text:
\begin{document}

\title{Statistics of linear  families  of  smooth functions on knots}

\date{Started  June 2, 2010. Completed  on June 4, 2010.
Last modified on {\today}. }

\author{Liviu I. Nicolaescu}

\address{Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556-4618.}
\email{nicolaescu.1@nd.edu}
\urladdr{\url{http://www.nd.edu/~lnicolae/}}

\begin{abstract}   Given a   knot $K$ in an Euclidean space $\bR^n$,  and   a     finite dimensional subspace $\bsV\subset C^\infty(K)$,  we    express the expected number of critical  points of  a random function in $\bsV$in terms of  an integral-geometric invariant of $K$ and $\bsV$.  When $\bsV$  consists of  the restrictions   to $K$ of homogeneous polynomials  of degree $\ell$  on $\bR^n$,  this invariant  takes the form  of total curvature of  a certain immersion of $K$.  In particular, when $K$ is the unit circle in $\bR^2$ centered at the origin,  then the expected number of critical points of the restriction to $K$ of a random  homogeneous polynomial  of  degree $\ell$  is $2\sqrt{3\ell-2}$, and the expected number of critical points on $K$ of a random  trigonometric polynomial of degree $k$ is approximately $1.549 k$.  \end{abstract}

\maketitle

\begin{center}
\textit{To the memory of my mathematical  hero, Vladimir Igorevich Arnold}
\end{center}

\tableofcontents

\section*{Introduction}
\setcounter{equation}{0}

A celebrated   result of F\'{a}ry and Milnor \cite{Fa, Mil} states that the expected  number of critical points  of the restriction  to a knot $K\hra \bR^3$ of a  random linear map  $h:\bR^3\ra \bR$  is equal to an integral-geometric  invariant of the knot, namely, its (suitably normalized) total curvature.     It is natural to ask how this  result  changes if, instead of  random linear maps, we look  at  random  homogeneous polynomials $P:\bR^3\ra \bR$ of a given degree $\ell$. It is convenient to investigate an even more general situation.

Suppose that $\bsU$ is an oriented   real Euclidean space  of dimension $n$  and $K\hra \bsU$ is a knot in $\bsU$, i.e., an smoothly embedded $S^1$.  We assume that $0\not\in K$.  For any positive integer $\ell$ we denote by $\eQ_\ell(\bsU)$ the space of symmetric $\ell$-linear forms  on $\bsU$. Each such form  defines a polynomial  function
\[
P_\Phi:\bsU\ra \bR,\;\; P_\Phi(\bu)=\Phi(\underbrace{\bu,\dotsc,\bu}_\ell).
\]
We denote by $f_\Phi$ the restriction     of $P_\Phi$ to the knot $K$.  We  will refer to such functions as   \emph{polynomial functions on $K$ of  degree  $\ell$}. For  random $\Phi$, the function   $f_\Phi$ is Morse,  and we denote by $\mu_K^\ell(\Phi)$ its number of critical points.      We can regard $\mu_K(\Phi)$ as a random variable and  ask how  is  its expectation  related to the global geometry of $K$.

To describe this relationship,  denote by $\eS^\ell$ the unit sphere in $\eQ_\ell(\bsU)$ with respect to the natural metric $\lan-,-\ran_\ell$ induced by the Euclidean metric on  $\bsU$. The expected  number of  critical points  of $f_\Phi$ is the real number $\mu_K^\ell$ defined by
\[
\mu_K^\ell=\frac{1}{{\rm area}\,(\eS^\ell)} \int_{\eS^\ell} \mu_K^\ell(\Phi)\,dS(\Phi).
\]
In  Theorem \ref{th: vero} we express  $\mu_K^\ell$ as an integral geometric invariant of $K$.     This invariant is  described in terms of the Veronese  map $\eV_\ell :\bsU\ra \eQ_\ell(\bsU)$  uniquely determined by the equality
\[
\bigl\lan\,\Phi, \eV_\ell(\bu)\,\bigr\ran_\ell=\Phi(\underbrace{\bu,\dotsc,\bu}_\ell),\;\;\forall \bu\in\bsU,\;\;\Phi\in\eQ_\ell(\bsU).
\]
The restriction of the Veronese map to $K$ produces an immersion
\[
\eV_\ell: K\ra \eQ_\ell(\bsU)
\]
that we called the \emph{$\ell$-th Veronese immersion} of $K$.    In Theorem \ref{th: vero} we show that $\mu_K^\ell$ is the total curvature of the $\ell$-th Veronese immersion.       The case $\ell=1$ of this theorem is the celebrated  result of F\'{a}ry and Milnor \cite{Fa, Mil}.

As an application we compute $\mu_K^\ell$, when $K$ is the unit circle in $\bR^2$ centered at the origin. More precisely, we show that  in this case $\mu^\ell_K=2\sqrt{3\ell-2}$. In other words,  the  expected  number of critical points  of a polynomial function on $K$ of degree $\ell$ is $2\sqrt{3\ell-2}$.

We obtain Theorem  \ref{th: vero} as a special case of  Theorem \ref{th: main}     that describes the   expected    number of critical points  of a random  function belonging to a fixed finite dimensional subspace $\bsV\subset C^\infty(K)$ satisfying a nondegeneracy condition (\ref{eq: nondeg}).

Theorem \ref{th: main} has another interesting consequence. More precisely, in Theorem \ref{th: trig} we show that the expected number of critical points on the unit circle of a random     trigonometric polynomial of degree $n$ is
\[
T_n=2\sqrt{\frac{3\beta_5(n+1)}{5\beta_3(n+1)}},
\]
where $\beta_k$ denotes the $k$-th Bernoulli polynomial. In particular
\[
T_n\sim 2n\sqrt{\frac{3}{5}} \;\;\mbox{as $n\ra \infty$}.
\]

\medskip

\noindent \ding{45} {\bf Notations.} We will  denote by $\bsi_n$ the ``area'' of the  round  $n$-dimensional sphere $S^n$ of radius $1$, and by $\bom_n$  the  ``volume'' of the  unit ball in $\bR^n$.  These quantities are uniquely determined by the equalities (see \cite[Ex. 9.1.11]{N1})
\[
\bsi_{n-1}=n\bom_n,\;\;\bom_n=\frac{\Gamma(1/2)^n}{\Gamma(1+n/2)},\;\;\Gamma(1/2)=\sqrt{\pi},
\]
where $\Gamma$  is Euler's Gamma function.

\section{An   abstract result}
\label{s: 1}
\setcounter{equation}{0}

Let $\bsU$ and $\bsV$ be a real, oriented Euclidean space of dimension $n$.  We denote by $(-,-)_\bsU$ the inner product in $\bsU$ and by $S(\bsU)$ the unit sphere in $\bsU$. Suppose that  $K\hra \bsU$ is a smooth knot in $\bsU$, i.e.,   the image of a smooth embedding  $S^1\hra \bsU$.  Fix  an arclength parametrization of $K$
 \[
 s\mapsto \bx(s),\;\; 0\leq s\leq L,
 \]
 where $L$ denotes the length of $K$.

 Assume that we are given a  finite dimensional subspace $\bsV\subset  C^\infty(\bsU)$ of dimension  $N$ satisfying the nondegeneracy condition
 \begin{equation}
 \forall s\in [0,L],\;\;\exists  \bv\in\bsV:\;\; d_{\bx(s)}\bv (\bx'(s))\neq 0,
 \label{eq: nondeg}
 \end{equation}
 where $d_\bx \bv $ denotes the differential of the function  $\bv$ at $\bx$.   We fix an  Euclidean inner product $(-,-)_\bsV$ on $\bsV$ and we denote by $S(\bsV)$ the unit sphere in $\bsV$ centered at the origin.

