TITLE: Enthalpy in Euler's Equations: Does Isentropic Flow Imply Homentropic Flow?
QUESTION [1 upvotes]: In the derivation of the Euler's equation of motion for ideal fluids by Landau and Lifshitz, I am able to follow the derivation of the following equation
$$ \frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (\textbf{v} \cdot \nabla) \textbf{v} = - \frac{\nabla p}{\rho} + \textbf{g}$$
This is assuming there are no viscous forces present. Furthermore if we assume there is no heat input to the system, then the process is fully isentropic, and this is expressed via the equation:
$$ \frac{Ds}{Dt} = \frac{\partial s}{\partial t} + \textbf{v} \cdot \nabla s = 0$$
This is clear to me. However, introducing thermodynamic quantities and potentials into the game is what has me confused. Namely, when we introduce the specific volume $V$, specific entropy $s$, specific enthalpy $w$, specific internal energy $u$, we promote them to  fields (i.e. functions of $\textbf{x}$ and $t$ ) and  then assume that they locally obey the usual thermodynamic equations:
$$ dw = T ds + V dp $$
$$ du = T ds - p dV $$
The notation of the differential $d$ sort of confuses me. What I think the above equation implies is three seperate equations, e.g. for the enthalpy equation is given by:
$$ \frac{Dw}{Dt}  = T \frac{Ds}{Dt} + V \frac{Dp}{Dt}$$
$$ \frac{\partial w}{\partial t}  = T \frac{\partial s}{\partial t} + V \frac{\partial p}{\partial t}$$
$$ \nabla w  = T \nabla s + V \nabla p$$
and I believe this is so because the third equation above appears on page 9 in Landau and Lifshitz. Now Landau and Lifshitz makes the claim that isentropic condition $\frac{Ds}{Dt} =0$ implies that $\nabla w = - \frac{1}{\rho} \nabla p$. But in order for this to be true, per above we need $\nabla s = 0$, which is not the same thing as $\frac{Ds}{Dt} =0$.
We need the equation $\nabla w = V \nabla p = \frac{1}{\rho} \nabla p $ in order to write as Euler's equation as
$$ \frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (\textbf{v} \cdot \nabla) \textbf{v} = - \nabla w+ \textbf{g}$$
So I guess my main questions are : where is the logic wrong here? Is my interpretation of the equation with the differentials  wrong? How does $\frac{Ds}{Dt} = 0$ imply that $\nabla w = V \nabla p$?

REPLY [1 votes]: You're right with your interpretation of the differentials, from differantials to differential operators acting on fields.
You're right that in continuum mechanics, we are promoting thermodynamic variables to fields, since we often assume local thermodynamic equilibrium, i.e. time to reach thermodynamic equilibrium is much faster than characteristic time of fludi dynamics processes.
The last step you're missing, and Landau is not telling you explicitly is that

$\dfrac{D s }{D t} = 0$ and
uniform conditions at far field,  $s(\mathbf{r}_{\infty, inflow}) = \overline{s}$, and
no shock occurring in the domain of interest (since shocks produce entropy, even though you can't approach them with differential equations, if you model them as discontinuities in a inviscid model),

imply that entropy is uniform across the whole domain of interest $s(\mathbf{r}) = \overline{s}$.
You can get a "qualitative proof" of the statement above thinking at the meaning of the material derivative, i.e. the rare of change of the quantity involved as it's seen by a material particle. If every material particle entering the domain of interest has initial entropy $\overline{s}$ (assumption 2.), and there is no entropy production (assumption 3., no shock), all the fluid particles in the domain have the same entropy, and thus the space derivatives are zero as well.


text:
{\bf Problem.} A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length 7. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside of the circle. If $\angle{APD}=60^\circ$ and $BP=8$, then $r^2=$
$\text{(A) } 70\quad \text{(B) } 71\quad \text{(C) } 72\quad \text{(D) } 73\quad \text{(E) } 74$

{\bf Level.} Level 5
{\bf Type.} Geometry
{\bf Solution.} [asy]  import olympiad;  import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); draw(circ); draw(A--P--D); label('$A$', A, N); label('$D$', D, S); label('$C$', C, SE); label('$B$', B, NE); label('$P$', P, E); label('$60^\circ$', P, 2 * (dir(P--A) + dir(P--D))); label('$10$', A--B, S); label('$8$', B--P, NE); label('$7$', C--D, N);  [/asy]
Applying Power of a Point on $P$, we find that $PC=9$ and thus $PD=16$. Observing that $PD=2BP$ and that $\angle BPD=60^{\circ}$, we conclude that $BPD$ is a $30-60-90$ right triangle with right angle at $B$. Thus, $BD=8\sqrt{3}$ and triangle $ABD$ is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem $AD=2r=2\sqrt{73}$. From here we see that $r^2=\boxed{73}$.

set_name: MATH


TITLE: Convert a equation with fractions into whole numbers
QUESTION [1 upvotes]: So I have this equation:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So this is a really easy problem, I could just multiply
$$\frac{2}{3}*\frac{3}{3} = \frac{6}{9}$$
Then subtract
$$\frac{6}{9}a^2 - \frac{4}{9}a^2 = \frac{2}{9}a^2=8a$$
$$36a=a^2$$
$$36=a$$
However, I want to solve the equation by getting rid of the fractions right at the beginning:
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
So I thought it'd be much simpler if I could get rid of these fractions by multiplying everything by a single value. Therefore, I thought what value can I multiply 2 and 4 so it gives me a divisible value by 3 and 9? It took me some time but I came up with 9
$$\frac{2}{3}a^2-\frac{4}{9}a^2 = 8a$$
$$9(\frac{2}{3}a^2-\frac{4}{9}a^2) = 9(8a)$$
$$6a^2-9a^2 = 8a$$
My question is if there is an easier way to find this value that when multiplied it eliminates the fractions? It took a few valuable seconds to figure out it was 9, so I was wondering if this process has a name or any rules to find it quicker?

REPLY [0 votes]: Method 1:  $\frac ab x + \frac cd x = gy$
Multiple by both $b$ and $d$.  Multiplying by $bd$ will always eleminate everything.  so $bd\frac abx + bd \frac cd x = adx + bc x = bdgy$.
Ex: $\frac 23 a^2 - \frac 49 a^2 = 8a$
$3*9\frac 23 a^2 -3*9\frac 49 a^2 = 3*9*8a$
$18a^2 - 12a^2 = 216 a$
But notice that was over kill. Every term is a multiple of 3.
Method 2: Do it one term at a time:
$\frac ab x + \frac cd x = gy$
$ ax + \frac {bc}dx = by$
$ dax + bcx = bdy$.
Ex:
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$3\frac 23 a^2 - 3\frac 49a^2 = 2a^2 - \frac 43 a^2 = 3*8a = 24a$
$3*2a^2 - 3\frac 43 a^2 = 6a^2 - 4a^2 = 3*24a = 72a$
Notice in this case you avoided the overkill by factoring out the excessive $3$ as it appeared.  Had you started with $9$ it wouldn't have come up at all.
$\frac 23 a^2 - \frac 49 a^2 = 8a$
$9*\frac 23 a^2 - 9*\frac 49 a^2 = 9*8a$
$3*2a^2 - 4a^2 = 72a$.
Method 3:  The conclusion we can reach. Multiply by the least common multiple.
$\frac ab x + \frac cd x = gy$
$\text{lcm}(bd)\frac ab x + \text{lcm}(bd) \frac cd = \text{lcm}(bd)gy$
$d'a x + b'c x = \text{lcm}(bd) gy$.
Example: $\frac 23 a^2 - \frac 49 a^2 = 8a$.  $3=3$ and $9 = 3^2$ so $\text{lcm}(3,9) = 9$.
$9 \frac 23 a^2 + 9\frac 49 a^2 = 9*8a$
$6a^2 - 4a^2 = 9*8a$.
But notice:  You not only should get rid of the denominaters you should also get rid of the common factor two.
Final  $\frac ab x + \frac cd x = \frac gh y$.  Multiply both sides by $\frac {\text{lcm}(b,d,h)}{\gcd(a,c,g)}$
So for $\frac 23 a^2 - \frac 49 a^2 = 8a$.  $\text{lcm}(3,9) = 3$ and $\gcd(2,4,8) = 2$.
So multiply everything by $\frac 92$.
$\frac 92\frac 23 a^2 - \frac 92\frac 29 a^2 = \frac 928a$
$3a^2 - 2a^2 = 36a$.


TITLE: How large is the set of unimodular lattices whose sucesssive minima cannot be attained by a basis of lattice?
QUESTION [4 upvotes]: Recall that the $i$-th successive minimum of $L\in \mathcal L$ (space of full rank lattices in $\mathbb R^d$), denoted $\lambda_i(L)$ is the infimum of the radii of the balls containing $i$-linearly independent vectors in $L$. All norms are Euclidean 2-norm.
Let $\Lambda$ be a unimodular lattice in $\mathbb R^d$. For $d\le 4$, it is possible to construct a basis $\{v_1,\dots, v_d\}$ of $\Lambda$ such that $\lambda_i(\Lambda)=\|v_i\|$ for any $i$.
If $d>4$, then the example $\frac{1}{2^{1/d}}L_0$, where $L_0:=Span_{\mathbb Z} \{e_1,\dots, e_{d-1},\frac{1}{2}(e_1+\cdots+e_d) \}$ shows it may be impossible to find such a basis. The coefficients $\frac{1}{2^{1/d}}$ is used to scale $L_0$ to a unimodular lattice.
It seems to me that such examples are really rare, but how rare?
Consider the set $S$ of unimodular lattices whose sucesssive minima cannot be attained by a basis of lattice. Recall the space of unimodular lattices are be identified with the homogeneous space $SL(d,\mathbb R)/SL(d,\mathbb Z)$. Can we see that $S$ is contained in (countable union) of submanifold/subvariety of dimension strictly lower than the full dimension? Or show it has Haar measure zero?
If it does not have measure zero, can we give an estimate of its measure?
(1) To begin with can anyone construct any examples different from this $\frac{1}{2^{1/d}}L_0$? (As Alison commented below small perturbation of this $\frac{1}{2^{1/d}}L_0$ example may work)
(2) Can any one construct any examples significantly different from this $\frac{1}{2^{1/d}}L_0$. For example, with $\lambda_1(\Lambda)$ close to zero? Such lattices are often called "unbounded".

REPLY [1 votes]: To begin with can anyone construct any examples different from this $(1/\sqrt[d]{2})L_0$?

My (somewhat imperfect) undertsanding of this is that the answer to this question is no.
Questions related to this are studied in various works by Martinet (and others), see for example this and this, though this second paper is more relevant.
The setup is as follows.
One starts with a lattice $\Lambda$.
Then one considers its Minkowskian Sublattice $\Lambda'$, which has as its basis some set of linearly independent representatives of length $\lambda_i(\Lambda)$.
One can then consider the quotient $\Lambda / \Lambda'$, and try to classify possible pairs $(\Lambda, \Lambda')$.
This is done by letting $d\mathbb{Z}$ be the annihilator of $\Lambda/\Lambda'$, and then writing
$$\Lambda = \mathsf{span}_{\mathbb{Z}}\langle \Lambda, f_1,\dots, f_k\rangle$$
for $f_i = \frac{\sum_i \alpha_i^{(j)} e_i}{d}$, and $\alpha^{(j)} = (\alpha_1^{(j)},\dots, \alpha_n^{(j)})\in(\mathbb{Z}/d\mathbb{Z})^n$ certain words of a $\mathbb{Z}/d\mathbb{Z}$-linear code.
In this way, the example of $L_0$ corresponds to applying construction A to the (binary) repetition code $[1,1,\dots,1]$.
But one can obtain other pairs $(\Lambda, \Lambda')$ (starting in dimension 9) by instead lifting other codes.
The second link classifies the codes that can appear (in small dimensions).
Each code leads to a (somewhat) different analogue of $L_0$, though the linked work does not contain unbounded lattices.
Instead it contains the comment

It should be noted that the two mentioned applications make use only
of results for well rounded lattices, that is, for lattices with minimal vectors spanning $E$. In other words, these lattices have equal successive minima $\lambda_1(L),\dots, \lambda_n(L)$. Indeed, a deformation argument (see [Mar01, Theorem 1.5]) shows that all codes can be realized using well rounded lattices.

