On a family of symmetric matrices

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$   Suppose that $w: \bR\to[0,\infty)$ is an integrable function.  Consider its   Fourier transform $\newcommand{\ii}{\boldsymbol{i}}$

$$ \widehat{w}(\theta)=\int_{\bR}  e^{-\ii x\theta}w(\theta) dx. $$

 For any $\vec{\theta}\in\bR^n$ we form the  complex Hermitian  $n\times n$ matrix

$$A_w(\vec{\theta})= \bigl(\; a_{ij}(\vec{\theta})\;)_{1\leq i,j\leq n},\;\; a_{ij}(\vec{\theta})=\widehat{w}(\theta_i-\theta_j) .$$

Observe that  for any $\vec{z}\in\mathbb{C}^n$  we have $\newcommand{\bC}{\mathbb{C}}$

$$ \bigl(\; A_w(\vec{\theta})\vec{z},\vec{z}\;\bigr)=\sum_{i,j} \widehat{w}(\theta_i-\theta_j) z_i\bar{z}_j =\int_{\bR}  | T_{\vec{z}}(x,\vec{\theta})|^2 w(x) dx, $$

where  $T_{\vec{z}}( x)$ is  is the trigonometric polynomial $\newcommand{\vez}{\vec{z}}$

$$T_{\vez}(x,\vec{\theta})= \sum_j z_j e^{\ii \theta_j x}. $$


We denote by $(-,-)_w$ the inner product

$$ (f,g)_w=\int_{\bR} f(x) \bar{g(x)} w(x) dx,\;\;f,g:\bR\to \bC. $$

We see that $A_w(\vec{\theta})$  is the Gramm-Schmidt matrix

$$ a_{ij}(\vec{\theta})= (E_{\theta_i}, E_{\theta_j})_w,\;\;  E_\theta(x)=e^{\ii\theta x}. $$


We see that $\sqrt{\;\det A_w(\vec{\theta})\;}$ is equal to the $n$-dimensional volume   of the parallelepiped   $P(\vec{\theta})=L^2(\bR, wdx)$ spanned  by the  functions $E_{\theta_1},\dotsc, E_{\theta_n}$. We observe  that if these exponentials are  linearly  dependent,  then this volume is  zero.  Here is a first elementary result.


Lemma  1.   The exponentials  $E_{\theta_1},\dotsc, E_{\theta_n}$ are linearly dependent  (over $\bC$) if and only if $\theta_j=\theta_k$ for  some $j\neq k$.


Proof.    Suppose that

$$\sum_{j=1}^n z_j E_{\theta_j}(x)=0,\;\;\forall x\in \bR. $$

Then for any $f\in \eS(\bR)$ we have

$$\sum_{j=1}^n z_j E_{\theta_j}(x)f(x)=0,\;\;\forall x\in \bR. $$


By taking the Fourier Transform of the last equality we deduce

$$ \sum_{j=1}^n z_j \widehat{f}(\theta-\theta_j) =0.  \label{1}\tag{1}$$

If we now choose $\newcommand{\ve}{{\varepsilon}}$ a  family $f_\ve(x)\in\eS(\bR)$ such that, as $\ve\searrow 0$,  $\widehat{f}_\ve(\theta)\to\delta(\theta)=$ the Dirac  delta function concentrated at $0$, we  deduce  from (\ref{1})  that


$$\sum_{j=1}^n z_j\delta(\theta-\theta_j)=0. \tag{2}\label{2} $$

 Clearly this can happen if and only if  $\theta_j=\theta_k$ for  some $j\neq k$.  q.e.d.


If we set

$$ \Delta(\vec{\theta}) :=\prod_{1\leq j<k\leq n} (\theta_k-\theta_j), $$

then we deduce from the above lemma that

$$ \det A_w(\vec{\theta})= 0 \Leftrightarrow  \Delta(\vec{\theta})=0. $$

A more precise statement is true.


Theorem 2.  For any integrable weight $w:\bR\to [0,\infty)$  such that $\int_{\bR} w(x) dx >0$ there  exists a  constant $C=C(w)>0$ such that for any $\theta_1,\dotsc, \theta_n\in [-1,1]$ we have

$$ \frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det  A_w(\vec{\theta}). \tag{E}\label{E}$$


Proof.          We regard $A_w(\vec{\theta})$ as a hermitian operator

$$ A_w(\vec{\theta}):\bC^n\to \bC^n. $$

We denote by $\lambda_1(\vec{\theta})\leq \cdots \leq \lambda_n(\vec{\theta}) $ its eigenvalues so that

$$\det A_w(\vec{\theta})=\prod_{j=1}^n \lambda_j(\vec{\theta}) \tag{Det}\label{D}. $$

Observe that $\newcommand{\Lra}{\Leftrightarrow}$ $\newcommand{\eO}{\mathscr{O}}$

$$\vec{z}\in \ker A(\vec{\theta}) \Lra  \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in{\rm supp}\; w \Lra   \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in\bR. \tag{Ker}\label{K}$$

We want to give a  more precise description    of $\ker A_w(\vec{\theta})$.    Set

$$ I_n:=\{1,\dotsc, n\},\;\; \Phi_{\vec{\theta}}=\{ \theta_1,\dotsc,\theta_n\}\subset \bR. $$


We want to emphasize that $\Phi{\vec{\theta}}$ is not a multi-set so that $\#\Phi(\vec{\theta})\leq n.$  $\newcommand{\vet}{{\vec{\theta}}}$.

 Example 3.  For example  with $n=6$ and $\vet=(1,2,3,2,2,4)$ we have

$$ \Phi_\vet=\Phi_{(1,2,3,2,2,4)}=\{1,2,3,4\}. $$

---

For $\newcommand{\vfi}{{\varphi}}$ $\vfi\in\Phi_\vet$ we set


$$ J_\vfi=\bigl\{ j\in I_n;\;\; \theta_j=\vfi\;\bigr\}. $$

In the example above for $ \vet=(1,2,3,2,2,4)$ and $\vfi=2$ we have $J_\vfi=\{2,4,5\}$.   $\newcommand{\vez}{\vec{z}}$  For $J\subset I_n$ we set


$$S_J:\bC^n\to \bC,\;\;S_J(\vez)=\sum_{j\in J} z_. $$

In particular, for  any $\vfi\in\Phi_\vet$ we define

$$S_\vfi:\bC^n\to \bC,\;\; S_{\vfi}(\vec{z})=S_{J_\vfi}(\vez)=\sum_{j\in J_\vfi} z_j. $$

We deduce

$$  \sum_{j\in I_n} z_jE_{\theta_j}=\sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi. $$

Using (\ref{K}) we deduce

$$ \vez\in\ker A(\vet)\Lra \sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi\Lra S_\vfi(\vez)=0,\;\;\forall \vfi\in \Phi_\vet . \tag{3}\label{3}$$

In particular we deduce

$$\dim \ker A(\vet)=n-\#\Phi_\vet. $$


Step 1.   Assume  that $w$ has compact support so that $\widehat{w}(\theta)$  is real analytic over $\bR$.   We will show  that we have the two-sided estimate

$$ \frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det  A_w(\vec{\theta}) \leq C |\Delta(\vec{\theta})|^2. \tag{$E_*$}\label{Es} $$

 In this case  $\det A_w(\vet)$ is real analytic and symmetric in the variables $\theta_1,\dotsc, \theta_n$  and vanishes   if and only if $\theta_j=\theta_k$ for some $j=k$.  Thus $\det A_w(\vet)$ has a  Taylor series expansion (near $\vet=0$)


$$\det A_w(\vet)= \sum_{\ell\geq 0} P_\ell(\vet), $$

where $P_\ell(\vet)$ is a  symmetric polynomial  in $\vet$ that vanishes   when $\theta_j=\theta_k$ for some $j\neq k$. Symmetric  polynomials of this type  have the form,


$$\Delta(\vet)^{2N} \cdot Q(\vet) $$

where $N$ is some positive integer  and $Q$ is a symmetric polynomial.  We deduce from  the \Lojasewicz inequality  for  subanalytic functions that there exists  $C=C(w)>0$, a positive integer $N$  and a rational number and $r>0$ such that


$$ \frac{1}{C} |\Delta(\vet)|^{r}\leq   \det A_w(\vet) \leq C \Delta(\vet)^{2N},\;\;\forall |\vet|\leq 2\pi. \tag{4} \label{4} $$

We want to show  that in (\ref{4})  we have $2N=r=2$.   We argue by contradiction, namely we assume that $r\neq 2$ or $N\neq 1$.      Let

$$\vet(t)= (0, t, \theta_3, \dotsc, \theta_n), \;\; 0\leq |t| < \theta_3<\cdots < \theta_n. $$

Set $A_w(t)=A_w\bigl(\,\vet(t)\;\bigr)$.   Denote its eigenvalues by

$$0\leq \lambda_1(t)\leq \lambda_2(t)\cdots \leq \lambda_n(t). $$

The eigenvalues are so arranged so that the functions $\lambda_k(t)$ are real analytic for $t$ in a neighborhood of $0$. We deduce from (\ref{3}) that $\ker A_w(0)$ is one dimensional so that $\lambda_1(0) =0$,  $\lambda_k(0)>0$, $\forall  k>1$. Hence

$$ \det  A_w(t) \sim \lambda_1(t) \prod_{k=2}^n \lambda_k(0)\;\;\mbox{as $t\searrow 0$}.  \tag{5}\label{5} $$

On the other hand

$$\Delta(\vet(t))^2 \sim Zt^2  \;\;\mbox{as}\;\; t\searrow 0 $$

for some positive constant $Z$.  Using this estimate in (\ref{4}) we deduce $r=2N$.  On the other hand, using the above estimate in (\ref{5}) we deduce

$$ \lambda_1(t) \sim Z_1 t^{2N} \;\;\mbox{as}\;\;t\searrow 0. \tag{6}\label{6}, $$

for another positive constant $Z_1$.

The   kernel of $A_w(0)$ is spanned by the  unit vector

$$ \vez(0)= (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0,\dotsc 0). $$

We can find a real analytic family of vectors $t\mapsto \vec{z}(t)$ satisfying

$$|\vez(t)|=1,\;\; A_w(t) \vez(t)=\lambda_1(t)\vez(t),\;\;\lim_{t\to 0}\vez(t)=\vez(0). $$

In particular, we deduce

$$ \dot{A}_w(0)\vez(0)+A_w(0)\dot{\vez}(0)=\dot{\lambda}_1(0)\vez(0)+\lambda_1(0)\dot{\vez}(0)=0. $$

A simple computation  shows that $\dot{A}_w(0) \vez(0)=0$ so we deduce  $A_w(0)\dot{\vez}(0)=0$.  This shows that

$$\dot{z}_1(0)+\dot{z}_2(0)=0,\;\;\dot{z}_k(0)=0,\;\;\forall k>2. $$

$$ \lambda_1(t)= (A_w(t) \vez(t),\vez(t))= \int_{\bR}   \Bigl| \;\underbrace{\sum_{j=1}^n z_j(t) e^{\theta_j(t) x}}_{=:f_t(x)}\;\Bigr|^2 w(x) dx. $$

Observe that

$$f_t(x):= \sum_{j=1}^n z_j(t) e^{\theta_j(t) x}= \frac{1}{\sqrt{2}}(1-e^{\ii t x}) +\sum_{j=1}^k \ve_j(t) e^{\ii\theta_j(t) x},\;\;\ve_j(t)=z_j(t)-z_j(0). $$

We   deduce that

$$  \lim_{t\to 0} \frac{1}{t}f_t(x) = -\frac{\ii x}{\sqrt{2}} + \sum_{k=1}^n \dot{z}_k(0)= -\frac{\ii x}{\sqrt{2}}\tag{7}\label{7}$$

uniformly  for  $x$   on compacts. Since  $w$ has compact support  we deduce that (\ref{7}) holds for uniformly for $x$ in the support of $w$.  We deduce that

$$ \lambda_1(t)\sim \frac{1}{2}\;\underbrace{\left(\int_{\bR}  x^2 w(x)dx \right)}_{=\widehat{w}''(0)}\;t^2\;\;\mbox{as $t\to 0$}. $$

Using the last equality in (\ref{6}) we obtain $2N=2$ which proves  (\ref{Es}) .


Step 2.    We will show that if (\ref{E}) holds for $w_0$ and $w_1(x) \geq w_0(x)$,  $\forall x$,  then (\ref{E}) holds for  $w_1$ as well.       For any weight $w$  and any $\vet$ such that the $\Delta(\vet)\neq 0$ consider the ellipsoid

$$ \Sigma_w:=\bigl\{\vez\in\bC^n;\;\; (A_w\vez,\vez)\leq 1\bigr\}. $$

Then


$$ {\rm vol}\, \bigl(\;\Sigma_w(\vet)\;\bigr)=\frac{\pi^n}{n!\det A_w(\vet)}. $$

Observe that if $ w_0\leq w_1$ then $\Sigma_{w_0}(\vet)\subset \Sigma_{w_1}(\vet)$ and we  deduce

$$ \det A_{w_0}(\vet) \leq \det A_{w_1}(\vet). $$

This proves our claim.

Step 3. We show that (\ref{E}) holds for any integrable weight.   At  least one of the level sets $\{w\geq \ve\}$, $\ve>0$ is nonempty.  We can find a compact set of nonzero measure  $K \subset \{w\geq \ve \}$. Now define $w_0=I_{K}$. Clearly $I_K\leq w$.   From  Step 1 we know that (\ref{E}) holds for $w_0$. Invoking Step 2 we deduce that (\ref{E}) holds for  $w$.  Q.E.D.

The half-life of a theorem, or Arnold's principle at work - MathOverflow

This is a very interesting thread.


The half-life of a theorem, or Arnold's principle at work - MathOverflow

Supersymmetry in doubt

About a dozen of years ago, at a Great Lakes Conference dinner at Northwestern  I asked  Witten  what parts of high energy physics he thinks will be  confirmed  experimentally in our lifetime.  Super-symmetry was one of the first things he mentioned. Now comes  this news from the Large Hadron Collider casting some doubt on the supersymmetry premise.

