## Tuesday, November 27, 2012

### FRANK FEST: Workshop on High Dimensional Topology @ ND 2012

This is a conference in honor of Frank Connolly who is retiring this semester.

Workshop on High Dimensional Topology @ ND 2012

## Wednesday, November 21, 2012

### Sharp nondegeneracy estimates for a family of random Fourier series


Suppose that $w\in \eS(\bR)$ is an even, nonnegative  Schwartz  function.  Assume that $w\not\equiv 0$. $\newcommand{\hw}{\widehat{w}}$ We denote by $\hw(t)$   its Fourier transform

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

For $n\in \bZ$ we  set $\newcommand{\be}{\boldsymbol{e}}$

$$\be_n(\theta) :=\frac{1}{\sqrt{\pi}}\begin{cases} \frac{1}{\sqrt{2}}, & n=0,\\ \sin n \theta , & n<0,\\ \cos n\theta , & n>0. \end{cases}$$

Observe that  the collection $\lbrace \be_n(\theta)\rbrace_{n\in\bZ}$ is an orthonormal  basis of $L^2(\bR/2\pi\bZ)$.   $\newcommand{\bT}{\mathbb{T}}$      For any positive integer $N$ we  denote by $\bT^N$ the $N$-dimensional torus

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

Consider the random Fourier series

$$f_\ve(\theta)=\sum_{n\in \bZ} \sqrt{w(\ve n) } c_n \be_n(\theta),$$


$$\eE^{\ve}:\bT^1\times \bT^1\to \bR,\;\;\eE^\ve (\theta,\vfi)=\bsE\bigl( f_\ve(\theta)\cdot f_\ve(\vfi)) =\sum_{n\in \bZ} w(\ve n) \be_n(\theta)\be_n(\vfi)$$

$$=\frac{1}{2\pi} w(0)+\frac{1}{\pi}\sum_{n>0}w(\ve n)\cos n(\theta-\vfi)=\frac{1}{2\pi}\sum_{n\in\bZ}w(\ve n) e^{\ii n(\theta-\vfi)}=W_\ve(\theta-\vfi). \tag{1}\label{1}$$

Poisson formula.   For any $\phi\in \eS(\bR)$  and any $c\in\bR\setminus 0$ we have

$$\frac{2\pi}{c}\sum_{n\in\bZ} \phi\Bigl(\frac{2\pi n}{c}\Bigr)= \sum_{\nu\in\bZ} \widehat{\phi}(n c).$$

Suppose $\phi\in\eS(\bR)$ and $c$ are such that

$$\phi\Bigl(\;\frac{2\pi n}{c}\;\Bigr)= w(\ve n) e^{\ii n(\theta-\vfi)} .$$

If we formally replace $n =\frac{c x}{2\pi}$ we deduce from the above equality that

$$\phi(x)= w\Bigl(\frac{\ve c x}{2\pi}\Bigr) e^{\ii\frac{c(\theta-\vfi)x}{2\pi}}=w(ax)e^{\ii b x}, \;\; a:=\frac{\ve c}{2\pi},\;\;b :=\frac{c(\theta-\vfi)}{2\pi}.$$

Then

$$\widehat{\phi}(t) =\int_{\bR} e^{-\ii tx} w(ax) e^{\ii bx} dx = \frac{1}{a}\int_{\bR} e^{-\ii \frac{t-b}{a}x} w(y) dy = \frac{1}{a}\hw\Bigl( \frac{t-b}{a}\Bigr).$$

We now set $c:=2\pi$ so that $a=\ve$, $b=(\theta-\vfi)$. Using The Poisson formula in (\ref{1}) we deduce

$$W_\ve(\theta-\vfi)=\eE^\ve(\theta,\vfi) =\frac{1}{2\pi\ve} \sum_{n\in\bZ} \hw\Bigl(\frac{2\pi n-(\theta-\vfi)}{\ve}\Bigr) . \tag{2}\label{2}$$

Now consider the  random function

$$F_\ve:\bT^N\to \bR,\;\; F_\ve(\vec{\theta}) = \sum_{j=1}^n f_\ve(\theta_j).$$

The  correlation kernel of this  random function is

$$\eE_N^\ve(\vec{\theta},\vec{\vfi}) =\sum_{1\leq j,k\leq N} \eE^\ve(\theta_j-\vfi_k).$$

The differential of $F_\ve$ at a point $\vec{t}\in\bT^N$ is a Gaussian  random vector with covariance matrix $\newcommand{\pa}{\partial}$

$$S^\ve(\vec{t})= \Bigl( S^\ve_{jk}(\vec{t})\;\Bigr)_{1\leq j,k\leq N},\;\; S^\ve_{jk}(\vec{t})= \frac{\pa^2}{\pa\theta_j\pa \vfi_k} \eE_N^\ve\bigl(\;\vec{\theta},\vec{\vfi}\;\bigr)|_{\vec{\theta}=\vec{\vfi}=\vec{t}}=-W_\ve''(t_j-t_k)=\frac{1}{2\pi}\sum_{n\in\bZ} n^2w(\ve n) e^{\ii n(t_j-t_k)}\tag{3}\label{3}.$$

Definition. We say that $\vec{t}\in\bT^n$ is  nondegenerate if $t_j-t_k\in\bR\setminus 2\pi\bZ$, $\forall j\neq k$. We denote by $\bT^N_*$ the collection of nondgenerate  points in $\bT^N$.

$$\ast\ast\ast$$

We have the following result similar to the one in our   previous post.