As their primary goal is to classify the codes which can arise from the above construction, they reduce their analysis to the case of well-rounded lattices (which are not unbounded in your sense).
None of this is useful for your overall of quantifying how "large" this set of lattices is, but may be useful for additional explicit examples of analogues of $L_0$.


TITLE: Classify all groups with order $pq$, $p$ is not equal to $q$
QUESTION [2 upvotes]: I am reading the following material:
http://www.math.ucla.edu/~iacoley/hw/algqual.pdf
. At page 5, there is a proof classifying all groups with order $pq$(they are all primes and $p$ is not equal to $q$). I am lost in the last two sentence, which claims the semidirect product are all isomorphic. Could someone help me to understand it? Thanks!

REPLY [4 votes]: I'll try my best to explain both sentences in turn. Apologies in advance if I get anything wrong.

For two non-trivial homomorphisms $\varphi, \psi : P \rightarrow \text{Aut}Q$, we can construct an isomorphism of
  $Q \rtimes_\varphi P$ and $Q \rtimes_\psi P$ via an automorphism of $P$, since $P$ is cyclic.

Since $P$ has a prime order $p$, all of its non-trivial elements are generators. This gives us precisely $p - 1$ unique automorphisms of $P$, since what matters is where the automorphism sends the generators. Specifically, one particular generator can be sent to any of the $p - 1$ generators in $P$.
Take a generator $g \in P$. Then, we know $\varphi(g) = a_\varphi$ and $\psi(g) = a_\psi$ for $a_\varphi, a_\psi \in \text{Aut}Q$. If $a_\phi = a_\psi$, then things are straightforward, so let's suppose $a_\phi \neq a_\psi$. Then, we take an automorphism $\alpha : P \rightarrow P$ such that $\psi(\alpha(g)) = a_\varphi = \varphi(g)$.
We are guaranteed such an $\alpha$ exists since the images of $\varphi$ and $\psi$ are the same unique subgroup of $\text{Aut}Q$ of order $p$, which I will call $\text{Aut}Q^*$. Since $|P| = |\text{Aut}Q^*|$, $\varphi$ and $\psi$ are surjective onto $\text{Aut}Q^*$, and so for any $a \in \text{Aut}Q^*$, there exist $g,g' \in P$ such that $\varphi(g) = a$ and $\psi(g') = a$. We are simply finding the automorphism $\alpha$ where $\alpha(g) = g'$, and thus $\psi(\alpha) = \varphi$.
Take a mapping $\beta : Q \rtimes_\varphi P \rightarrow Q \rtimes_\psi P$ given by:
$$\beta : (q,p) \rightarrow (q,\alpha(p))$$
Suppose $\beta((q_1,p_1)) = \beta((q_2,p_2))$. Then, $(q_1,\alpha(p_1)) = (q_2,\alpha(p_2))$, which gives us $q_1 = q_2$ and $\alpha(p_1) = \alpha(p_2)$; apply the inverse of $\alpha$ to obtain $p_1 = p_2$, so $\beta$ is injective.
Take $(q,p) \in Q \rtimes_\psi P$. Then, we clearly have $\beta((q,\alpha^{-1}(p))) = (q,p)$, so $\beta$ is surjective (and therefore bijective). Furthermore, we have:
$$\begin{equation}
\begin{split}
\beta((q_1,p_1) \times_\varphi (q_2,p_2)) & = \beta((q_1\varphi(p_1)(q_2),p_1p_2)) \\ & = (q_1\varphi(p_1)(q_2),\alpha(p_1p_2)) \\ & = (q_1\psi(\alpha(p_1))(q_2),\alpha(p_1)\alpha(p_2)) \\ & = (q_1,\alpha(p_1)) \times_\psi (q_2,\alpha(p_2))) \\ & = \beta((q_1,p_1)) \times_\psi\beta((q_2,p_2))
\end{split}
\end{equation}$$
where $\times_\psi$ and $\times_\varphi$ are the operations of $Q \rtimes_\psi P$ and $Q \rtimes_\varphi P$, respectively; thus, $\beta$ is a homomorphism. Therefore, since $\beta$ is bijective, it is also an isomorphism, and so $Q \rtimes_\varphi P \cong Q \rtimes_\psi P$. This applies to all such pairs of semidirect products with nontrivial homomorphisms $\varphi, \psi$, so there is only one (up to isomorphism) such semidirect product.

Hence there are two nonisomorphic groups of order $pq$, which correspond to a trivial $\varphi$ and a nontrivial $\varphi$.

Combined with what was found earlier in the solution, we have two total cases:

If $p \nmid q - 1$, then $G \cong \mathbb{Z}_{pq}$.
If $p \mid q - 1$, then $G \cong Q \rtimes_\varphi P$ for a normal $q$-Sylow subgroup $Q$, a $p$-Sylow subgroup $P$, and a homomorphism $\varphi : P \rightarrow \text{Aut}Q$.

Thus, we have classified groups of order $pq$ into two nonisomorphic groups up to isomorphism.


TITLE: Motivation for Application of Dedekind-Hasse Norm
QUESTION [1 upvotes]: I am reading about how one can use the Dedekind-Hasse norm to show that $\textbf{Z}[(1+\sqrt{-19})/2]$ is a Principal Ideal Domain. I understand the proof outlined in the link I gave, but I have a question about step C. of part ii. at the bottom of page 3. They essentially define a bunch of new variables, and it indeed turns out that such definitions do work out in the proof, which I can understand.
However, I don't really know how I would be able to come up with something like that; it really does seem like they just pulled these out of nowhere.
I ended up deriving that
$$N\left(\frac\alpha\beta s-t\right)=\frac{a^2s^2-2acst+19b^2+c^2t^2}{c^2}=\frac{(as-ct)^2+19b^2}{c^2}$$
but I am not sure how to proceed to what they ended up getting.
Can someone shed light on this?

REPLY [1 votes]: we have $\frac{\alpha}{\beta}=\frac{a+b\delta}{c}$, $s=y+x\delta$, $t=q-z\delta$,  $ax+by+cz=1$, and $ay+\delta^2 bx=cq+r$, where $\delta=\sqrt{-19}$.
Now
\begin{align*}
\frac{\alpha}{\beta}s-t& = \left(\frac{a+b\delta}{c}\right)(y+x\delta)-q+z\delta\\
&=\left(\frac{ay+bx\delta^2}{c}-q\right)+\left(\frac{ax+by}{c}+z\right)\delta\\
&=\left(\frac{cq+r}{c}-q\right)+\left(\frac{1-cz}{c}+z\right)\delta\\
&=\left(\frac{r}{c}\right)+\left(\frac{1}{c}\right)\delta.
\end{align*}
Now
$$N\left(\frac{\alpha}{\beta}s-t\right)=\left(\frac{r}{c}\right)^2+19\left(\frac{1}{c}\right)^2=\frac{r^2+19}{c^2}$$


text:
\begin{document}

\title{Feynman graphs and renormalization \\ in quantum diffusion
\footnote{Talk given by M. Salmhofer at the conference in honour of
Wolfhart Zimmermann's 80th birthday, Ringberg Castle, February 3--6, 2008}}

\author{L\'aszl\'o Erd\H os
\\[1ex]
\small
Mathematisches Institut, Universit\" at M\" unchen,
Theresienstr. 39, D-80333 M\" unchen\\
\small
\texttt{lerdos@mathematik.uni-muenchen.de}
\\[2ex]
Manfred Salmhofer
\\[1ex]
\small
Institut f\" ur Theoretische Physik, Universit\" at Leipzig,
Postfach 100920, 04009~Leipzig\\
\small
and Max--Planck--Institut f\" ur Mathematik, Inselstr.\  22,
D-04103 Leipzig \\
\small
\texttt{salmhofer@itp.uni-leipzig.de}
\\[2ex]
Horng--Tzer Yau
\\[1ex]
\small
Mathematics Department, Harvard University,
Cambridge, MA 02138, USA
\\
\small
\texttt{htyau@math.harvard.edu}
}

\maketitle
\begin{abstract}
We review our proof that in a scaling limit, the time evolution
of a quantum particle in a static random environment leads to
a diffusion equation.  In particular, we discuss the role of
Feynman graph expansions and of renormalization.
\end{abstract}

\section{Introduction}
The emergence of irreversibility from reversible dynamics
in large systems has been one of the fundamental questions
in science since the days of Maxwell and Boltzmann.
The famous debate about the statistical character
of the second law of thermodynamics and the related controversy
about Boltzmann's {\em Sto\ss zahlansatz} in the derivation of
his transport equation has been very fruitful for physics and
mathematics. After Lanford's rigorous justification of the Boltzmann
equation for a classical many--particle system at short kinetic time
scales\cite{Lanford}, the mathematical justification of the Boltzmann equation
at longer timescales has remained a challenge up to the present time.
The analogous statement for quantum systems remains open
even at the short kinetic timescale.

A related important question is to understand how Brownian
motion emerges as an effective law from time-reversal-invariant
microscopic physical laws,
as given by a Hamiltonian system or the Schr\" odinger equation.
Kesten-Papanicolaou\cite{KP} proved that the velocity distribution
of a classical particle moving in an environment consisting of
random scatterers (i.e., Lorenz gas with random scatterers)
converges to a Brownian motion in a weak coupling limit
in dimensions $d\ge 3$.
In this model the bath of light particles whose fluctuations lead to
the Brownian motion of the observed particle is
replaced with random static impurities.
A similar result was obtained in $d=2$ dimensions\cite{DGL2}.
Recently\cite{KR}, the same evolution was controlled
on a longer time scale and the position process was proven
to converge to Brownian motion as well.
Bunimovich and Sinai\cite{BS} proved the convergence of the
periodic Lorenz gas with a hard core interaction to a Brownian motion.
In this model the only source of randomness is the distribution
of the initial condition.
Finally, D\"urr, Goldstein and Lebowitz\cite{DGL1} proved that
the velocity process of a
heavy particle in a light ideal gas, which is
a model with a dynamical environment, converges to the
Ornstein-Uhlenbeck process.

Although Brownian motion was discovered and first studied
theoretically in the context of classical dynamics,
it also describes the motion of
a quantum particle in a random environment, on a timescale that is long
compared to the standard kinetic timescale\cite{ESY1,ESY2,ESY3}.
In the following we describe
this result and the strategy of the proof in a bit more detail.  Besides the motivation
discussed above, the random Schr\" odinger operator that we study is also
the standard model for transport of electrons in metals with impurities,
which plays a central role in the theory of the metal--insulator transition\cite{Lee,VW}.
The outstanding open mathematical question in this area is the proof
of the extended states conjecture, stating that in dimensions $ d \ge 3$,
at weak disorder, the spectrum of such Hamiltonians is absolutely continuous.
Despite much effort, this conjecture has up to now only been proven\cite{Kl,ASW,FHS}
on the Bethe lattice, which can be interpreted as the case $d = \infty$.
In a system with a magnetic field, the existence of dynamical delocalization
at certain energies near the Landau levels has been proven recently\cite{GKS}.