BBC News - Popular physics theory running out of hiding places

A word of caution though.   About a year ago people thought that the Large Hadron Collider detected a particle traveling faster than the speed of light.


 News - Popular physics theory running out of hiding places

More about this SUSY injury at the Not Even Wrong blog.

Influence By Degree, Episode 1

An interesting BBC documentary on how private donors  influence academic life.

BBC World Service - The Documentary, Influence By Degree, Episode 1

Tim Gowers recovering after a heart intervention

I'll let the Great Man tell the story.

Analele Stiintifice Iasi

I thought I ought to do some advertising for the math journal published by my Alma Mater.  I am talking of course of the  Analele Stiintifice ale Universitatii  "Al.I. Cuza" Iasi. Sectia Matematica.

Separable random functions

$\newcommand{\si}{\sigma}$ $\newcommand{\es}{\mathscr{S}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bsT}{\boldsymbol{T}}$ First, what is a random function? To  define it we need a parameter space $\bsT$, $\newcommand{\eS}{\mathscr{S}}$ a  probability space $(\Omega, \eS, P)$, and a target space $X$. Roughly speaking  a  random function $\bsT\to X$ is defined to be a choice of probability measure (and underlying $\si$-algebra of events)  on $X^{\bsT}=$ the space of functions $\bsT\to X$.


In applications $X$ is a metric space and $\bsT$ is a  locally closed subset of some Euclidean space $\bR^N$. (Example to keep in mind:  $\bsT$ an open subset of $\bR^N$ or $\bsT$ a properly embedded submanifold of $\bR^N$. Often $X$ is a vector space.)

A random function on $\bsT$ is then a function

$$ f:\bsT\times\Omega\to X,\;\; \bsT\times\Omega\ni (t,\omega)\mapsto f(t,\omega) \in X, $$

such that, for any $t\in\bsT$,   the   correspondence

$$ \Omega\ni \omega\mapsto f_t(\omega) :=f(t,\omega)\in X $$

is measurable  with respected to the $\si$-algebra of Borel subsets  of $X$. In other words,  a random function on $\bsT$ is a family of random variables (on the same probability space) parameterized by $\bsT$.


Observe that we have a natural  map $\Phi: \Omega\to X^{\bsT}$,

$$\Omega\ni \omega\mapsto f_\omega\in X^\bsT,\;\;f_\omega(t)=f(t,\omega). $$

The pushforward via $\Phi$ of $(\eS,P)$ induces  structure of probability space on $X^{\bsT}$.  The functions $f_\omega$, $\omega\in \Omega$ are called the   sample functions  of the given random function.

Let us observe that  there are certain  properties of functions which a priori may not  measurable subsets of $\Omega$.  For example the set of $\omega$'s such that $f_\omega$ is continuous on $\bsT$   may not be measurable if $\bsT$   is uncountable.  To deal  with such issues  we will restrict our attention to certain  classes of  random functions, namely the  separable ones.

Definition 1.   Suppose that  $\bsT$ is a locally closed subset  of $\bR^N$ and $X$ is a Polish space, i.e., a  complete, separable  metric space. Fix a countable, dense subset $S\subset \bsT$. A random function $f:\bsT\times\Omega\to  X$ is called $S$-separable    if  there exists a negligible subset $N\subset  \Omega$, with the following property: for any closed subset $F\subset X$, any open subset $U\subset \bsT$ the  symmetric  difference of the sets

$$ \Omega(U,F):=\Bigl\{  \omega\in \Omega;\;\;  f_\omega(t)\in F,\forall t\in U\;\Bigr\},\;\;\Omega_S(U, F):=\Bigl\{  \omega\in \Omega;\;\; f_\omega(t)\in F,\;\;\forall t\in U\cap S\;\Bigr\} \tag{1}\label{1}$$

 is a subset of $N$, i.e.,

$$\Omega(U,S)\setminus \Omega_S(U,S),\;\;\Omega_S(U,F)\setminus \Omega(U,F)\subset N. $$


Definition 2.  Let $\bsT$ and $X$ be as in  Definition 1.  A random function  $g: \bsT\times \Omega\to X$ is called a  version of the  random function $f:\bsT\times \Omega\to X$  if

$$ P(g_t=f_t)=1,\;\;\forall t\in\bsT. $$


Let me   give an application of separability.      We say that a random function $g:\bsT\times \Omega\to X$ is  a.s.  continuous if

$$ P\bigl(\;\lbrace \omega;\;\; f_\omega: \bsT\to X\;\;\mbox{is continuous} \rbrace\;\bigr)=1. $$


Proposition 3.   Suppose that $f$ is an $S$-separable  random function $\bsT\times \Omega\to \bR$ and $g$ is a version of $f$.  If $g$ is a.s. continuous, then
\[
P\bigl(\lbrace \omega; \;\;g_\omega=f_\omega\rbrace\bigr)=1.
\]
In particular,  $f$ is a.s. continuous.

Proof.    Consider the set $N\subset \Omega$ in the definition of $S$-separability of $f$.   Define


$$\Omega_*:=\bigl\lbrace \;\omega\in \Omega\setminus N;\;\;g_\omega\;\mbox{is continuous},\;\;g(s,\omega)=f(s,\omega),\;\;\forall s\in S\;\bigr\rbrace. $$

Observe that $P(\Omega_*)=1$.   We will prove that
\[
g_\omega(t)=f_\omega(t),\;\;\forall \omega\in \Omega_*,\;\; t\in\bsT.\tag{$\ast$}\label{ast}
\]

 Fix an open set $U\subset \bsT$. For any $\omega\in\Omega_*$ set

\[
M_\omega(U, S):=  \sup_{t\in S\cap U} f_\omega(t).
\]
Invoking the definition of separability with $F=(-\infty, M_\omega(U, S)]$ we deduce  that
\[
f_\omega(t)\leq M_\omega(U,S),\;\;\forall t\in U,
\]
so that
\[
\sup_{t\in U} f_\omega(t)\leq \sup_{t\in S\cap U} f_\omega(t)\leq \sup_{t\in U} f_\omega(t).
\]
In other words, for any open set $U\subset\bsT$ we have
\[
\sup_{t\in U}f_\omega(t)=\sup_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{2}\label{2}.
\]
A variation of the above argument shows
\[
\inf_{t\in U}f_\omega(t)=\inf_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{3}\label{3}.
\]
Now let $\omega\in\Omega_*$,  $t_0\in T$. Given   $\newcommand{\ve}{\varepsilon}$ $\ve>0$, choose  an neighborhood $U=U(\ve \omega)$ of $t_0$ such that
\[
 g_\omega(t_0)-\ve\leq g_\omega(t)\leq g_\omega(t_0)+\ve,\;\;\forall t\in U(\ve,\omega).
\]
Since $g_\omega(t)=f_\omega(t)$ for $t\in S\cap U$ we deduce from (\ref{2}) and (\ref{3}) that
\[
g_\omega(t_0)-\ve \leq \inf_{t\in U(\ve,\omega)} f(t) \leq \sup_{t\in U(\ve,\omega)} f_\omega(t)\leq g_\omega(t_0)+\ve.
\]
In particular, we deduce
\[
g_\omega(t_0)-\ve \leq f_\omega(t_0)\leq g_\omega(t_0)+\ve,\;\;\forall \ve>0.
\]
This proves (\ref{ast}).     Q.E.D.


We have the following result.

Theorem 4.  Suppose that $f:\bsT\times \Omega\to X$ is a random function, where $\bsT$ are Polish spaces. If $X$ is compact, then  $f$ admits a separable version.


Proof.     We follow the approach in Gikhman-Skhorohod. $\DeclareMathOperator{\cl}{\mathbf{cl}}$  Fix a countable dense subset $S\subset \bsT$. $\newcommand{\eV}{\mathscr{V}}$ Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$ and with rational radii. For any  $\omega\in\Omega$  and any  open set $U\subset \bsT$ we set

$$ R(U,\omega):=\cl\bigl\lbrace f_\omega(t);\;\;t\in S\cap U\,\bigr\rbrace, $$

$$ R(t,\omega)=\bigcap_{t\in V\in\eV} R(V,\omega), $$

where $\cl$ stands for the closure of a set.   Observe that $R(t,\omega)\neq\emptyset$ because it is the intersection of a family of compact sets such that any finitely many sets in the family have nonempty intersection.


Lemma 5.  The following statements are equivalent.

(a) The random function  $f$ is $S$-separable.

(b) There exists $N\subset \Omega$ such that $P(N)=0$ and for any $\omega\in\Omega\setminus N$ and any $t\in\bsT$ we have $f_\omega(t)\in R(t,\omega)$.

Proof of the lemma.  (a) $\Rightarrow$ (b)   We know that $f$ is $S$-separable. Choose $N$ as in the definition of $S$-separability.   Fix $\omega_0\in \Omega\setminus N$ and $t_0\in \bsT$.   For any ball $V\in \eV$ that contains $t_0$  we have

\[
\bigl\lbrace \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in S\cap B\;\bigr\}=\bigl\lbrace\; \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in B\;\bigr\rbrace.
\]
Observe that $\omega_0$ belongs to the set in the left-hand-side of the above equality and so it must belong to the set in the right-hand-side.  Hence

\[
f_{\omega_0}(t)\in R(B,\omega_0),\;\;\forall t\in B
\]
and therefore $f_{\omega_0}(t_0)\in R(B,\omega_0)$ for any $B\in\eV$ that contains $t_0$. Thus $f_{\omega_0}(t_0)\in R(t_0,\omega_0)$ which finishes the proof of the implication (a) $\Rightarrow$ (b).


(b) $\Rightarrow$ (a)   Set $\Omega_*=\Omega\setminus N$.  Suppose that $F\subset X$ is closed.  For any $B\in\eV$  and $\omega\in \Omega_*$ we have

$$ f_\omega(t)\in F\;\;\forall t\in S\cap B \Leftrightarrow F\supset R(B,\omega). $$

Since $R(t,\omega)\subset R(B,\omega)$ for any $t\in B$ we deduce  that

$$ \Omega(B,F):=\bigl\{ \omega\in \Omega_*;\;\; f_\omega(t)\in F\;\;\forall t\in B\;\bigr\} =\bigl\{ \omega\in \Omega_*;\;\; f+\omega(t)\in F\;\;\forall t\in S\cap B\;\bigr\}=\Omega_S(B,F). $$


If $U$  an open set then we can write $U$ as a countable union of balls in $\eV$

$$U=\bigcup_n B_n. $$

Then

$$ \Omega(U,F)=\bigcap_n \Omega(B_n, F)=\bigcap_n\Omega_S(B_n, F)= \Omega_S(U,F). $$

This finishes the proof of the lemma.  q.e.d.



Lemma 6.  For any  Borel set $B\subset X$ there exists a countable subset $C_B\subset \bsT$ such that  for any $t\in\bsT$ the set

$$N(t, B):=\bigl\{ \omega\in \Omega; \;\;f_\omega(\tau)\in B,\;\;\forall \tau\in C_B,\;\;f_\omega(t)\in \bsT\setminus B\;\bigr\} $$

has probability $0$.


Proof of the lemma.   We construct $C_B$ recursively.  Choose $\tau_1\in\bsT$ arbitrarily and set $C_B^1:=\{\tau_1\}$.   Suppose that we have  constructed  $C_B^k=\{\tau_1,\dotsc,\tau_k\}$.    Set



$$ N_k(t):=\bigl\{\omega;\;\; g_\omega(\tau)\in B\;\;\forall \tau\in C_B^k,\;\;f_\omega(t)\in\Omega\setminus B\;\bigl\},\;\;p_k=\sup_{t\in\bsT} P\bigl(\;N_k(t)\;\bigr). $$

Observe that  $p_1\geq p_2\geq \cdots \geq p_k$. If $p_k=0$   we stop and we set $C_B:=C_B^k$.

If this is not the case, there exists $\tau_{k+1}\in \bsT$ such that

$$P\bigl(N_k(\tau_{k+1}\bigr)\geq \frac{1}{2}p_k. $$

Set $ C_B^{k+1}:=C_B^k\cup\{\tau_{k+1}\}$.  Observe that the events $N_1(\tau_2),\dotsc , N_k(\tau_{k+1})$ are mutually  exclusive  and thus

$$ 1\geq \sum_{j=1}^k P(N_j(\tau_{j+1})) \geq \frac{1}{2}\sum_{j=1}^k p_{j+1}. $$

Hence $\lim_{n\to\infty} p_n=0$. Now  set

$$N(t, B):=\bigcap_{k\geq 1} N_k(t). $$
                                                                                                                                                          q.e.d.


Lemma 7.   $\newcommand{\eB}{\mathscr{B}}$ Suppose that $\eB_0$ is a countable family  of Borel subsets of $X$ and $\eB$  is the family obtained by taking the intersections of  all the subfamilies of $\eB_0$.  Then there exists a countable subset $C\subset \bsT$, and for each $t$ a subset $N(t)$ of probability zero  such that for any $B\in \eB$ we have

$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t). $$

Proof.   For any $t\in \bsT$ we define

$$ C:=\bigcup_{B\in\eB_0}C_B, \;\; N(t) :=\bigcup_{B\in\eB_0} N(t,B), $$

where $C_B$  and $N(t,B)$ are  constructed as in Lemma 6. Clearly $C$ is countable.

If $B'\in\eB$ and $B\in\eB_0$ are such that $B\supset B'$, then

$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t,B)\subset N(t). $$

If

$$ B'=\bigcap_{k\geq 1} B_k,\;\;B_k\in\eB_0\;\;\forall k, $$

then


$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B'\;\bigr\}\subset \bigcup_{k\geq 1} \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B_k\;\bigr\}\subset \bigcup_{k\geq 1} N(t, B_k)\subset N(t). $$
                                                                                                                                                     q.e.d.