 As explained in \cite[\S 1.2]{N1}, the  condition (\ref{eq: nondeg}) implies that for generic $\bv\in\bsV$, the restriction of the function $\bv$ to   $K$ is a Morse function. We denote  by $\mu_K(\bv)$  its number of critical points.      Note that for any  nonzero scalar $t\in\bR$ the restriction of $t\bv$ to $K$ is Morse if and only if    the restriction of $\bv$ to $K$ is such.     Thus, it suffices to concentrate on functions $\bv\in S(\bsV)$.   The main goal    of this section  is the computation of the expected number of critical points of a random function in $S(\bsV)$, i.e., the quantity
 \[
 \mu_K^\bsV:=\frac{1}{{\rm area}\,\bigl(\,S(\bsV)\,\bigr)}\int_{S(\bsV)} \mu_K(\bv)\,|dS(\bv)|=\frac{1}|{\bsi_{N-1} \bigl(\,S(\bsV)\,\bigr)}\int_{S(\bsV)} \mu_K(\bv)\,|dS(\bv)|,
 \]
where  $|dS|$  is the ``area'' density on   $S(\bsV)$.

 To describe the result we need  to introduce the main  characters.  Set $\bsV\dual:=\Hom(\bsV,\bR)$ and observe that we have a smooth map $\xi: K\ra \bsV\dual$ that associates to   each point $\bx(s)\in K$ the linear map
 \begin{equation}
 \xi_s:\bsV\ra \bR,\;\ \; \bv\mapsto d_{\bx(s)}\bv(\bx'(s))\in \bR.
 \label{eq: xi}
 \end{equation}
 Using the metric induced isomorphism $\bsV\dual\ra \bsV$ we obtain a dual map
 \begin{equation}
 \xi^\dag: K\ra \bsV.
 \label{eq: xi-d}
 \end{equation}
 The nondegeneracy condition (\ref{eq: nondeg}) implies  that $\xi^\dag_s\neq 0$,  $\forall s$. We  can thus define a
 smooth map
 \[
 \bnu: K\ra S(\bsV),\;\;\bnu(s):=\frac{1}{|\xi^\dag_s|}\xi_s.
 \]
 \begin{theorem} Let $\bsU$, $K$ and $\bsV$ be as above then
 \[
 \mu_K^\bsV:=\frac{1}{\bsi_{N-1}}\int_{S(\bsV)} \mu_K(\bv)\, |dS(\bv)|= \frac{1}{\pi}\int_K |\nu'(s)|\, |ds|.
 \]
 \label{th: main}
 \end{theorem}

\begin{proof} Set
 \[
 E_\bnu:=\bigl\{\, (\bx,\bv)\in K\times S(\bsV);\;\; \bnu(\bx)\perp \bv \,\bigr\}.
 \]
 Note that $E_\bnu$ can be  alternatively defined as the zero set of the function
 \[
 F: K\times S(\bsV)\ra \bR,\;\;F(\bx,\bv)=(\bnu(x),\bv)_\bsV,
 \]
 and  the differential of $F$ is nonzero along the level set  $\{F=0\}$. This shows that $E_\bnu$ is a smooth submanifold of  $K\times S(\bsV)$ and
 \[
 \dim E_\bnu=\dim S(\bsV)= N-1.
 \]
 We    denote by $g_E$ the  metric on $E_\bnu$ induced  by the natural metric of  $K\times S(\bsV)$, and by $|dV_E|$ the associated     volume density. We have two natural (left and right) smooth maps
 \[
 K\stackrel{\lambda}{\longleftarrow} E_{\bnu}\stackrel{\rho}{\Lra} \bsS(V).
 \]
 The fiber of $\lambda$ over $\bx\in K$ is the Equator
 \[
 E_\bx:=\bigl\{ \bv\in S(\bsV);\;\; \bv\perp \bnu(x)\,\bigr\},
 \]
 while the fibers of $\rho$ are generically finite.  Tautologically, the restriction of $\lambda$ to any fiber of $\rho$ is injective.         More importantly, for any $\bv\in S(\bsV)$ the subset $\lambda\bigl(\rho^{-1}(\bv))\subset K$ is the set of critical points of  the restriction of $\bv$ to $K$.   Hence,  for generic $\bv\in\bsV$ we have
 \[
 \mu_K(\bv)=\#\rho^{-1}(\bv).
 \]
 We  have thus reduced the problem to computing  the average  number  of points  in the fibers of $\rho$ in terms of  integral-geometric invariants of the map $\bnu$.      We will achieve this in (\ref{eq: av4}).

   The area   formula  (see \cite[\S 3.2]{Feder} or \cite[\S 5.1]{KP}) implies  that
\begin{equation}
\int_{S(\bsV)} \# \rho^{-1}(\bw) |dS(\bv)|=\int_E J_\rho(x,\bv) |dV_E(x,\bv)|,
\label{eq: av}
\end{equation}
where   the nonnegative  function $J_\rho$ is the Jacobian of $\rho$ defind by the equality
\[
\rho^*|dS|=J_\rho \cdot |dV_E|.
\]
To compute the integral in the right-hand side of (\ref{eq: av}) we need a  more explicit description of the geometry of $E_\bnu$.

Let $(\bx_0,\bv_0)\in E_\bnu$.  Fix an orthonormal frame $e_1,\dotsc,  e_{N-2}$ of the tangent space of $E_{\bx_0}$ at $\bv_0$.     Assume $\bx_0=\bx(s_0)$.  Then  the tangent space  of $E_\bnu$ at $(\bx_0,\bv_0)$ consists of tangent vectors
\[
\dot{s}\bx'(s_0)\oplus \dot{\bv}\in T_{\bx_0}K\oplus T_{\bv_0}S(\bsV),\;\;\dot{s}\in\bR,
\]
such that
\begin{equation}
\dot{s}(\bnu'(s_0),\bv_0)_\bsV+ (\nu(s_0),\dot{\bv})_\bV=0.
\label{eq: tan}
\end{equation}
 Define
 \[
 e_0= \bx'(s) -(\bnu'(s_0),\bv_0)_\bsV \bnu(s_0).
 \]
For simplicity we set
\[
\mu(\bx_0, \bv_0):=\bigl|\, (\bnu'(s_0),\bv_0)_\bsV\,\bigr|.
\]
The equality (\ref{eq: tan}) implies that the collection
 \[
 e_0, e_1,\dotsc,  e_{N-2}
 \]
 is an orthogonal  basis of $T_{(\bx_0,\bv_0)}E_{\bnu}$. Moreover, the length of $e_0$ is
 \[
 |e_0|=\sqrt{ 1+\mu(\bx_0,\bv_0)^2}.
 \]
 If we denote by $dS_\bx$ the  area form on $E_\bx$ then we see that at $(\bx_0, \bv_0)$ we have
 \[
 ds\wedge dS_\bx(e_0, e_1,\dotsc, e_{N-2})= \pm 1.
 \]
 Hence at $(\bx_0,\bv_0)$ we have
 \begin{equation}
 |dV_E|=|e_0|\cdot |ds\wedge dS_\bx|=\sqrt{ 1+\mu(\bx_0,\bv_0)^2}\bigl|ds\wedge dS_\bx|.
\label{eq: vol}
\end{equation}
The differential of  $\rho$ at $(\bx_0, \bv_0)$ is the linear map
 \[
 \rho_*: T_{(\bx_0,\bv_0)}E_\bnu\ra T_{\bv_0}S(\bsV)
 \]
 given by
 \[
 e_0\mapsto  (\bnu'(s_0),\bv_0)_\bsV\,\cdot\,  \bnu(s_0),\;\; e_i\mapsto e_i,\;\;1\leq i\leq N-2.
 \]
 We conclude that
 \begin{equation}
 J_\rho(\bx_0,\bv_0)=\frac{\mu(\bx_0,\bv_0)}{\sqrt{ 1+\mu(\bx_0,\bv_0)^2}}.
 \label{eq: jac}
 \end{equation}
 Using (\ref{eq: vol}), (\ref{eq: jac}) and the co-area formula  for the map $\lambda$ \cite[Prop. 9.1.8]{N0} we deduce
 \begin{equation}
 \int_E J_\rho(x,\bv) |dV_E(x,\bv)|=\int_K \underbrace{\left(\int_{E_\bx} \mu(\bx, \bv) |dS_\bx(\bv)|\,\right)}_{=:J(\bx)} |ds(\bx)|.
 \label{eq: av1}
 \end{equation}

