Friday, July 27, 2012

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.

Wednesday, July 25, 2012

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

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

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
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

$$ \left[\begin{array}{cc}
 z & \bar{u} \\
u & -z
\right] \cdot \left[
\right]. $$

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

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


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

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


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

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

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

Tuesday, July 24, 2012

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

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


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

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)\\

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.

Friday, July 20, 2012

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).

Thursday, July 19, 2012

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)

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


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

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