The proof of Theorem 4 is now within reach.  Suppose that $S$ is a countable and dense  set of points in $\bsT$ and $D$ is a countable dense subset of $X$.  Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$. Denote by $\eB_0=\eB_0(D)$ the collection of open balls with  in $X$ of rational radii centered at points in $D$.  As in Lemma 7, denote by $\eB$ the collection of sets obtained by taking intersections of arbitrary families in   $\eB_0$. Clearly $\eB$ contains all the closed subsets of $X$.



Fix a ball $V\in \eV$.  Lemma 7  applied to the restriction of $f$ to $V$ implies the existence of a countable set

$$C(V)\subset V$$
  
and  of a family of  negligible sets

$$N_V(t)\subset \Omega,\;\;t\in V $$

such that  for any $B\in eB$

$$\{ \omega;\;\;  f_\omega(\tau)\in B,\;\;\forall \tau\in C,\;\;f_\omega(t)\in V\setminus B\;\bigr\}\subset N_V(t). $$

Set

$$C=\bigcup_{V\in\eV}C(V),$$

while for $t\in  \bsT$ we set

$$ N(t):=\bigcup_{\eV\niV\ni t}N_V(t).

Clearly $C$ is both countable and dense in $\bsT$. We can now construct a $C$-separable version  $\tilde{f}$ of $f$. Define

• $\tilde{f}_\omega(t)= f_\omega(t)$  if $t\in C$ or $\omega\not\in N(t)$
• If $\omega\in N(t)$ and $t\in \bsT\setminus C$ we assign  $\tilde{f}^V_\omega(t)$  an arbitrary value in $R(t,\omega)$.

By construction $\tilde{f}$ is a version of $f$ because for any $t\in \bsT$

$$\{\omega;\;\;\tilde{f}_\omega(t)\neq f_\omega(t)\;\}\subset N(t). $$

Since $f_\omega(\tau)=\tilde{f}_\omega(\tau)$ for any $\tau\in C$, $\omega\in \Omega$ sets $R(t,\omega)$, defined as  as in Lemma 5, are the same for both   functions $\tilde{f}$ and $f$. By construction $\tilde{f}_\omega(t)\in R(t,\omega)$,   $\forall t,\omega$.    Q.E.D.

On convolutions

Somebody on MathOverlow  asked for some intuition behind the operation of convolution of two functions. Here is my take on this.  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$


Suppose we are given a function $f:\bR\to \bR$. Discretize  the real axis and think of it as  the collection of point $\Lambda_\hbar:=\hbar \bZ$, where $\hbar>0$ is a small number.  We can then approximate $f$ with its restriction  $f^\hbar:=f|_{\Lambda_\hbar}$. This  is   determined by its generating function, i.e., the   formal power series $\newcommand{\ii}{\boldsymbol{i}}$

$$G^\hbar_f(t)=\sum_{n\in\bZ}f(n\hbar)t^n\in \bR[[t,t^{-1}]]. $$

Then

$$G^\hbar_{f_0\ast f_1}(t)= G^\hbar_{f_0}(t)\cdot G^\hbar_{f_1}(t).\tag{1} \label{1} $$

Observe that if  we   set $t=e^{-\ii\xi \hbar}$, then

$$G^\hbar_f(t)=\sum_{x\in\Lambda_\hbar} f(x) e^{-\ii \xi x}. $$

Moreover

$$ \hbar G^\hbar_f(e^{-\ii\xi \hbar})=\sum _{n\in \bZ} \hbar f(n\hbar) e^{-\ii\xi(n\hbar)}, \tag{2}\label{2}$$

and the expression in the right hand sum is  a "Riemann sum"  approximating

$$\int_{\bR} f(x)^{-\ii\xi x} dx. $$

Above we recognize the Fourier transform of $f$. If we let $\hbar\to 0$  in (\ref{2}) and we use (\ref{1})  we obtain the wellknown fact that the Fourier transform  maps the convolution to the usual pointwise product of functions. (The fact that this rather careless passing to the limit  can be rigorous is what the Poisson formula is all about.)

The above argument shows that we can regard $\hbar G_f^\hbar(1)$ as an approximation for $\int_{\bR} f(x) dx$.





Denote by $\delta(x)$ the Delta function concentrated at $0$. The Delta function concentrated at $x_0$ is then $\delta(x-x_0)$. What could be the generating function of $\delta(x)$, $G_\delta^\hbar$?  First, we know that $\delta(x)=0$, $\forall x\neq 0$ so that


$$G_\delta^\hbar(t) =ct^0=c. $$

The constant $c$ can be determined from the equality 

$$ 1= \int_{\bR} \delta(x) dx=\hbar G_\delta^\hbar(1)=\hbar c$$

Hence $\hbar G_\delta^\hbar(1)=1$.  Similarly

$$ G^\hbar_{\delta(\cdot-n\hbar)} =\frac{1}{\hbar} t^n. $$

In particular, the discretization $\delta^\hbar(x-n\hbar)$ of $\delta(x-n\hbar)$ is the  function $\Lambda_\hbar\to \bR$ with value $\frac{1}{\hbar}$ at $x=n\hbar$ and $0$  elsewhere.

Putting together all of the above we obtain an equivalemn description for the  generating functon af a function $f:\Lambda_\hbar\to\bR$. More precisely

$$ G^\hbar_f(t)=\hbar\sum_{\lambda\in\Lambda_\hbar}f(\lambda) G^\hbar_{\delta(\cdot-\lambda)}(t). $$
In other words

$$f^\hbar= \hbar\sum_{\lambda\in\Lambda_\hbar} f(\lambda)\delta^\hbar_\lambda,\;\;\delta^\hbar_\lambda(\cdot):=\delta^\hbar(\cdot-\lambda). \tag{3}\label{3}$$


The last  equality  suggests an interpretation for the generating function as an algebraic encoding of the fact that $f:\Lambda_\hbar\to\bR$ is a superposition of $\delta$ functions  concentrated along the points of the lattice $\Lambda_\hbar$. The  factor $\hbar$ in (\ref{3}) is a discretization of the infinitesimal $dx$, which indicates that $\hbar\delta^\hbar_\lambda$  should be viewed as a measure.    Observe that

$$(\hbar\delta^\hbar_\lambda)\ast (\hbar\delta^\hbar_\mu)=\hbar\delta^\hbar_{\lambda+\mu}. \tag{4}\label{4}$$

Midwest Dynamical Systems Seminar - Department of Mathematics - University of Notre Dame

Midwest Dynamical Systems Seminar - Department of Mathematics - University of Notre Dame

Mathgen paper accepted! | That's Mathematics!

 I've just found out from a colleague of this site MathGen  which randomly generates math paper. Apparently  one such paper has recently been accepted for publication by an Open Access journal, you know, the kind where you pay to have your  paper publishe.  More details  at this site
 Mathgen paper accepted! | That's Mathematics!

Here is the paper MathGen  produced on my behalf.

Journal of Gokova Geometry and Topology

 I thought that you, yes you the guy reading these lines,    should have a look at this    journal of geometry and topology, Journal of Gokova Geometry Topology. It has  a great editorial board and it looks for   great papers to publish.

GmailTeX

 If you wanted to e-mail math formulas and did  not know how, try the link below.

GmailTeX

On an integral geometric formula

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bsV}{{\boldsymbol{V}}}$ $\DeclareMathOperator{\Graffr}{\mathbf{Graff}^c}$  $\newcommand{\be}{\boldsymbol{e}}$ $\newcommand{\bv}{\boldsymbol{v}}$   $\DeclareMathOperator{\Grr}{\mathbf{Gr}^c}$ $\newcommand{\Gr}{\mathbf{Gr}}$ $\newcommand{\Graff}{\mathbf{Graff}}$

Suppose that $\bsV$ is a finite dimensional real Euclidean  space, $M\subset \bsV$  is a smooth compact submanifold of dimension $m$ and codimension $r$ and we set


$$ N:=\dim \bsV=m+r. $$

  For any  nonnegative integer $c\leq \dim \bsV$ we denote by $\Graff^c(\bsV)$ the  Grassmannian  of  affine subspaces of $\bsV$ of codimension $c$,   by $\Gr^c(\bsV)$ the Grassmannian of  codimension $c$ vector subspaces of $\bsV$. We set $\Gr_k(\bsV):=\Gr^{N-k}(\bsV)$.

The   codimension $c$ Radon transform of a smooth function $f: M\to  \bR$  is a function

$$ \widehat{f}:\Graff^c(\bsV)\to\bR , $$

such that $\newcommand{\eH}{\mathfrak{H}}$

$$\widehat{f}(S) =\int_{S\cap M} f(x) d\eH^{m-c}(x),  \;\; \forall S\in \Graff^c(M), \label{r}\tag{R}$$

where $d\eH^{m-c}$ denotes the $(m-c)$-dimensional Hausdorff measure.   If $c\leq \dim M$ then a generic   affine plane $S\in\Graff^c(\bsV)$ intersects  $M$ transversally in which case the Hausdorff measure in (\ref{r}) is the usual Lebesgue measure induced my the  natural Riemann metric on $S\cap M$.

I want  to explain how to  recover the integral of $f$ over $M$ from its Radon transform.



Observe that we have an incidence set $\newcommand{\eI}{\mathscr{I}}$

$$\eI^c(\bsV) :=\Bigl\{ (\bv, S)\in \bsV\times \Graffr(\bsV);\;\; \bv\in S\;\Bigr\} $$

equipped with  natural projections

$$ \bsV\stackrel{\lambda}{\leftarrow}\eI^c(\bsV)\stackrel{\rho}{\to}\Graffr(\bsV).\label{F}\tag{F} $$


For any subset $X\subset \bsV$ we define

$$\eI^c(X):=\lambda^{-1}(M)\subset \eI^r(X),\;\; \Graffr(X)=\rho\Bigl(\;\eI^r(X)\;\Bigr). $$

Note that

$$ \Graffr(X)=\Bigl\{ S\in \Graffr(\bsV);\;\; S\cap X\neq \emptyset\;\Bigr\} $$

and for any $\bv\in\bsV$ we have

$$ \lambda^{-1}(\bv) =\bigl\{ \bv+S;\;\;S\in \Grr(\bsV)\;\bigr\}=\Graffr(\bv)\subset \Graffr(\bsV). $$


Observe that  $\eI^c(V)\to \bsV$ is a smooth fiber bundle  with fiber  $\Gr^r(\bsV)$. In particular,  $\eI^c(M)\to M$ is the bundle obtained  by restricting to the submanifold $M$.  Its fiber is also $\Gr^c(\bsV)$.

At this point I need to recall some  basic facts described in great detail in Sections 9.1.2, 9.1.3 of  Lectures on the Geometry of Manifolds.

 The Grassmannain $\Gr^c(\bsV)$ is equipped with a canonical $O(\bsV)$-invariant metric  with volume  density $|d\gamma^c_\bsV|$ with total volume $\newcommand{\sbinom}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}}$

$$\int_{\Gr^c(\bsV)} |d\gamma_\bsV^c(L)|=\sbinom{N}{c}, $$

where $\sbinom{N}{c}$ is defined  in   equation (9.1.66) of the Lectures.

Now observe that  we have a natural projection $\pi: \Graff^c(\bsV)\to \Gr^c(\bsV)$ that associates  to each affine  plane its translate through the origin.    A plane $S\in\Graff^c(\bsV)$ intersects the orthogonal complement  of $\pi(S)$ in a unique point $C(S)=S\cap \pi(S)^\perp$.   We obtain a an embeding

$$ \Gr^c(\bsV)\ni S\mapsto \bigl(\;C(S), \pi(S)\;\bigr)\in \bsV\times\Gr^c(\bsV),\;\;C(S)\perp \pi(S), $$
$\newcommand{\eQ}{\mathfrak{Q}}$
and we  will regard  $\Graff^c(\bsV)$ as a submanifold of $\bsV\times \Gr^c(\bsV)$.  As such,     it becomes  the total space of  a vector bundle $\eQ_c\to\Gr^c(\bsV)$, in fact a subbundle of the trivial bundle $\bsV\times  \Gr^c(\bsV)\to\Gr^c(\bsV)$.   The orthogonal complement  $\eQ_c^\perp$ of this bundle is the tautological vector bundle $\newcommand{\eU}{\mathscr{U}}$  ${\eU}^c\to\Gr^c(\bsV)$.  In particular

$$\dim\Gr^c(\bsV)= c(N-c)+  c. $$

Along $\Graff^c(\bsV)$ we have a canonical vector bundle,  the  vertical bundle $VT\Graff^c(\bsV)\subset T\Graff^c(\bsV)$   consisting of the kernels of $d\pi$, i.e., vectors tangent to the fibers of $\pi$. The  vertical bundle is equipped with a natural density  $|d\bv|_c$ which when restricted to a fiber of $\pi^{-1}(L)$  induces the natural volume form on the fiber $L^\perp$ viewed as a vector subspace of $\bsV$. As in Section 9.1.3 of the Lectures we define a product  density $|d\tilde{\gamma}^c|=|d\tilde{\gamma}_\bsV^c|$ on $ \Graff^c(\bsV)$,

$$|d\tilde{\gamma}_\bsV^c|= |d\bv|_c\times \pi^*|d\gamma_\bsV^{c}| $$

Alternatively, the vector bundle $\eQ_c$, as a subbundle of the trivial bundle $\bsV\times \Gr^c(\bsV)\to\Gr^c(\bsV)$  is equipped with a natural metric connection. The  horizontal subbundle   $HT\eQ_c\subset T\eQ_c$  is isomorphic to $\pi^* T\Gr^c(\bsV)$ and thus comes equipped with a natural metric.    The  vertical subbundle $VT\eQ_c=VT\Graff^c(\bsV)$ is also equipped with a  natural  metric and in this fashion we obtain a metric on $\Graff^c(\bsV)=\eQ_c$. The density $|d\tilde{\gamma}^c_\bsV|$  is the volume density defined by this metric.


Suppose now that $c\leq m=\dim M$.   We denote by $\Graff^c_*(M)$ the subset of $\Graff^c(M)$ consisting of affine planes that intersect $M$ transversally.    This is an open subset of $\Graff^c(M)$.  The condition $c\leq m$ implies that this set is nonempty.  (For $c=1$ this follows from the fact that the restriction to $M$ of a generic linear function is a Morse function. Then look at iterated slicing by hyperplanes.)