To proceed further we need the following elementary result.

 \begin{lemma} Suppose $\bsW$ is an $(m+1)$-dimensional oriented real Euclidean space with inner product $(-,-)$, and $\bv_0\in\bsW$. Denote by $S(\bsW)$ the unit  sphere in   $\bsE$, and by $|dS|$ the ``area'' density on $S(\bsW)$.  Then
 \[
I(\br_0) =\int_{S(\bsW)} |(\bv_0,\bw)| |dS(\bw)|=\bom_m|\bv_0|.
 \]
 \label{lemma: cos}
 \end{lemma}

 \begin{proof} Fix an orthonormal basis $(e_0,e_1,\dotsc, e_m)$ of $\bsW$ and denote by $(w_0,\dotsc, w_m)$ the resulting coordinates.   Observe that for any  orthogonal transformation $T: \bsW\ra \bsW$ we have $I(T\bv_0)=I(\bv_0)$ so that $I(\br_0)$ depends only on the length of $\bv_0$.  Thus, after an orthogonal transformation,  we can assume  $\bv_0= ce_0$. We can then write
 \[
 I(ce_0)=|c|\int_{S(\bsW)}  |(e_0,\bw)| |dS(\bw)|= |c|\int_{S(\bsW)}  |w_0| |dS(\bw)|.
 \]
 Denote by $S_+(\bsW)$ the hemisphere $w_0>0$. We have
 \[
 I(ce_0)= 2|c|\int_{S_+(\bsW)}  w_0 |dS(\bw)|.
 \]
 The upper  hemisphere $S_+(\bsW)$ is the graph of the map
 \[
 w_0= \sqrt{1-r^2},\;\;r :=  \sqrt{w_1^2+\cdots +w_m^2}<1.
 \]
 Observe that
 \[
 |\nabla w_0|=\frac{r^2}{1-r^2}\;\; \mbox{and}\;\; |dS(\bw)|=\frac{1}{\sqrt{1-r^2}} |dV_m|=\frac{1}{w_0}|dV_m|,
 \]
 where $|dV_m|$ denotes the Euclidean volume density on the $m$-dimensional Euclidean space with coordinates $(w_1,\dotsc, w_m)$. We deduce
 \[
 I(ce_0)= 2|c|\int_{r<1} |dV_m|=2|c|\bom_m.
 \]
 \end{proof}

  Using Lemma \ref{lemma: cos}  in the special case when $m+1=N-1$, $\bsW$ is the  hyperplane containing the Equator $E_{\bx_0}$, $\bx_0=\bx(s_0)\in K$,  and $\bv_0=\bnu'(s_0)$ we deduce
  \[
  J(\bx_0)=I(\bnu'(s_0))=2|\bnu'(s_0)|\bom_{N-2}.
  \]
  From (\ref{eq: av}) we now deduce
  \begin{equation}
  \frac{1}{\bsi_{N-1}}\int_{S(\bsV)} \# \rho^{-1}(\bv) |dS(\bv)|=\frac{2\bom_{N-2}}{\bsi_{N-1}} \int_K |\bnu'(s)|  |ds|.
  \label{eq: av2}
  \end{equation}
  Using the   equalities
  \[
  \bsi_{N-1}= N\bom_N,\;\;\bom_m=\frac{\Gamma(1/2)^m}{\Gamma(1+m/2)}.
  \]
  we deduce
  \[
  \frac{2\bom_{N-2}}{\bsi_{N-1}}=\frac{2}{N\Gamma(1/2)^2} \frac{\Gamma(1+N/2)}{\Gamma(N/2)}=\frac{1}{\pi},
  \]
  where at the last step we have use the  classical identities
  \[
  \Gamma(1+N/2)=\frac{N}{2}\Gamma(N/2),\;\;\Gamma(1/2)=\sqrt{\pi}.
  \]
  We have thus proved
  \begin{equation}
   \frac{1}{\bsi_{N-1}}\int_{S(\bsV)} \# \rho^{-1}(\bv) |dS(\bv)|=\frac{1}{\pi}\int_K |\bnu'(s)|  |ds|.
   \label{eq: av4}
   \end{equation}
   \end{proof}

   \section{Polynomial  Morse functions on knots}
   \setcounter{equation}{0}

  Let $\bsU$ and $K\hra\bsU$ as above.   We  want to apply  the  results in the previous section to a special choice of $\bsV$.  Fix a positive integer $\ell$.   Denote by $\eQ_\ell(\bsU)$ the space of  symmetric $\ell$-linear  forms on $\bsU$, or equivalently, the space of homogeneous polynomials on $\bsU$ of degree $\ell$.   This will be our choice of subspace $\bsV\subset C^\infty(\bsU)$. The nondegeneracy assumption (\ref{eq: nondeg}) translates into the condition $0\not\in K$. Moreover,
  \[
  \dim\eQ_\ell(\bsU)=\binom{n+\ell-1}{\ell}.
  \]
  The   Euclidean metric on $\bsU$ induces an inner product $\lan-,-\ran_\ell$ on $\eQ_\ell(\bsU)$. Denote by $\eS^\ell$ the unit sphere in $\eQ_\ell(\bsU)$. For any $\Phi\in \eS^\ell$ we obtain a function
 \[
 f_\Phi:K\ra \bR,\;\; f_\Phi(\bx)=\Phi(\underbrace{\bx,\dotsc, \bx}_k).
 \]
 The function $f_\Phi$ is a Morse function for almost all $\Phi\in \eS^\ell$. We denote by $\mu_K(\Phi)$  the number of critical points of   $f_P$.   The main goal of this section is to describe the average
 \[
 \mu_K^\ell= \frac{1}{\bsi_{N_\ell-1}} \int_{\eS^\ell} \mu^\ell_K(\Phi) \,|dS(\Phi)|
 \]
 in terms of  integral-geometric  invariants of $K$.  We will achieve this   by reducing the problem to the situation investigated in the previous section.

 We begin with a linear algebra digression.    The $\ell$-th Veronese map is the linear map
 \[
 \eV=\eV_\ell: \underbrace{\bsU\otimes\cdots\otimes\bsU}_\ell\ra \eQ_\ell(\bsU),
 \]
 uniquely determined by the requirement
 \begin{equation}
\bigl\lan\, \Phi,\eV(\bu_1,\dotsc, \bu_\ell)\,\bigr\ran_\ell=\Phi(\bu_1,\dotsc,\bu_\ell),,\;\;\forall \bu_1,\dotsc,\bu_\ell \in\bsU,\;\;\Phi\in \eQ_\ell(\bsU).
 \label{eq: ver0}
 \end{equation}
 Note that for any smooth paths
 \[
 t\mapsto \bu_i(t)\in    \bsU,\;\;i=1,\dotsc, \ell,
 \]
 we have
 \[
 \frac{d}{dt}\eV(\bu_1,\dotsc,\bu_\ell)=\eV(\dot{\bu}_1,\bu_2,\dotsc,\bu_\ell)+\cdots +\eV(\bu_1,\dotsc,\bu_{\ell-1}, \dot{\bu}_\ell),
 \]
 where a  dot indicates $t$-differentiation.    For any  $\bu\in  \bsU$ we set
 \[
 M_\bu:=\eV(\underbrace{\bu,\dotsc,\bu}_\ell).
 \]
 With these notations we  deduce that
 \[
 f_\Phi(\bx)=\lan \Phi, M_\bx\ran_\ell,\;\;\forall \bx\in K,\;\;\Phi\in\eS^\ell.
 \]
 We deduce that $\bx_0=\bx(s_0)$ is a critical point of $f_\Phi$ if
 \[
 \lan \Phi, M'_{\bx(s_0)}\ran_\ell=0,
 \]
  where a prime $'$ indicates $s$-differentiation.  Observe  that
 \[
 M'_{\bx(s)}=\ell \eV(\bx'(s),\underbrace{\bx(s),\dotsc,\bx(s)}_{\ell-1}),
 \]
 Since $\bx(s),\bx'(s)\neq 0$, $\forall s$ we deduce
 \begin{equation}
 M'_{\bx(s)}\neq 0,\;\;\forall s.
 \label{eq: immer}
 \end{equation}
Define
 \[
\bnu_\ell:K\ra \eS^\ell,\;\; \bnu(s) =\frac{1}{|M'_{\bx(s)}|}M'_{\bx(s)}.
 \]
Using (\ref{eq: av4}) we deduce
 \begin{equation}
 \mu^\ell_K=\frac{1}{\pi} \int_K |\bnu_\ell'(s)| |ds|.
 \label{eq: av5}
 \end{equation}
 To formulate this in a more geometric fashion observe that (\ref{eq: immer}) implies that the map
 \begin{equation}
 K\ni\bx\mapsto \eV(\underbrace{\bx,\dotsc,\bx})\in\eQ_\ell
 \label{eq: vero}
 \end{equation}
 is an immersion.   We will refer to it  as the \emph{$\ell$-th Veronese immersion}  of  $K$.   For every $s$ the unit vector $\bnu_\ell(s)$ is the unit tangent vector field along this  Veronese embedding.   The integral    in the right-hand side of (\ref{eq: av5})  is then precisely the total curvature of the $\ell$-th Veronese immersion of $K$. We denote it by $\tau_\ell(K)$.  We have thus proved the following result.

