TITLE: Tensor with multi-rank $(1,1,1)$
QUESTION [0 upvotes]: I want to Show that a $2 \times 2 \times 2$ tensor cannot have multi-rank $(1, 1, 2)$ and it has rank $1$ if and only
if it has multi-rank $(1, 1, 1)$?

REPLY [0 votes]: If your tensor $T$ has multirank $(1,1,2)$ then it can be written in the form $a \otimes B$, where $B$ has multirank $(1,2)$. Then we can write the original tensor as $a \otimes b \otimes c$, where $c$ has rank $2$. But that is nonsense, $c$ is a vector in a vector space, so it has rank $1$ (unless $c=0$, but in that case your original tensor is zero).
If $T$ has rank $1$, then $T = a \otimes b \otimes c$ has multirank $(1,1,1)$.
If $T$ has multirank $(1,1,1)$, then $T = a \otimes B$ where $B$ has multirank $(1,1)$. So then $B = b \otimes c$. So $T = a \otimes b \otimes c$, and $T$ has rank $1$.


TITLE: Why does the equality $\int \mathbf{r}\times \nabla f d\nu=0$ holds?
QUESTION [0 upvotes]: My question comes from Problem 4. Classical laws for angular motion of  Practical Quantum Mechanics.
And the Prob.4 is that,

To show that the classical relation between angular momentum $\mathbf{L}=\mathbf{r}\times \mathbf{p}$ and torque $\mathbf{T}=\mathbf{r}\times\mathbf{F}$(where $\mathbf{p}$ stands for linear momentum and $\mathbf{F}$ for force),
  $$
\frac{d\mathbf{L}}{dt}=\mathbf{T},
$$
  still holds for the space averages in quantum mechanics.

In the solution, the author applied one equation, i.e.
$$
\iiint (\mathbf{r}\times \nabla f) d\nu=0
$$
where $\mathbf{r}=x\hat{i}+y\hat j+z\hat k$, and he said that the equality is a "general vector rule". However, I couldn't find this equality anywhere(include vector analysis tetbooks, wikipedia and this site).
I have also try to prove that equality holds. According the rule $\nabla\times (f\mathbf{A})=f\nabla\times \mathbf{A}+\nabla f \times \mathbf{A}$, I got
$$
\mathbf{r}\times \nabla f=-\nabla\times (f\mathbf{r})
$$
but I don't know what should I do to prove that the integration is zero.
So, I need help.

REPLY [1 votes]: Any volume integral of curl $\mathbf A$:
$$
\int_V \nabla\times \mathbf A \,dV
$$
can be calculated also as surface integral of $\mathbf A$:
$$
\oint_\Sigma d\boldsymbol\Sigma\times\mathbf{A};
$$
here $\Sigma$ is boundary of the region and $d\boldsymbol \Sigma = \mathbf nd\Sigma$ is vector whose magnitude is that of area $d \Sigma$ has direction of normal to the surface and is oriented outwards.
If magnitude of $\mathbf n \times \mathbf A$ is zero at every point of some closed boundary surface, surface integral over this boundary will be zero.
Whether this is so depends on the physical situation considered. For example, it is true if $\mathbf A$ is electrostatic field and the boundary is in metal, since field vanishes in metal. It isn't true if $\mathbf A$ is magnetic field of simple circuit and boundary is in vacuum, since the field vanishes nowhere; it only decays with distance.
If the decay is fast enough though, the above integral may have limit 0 as the boundary is expanded to infinity.
Your integral has this form if we take $\mathbf A=f\mathbf r$. So, the condition of validity of the relation is either 1) $f$ vanishes on some closed boundary surface, or 2) $f$ decays with distance rapidly enough.


TITLE: What effect does rotating the coordinate axes have on a given vector? (rotation matrices)
QUESTION [1 upvotes]: Suppose I have a vector $\vec{B}$ represented in the $x, y, z$ coordinate system. If I were to rotate my coordinate axes by applying the rotation matrices $R_x(\theta)$ and $R_z(\theta)$ to the $x$ and $z$ axes respectively, how do I compensate for these changes in my $\vec{B}$? What rotations do I have to apply to $\vec{B}$ so it can be represented in the new coordinate system?
P.S: The matrices I'm talking about are these. I know the relevant angles for each of these in my problem.

REPLY [0 votes]: If your coordinate system changes according to some transformation, then the coordinates of a fixed vector transform according to the inverse of the transformation. In this case, our transformations are simple rotations, and the inverse of a rotation is simply the rotation around the same axis but with the opposite angle. In symbols: $R^{-1}(\theta) = R(-\theta)$. Therefore, if we apply $R_x$ followed by $R_z$ (note that the order matters) to the original basis, we find that the coordinate vector of $\vec{B}$ in the new basis is given by $$(R_z(\theta_2)R_x(\theta_1))^{-1}\vec{B} = R_x^{-1}(\theta_1)R_z^{-1}(\theta_2)\vec{B} = R_x(-\theta_1)R_z(-\theta_2)\vec{B}.$$


TITLE: Incorrect radius of ellipsoid
QUESTION [1 upvotes]: I've seen the correct way of finding $r(\theta,\phi)$ for the purposes of integrating for area, but that left me wondering why we can't just use:
$$
x = a \cdot \cos{\theta} \cdot \sin{\phi} \\
y = b \cdot \sin{\theta} \cdot \sin{\phi} \\
z = c \cdot \cos{\phi}
$$
and then say that:
$$
r(\theta,\phi) = \sqrt{ x^{2} + y^{2} + z^{2}} \\
r(\theta,\phi) = \sqrt{ (a \cdot \cos{\theta} \cdot \sin{\phi})^{2} + (b \cdot \sin{\theta} \cdot \sin{\phi})^{2} + (c \cdot \cos{\phi})^{2}}
$$
Numerically computing A = $\int_{0}^{2\pi}\int_{0}^{\pi}r(\theta,\phi)^{2}\sin(\phi)d\phi d\theta$ for area returns an incorrect value. Obviously something is wrong, but what?

REPLY [0 votes]: $\newcommand{xphi}{\underline\phi}\newcommand{xtheta}{\underline\theta}$
The formula in the question is a formula that generates points on the ellipsoid from arbitrary parameters $\xphi$ and $\xtheta$ as follows:
First, consider a point $P$ on the unit sphere with spherical coordinates
$(r,\phi,\theta) = (1, \xphi,\xtheta).$
Then stretch (or compress) the sphere along the $x,$ $y,$ and $z$ axes by factors $a,$ $b,$ and $c$ respectively.
The image of $P$ under this transformation is a point $P'$ on the ellipsoid with Cartesian coordinates
$$(x,y,z) = (a \cos{\xtheta} \sin{\xphi},
 b \sin{\xtheta} \sin{\xphi},
 c \cos{\xphi}).$$

The result of the transformation is typically that the point $P'$ is not only at a different distance from the origin than $P$ was, but is also not on the same radial from the origin as $P$ was. That is, the distortion of the sphere into the ellipsoid tends to change both the distance and direction to each of the points on the surface.
If we now take the spherical coordinates $(r,\phi,\theta)$ of the transformed point $P',$ we will (typically) find that $r \neq 1,$ $\phi \neq \xphi,$
and $\theta \neq \xtheta.$
In short, although the formulas in the question use symbols $\phi$ and $\theta$ that often represent spherical coordinates, those symbols are not spherical coordinates in those formulas.
Because the parametric formulas do not use spherical coordinates, you will get incorrect results if you pretend that the parameters are spherical coordinates and integrate over them as if they were spherical coordinates.
The reason I rewrote the formulas from the question using $\xphi$ rather than $\phi$ and $\xtheta$ rather than $\theta$ was in order to be able to clearly distinguish them from the usual symbols for spherical coordinates while keeping some indication of how my formulas over these parameters correspond to the formulas in the question.
Note that
the article in Wolfram Mathworld
goes even further in relabeling the parameters: that article uses $u$ where you used $\theta$ and $v$ where you used $\phi.$


TITLE: Basic question with tiling tiles in a rectangle
QUESTION [0 upvotes]: Find the number of ways of tiling a 2 by 10 rectangle with 1 by 2 tiles which can be placed in any orientation.
I’m just not sure how to approach it begin counting it

