Hope to add bits of mathematics or mathematical news I find interesting.
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
$$\bsi_x :=\left[\begin{array}{cc} 0 & 1\\
1 & 0
\end{array}
\right],\;\;\bsi_y:=\left[
\begin{array}{cc} 0 & -\ii \\
\ii & 0
\end{array}
\right],\;\;\bsi_z:=\left[
\begin{array}{cc} 1 &0\\
0 & -1
\end{array}
\right]. $$
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
$$ \frac{\ii}{2\pi}\int_{\bC} \Omega =-1. $$
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
$$ \frac{\ii}{2\pi}\int_{\bC} \Omega =-1. $$
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
$$\widehat{\phi}(\xi)=\widehat{v}\Bigl(\;\xi+\frac{1}{\ve}\theta\;\Bigr). $$
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.
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.
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).
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)
$$ F^+_\nu(\xi, c)= -\frac{\sqrt{\pi}}{2}\left(\frac{2c}{\xi}\right)^\nu\Gamma\Bigl(\nu+\frac{1}{2}\Bigr)\bigl(\;Y_{-\nu}(c \xi ) + \ii J_{-\nu}(c \xi )\;\bigr), $$
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.
$$ F^+_\nu(\xi, c) :=\int_c^\infty (x^2-c^2)^{ \nu-\frac{1}{2} } e^{-\ii\xi x} dx,$$
$$ F^-_\nu(\xi,c) :=\int_{-\infty}^{-c} (x^2-c^2)^{\nu-\frac{1}{2}} e^{-\ii\xi x} dx,$$
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.
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