 \begin{theorem} Let  $K\subset \bsU$ be a knot in the Euclidean space $\bsU$ such that $0\not \in K$. Then for any positive integer $\ell$ we have
 \[
 \mu_K^\ell=\frac{1}{\pi}\tau_\ell(K),
 \]
 where $\tau_\ell(K)$ denotes  the  total curvature of the $\ell$-th Veronese immersion  of $K$ defined by (\ref{eq: vero}).\qed
 \label{th: vero}
 \end{theorem}

 \begin{remark}  (a)The  case $\ell=1$ in the above theorem  is precisely  the celebrated result of  F\'{a}ry and Milnor, \cite{Fa, Mil}.

 (b) The arguments in the proof   of  Theorem \ref{th: vero} show something more. More precisely, if $X\subset \bsU$ is a compact submanifold of $\bsU$, then  the  expected number of critical points  of $P_\Phi|_X$ is  equal to the expected number of critical points  of the function $\lambda \circ \eV_\ell|_X$, where $\lambda:\eQ_\ell\ra \bR$ is a random  linear function of norm $1$.  The arguments in Chern and Lashof, \cite{CL}, show that this number is the total curvature of the $\ell$-th Veronese immersion $\eV_\ell:X\ra \eQ_\ell(\bsU)$. \qed
 \end{remark}

 \section{An application}
 \setcounter{equation}{0}

 We want to compute the  higher total curvatures $\tau_\ell(K)$ when $K$ is the unit circle in $\bsU=\bR^2$ centered at the origin.   Denote by $K_\ell$ the $\ell$-th  Veronese  immersion of $K$.  We begin by providing a more explicit description of  the  Veronese map
 \[
 \eV_\ell: \underbrace{\bsU\otimes \cdots\otimes\bsU}_\ell\ra \eQ_\ell(\bsU)
 \]
 For any
 $\ell$-linear form
 \[
 \Psi: \bsU\times\cdots\times \bsU\ra \bR
 \]
  we define its symmetrization $\boldsymbol{\lan}\,\Psi\,\boldsymbol{\ran}\in\eQ_\ell(\bsU)$ by
  \[
  \boldsymbol{\lan}\,\Psi\,\boldsymbol{\ran}(\bu_1,\dotsc,\bu_\ell):=\frac{1}{\ell!}\sum_{\si\in\mathfrak{S}_\ell} \Psi(\bu_{\si(1)},\dotsc,\bu_{\si(\ell)}),
  \]
  where $\mathfrak{S}_\ell$ denotes the group of permutations of $\{1,\dotsc,\ell\}$. If $\bu_1,\dotsc, \bu_\ell\in \bsU$, we denote by $\bu_i^\dag$ the linear functionals
\[
\bu_i^\dag:\bsU\ra \bR,\;\;\bu_i^\dag(\bu)=(\bu_i,\bu)_\bsU.
\]
Then
\[
\eV_\ell(\bu_1,\dotsc,\bu_\ell)= \boldsymbol{\bigl\lan}\,\bu_1^\dag\otimes\cdots\otimes\bu_\ell^\dag\,\boldsymbol{\bigr\ran}.
\]
Denote by $(e_1,e_2)$ the canonical orthonormal  basis of $\bR^2$  and by $(e^1,e^2)$ its dual basis.    An \emph{orthonormal basis} of $\eQ_\ell$ is given by the monomials
\[
\Phi_j= \binom{\ell}{j}^{1/2}\boldsymbol{\Bigl\lan}\, (e^1)^{\otimes j}\otimes (e^2)^{\otimes (\ell-j)}\,\boldsymbol{\Bigr\ran},\;\;0\leq j\leq \ell.
\]
For any $\bx\in\bsU$ we have
\[
\eV_\ell(\bx)= \sum_{j=0}^\ell V_j(\bx)\Phi_j,
\]
where
\[
V_j(x)= \lan \eV_\ell(\bx),\Phi_j\ran_\ell =\Phi_j(\bx,\dotsc,\bx)= \binom{\ell}{j}^{1/2}  x_1^jx_2^{\ell-j}.
\]
The map $\eV_\ell$ has an important invariance property.  Observe that  we have a linear    right action of  $SO(\bsU)$ on $\eQ_\ell$
\[
\eQ_\ell(\bsU)\times SO(\bsU)\ni (\Phi,R)\mapsto \eT_R\Phi\in \eQ_\ell(\bsU),
\]
where $\eT_R:\eQ_\ell(\bsU)\ra \eQ_\ell(\bsU)$ is the \emph{isometry} defined  by
\[
(\eT_R\Phi) (\bu_1,\dotsc, \bu_\ell)=\Phi(R\bu_1,\dotsc, R\bu_\ell),
\]
for all $\Phi\in\eQ_\ell$ and  $\bu_1,\dotsc,\bu_\ell\in \bsU$.   The   equality (\ref{eq: ver0}) implies that  for any $\bu_1,\dotsc,\bu_\ell\in\bsU$ we have
\begin{equation}
\eV_\ell(R\bu_1,\dotsc,R\bu_\ell)=\eT_R^\dag \bigl(\eV_\ell(\bu_1,\dotsc,\bu_\ell)\,\bigr),
\label{eq:  action}
\end{equation}
where $\eT_R^\dag:\eQ_\ell(\bsU)\ra \eQ_\ell(\bsU)$ denotes the  adjoint of $\eT_R$.