REPLY [0 votes]: I am not going to be providing a full solution but just a hint;
let $a_n$ denote number of ways to cover a $2*n$ rectangle with given tiles.
I'd suggest you to draw and see how $a_n$ = $a_{n-1}$ + $a_{n-2}$. (HINT; you can place the $2*1$ tiles in two ways, horizontally and vertically) and then find $a_0$ , $a_1$ and $a_2$ to get the rest or study some methods to solve recusrsion


text:
\begin{document}
\title {On the Dynamics of Crystal Electrons,\\
        high Momentum Regime}

\author{Joachim Asch${}^\ast$\and Fran\c cois Bentosela${}^\ast$\and
Pierre Duclos\thanks{CPT-CNRS, Luminy Case 907,  F-13288 Marseille
Cedex 9, France. e-mail: user@cpt.univ-mrs.fr}
\and  Gheorghe Nenciu
\thanks{{Dept.Theor.Phys. University of Bucharest,
P.O.Box MG 11, 76900- Bucharest, Romania. e-mail: nenciu@barutu.fizica.unibuc.ro}} }
\date{26.1.2000}

\maketitle

\begin{abstract}We study the quantum dynamics generated by
$H^{SW}=-{d^2\over dx^2}+V-x$ with  $V$ a real periodic
function of weak regularity. We prove that the continuous spectrum of
$H^{SW}$ is never empty, and furthermore that for $V$ small enough
there are no bound states.
 \end{abstract}

 \section{Introduction}
Consider for $\alpha\ge0$   potentials of the form
\begin{assert}
$V:\bR\to\bR, V(x+\gamma)=V(x)\quad(x\in\bR,\gamma\in2\pi\bZ)$
with
$$\Vert V\Vert_\alpha^2:=\sum_{n\in\bZ}\vert\widehat
V(n)\vert^2(1+n^2)^\alpha<\infty$$
 where  $\widehat V(n):={1\over \sqrt{2\pi}}\int_0^{2\pi}e^{-inx}
V(x)\ dx$.
\end{assert}
The Stark Wannier Hamiltonian is the
selfadjoint operator
$$H^{SW}:=-{d^2\over dx^2}-x+V$$
defined in $L^2(\bR)$ by extension from the core
$C_0^\infty(\bR)$; this is a corollary of the Faris Lavine
theorem, see \cite{reedsimo2}.

We shall prove that there are always propagating states and that
small potentials with some regularity
cannot bind:
\begin{theorem}\label{cont}
\begin{itemize}
\item[(i)] Let $\alpha>0$ then
\[\sigma_{cont}(H^{SW})\neq\emptyset;\]
\item[(ii)] for $\alpha>1/2$ there is a $c>0$ such that for ${\Vert
V
\Vert_\alpha}<c$
\[\sigma_{pp}(H^{SW})=\emptyset.\]
\end{itemize}

\end{theorem}

This theorem is a consequence of our main result, Theorem
\ref{main}, which asserts that the probability to be
accelerated in the future grows with the momentum of the
initial state.

Remark that our results are stated  in terms of the Stark--Wannier
problem  but apply as well to the problem of driven quantum rings,
see \cite{avronemi}.

The dynamics of crystal electrons described by the present model
have been studied since \cite{wannier} both in mathematics and
physics literature, see \cite{nenc91} for a review. The general
problem is to  understand how the reflections at the band edges
accumulate to localize the electron or to create resonances; it is
far from being settled. Our contribution is to the question: how do
spectral properties change when
$\alpha$ is diminishing? We refer to \cite{ao} for a physical
discussion of this theme. Answers for two extreme cases are known: if
$\alpha>5/2$ then the spectrum of $H^{SW}$ is absolutely continuous
see \cite{6} and \cite{shab, brie} for generalizations; on the
other hand there are models (corresponding to $\alpha<0$) for which
the spectrum has no absolute continuous component,
\cite{nenc2,joye} or is even pure point \cite{aschduclexne}. If $V$
is analytic one has certain informations on existence and width of
resonances, \cite{grecsacc, aschbrie}.

The paper is organized as follows: in the next section we state our
main result, Theorem \ref{main}, precisely and infer Theorem
\ref{cont}. The third section contains the dynamical information:
the proof of Theorem \ref{main} organized in several subsections.

\section{Spectral properties}
Denote the free Stark Hamiltonian by $H^{SW}_0=-\Delta-x$; by
$D:=-i\partial_x$ the momentum operator which is
selfadjoint on $H^1(\bR)$; $\chi$ is the binary fonction with
values $\chi(True)=1, \chi(False)=0$;
$\chi(D-t\in\lbrack a,b\rbrack)$ is the cutoff function in
Fourierspace of the interval
$\lbrack t+a,t+b\rbrack$;
$cte.$ a generic constant which may change from line to line;
\[\langle
n\rangle:=\left\lbrace\begin{array}{ll}1&n\le0\\
n&n>0\end{array}\right..\]

The main theorem is:
\begin{theorem}\label{main}
Let $\alpha>0, M>0$. There exists a $c=c_{\alpha,M}>0$ such that for
$V$ with $\Vert V\Vert_\alpha\le M$ and for all
$t\in\bR, n\in\bZ$ it holds:
\begin{itemize}
\item[(i)] \beqno&&\Vert\chi(D-t\in\lbrack n,
n+1))(e^{-iH^{SW}t}-e^{-iH^{SW}_0t})\Vert\le\\&& c_{\alpha,M}{\Vert
V\Vert_\alpha\over
\langle sign(t) n\rangle^{min\lbrace1,\alpha\rbrace}};\eeqno

\item[(ii)] for $\psi\in L^2(\bR)$ this implies:
\beqn{}&&\Vert\chi(D-t\in\lbrack n,
n+1))e^{-iH^{SW}t}\psi\Vert\ge\nonumber\\ &&\Vert\chi(D\in\lbrack
n,n+1))\psi\Vert- c_{\alpha,M}{\Vert V\Vert_\alpha\over
\langle sign(t)
n\rangle^{min\lbrace1,\alpha\rbrace}}\Vert\psi\Vert.\label{HM}\eeqn
\end{itemize}
\end{theorem}

\begin{remark}
The dynamical meaning of (ii) is that a large part of a state whose
initial momentum is high enough is accelerated, i.e.: For any
$\varepsilon>0$ there is an
$n$ such that for
$\psi=\chi(D\in\lbrack n,n+1))\psi, \Vert\psi\Vert=1$ it holds
\[\Vert\chi(D-t\in\lbrack
n,n+1))e^{-iH^{SW}t}\psi\Vert\ge1-\varepsilon\qquad(t>0).\]
\end{remark}
We show now that the result on the spectrum follows from this:

\medskip
{\bf Proof of Theorem \ref{cont}.}
For a state $\psi$  in the pure point spectral subspace of
$H^{SW}$ it holds:
\[\lim_{R\to\infty}\sup_{\pm t\ge0}\Vert\chi(D\ge
R)e^{-iH^{SW}t}\psi\Vert=0,\]
see, for exemple \cite{enssvese}. This implies for $n\in\bZ$
\beq\label{BS}\lim_{t\to\pm\infty}\Vert\chi(D-t\in\lbrack
n,n+1))e^{-iH^{SW}t}\psi\Vert=0.\eeq
{\it ad (i):}  Consider  the initial state
$\psi=\cF^{-1}\chi(p\in\lbrack n,n+1))$ where $\cF^{-1}$ denotes the
inverse unitary Fourier transform. By the inequality (\ref{HM}) it
holds for
$t>0,n$ large enough
\beqno\Vert\chi(D-t\in\lbrack
n,n+1))e^{-iH^{SW}t}\psi\Vert\ge1 -c{\Vert
V\Vert_\alpha\over
\langle sign(t)
n\rangle^{min\lbrace1,\alpha\rbrace}}\ge{1\over2}\eeqno  which
contradicts (\ref{BS}). So $\psi$ has a part in the continuous
subspace of $H^{SW}$.