Proposition 1.  There exists $\ve_0=\ve_0(w,N)>0$ such that if $\ve \in (0,\ve_0)$ and  $\vec{t}\in \bT^N$ is nondegenerate, then the  matrix $S^\ve(\vec{t})$ is positive  definite.

Proof.     Set

$$Z_\ve:=\bigl\{ n\in\bZ;\;\;w(\ve n)\neq 0\;\bigr\}.$$

Consider the space $H_\ve$ consisting of functions $\newcommand{\bC}{\mathbb{C}}$

$$u: Z_\ve \to \bC,\;\;\sum_{n\in Z_\ve} |u(n)|^2 n^2 w(\ve n) <\infty.$$

This is a separable  Hilbert space with inner product

$$(u,v)_\ve= \frac{1}{2\pi} \sum_{n\in\bZ} u(n)\cdot \overline{v(n)}\; n^2w(\ve n).$$

We denote by $\Vert-\Vert_\ve$ the associated norm.

For $t\in\bT^1$ consider the truncated character $\chi^\ve_t:Z_\ve\to \bT^1$, $\chi^\ve_t(n)=e^{\ii tn}$.  For $\vec{z}\in \bC^N$ $\newcommand{\vez}{{\vec{z}}}$  and $\vec{t}\in \bT^N$ consider $T_{\vez,\vec{t}}\in H_\ve$

$$T_{\vez,\vec{t}}(n)=\sum_{j=1}^n z_j \chi^\ve _{t_j}(n)=\sum_{j=1}^N z_j e^{\ii t_j n},\;\;n\in Z_\ve.$$

From the equality (\ref{3}) we deduce that

$$\sum_{j,k=1}^n S^\ve_{jk}(\vec{t}) z_j\bar{z}_k = \Vert T_{\vez,\vec{t}}\Vert_\ve^2.$$

Thus, the matrix $S^\ve(\vec{t})$ has a kernel if and only if the truncated characters $\chi^\ve_{t_1},\dotsc, \chi^\ve_{t_N}$ are linearly dependent.  We show that this is not possible if  $\vec{t}$ is  nondegenerate and  $\ve$ is sufficiently small.

Fix  $\ve_0=\ve_0(N,w)$ such that  if $\ve<\ve_0$ the support   of $x\mapsto w(\ve x)$ contains a long  interval of the form $[\nu_\ve, \nu_\ve+N-1]$, for some integer $\nu_\ve>0$. (Recall that $w$ is even.) In other words $\nu_\ve,\nu_\ve+1,\cdots,\nu_\ve+N-1\in Z_\ve$.

Let $\ve\in (0,\ve_0)$  and suppose that  $\vez\in\bC^N\setminus 0$ and $\vec{t}\in\bT^N$ are such that  such that  $T_{\vez,\vec{t}}=0$.   Thus

$$(T_\vez, u)_\ve =0,\;\;\forall u\in H_\ve$$

For any $m\in Z_\ve$ consider the Dirac function $\delta_m: Z_\ve\to \bC$,  $\delta_m(n)=\delta_{mn}=$ the Kronecker  delta.

We deduce that for any $m=\nu_\ve,\nu_\ve+1,\dotsc, \nu_\ve+N-1$ we have

$$0 = (T_\vez, \delta_m)_\ve=m^2w(\ve m) \sum_{j=1}^N z_j e^{\theta_j m} .$$

This can happen if and only if

$$0= \det\left[ \begin{array}{cccc} e^{\ii \nu_\ve t_1} & e^{\ii \nu_\ve t_2} & \cdots & e^{\nu_\ve t_N}\\ e^{\ii(\nu_\ve+1)t_1} & e^{\ii(\nu_\ve+1)t_2} & \cdots & e^{\ii (\nu_\ve+1) t_N}\\ \vdots & \vdots &\vdots &\vdots\\ e^{\ii(\nu_\ve+N-1)t_1} & e^{\ii(\nu_\ve+N-1)t_2} &\cdots & e^{\ii(\nu_\ve+N-1)t_N} \end{array} \right] = e^{\ii\nu_\ve(t_1+\cdots +t_N)} \prod_{j<k} \Bigl( e^{\ii t_k}-e^{\ii t_j}\Bigr) .$$

This shows that  $S^\ve(\vec{t})$ has a kernel if and only of $\vec{t}$ is degenerate.    Q.E.D.