Consider the   standard parametrization $\theta\mapsto\bx(\theta)=(\cos\theta,\sin\theta)$ of the unit circle in $\bsU$.  The $\ell$-th  Veronese immersion of the unit circle is given by the map
\[
V: [0,2\pi]\ni\theta \mapsto \bigl(\, V_0(\theta),\dotsc, V_\ell(\theta)\,\bigr)\in \bR^{\ell+1}
\]
\[
V_j(\theta)=\binom{\ell}{j}^{1/2}(\cos\theta)^j(\sin\theta)^{(\ell-j)},\;\;0\leq j\leq \ell.
\]

We have
\[
V'(\theta)=\ell \eV_\ell\bigl(\,\bx'(\theta),\bx(\theta), \dotsc, \bx(\theta)\,\bigr),,\;\;\bx''(\theta)=-\bx(\theta).
\]
and
\[
V''(\theta)=\ell \eV_\ell\bigl(\,\bx''(\theta),\bx(\theta), \dotsc, \bx(\theta)\,\bigr)+ \ell(\ell-1) \eV_\ell\bigl(\,\bx'(\theta),\bx'(\theta), \bx(\theta),\dotsc, \bx(\theta)\,\bigr).
\]
If we denote by $R_\theta$ the counterclockwise rotation of $\bR^2$ of angle $\theta$, then
\[
\bx(\theta)= R_\theta\bx(0),\;\;\bx'(\theta)= R_\theta \bx'(0).
\]
Using (\ref{eq: action}) we deduce that
\[
V'(\theta)=\eT_{R_\theta}^\dag V'(0),\;\; V''(\theta)= \eT_{R_\theta}^\dag  V''(\theta).
\]
In particular, since the map  $\eT_R^\dag:\eQ_\ell(\bsU)\ra \eQ_\ell(\bsU)$ is an isometry we deduce
\[
|V'(\theta)|=|V'(0)|,\;\;|V''(\theta)|= |V''(0)|,\;\;\forall \theta.
\]
Observe that for $0<j <\ell$ we have
\[
V'_j(\theta)=\binom{\ell}{j}^{1/2}\left( -j(\cos\theta)^{(j-1)}(\sin\theta)^{(\ell-j+1)}+(\ell-j)(\cos\theta)^{(j+1)}(\sin\theta)^{(\ell-j-1)}\,\right)
\]
\[
=\binom{\ell}{j}^{1/2}(\cos\theta)^{(j-1)}(\sin\theta)^{(\ell-j-1)}\bigl(\,-j\sin^2\theta+(\ell-j)\cos^2\theta\,\bigr)
\]
\[
=\binom{\ell}{j}^{1/2}(\cos\theta)^{(j-1)}(\sin\theta)^{(\ell-j-1)}\bigl(-j +\ell\cos^2\theta).
\]
Next we observe that
\[
V_\ell'(\theta)= -\ell (\cos\theta)^{\ell-1}\sin\theta,\;\; V_0'(\theta)=\ell(\sin\theta)^{\ell-1}\cos\theta.
\]
This shows that for $\theta=0$ only the component $V'_{\ell-1}(0)$ is non trivial and it is equal to $\ell^{1/2}$. This proves that
\[
|V'(\theta)|=\ell^{1/2}.
\]
Thus the arclength parameter along the  curve $\theta\mapsto V(\theta)$ is $s=\ell^{1/2}\theta$, the length of $K_\ell$ is $2\pi\ell^{1/2}$  and
\[
\frac{d}{ds} = \ell^{-1/2}\frac{d}{d\theta}.
\]
Next we compute $|V''(0)|$.   If $\ell-j-1\geq 2$, i.e. $j\leq \ell-3$ we have $V_j''(0)=0$.
\[
V_\ell''(0)=-\ell,  \;\; V_{\ell-1}''(0)=0,\;\;V''_{\ell-2}=2\binom{\ell}{2}^{1/2}.
\]
Hence
\[
\left|\frac{d^2}{d\theta^2}V(\theta)\right|^2= \ell^2 +4\binom{\ell}{2} = 3\ell^2-2\ell \Rightarrow \left|\frac{d^2}{ds^2}V(s)\right|^2=\frac{3\ell-2}{\ell}.
\]
Thus
\[
\tau_\ell(K)= \frac{1}{\pi} \int_0^{2\pi\ell^{1/2} } \Biggl|\frac{d^2}{ds^2} V(s)\,\Biggr| ds= 2\sqrt{3\ell-2}.
\]
We have thus proved the following result.

\begin{theorem} If $K$ is the unit circle in  $\bR^2$ centered at the origin, then
\[
\mu_K^\ell=2\sqrt{3\ell-2},\;\;\forall \ell\geq 1.\proofend
\]
\label{th:  typical}
\end{theorem}

\begin{remark} (a) The case $\ell=1$ of the above theorem is   obvious.  For the case $\ell=2$, i.e., the equality $\mu_K^2=4$, we can give two alternate proofs.     For the first proof we  observe that
\[
\eV_2(\cos\theta,\sin\theta) =\bigl(\, x\theta),y(\theta),z(\theta)\,\bigr) (\sin^2\theta, \sqrt{2}\sin\theta \cos\theta,\cos^2\theta).
\]
Observe that  the image of the   $2$nd Veronese immersion is the intersection of the sphere $x^2+y^2+z^2=1$ with the plane $x+z=1$ which is a circle of radius $\frac{1}{\sqrt{2}}$.  Moreover, the Veronese  immersion double covers this circle. These facts imply easily that $\mu_K^2=4$.

For the second proof  we observe that the typical pencil of plane conics
\[
C_t=\{ ax^2+2bxy+c^2y^2=t\}
\]
 has  four points of tangency with the  unit circle.

 \begin{figure}[ht]
\centering{\includegraphics[height=2.5in,width=2.5in]{vero.eps}}
\caption{\sl The  second Veronese embedding of the circle of radius $1$ with center $(120,0)$.}
\label{fig: 1}
\end{figure}

 (b) The invariant $\mu_K^\ell$ is invariant under rotations, but not under translations. For example if $K$ is the    circle of radius $1$ in the plane centered at the point $(3,0)$. Then the second Veronese immersion of $K$ is given by
 \[
 [0,2\pi]\ni t\mapsto \bigl(\, (3+\cos t)^2, \sqrt{2}(3+\cos t)\sin t, \sin^2 t\,\bigr)\in\bR^3.
 \]
 \emph{MAPLE} aided computation shows that its pointwise curvature is
 \[
\kappa(t)= \sqrt {-{\frac {108\, \left( \cos \left( t \right)  \right) ^{3}-54\,
 \left( \cos \left( t \right)  \right) ^{2}-360\,\cos \left( t
 \right) -238}{ -2\left( 9\, \left( \cos \left( t \right)  \right) ^{2}-
19-6\,\cos \left( t \right)  \right) ^{3}}}},
\]
while its total curvature is
\[
\mu_K^2\approx 2.065 < 2\sqrt{3\cdot 2-2}=4.
\]
Numerical experimentation suggest  that if $K(d)$ is the circle of radius $1$ centered at the point $(d,0)$ then
\[
\lim_{d\ra \infty} \mu^2_{K(d)} =2.
\]
In view of Fenchel theorem, this would indicate that  for $d$ large  the second  Veronese  embedding of $K(d)$ is very close to a planar convex curve.   The \emph{MAPLE} generated   Figure \ref{fig: 1}   seems to confirm this, since the second Veronese immersion   is contained in a box  of dimensions $1\times a\times b$, where $a>300$ and $b>400$.\qed

 \end{remark}

\section{Critical points of trigonometric polynomials}
\setcounter{equation}{0}

As another application of Theorem \ref{th: main}, we want to compute the expected number of critical points on the unit circle of a trigonometric polynomial of degree $\leq n$.