\noindent{\it ad (ii):}
For $\psi$ in the pure point subspace take the limits $t\to sign (n)\
\infty$
 in  (\ref{HM}); the equality (\ref{BS}) implies
\[\Vert\chi(D\in\lbrack n,n+1))\psi\Vert\le c_{\alpha, M}{\Vert
V\Vert_\alpha\over\langle\vert
n\vert\rangle^{min\lbrace1,\alpha\rbrace}}\Vert\psi\Vert
\qquad(n\in\bZ)\]
which leads to
\[\Vert\psi\Vert^2=\sum_{n\in\bZ}\Vert\chi(D\in\lbrack n,
n+1))\psi\Vert^2\le c_{\alpha, M}\Vert
V\Vert_\alpha^2\sum_{n\in\bZ}{1\over\langle\vert
n\vert\rangle^{min\lbrace2,2\alpha\rbrace}}\Vert\psi\Vert^2\] which
is a contradiction for
$\Vert V\Vert_\alpha$ small enough and $\alpha>1/2$; so there are
no bound states.
\hfill$\Box$

\section{Dynamics}
To prove Theorem \ref{main} we decompose the operator $H^{SW}$
in the Bloch representation.  Denote by
$\bV$ the convolution operator
$\bV\psi(n)={1\over\sqrt{2\pi}}\sum_{m\in\bZ}\widehat
V(n-m)\psi(m)$ in $L^2(\bZ)$.

$V$ is real, so it holds: $\overline{\widehat
V(n)}=\widehat V(-n)$. We can always substract a constant from
$H^{SW}$ so we suppose that $\widehat V(0)=0$.

Consider in
\[L^2(\lbrack 0,1), dk; L^2(\bZ))\simeq\int_{\lbrack 0,1)}^\oplus
L^2(\bZ) \ dk\]
the time dependent operator
\[H(t)\psi(k,n)=(n+k+t)^2\psi(k,n)+\bV\psi(k,n).\]
$H_0(t)$ denotes this operator for $V=0$ which has constant domain
$H^2(\bZ)$.
$\bV$ is  bounded from $H^2(\bZ)$ to $L^2(\bZ)$ for $\alpha\ge0$
so the propagator  $\bU$ generated by $H(t)$ is well defined in
the strong sense. Its relation to the Wannier Stark propagator is:
\beq \bU(t)=\cB e^{-itx}e^{-iH^{SW}t}\cB^{-1}\label{BR}\eeq
where $\cB$ is the Fourier-Bloch transformation
\[\cB\psi(k,n)={1\over\sqrt{2\pi}}\int_0^{2\pi}e^{-inx}
\left\{\sum_{\gamma\in2\pi\bZ}
e^{-ik(x+\gamma)}\psi(x+\gamma)\right\}\ dx\]
which maps $L^2(\bR)$ unitarily onto $L^2(\lbrack 0,1), dk;
L^2(\bZ))$.

 Remark that $\cB\psi(k,n)=\cF\psi(k+n)$. We denote the
quantities for the case $V=0$  by a subscript $0$.

The following statement is a reformulation of Theorem \ref{main}
in this representation.

Denote $P_n$ the projection on the the site $n$
in $L^2(\bZ)$ then:
\begin{theorem}\label{tech}
For $\alpha>0, M>0$ there exists $c=c_{\alpha,M}>0$  such that for
$V$ with $\Vert V\Vert_\alpha\le M$ it holds in $\int_{\lbrack
0,1)}^\oplus L^2(\bZ) \ dk$ for $t\in\bR, n\in\bZ$:
\beqno {}&& \Vert P_n(\bU(t)-\bU_0(t))\Vert_\oplus=\\
&&\Vert\int_0^t P_n\bU^\ast_0\bV \bU\Vert_\oplus\le
c_{\alpha,M}{\Vert V\Vert_\alpha\over
\langle sign(t) n\rangle^{min\lbrace1,\alpha\rbrace}}.\eeqno
\end{theorem}
We first prove that this is indeed equivalent to Theorem \ref{main}:

{\bf Proof of Theorem \ref{main}.}
By the identity
(\ref{BR}) it holds:
\beqno{}&&\Vert\chi(D-t\in\lbrack
n,n+1))(e^{-iH^{SW}t}-e^{-iH^{SW}_0t})\Vert_{L^2(\bR)}=\\
{}&&\Vert\cB e^{-itx}\chi(D-t\in\lbrack
n,n+1))e^{itx}\cB^{-1}\cB e^{-itx}(e^{-iH^{SW}t}-e^{-iH^{SW}_0t})
\cB^{-1}\Vert_\oplus=\\
&&\Vert\chi(.+k\in\lbrack n,n+1))(\bU(t)-\bU_0(t))\Vert_\oplus=\\
&&\Vert P_n(\bU(t)-\bU_0(t))\Vert_\oplus.\eeqno
so part (i) is equivalent to Theorem \ref{tech}. To see part (ii)
observe that
\[e^{iH^{SW}_0t}\chi(D-t\in\lbrack
n,n+1))e^{-iH^{SW}_0t}=\chi(D\in\lbrack n,n+1))\]
so by (i) and the unitarity of $e^{iH^{SW}_0t}$:
\beqno{\ }&&\Vert\chi(D-t\in\lbrack n,
n+1))e^{-iH^{SW}t}\psi\Vert\ge\\
&&\Vert\chi(D-t\in\lbrack n,
n+1))e^{-iH^{SW}_0t}\psi\Vert- c_{\alpha,M}{\Vert
V\Vert_\alpha\over
\langle sign(t)
n\rangle^{min\lbrace1,\alpha\rbrace}}\Vert\psi\Vert=\\
&&\Vert\chi(D\in\lbrack n, n+1))\psi\Vert- c_{\alpha,M}{\Vert
V\Vert_\alpha\over
\langle sign(t)
n\rangle^{min\lbrace1,\alpha\rbrace}}\Vert\psi\Vert.\eeqno
 \hfill$\Box$

In the rest of the paper we prove Theorem \ref{tech}.

\subsection{\bf Proof of Theorem \ref{tech}}
 $\bU(t)=\int^\oplus U^k(t)\ dk$. For $\psi\in H^2(\bZ)$,
$\Omega^k:={U_0^k}^\ast U^k$ it holds
\beqno{}&&\Vert P_n(U^k(t)-U^k_0(t))\psi\Vert_{L^2(\bZ)}=
\Vert P_n(\Omega^k(t)-\bI)\psi\Vert\le\\
&&\Vert\int_0^tP_n {U_0^k}^\ast\bV U_0^k\Omega^k\psi\Vert.\eeqno
Remark that
\[U^k(t,s)=U^0(t+{k},s+{k})\]
so it is sufficient to estimate
\beq\label{INT}\Vert\int_0^t P_n{U_0^0}^\ast\bV
U_0^0\Omega^0\psi\Vert.\eeq
We shall give the argument for nonnegative integers $n$ and
$t>0$. For $n<0, t<0$ the result then follows from
time reversal, i.e.: because for
$\bU(-t)=T\bU(t)T^{-1}$ with $T\psi(n):=\overline\psi(-n)$ it
holds:
$\Vert P_{-n}(\bU(-t)-\bU_0(-t))\Vert=\Vert
P_n(\bU(t)-\bU_0(t))\Vert$. For the remaining cases we shall give
an argument later.

In the sequel we shall drop the superscript $0$. We shall also
suppress the states $\psi$; the norm estimates below are to be
understood as uniform estimates  proven on the dense set
$H^2(\bZ)$ and valid by extension in the operator norm.

Rapid oscillations are responsible for the smallness of (\ref{INT}).
Remark firstly that with
\[E_n(t):=(n+t)^2 \hbox{ it holds }\]
\[U_0(t)P_n=e^{-i\int_0^tE_n}P_n,\]
secondly that in every time interval
\[I_l:={l\over2}+\lbrack-{1\over4},{1\over4})
\qquad(l\in\bN)\]
there is exactly one degeneracy, namely
\[E_n\left({l\over 2}\right)=E_{-n-l}\left({l\over 2}\right)\]
corresponding to some point of stationary phase of
$e^{-i\int_0^t(E_n-E_{-n-l})}$. Denote
\[\bP_{n,l}:=P_n+P_{-n-l}, \quad \bP_{n,l}^\bot:=\bI-\bP_{n,l}.\]
Heuristically, the contribution
\beq\label{TWOINT}
\Vert\int_{I_l}P_nU_0^\ast\bV U_0\Omega\Vert\le
\Vert\int_{I_l}P_nU_0^\ast\bV \bP_{n,l}U_0\Omega\Vert+
\Vert\int_{I_l}P_nU_0^\ast\bV \bP_{n,l}^\bot U_0\Omega\Vert\eeq
to (\ref{INT}) is small for two reasons; loosely speaking: the first
term describes the probability that a reflection from momentum $n$ to
$-n-l$ takes place, i.e. that the electron behaves adiabatically.
This are less likely for $n$ large and $\widehat V$ small.
The second term
describes transitions to the other states which is less probable
for $n$ large because the energetic distance to these states grows
like $\vert2n+l\vert$.