\section{The problem and the main result}
We consider random Schr\" odinger operators,
both on a lattice and in the continuum, in $d \ge 3$
dimensions. In this presentation, we focus on the case $d=3$.
The time evolution of the {\em Anderson Model} (AM) is generated by
\begin{equation}
\I \sfrac{\del}{\del t} \psi(t) = H \psi (t)
, \;
\psi (0) = \psi_0
\;
\mbox{ with }
\;
H =  -\frac12 \Delta + \lambda V_\omega
\;
\mbox{ on }
\ell^2 (\bZ^d)
\end{equation}
where $-\Delta$ is the standard discrete Laplacian
and the potential is given by
$ V (x) = \sum_{a \in \bZ^d} V_{a} (x)$, with $ V_a(x) = v_a \delta_{x,a}$,
and $v_a$  independent identically distributed (i.i.d.) random variables.
We assume that
$m_k = \bE \left( v_a ^k \right)$ satisfies
\begin{equation}\label{momentcondition}
\forall i \le 2d:
m_i < \infty , \qquad
m_1=m_3=m_5=0, \quad m_2 = 1.
\end{equation}
The continuum analogue of this model is the {\em Quantum Lorentz Model} (QLM),
where $H =  -\frac12 \Delta + \lambda V_\omega$  on $L^2 (\bR^d)$,
with $\Delta $ the standard Laplacian, $V_\omega (x) = \int_{\bR^d} B(x-y)\dd \mu_\omega(y)$,
where $B$ is a fixed spherically symmetric Schwarz function with $0 \in $ supp $\hat B$,
$\mu_\omega$ is a Poisson point process on $\bR^d$ with
homogeneous unit density and i.i.d.\ random masses:
\begin{equation}
\mu_\omega = \sum_{\gamma=1}^\infty v_\gamma(\omega) \delta_{y_\gamma(\omega)} .
\end{equation}
$\{ y_\gamma(\omega)\}$ is Poisson,
independent of the weights $\{ v_\gamma(\omega)\}$.
Again, $m_k:=\bE_v \, v_\gamma^{k}$  is assumed to satisfy
(\ref{momentcondition}).

Suppose the initial state is localized, i.e.\ $\hat \psi_0$ is smooth.
How does the solution $\psi(t) = \E^{-\I t H} \psi_0$ behave
for large $t$ ?
If $\lambda =0$, the time evolution is easily calculated in
Fourier space: $\hat\psi (t,k) = \E^{-\I t e(k)} \hat \psi_0 (k) $,
with $e(k) = k^2/2$ (QLM) or $e(k) = \sum_{i=1}^d (1- \cos k_i)$ (AM).
It is equally easy to see that the motion is ballistic, i.e.\
\begin{equation}
\langle X^2\rangle_{t} = \langle \psi(t) , X^2 \psi(t) \rangle
\sim t^2 .
\end{equation}
If $\lambda \ne 0$, one expects either localization, $ \langle X^2\rangle_{t}  = O(1)$ for all $t$,
or diffusive behaviour (extended states),
$\langle X^2\rangle_{t}  =  O(t)$, depending on $\lambda$ and $\hat \psi_0$.
The localized behaviour corresponds to dense pure point spectrum at almost every energy;
this was proven for large disorder\cite{FS,AM} and away from the spectrum of the
Laplacian. Extended states correspond to absolutely continuous spectrum.
As mentioned, the latter has been proven\cite{Kl,ASW,FHS} on the Cayley tree
for small $\lambda > 0$.
At this time there is no proof of existence of extended states in $d=3$.
For a simpler case, namely that of randomness with a decaying envelopping function,
i.e.\  $V_\omega (x) = \omega_x h(x)$, $\omega_x$ i.i.d., $h$ fixed,  there is a proof\cite{RS,B}
that for $\eta > \frac12 $ and $h (x) \sim |x|^{-\eta}$ as $|x| \to \infty$,
$H = - \Delta + V_\omega$ has absolutely continuous spectrum.

Our result is formulated in terms of the {\em Wigner function}
\begin{equation}
W_\psi (x,v)
=
\int \dd y \;
\E^{\I vy}
\overline{\psi (x+\frac{y}{2})} \,
\psi (x-\frac{y}{2})
\end{equation}
which can be thought of as an analogue of a phase space density (but can become negative).
Its marginals are
$
\int W_\psi (x,v)  \dd x
=
|\hat \psi (v)|^2
$ and
$
\int W_\psi (x,v)  \dd v
=
|\psi (x)|^2
$.
Moreover,
\begin{equation}\label{hat-wig}
\hat W_\psi (\xi,v) = \int \dd x\; \E^{-\I x \xi} W_\psi (x,v)
=
\overline{\hat \psi (v - \xi/2)}\, \hat \psi (v+\xi/2) .
\end{equation}
On the lattice, one has to
modify the definition of the Wigner transform slightly\cite{ESY3}.

The {\em kinetic scaling} is given by
\begin{equation}
\epkin = \lambda^2, \qquad \cT = \epkin t, \qquad \cX= \epkin x,
\end{equation}
i.e.\ the microscopic time and space variables both become of order $\lambda^{-2}$,
so that velocities remain unscaled.

\begin{theorem}
\begin{equation}
\bE W^\epkin_{\psi(\cT \epkin^{-1})}  (\cX,\cV)
\gtoas{\epkin \to 0} F(\cX,\cV,\cT),
\end{equation}
$F$ the solution of the {\em linear Boltzmann equation}
\begin{eqnarray}
& &\frac{\del}{\del \cT} F(\cX,\cV,\cT)
+
(\nabla e) (\cV) \cdot \nabla_{\cX} F(\cX,\cV,\cT)
\nonumber\\
&&=
2\pi \int \dd \cU \;
\delta(e(\cU)-e(\cV)) \;
\abs{\hat B(\cU - \cV)}^2 \;
\left[
F(\cX,\cU,\cT) - F(\cX,\cV,\cT)
\right] .
\end{eqnarray}
\end{theorem}

\noindent
This theorem was first proven for the continuum for small
time $\cT$ \cite{Sp1}, then for arbitrary time\cite{EY},
and later extended to the lattice case\cite{Ch}.

The {\em diffusive scaling} is defined by
\begin{equation}
\eplo = \lambda^{2+\kappa/2},
\qquad
X = \eplo x,
\qquad
T= \eplo \lambda^{\kappa/2} t = \lambda^{\kappa+2} t .
\end{equation}
This is long compared to the kinetic timescale:
the kinetic variables $\cX $ and $\cT$ diverge as $\lambda \to 0$
when $X$ and $T$ are kept fixed,
\begin{equation}
\cX = \lambda^{-\kappa/2} X ,
\qquad
\cT = \lambda^{-\kappa} T.
\end{equation}
A first hint at diffusion is that under this scaling $\cX^2 / \cT =X^2/T$
is independent of $\lambda$. The result for the Anderson model is

\begin{theorem}
Let $d=3$, $\psi_0 \in \ell^2(\bZ^3)$ and $\psi(t)$ be the solution to the
random Schr\" odinger equation with initial condition $\psi_0$.
If  $\kappa > 0$ is small enough
and $\eplo = \lambda^{2+\kappa/2}$,
then in the limit $\lambda  \to 0$,
$\bE W^\eplo_{\psi(\lambda^{-2-\kappa }T)} $ converges
weakly to the solution $f$ of a heat equation.

More precisely: denote
$
\langle F\rangle_E
=
\Phi(E)^{-1}
\int \dd v\; F(v) \delta(E-e(v)) ,
$
where $ \Phi(E) = \int \dd v\; \delta(E-e(v))$.
Let $E \in (0,3)$  and
$
D_{ij} (E)
=
\frac{1}{2\pi \Phi(E)}
\langle
\nabla_i e\; \nabla_j e
\rangle_E,
$
and let $f$ be the solution of the heat equation
\begin{eqnarray}
\frac{\del}{\del T} f(T,X,E)
&=&
\nabla_X \cdot D(E) \nabla_X\;  f(T,X,E)
\\
f(0,X,E)
&=&
\delta(X) \;
\langle
|\hat\psi_0|^2 \rangle_E \; .
\end{eqnarray}
Let $\cO(x, v)$ be a Schwartz function on $\bR^d\times \bR^d/2\pi \bZ^d$. Then
\begin{eqnarray}
&&
\lim_{\eplo \to 0}
\sum\nolimits_{X \in (\eplo\bZ/2)^d}
\int \! \dd v \; \cO(X, v)
\; \bE
 W^\eplo_{\psi(\lambda^{-\kappa-2} T)} (X, v)
\nonumber\\
&&
\qquad = \int_{\bR^d} \dd X \int \!\! \dd v \; \cO(X, v) \; f(T, X, e(v))  .
\end{eqnarray}
The limit is uniform on $[0,T_0]$ for any $T_0 > 0$.
\end{theorem}

\noindent
We discuss some of the ideas in the proof of this theorem in Section \ref{proofsec}.

\medskip\noindent
If $\hat \psi_0 \in C^1$
and $\lambda $ is small enough, we have the more detailed error estimate
\begin{eqnarray}
&&
\int \dd v \int \dd \xi \; \hat \cO (\xi,v) \;
\bE \hat W^\eplo_{\psi (\lambda^{-2-\kappa} T)} (\xi,v)
\\
&=&
\int \dd \xi \int \Phi (E) \dd E \;
\E^{- \frac{T}{2} \langle \xi, \; D(E) \xi \rangle_E}
\langle \hat \cO (\xi, \cdot ) \rangle_E\;
\langle \hat W_{\psi_0} (\eplo \xi, \cdot )\rangle _E
+
o(1) .
\nonumber
\end{eqnarray}
The Boltzmann equation also gives the same diffusion equation
in the long time limit, but it was itself derived from the
quantum mechanical time evolution only for shorter timescales.
The main difficulty in the proof is to deal with contributions that
vanish for $\lambda \to 0$ under kinetic scaling, but that become
important under the above--defined diffusive scaling.
More technically speaking, in the Feynman expansion done
to analyze the time evolution, most of these terms would even diverge
under diffusive scaling if we did not renormalize the propagator.

The allowed values of $\kappa $ are in an interval  $[0,\kappa_0)$,
where $\kappa_0$ is a universal constant.
For technical  reasons, $\kappa_0 $ has to be chosen very small
in the proof. Heuristically, i.e.\ ignoring many of the technical
complications and assuming optimal bounds, one would expect
the remainder of the renormalized Feynman graph expansion to vanish up to
$\kappa_0=2$, and to diverge for $\kappa_0 > 2$.

The diffusive scaling leads to a diffusion on the energy shells.
A diffusion mixing energy shells is expected to start at $t = \lambda^{-4}$.

An intuitive way of interpreting the expansion described below
is as an expansion in the number $N$ of collisions of the particle
with the randomly (but statically) arranged obstacles represented
by the potential. As compared to the previous results\cite{EY,Ch},
the main new feature here is that under diffusive scaling,
the effective number of collisions of the  particle diverges.
That is, not only is it necessary to expand to an order $N$ that
diverges as $\lambda \to 0$, but also the main contribution
does not come from terms with a finite number of collisions.