Set

$$ \eI^c_*(M)= \rho^{-1}\bigl(\;\Graff^c_*(M)\;\bigr)\subset \eI_M $$

The fiber of  $\rho:\eI_*^c(M)\to \Graff^c_*(M)$ over $S\in \Graff^c_*(M)$ is the submanifold $S\cap M$ which is equipped with a metric density.  We obtain a density on $\eI^c_*(M)$

$$ |d\nu^c_M|= |dV_{S\cap M}|\times \rho^*|d\tilde{\gamma}^c|. \tag{$\nu^c$}\label{nu}$$

If $f: M\to\bR$ is a smooth function, then

$$\int_{\eI^c_*(M)}\lambda^*(f) |d\nu^c_M|=\int_{\Graff^c_*(M)}\left(\int_{S\cap M} f|dV_{S\cap M}\right) |d\tilde{\gamma}^c(S)|. \label{1}\tag{1} $$

For any  vector subspace $U\subset \bsV)$ we denote by $\Gr^c(\bsV)_U$ the set consisiting of subspaces $L\in\Gr^c(\bsV)$ that intersect $U$ transversely.


We now want to integrate $\lambda^*(f)$ along the fibers of $\lambda :\eI^c_*(M)\to M$.  For any  vector subspace $U\subset \bsV)$ we denote by $\Gr^c(\bsV)_U$ the set consisting of subspaces $L\in\Gr^c(\bsV)$ that intersect $U$ transversely.

The fiber of this map over a point $x\in M$ is an open  subset of $x+\Gr^c(\bsV)_{T_xM}\subset \Graff^c(\bsV)$ with negligible complement.    The density $|d\nu^c_M|$ on $\eI^c_*(M)$ induces  a density

$$ |d\nu^c_x|=|d\nu^M|/\lambda^*|dV_M| $$


on each fiber $\lambda^{-1}(x)$ and we deduce

$$ \int_{\eI^c_*(M)} \lambda^* f|d\nu^c(M)|= \int_M\left(\int_{\lambda^{-1}(x)}|d\nu^c_x|\right) f(x)|dV_N(x)|. \label{2}\tag{2} $$

The density $|d\nu^c(x)|$ is  the restriction of a density $|d\bar{\nu}^c_x|$ on $\Gr^c(\bsV)_{T_xM}$. In fact, a reasoning similar to the one   in the proof  of Lemma 9.3.21 in the Lectures implies that  for any $U\in\Gr_m(\bsV)$ there exists a canonical density $|d\bar{\nu}^c_U|$ on $\Gr^c(\bsV)_U$ such that

$$ T_*|d\nu^c_U|=|d\bar{\nu^c}_{T(U)}|,\;\;\forall T\in O(\bsV),\;\;U\in \Gr_m(\bsV), \label{3}\tag{3} $$


$$ |d\bar{\nu}^c_x|=|d\bar{\nu}^c_{T_xM}|,\;\;\forall x\in M. \label{4}\tag{4}$$


 Using (\ref{3}) (\ref{4}) in (\ref{2}) we deduce that there  exists a constant $Z=Z(N,m,c)$ that depends only on $N,m,c$ such that

$$ Z(N,m,c)=\int_{\nu^{-1}(x)} |d\bar{\nu}^c_x|,\;\;forall x\in M. $$

Using this in (\ref{2})  we conclude from (\ref{1}) that

$$ Z(N,m,c)\int_{M}f(x)\; |dV_M(x)| =\int_{\Graff^c(\bsV)}\left(\int_{S\cap M} f(x)|dV_{S\cap M}|\right) |d\tilde{\gamma}^c_\bsV(S)|.\label{5}\tag{5} $$




To find the constant $Z(N,m,c)$ we choose $M$ and $f$ judiciously.  We let $M=\Sigma^m$, the unit $m$-dimensional sphere contained in some $(m+1)$-dimensional subspace of $\bsV$.  Then, we let $f\equiv 1$. We deduce from (\ref{5}) that

$$ Z(N,m,c)=\frac{1}{{\rm vol}\;(\Sigma^m)} \int_{\Graff^c(\bsV)} {\rm vol}\,(S\cap \Sigma^m)\;|d\tilde{\gamma}^c_\bsV(S)|.\label{6}\tag{6} $$


Using the Crofton formula in Theorem 9.3.34 in the Lectures  in the special case $p=m-c$ we deduce


$$Z(N,m,c)=\sbinom{m}{c}. $$

Remark. 1        Consider the Radon transform


$$ C_0^\infty(\bsV)\ni f\mapsto  \widehat{f}\in C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr), \;\; \widehat{f}(S)=\int_S f(x)|dV_S(x)|,\;\;\forall S\in \Graff^c(\bsV). $$

Observe that $\widehat{f}$ has compact support.   Indeed, if  the support of $f$ is contained in a ball of radius $R$, then for any affine plane $S\in \Graff^c(\bsV)$ such that ${\rm dist}\,(0,S)>R$ we have $\widehat{f}(S)=0$.

Consider the dual Radon transform  $\newcommand{\vfi}{\varphi}$


$$ C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr)\ni \vfi\mapsto \check{\vfi}\in C^\infty(\bsV),\;\;\check{\vfi}(x)=\int_{\Gr^c(\bsV)} \vfi(x+L)\;|d\gamma^c(L)|,\;\;\forall x\in \bsV. $$



Consider the fundamental double fibration  (\ref{F}). Given $f\in C_0^\infty(\bsV)$, $\vfi\in C^\infty_0\bigl(\;\Graff^c(\bsV)\;\bigr)$ we obtain a function 


$$ \Phi=\lambda^*(f)\cdot \rho^*(\vfi)\in C_0^\infty(\bsV) $$

Arguing as above, with $M=\bsV$  we  observe that $\Graff^c_*(\bsV)=\Graff^c(\bsV)$ and we obtain as in (\ref{nu}) a density  $|\nu^c_\bsV|$ on $\eI^c_*(\bsV)=\eI^c(\bsV)$.   Denote by $\rho_*\Phi |d\nu^c_\bsV|$ the   pushfoward  of the density $\Phi|d\nu^c_\bsV$. It is a density on $\Graff^c(\bsV)$  and we have the  Fubini formula (coarea formula)

$$ \int_{\eI^c(\bsV)} \Phi(x,S) |d\nu^c_\bsV(x,S)|=\int_{\Graff^c(\bsV)}\rho_*\Phi|d\ni^c_\bsV|\label{7}\tag{7} $$

Similarly, we obtain

$$ \int_{\eI^c(\bsV)} \Phi(x,S) |d\nu^c_\bsV(x,S)|=\int_{\bsV} \lambda_*\Phi |d\nu^c_\bsV|(x).\label{8}\tag{8} $$

From the construction  of $|d\nu^c_\bsV|$ we deduce immediately that

$$ \rho_*\Phi|d\nu^c_\bsV|(S)=  \widehat{f}(S) \vfi(S) |d\tilde{\gamma}^c|(S). $$


From the definitions of $|d\nu^c_\bsV|$, $|d\gamma^c_\bsV|$ and  $|d\tilde{\gamma}^c_\bsV|$ it follows easily that

$$ \lambda_*\Phi |d\nu^c_\bsV|(x) =  f(x)\check{\vfi}(x)|dx| $$

Using the last equalities in (\ref{7}) and (\ref{8})  we deduce

$$ \int_{\bsV} f(x)\check{\vfi}(x)=\int_{\Graff^c(\bsV)} \widehat{f}(S)\vfi(S) |d\tilde{\Gamma}^c(S)|. \tag{D}  \label{d} $$

The equality (\ref{d}) shows that the operations $f\mapsto \widehat{f}$ and $\vfi\mapsto \widehat{\vfi}$ are indeed dual to each other.   Note also that if we set  $\vfi\equiv 1 $ in (\ref{d}) then

$$\check{\vfi}(x)={\rm vol}\,\bigl(\;\Gr^c(\bsV)\;\bigr)=\sbinom{N}{c} $$

and in this case we reobtain (\ref{5}) in the special case $M=\bsV$.  The  equality (\ref{d}) is  important for another reason.

Denote by $C_0^{-\infty}(\bsV)$ the space of generalized functions with compact supports, then we can extend the Radon transform to such objects. If $u\in C_0^{-\infty}(\bsV)$ then we define its   Radon transform $\widehat{u}$ to be the compactly supported  generalized density on $\Graff^c(\bsV)$ defined by the equality

$$ \langle \widehat{u},\vfi\rangle=\langle u,\check{\vfi}\rangle ,\;\;\forall \vfi\in C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr). $$

If $M$ is a compact submanifold of $\bsV$, then we get a  Dirac-type  generalized function $\delta_M$ on $\bsV$   defined by integration along $M$ with respect to the volume density on $M$ determined by the induced metric.   Then

$$ \langle\widehat{\delta}_M,\vfi\rangle =\int_M  \check{\vfi}(x) |dV_M(x)|,\;\;\forall  C^\infty\bigl(\;\Graff^c(\bsV)\;\bigr). $$

The  generalized function $\widehat{\delta}_M$ is represented by a locally integrable function

$$\widehat{\delta}_M(S) =\eH^{m-c}(M\cap S),\;\;\forall S\in \Graff^c(M). $$

An introduction to the Laplace method in the asymptotics of integrals

$\newcommand{\bR}{\mathbb{R}}$ There are many sources explaining   this old technique of   determining the asymptotic behavior of certain integrals depending on a small parameter $\hbar$.     However, in applications, the integrals do not quite fit  the setup described in most books I have consulted  and I thought it would be nice  to  present   the general strategy. What follows is folklore,  and  even not  the most  general possible result, but I took great pain to highlight  the  key features one should look   for when attempting to use the Laplace technique in a concrete case.

Consider an interval $(a,b)\subset \bR$  and  a family of  $C^2$-functions

$$\phi_\hbar: (a,b)\to \bR,\;\;\hbar>0,  $$

where the interval $(a,b)$ could be finite, or infinite. We are interested in the behavior of the integral

$$I_\hbar:=\int_a^b e^{-\phi_\hbar(t)} dt $$

as $\hbar \searrow 0$ given that the functions  $\phi_h$ satisfy certain conditions



$\mathbf{C}_1$   For any $\hbar>0$ the function $\phi_\hbar$ has a unique critical point $\tau=\tau(\hbar)\in (a,b)$. Moreover, $\phi_\hbar''(\tau)>0$.  In other words, $\tau$ is a nondegenerate local minimum, and the uniqueness  assumption implies that $\phi_\hbar$ achieves its absolute minimum at $\tau$. We set $\newcommand{\si}{\sigma}$

$$ \si=\si(\hbar):=    \frac{1}{\sqrt{\phi_\hbar''(\tau)}}. $$

$\mathbf{C}_2$ The numers $\tau(\hbar)$ and $\si(\hbar)$ satisfy the conditions

$$ \lim_{\hbar\to 0}\frac{\tau(\hbar)-a}{\si(\hbar)}=\lim_{\hbar\to 0}\frac{ b-\tau(\hbar)}{\si(\hbar)}=\infty. $$


$\mathbf{C}_3$

$$\lim_{\hbar\to 0}\bigl(\,\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\,\bigr)=\frac{x^2}{2},\;\;\forall x\in\bR $$

$\mathbf{C}_4$ There exists $u:\bR\to \bR$ such that

$$\int_{\bR}e^{-u(x)} dx <\infty\;\;\mbox{and}\;\;\phi_\hbar(\tau+ \si x)-\phi_\hbar(\tau)\geq u(x),\;\;\forall \hbar,\;\;x\in J(\hbar). $$

Then, under the assumptions $\mathbf{C}_1,\dotsc,\mathbf{C}_4$  we have

$$ I_\hbar \sim\sqrt{2\pi} \si e^{-\phi_\hbar(\tau)}\;\;\mbox{as $\hbar\to 0$}. \label{A}\tag{A}  $$

Proof of (\ref{A}). We make the change of variables $t=\tau+\si x$ in the integral $I_\hbar$  to conclude that

$$I_\hbar=\si e^{-\phi_\hbar(\tau)}\int_{J(\hbar)} e^{-\bigl(\;\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\;\bigr)} dx, $$

where

$$ J(\hbar)= \Bigl[\frac{a-\tau}{\si},\frac{b-\tau}{\si}\Bigr]. $$

The condition $\mathbf{C}_2$ implies that the intervals  $J(\hbar)$ expand to $\bR$ as $\hbar \to 0$    Set $\newcommand{\vfih}{\varphi_\hbar}$

$$\vfih(x): =\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau). $$

 $\mathbf{C}_3$ implies that

$$\vfih(x)\to\frac{x^2}{2}\;\;\mbox{as $\hbar\to 0$} ,\;\;\forall x\in\bR. $$


We can now use $\mathbf{C}_4$ to invoke the dominated convergence theorem  and conclude that

$$\lim_{\hbar\to 0} \int_{J(\hbar)} e^{-\vfih(x)} dx=\int_{\bR}e^{-\frac{x^2}{2}}=\sqrt{2\pi}. $$

This completes the proof  of (\ref{A}).


Remark 1.   (a) Often in applications each of the functions $\phi_\hbar$ is convex.   In such cases the bound  $\mathbf{C}_4$ is a consequence of   the bound

$$ \bigl|\;\phi'_\hbar(\tau\pm \si)\;\bigr|= O\Bigl(\frac{1}{\si}\Bigr) \;\;\mbox{as $\hbar\to 0$}\label{B}\tag{B}. $$

Indeed,  $\vfih$ is convex and thus its graph its situated above  either of the tangent lines at $x=\pm 1$.    Thus

$$ \vfih(x) \geq \max\Bigl\{ \vfih'(1)(x-1)+\vfih(1),\;\;\vfih'(-1)(x+1) +\vfih(-1)\Bigr\}. $$

Observing  that

$$\vfih(\pm 1)>\vfih(0)=0,\;\;0< \pm \vfih'(\pm 1)= \pm \si\phi_\hbar(\tau\pm \si)=O(1), $$

  we deduce that in $\mathbf{C}_4$ we can choose $u(x)$ of the form $u(x)=C(|x|-1)$ for some positive constant $C$.