We shall now proceed to the proof according to this intuition. We
first treat the reflection to $-n-l$:
\begin{lemma}\label{cross}
There is a numerical constant $c>0$, independent of $V$, such that
for all $\alpha<\beta\in I_l$
\[\Vert\int_{\alpha}^\beta P_nU_0^\ast\bV U_0\bP_{n,l}\Omega\Vert\le
c\sqrt{1+\Vert V\Vert_0}{\vert\widehat V(2n+l)\vert\over
\sqrt{\ 2n+l}}\]
\end{lemma}

{\bf Proof.} We supposed that $\widehat V(0)=0$ so $P_n\bV P_n=0$;
with  the notation
\[\varphi(t):=\int_0^t(E_n-E_{-n-l})=\int_0^t(2n+l)(-l+2\tau)\
d\tau;
\quad\bV_{n,l}:=P_n\bV P_{-n-l};\]
we shall  estimate
\[\Vert\int_{I_l}e^{i\varphi }\bV_{n,l}\Omega\Vert.\]
It will be clear that the reasoning holds uniformly if we
integrate only on $\lbrack\alpha, \beta\rbrack\subset I_l$. This
estimate is done by a stationary phase calculation in the spirit of
\cite{bornfock}.

Decompose for an $a\in(0,1/2)$
\[I_l=I_l\setminus\left\lbrace{l\over2}+\lbrack-{a\over2},
{a\over2}\rbrack\right
\rbrace
\bigcup\left\lbrace{l\over2}+\lbrack-{a\over2},{a\over2}\rbrack
\right\rbrace =:I_l^{NS}\cup I_l^{S}.\]
Then as $\Vert\Omega\Vert=1$:
\[\Vert\int_{I_l^S}e^{i\varphi}\bV_{n,l}\Omega\Vert\le{a}\Vert\bV_{n,l}\Vert.\]
On the other hand an integration by parts of
$(\partial_t e^{i\varphi})\bV_{n,l}\Omega/i\dot\varphi$ yields
\beqno{}&&\left.\left\Vert\int_{I_l^{NS}}e^{i\varphi}\bV_{n,l}\Omega
\right\Vert\le
\left\Vert{\bV_{n,l}\over\dot\varphi}\right\Vert
\right\arrowvert_{\partial
I_l^{NS}}+\int_{I_l^{NS}}\left\Vert{\bV_{n,l}\dot\Omega\over\dot
\varphi}
\right\Vert+\left\Vert{\varphi^{\prime\prime}\bV_{n,l}\Omega\over\dot
\varphi^2}\right\Vert.\eeqno
Now observe that $\vert\dot\varphi(t)\vert=\vert 2n+l\vert\vert
-l+2t\vert\ge \vert 2n+l\vert a$, and that
$i\dot\Omega=U_0^\ast\bV U_0\Omega$, so the term on the last
expression is smaller than
\beqno{}&&4{\Vert \bV_{n,l}\Vert\over a\vert
2n+l\vert}+{\Vert \bV_{n,l}\Vert\over \vert
2n+l\vert}\int_{I_l^{NS}}\left({\Vert
P_{-n-l}\bV\Vert\over\vert
l-2t\vert}+{\vert
2\vert\over\vert
l-2t\vert^2}\right)\ dt\\
{}&&\le cte.{\Vert \bV_{n,l}\Vert\over a\vert
2n+l\vert}(1+\Vert V\Vert_0)\eeqno where we have used that
$\Vert P_{-n-l}\bV\Vert\le\Vert V\Vert_0$.
Thus for $a\in(0,1/2)$:
\[\left\Vert\int_{I_l}P_nU_0^\ast\bV U_0\bP_{n,l}\Omega\right\Vert\le
a\Vert\bV_{nl}\Vert+{1\over a}cte.{\Vert \bV_{n,l}\Vert\over \vert
2n+l\vert}(1+\Vert V\Vert_0)\]
The minimum of $a\alpha+\beta/a$ for positive $a$ is
$2\sqrt{\alpha\beta}$, thus
\[\left\Vert\int_{I_l}P_nU_0^\ast\bV\bP_{n,l}U_0\Omega\right\Vert\le
c{\Vert\bV_{n,l}\Vert\over
\sqrt{\vert 2n+l\vert}}\sqrt{1+\Vert V\Vert_0}\]

which implies our assertion as
$\Vert\bV_{n,l}\Vert\le{1\over\sqrt{2 \pi}}\vert
\widehat V(2n+l)\vert$.

\hfill$\Box$

\bigskip
The other levels are separated by large gaps. We start the proof
that the transition probability to them is small  with a double
integration by parts lemma.

We have $H_0(t)P_n=E_n(t)P_n$
 Denote by
\[\widehat R_l(t):=(H_0(t)-E_n(t))^{-1}\bP_{n,l}^\bot\]
the reduced resolvent. The Friedrichs
twiddle operation is fundamental in adiabatic theories. The version
needed here is defined for an operator $A$ on $L^2(\bZ)$ by
\[\widetilde{A}_l:=P_nA\widehat R_l.\]

\begin{lemma}\label{IBP}
For $\beta>\alpha$ it holds on $H^2(\bZ)$
\beqno{} &&\int_\alpha^\beta U_0^\ast P_n\bV \bP_{n,l}^\bot
U_0\Omega=\\ {}&&\int_\alpha^\beta
U_0^\ast\left({\widetilde{\widetilde
\bV_l \bV}}_l\bV+i\dot{\widetilde{\widetilde
\bV_l \bV}_l}-\widetilde\bV_l\bV
\bP_{n,l}-i\dot{\widetilde\bV}_l\right)U_0\Omega\\  {}&&
+\left.iU_0^\ast\left(\widetilde\bV_l-
\widetilde{\widetilde\bV_l\bV}_l\right)
U_0\Omega\right\arrowvert_\alpha^\beta\eeqno
\end{lemma}

{\bf Proof.} The twiddle operation is an inverse commutator.
A dot $\dot{}$ or a prime ${}^\prime$ denotes differentiation. It
holds:
\[i\partial_t U_0^\ast\widetilde{\bV}_lU_0=
U_0^\ast (\lbrack
\widetilde{\bV}_l, H_0-E_n\rbrack+i\dot{\widetilde{\bV}_l})U_0=
U_0^\ast( P_n\bV \bP_{n,l}^\bot+i\dot{\widetilde{\bV}_l})
U_0\]
so an integration by parts and $i\partial_t\Omega=U_0^\ast\bV
U_0\Omega$ yield
\beqno{} &&\int_\alpha^\beta U_0^\ast P_n\bV \bP_{n,l}^\bot
U_0\Omega=\\ {}&&-\int_\alpha^\beta
U_0^\ast\left({\widetilde\bV_l
}\bV+i\dot{\widetilde\bV}_l\right)U_0\Omega
+\left.iU_0^\ast\widetilde\bV_l
U_0\Omega\right\arrowvert_\alpha^\beta.\eeqno
The decomposition $\widetilde\bV_l\bV=\widetilde\bV_l\bV\bP_{n,l}+
\widetilde\bV_l\bV\bP_{n,l}^\bot$, the identity
\[i\partial_tU_0^\ast\widetilde{\widetilde\bV_l\bV}_lU_0=
U_0^\ast(\widetilde\bV_l\bV\bP_{n,l}^\bot+
i\dot{\widetilde{\widetilde\bV_l\bV}_l})U_0\]
and a second integration by parts imply
\beqno{} &&-\int_\alpha^\beta U_0^\ast
\widetilde{\widetilde\bV_l\bV}_lP_{n,l}^\bot U_0\Omega=\\
{}&&\int_\alpha^\beta
U_0^\ast\left(\widetilde{\widetilde\bV_l\bV}_l\bV+
i\dot{\widetilde{\widetilde\bV_l\bV}_l}\right)U_0\Omega
-\left.iU_0^\ast\widetilde{\widetilde\bV_l\bV}_l
U_0\Omega\right\arrowvert_\alpha^\beta.\eeqno
Thus the assertion is proved.
\hfill$\Box$

Concerning the second contribution to (\ref{TWOINT}) we shall now
proceed to estimate the different terms of
\beq\label{off}\Vert\int_{I_l}P_nU_0^\ast\bV \bP_{n,l}^\bot
U_0\Omega\Vert\eeq
 defined by Lemma \ref{IBP} one after the other.