\section{Collision histories, Feynman graphs, and ladders}\label{proofsec}
We discuss some of the ideas of the proof for the example
of the Anderson model,  i.e.\ the lattice situation.
For the detailed bounds of Feynman graphs, the lattice
leads to a number of complications\cite{ESY3}, but for the presentation
it is easier.

\subsection{Collision histories}
Let us start with a formal time--ordered expansion, setting
$H_0 = - \frac12 \Delta$ and expanding in $\lambda V$.
Then
$\psi(t) = \E^{-\I t H} \psi_0 = \sum_{n \ge 0} \psi^{(n)} (t)$
with
\begin{eqnarray}
 \psi^{(n)} (t)
&=&
(-\I \lambda)^n
\int \dd \mu_{n+1} (s)
\E^{-\I s_{n} H_0} V \E^{-\I s_{n} H_0} \ldots
V \E^{-\I s_{0} H_0} \psi_0
\end{eqnarray}
where $s= (s_0, \ldots, s_n)$ and
\begin{equation}\label{mudef}
\dd \mu_{n+1} (s)
=
\int\limits_{[0,\infty)^{n+1}}
\dd s_0 \ldots \dd s_n \;
\delta \left( t - \sum_{j=0}^{n} s_j\right) .
\end{equation}
Because $V=\sum_{a\in \bZ^d} V_a$,  it is natural to split
each $\psi^{(n)}$ further,
$\psi^{(n)} (t) = \sum_{\ba_n} \psi^{(n)}_{\ba_n} (t)$.
Every sequence  of obstacle labels
$\ba_n = (a_1,\ldots,a_n) \in (\bZ^d)^n$
represents a {\em collision history},
and for $k \in \{1, \ldots, n-1\}$, the time variables $s_k$ in (\ref{mudef})
are the time differences between two subsequent collisions.
The delta function in (\ref{mudef})
enforces the constraint that these time differences, together with the
propagation times $s_0$ before the first collision and $s_{n}$ after the
last one, add up to the total time $t$. We shall discuss convergence
questions about this expansion later.

Our detailed analysis takes place in momentum space, where each
$V$ acts as a convolution operator, so that
\begin{eqnarray}\label{psimom}
\hat \psi_n (t,p_n)
\!=\!
(\mbox{-}\I)^n\!
\int \dd \mu_{_{n+1}} \!(s)\!
\int \!
\pli_{j=0}^{n-1} \sfrac{\dd^d p_j }{(2\pi)^d} \, \E^{-\I s_j e(p_j)} \!
\pli_{j=1}^n \hat V (p_j \mbox{-} p_{j-1})
\hat \psi_0 (p_0) .
\end{eqnarray}
Very schematically, one can represent this
as follows, where each of the dashed lines represents
a factor $\lambda V$ and each of the full lines
gets a phase factor $ \E^{-\I s_j e(p_j)}$.
\begin{center}
    \includegraphics[height=10mm]{befav.eps}
\end{center}
As the convolution formula shows, one can define a momentum flow
in this graph, where the momentum change $p_j - p_{j+1}$
flows away through the dashed line. Before the disorder average,
there is no translation invariance in the system, so every scattering
at an obstacle changes the momentum of the particle.

\subsection{Disorder average and graphs}

Recalling (\ref{hat-wig}), we have
\begin{equation}
\bE \left[\hat W_{\psi(t)} (\xi,v)\right]
=
\sum_{n,n'} \sum_{\ba_n,\ba'_{n'}}
\bE \left[
\overline{\hat \psi^{(n)}_{\ba_n} (t,v - \xi/2)}
\hat \psi^{(n')}_{\ba'_{n'}} (t,v+\xi/2)
\right] .
\end{equation}
Note that there are now two, a priori independent, collision histories,
one for $\psi$ and one for $\bar \psi$. It will be part of the proof
to show that, in the scaling limit we consider, the only contributions
after self--energy renormalization come from the so-called ladder graphs,
where the two collision histories are identical: $n=n'$  and ${\bf a}_n = {\bf a}'_n$.

Because the disorder is i.i.d., translation invariance holds for the
average, which means that momentum conservation also holds
for the dashed lines, which for the Anderson model simply
correspond to a factor $\lambda^2$, since the second moment
of the disorder was normalized to $1$ in (\ref{momentcondition}).

The result can be represented as a graph built of two particle
lines, particle--disorder vertices, which are joined by disorder lines,
and, if the randomness is non-Gaussian,
disorder-disorder vertices, which correspond to the higher
moments of the disorder distribution. An example is
\begin{center}
    \includegraphics[height=20mm]{aftav.eps}
\end{center}
Particle lines get propagators $\E^{-\I s_j e(p_j)}$, interaction lines
give factors $\lambda^2$, and the disorder-disorder vertex
of degree four corresponds to a factor $m_4 \lambda^4$.

It is clear that in the way the expansion was introduced above,
one really needs the assumption that arbitrary moments, not just the first
$2d$ ones, exist.  The expansion employed in the true proof contains a
stopping rule which avoids high moments, but we shall not discuss this
here in more detail. In fact, we shall in the following assume
for simplicity that the disorder is  Gaussian, so that there are no
vertices of higher degree for the dashed lines, and the average just
corresponds to a  pairing of interaction lines.

An example of a pairing is as follows
\begin{center}
    \includegraphics[height=20mm]{recpairing.eps}
\end{center}
Note that here, there is  a crossing of the two pairing lines
in the graphical representation, but there are no vertices in which more than
one interaction line enters.

A special class of pairings are the {\em up--down pairings},
where $n=n'$ and the pairing corresponds to a permutation $\si \in \cS_n$:
\begin{center}
    \includegraphics[height=20mm]{perm2.eps}
\end{center}

The most important term turns out to be the {\em ladder graph},
corresponding to $\si = \mbox{ id:} $

\begin{center}
    \includegraphics[height=20mm]{ladder.eps}
\end{center}

\subsection{Graph bounds}
In the following, we give a brief discussion of bounds of the
contributions of individual graphs,  restricting to up--down pairings.
If one takes a bound in the representation (\ref{psimom}),
each phase factor is replaced by $1$.
This leads to a bound of order $(\lambda t)^n/n!$ (where the $n!$ comes from
the time ordering implied by the delta function in (\ref{mudef})),
which does not even allow to consider the kinetic scaling where
$\lambda^2 t$ is fixed.  For this reason, the following
{\em propagator representation} is useful. Let $\eta > 0$.
Then, inserting the Fourier representation of the delta function,
\begin{eqnarray}
&&
\int_{[0,\infty)^{n+1}} \dd^{n+1} s\;
\delta \left( t - \sum_{j=1}^{n+1} s_j \right)
\prod\limits_{j=1}^{n+1}
\E^{-\I s_j e(p_j)}
\nonumber
\\
&=&
\E^{t\eta}
\int_{[0,\infty)^{n+1}} \dd^{n+1} s\;
\delta \left( t - \sum_{j=1}^{n+1} s_j \right)
\prod\limits_{j=1}^{n+1}
\E^{-\I s_j (e(p_j) - \I \eta)}
\nonumber
\\
&=&
\E^{t\eta}
\int \sfrac{\dd\alpha}{2\pi} \;
\E^{-\I t \alpha}
\int_{[0,\infty)^{n+1}} \dd^{n+1} s\;
\prod\limits_{j=1}^{n+1}
\E^{-\I s_j (\alpha - e(p_j) + \I \eta)}
\nonumber
\\
&=&
\I^{-n}
\E^{t\eta}
\int \sfrac{\dd\alpha}{2\pi} \;
\E^{-\I \alpha t}
\prod\limits_{j=1}^{n+1}
\frac{1}{\alpha - e(p_j) + \I \eta} \; .
\end{eqnarray}
It is convenient to choose $\eta = t^{-1}$.

The contribution of a permutation $\sigma \in \cS_n$,
corresponding to an up--down pairing graph $\Gamma_\sigma$,
to $\langle \hat \cO , \hat W^{\veps}_{\psi} \rangle $ is
\begin{eqnarray}
Val(\Ga_\si)
&=&
\lambda^{2n} \; \E^{2t\eta}
\int \sfrac{\dd \alpha \, \dd \beta}{(2\pi)^2}\;
\E^{\I (\beta - \alpha) t}
\nonumber
\\
&&
\int \dd \xi \int \pli_{j=0}^n \sfrac{\dd^d p_j }{(2\pi)^d}\int \pli_{k=0}^n \sfrac{\dd^d q_k }{(2\pi)^d} \;\;\;
\hat\cO (\xi,p_n) \hat W^{\veps}_{\psi_0} (\xi,p_0)
\nonumber
\\
&&
\pli_{j=0}^n
\frac{1}{\beta - \overline{\omega(q_j - \frac{\veps \xi}{2})} - \I \eta}\;
\frac{1}{\alpha - \omega(p_j - \frac{\veps \xi}{2}) - \I \eta}
\nonumber
\\
&&
\pli_{j=1}^n \delta \left( p_j - p_{j-1} - (q_{\si(j)} - q_{\si(j) -1}) \right) .
\end{eqnarray}
At the moment, $\omega (p) = e(p) \in \bR$; later,
$\omega$ will change under renormalization
and become complex.

A simple Schwarz inequality separating the dependence on the $p_i$ and
that on the $q_i$ implies that for all $\si$
\begin{equation}
\abs{Val(\Ga_\si)} \le Val(\Ga_{id}) .
\end{equation}
The ladder is easy to calculate at $\xi = 0$, and
a ladder of length $n$
is of order $\frac{1}{n!}(\lambda^2 t)^{n} = \frac{1}{n!} \cT^n$.

A crucial observation is that the values of graphs with {\em crossings}
get inverse powers of $t$, as compared to the ladder.
This follows from the bound
\begin{equation}
\int \dd p\;
\frac{1}{|\alpha - \omega(p) + \I \eta|}\;
\frac{1}{|\beta - \overline{\omega(\pm p + q)} - \I \eta|}
\le
C |\log \eta|^3
\frac{\eta^{-b}}{\tnorm{q} + \eta}
\end{equation}
($b=0$ for the continuum; $1/2 \le b \le 3/4$ on the lattice).
$\tnorm{p} = |p|$ {in the continuum},
$\tnorm{p} = \min\{ |p-v|: v_i \in \{0, \pm \pi\}\}$ on the lattice.
Again, here $\omega(p) = e(p)$.
This motivates why the ladder graph gives the dominant contribution
under kinetic scaling.
However, the number of graphs goes like $n!$, which cancels the $1/n!$,
hence expanding to infinite order one gets majorants by geometric series,
which converge  only on very short kinetic timescales $\cT$.
This is the reason for the restriction to small kinetic timescales
in the first proof\cite{Sp1} of the Boltzmann equation for the QLM.