(b) Both conditions $\mathbf{C}_3$ and $\mathbf{C}_4$ would follow immediately if one can  prove that there exists  a $C^1$-function

$$\Psi:[0,\infty)\times \bR\to \bR,\;\;(\hbar,x)\mapsto \Psi(\hbar,x), $$

such that

$$\Psi(0,x)=\frac{x^2}{2},\;\;\Psi(\hbar,x)=\vfih(x),\;\;\forall x\in J(\hbar),\;\;\forall \hbar >0. $$



(c) The esence of the above results is that, under appropriate assumptions, we can replace $\phi_\hbar(t)$ with its $2$-nd order jet at $\tau$

$$ \phi_\hbar(t)\approx \phi_\hbar(\tau)+\frac{(t-\tau)^2}{2\si^2} $$ 

and deduce that

$$ I_\hbar\sim \int_a^b e^{-\phi_\hbar(\tau)-\frac{(t-\tau)^2}{2\si^2}} dt,\;\;\mbox{as $\hbar \to 0$} $$

Example 1. Let me illustrate  how the above strategy works  in the  classical situation described in all the books on asymptotics of integrals. Consider a $C^2$ convex function $\phi:(-a, a)\to\bR$ with a unique minimum at $\tau =0$ and such that $\phi''(0)>0$.  Set $\phi_\hbar(t)=\frac{1}{\hbar}\phi(t)$ so that

$$I_\hbar=\int_{-a}^ae^{-\frac{1}{\hbar}\phi(t)} dt. $$

In this case

$$\tau=0,\;\; \si =\sqrt{\frac{\hbar}{\phi''(0)}}. $$

The conditions $\mathbf{C}_1,\mathbf{C}_2$  are obviously satisfied.      As for $\mathbf{C}_3$ we observe that in this case we have

$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)=\frac{1}{\hbar}\bigl(\,\phi(\si x)-\phi(0)\;\bigr) = \frac{1}{\hbar}\Bigl(\;\phi'(0)\si x+ \frac{\phi''(0)}{2}\si^2 x^2+ o(\si^2)\;\Bigr)= \frac{x^2}{2}+ o(1). $$

Condition (\ref{B}) is also satisfied since

$$\vfih(\pm 1)=\frac{1}{\sqrt{\hbar\phi''(0)}}\phi'\Bigl(\pm \sqrt{\frac{\hbar}{\phi''(0)}}\Bigr) \to \pm 1\;\;\mbox{as $\hbar\to 0$}. $$

Hence we conclude that

$$\int_{-a}^a e^{-\frac{1}{\hbar}\phi(t)} dt \sim\sqrt{\frac{2\pi\hbar}{\phi''(0)}}\;\;\mbox{as $\hbar\to  0$}. $$


Example 2.     Consider  the integral

$$ \Gamma(\lambda +1)=\int_0^\infty  t^\lambda e^{-t}  dt,\;\;\lambda \to  \infty. $$

Observing  that it has the form $I_\hbar$ where $a=0$, $b=\infty$, $\hbar=\frac{1}{\lambda}$  and

$$\phi_\hbar(t)= t-\lambda \log t. $$

In this case we have

$$\phi'_\hbar(t)=1-\frac{\lambda}{t},\;\; \phi_\hbar(t)=\frac{\lambda}{t^2}, \;\; \tau(\lambda)=\lambda,\;\;\si(\lambda)=\frac{1}{\sqrt{\lambda}}. $$

thus, the conditions ($\mathbf{C}_1$) and ($\mathbf{C}_2$) are   satisfied.   To  verify ($\mathbf{C}_3$) observe   that

$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)= (\lambda +\sqrt{\lambda}x)-\lambda\log(\lambda+\sqrt{\lambda}x)- \lambda -\lambda \log \lambda $$

$$=\sqrt{\lambda} x-\lambda\log\Bigl(1+\frac{x}{\sqrt{\lambda}}\Bigr). $$

The  condition $\mathbf{C}_3$ now follows from the  Taylor expansion of $\log(1+s)$ at $s=0$. To prove  $\mathbf{C}_4$ we observe that in this case $\phi_\hbar$ is continuous, so it suffices to check (\ref{B}), i.e., $\vfih(\pm 1)=O(1)$.  In this case we have

$$\vfih'(\pm1)=\sqrt{\lambda}-\lambda\log\Bigl(1\pm \frac{1}{\sqrt{\lambda}}\Bigr), $$

and (\ref{B}) follows by using the Taylor expansion of $\log(1+s)$ at $s=0$.  In this case

$$e^{-\phi_\hbar(\tau)}=\lambda^\lambda e^{-\lambda}$$

and we deduce

$$ \Gamma(\lambda+1)\sim \sqrt{2\pi}\lambda^{\lambda-\frac{1}{2}}e^{-\lambda}\;\;\mbox{as $\lambda\to\infty$}. $$

Example 3.  Suppose that $w:[0,\infty)\to\bR$ is a smooth function such that

$$w(t), \;w'(t),\;\;w''(t) >0,\;\;\forall t>T>1. $$

Then

$$\mu_\lambda=\int_0^\infty t^\lambda e^{-w(t)} dt <\infty,\;\;\forall \lambda >0$$

and  I would like to investigate the behavior of $\mu_\lambda$ as $\lambda\to \infty$.  The quantitites $\mu_k$, $k\in\mathbb{Z}_{\geq 0}$, are the moments of the measure $e^{-w(t)}dt$  on the positive semiaxis.  Note that

$$\mu_\lambda=\int_0^T t^\lambda e^{-w(t)} dt+\int_T^\infty t^\lambda  e^{-w(t)} dt. $$

Observe that

$$ \int_T^\infty t^\lambda e^{-w(t)} dt \geq T^\lambda\int_T^\infty e^{-w(t)} dt, $$

while

$$ T^{-\lambda} \int_0^T t^\lambda e^{-w(t)} dt= \int_0^T \left(\frac{t}{T}\right)^\lambda e^{-w(t)} dt \to 0\;\;\mbox{as $\lambda\to \infty$}. $$

Thus, as $\lambda \to \infty$ we have

$$\mu_\lambda\sim I_\lambda:=\int_T^\infty t^\lambda e^{-w(t)}  dt. $$

Observe that

$$I_\lambda =\int_T^\infty e^{-\phi_\lambda(t)} dt,\;\;\phi_\lambda(t)=w(t)-\lambda\log t. $$

We will show that the Laplace method is applicable in this case if we  assume that  we have an asymptotic estimate $\newcommand{\bZ}{\mathbb{Z}}$

$$ tw'(t)\sim A t^\alpha(\log t)^p\;\;\mbox{as $t\to\infty$},\;\; A>0\;\;,\alpha>1,\;\;p\in {\bZ}_{\geq 0}\tag{$\ast$} \label{ast} $$

which is twice differentiable.

 The critical points of $\phi_\lambda$ are solutions of the equality

$$\lambda=tw'(t). $$

Since  $tw'(t)$ is increasing and     $tw'(t)\to\infty$ as $t\to \infty$  we deduce that the above equation  has a unique solution $\tau=\tau(\lambda)$  for $\lambda \gg 0$.    The correspondence  $\lambda\to \tau(\lambda)$ is smooth  and $\tau(\lambda)\to \infty$ as $\lambda\to \infty$.   This proves $\mathbf{C_1}$. In view of (\ref{ast}) we deduce that

$$\lambda(\tau)\sim At^\alpha(\log \tau)^p \;\;\mbox{as $\tau\to \infty$}. $$

Observe that

$$\phi_\lambda''(\tau)=w''(\tau)+\frac{\lambda}{\tau^2} = w''(\tau)+\frac{w'(\tau)}{\tau}=\frac{\tau w''(\tau)+w'(\tau)}{\tau}. $$

Hence

$$\si =\sqrt{\frac{\tau}{\tau w''(\tau)+w'(\tau)}}=\sqrt{\frac{\tau}{\lambda'(\tau)},}\;\;\frac{\tau}{\si}= \sqrt{\tau \lambda'(\tau)}=\sqrt{\tau^2w''(\tau) +\tau w'(\tau)}\to \infty. $$

This proves  $\mathbf{C_2}$. Observe

$$ \phi_\lambda(\tau+\si x)-\phi_\lambda(\tau)= w(\tau+ \si x)-w(\tau) -\lambda\log \Bigl(1+\frac{\si}{\tau}x \Bigr) = w(\tau+ \si x)-w(\tau) -\tau w'(\tau)\log \Bigl(1+\frac{\si}{\tau}x \Bigr) $$

$$=\si^2\left( w''(\tau) +\frac{w'(\tau)}{\tau}\right)\frac{x^2}{2} +\frac{\si^3x^3}{3!}w^{(3)}(\theta)  + O\left(\frac{\si^3w'(\tau) x^3}{\tau^2}\right),$$

 for some $\theta=\theta(\tau,x)\in [\tau,\tau+\si x]$. Now observe that

$$ \frac{\si^3 w'(\tau)}{\tau^2}=\frac{\si}{\tau}  \frac{\si^2w'(\tau)}{\tau}=\frac{\si}{\tau}\frac{w'(\tau)}{\tau w''(\tau)+ w'(\tau)}\leq \frac{\si}{\tau}\to 0,\;\;\mbox{as $\tau\to \infty$}. \tag{1}\label{2}$$

To verify $\mathbf{C}_3$ we need to prove that

$$\lim_{t\to \infty}\si^3w^{(3)}(\theta)=0.\tag{2} \label{1} $$

It is time to use (\ref{ast}). We have

$$w'(t)+ tw''(t) \sim   A\alpha t^{\alpha-1}(\log t)^p+ Apt^{\alpha-1} (\log t)^{p-1}, $$


$$2w''(t) +tw^{(3)}(t)\sim A\alpha(\alpha-1)t^{\alpha-2}(\log t)^p+A\alpha pt^{\alpha-2}(\log t)^{p-1}+Ap(\alpha-1) t^{\alpha-2} (\log t)^{p-1}+ Ap(p-1)t^{\alpha-2}(\log t)^{p-2}. $$

Now observe that

$$ t^3w^{(3)}(t)\sim -2t^2 w''(t)+A\alpha(\alpha-1)t^{\alpha}(\log t)^p+Ap t^\alpha(2\alpha-1) (\log t)^{p-1}+ Ap(p-1)t^{\alpha}(\log t)^{p-2}, $$

and

$$ -2t^2w''(t)\sim  2tw'(t) -2A\alpha t^{\alpha}(\log t)^p-2 Apt^{\alpha} (\log t)^{p-1} = -2A(\alpha-1)t^\alpha(\log t)^p -2Apt^\alpha (\log t)^{p-1}.\tag{3}\label{3}$$

Hence
$$t^3w^{(3)} (t) \sim  At^{\alpha}\Bigl(\;(\alpha-1)(\alpha-2)(\log t)^p + p(2\alpha-3)(\log t)^{p-1}+ p(p-1)(\log t)^{p-2}\;\Bigr).$$
Also

$$\frac{\si}{\tau}=\frac{1}{\sqrt{\tau\lambda'(\tau)}} \sim \frac{1}{\sqrt{A\alpha}}\tau^{-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}},\;\;\si \sim \frac{1}{\sqrt{A\alpha}}\tau^{1-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}}. \tag{4}\label{4}$$


We distinguish three cases.

 Case 1. $\alpha \neq 2$  In this case $\si\to 0$ and  $\newcommand{\ve}{\varepsilon}$


$$\theta= \tau(1 +c\frac{\si}{\tau}x) $$

$$w^{3}(\theta) =Ct^{\alpha-3} (\log t)^p\bigl( \;1+\ve(t)\;\bigr), $$

where $C=A(\alpha-1)(\alpha-2)\neq 0$ and $\ve(t)\to 0$ as $t\to \infty$. Then


$$ \si^3w^{(3)}(\theta) \sim\frac{C}{\sqrt{A\alpha}} t^{-\frac{\alpha}{2}} (\log t)^{-\frac{p}{2}} \to 0. $$

This proves (\ref{1}).

Case 2.   $\alpha=2$ $p=1$ In this case (\ref{ast})  implies

$$\lim_{t\to\infty}w^{(3)}(t)=0,\;\;\si(t)=O(1)\;\;\mbox{as $t\to\infty$}. $$

which clearly implies (\ref{1}).

Case 3.  $\alpha=2, p>0$ Proceed as in Case 1.

Finally, we  want to check $\mathbf{C}_4$. In our case $\phi_h$ is convex and we will verify (\ref{B}).  We have

$$\vfih(x) = \frac{d}{dx}\Bigl( w(\tau+\si x) -w(\tau)-\tau w'(\tau)\log\Bigl(1+\frac{\si}{\tau}x\Bigr)\;\Bigr) $$

$$=\si w'(\tau+\si x)-\frac{\si w'(\tau)}{1+\frac{\si}{\tau}x}=\si w'(\tau)\left(1-\frac{1}{1+\frac{\si}{\tau}x}\right)+\frac{1}{2}\si^2x^2w''(\theta)=\frac{\si^2 w'(\tau)}{\tau}x^2+ \frac{1}{2}\si^2x^2w''(\theta)+ O\left( \frac{\si^3w'(\tau)}{\tau^2} x^2\right), $$

for some  $\theta\in [\tau,\tau+\si]$.    As in the proof of (\ref{2}) we deduce that

$$ \frac{\si^2w'(\tau)}{\tau}=O(1),\;\; \frac{\si^3w'(\tau)}{\tau^2}=o(1). $$

We only need to verify that

$$ \si^2w''(\theta)=O(1). $$

This follows from  (\ref{3}) and (\ref{4}}).

A proof of the abc conjecture was recently posted

A few weeks ago (August 2012)  Shinichi Mochizuki has  posted  a proof of the   ABC conjecture.  He proves  a more general result which  has many nontrivial consequences, one being the  ABC conjecture.