Let $K$ denote the     unit circle in $\bR^2$ centered at the origin and denote by $\bsV_n$ the vector space of trigonometric polynomials of degree $\leq n$ on $K$, i.e.,     functions $f:K\ra \bR$ of the form
\[
f(\theta)= a_0+\sum_{k=1}^n  \bigl(\,a_k\cos k\theta+b_k\sin k\theta\,\bigr),
\]
equipped with the inner product
\[
(f,g)_\bsV= \frac{1}{2\pi}\int_0^{2\pi}  f(\theta) g(\theta) d\theta.
\]
Thus
\[
\left| a_0+\sum_{k=1}^n  \bigl(\,a_k\cos k\theta+b_k\sin k\theta\,\bigr)\,\right|^2_\bsV= a_0^2 + \frac{1}{2}\sum_{k=1}^n (a_k^2+b_k^2).
\]
We  obtain  an orthonormal basis of $\bsV_n$
\[
\be_0=1,\;\;\be_{k}=\sqrt{2} \cos k\theta,\;\;\bsf_k= \sqrt{2} \sin k\theta ,\;\;k=1,\dotsc, n.
\]
The function
\[
\xi: K\ra \bsV_n,\;\;[0,2\pi] \ni s \mapsto \xi_s^\dag
\]
defined by  (\ref{eq: xi}) +(\ref{eq: xi-d}) is given in this case by  the map
\[
s\mapsto \xi_s^\dag= \sqrt{2}\sum_{k=1}^n k\bigl(-(\sin ks)\be_k +(\cos ks)\bsf_k\,\bigr).
\]
Observe that
\[
|\xi_s^\dag|^2= 2\sum_{k=1}^n k^2.
\]
We set
\[
\bnu(s):=\frac{1}{|\xi_s^\dag|}\xi_s^\dag=\left(\sum_{k=1}^n k^2\right)^{-1/2} \sum_{k=1}^n k\bigl(-(\sin ks)\be_k +(\cos ks)\bsf_k\,\bigr).
\]
Then
\[
\bnu'(s)= \left(\sum_{k=1}^n k^2\right)^{-1/2} \sum_{k=1}^n k^2\bigl(\, (\cos ks)\be_k -(\sin ks)\bsf_k\,\bigr),
\]
so that
\[
|\bnu'(s)|= \left(\sum_{k=1}^n k^2\right)^{-1/2}\cdot \left(\sum_{k=1}^n k^4\right)^{1/2}
\]
Using the classical identity
\[
\sum_{k=1}^n k^j =\frac{1}{j+1}\beta_{j+1}(n+1),
\]
where $\beta_\ell$ is the $\ell$-th Bernoulli polynomial   we deduce
\begin{equation}
|\bnu'(s)|=\sqrt{\frac{3\beta_5(n+1)}{5\beta_3(n+1)}},
\label{eq: trig}
\end{equation}
where
\[
\beta_3(x)= x^3-\frac{3}{2}x^2+\frac{1}{2}x,\;\;\beta_5(x)=x^5-\frac{5}{2}x^4 +\frac{5}{3}x^3-\frac{1}{6}x.
\]
The equality (\ref{eq: trig}) and Theorem \ref{th: main} imply the following result.

\begin{theorem} The expected number of critical points on the unit circle of a  trigonometric polynomial of degree $\leq n$ is
\[
T_n=2 \sqrt{\frac{3\beta_5(n+1)}{5\beta_3(n+1)}}=2\sqrt{\frac{\sum_{k=1}^nk^4}{\sum_{k=1}^nk^2}}.
\]
In particular
\[
T_n \sim 2n\sqrt{\frac{3}{5}} \;\;\mbox{as $n\ra \infty$}.\proofend
\]
\label{th: trig}
\end{theorem}

\begin{remark}     The above theorem  indicates  that for large $k$, a trigonometric  polynomial  of degree $k$  is expected to have approximately  $ 2\sqrt{\frac{3}{5}} k\approx 1.549 k$ critical points.  In particular, it will likely have less that $2k$ critical points on the unit circle.

\begin{figure}[ht]
\centering{\includegraphics[height=2.5in,width=2.5in]{trig.eps}}
\caption{\sl A ``random'' trigonometric polynomial of degree $4$.}
\label{fig: trig}
\end{figure}

In the \emph{MAPLE} generated Figure  \ref{fig: trig} we depicted the graph of  a ``random''  trigonometric polynomial of degree $4$
\[
1.2\cos t  + 2.35\sin t  - 3.17\cos 2t + 2.71\sin 2t + 1.53\cos 3t  - 4.17\sin 3t
 \]
 \[
 +\cos 4t  - 1.15\sin 4t .
\]
 We notice  that it has  $6$ critical points, which is very close to the  expected value $T_4\approx 6.87$. \qed \end{remark}


TITLE: Solving for a single variable in a quadratic system
QUESTION [0 upvotes]: I have a quadratic system of $n$ equations that looks like:
$$ A_{ij}x_{j}y + B_{ij}x_{j}=0 $$
For $i=0...n$, where $A_{i,j}$ and $B_{ij}$ are integer matrices and sums over $j$ are implied. Is there some efficient way to solve this system for y only, without having to implicitly solve all $x_{j}$?
(Bonus question: what if the system looks like $ A_{ijk}x_{j}y_{k} + B_{ij}x_{j}=0 $?)

REPLY [1 votes]: Your problem is a texbook example of "generalized eigenvalue problem". It is a standard problem implemented in numerical libraries such as lapack.
The routines that compute solutions of the eigenproblem may be used to yield only the eigenspectrum without computing eigenvectors that is $x.$ They are also quicker in that case.
This is a reference:
http://www.netlib.org/lapack/lug/node35.html


TITLE: Divergence/convergence of an integral $\int_{-\infty}^{+\infty} (1-e^{ix}) |x|^{-n} dx$
QUESTION [0 upvotes]: I am told that the following integral converges for $1<n<3$.
$$
\int_{-\infty}^{+\infty} (1-e^{ix}) |x|^{-n} dx
$$
I am a bit baffled. Anyone with a clue or where to start with this in order to evaluate it ?

REPLY [0 votes]: I think the convergence of the integral is covered fine in the other answers, so just let me give you a non-rigorous (depending on your beliefs) way of calculating its value.
With the definition of Fourier transform $\hat f(\xi)=\int_{-\infty}^{+\infty}e^{-ix\xi}f(x)\,dx$ your integral can be seen to be $\hat{f}(0)-\hat{f}(-1)$ (if we agree to split the integral!), where $f(x)=|x|^{-n}$. It is known (see for example the book on classical Fourier analysis by Grafakos) that the Fourier transform of $f$ is
$$
\hat{f}(\xi)=2|\xi|^{n-1}\Gamma(1-n)\sin(n\pi/2).
$$
This gives that your integral equals
$$
\hat{f}(0)-\hat{f}(-1)=0-2\Gamma(1-n)\sin(n\pi/2)=-2\Gamma(1-n)\sin(n\pi/2).
$$


TITLE: How to find the limit of a piecewise function that includes a "otherwise"?
QUESTION [0 upvotes]: I have to see if the limit of the sequence $\{y_n\}$ converges to $0$.
$$y_n=\left\{ \begin{align}
1 & \textrm{ if } n \textrm{ is divisible by } 500 \\
\tfrac{1}{n} & \textrm{ otherwise }
\end{align} \right.$$
I want to say no it does not because it either will converge to $1$ or $0$. What is throwing me off is how to verify this by showing work. I understand how to do piece wise functions when given conditions such as $x \leq 5$ and $x>5$ (for example). I do not know how to explain or go about solving this with "otherwise".
Help?

REPLY [2 votes]: If the sequence converges to a limit then all of its subsequences converge to the same limit. Consider the subsequence of multiples of $500$ and odd multiples of $250$, and you are done !


TITLE: Are these statements about reducible polynomials correct?
QUESTION [0 upvotes]: Gauss's Lemma: If an integer polynomial is reducible in $\mathbb Q[x]$, then it is reducible in $\mathbb Z[x]$.

It is obvious that if an integer polynomial is reducible in $\mathbb Z[x]$, then it is reducible in $\mathbb Q[x]$.
So: An integer polynomial is reducible in $\mathbb Q[x]$ if and only if it is reducible in $\mathbb Z[x]$.
A primitive polynomial is irreducible in $\mathbb Z[x]$ if and only if it is irreducible in $\mathbb Q[x]$

REPLY [2 votes]: It is obvious that if an integer polynomial is reducible in $\mathbb Z[x]$, then it is reducible in $\mathbb Q[x]$.

This may not hold if the polynomial is not primitive. Consider $2 \cdot x$, since $2$ is not a unit in $\Bbb{Z}[x]$, $2x$ is reducible in $\mathbb{Z}[x]$, but it is not reducible in $\Bbb{Q}[x]$.
Note that irreducible polynomials in $\Bbb{Z}[x]$ are always either constant or primitive, so you can also say that every irreducible polynomial in $\Bbb{Z}[x]$ is irreducible or a unit in $\Bbb{Q}[x]$.