\subsection{Expansions to finite order and remainder terms}

Major progress\cite{EY} came from the realization that
one can do an expansion to finite order with an efficient
remainder estimate.
A natural way to generate a finite--order expansion is the {\em Duhamel formula}
\begin{equation}
\psi(t) = \E^{-\I t H} \psi_0
=
\E^{-\I t H_0} \psi_0 + \int_0^t \dd s\;
\E^{-\I (t-s) H} \lambda V \E^{-\I s H_0} \psi_0 .
\end{equation}
Iteration gives
\begin{equation}
\psi(t) = \sum_{n=0}^{N-1} \psi^{(n)} (t) + \Psi_N (t),
\end{equation}
where
\begin{equation}
\Psi_N (t) = (-\I) \int_0^t \dd s\; \E^{-\I (t-s) H} \lambda V \psi^{(N-1)} (s)
\end{equation}
and
\begin{equation}
\psi^{(n)} (t)
=
(-\I \lambda)^n
\int \dd \mu_{n+1} (s)
\E^{-\I s_{n} H_0} V \ldots V\; \E^{-\I s_0 H_0} \psi_0 .
\end{equation}
An alternative way of looking at this is via
its relation to the resolvent formula
\begin{equation}
R_z
=
R_z^{(0)}
+ R_z \lambda V R_z^{(0)}
\end{equation}
where $R_z = (z-H)^{-1}$ and $R_z^{(0)} = (z-H_0)^{-1}$.
Iteration of the resolvent equation and using the Fourier
transform gives the above propagator representation directly.
The Duhamel formula is obtained via
\begin{equation}
\E^{-\I t H }
=
- \E^{t \eta}
\int
\sfrac{\dd \alpha}{2\pi \I} \;
\E^{- \I \alpha t} \;
R_{\alpha + \I \eta} .
\end{equation}

\noindent
The second crucial ingredient is that one can use
the unitarity of the full time evolution to reduce all terms
to ones where no $H$ appears in the
time evolution any more:
\begin{eqnarray}
\norm{\Psi_N (t)}
&\le &
\int_0^t \dd s\; \norm{\E^{-\I (t-s) H} \lambda V \psi^{(N-1)} (s)}
\le
\int_0^t \dd s\; \norm{ \lambda V \psi^{(N-1)} (s)} .
\end{eqnarray}
Thus
\begin{equation}
\norm{\Psi_N (t)} ^2
\le t\;  |\lambda|^2
\int_0^t \dd s\; \norm{ V \psi^{(N-1)} (s)}^2.
\end{equation}
The remaining integral over $s$ effectively gives a factor $t$,
which is the price to pay for this unitarity bound.
To control this factor, one needs exhibit more factors $t^{-1}$ in graphs
with several independent crossings, and treat graphs with
only one crossing explicitly (in the resolvent iteration, the
unitarity bound would be replaced by $\norm{R_{\alpha + \I \eta} } \le \eta^{-1}$).

By a Schwarz inequality, one can see that the Wigner transform is
continuous in $L^2$ norm:
\begin{equation}
\abs{\bE
\left(
\langle \hat \cO , \hat W^\veps_{\psi_1} \rangle
-
\langle \hat \cO , \hat W^\veps_{\psi_2} \rangle
\right) }
\le
C
\int\dd \xi \; \sup\limits_{v} \abs{\hat \cO (\xi,v)} \;
\sqrt{
\bE \norm{\psi_1}^2 \;
\bE \norm{\psi_1-\psi_2}^2
} .
\end{equation}
Thus the unitarity bound can also be used for the Wigner transform.
The proof  of the Boltzmann equation\cite{EY} on an arbitrarily large
kinetic timescale $\cT$  uses an expansion up to order
$N \sim \log t$.
The ladder terms give the gain  term in the Boltzmann equation.
The lowest order self--energy  correction gives the loss term
in the Boltzmann equation.  It corresponds to the ``gate'' graph
\begin{center}
    \includegraphics[height=15mm]{gate.eps}
\end{center}

\subsection{Long time scale: renormalization}
Because the ladder with $n$ rungs is of order $ (\lambda^2 t)^n / n! $,
it diverges under diffusive scaling, and so do other graphs.
To increase the time beyond $\lambda^{-2}$, we need to do
a renormalization. Formally, one can think of this as a resummation
of the gate diagrams, which are of self--energy type, but this geometric
series converges only for small $\lambda^2 t$. A way to avoid such
formal resummations is to change the way $H$ is split into a ``free''
and an interaction part, i.e., expand around a different $H_0$.
For $\veps > 0 $ set
\begin{equation}
\Theta_\veps (\alpha) = \int \dd q \frac{1}{\alpha - e(q) + \I \veps} \; .
\end{equation}
This is the value of the gate diagram at energy $\alpha$ in the Anderson model
(in the QLM, the integrand contains an additional factor from the interaction function).
The limit $\Theta (\alpha) = \lim_{\veps \to 0+} \Theta_\veps (\alpha)$ exists
and is H\" older continuous\cite{ESY1} in $\alpha$ of order $1/2$.
Let
\begin{equation}
\theta(p) = \Theta (e(p)) .
\end{equation}
The idea is now to put $\lambda^2 \theta (p)$ as a counterterm,
which by construction subtracts every insertion of a gate diagram at the point
$\alpha = e(p)$ where the particle propagator is singular. Because
$\alpha$ and $\eta$ appear only as auxiliary quantities in the expansion,
it was necessary to take $\veps \to 0$ above and define $\theta$ in an
$\alpha$--independent way.

The counterterm is added and subtracted so that the Hamiltonian does not change:
let $\omega (p) = e(p) + \lambda^2 \theta(p) $ and decompose
\begin{equation}
H= \omega (P) + U, \qquad U = \lambda V - \lambda^2 \theta(P)
\end{equation}
(where $P$ denotes the momentum operator).
The function $\omega$ can be thought of as a new dispersion relation of
energy as a function of momentum. However, $\omega$ also
has a negative imaginary part, roughly of order $\lambda^2$.
More precisely, for $d \ge 3$ there is $c >0$ such that
\begin{equation}
\mbox{ Im } \omega (p) \le - c \lambda^2 \tnorm{p}^{d-2} .
\end{equation}
Thus $H_0$ is no longer selfadjoint. However, the negative sign
of Im $\omega$ implies that the resolvent $R_{\alpha + \I \eta}$
is still well-defined, since the imaginary parts add up with the same
sign. Correspondingly, the time evolution operator $\E^{-\I s H_0}$ is
no longer unitary but it remains bounded for $s \ge 0$.
Both the Duhamel and the resolvent iteration are thus well-defined.
Besides the new propagator $(\alpha + \I \eta - \omega (p))^{-1}$,
the important change is that every factor $U$ now also contains
a counterterm insertion $- \lambda^2 \theta (p)$.
The point about these iterations is that they can be stopped (or even
modified) after every expansion step. It is thus clear that one can
group the counterterms that appear in the expansion together
with the gates that get created when taking the average over the
disorder.  The cancellation among these two terms provides a small
factor that makes such terms vanish in the diffusive scaling limit.
Moreover, it is clear that one can implement rules for
stopping the expansion independently of the subsequent disorder average.
In particular, because the randomness is i.i.d., one can avoid moments
beyond the power $2d$ by stopping the expansion when a given site
has appeared in the collision history $d$ times.
The terms to which no such repetition or renormalization cancellation
applies are expanded up to order
$n \sim \lambda^2 t  \lambda^{-\delta} \sim \lambda^{-\kappa-\delta} $,
where $\delta > 0$ depends on $\kappa$. The intuition behind this is that
certain graphs with $n \sim \lambda^2 t \sim \lambda^{-\kappa}$ give the
main contribution, and expanding up to an order that is $\lambda^{-\delta}$
higher leads again to small factors.

The imaginary part of $\omega$ gives effectively a regularization
$O(\lambda^2)$ instead of $O(\eta)$ for the denominators,
which changes the values of all diagrams significantly.
In particular, the integral for one rung of the ladder becomes
\begin{equation}
 \int \sfrac {\lambda^2\; \dd p}{ (\alpha - \overline{\omega(p+r)}
      -i\eta)
     \; (\beta - \omega(p-r)  +i\eta)} \;
=
 1+ C_0\lambda^{1- O(\kappa)}\; .
\end{equation}
where $C_0$ is a constant.
Thus with this renormalization, the ladders become of order 1,
so that one can go beyond kinetic scaling.
Indeed, in the language of Feynman graphs, the main result
can be stated informally as

\begin{quote}
After renormalization, the sum of the ladder graphs for the Wigner
transform converges to a solution of the heat equation in the
diffusive scaling limit.
\end{quote}

The precise statements are Theorems 5.1, 5.2, and 5.3 in Ref.\cite{ESY1}.
They involve in particular proving that the terms which do not
correspond to pure up--down pairings vanish in the limit,
and dealing with a number of technical complications which
arise from the fact that one has to do an expansion to a finite order.

\subsection{The key estimate for controlling combinatorics}
We have had to leave out almost all technical details
to avoid overloading the presentation,
but we should like to at least mention
the heart of the proof  here at the end,
to clarify the main ideas about the Feynman graph expansion.

Focusing on up-down pairings,
we have to deal with a combinatorial problem of
bounding the sum over the $n!$ permutations $\sigma \in \cS_n$.
As mentioned, with an expansion to infinite order,
one cannot get beyond the kinetic scaling because of this
factor $n!$. The control of the remainders is done here
by choosing  an appropriate stopping $n$ for the expansion
and by ``beating down the combinatorics by power counting''.
That is, we prove exponential
suppression of the values of Feynman graphs in the
number of crossings they have, that is, loosely speaking,
in their complexity.

The precise notion capturing the complexity of a permutation
$\sigma \in \cS_n$ is its degree $d(\sigma)$, defined as
the number of non--ladder and non--antiladder indices.
Essentially, the ladder indices are those for which $\sigma (i+1) = \sigma (i) + 1$,
and the antiladder indices are those for which $\sigma(i+1) = \sigma (i) -1$.

\begin{theorem}
Let $\Ga_\si$ be the Feynman graph corresponding to
$\sigma$. There is $\gamma > 0$ such that for all $\sigma$
\begin{equation}\label{combest}
\abs{Val(\Ga_\si)} \le C \lambda^{\gamma d(\sigma)} .
\end{equation}
\end{theorem}

\noindent
This theorem is proven using a special integration algorithm for
bounding the values of large Feynman graphs\cite{ESY1}.

The number of permutations with degree $D$ is
\begin{equation}
\cN_{n,D}
=
|\{ \sigma \in \cS_n : d(\sigma) =D \}|
\le 2 (2n)^D .
\end{equation}
Expanding up to $n = O(\lambda^{-\kappa - \delta})$, $\delta >0$, we have
by (\ref{combest}),
{if $\gamma - \kappa - \delta > 0$},
\begin{equation}
\sum_{{\sigma \in \cS_n \atop d(\sigma) \ge D}}
\lambda^{\gamma  d(\sigma)}
=
\sum_{d=D}^k \lambda^{\gamma d} \cN_{n,d}
\le
2
\sum_{d=D}^k (2\lambda)^{d(\gamma - \kappa - \delta)}
\le
O(\lambda^{D (\gamma - \kappa - \delta)}).
\end{equation}
Thus the contribution from the sum of all terms with degree $D \ge 2$
is small if $\gamma - \kappa - \delta > 0$, hence the essential restriction for
the value of $\kappa $ is that of $\gamma$. As mentioned, one would
hope to get close to $\gamma=2$ in (\ref{combest}),
but $\gamma $ has to be chosen smaller for technical reasons.

\section{Conclusion}
We have shown that, for random Schr\" odinger operators
with a weak static disorder
the quantum mechanical time evolution can be controlled
on large space and time scales where a diffusion equation
governs the behavior. The Schr\"odinger evolution is time--reversible --
yet irreversibility on large scales emerges. This
apparent controversy is resolved by noting that along the scaling limit
microscopic degrees of freedom have been effectively integrated out.