I am not a number theorist  and I have no  chance of understanding Mochizuki's arguments (whichh seem algebraic geometric in nature)   and I wonder if there is "official" confirmation of their validity.

Update Sept. 5, 2012  A few more details about this announcement  can be found  here.

On the Banchoff-Chmutov surfaces

This discusses a  question posed on MathOverflow   by Leon Lampret.


Denote by $T_n$ the  Chebyshev polynomial of the first kind and degree $n$ uniquely determined by the  equality $\newcommand{\bR}{\mathbb{R}}$

$$ T_n(\cos t)=\cos nt,\;\;\forall t\in\bR. $$


Denote by $U_n$ the  Chebyshev polynomial of degree $n$ and of the second kind uniquely determined  by the equality

$$ U_n(\cos t)=\frac{\sin (n+1) t}{\sin t},\;\;t\in\bR. $$

They are related by two equalities

$$T_n'= n U_{n-1}, \;\; T_n(x)^2 -(x^2-1)U_{n-1}(x)^2=1. \tag{1}$$


The polynomial $T_n$ has $n$ distinct real zeros located in $(-1,1)$ and thus, by Rolle's theorem, all its critical points are located in $(-1,1)$.


The polynomial $T_n$ is a solution of the second order linear differential equation

$$ (1-x^2)y''-xy'+n^2 y=0, $$

which shows that all the critical points of $T_n$ are nondegenerate.

The  Banchoff-Chmutov surface   $Z_n$ is defined by

$$Z_n:=\Bigl\lbrace (x,y,z)\in\bR^3;\;\; \underbrace{T_n(x)+T_n(y)+ T_n(z)}_{=: f_n(x,y,z)} =0\;\bigr\rbrace. $$


Remark.  (a)   $Z_n$ is a smooth submanifold of $\bR^3$. To see this, we  rely on the implicit function theorem. Observe that if  $df_n(x_0,y_0,z_0)=0$, then

$$ T_n'(x_0)=T_n'(y_0)=T_n'(z_0) =0. $$

In particular (1) implies

$$U_{n-1}(x_0)=U_{n-1}(y_0)=U_{n-1}(z_0)=0. $$

Invoking (1) again we deduce

$$T_n(x_0)=T_n(y_0)=T_n(z_0)=1.$$

This  shows that $(x_0,y_0,z_0)\not\in Z_n$.

(b) If $n$ is even, then $Z_n$ is compact. Indeed, in this case $T_n$ is an even polynomial and

$$\lim_{|x|\to\infty} T_n(x)=\infty. $$

This implies that $Z_n$ is  bounded, thus compact since it is obviously closed.


Assume that $n$ is   even  and consider the function

$$f: Z_n\to \bR,\;\; h(x,y,z)= z. $$

The critical points of  $h$.  A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff the gradient of $f_n$ points in the $z$-direction, i.e.,

$$  T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y) $$

Now the critical points  of $T_n$ are all located in the interval $[-1,1]$ and can be  easily determined from the defining equality

$$ T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A} $$

so that

$$ T_n'(\cos t) = n\frac{\sin nt}{\sin t} $$

This  nails the critical points  of $T_n$ to

$$u_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$

Note that

$$ T_n(u_k)= \cos k\pi=(-1)^k $$

so that the critical points of $h$ on the surface $Z_n$  are


$$\bigl\lbrace (u_j,u_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace.  $$

Now we need to count the solutions of the equations


$$T_n(x)=0,\;\pm 2. $$

The equation $T_n(x)=0$ has $n$ solutions, all situated  in $[-1,1]$.

On the interval $[-1,1]$  we deduce from  (A) that $|T_n|\leq 1$.  The polynomial $T_n$ is even and is increasing on $[1,\infty)$.  (It is increasing since $T_n'(x)\neq 0$ for $x\geq 1$.)We conclude that the equation $T_n(x)=-2$ has no solutions, while the equality $T_n(x)=2$ has two solutions.   Thus the critical set of $h$  splits  into three parts

$$ C_0=\big\lbrace\, (u_j,u_k,z);\;\;1\leq j\leq n-1,\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\,\big\rbrace, $$

$$ C_2^+=\big\lbrace \,(u_j,u_k,z);\;\; j,k\in [1,n-1]\cap 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\,\big\rbrace, $$


$$ C_2^-= \lbrace (u_j,u_k,z);\;\;j,k\in [1,n-1]\cap 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace. $$


From the above discussion we deduce  that the points in $C_2^-$ are minima and the points in $C_2^+$ are maxima.


The function $h$ is a Morse function.         Note that $h$  is defined implicitly, by solving for $z$ in the  nonlinear equation

$$  T_n(x)+ T_n(y) + T_n(z)=0. \tag{2} $$

Suppose that $(x_0,y_0,z_0)$ is a critical point of $h$,  $x_0=u_j$, $y_0=u_k$. Differentiating (2) near this critical point we deduce $\newcommand{\pa}{\partial}$

$$ \frac{\pa z}{\pa x}T_n'(z) +T_n'(x) =0,\;\;  \frac{\pa z}{\pa y} T_n'(z) + T_n'(y)=0. $$

 Differentiating the above again we deduce that

$$\frac{\pa^2 z}{\pa x\pa y}|_{(u_j,u_k)}=0,  \tag{3}$$

$$\frac{\pa^2 z}{\pa x^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_j)= 0,\;\; \frac{\pa^2 z}{\pa y^2}|_{(u_j,u_k)} T_n'(z_0) +T_n''(u_k)= 0. $$

Now let us  observe that $T_n(z_0)=0$ or $T_n(z_0)=2$.  In the first case $T_n'(z_0)\neq 0$ because $T_n$ has only simple zeros. In the second case $T_n'(z_0)\neq 0$ because  in this case $|z_0|>1$ and $T_n$ has no critical points outside $(-1,1)$.  Hence

$$\frac{\pa ^2 z}{\pa x^2}|_{(u_j,u_k)}= -\frac{T_n''(u_j)}{T_n'(z_0)},\;\;\frac{\pa ^2 z}{\pa y^2}|_{(u_j,u_k)}= -\frac{T_n''(u_k)}{T_n'(z_0)}. \tag{4} $$


The point $u_k$ is a local minimum for $T_n$ if $k$ is odd and a local maximum if $k$ is even.  Moreover, these are nondegenerate critical points  of $T_n$.


This proves that all the critical points of $h$ are nondegenerate. Moreover, if $(u_j, u_k)\in C_0$ then the numbers  $T_n''(u_j)$ and $T_n''(u_k)$ have opposite signs  and invoking (3) and (4) we deduce that  this point is a saddle point.



Thus  the Euler characteristic of $Z_n$ is

$$ \chi(Z_n)= {\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0. $$




Now  observe that

$$ {\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$

$$ {\rm card} \; C_0 = n\Bigl( \frac{n(n-2)}{4}+\frac{n(n-2)}{4}\Bigr)=\frac{n^2(n-2)}{2}. $$

Thus the Euler characteristic of $Z_n$  is

$$\chi(Z_n)=\frac{n^2(3-n)}{2}. \tag{E} $$

Here are  images of $Z_2, Z_4, Z_6$, courtesy of  StackExchange (hat tip to Igor Rivin)

Z_2

Z_6

Z_4

The above computations do not  explain whether $Z_n$ is connected or not.  To check that it suffices to look at the critical values  of the above  function corresponding to saddle points. These critical values are the zeros $\zeta_1<\dotsc <\zeta_n$ of $T_n$. The level set

$$ Z_n\cap \lbrace z=\zeta_k\rbrace $$

is the algebraic curve

$$ T_n(x)+T_n(y)=0. \tag{C} $$

This forces $|x|,|y|\leq 1$ because $T_n(x)> 1$ for $|x|> 1$ and $|T_n(x)\leq 1$ for $|x|\leq 1$.  We can use the  *homeomorphism*

$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1] $$

to give an alternate description to (C).  It is  the singular curve  inside  the square $[0,\pi]\times [0,\pi]$ with coordinates $(s,t)$ described by

$$\cos ns+ \cos nt =0.$$

This can be easily visualized as the intersection of the square with the  grid

$$ s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n} $$

which is  connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.


Example.    The equality (E) predicts that $\chi(Z_6)=-54$. Let us  verify this directly.    Here is a more detailed image of $Z_6$.

Z_6

We can give an alternate description of $Z_6$ as follows. Consider the $1$-dimensional simplicial complex $C\subset \bR^3$ depicted below

 The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhoof $T$ of $C$ in $\bR^3$ and thus

$$\chi(Z_6)=\chi(\pa T)=2\chi(T)=2\chi(C). $$

Thus formula (E) predicts that

$$\chi(C)= -27.$$

Let us verify this directly.   The vertex set of $C$  consists of

• 8 Green vertices of degree 3.
• 12 Red vertices of degree 4.
• 6 Blue vertices of degree 5.
• 1 Black vertex of degree 6.

The number $V$ of vertices of $C$ is thus

$$V= 8+12+6+1=27. $$

The total number $E$ of edges of $C$   is half the sum of degrees of vertices so that

$$ E=\frac{1}{2}( 3\times 8+ 4\times 12+ 5\times 6+ 6\times 1)= \frac{1}{2} (24+48+30+6)=54. $$

Hence 

$$\chi(C)= 27-54 =-27. $$

Bill Thurston has passed away

I have found out that Bill Thurston has passed away last night after a fight with melanoma.            This is  a premature closing of an amazing and transformative  chapter in the history of mathematics.      Like all the greats,    there won't  be anyone like  Thurston.     Rest in peace!

Invariants of symmetric matrices

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{SO}{SO}$  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\eQ}{\mathscr{Q}}$ $\DeclareMathOperator{\tr}{tr}$

Fix an integer $m>2$, and denote by $\eS_m$  the  space of real, symmetric   $m\times m$ matrices. The group  $\SO(m)$ of  orthogonal transformations  of $\bR^m$ acts on $\eS_m$ by conjugation. Fix a unit vector  $\eta\in\bR^m$. We obtain a subgroup   $\SO(m-1)$ of $\SO(m)$ consisting of transformations that fix the vector $\eta$.   Denote by $\eQ_m$ the space  of   $\SO(m-1)$-invariant  homogeneous quadratic polynomials  on $\eS_m$.   Then  $\eQ_m$ is spanned by the quadratic polynomials

$$   A\mapsto \tr A^2,\;\; A\mapsto (\tr A)^2, \;\;A\mapsto |A\eta|^2, $$

$$ A\mapsto (A\eta,\eta)^2,\;\; A\mapsto (\tr A)(A\eta,\eta), $$

where   $(-,-)$ denotes the canonical inner product on $\bR^m$.

There is   a simple proof  of this fact due to Robert Bryant.

A simple homotopic trick

Here is a simple fact, which seems to be well known to   homotopy theorists.  It might come in handy.

 Suppose that  we have two fibrations

$$ f:Y\to B,\;\; \pi:E\to B, $$


such that  $E$ is contractible.  The pullback to $Y$ of the fibration $\pi: E\to B$  via the map $f$ is a new fibration $g: X\to Y$.   Then the homotopy fiber  $Z$ of $Y\to B$ is homotopic to  $X$.


 Indeed we have a fibration $F:X\to E$ with t homotopy fiber $Z$.  Since $E$ is contractible we  deduce that $X$  is homotopic to $Z$.

This is particularly useful when $X\to Y$ is a principal  $G$-bundle classified by a map $f: Y\to BG$.  The map $f$ can be assumed to be a fibration. The homotopy fiber of $f$ is then $X$.   From the Leray-Serre spectral sequence we obtain a spectral sequnce  converging to the cohomology of $Y$ with $E_2$-term

$$E_2^{p,q}= H^p\bigl(\;BG, \tilde{H}^q(X)\;\bigr), $$

where $\tilde{H}^q(X)$ denotes a local system on $BG$ with stalk $H^q(X)$.

A wonderful math video archive

CUNY has a wonderful video archive going back to the 80s.     What a great idea!  You get to see some of the classics that are no longer among  the living. Hat tip to Richard Hind!

Injective modules are hard to deal with

This is related to a Mathoverflow question. The proofs of existence of injective  resolutions require the axiom of choice, in one form or another. Translation:  these proofs  are not constructive, so there are no general algorithms for producing such objects.  This becomes a painful issue in concrete  situations.     This  has similarities with another famous existence result, the Hahn-Banach theorem  which postulates the existence of  continuous linear functionals with certain properties. It is particularly useful for existence theorems for PDE's. Unfortunately it gives you no guide for finding those solutions.

Some Berry phase computations

 $\newcommand{\bp}{{\boldsymbol{p}}}$ $\newcommand{\bq}{\boldsymbol{q}}$  $\newcommand{\bsi}{\boldsymbol{\sigma}}$ $\newcommand{\pa}{\partial}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ii}{\boldsymbol{i}}$  $\newcommand{\bsS}{\boldsymbol{S}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$

I'm attending this seminar on topological insulators. I still do not understand the physics, but here is some mathematics.