In the following lemma we collect facts that shall be used
frequently and often without comment:

\begin{lemma}\label{estimates}
Let $a\neq b\in\bZ$.
For $\alpha, \beta>0$ there is a $cte.>0$ such that it holds
\[\sup_{j\in\bZ\setminus\lbrace a,
b\rbrace}{1\over \vert j-a\vert^\alpha\vert
j-b\vert^\beta}=\sup_{j<{a+b\over2}}\ldots+
\sup_{j\ge{a+b\over2}}\ldots\le cte. {1\over \vert
a-b\vert^{{min\lbrace \alpha, \beta\rbrace}}};\]
and for $\alpha, \beta>1$:
\[\sum_{j\in\bZ\setminus\lbrace a, b\rbrace}{1\over\vert
j-a\vert^\alpha}{1\over
\vert j-b\vert^\beta}\le\sum_{j<{a+b\over2}}\ldots+
\sum_{j\ge{a+b\over2}}\ldots\le
cte.{1\over\vert a-b\vert^{min\lbrace \alpha,\beta\rbrace}};\]
let $A$ be the operator on $L^2(\bZ)$ whose kernel is
$A(i,j)=f(i)g(i-j)$ for some $f,g\in L^2(\bZ)$, then
\[\Vert A\Vert=\sup_{\Vert
\varphi\Vert=\Vert\psi\Vert=1}\vert\langle\varphi,
A\psi\rangle\vert\le\Vert f\Vert_{L^2(\bZ)}\Vert
g\Vert_{L^2(\bZ)}.\]
\end{lemma}

The smallness of the terms in Lemma (\ref{IBP}) results from the
presence of the reduced resolvent; we shall use that for $m\neq n,
m\neq -n-l, t\in I_l$ it holds as
$\vert m+n+l\vert\ge1$ and so $\vert
m+n+l+\alpha\vert\ge{1\over2}\vert m+n+l\vert$ for
$\alpha<{1\over2}$:
\[\inf_{t\in I_l}\vert E_m(t)-E_n(t)\vert=
\inf_{I_l}\vert(m-n)(m+n+2t)\vert\ge{1\over 2}\vert
m-n\vert\vert m+n+l\vert\]
and as in  Lemma (\ref{estimates}):
\[\inf_{m\in\bZ\setminus\lbrace n, -n-l\rbrace}\inf_{t\in I_l}\vert
E_m(t)-E_n(t)\vert\ge cte.\vert n-(-n-l)\vert=cte.\vert
2n+l\vert.\]

The first relevant term in the integrand of Lemma (\ref{IBP}) is
$\widetilde{\widetilde\bV_l\bV}_l\bV=P_n\bV\widehat{R}_l\bV\widehat
R_l\bV$. Now
\[P_n\bV(i,j)=\delta_{ni}\widehat V(n-j);\quad\vert\widehat
R_l\bV(i,j)\vert=\chi(i\neq n)\chi(i\neq -n-l){\widehat V(i-j)\over
E_i-E_n}.\]
By the third point of Lemma (\ref{estimates}) we get
\[\Vert\widehat R_l\bV\Vert\le\Vert
V\Vert_0\left(\sum_{\bZ\setminus\lbrace
n,-n-l\rbrace}\left\vert{2\over
(i-n)(i+n+l)}\right\vert^2\right)^{1/2}\le cte.\Vert
V\Vert_0{1\over\vert2n+l\vert}\]
so
\beq\label{A}\sup_{t\in I_l}
\Vert\widetilde{\widetilde\bV_l\bV}_l\bV\Vert\le\sup_{I_l}\Vert
P_n\bV\Vert\Vert\widehat R_l\bV\Vert^2\le cte.\Vert
V\Vert^3_0{1\over\vert 2n+l\vert^2}.\eeq
For the second term it holds
\[\Vert\dot{\widetilde{\widetilde\bV_l\bV}_l}\Vert=
\Vert(P_n\bV\widehat R_l\bV\widehat
R_l)^\prime\Vert\le2\Vert\bV\dot{\widehat R_l}\Vert\Vert\bV\widehat
R_l\Vert\]
so for $j\in\{n,-n-l\}$:
\[\sqrt{2\pi}\bV\dot{\widehat R_l}(i,j)={\widehat V(i-j)(\dot
E_j-\dot E_n)\over(E_j-E_n)^2}={2\widehat
V(i-j)\over(j-n)(j+n+2t)^2}\] so
\[\Vert\bV\dot{\widehat R_l}\Vert\le
cte.\Vert\bV\Vert_0{1\over\vert2n+l\vert}\]
and together with the estimate of $\bV\widehat R_l$ above it results:
\beq\label{B}\sup_{t\in
I_l}\Vert\dot{\widetilde{\widetilde\bV_l\bV}_l}\Vert\le cte.\Vert
V\Vert_0^2{1\over\vert2n+l\vert^2}.\eeq
 Next we discuss
$\dot{\widetilde\bV}_l=P_n\bV\dot{\widehat R_l}$.
\[\sqrt{2\pi}\dot{\widetilde\bV}_l(i,j)=2\delta_{in}{\widehat
V(i-j)\over(j-n)(j+n+2t)^2} {\vert i-j\vert^\alpha\over\vert
i-j\vert^\alpha}\] so:
\beq\label{C}\sup_{I_l}\Vert\dot{\widetilde\bV}_l\Vert\le
cte.\Vert V\Vert_\alpha{1\over\vert2n+l
\vert^{min{\lbrace1+\alpha,2\rbrace}}}.\eeq
The next contribution to (\ref{off}) comes from
\[\widetilde\bV_l\bV \bP_{n,l}=P_n\bV\widehat R_l\bV
P_n+P_n\bV\widehat R_l
\bV P_{-n-l}.\]
 The kernel of the
second term is
\[{2\pi}P_n\bV\widehat R_l\bV
P_{-n-l}(i,j)=\delta_{in}\delta_{j(-n-l)}
\sum_{\bZ\setminus\lbrace n,-n-l\rbrace}{\widehat V(n-m)\widehat
V(m+n+l)\over(m-n)(m+n+2t)}\]
so
\beqn&&\sup_{t\in I_l}\Vert P_n\bV\widehat R_l\bV
P_{-n-l}\Vert\nonumber\\ &&\le cte.
\sum_{\bZ\setminus\lbrace n,-n-l\rbrace}\left\vert{\widehat
V(n-m)\widehat V(m+n+l)\over(m-n)(m+n+l)}\right\vert{\vert
m-n\vert^\alpha\vert m+n+l\vert^\alpha\over\vert m-n\vert^\alpha\vert
m+n+l\vert^\alpha}\nonumber\\
&&\le cte.\Vert
V\Vert_\alpha^2{1\over\vert2n+l\vert^{1+\alpha}}\label{D}\eeqn
by the
Cauchy-Schwartz inequality and Lemma (\ref{estimates}).

The other term is more difficult, it is the part which is first
scattered out of the state $n$, then back to it:
\[{2\pi}P_n\bV\widehat R_l\bV P_n(i,j)=
\delta_{in}\delta_{jn}\sum_{\bZ\setminus\lbrace
n,-n-l\rbrace}{\vert\widehat V(n-m)\vert^2\over(m-n)(m+n+2t)}.\]
Changing coordinates we obtain
\beqno &&{2\pi}P_n\bV\widehat R_l\bV P_n(i,j)=
\delta_{in}\delta_{jn}\sum_{p\in\bZ\setminus\lbrace
0,-(2n+l)\rbrace}{\vert\widehat V(p)\vert^2\over p(p+2n+2t)}=\\
&&{\vert\widehat V(2n+l)\vert^2\over
(2n+l)(4n+l+2t)}+\delta_{in}\delta_{jn}\sum_{p\in\bZ\setminus\lbrace
0,\pm(2n+l)\rbrace}{\vert\widehat V(p)\vert^2\over p(p+2n+2t)}=\\
&&{\vert\widehat V(2n+l)\vert^2\over
(2n+l)(4n+l+2t)}+\delta_{in}\delta_{jn}\sum_{p\in\bZ\setminus\lbrace
0,\pm(2n+l)\rbrace}-{\vert\widehat V(p)\vert^2\over p^2-(2n+2t)^2}
\eeqno
because only
$-{\vert\widehat V(p)\vert^2\over p^2-(2n+2t)^2}$,
the symmetric part of ${\vert\widehat V(p)\vert^2\over
p(p+2n+2t)}$, contributes to the symmetric sum. Physically speaking
it is destructive interference of the contributions of the
states $m=n\pm p$ which is at work here.