Although the expansion methods we use bear some resemblance to those
of constructive quantum field theory, there are also a few
noteworthy differences.
First, because we analyze the time evolution at real time,
the (near--)singularities of the propagators
are located on hypersurfaces, and not at points, as would be the case
in Euclidean field theories.
The singularity structure is to some extent similar to that in
real time Fermi surface problems, although  there is no
fixed Fermi surface here -- the integrals
over $\alpha$ and $\beta$ ``test'' all possible level
sets of the function $e(p)$, and this leads to a number
of serious complications.
Second, we are able to control the combinatorics of a
straightforward Feynman graph expansion in momentum space,
while the analysis in constructive field theory (to our knowledge,
always) needs to be done by cluster expansions in position space
to avoid the divergence of an infinite series of Feynman graphs.
The reason for this is twofold:
the unitarity bound allows us to do an expansion to a finite order,
and our strong improvement (\ref{combest}) over standard power counting bounds allows us to push this order so high that we can reach the scale
where diffusion sets in, while still retaining control of the remainders.

The genuine challenge is to show diffusion without taking scaling limits,
i.e. for a fixed (small) disorder $\lambda$ and for any time independent
of $\lambda$. With expansion techniques, this would require
to renormalize not only the self--energy to arbitrary order but
also the four--point functions. Refining the self--energy renormalization
poses no fundamental difficulty. The correct renormalization of all
four--point functions in this problem, however, remains a widely open problem.

\bibliographystyle{ws-procs9x6}

config: arxiv


TITLE: A string is chosen uniformly at random from the set of all binary strings of length n. Show that the expected number of zeroes is n/2
QUESTION [0 upvotes]: This was a problem on a past exam.
A string is chosen uniformly at random from the set of all binary strings of length $n$. Show that the expected number of zeroes in the string is $\frac{n}{2}$.
Can someone please provide a proof?

REPLY [2 votes]: Let $X$ be the number of zeros in a string. Then $P(X=k)=2^{-n}\binom{n}{k}$ for $0\leq k\leq n$. The expected number of zeroes in the string equals
$$
\begin{align}
\sum_{k=0}^nk P(X=k)=\sum_{k=0}^n k \frac{1}{2^n}\binom{n}{k}&=2^{-n}
\sum_{k=1}^n\binom{n}{k}\binom{k}{1}\\&=2^{-n}
\sum_{k=1}^nn\binom{n-1}{k-1}\tag{0}\\
&=n2^{-n}\sum_{u=0}^{n-1}\binom{n-1}{u}\\
&=n2^{-n}2^{n-1}\tag{1}\\
&=n/2
\end{align}
$$
where in (0) we used the fact that
$$
\binom{n}{k}\binom{k}{1}=\binom{n}{1}\binom{n-1}{k-1}
$$
and in (1) we used the binomial theorem.


TITLE: Show that a subspace of $S \times S$ is not separable
QUESTION [1 upvotes]: I'm trying to show that on $S \times S$ where $S$ is the Sorgenfrey line, the subspace $Y=\{(-x,x)\}$ is discrete so I can show that $Y$ isn't separable. I'm taking $u=[-x,a) \times [x,b)$ such that $-x<a,~ x<b$ and saying $u \cap Y=\{(-x,x)\}~ \forall x \in \mathbb{R} $. $S$ is $T_1$ so $u \cap Y$ is closed. Then $Y$ is discrete and the only dense set of a discrete set is the set of all points from that set, so there is no countable subset of $Y$ dense in $Y$, so $Y$ isn't separable.
Anything I'm missing? How should I state the last part? Can I use a similar method to show $S \times S$ isn't Lindelöf?

REPLY [1 votes]: You want to show that singletons are open in order to prove that $Y=\{(-x,x):x\in\Bbb R\}$ is discrete. Then from the uncountability of $Y$ you have that $Y$ is not separable and hence $S\times S$ is not separable (because a subspace of a separable space is separable).
You don't need to know anything about $T_1$; the definitions will suffice. The set $u=[-x,a)\times[x,b)$ (picking any $a>-x,b>x$) is open by definition of the product topology and the Sorgenfrey line, so $u\cap Y=\{(-x,x)\}$ is open by definition of a subspace. Thus singletons are open.


text:
\begin{document}

\author{Matthias Gundlach
\\ Institut f\"ur Dynamische Systeme, Universit\"at Bremen\\
Postfach 330 440, 28334 Bremen, Germany
\and Andrei Khrennikov\footnote{This research was supported by the
  visiting professor fellowship at Bremen University.} \& Karl-Olof Lindahl
\footnote{This research was supported by the grant
''Strategical investigations'' of V\"axj\"o University.}\\
School of Mathematics and Systems Engineering\\
V\"{a}xj\"{o} University, 351 95, V\"{a}xj\"{o}, Sweden}
\title{On ergodic behavior of $p$-adic dynamical systems
\footnote{This note was published in \emph{Infinite Dimensional Analysis}, Vol. 4,
No. 4  (2001) 569--577.}}

\date{\empty}
\maketitle

\begin{abstract}
Monomial mappings, $x\mapsto x^n$, are topologically transitive and
ergodic with respect to Haar measure on the unit circle in the
complex plane.
In this paper we obtain an anologous result for monomial dynamical systems
over $p-$adic
numbers. The process is,
however, not straightforward. The result will depend on the natural number
$n$. Moreover,
in the $p-$adic case we never have ergodicity on the unit circle, but on
the circles
around the point $1$.
\end{abstract}

\section{Introduction}

Investgations in $p-$adic quantum physics
  \cite{VV} -- \cite{Kh1} (especially string theory   \cite{VV},
  \cite{Freund}, \cite{Vladimirov1})
stimulated an increasing interest in studying $p-$adic dynamical systems,
see for example \cite{Thiran et al} -- \cite{Bosio}.
Some steps
in this direction \cite{Kh1}, \cite{Albeverio3}
demostrated that even the simplest (monomial) discrete dynamical
systems over the fields of $p-$adic numbers $\mathbb{Q}_p$ have quite complex
behavior. This behavior depends crucially on the prime number $p>1$ (which
determines
$\mathbb{Q}_p$). By varying $p$ we can transform attractors into centers of
Siegel discs
and vice versa. The number of cycles and their lengths also depend
crucially on $p$ \cite{Pezeda}, \cite{Nilsson}. Other aspects (such as recurrence,
renormalization and conjugacy)
of monomials and the dynamics of the endomorphisms of the multiplicative group in general was
investigated by Arrowsmith and Vivaldi \cite{Arrowsmith Vivaldi2}.

Some applications of discrete $p-$adic dynamical systems to cognitive
sciences
and neural networks were considered in \cite{Kh1},
\cite{Albeverio4}. Some of these cognitive models are described
by random dynamical systems in the fields $p-$adic numbers, see
\cite{Dubischar}, \cite{Li1}.
In this note we study ergodicity of monomial $p-$adic dynamical
systems on spheres.
For a system $\psi_n(x)=x^n$, $n=2,3,...,$ the result depends crucially on
the relation
between $n$ and $p$. Our proof is essentialy based on $p-$adic analysis
(analytic mappings),
\cite{Schikhof}. We remark that the corresponding fact for the field of
complex numbers,
$\psi_n: \mathbb{C}\rightarrow \mathbb{C}$, for the sphere
$\left|z\right|=1$, is rather
trivial, see \cite{Walters}.

\section{$p$-adic numbers}\label{p-adic numbers}

The system of $p$-adic numbers $\mathbb{Q}_p$ was constructed by K. Hensel
in the 1890s.

The field of real numbers $\mathbb{R}$ is constructed as the completion of the
field of rational
numbers $\mathbb{Q}_p$ with respect to the metric $\rho(x,y)$ $=$ $\vert x - y
\vert$, where $\vert
\cdot\vert$ is the usual valuation of distance given by the absolute value.
The fields of $p$-adic
numbers $\mathbb{Q}_p$  are constructed in a corresponding way, but  using
other  valuations.
For a prime number $p$, the $p$-adic valuation $\vert \cdot\vert_p $ is
defined in the following
way. First we define it for natural numbers. Every natural number $n$ can
be represented as the
product of prime  numbers, $n$ $=$ $2^{r_2}3^{r_3} \cdots p^{r_p} \cdots$,
and we define
$\vert n\vert_p$ $=$ $p^{-r_p}$, writing $\vert 0 \vert_p$ $=0$  and
$\vert -n\vert_p$ $=$
$\vert n \vert_p$.  We then extend the definition of the $p$-adic valuation
$\vert\cdot\vert_p$
to all rational numbers by setting
\[
\vert n/m\vert_p= \vert n\vert_p/\vert m\vert_p
\]
for $m$ $\not=$ $0$. The completion of $\mathbb{Q}_p$ with respect to the
metric
$\rho_p (x,y)$ $=$
$\vert x- y\vert_p$ is the locally compact field of $p$-adic numbers
$\mathbb{Q}_p$.

The number fields $\mathbb{R}$ and $\mathbb{Q}_p$ are unique in a
 sense, since by Ostrovsky's
theorem (\cite{Schikhof}), $\vert \cdot \vert$ and  $\vert\cdot\vert_p$ are the
only possible  valuations
on $\mathbb{Q}$, but have quite distinctive properties. The field of real
numbers $\mathbb{R}$ with its
usual valuation satisfies $\vert n\vert$ $=$ $n$ $\to$ $\infty$ for
valuations of natural numbers $n$
and is said to be {\it Archimedean.\/} By a well know theorem of number
theory \cite{Schikhof}, the only
complete Archimedean fields are those of the real and the complex numbers.
In contrast, the fields
of $p$-adic numbers, which satisfy $\vert n\vert_p$ $\leq$ $1$ for all $n$
$\in\mathbb{N}$, are
examples of {\it non-Archimedean\/} fields.

Unlike the absolute value distance $\vert \cdot \vert$, the $p$-adic
valuation satisfies the strong
triangle inequality
\begin{equation}\label{str}
|x+y|_p \leq  \max[|x|_p,|y|_p],  \quad x,y \in \mathbb{Q}_p,
\end{equation}
with equality in the case that $|x|_p\neq |y|_p$.

Write $B_r(a)$ $=$ $\{x\in \mathbb{Q}_p: |x -a|_p \leq r\}$ and $B_r^-(a)$ $=$
$\{x\in \mathbb{Q}_p:
|x -a|_p < r\}$ where $r$ $=$ $p^n$ and $n$ $=$ $0$, $\pm 1$, $\pm 2$,
$\ldots$.
These are the ``closed'' and ``open''  balls in $\mathbb{Q}_p$ while the sets
$S _r(a)$ $=$ $\{x \in K: |x -a|_p = r \}$ are the spheres in $\mathbb{Q}_p$
of such radii $r$.
Any $p$-adic
ball $B_r(0)$ is an additive subgroup of $\mathbb{Q}_p$, while the ball
$B_1(0)$ is also
a ring, which is called the {\it ring of $p$-adic integers} and is denoted
by $\mathbb{Z}_p$.

As a consequence of the strong triangle property (\ref{str}) every element
$b$ of a ball $B_r(a)$ is the center of this
ball in the sense that $B_r(b)=B_r(a)$ for every $b\in B_r(a)$.

Each $p-$adic number has a unique expansion of the form
$\sum_{n=M}^{\infty}a_np^n$, where $M$ is an integer and  $a_n\in \{0, ...,p-1\}$.
The $p-$adic integers $\mathbb{Z}_p$, are $p-$adic
numbers with expansions $\sum_{n=0}^{\infty}a_np^n$.
The absolute value of a $p-$adic number is $p^{-k}$ if and only if $k$ is the
first
digit in the expansion which differs from zero. Thus
$\left|x-y\right|_p\leq p^{-k}$ if and only if
$x\equiv y\mod p^k$.

