

REPLY [4 votes]: Once you have a method to estimate the top Lyapunov exponent, you can use the action on wedge products to estimate the other exponents (This should be clear if you examine the proof of Oseledet's theorem.) Estimating the top exponent is difficult, in general. Doing it via the definition is very slow as you observed. More efficient approaches are usually based on Furstenberg's foundational paper [1] that preceded Oseledets' work. The key is to estimate the stationary measure on projective space constructed in [1]. See also [2].  Sometimes this measure is discrete which allows for efficient computation [3]. A sophisticated complex analysis approach is in [4].
[1] Furstenberg, Harry. "Noncommuting random products." Transactions of the American Mathematical Society 108, no. 3 (1963): 377-428.
[2] Gol'dsheid, I. Ya, and Grigorii Aleksandrovich Margulis. "Lyapunov indices of a product of random matrices." Russian mathematical surveys 44, no. 5 (1989): 11-71.
[3] Kenyon, Richard, and Yuval Peres. "Intersecting random translates of invariant Cantor sets." Inventiones mathematicae 104, no. 1 (1991): 601-629.
[4] Pollicott, Mark. "Maximal Lyapunov exponents for random matrix products." Inventiones mathematicae 181, no. 1 (2010): 209-226.


TITLE: Norm of Integral Operator on $E=\{u \in C[0,1]\ ,\ u(0) = 0\}$
QUESTION [0 upvotes]: There are similar question but the characterization of the space $E$ that I have gives me problem in computing the actual norm.
Let $E=\{u \in C[0,1]\ ,\ u(0) = 0\}$ with the usual $\parallel \cdot\parallel_\infty$-norm.
Consider the linear functional $f$ defined as: $$f(u)=\int_0^1u(t)dt, \ \forall u\in E\ .$$
$\cdot$Show that  $f ∈E^⋆$ and compute $∥f∥_{E^⋆}$.
I can show that $f ∈E^⋆$ but I have troubles computing the actual norm of $f$.
$f ∈E^⋆$ iff $\ \ \ $ $f\ : \ E$ -> $\mathbb{R}$ and  $f$ is linear and continuous. Linearity follows easily from the properties of the integral. In order to show continuity, we show that $f$ is bounded since for linear operator boundedness is equivalent to continuity.
$|f(u)| = |\int_0^1 u(t)dt\ |\ \leq\ \int_0^1 |u(t)|dt\ \leq ||u||_\infty$, so taking the $\sup_{||u||\leq 1}$ we get that $||f||\leq 1$. Hence I would guess that: $$||f||=1$$
since we have that: $\forall u \in E, |f(u)|\leq||f||\cdot||u||_\infty$.
Now I should find some $u\in E$ such that $||u||_\infty = 1$ and $|f(u)|=1$ and the trick would be done, but I can not find such function.
Any suggestion?

REPLY [0 votes]: It is not necessary to find $u\in E$ such that $|f(u)|/||u||_\infty=1$. It is enough to find a sequence $u_n$ such that $|f(u_n)|/||u_n||_\infty\to1$.
For this take the function(s) $u_n(x)=1$ for $x\geq1/n$ and $u_n(x)=nx$ for $x\in[0,1/n]$.
We can see that for this sequence
$$|f(u_n)|/||u_n||_{\infty}=|\int_{0}^{1}u_n(x)dx|\geq|\int_{1/n}^{1}u_n(x)dx|=1-\frac{1}{n}\to1.$$