TITLE: 2-NEXPTIME-complete problems
QUESTION [10 upvotes]: We have a problem and we found an algorithm that appear to be 2-nexptime.
I would like to find known 2-nexptime-complete problems in order to find a lower bound.
I found in literature mainly two such problems:

whether PCP as a solution of size less than $2^{2^n}$
and the tilling problem for a square of size $2^{2^n}$

However I was unable to encode these problems in mine. So I would like to know other 2-NEXPTIME-complete problems, first to have more intuition on this class, and second in the better case prove a lower bound.
I do not provide here the problem on purpose in order to have a broad overview of 2-NEXPTIME.
Thanks

REPLY [4 votes]: The obvious N2Exp problem is of course the word acceptance problem for 2exp time bounded nondeterministic Turing machines. Using this might be as hard/easy as 2exp tiling, because the simulation of such a Turing machine computation in essence also requires you to define a double-exponentially large grid (2exp many configurations of memory tapes of length 2exp each) that is then filled in a non-deterministic way. In practice, showing N2Exp lower bounds often boils down to constructing such a grid (and making sure that it is not a tree or something else of insufficient structure). The "N" (nondeterminism) is often an inherent part of the problem and not so difficult to get once you have a large enough grid (if not, one would maybe shoot for 2exp at first).
Another practical N2ExpTime-complete problem is reasoning in expressive description logics (DL). In particular, the DL $\mathcal{SROIQ}$ that is underlying the W3C OWL 2 Web Ontology Language standard is N2ExpTime-complete (Yevgeny Kazakov: RIQ and SROIQ Are Harder than SHOIQ. KR 2008: 274-284). Now this is probably not a problem you want to use in reductions, since the definition of the logic is a bit unwieldy because of its many features. The actual lower bound proof for $\mathcal{SROIQ}$ has also been done by reduction to 2-exp tiling. However, depending on your problem, the construction given for $\mathcal{SROIQ}$ could be inspiring to see how to craft such large grids.
The tiling also shows another general pattern: N2Exp really is like NP, you just need to find a way to encode even larger problem instances very efficiently. In principle, you could try to scale up any NP problem. The reason why tiling is nice is that you only need to scale the size of the grid in this case (which is rather uniform).
On the other hand, if your problem is possibly only 2ExpTime-complete, then you could get away with an exponentially space-bounded alternating Turing machine simulation. If you have troubles building a 2exp grid, but you can get to exponential sizes, then this might be worth a try.


TITLE: Unrecognized Quantifiers
QUESTION [1 upvotes]: Which quantifiers are represented by the first two symbols ($\iota$ and $\mathcal{A}$) in the set in which Q belongs. I could not find a reference on the Quantifier Wikipedia article.

(Source: Introductory slide on Lambda Calculus (slide 32) in a slide-deck on grammars for parsing natural language sentences.)
Here's an example where the first one is used; might be helpful to pin-point which quantifier it represents in case the author used non-standard symbols.

(Source: Slide 45 in the same slide-deck.)

REPLY [6 votes]: The iota ($\iota$) and script A ($\mathcal{A}$) are determiners, not quantifiers in the usual sense; they’re described later, in the slides titled Definite Determiners and Indefinite Determiners. The expression $\iota x.F(x)$ is a term representing the unique entity $x$ such that $F(x)$ is true when such an entity exists and is undefined otherwise; see also this web page. $\mathcal{A}$ is the indefinite counterpart of $\iota$: $\mathcal{A}x.F(x)$ selects any entity $x$ such that $F(x)$ is true. Thus, $\iota$ functions somewhat like the definite article the, while $\mathscr{A}$ functions rather like the indefinite article a(n). The examples in the relevant slides are fairly clear.


TITLE: Prove that the number of answers for $|a_1|+|a_2|+...+|a_k| \le n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n| \le k$.
QUESTION [7 upvotes]: I'm trying to solve this problem:

Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|a_1|+|a_2|+...+|a_n|≤k$. all $a_i$ is an integer number.

I have a solution for non negative $a_i$:
We convert $|a_1|+|a_2|+...+|a_k|≤n$ to $a_1+a_2+...+a_k+c=n$ and convert $|a_1|+|a_2|+...+|a_n|≤k$ to $a_1+a_2+...+a_n+c=k$.
By bars and stars theorem:
Number of answers for $a_1+a_2+...+a_k+c=n$ is equal to $\binom{n+k}{k}$ and number of answers for $a_1+a_2+...+a_n+c=k$ is equal to $\binom{n+k}{n}$. That $\binom{n+k}{k} = \binom{n+k}{n}$.
but i don't know how to prove for negative integers.

REPLY [0 votes]: I just saw this question as it came back to the front page. I wrote this answer a couple of months ago, which proves that this number is
$$
\sum_{k=0}^n2^k\binom{n}{k}\binom{m}{k}
$$
which is also shown in answers to this question. My answer starts with a recurrence (like Mike Earnest's answer) and solves it by induction.


TITLE: On symmetric expressions of polynomials.
QUESTION [1 upvotes]: $\alpha, \beta,  \gamma$ are the roots of equation $x^3+px^2+qx+r=0$ . Find $$\sum{\frac{\alpha^2 + \beta^2}{\alpha+\beta}}.$$

I can solve the symmetric expression by :
$$\frac{\alpha^2 + \beta^2}{\alpha+\beta} = (\alpha+\beta) -\frac{2\alpha \beta}{(\alpha+\beta) }.$$
Then,  advance forward.  But I wanted to know,  if there is any other simpler method.  I've seen such symmetric expressions solved by multiplying the inner terms and simpler terms pop out . But here I cannot...
Please help. Thanks.

REPLY [2 votes]: (Following up on my comment, too long to post as one.)

$\frac{(\alpha) ^2+(\beta) ^2}{\alpha+\beta} = (\alpha+\beta) -\frac{2\alpha \beta}{(\alpha+\beta) }$

Also using that $\,\alpha\beta=-r /\gamma\,$ and $\,\alpha+\beta=-p-\gamma\,$:
$$
A = \sum \frac{\alpha ^2+\beta ^2}{\alpha+\beta} = \sum \left((\alpha+\beta) -\frac{2\alpha \beta}{\alpha+\beta} \right) = 2 \underbrace{\,\sum \alpha\,}_{=\,-p} + 2r \underbrace{\,\sum \frac{1}{\alpha(p+\alpha)}\,}_{= \,B}
$$
Since $\,\dfrac{1}{\alpha(p+\alpha)}=\dfrac{1}{p}\left(\dfrac{1}{\alpha}-\dfrac{1}{p+\alpha}\right)\,$:
$$
B = \frac{1}{p}\left(\underbrace{\,\sum \frac{1}{\alpha}\,}_{=C} - \underbrace{\,\sum \frac{1}{p+\alpha}\,}_{= D}\right)
$$
$\dfrac{1}{\alpha}\,$ are the roots of the reciprocal polynomial $\,x^3 f\left(\frac{1}{x}\right) = 1+px+qx^2+rx^3\,$, so:
$$
C = \sum \frac{1}{\alpha} = -\frac{q}{r}
$$
$p+\alpha$ are the roots of $\,f(x-p) = x^3 - 2 p x^2 + (p^2 + q) x - p q + r\,$, so:
$$
D = \sum \frac{1}{p+\alpha}=-\frac{p^2 + q}{- p q + r}
$$
What's left is to piece it all back together, and also cover the special cases like $\,r=0\,$.


TITLE: Analytically continuing a function defined by an integral to a larger region
QUESTION [2 upvotes]: Let's say that for certain restrictions on the complex parameters $\alpha$ and $\beta$ , you determine that  $$\int_{a}^{b} f(x,\alpha, \beta) \, dx = g(\alpha,\beta).$$
But you can show that $ \int_{a}^{b} f(x,\alpha,\beta) \, dx$ converges for values of $\beta$ in a larger region.
Can you conclude that $\int_{a}^{b} f(x,\alpha,\beta) \, dx = g(\alpha,\beta)$ in this larger region if $g(\alpha,\beta)$ is analytic with respect to $\beta$ in the larger region?
For example,  if $\text{Re}(\beta) > \text{Re}(\alpha) >-1$,  one can use contour integration to show that  $$\int_{0}^{\frac{\pi}{2}} \cos^{\alpha}(x) \cos (\beta x) \, dx = \frac{\pi}{2^{\alpha+1}} \frac{\Gamma(\alpha+1)}{\Gamma(\frac{\alpha}{2} + \frac{\beta}{2} +1)\Gamma(\frac{\alpha}{2} - \frac{\beta}{2} +1)}. $$
I want to argue that this equation holds for all complex values of $\beta$.