 Now for $\alpha>0$
\beqno&&\sup_{t\in I_l}\sum_{\bZ\setminus\lbrace
0,\pm(2n+l)\rbrace}{\vert\widehat V(p)\vert^2\over
\vert p^2-(2n+2t)^2\vert}{p^{2\alpha}\over p^{2\alpha}}\le\\
&&\Vert
V\Vert_\alpha^2\sup_{p,t}{1\over \vert p^{2\alpha}\vert\vert
p^2-(2n+2t)^2\vert}
\le\Vert V\Vert_\alpha^2
{cte.\over\vert2n+l\vert^{min\lbrace1+2\alpha,2\rbrace}};\eeqno
to see this recall that for $p\neq\pm(2n+l)$
\[\inf_{t\in I_l}\vert p^2-(2n+t)^2\vert\ge{1\over4}\vert
p^2-(2n+l)^2\vert,\]
and take the supremum over $\vert p\vert<{2n+l\over2}$ and $\pm
p\ge{\vert 2n+l\vert\over2}$ separately.

Furthermore
\[{\vert\widehat
V(2n+l)\vert^2\over\vert(2n+l)(2n+l+2n+2t)\vert}\le\Vert
V\Vert_0^2{cte.\over\vert2n+l\vert^2},\]
so the estimate for the backscattering term is
\beqn\label{E}\sup_{t\in I_l}\Vert P_n\bV\widehat
R_l\bV P_n\Vert&\le&cte. \sup_{t\in
I_l}\left\vert\sum_{\bZ\setminus\lbrace n,
-n-l\rbrace}{\vert\widehat
V(n-m)\vert^2\over(m-n)(m+n+2t)}\right\vert\nonumber\\
&\le&cte. \Vert
V\Vert_\alpha^2{1\over\vert2n+l\vert^
{min\lbrace1+2\alpha,2\rbrace}}.\eeqn
We are left with the boundary terms in (\ref{off}). We first discuss
\[\widetilde{\widetilde\bV_l\bV_l}
\left(t={l\over2}\pm{1\over4}\right).\]

$\widetilde{\widetilde\bV_l\bV_l}=P_n\bV_l\widehat
R_l\bV_l\widehat R_l$, so
\beq\label{F}\Vert\widetilde\bV_l\widetilde\bV_l({l\over2}\pm{1\over4})\Vert
\le\Vert\bV_l\widehat R_l({l\over2}\pm{1\over4})\Vert^2\le
cte.\Vert V\Vert_0^2{1\over\vert 2n+l\vert^2}\eeq
where we used the estimate which led to (\ref{A}).

For the other boundary term we have to be more careful:
Consider for $t_{l+1}:={l\over2}+{1\over4}$
\beqno&&
\sqrt{2\pi}\left(\widetilde\bV_l(t_{l+1})-\widetilde\bV_{l+1}(t_{l+1})\right)(i,j)=\\
&&P_n\bV(H-E_n)^{-1}(t_{l+1})(P_{-n-(l+1)}-P_{-n-l})(i,j)=\\
&&\delta_{in}\delta_{j(-n-l)}{\widehat
V(2n+l)\over{3/2}(2n+l+1)}-\delta_{j(-n-(l+1))}{\widehat
V(2n+l+1)\over{1/2}(2n+l)}.\eeqno
So
\beq\label{G}\Vert\widetilde\bV_l(t_{l+1})-\widetilde\bV_{l+1}(t_{l+1})\Vert\le
cte.\left({\vert\widehat
V(2n+l)\vert\over\vert2n+l\vert}+{\vert\widehat
V(2n+l+1)\vert\over\vert 2n+l+1\vert}\right).\eeq
Furthermore it holds for $x\in\lbrack-{1\over4},{1\over4})$:
\[\sqrt{2\pi}\widetilde\bV_l\left({l\over2}+x\right)(i,j)=\delta_{in}
{\widehat V(n-j)\over(j-n)(j+n+l+2x)}\chi(j\neq
n)\chi(j\neq-n-l),\]
so for $\alpha>0$
\beqn\label{H}\Vert\widetilde\bV_l({l\over2}+x)\Vert^2&\le&
cte.\sum_{\bZ\setminus\lbrace n, -n-l\rbrace}{\vert\widehat
V(n-j)\vert^2\over\vert(j-n)(j+n+l+2x)\vert^2}\nonumber\\
&\le&\Vert V\Vert^2_0
{cte.\over\vert2n+l\vert^{2}}.\eeqn

 With these observations we
have finished the proof of Theorem (\ref{main}). We assemble the
argument.

We have
\beqno&&\Vert\int_0^t P_nU_0^\ast\bV U\Vert_{L^2(\bZ)}=\\
&&\Vert\int_0^{1\over4} \dots\Vert+
\sum_{l=1}^{L-1}\Vert\int_{I_l} \dots\Vert+
\Vert\int_{{L\over2}-{1\over4}}^{t} \dots\Vert\le\\
&&\sum_{l=0}^\infty\Vert\int_{I_l} \dots\Vert\le\\
&&\sum_{l=0}^\infty\Vert\int_{I_l}
\dots\bP_{n,l}\Vert+\sum_{l=0}^\infty\Vert\int_{I_l}
\dots\bP_{n,l}^\bot\Vert\eeqno
where $L$ is the integer such that
$t\in\left\lbrack{L\over2}-{1\over4},{L\over2}+{1\over4}\right)$;
$I_0:=\left\lbrack0,{1\over4}\right)$; for the index $L$ we
have redefined
$I_L:=\left\lbrack{L\over2}-{1\over4},t\right)$.

Remark that for $n>0, \beta>1$
\[\sum_{l=0}^\infty{1\over\langle2n+l\rangle^\beta}\le
{cte.\over\langle n\rangle^{\beta-1}}.\]

So by Lemma (\ref{cross})  it holds
\beqno&&\sum_{l=0}^\infty\Vert\int_{I_l}
\dots\bP_{n,l}\Vert\le cte.\sqrt{1+\Vert V\Vert_0}\sum
{\vert\widehat V(2n+l)\vert\over\sqrt{\langle 2n+l\rangle}}\\
&&\le cte.\sqrt{1+\Vert V\Vert_0}\Vert
V\Vert_\alpha\left(\sum{1\over\langle 2n+l\rangle^{2\alpha+1}}
\right)^{1/2}\\
&&\le cte.\Vert V\Vert_\alpha{1\over \langle n\rangle^\alpha}.\eeqno
By Lemma (\ref{IBP}) and the estimates (\ref{A}, \ref{B}, \ref{C},
\ref{D}, \ref{E}, \ref{F}, \ref{G}, \ref{H}) we have
\beqno&&\sum_{l=0}^\infty\Vert\int_{I_l}
\dots\bP_{n,l}^\bot\Vert\le\\
&&\sum \sup_{I_l}\left(\Vert{\widetilde{\widetilde
\bV_l \bV}}_l\bV\Vert+\Vert\dot{\widetilde{\widetilde
\bV_l \bV}_l}\Vert+\Vert\widetilde\bV_l\bV
\bP_{n,l}\Vert+\Vert\dot{\widetilde\bV}_l\Vert\right)+\\
&&\sum
\left(\Vert\widetilde{\widetilde\bV_l\bV}_l(t_{l+1})\Vert+\Vert
\widetilde{\widetilde\bV_l\bV}_l(t_l)\Vert\right)+\\
&&\sum\left(\Vert\widetilde \bV_{l}(t_{l+1})-\widetilde
\bV_{l+1}(t_{l+1})\Vert\right)+\Vert\widetilde
\bV_0(t_0)\Vert+\Vert\widetilde
\bV_L(t_L)\Vert\\
&&\le c_{\alpha, M}\Vert V\Vert_\alpha {1\over
\langle n\rangle^{min\lbrace1,\alpha\rbrace}}.
\eeqno
This proves the case $sign(n t)>0$. From our calculations it is
clear that for
$sign(n t)<0$ all the estimates give a bound proportional to
$\Vert V_\alpha\Vert$ and no decay in $n$. Thus the proof of
Theorem (\ref{main}) is finished.
\hfill$\Box$

\section{Acknowledgements}

This paper was partly written during the visits of G. Nenciu at CPT Marseille
and of F. Bentosela and P.Duclos at Univ. of Bucharest. Financial support
of CPT Marseille and the Romanian Ministry of Education (grant CNCSU 13C)
is hereby acknowledged.

config: arxiv


TITLE: maximal algebraically independent sets in ring extensions
QUESTION [2 upvotes]: Let $E/K$ be a field extension. It is a well known fact that all maximal subsets $A \subset E$ consisting of algebraically independent elements over $K$ have the same cardinality (which is by definition the transcendence degree of $E$ over $K$).
I suspect that this does not hold for arbitrary $K$-algebras $E$ (not even for finitely generated ones), but I don't know any counterexample. Can you name some preferably simple example where the statement doesn't hold?
Thank you in advance!