Consider the Pauli matrices

$$\bsi_x :=\left[\begin{array}{cc} 0 & 1\\
1 & 0
\end{array}
\right],\;\;\bsi_y:=\left[
\begin{array}{cc} 0 & -\ii \\
\ii & 0
\end{array}
\right],\;\;\bsi_z:=\left[
\begin{array}{cc} 1 &0\\
0 & -1
\end{array}
\right]. $$

For $\bp =(x,y,z)\in \bR^3$ we set

$$\bsi(\bp) = x\bsi_x +y\bsi_y+ z\bsi_z=\left[\begin{array}{cc} z & x-\ii y\\
x+\ii y & -z
\end{array}
\right].$$
The family of hermitian  matrices $\bsi(\bp)$ satisfy the  Clifford identities

$$\bsi(\bp)\cdot\bsi(\bq) +\bsi(\bq)\cdot \bsi(\bp) =2\bp\cdot \bq,\;\;\forall \bp,\bq\in\bR^3. $$

In particular

$$\bsi(\bp)^2= |\bp|^2,\;\;\forall\bp\in\bR^3, $$

so that the eigenvalues  of $\bsi(\bp)$ are $\pm |\bp|$.    Consider the unit sphere

$$\bsS^2:=\bigl\lbrace\;\bp\in\bR^3;\;\;|\bp|=1\;\bigr\rbrace. $$

We obtain a family  of  hermitian matrices $\lbrace\bsi(\bp)\rbrace_{\bp\in\bsS^2}$ with eigenvalues $\pm 1$.  Denote by $V_{\bp}$ the $1$-eigenspace      of the matrix $\bsi(\bp)$,  $\bp\in \bsS^2$.   Suppose $\bp=(x,y,z)$ is not the South Pole $P_-=(0,0,-1)$, i.e.,   $z\neq -1$. We set $u=x+\ii y$.  To find a basis  of $V_\bp$ we  need to solve the system

$$ \left[\begin{array}{cc}
 z & \bar{u} \\
u & -z
\end{array}
\right] \cdot \left[
\begin{array}{c}
z_1\\
z_2
\end{array}
\right]=\left[\begin{array}{c}
z_1\\
z_2
\end{array}
\right]. $$

We deduce that

$$z_2= \frac{u}{1+z} z_1.$$

The stereographic projection from the  South Pole  is the map

$$\zeta: \bsS^2\setminus P_-\to\bR^2=\bC$$

that associates to each  point $\bp\in\bsS^2\setminus  P_-$ the  intersection of the line $P_-\bp$ with  the coordinate plane $z=0$. Concretely, if $\bp=(x,y,z)$, then

$$\zeta(\bp)= \frac{u}{1+z}. $$

We deduce that $V_\bp$ is spanned by  the vector

$$\vec{z}(\bp)= (1,\zeta(\bp)). $$

We write $\zeta=\zeta_1+\ii \zeta_2$ so that we can use $(\zeta_1,\zeta_2)$ as coordinates on $\bsS^2\setminus P_-$.  Consider the normalized vector

$$|\Psi_\bp\ran :=\frac{1}{|\vec{z}(\bp)|} \vec{z}(\bp)= \frac{1}{\sqrt{1+|\zeta|^2}}(1,\zeta)$$.


Set

$$ G(\zeta):= \sqrt{1+|\zeta|^2}. $$  Note  that

$$d\left(\frac{1}{G}\right) =-\frac{dG}{G^2}.$$


The Berry connection  $\nabla$ is obtained from the equality

$$\nabla|\Psi_\bp\ran = |\Psi_\bp\ran\lan \Psi_\bp|d\Psi_\bp\ran. $$


Observe that

$$ d|\Psi_\bp\ran = -\frac{dG}{G^2}(1,\zeta)+\frac{1}{G} (0,d\zeta),$$

$$ \lan \Psi_\bp |d\Psi_\bp\ran =-\frac{(1+|\zeta|^2)dG}{G^3}+\frac{1}{G^2}\bar{\zeta} d\zeta $$

$$= -\frac{dG}{G} +\frac{1}{G^2} \bar{\zeta} d\zeta=\underbrace{-d\log G+\frac{1}{1+|\zeta|^2}\bar{\zeta} d\zeta}_{=:\omega}. $$

The  $1$-form associate to the  Berry connection is the above $\omega$. The curvature of the Berry connection is

$$\Omega= d\omega = -\frac{d|\zeta|^2}{(1+|\zeta|^2)^2}\bar{\zeta}d\zeta + \frac{1}{1+|\zeta|^2}d\bar{\zeta}\wedge d\zeta=-\frac{|\zeta|^2}{(1+|\zeta|^2)^2} d\bar{\zeta}\wedge d\zeta +\frac{1}{1+|\zeta|^2} d\bar{\zeta}\wedge d\zeta$$

$$ =\frac{1}{(1+|\zeta|^2)}  d\bar{\zeta}\wedge d\zeta =\frac{2\ii}{(1+|\zeta|^2)^2} d\zeta_1\wedge d\zeta_2. $$

To compute the integral of $\Omega$ we use polar coordinates, $\zeta=r e^{\ii\theta}$ so that

$$d\zeta_1\wedge d\zeta_2 =rdrd\theta$$

and

$$\int_{\bC}\Omega=\ii \int_0^{2\pi}d\theta\int_0^\infty \frac{2rdr}{(1+r^2)^2} dr=\ii\int_0^{2\pi} d\theta\int_0^\infty \frac{d(1+r^2)}{(1+r^2)^2} =2\pi \ii. $$


The Chern form of the Berry connection is $\frac{\ii}{2\pi} \Omega$ and the first Chern number is

$$ \frac{\ii}{2\pi}\int_{\bC} \Omega =-1. $$

Random functions on tori

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bC}{\mathbb{C}}$  $\newcommand{\bT}{\mathbb{T}}$ $\newcommand{\eE}{\mathscr{E}}$ $\newcommand{\ve}{{\varepsilon}}$ $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$ $\newcommand{\bu}{\boldsymbol{u}}$ $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\var}{\boldsymbol{var}}$ $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bsU}{\boldsymbol{U}}$  $\newcommand{\vfi}{\varphi}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\teE}{\widetilde{\mathscr{E}}} $ $\newcommand{\pa}{\partial}$  $\DeclareMathOperator{\Hess}{\boldsymbol{Hess}}$ $\DeclareMathOperator{\diag}{Diag}$ $\newcommand{\one}{\boldsymbol{1}}$





Consider the $m$-dimensional torus

$$\bT^m:=\bR^m/(2\pi\bZ)^m $$

equipped with the  flat metric

$$g:=\sum_{j=1}^m (d\theta^j)^2. $$

It has volume ${\rm vol}_g(\bT^m) =(2\pi)^m.$   The eigenvalues of the corresponding  Laplacian are

$$|\vec{k}|^2,\;\;\vec{k}=(k_1,\dotsc, k_m)\in\bZ^m. $$

 For $\vec{\theta}=(\theta^1,\dotsc, \theta^m) \in\bR^m$ and $\vec{k}\in\bZ^m$ we set

$$ \lan\vec{k},\vec{\theta}\ran =\sum_jk_k\theta^j. $$

Denote by $\prec$ the lexicographic order on $\bR^m$.  An  orthonormal basis of $L^2(\bT^m)$ is given by the  functions $(\Psi_{\vec{k}})_{\vec{k}\in\bZ^m}$, where


$$ \Psi_{\vec{0}}(\vec{\theta}) =\frac{1}{(2\pi)^{\frac{m}{2}}}$$,


$$\Psi_{\vec{k}}(\vec{\theta})=\frac{\sqrt{2}}{(2\pi)^{m/2}} \sin\lan \vec{k},\vec{\theta}\ran, \;\;\vec{k}\succ\vec{0}, $$


$$ \Psi_{\vec{k}}(\vec{\theta})=\frac{\sqrt{2}}{(2\pi)^{m/2}} \cos\lan\vec{k},\vec{\theta}\ran,\;\;\vec{k}\prec\vec{0}. $$



Fix a  nonnegative Schwartz function $w\in \eS(\bR)$, set $w_\ve(t)=w(\ve t)$ and consider the random  function

$$ \bu_\ve(\vec{\theta})=\sum_{\vec{k}\in\bZ^m} X_{\vec{k}}\Psi_{\vec{k}}(\vec{\theta}), $$

where $X_{\vec{k}}$ are independent  Gaussian random variables with mean $0$ and variances

$$\var(X_{\vec{k}})= w(\ve|\vec{k}|). $$

We denote by $N(\bu_\ve)$ the number of critical points  of $\bu_\ve$ and  by $N_\ve$ its expectation

$$ N_\ve =\bsE\Bigl(\; N(\bu_\ve)\;\Bigr). $$

A simple computation shows that the covariance kernel of this random function is

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)= \frac{1}{(2\pi)^m}\sum_{\vec{k}\in\bZ^m }w(\ve|\vec{k}|)e^{-\ii\lan\vec{k}, \vec{\theta}_2-\vec{\theta_1}\ran}. $$

Set $\vec{\theta}:=\vec{\theta}_2-\vec{\theta}_1$ and define $\phi:\bR^m\to\bC$ by

$$\phi(\vec{x})=e^{-\ii\lan\vec{x},\frac{1}{\ve}\vec{\theta}\ran}  w(|\vec{x}|). $$

We deduce that

$$ \eE^\ve(\vec{\theta}_1,\vec{\theta}_2)=\frac{1}{(2\pi)^m}\sum_{\vec{k}\in\bZ^m}  \phi(\ve\vec{k}). $$
Using Poisson formula we deduce  that for any $a>0$ we have

$$\sum_{\vec{k}\in\bZ^m}\phi\left(\frac{2\pi}{a}\vec{k}\right)=\left(\frac{a}{2\pi}\right)^m \sum_{\vec{\nu}\in\bZ^m}\widehat{\phi}(a\vec{\nu}), $$

where  for any $f\in\eS(\bR^m)$ we denote by $\widehat{f}(\xi)$ its Fourier transform

$$\widehat{f}(\xi)=\int_{\bR^m} e^{-\ii\lan\xi,\vec{x}\ran} f(\vec{x})|d\vec{x}|. $$

If we let  $\frac{2\pi}{a}=\ve$, then we deduce

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)=\frac{1}{(2\pi\ve)^m} \sum_{\vec{\nu}\in\bZ^m}\widehat{\phi}\left(\frac{2\pi}{\ve}\vec{\nu}\right). $$


Let $v:\bR^m\to \bR$,  $v(\vec{x})=w(|\vec{x}|) $. Then

$$\widehat{\phi}(\xi)=\widehat{v}\Bigl(\;\xi+\frac{1}{\ve}\theta\;\Bigr). $$

Hence

$$\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)= \frac{1}{(2\pi\ve)^m}\sum_{\vec{\nu}\in\bZ^m}\widehat{v}\left(\frac{1}{\ve}\vec{\theta}+\frac{2\pi}{\ve}\vec{\nu}\right). $$

Now observe that if $|\theta| \ll 2\pi$, then for $\vec{\nu}\in\bZ^m\setminus 0$ then for any $N>0$ there exists a constant  $C_N>0$ such that

$$ \left|\widehat{v}\left(\frac{1}{\ve}\vec{\theta}+\frac{2\pi}{\ve}\vec{\nu}\right)\right|\leq   C_N\ve^N|\nu|^{-N}. $$

We deduce that

$$ \eE^\ve(\vec{\theta}_1,\vec{\theta}_2) = \frac{1}{(2\pi\ve)^m}\left(\widehat{v}\left(\; \frac{1}{\ve}\vec{\theta}\;\right)+O\bigl(\; \ve^N\;\bigr)\;\right),\;\;\forall N>0. $$

The last asymptotic expansion can be  differentiated with respect to $\vec{\theta}_1$ and $\vec{\theta}_2$.

Now define the random function

$$\bsU_\ve:\bT^m\times \bT^m\to \bR,\;\;\bsU_\ve(\vec{\theta},\vec{\vfi})=\bu_\ve(\vec{\theta})+\bu_\ve(\vec{\vfi}). $$

We denote  by $N(\bsU_\ve)$ the number of critical points of $\bsU_\ve$ situated  outside the diagonal. Note that

$$ N(\bsU_\ve)= N(\bu_\ve)^2-N(\bu_\ve). $$

We would like to  understand the behavior of the expectation of $N(\bsU_\ve)$ as $\ve\searrow 0$.

The covariance kernel  of $\bsU_\ve$ is the function

$$ \widetilde{\eE}^\ve(\vec{\theta}_1,\vec{\vfi}_1; \vec{\theta}_2,\vec{\vfi}_2) =\eE^\ve(\vec{\theta}_1,\vec{\theta}_2)+\eE^\ve(\vec{\theta}_1,\vec{\vfi}_2)+\eE^\ve(\vec{\vfi}_1,\vec{\theta}_2)+\eE^\ve(\vec{\vfi}_1,\vec{\vfi}_2)$$

$$= \frac{1}{(2\pi\ve)^m}\Bigl( \;\widehat{v}(\;\ve^{-1}(\vec{\theta}_2-\vec{\theta}_1)\;)+ \widehat{v}(\ve^{-1}(\vec{\vfi}_2-\vec{\theta}_1)\;)\\
+\widehat{v}(\;\ve^{-1}(\vec{\theta}_2-\vec{\vfi}_1)\;) +\widehat{v}(\;\ve^{-1}(\vec{\vfi}_2-\vec{\vfi}_1)\;)+O(\ve^\infty)\;\Bigr). $$

Let us introduce the notation

$$\Theta:=(\vec{\theta},\vec{\vfi})\in\bT^m\times\bT^m , d(\Theta):=\vec{\vfi}-\vec{\theta}. $$


We  need to understand the quantities

$$\pa^\alpha_{\Theta_1}\pa^\beta_{\Theta_2}\teE^\ve(\Theta_1,\Theta_2)_{\Theta_1=\Theta_2=\Theta}=\bsE\bigl(\;\pa^\alpha_\Theta\bsU_\ve(\Theta)\cdot\pa^\beta_\Theta\bsU_\ve(\Theta)\;\bigr). $$

Note that $\widehat{v}(\xi)$ is radially symmetric, in fact it  can be written as $f(|\xi|^2)$  for some smooth function $f$. Indeed, we have  (see  Michael Taylor's notes;  he uses a different normalization for the Fourier transform.)


$$\widehat{v}(\xi)=\int_{\bR^m} v(|\vec{x}|) e^{-\ii\lan\xi,\vec{x}\ran}  =(2\pi)^{\frac{m}{2}}|\xi|^{1-\frac{m}{2}}\int_0^\infty v(r) J_{\frac{m}{2}-1}(r|\xi|) dr,$$

where $J_\nu$ denotes the Bessel function of first type and order $\nu$. For  any multi-indices $\alpha,\beta$ we have



$$(2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\theta}_2}\teE^\ve(\Theta,\Theta)=  \ve^{-m-|\alpha|-|\beta|} \Bigl(\; (-1)^{|\alpha|}\pa^{\alpha+\beta}_\xi \widehat{v}(0) + O(\ve^\infty)\,\Bigr), \tag{1}$$


$$ (2\pi)^m\pa^\alpha_{\vec{\vfi}_1}\pa^\beta_{\vec{\vfi}_2}\teE^\ve(\Theta,\Theta)= \ve^{-m-|\alpha|-|\beta|}\Bigl( (-1)^{|\alpha|} \pa^{\alpha+\beta}_\xi\widehat{v}(0) +O(\ve^{\infty})\;\Bigr). \tag{2}$$