REPLY [2 votes]: The Gamma function has simple poles on $-\mathbb{N}$, and is holomorphic and nonzero everywhere else. So your right hand side is an entire function with respect to $\beta$, for $\alpha$ fixed.
On the left-hand side, still for a fixed $\alpha$, the integrand is clearly entire with respect to $\beta$ for each $t$. Integrating with respect to $x$ yields an entire function here. Indeed, the partial derivative with respect to $\beta$ is locally uniformly (in $\beta$) dominated by an integrable function (no longer depending on $\beta$) over $x\in[0,\pi/2]$. So holomorphy follows by dominated convergence. We could avoid DCT, but it does not make the argument shorter. Note also that, unlike what you said in the comments, we need to be careful: if $g(z)=\int f(t,z)dt$ converges on some open set $\Omega$ in $\mathbb{C}$, and if $z\longmapsto f(t,z)$ is holomorphic on $\omega$ for each $t$, it does not necessarily imply that $g$ is holomorphic. See the example below.
So you have two entire functions which coincide on $(\alpha,+\infty)$. By the principle of isolated zeroes, they must coincide on the whole of $\mathbb{C}$. So your formula holds for all $\alpha>-1$ and all $\beta\in\mathbb{C}$.
Note: I couldn't find a wikipedia reference for the principle of isolated zeroes. So I suspect it is not the most popular terminology. What I mean is: the zeroes of a nonconstant holomorphic function on a domain are isolated. Conversely: if the zero set of a holomorphic function on a domain has a limit point in this domain, then the function is constant equal to zero on the whole domain.
Example: Here is why we need to be careful before claiming the left hand side is holomorphic. Consider $g(z):=\int_0^1\sin(z/t)dt$. This is well-defined on $\mathbb{C}$, and $g(0)=0$. Now consider
$$\frac{g(1/n)}{1/n}=\int_0^1n\sin (1/(nt))dt.$$
If $g$ was holomorphic at $0$, this would converge when $n$ tends to $+\infty$. Now recall $\sin x\geq 2x/\pi$ on $[0,\pi/2]$. Hence $n\sin (1/(nt))\geq  2/(\pi t)$ on $[2/(\pi n),1]$. Hence
$$
\frac{g(1/n)}{1/n}\geq \frac{2}{\pi}\int_{\frac{2}{\pi n}}^1\frac{1}{t}dt=\frac{2}{\pi}\log\left(\frac{\pi n}{2}\right)\longrightarrow +\infty.
$$
So $g$ is not holomorphic at $0$.


TITLE: Unknot: embedded vs immersed bounding disk
QUESTION [2 upvotes]: Suppose that we have a knot $K\subset \mathbb{S}^3$. If $K$ bounds* an embedded 2-disk then $K$ is the unknot.
But what happens if $K$ bounds an immersed 2-disk?
The immersed disk generically will have two kind of singularities, double points and triple points.  The double points will be closed curves while the triple points a finite number of points.

1) What is an example when $K$ bounds an immersed disk but not an embedded one?
2) Can we remove the double point singularities?
3)Are the triple points related to the genus of a Seifert surface for $K$?

*Formally, $K$ bounds an embedded (immersed) 2-disk  if $\exists  f:(\mathbb{D}^2,\partial \mathbb{D}^2)\to (\mathbb{S}^3,K)$  that is a smooth embedding (immersion). We can assume that in both cases $f|_{\partial \mathbb{D}^2}$ is a diffeomorphism onto the image if necessary.

REPLY [3 votes]: Every (smooth) knot in $S^3$ bounds an immersed smooth disk in $S^3$, see e.g.

Bruce Trace, A general position theorem for surfaces in Euclidean 4-space. Combinatorial methods in topology and algebraic geometry (Rochester, N.Y., 1982), 123–137, Contemp. Math., 44, Amer. Math. Soc., Providence, RI, 1985.

If A knot $K\subset S^3$ is a "ribbon knot" if it bounds an immersed disk in $S^3$ with only double singularities. There exist nontrivial ribbon knots in $S^3$, in particular, you cannot remove their double intersections. In particular, there are knots which bound immersed disks with double self-intersections, which do not bound embedded disks.

Here is an example taken from Wikipedia:
Every ribbon knot bounds a smooth embedded disk in $B^4$, the unit 4-ball. In other words, every ribbon knot is "slice." It is a famous open problem if every slice knot is ribbon.
In general, triple self-intersections  are more directly related to the 4-ball genus rather than the Seifert genus of a knot.


TITLE: Determinant of identity minus adjacency matrix
QUESTION [0 upvotes]: Let $M$ be the adjacency matrix of a directed graph $G$. Is there any known relation between $\det(\textrm{id}-M)$ and the cycles of $G$?
It is easy to see that if $G$ is acyclic then this determinant is $1$ (because we can take $M$ to be strictly upper triangular). What does this measure in general?
Somewhat related: How to tell if a directed graph is acyclic from the adjacency matrix

Using sage I computed these determinants for all the 9608 directed graphs on $5$ vertices and I found that we have $\det(I-M)=1$ iff $G$ is acyclic.
Moreover this is the list of possible determinants together with the number of graphs with such determinant:
[(-48, 1), (-40, 1), (-36, 2), (-32, 6), (-30, 3), (-28, 9), (-27, 1),
(-26, 4), (-25, 4), (-24, 36), (-23, 4), (-22, 18), (-21, 9), (-20, 49),
(-19, 12), (-18, 75), (-17, 23), (-16, 144), (-15, 76), (-14, 124),
(-13, 69), (-12, 361), (-11, 116), (-10, 339), (-9, 290), (-8, 676),
(-7, 294), (-6, 917), (-5, 500), (-4, 1195), (-3, 889), (-2, 1144), (-1,
749), (0, 1166), (1, 302)]

REPLY [1 votes]: According to my calculations in sage, there are 11 graphs (out 156) on six vertices such that $\det(I-M)=1$ and exactly two of these eleven are trees. This shows that is is going to be very difficult to determine information about cycles from the value of $\det(I-M)$. (There is nothing special about the value '1' here, for example example there are 35 graphs with $\det(I-M)=0$ and two of these are trees.)
Note that, trees aside, no useful combinatorial interpretation of $\det(M)$ is known, and there seems to be little reason to expect $\det(I-M)$ to work better.


TITLE: Spacetime with a point defect
QUESTION [0 upvotes]: What is the metric for a spacetime with a point defect? Spacetime metric with line defects are well-known, they are basically Cosmic strings. Is anyone aware of an example for spacetime with point defects?

REPLY [2 votes]: I suspect that what OP is looking for is the spacetime of global monopole, a solution of GR with a deficit of solid angle:
$$
 ds^2=-dt^2+\frac{dr^2}{\alpha ^2}+r^2d\theta ^2+r^2\sin ^2\theta d\varphi ^2,\tag{*}
$$
where $\alpha<1$. Metric of this type arises as a solution of Einstein field equation with the matter in the form of triplet of scalar fields with a “mexican hat”–type potential with global $SO(3)$ symmetry. This global symmetry is broken by via a “hedgehog”-like ansats. Near the origin the metric actually has “microstructure”, so rather than conical singularity of $(*)$ the curvature remains bounded, plus there is also a small negative mass term.
Reference:

Barriola, M., & Vilenkin, A. (1989). Gravitational field of a global monopole. Physical Review Letters, 63(4), 341, doi:10.1103/PhysRevLett.63.341.