REPLY [2 votes]: Let $E=K[X,Y,Z]/(XY,XZ)$. Then $x$ (the residue class of $X$) is a maximal algebraically independent subset, and $y,z$ are also algebraically independent.


TITLE: What is the underlying function?
QUESTION [1 upvotes]: I have some value pairs. They are inversely proportional. I want to have a formula to get the second value from the first value.
Speed Delay
10    0.142
15    0.087
20    0.064
25    0.049
30    0.036
37    0.03
45    0.017
68    0.016

I guess there have to be an offset and a factor. Is it necessary to have min and max values?

REPLY [0 votes]: Based on plugging the data points into Logger Pro, a graphing software, I used a curve fit for  an natural exponent function and got:
$$y = A\cdot exp(-Cx) + B \rightarrow y = .3195\cdot exp(-.09558x) + .01664 $$where $x$ is the speed and $y$ is the delay and $exp$ is shorthand for $e^\text{something}$.
However, as you can see below, obviously not all of your data points fit the given curve. Perhaps your calculation for a formula in the comment above may suit you better depending on your needs. Note that I did play around with other functions within the pre-loaded ones (there's a lot!) in Logger Pro in hopes of finding a better curve fit, to no avail.


TITLE: A geometric sequence of measurable functions with $|f_n|\leq 1$ on a measure space $(X,\mathcal M, \mu)$.
QUESTION [3 upvotes]: Given a measurable function with $|f|< 1$ on a measure space $(X,\mathcal M, \mu)$, does the limit exist:
$$\lim_n \int 1 + f + f^2 + f^3 + \cdots  + f^n \,d\mu$$
Some consequences up front:
$f^n \in L^1$ for all $n\geq 1$.
If $\sum_0^n f^i(x) = S_n(x)$, then
My conclusions:
For each $x$, $S_n(x) \rightarrow \frac{1}{1-f}$ since $|f(x)|<1$.
If there exists $\epsilon>0$ such that $|f|\leq 1-\epsilon$, then $|\frac{1}{1-f}| \leq \epsilon^{-1}$, which, since $\mu(X)<\infty$, means that $|S_n|$ is dominated by an $L^1$ function. Thus $\lim_n \int S_n = \int \lim S_n = \int \frac{1}{1-f} < \infty$.
If on the other hand there does not exist such an $\epsilon$, we have for $\epsilon>0$ the set $\{|f(x)|>1- \epsilon\}$ is of positive measure. If $|f|>1-\epsilon$, then $\int |1 + f + \cdots + f^n| \geq \int_{|f|>1-\epsilon} |\frac{1-f^{n+1}}{1-f}| \geq \int_{|f|>1-\epsilon} \frac{1-|f|^{n+1}}{\epsilon} \rightarrow \mu(|f|>1-\epsilon)\epsilon^{-1}$. But this doesn't help me prove that the limit is infinity. I'm trying to show the limit always exists - even in the infinite case.

REPLY [1 votes]: If $f$ is real-valued, splitting it into its positive and negative part solves the problem. Let
$$g(x) = \max \{ 0, f(x)\};\quad h(x) = \max \{0, -f(x)\}.$$
Then $f(x) = g(x)-h(x)$, and at each point of $X$, at most one of $g$ and $h$ takes a value $\neq 0$. Then
$$f(x)^n = g(x)^n + (-1)^n h(x)^n,$$
and both $g^n$ and $h^n$ monotonically converge to $0$, so
$$G_n = \int_X g^n\,d\mu \searrow 0;\quad H_n = \int_X h^n\,d\mu \searrow 0.$$
The series
$$\sum_{n=0}^\infty (-1)^n H_n$$
is (conditionally, probably) convergent by the alternating series criterion. The series
$$\sum_{n=0}^\infty G_n$$
converges either to a finite limit, or to $+\infty$. Thus
$$\int_X S_n\,d\mu = \sum_{k=0}^n G_k + \sum_{k=0}^n (-1)^k H_k$$
converges, either to a finite limit, or to $+\infty$.


TITLE: Correct notation when solving a double integral
QUESTION [0 upvotes]: I am wondering if isolating the x-terms like I show below is correct when solving a double integral, or should I completely solve the first integral ($dy$) with the x-terms beside it before I can move to the next integral.
I know that this is correct
$$
\int_{0}^{1} \int_{-3}^{3} \frac{xy^2}{x^2+1} \ dy \ dx = \left( \int_{0}^{1}  \frac{x}{x^2+1} \ dx \right) \left( \int_{-3}^{3} y^2 \ dy \right)
$$
but is this also correct?
$$
\int_{0}^{1} \int_{-3}^{3} \frac{xy^2}{x^2+1} \ dy \ dx = \int_{0}^{1} \left[ \left( \frac{x}{x^2+1} \right) \int_{-3}^{3} y^2 \ dy \right] \ dx = \cdots
$$
Thank you.

REPLY [0 votes]: Yes. I worked out the integration and all three turn out as $9\ln 2$.
However, do not think about applying this to double integrals in general.


TITLE: Find the limiting distribution of $Z_n=n[1-F(Y_n)]$
QUESTION [2 upvotes]: Let $Y_n$ denote the maximum (the last order statistic) of a random sample of size $n$ from a distribution of the continuous type that has c.d.f $F(x)$ and pdf $f(x) = F'(x)$. Find the limiting distribution of $Z_n=n[1-F(Y_n)]$
I dont have the solution to this but the answer my teacher gave us is that the limiting distribution of $Z_n$ should be exponential with mean $1$. If you define $Z_n=n[1-Y_n]$ then that limiting distribution is exponential with mean $1$ but I dont know how to work with $F(Y_n)$

REPLY [1 votes]: This is essentially the same idea given by  @user365239, but stated a little differently.
Let $X_1,\dots,X_n$ be IID with DF $F$. Let $Y_n =\max_{j=1\dots n} X_j$.
Let $Z_j = n(1-F(X_j))$. Then $Z_1,\dots,Z_n$ are IID, and observe that
$$n(1-F(Y_n)) = \min_{j=1\dots n}Z_j.$$
Also, since  $F(X_j)\sim U[0,1]$, we also have  $1-F(X_j)\sim U[0,1]$, and so  $Z_j\sim U[0,n]$.
From this we obtain:
$$ P( \min_{j=1...n} Z_n > x) = P( U[0,n]>x)^n = (1-x/n)^n \to e^{-x},$$
where the second equality holds when $n>x$.


TITLE: Pigeon Hole Principle Algorithm
QUESTION [1 upvotes]: The “pigeonhole principle” states that if n+1 objects (e.g., pigeons) are to be distributed into n holes then some hole must contain at least two objects. This observation is obvious but useful.
Employ the pigeonhole principle to prove the following:
Claim: Let G be an undirected graph with at least two vertices. Then there exist distinct vertices v and w in G that have the same degree.
Thank you so much for all your help, a couple of problems on this homework were unlike anything we've done in class thus far. I think they're supposed to be an easy concept questions but I'm struggling.

REPLY [2 votes]: It is tacitly assumed that the graph $G$ is finite and contains no loops or double edges. Otherwise it is easy to give counterexamples.
Let $v_0$ be the number of isolated vertices,  and denote by $v'$ the number of vertices of  degree $\geq1$. When $v_0=2$ we are done. Otherwise $v'\geq1$, and $d_\max$, the maximal occurring degree, is $\geq1$. Since a vertex of maximal degree is connected to  $d_\max$ other nonisolated vertices we have
$$v'\geq d_\max+1\ .$$
The $v'$ vertices of degree $\geq1$ can have degrees from $1$ to $d_\max$ inclusive. As   $v'>d_\max$ we can conclude by the pigeon hole principle that two of them have to have the same degree.