Let $(m,n)$ denote the greatest common divisor of $m$ and $n$.
We say that $n$ is a (multplicative) \emph{unit} (with respect to the
prime number $p$) iff $(n,p)=1$. Let $G_{p^l}$, $l\geq 1$, be the
multiplicative group of units  in the residue field modulo $p^l$.
Let us by $\langle n\rangle =\{n^N:\quad N\in\mathbb{N}\}$ denote the set
\emph{generated} by $n$.

Let $\psi_n$ be a (monomial) mapping on $\mathbb{Z}_p$ taking $x$ to $x^n$.
Then all spheres
$S_{p^{-l}}(1)$ are
$\psi _n$-invariant iff $n$ is a multplicative unit.
This is a consequence of the following result in $p-$adic analysis.
A proof is given in Appendix \ref{lemma}.

\begin{lemma}
\label{lemma 1}Let $x,y\in S_1(0)$ and suppose $\left| x-y\right| _p<1.$
Then for all natural numbers $n,$
\begin{equation}
\left| x^n-y^n\right| _p\leq \left| n\right| _p\left| x-y\right| _p,
\label{fundamental-inequality}
\end{equation}
with equality for $p>2$. Moreover equality also holds for $p=2$ if $n$
is odd.
\end{lemma}

\vspace{1.5ex}
\noindent
In particular, $\psi_n$ is an isometry on $S_{p^{-l}}(1)$ if and only if $(n,p)=1$.

Therefore we will henceforth
assume that $n$ is a
unit.  Also note that, as a consequence, $S_{p^{-l}}(1)$ is not a group under
multiplication. Thus our investigations are not about the dynamics on a
compact (abelian) group.

\section{Minimality}

Let us consider the dynamical system $x\mapsto x^n$ on spheres
$S_{p^{-l}}(1)$.
The result depends crucially on the following well known result from group
theory.

\begin{lemma}\label{generation}
Let $p>2$ and $l$ be any natural number, then the natural number $n$ is a
generator of
$G_{p^l}$ if and only if $n$ is a generator of $G_{p^2}$. $G_{2^l}$ is
noncyclic for
$l\geq 3$.
\end{lemma}

\vspace{1.5ex}
\noindent
Recall that a dynamical system given by a continuous transformation $\psi$ on
a compact metric space $X$ is called \emph{topologically transitive} if there
exists
a dense orbit $\{\psi^n(x): n\in\mathbb{N}\}$ in $X$, and (one-sided)
\emph{minimal}, if all orbits for $\psi$ in $X$ are dense. For the case of
monomial systems $x\mapsto x^n$ on spheres $S_{p^{-l}}(1)$ topological
transitivity
means the existens of an $x\in S_{p^{-l}}(1)$ s.t. each $y\in S_{p^{-l}}(1)$
is a limit point in the orbit of $x$, i.e. can be represented as
\begin{equation}\label{limit point}
y=\lim_{k\rightarrow\infty} x^{n^{N_k}},
\end{equation}
for some sequence $\{N_k\}$, while minimality means that such a property holds
for any $x\in S_{p^{-l}}(1)$. Our investigations are based on the following
theorem.

  \begin{theorem}
    For $p\neq 2$ the set $\langle n\rangle$ is dense in
    $S_1(0)$ if and only if $n$ is a generator of $G_{p^2}$.
  \end{theorem}
    {\noindent \bf Proof. }
      We have to show that for every $\epsilon >0$ and
      every $x\in S_1(0)$ there is a $y\in \langle n\rangle$ such
      that $\left|x-y\right|_p<\epsilon$ if and only if $n$ is a generator of $G_{p^2}$.
      First, assume that $n$ is a generator of $G_{p^2}$.
      Let $\epsilon>0$ and $x\in S_1(0)$ be
      arbitrary. Because of the discreteness of the $p-$adic metric we can assume that
      $\epsilon=p^{-k}$ for some natural number $k$. But (according to Lemma \ref{generation})
      if $n$ is a generator of $G_{p^2}$,
      then $n$ is also a generator of $G_{p^l}$ for every natural number $l$ (and $p\neq 2$)
      and especially
      for $l=k$. Consequently there is an $N$ such that $n^N=x\mod p^k$. From the
      definition of the $p-$adic metric we have that $\left|x-y\right|_p<p^{-k}$ if and only if
      $x\equiv y\mod p^k$. Hence  $\left|x-n^N\right|_p<p^{-k}$. By reversing the arguments above,
      we see that the converse is also true.
    $\square$

Let us consider $p\not=2$ and for $x\in B_{p^{-1}}(1)$ the $p$-adic
exponential  function $t \mapsto x^t,$ see, for example~\cite{Schikhof}.
This function is well defined and continuous as a map from $\mathbb{Z}_p$ to
$\mathbb{Z}_p$.
In particular, for each $a \in \mathbb{Z}_p$, we have
\begin{equation}
\label{l1}
x^a=\lim_{k\rightarrow a} x^k, \; \; k\in\mathbb{N}.
\end{equation}
We shall also use properties of the $p$-adic logarithmic function, see, for
example \cite{Schikhof}.
Let $ z \in B_{p^{-1}}(1).$  Then $\log z$ is well defined.
For $z=1+\lambda$ with $\left|\lambda \right|_p \leq 1/p$, we have:
\begin{equation}
\label{l2}
 \log{z}=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}\Delta^k}{k}=\lambda
(1+\lambda \Delta_\lambda),
      \quad \left|\Delta_\lambda \right|_p\leq 1.
\end{equation}

\vspace{1.5ex}
\noindent
By using (\ref{l2}) we obtain that  $\log : B_{p^{-1}}(1) \to B_{p^{-1}}(0)$ is
an isometry:
\begin{equation}
\label{l3}
\vert \log x_1 - \log x_2 \vert_p= \vert x_1 - x_2\vert_p,\; \;  x_1, x_2
\in B_{1/p}(1) \; .
\end{equation}

\begin{lemma} Let $x \in B_{p^{-1}}(1), x \not = 1, a \in \mathbb{Z}_p$ and
let $\{ m_k \}$ be a sequence of natural numbers. If $x^{m_k} \to x^a,
k \to \infty,$ then $m_k \to a$ as $k \to \infty,$ in $\mathbb{Z}_p$.
\end{lemma}

\vspace{1.5ex}
\noindent
This is a consequence of the isometric property of $\log$.

\vspace{1.5ex}

  \begin{theorem}\label{theorem-minimal}
    Let $p\not=2$ and $l\geq 1$. Then, the monomial dynamical system
    $x\mapsto x^n$ is minimal on the circle $S_{p^{-l}}(1)$ if and only if
$n$ is a
    generator of $G_{p^2}$.
  \end{theorem}
{\noindent \bf Proof. }
Let $x \in S_{p^{-l}}(1).$
Consider the equation $x^a = y$.  We have that $a=\frac{\log x}{\log y}$.
As $\log: B_{p^{-1}}(1) \to B_{p^{-1}}(0)$ is an isometry, we
have $\log(S_{p^{-l}}(1)) = S_{p^{-l}}(0).$ Thus
$a= \frac{\log x}{\log y} \in S_1(0)$ and moreover, each
$a \in S_1(0)$ can be represented as $\frac{\log x}{\log y}$
for some $y \in S_{p^{-l}}(1)$.

Let $y$ be an arbitrary element of $S_{p^{-l}}(1)$ and
let $x^a=y$ for some $a\in S_1(0).$ By Theorem 3.1, if $n$ is a generator of
$G_{p^2},$
then each $a \in S_1(0)$ is a limit point of the sequence $\{ n^N
\}_{N=1}^\infty.$
Thus $a= \lim_{k\to\infty} n^{N_k}$ for some subsequence $\{N_k \}.$ By
using the continutity
of the exponential function we obtain (\ref{limit point}).

Suppose now that, for some $n,$ $x^{n^{N_k}} \to x^a.$ By Lemma 3.2 we
obtain that
$n^{N_k} \to a$ as $k \to \infty.$ If we have (\ref{limit point}) for all $y
\in S_{p^{-l}}(1),$
then each $a\in S_1(0)$ can be approximated by elements $n^N.$ In
particular,
all elements in the set $\{1,2,...,p-1,p+1,.., p^2-1 \}$ can be approximated modulo $p^2$.
Thus $n$ is a is a generator of $G_{p^2}$.
$\square$.

\vspace{1.5ex}
\noindent {\bf Example }In the case that $p=3$ we have that $\psi_n$ is
minimal if $n=2$,
$2$ is a generator of $U_{3^2}=\{1,2,4,5,7,8\}$. But for $n=4$ it is not;
$\langle 4\rangle\mod 3^2=\{1,4,7\}$. We can also see this by noting that
$S_{1/3}(1)=B_{1/9}(4)\cup B_{1/9}(7)$ and that $B_{1/9}(4)$ is invariant
under $\psi_4$.

\begin{corollary}
If $a$ is a fixed point of the monomial dynamical system $x\mapsto x^n$, then
this is minimal on $S_{p^{-l}}(a)$ if and only if $n$ is a generator of
$G_{p^2}$.
\end{corollary}

\vspace{1.5ex}
{\noindent \bf Proof. }
The assertion follows immediately from Theorem 3.2 by
topological conjugation via $x\mapsto x/a$,
$S_{p^{-l}}(a)\mapsto S_{p^{-l}}(1)$.
$\square$

\vspace{1.5ex}
\noindent {\bf Example }
Let $p=17$ and $n=3$. On $\mathbb{Q}_{17}$ there is a primitive $3$rd root of
unity, see for example \cite{Li1}.
Moreover, $3$ is also a generator of $G_{17^2}$. Therefore there exist
$n$th roots of unity different from $1$ around which the dynamics is
minimal.

\section{Unique ergodicity}

In the following we will show that the minimality of the monomial dynamical
system $\psi_n:$ $x\mapsto x^n$ on the sphere $S_{p^{-l}}(1)$ is equivalent
to its \emph{unique ergodicity}. The latter property means that there exists
a unique probability measure on $S_{p^{-l}}(1)$ and its Borel $\sigma-$algebra
which is invariant under $\psi_n$. We will see that this
measure is in fact the normalized restriction of the Haar measure on
$\mathbb{Z}_p$. Moreover, we will also see that the ergodicity with
respect to Haar measure of $\psi_n$ is also equivalent to its unique
ergodicity. We should point out that -- though many results are
analogous to the case of the (irrational) rotation on the circle, our
situation is quite different, in particular as we do not deal with
dynamics on topological subgroups.

\begin{lemma}\label{lemma-unique}
Assume that $\psi_n$ is minimal. Then, the Haar measure $m$ is the unique
$\psi_n-$invariant  measure on $S_{p^{-l}}(1)$.
\end{lemma}

{\noindent \bf Proof. }
First note that minimality of $\psi_n$ implies that $(n,p)=1$ and hence that $\psi_n$
is an isometry on $S_{p^{-l}}(1)$, see equation (\ref{fundamental-inequality}).
Then, as a consequence of Theorem 27.5 in \cite{Schikhof}, it follows that
$\psi_n(B_r(a))=B_r(\psi_n(a))$ for each ball $B_r(a)\subset S_{p^{-l}}(1)$.
Consequently, for every open set $U\neq\emptyset$ we must have that
$S_{p^{-l}}(1)=\cup_{N=0}^{\infty}\psi_n^N(U)$.
Then, it follows for a $\psi_n-$invariant measure $\mu$ that $\mu (U)>0$.

Moreover, we can split $S_{p^{-l}}(1)$ into disjoint balls of radii
$p^{-(l+k)}$, $k\geq 1$, on which $\psi_n$ acts as a permutation.
In fact, for each $k\geq 1$, $S_{p^{-l}}(1)$ is the union,
\begin{equation}\label{splitting}
S_{p^{-l}}(1)=\cup B_{p^{-(l+k)}}(1+b_lp^l+...+b_{l+k-1}p^{l+k-1}),
\end{equation}
where $b_i\in\{0,1,...,p-1\}$ and $b_l\neq 0$.