Remark.    Here is an alternate proof of  Proposition 1 that yields a bit more. The above proof shows that

$$S^\ve_{jk}(\vec{t})= (\chi_{t_j},\chi_{t_k})_\ve.$$

Suppose for simplicity that $0\in Z_\ve$, i.e.,    $w(0)>0$. Then for $\ve>0$ sufficiently small we have $1,\dotsc, N\in Z_\ve$.  Observe that

$$(\delta_j,\delta_k)_\ve= \frac{k^2w(\ve k)}{2\pi}\delta_{jk},\;\;j,k=1,\dotsc, N.$$

We have a Cauchy-Schwartz inequality

$$\Bigl|\; \bigl( \chi_{t_1}\wedge\cdots \chi_{t_n}, \delta_1\wedge \cdots \wedge \delta_N\;\bigr)_\ve\;\Bigr| \leq \bigl|\; \chi_{t_1}\wedge\cdots \wedge\chi_{t_n}\;\bigr|_\ve\cdot \bigl|\;\delta_1\wedge \cdots \wedge \delta_N\;\bigr|_\ve.$$

This translates to

$$\Bigl| \det\Bigl(\; (\chi_{t_j},\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N}\;\Bigr| \leq \sqrt{\det\Bigr( \; (\chi_{t_j},\chi_{t_k})_\ve\;\Bigr)_{1\leq i,j\leq N} } \cdot \sqrt{\det\Bigr( \; (\delta_j,\delta_k)_\ve\;\Bigr)_{1\leq i,j\leq N} },$$
or, equivalently

$$\prod_{j<k} \Bigl| e^{\ii t_j}-e^{\ii t_k}\Bigr|^2 \leq \frac{1}{(2\pi)^N}\Bigl(\prod_{j=1}^N j^2w(\ve j)\Bigr) \det S^\ve(\vec{t}). \tag{4} \label{4}$$

$$\ast\ast\ast$$

The basic question that interests me is the following: what  happens to $S^\ve(\vec{t})$ as $\ve\to 0$, and $\vec{t}$ is nondegenerate.

Observe that (\ref{2}) implies that

$$W_\ve''(t)=\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\frac{2\pi n-t}{\ve}\Bigr).$$

We make the change in variables $t=\ve \tau$, we set

$$C^\ve(\vec{\tau}) := S^\ve(\ve\vec{\tau})$$

and we deduce

$$C^\ve_{jk}(\vec{\tau})=\frac{1}{2\pi}\sum_{n\in\bZ}n^2w(\ve n) e^{\ii\ve n(\tau_j-\tau_k)} =-\frac{1}{2\pi\ve^3}\sum_{n\in\bZ}\hw''\Bigl(\tau_j-\tau_k-\frac{2\pi n}{\ve}\Bigr),\;\;0\leq \tau_j <\frac{2\pi}{\ve},\;\;j=1,\dotsc, N. \tag{5}\label{5}$$


$$\bde_k=\bde_k^\ve :=\frac{\sqrt{2\pi}}{k\sqrt{w(\ve k)}}\delta_k.$$

By construction,  the collection $\bde_1,\dotsc,\bde_N$ is an $(-,-)_\ve$-orthonormal basis of $\bsD_N$.

The  $(-,-)_\ve$-orthogonal projection $P_\ve=P_\ve(\vec{t}): X(\vec{t})\to \bsD_n$ is given by

$$P_\ve \chi_{t_j} =\sum_{k=1}^n (\chi_{t_j},\bde_k)_\ve \bde_k = \sum_{k=1}^N e^{\ii kt_j} \delta_k.$$

With respect to the natural bases $\chi_{t_1},\dotsc,\chi_{t_N}$ of $X(\vec{t})$ and $\delta_1,\dotsc, \delta_N$ of $\bsD_N$ the    projection is therefore given by the  Vandermonde matrix

$$V= V(\vec{t}),\;\; V_{kj}= e^{\ii k t_j},\;\; V=\left[\begin{array}{cccc} e^{\ii t_1} & e^{\ii t_2} &\cdots & e^{\ii t_N}\\ e^{2\ii t_1} & e^{2\ii t_2} & \cdots & e^{2\ii t_N}\\ \vdots &\vdots &\vdots &\vdots\\ e^{N\ii t_1} & e^{N\ii t_2} &\cdots & e^{N\ii t_N} \end{array} \right].$$

## Wednesday, November 14, 2012

### On a family of symmetric matrices


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

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

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


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

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

## Tuesday, November 13, 2012

### 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

### 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

## Friday, November 9, 2012

### Tim Gowers recovering after a heart intervention

I'll let the Great Man tell the story.

## Thursday, November 1, 2012

### 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.