The  main term of this asymptotics  is trivial if $|\alpha|+|\beta|$ is odd. Next


$$(2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\vfi}_2} \teE^\ve(\Theta,\Theta)= \ve^{-m-|\alpha|-|\beta}\Bigl( (-1)^{|\alpha|}\pa^{\alpha+\beta}_\xi\widehat{v}(\ve^{-1}d(\Theta) ) +O(\ve^\infty)\;\Bigr), \tag{3}$$


$$(2\pi)^m \pa^\alpha_{\vec{\vfi}_1}\pa^\beta_{\vec{\theta}_2}\teE^\ve(\Theta,\Theta) =\ve^{-m-|\alpha|-|\beta|} \Bigl(\;(-1)^{|\alpha|} \pa^{\alpha+\beta}_\xi \widehat{v}(\;-\ve^{-1}d(\Theta)\;)+ O(\ve^\infty)\;\Bigr)\\
=\ve^{-m-|\alpha|-|\beta|}\Bigl(\;(-1)^{|\beta|}\pa^{\alpha+\beta}_\xi\widehat{v}(\;\ve^{-1}d(\Theta)\;)+O(\ve^\infty)\;\Bigr).\tag{4}$$

For example  if $|\alpha|=2$ and $|\beta|=1$ we have

$$ (2\pi)^m\pa^\alpha_{\vec{\theta}_1}\pa^\beta_{\vec{\vfi}_1}\teE^\ve(\Theta,\Theta)= \ve^{-m-3}\Bigl(\;\pa^{\alpha+\beta}_\xi\widehat{v}(\;\ve^{-1}d(\Theta)\;)+O(\ve^\infty)\;\Bigr),\tag{3'} $$

Example 1.    Let us compute $\pa^\alpha_\xi f(|\xi|^2)$,  $|\alpha|\leq 4$.

We have

$$\pa_{\xi_i} f(|\xi|^2) = 2\xi_i f',\;\;\pa^2_{\xi_i\xi_j}f(|\xi|^2)= 2\delta_{ij} f' +4\xi_i\xi_j f'',$$

$$\pa^3_{\xi_i\xi_j\xi_k} f = 4\bigl(\; \delta_{ij}\xi_k+\delta_{ik}\xi_j+\delta_{jk}\xi_i\;\bigr)f''+8\xi_i\xi_j\xi_k f'''. $$

$$\pa^4_{\xi_i\xi_j\xi_k\xi_\ell} f(|\xi|^2)= 4\bigl(\;\delta_{ij}\delta_{k\ell}+\delta_{ik}\delta_{j\ell}+\delta_{jk}\delta_{i\ell}\;\bigr) f'' $$

$$+ 8\bigl(\; \delta_{ij}\xi_k\xi_\ell+\delta_{ik}\xi_j\xi_\ell+\delta_{jk}\xi_i\xi_\ell+\delta_{i\ell}\xi_j\xi_k+\delta_{j\ell}\xi_i\xi_k+\delta_{k\ell}\xi_i\xi_j\;\bigr) f''' +16\xi_i\xi_j\xi_k\xi_\ell f^{(4)}. $$

Example 2. Let's be more specific and  set $w(t)=e^{-t^2/2}$. Then $v(\vec{x})= e^{-|\vec{x}|^2/2}$, $f(s)=e^{-s}$ so that

$$\widehat{v}(\xi) = (2\pi)^{m} e^{-|\xi|^2/2}. $$


We can write

$$\widehat{v}(\xi) =(2\pi)^{m/2}\prod_{j=1}^m e^{-\xi_j^2/2}. $$

For any multi-index $\alpha=(\alpha_1,\dotsc, \alpha_m)$ we have


$$\pa^\alpha_\xi \widehat{v}(\xi) =(-1)^{|\alpha|}\underbrace{\left(\prod_{j=1}^m H_{\alpha_j}(\xi_j)\right) }_{=:H_\alpha(\xi)}\widehat{v}(\xi), $$

where $H_n$ denotes the $n$-th Hermite polynomial defined by

$$ \frac{d^n}{dx^n} e^{-x^2/2}= (-1)^nH_n(x) e^{-x^2/2}. $$


Let us point out that

$$ (-1)^{n+1}H_{n+1}(x) = (-1)^nH_n'(x) +(-1)^{n+1}xH_n(x), $$

$$ H_1(x)=x, \;\; H_2(x)=x^2-1,\;\;H_3(x) =x^3-3x,\;\; H_4(x)=x^4-6x^2 +3. $$



Observe that

$$\Hess\bsU_\ve(\Theta) =\Hess \bu_\ve(\vec{\theta})\oplus \Hess(\bu_\ve(\vec{\vfi}). $$



We need to understand the statistics of the following two random objects.

 $$d\bsU_\ve(\Theta),, $$


$$H_c(\Theta):= \bsE\Bigl(\;\Hess \bsU_\ve(\Theta)\;\bigr|\; d\bsU_\ve(\Theta)=0\;\Bigr),$$

as $d(\Theta)\to 0$, i.e.,  $\Theta$ approaches the diagonal in  $\bT^m\times \bT^m$. The covariance form $V_\ve$ of $d\bsU_\ve(\Theta)$ becomes singular as $d\to 0$.  Fortunately, something miraculous seems to be happening: as $d\to 0$  the Gaussian random variable  $H_c$ ecomes highly concentrated near the trivial matrix and in the limit it becomes the deterministic $0$-matrix. This leads to remarkable compensation is the Kac-Rice formula.

Excellent polynomial mappings

This  answers  a question of Lior Bary-Soroker on Mathoverflow.  $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\pa}{\partial}$

Suppose that $f:\bR^n\to \bR$ is a polynomial. For any $\vec{a}\in\bR^n$ we define

$$f_{\vec{a}}:\bR^n\to \bR,\;\; f_{\vec{a}}(\vec{x})=f(\vec{x})-\vec{a}\cdot\vec{x}. $$

We want to prove that  for generic $\vec{a}$ the function $f_{\vec{a}}$ is an excellent Morse function, i.e., all critical points are nondegenerate, and no two of them  are  on the same level set of $f_{\vec{a}}$.  We set

$$C(\vec{a})=\lbrace \vec{x}\in\bR^n;\;\;df_{\vec{a}}(\vec{x})=0\;\rbrace=\lbrace\;\vec{x}\in\bR^n;\;\;df(\vec{x})=\vec{a}\;\rbrace. $$

We say that a set $S\subset\bR^n$ is generic (in semialgebraic sense) if its complement is a semialgebraic set of dimension $<n$.  Sard's theorem implies that for generic $\vec{a}$ the critical set $C(\vec{a})$ is discrete. Being semialgebraic this implies that it is also finite.

Define $\newcommand{\eZ}{\mathscr{Z}}$

$$\eZ:=\bigl\lbrace\; (\vec{x},\vec{a})\in\bR^n\times\bR^n;\;\;df(\vec{x})=\vec{a}\;\bigr\rbrace. $$

The set $\eZ$ is semialgebraic. We denote by $\pi:\eZ\to\bR^n$ the projection

$$\eZ\ni (\vec{x},\vec{a})\to \vec{a}\in \bR^n. $$

For any $S\subset \bR^n$ we set $\eZ(S):=\pi^{-1}(S)$. There exists a generic set $G\subset \bR^n$ such that for any connected component  $A$ of $G$ the  induced  map $\pi:\eZ(A)\to A$ is a twice differentiable   covering.   Thus there exists a positive integer  $m=m(A)$ and  twice differentiable maps

$$ \vec{u}_1,\dotsc,\vec{u}_m:A\to \bR^n $$

such that

$$\vec{u}_i(\vec{a})\neq\vec{u}_j(\vec{a}), \;\;\forall i\neq j,  \;\;\vec{a}\in A, \tag{1}$$

  and the  set $\eZ(A)$ is the union of the graphs  of the maps $\vec{u}_i$. In other words, for any $\vec{a}\in A$ we have

$$C(\vec{a})=\lbrace\; \vec{u}_1(\vec{a}),\dotsc,\vec{u}_m(\vec{a})\;\rbrace. $$

We claim that $f_{\vec{a}}$ is excellent for generic  $\vec{a}\in A$. We argue by contradiction. Suppose that this is not the case. Then there exists a nonempty, connected  open subset $A_*\subset A$ and indices $i\neq j$ such that

$$ f( \vec{u}_i(\vec{a}) -f(\vec{u}_j(\vec{a}))=\vec{a}\cdot\bigl(\vec{u}_i(\vec{a})-\vec{u}_j(\vec{a})\;\bigr),\;\;\forall \vec{a}\in A_*. \tag{2}$$

We denote by $\pa_k f$ the $k$-th partial derivative of $f$ and we write

$$\vec{a}= (a^1,\dotsc, a^n),\;\;\vec{u}_i=(u_i^1,\dotsc, u_i^n). $$

Note that

$$ \pa_kf(\vec{u}_i(\vec{a}))=a^k=\pa_kf(\vec{u}_j(\vec{a})),\;\;\forall k=1,\dotsc, n.\tag{3}$$

 Derivating (2) with respect to the variable $a^\ell$  using the equality  (3) we deduce

$$\sum_k a^k\pa_{a^\ell}\bigl(u_i^k-u_j^k\bigr)= u^\ell_i(\vec{a})-u^\ell_j(\vec{a}) +\sum_k a^k\pa_{a^\ell}\bigl(u_i^k-u_j^k),\;\;\vec{a}\in A_*. $$


We deduce  that

$$ u^\ell_i(\vec{a})=u^\ell_j(\vec{a}),\;\;\forall\ell=1,\dotsc, n,\;\forall\vec{a}\in A_*. $$

This contradicts (1).

Some nagging Fourier transforms

In this  computer dominated  era,  the Tables of Integrals have lost   their  attraction. Fortunately, they are still  around,  and they  can get you out of many  jams.  I was looking  for  some  compact descriptions of   some integrals $\newcommand{\ii}{{\boldsymbol{i}}}$

$$ F^+_\nu(\xi, c) :=\int_c^\infty (x^2-c^2)^{ \nu-\frac{1}{2} } e^{-\ii\xi x} dx,$$

$$ F^-_\nu(\xi,c) :=\int_{-\infty}^{-c} (x^2-c^2)^{\nu-\frac{1}{2}} e^{-\ii\xi x} dx,$$


where $c>0$, $\xi>0$ and $\nu \in  (-\frac{1}{2},\frac{1}{2})$.  Fortunately, the venerable  Gradshteyn and Ryzik  (G& R)  had the answer.  (If you're younger than forty  it is very likely you haven't heard   of this   relique of the cold war.) Here is the beautiful answer (G & R, 6th Edition,  3.77.17, 3.771.9, page 436)


$$ F^+_\nu(\xi, c)= -\frac{\sqrt{\pi}}{2}\left(\frac{2c}{\xi}\right)^\nu\Gamma\Bigl(\nu+\frac{1}{2}\Bigr)\bigl(\;Y_{-\nu}(c \xi ) + \ii J_{-\nu}(c \xi )\;\bigr), $$


and 


$$ F_\nu^-(\xi, c) =-\frac{\sqrt{\pi}}{2}\left(\frac{2c}{\xi}\right)^\nu\Gamma(\Bigl(\nu+\frac{1}{2}\Bigr)\bigl(\; Y_{-\nu}(c\xi)-\ii J_{-\nu}(c\xi)\;\bigr), $$




where $J_\nu$  is the Bessel function of the first kind and  order $\nu$,


$$ J_\nu(z) := \frac{z^\nu}{2^\nu}\sum_{k=0}^\infty (-1)^k\frac{z^{2k}}{2^{2k}k!\Gamma(\nu+k+1)},\;\;|\arg z|<\pi, $$


and $Y_\nu(z)$ is the Bessel function of the second kind and order $\nu$,


$$ Y_\nu(z)=\frac{1}{\sin\nu\pi}\bigl( \;\cos\nu\pi J_\nu(z) -J_{-\nu}(z) \;\bigr).$$


The above expression makes sense  only for non-integral $\nu$. To obtain the expression for integral $\nu$ we let $\nu$  approach an integer in the above equality. See G & R  (6th Edition)  Section 8.4-8.5.


We deduce  that



$$-\ii F_\nu^+(\xi,c)+\ii\xi F^-_\nu(\xi,c)= -\sqrt{\pi}\Gamma\Bigl(\nu+\frac{1}{2}\Bigr)\xi\left( \frac{2c}{\xi} \right)^\nu J_{-\nu}(c\xi). $$

We can rewrite this  as

$$-\ii F_\nu^-(\xi,c)+\ii\xi F_\nu^-(\xi, c) =  Const(\nu) c^\nu \xi^{-\nu+1}J_{-\nu}(c\xi). $$

Define  $\newcommand{\pa}{\partial}$

$$\chi_\pm^a(x_+^2-c^2):=\frac{1}{\Gamma(a+1)}(x_\pm^2-c^2)^a_+, $$

$$A_\nu(x)=\pa_x\Bigl(\chi_+^{\nu-1/2}(x^2_+-c^2)-\chi_+^{\nu-1/2}(x_-^2-c^2)\;\Bigr). $$

We conclude that
 $$\widehat{A}_\nu(\xi)=-\sqrt{\pi}(2c)^\nu |\xi|^{1-\nu}J_{-\nu}(c\xi). $$

July 21, 2012. The term $\xi^{-\nu+1}J_{-\nu}(c\xi)$ is fortunately the kernel of the  modified Hankel transform discussed in these  notes of Michael Taylor.  This is good news for my project.

The fundamental law of statistical physics « Equilibrium Networks

The fundamental law of statistical physics « Equilibrium Networks