TITLE: Where does this time-dependent wavefunction of $\Psi(x,t)= \sum_{n=1}^\infty \psi_n(x)\exp(\frac{-in^2\pi^2\hbar t}{2mL^2})$ come from?
QUESTION [0 upvotes]: I was reading this blog post  on simulating the probability desnity of the Schrodinger equation but there was one equation that I could not quite understand.
Firstly, defining
$$V(x) = \begin{cases} 0 & 0\le x\le L \\ \infty & \mathrm{otherwise} \end{cases}$$
for the classic Particle in a Box. Next, solving the time-independent Schrodinger equation for $V(x)$ you get
$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
I understand up to here but the next equation is what messes me up. The post says:

The general time dependent solution can be written as a linear combination of the separable solutions along with their time dependent part,
$$\Psi(x,t)= \sum_{n=1}^\infty \psi_n(x)\, \exp\left({\frac{-in^2\pi^2\hbar t}{2mL^2}}\right)$$

I do not understand where the term $\exp(\frac{-in^2\pi^2\hbar t}{2mL^2})$ comes from as the blog doesn't explain the equation and I do not know what this term is even called. It says its a combination of the seprable solutions with their time dependent part. What is the time dependant part? Why is it summed?

REPLY [0 votes]: That exponential factor is time evolution factor.
$$|\psi(t)\rangle = e^{-i \hat{H} t/\hbar} |\psi(0)\rangle $$
Such exponential factor comes from unitary property of time evolution operator to maintain its normalization condition. Here, $H$ is Hamiltonian which has eigenvalues as energy.
Energy eigenvalues for given hard wall potential can be obtained from following process:
$$-\frac{\hbar^2}{2m} \frac{d\psi}{dx} = E \psi \; \rightarrow \; \psi \sim \sin kx$$
$$k= \sqrt{\frac{2mE}{\hbar^2}} = \frac{n \pi}{L}$$
$$E_n = \frac{n^2 \pi^2 \hbar}{2mL^2}$$
Above DE has solutions you written, and $k$ is quantized by boundary condition. If you put those energy eigenvalues into time evolution factor, then you can get desired solution.
Linear combination means any wave function in this system can be expanded with energy eigenstates. It's because $\sin nx/L$ has completeness. (Think Fourier series)


TITLE: Integral definition of e
QUESTION [4 upvotes]: I know that $e$ can be defined via a convergent series: $$ e = \sum_{n=0}^\infty {1\over n!}$$
Or as a limit: $$ e = \lim_{n \to \infty} { \left(1 + {1 \over n}\right)^n }$$
Or as the value which satisfies the condition:

For any real number $x$, $f(x) = {df\over{dx}}$

Is there an integral definition of $e$ ?
Something like: $$e = \int_0^\infty ...$$

REPLY [0 votes]: I think it's worth pointing out the link between the examples you proffered.
Note $e^t$ is the solution to the differential equation $dx/dt = x, x(0) = 1$. Thus, $x(1) = e$.
Now, try to approximate the solution to this ODE using the forward Euler method with step size $1/h$ to estimate $x(1)$. The first step yields $x(1/h) \approx x(0) + (1/h)x(0) = (1 + 1/h)x(0) = 1 + 1/h$. The second step yields $x(2/h) \approx x(1/h) + (1/h)(x(1/h) = (1+1/h)x(1/h) = (1+1/h)^2$. Taking $h$ steps yields $x(1) \approx (1 + 1/h)^h$. Using consistent approximations results, you get that, as the step size $1/h$ goes to $0$, $x(1) \rightarrow e$.
Now, note that $\frac{d^2x}{dt^2} = \frac{d}{dt} \left( dx/dt \right) = \frac{d}{dt} x$. (This follows just by plugging in the equality $dx/dt = x$.) Inducting, we get that $\frac{d^nx}{dt^n} = x$ for all $n \in \mathbb{N}$, and the convergent sum result is simply a direct application of Taylor's theorem.
Basically, my point is that these equivalent definitions are not coincidence, and hint at a lot of neat results.


text:
\section{Concluding remarks}
\label{sec:concludex}

From the brief discussion given above, one notes
that many of the fractional and multifractional Gaussian processes have similar local properties,
in particular the local self-similarity (or having FBM as the tangent process at a point).
The LRD (or SRD) character is carried over from the fractional process to the corresponding multifractional process.
Some related processes such as step FBM
\cite{Benassi00}
and mixed FBM
\cite{Cheridito01}
are not included. The step FBM can be regarded as a special case of MBM, with
$H(t)$ a piecewise linear function. Such a multiscale process can be used to model
anomalous transport phenomena such as single-file diffusion.\cite{LimTeo09d}
Mixed FBM is a linear combination of two or more independent FBM, and it can be used to model
retarding anomalous diffusion,\cite{EabLim12}
financial time series,\cite{Mishura08,Cheridito01} telecommunication,\cite{Filatova08}
etc. As far as applications of fractional and multifractional stochastic processes are concerned, it is possible to select from a variety of processes one that provides the best description of the system under study.

Finally, we remark that path integral formulation of fractional stochastic processes has recently attracted considerable interest from physicists as well as mathematicians.\cite{Wio13,Kleinert12,JanakiramanSebastian12,EabLim14}
In view of the fact that several candidate theories of quantum gravity
\cite{Modesto09,Ambjorn10,Calcagni10,Calcagni12}
share the idea that spacetime is multifractal,
one would expect path integral formulation of fractional and multifractional stochastic processes may
play an important role in physics, just like the case in Brownian motion.


TITLE: Infinite pro-$p$ group of finite solvable length and finite coclass
QUESTION [1 upvotes]: I was reading about infinite pro-$p$ groups of finite coclass from the book "The Structure of Groups of Prime Power Order" by Leedham-Green and McKay. I asked this question in math.stackExchange before posting it here; there were no answers there, I thought I might share it in mathoverflow. My advance apologies if anything is inappropriate.
I was thinking about solvability and I think it can be shown that if $G$ is a solvable group of solvable length $l$ then every subgroup and quotient of $G$ has solvable length at most $l$ (please correct me if I am wrong). My question is related to the "opposite" of this property.
My question is

Let $S$ be an infinite pro-$p$ group of finite coclass. Suppose there exists a non-negative integer $t$ such that the solvable length of each lower central series quotient $S/\gamma_i(S)$ is less than or equal to $l$ for all $i\ge t$. Then is it true that the $S$ is solvable with solvable length less than or equal to $l$?

To recall, the coclass of a finite $p$-group $G$ of order $p^n$ is defined as $n-c$ where $c$ is the nilpotency class of $G$. In case of infinite pro-$p$ groups, an infinite pro-$p$ group $S$ is said to be of finite coclass $r$ if its lower central series quotients $S/\gamma_i(S)$ are finite $p$-groups and $S/\gamma_i(S)$ has coclass $r$ for all $i\ge t$ for some $t\ge 0$.
Thanks in advance.

REPLY [3 votes]: The answer is yes, $S$ is an inverse limit of its lower central quotients. As these have bounded derived length, the same goes for the Cartesian product of these groups.
By the way, all pro-$p$ groups of finite coclass are solvable, that's Theorem D of the coclass theory.


TITLE: A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
QUESTION [12 upvotes]: The inequality
$$
e_n:=\left(1+\frac1n\right)^n\leq3-\frac1n,
$$
where $n\in\mathbb{N}_+$, is certainly true, because we know, how LHS is connected with $e$. The other argument is the standard proof of boundedness of $(e_n)$, which uses the binomial theorem.
Are there any more elementary proofs of this inequality?

REPLY [7 votes]: Since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have:
$$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\leq\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\tag{1}$$
hence:
$$ \left(1+\frac{1}{n}\right)^n \leq \exp\left(1-\frac{1}{2n+2}\right)\leq\frac{e}{1+\frac{1}{2n+2}}=\frac{2n+2}{2n+3}e \tag{2}$$
that is stronger than $ \left(1+\frac{1}{n}\right)^n \leq 3-\frac{1}{n}$ for any $n\geq 2$.