We now show that $\psi_n$ is a (transitive) permutation on the partition
(\ref{splitting}). Recall that  every element of a $p-$adic ball is
the center of that ball (see section \ref{p-adic numbers}), and as
pointed out above $\psi_n(B_r(a))=B_r(\psi_n(a))$. Consequently we
have for all positve integers $k$, $\psi_n^k(a)\in B_r(a)\Rightarrow \psi_n^k(B_r(a))=B_r(\psi_n^k(a))=B_r(a)$ so that
$\psi_n^{Nk}(a)\in B_r(a)$ for every natural number
$N$. Hence, for a minimal $\psi_n$ a point of a ball $B$ of the
partition (\ref{splitting}) must move to another ball in the partition.
Furthermore, the minimality of $\psi_n$
shows indeed that $\psi_n$ acts as
a permutation on balls. By invariance of $\mu$, all balls must have the
same positive measure. As this holds for any $k$, $\mu$ must be the
restriction of Haar measure $m$.
$\square$

\vspace{1.5ex}
\noindent
The arguments of the proof of Lemma \ref{lemma-unique} also show that Haar
measure is always $\psi_n-$invariant. Thus if $\psi_n$ is uniquely ergodic,
the unique invariant measure must be the Haar measure $m$. Under these
circumstances it is known (\cite[Theorem 6.17]{Walters}) that $\psi_n$ must
be minimal.

\begin{theorem}\label{theorem-minimal-unique}
The monomial dynamical system $\psi_n:$ $x\mapsto x^n$ on $S_{p^{-l}}(1)$ is
minimal if and only if it is uniquely ergodic in which case the unique
invariant measure is the Haar measure.
\end{theorem}

\vspace{1.5ex}
\noindent
Let us mention that unique ergodicity yields in particular the ergodicity of
the unique invariant measure, i.e. the Haar measure $m$, which means that
\begin{equation}
\frac {1}{N}\sum_{i=0}^{N-1}f(x^{n^i})\rightarrow\int\! f\, dm \text{ for all }
x\in S_{p^{-l}}(1),
\end{equation}
and all continous functions $f:\quad S_{p^{-l}}(1)\rightarrow \mathbb{R}$.

On the other hand the arguments of the proof of Lemma \ref{lemma-unique}, i.e.
 the fact that $\psi_n$ acts as a transitive permutation on each partition of
$S_{p^{-l}}(1)$ into disjoint balls if and only if
$\langle n\rangle =G_{p^2}$,  proves that if $n$ is not a generator of
$G_{p^2}$ then the system is not ergodic with respect to Haar measure. Consequently,
if $\psi_n$ is ergodic then $\langle n\rangle =G_{p^2}$ so that the system is
minimal by Theorem \ref{theorem-minimal},
and hence even uniquely ergodic by Theorem \ref{theorem-minimal-unique}. Since
 unique ergodicity implies ergodicity  one has the following.

\begin{theorem}
The monomial dynamical system $\psi_n:$ $x\mapsto x^n$ on $S_{p^{-l}}(1)$ is
ergodic with respect to Haar measure if and only if it is uniquely ergodic.
\end{theorem}

Even if the monomial dynamical system $\psi_n:$ $x\mapsto x^n$ on
$S_{p^{-l}}(1)$ is ergodic, it cannot be mixing, especially not
weak-mixing. This can be seen from the fact that an abstract dynamical
system is weak-mixing if and only if the product of such two systems
is ergodic. If we choose a function $f$ on $S_{p^{-l}}(1)$ and define
a function $F$ on $S_{p^{-l}}(1)\times S_{p^{-l}}(1)$ by
$F(x,y):=f(\log x/\log y)$ (which is well defined as $\log$ does not
vanish on $S_{p^{-l}}(1)$), we obtain a non-constant function
satisfying $F(\psi_n(x),\psi_n(y))=F(x,y)$. This shows (see
\cite[Theorem 1.6]{Walters}) that $\psi_n\times\psi_n$ is not ergodic,
and hence $\psi_n$ is not weak-mixing with respect to any invariant
measure, in particular not to the Haar measure.

\vspace{1.5ex}
\noindent
Let us consider the ergodicity of a perturbed system
\begin{equation}\label{perturbation}
\psi_q=x^n+q(x),
\end{equation}
for some polynomial $q$ such that $q(x)\equiv 0\mod p^{l+2}$,
($\left | q(x)\right |_p\leq p^{-(l+2)}$). Under this  condition, the sphere $S_{p^{-l}}(1)$ is invariant. For such a system to be ergodic
 it is necessary that $n$ is a generator of $G_{p^2}$. This follows from the
 fact that for each $x=1+a_lp^l+a_{l+1}p^{l+1}+...$  on $S_{p^{-l}}(1)$
 the condition on $q$ gives
\begin{equation}
\psi_q^N(x)\equiv 1+n^N(a_l+a_{l+1}p)p^l \mod p^{l+2}.
\end{equation}
Now, $\psi_q$ acts as a transitive permutation on the $p(p-1)$ balls of radius
$p^{-(l+2)}$ if and only if $\langle n\rangle =G_{p^2}$. Consequently, a
perturbation (\ref{perturbation}) cannot make a nonergodic system ergodic.
However, it is not clear at this point if this is true also if $|q(x)|=p^{-(l+1)}$.

\section{Acknowledgement}
We would like to thank P.A. Svensson at V\"axj\"o Univerity for consultation in algebra.

\appendix

\section{Appendix}\label{lemma}

\begin{lemma}\footnote{This result was obtained in \cite{Li1}.}
Let $x,y\in S_1(0)$ and suppose $\left| x-y\right| _p<1.$
Then for all natural numbers $n,$
\begin{equation}
\left| x^n-y^n\right| _p\leq \left| n\right| _p\left| x-y\right| _p,
\end{equation}
with equality for $p>2$. Moreover equality also holds for $p=2$ if $n$
is odd.
\end{lemma}

{\noindent \bf Proof. }First note that
\begin{eqnarray*}
\left| x^n-y^n\right| _p &=&\left| (y+x-y)^n-y^n\right| _p=\left|
\sum_{k=0}^n\binom nky^{n-k}(x-y)^k-y^n\right| _p \\
&=&\left| \sum_{k=1}^n\binom nky^{n-k}(x-y)^k\right| _p\leq \max_k\left|
\binom nky^{n-k}(x-y)^k\right| _p,
\end{eqnarray*}
where the inequality is a consequence of the strong triangle inequality (\ref{str}). We
will see that the maximum occurs for $k=1$ and that the corresponding term is
greater than every other term (except for the case where $p=2$ and $n$ is even).
By the multiplicative property
of the valuation, the valuation of the $k$th term in the sum can be
estimated according to
\begin{align*}
\left| \binom nky^{n-k}(x-y)^k\right| _p &=\left| \binom nk\right| _p\left|
y\right| _p^{n-k}\left| x-y\right| _p^k \\
&=\left| n(n-1)\cdot ...\cdot (n-k+1)\right| _p\left| x-y\right| _p\frac{
\left| x-y\right| _p^{k-1}}{\left| k!\right| _p} \\
&\leq \left| n\right| _p\left| x-y\right| _p\frac{\left| x-y\right| _p^{k-1}
}{\left| k!\right| _p}\\&=\left| \binom n1y^{n-1}(x-y)^1\right| _p\frac{\left|
x-y\right| _p^{k-1}}{\left| k!\right| _p},
\end{align*}
where the inequality holds because $\left| a\right| _p\leq 1$ for every
integer $a\in \mathbb{Z}$ (together with the multiplicative property of the
valuation). Note that strong inequality holds in the case that $p=2$ and $n$ is odd;
if $n$ is odd we have $\left| n\right| _2=1$ ($n$ is not divisible by $2$). Moreover
we can assume that $n>1$ (the lemma is automatically true for $n=1$) so that
$\left| n-1\right| _2<1$.

To estimate the latter expression, $\left| x-y\right| _p^{k-1}/\left| k!\right| _p$,
we use the fact that
$\left| k!\right| _p\geq 1/p^{k-1}$ with strong inequality for $p>2$,
see for example the paper \cite[Lemma 25.5]{Schikhof}.
Moreover $\left| x-y\right| _p\leq 1/p$ by the condition of the lemma.
Consequently,
\[
\left| x-y\right| _p^{k-1}\leq 1/p^{k-1}.
\]
Hence we have that $\left| x-y\right| _p^{k-1}/\left| k!\right| _p\leq 1$
with strong inequality when $p>2.$ Therefore, for every integer $k$ such
that $1\leq k\leq n,$ we have
\[
\left| \binom nky^{n-k}(x-y)^k\right| _p\leq \left| \binom
n1y^{n-1}(x-y)^1\right| _p,
\]
with strong inequality for all $p$ and $n$ except the case  when $p=2$ and $n$ is even.
Hence, as a consequence of the strong triangle inequality (\ref{str}) we have that
\[
\left| x^n-y^n\right| _p\leq \left| \binom n1y^{n-1}(x-y)^1\right| _p=\left|
n\right| _p\left| x-y\right| _p,
\]
with equality for $p>2$ and, if $n$ is odd, also for $p=2$.
$\Box $


TITLE: Explicit construction of a finite field with $p^n$ elements
QUESTION [3 upvotes]: I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:
Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.
Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.
I don't understand the following points:

Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist?
Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?

Thanks in advance for any help!

REPLY [3 votes]: The existence of an irreducible polynomial can be shown by counting arguments.  For example, a quadratic is either irreducible, or the product of two linear polynomials.  We can restrict everything to be monic.  We now how many monic linear polynomials there are, and we know how many monic quadratic polynomials there are.  So we can count the product of two monic linear polynomials, and see that there are still more monic quadratics that must therefore be irreducible.  Note that this even allows you to calculate the exact number of irreducible polynomials.
This can be generalized to higher degrees; for example if a quadratic is not irreducible, we can factor it into a product of irreducible cubics, quadratics, and linear factors in certain predictable ways.  Lidl and Niederreiter use the Moebius function to give a formula for the number of irreducible polynomials of any given degree over a finite field.
For the second question, consider the elements of $\mathbb{F}_{q}[x]/(f(X))$; they have as representatives the polynomials of degree at most $n-1$, since a polynomial of degree $n$ or larger can have multiples of $f(x)$ subtracted from it to reduce the degree.  Counting these polynomials gives the order of the new field.


TITLE: Why does symmetry spontaneously break?
QUESTION [2 upvotes]: Why does symmetry spontaneously break? For example,in the case of the mexican hat potential, as far as I can tell, a particle that is at the unstable equilibrium ought to be able to continue staying at that equilibrium. Why would the blue particle move down?
Isn't it unphsical to say that the symmetry is broken?
What are the axioms of gauge theory that allow for spontaneous symmetry breaking?

REPLY [1 votes]: There are two things happening here.
SSB refers to cases where the symmetry of a Lagrangian or Hamiltonian does not match the symmetry of its ground states.
Taking the Hamiltonian case, you imagine a unitary operator $U$ that expresses a symmetry of the system, such that $U H U^\dagger = H$ and therefore $[U, H] = 0$ so that $U + U^\dagger$ and $i U - i U^\dagger$ must be observables that are conserved. We say that this symmetry is spontaneously broken when there is a family of ground-states $|g_\phi\rangle$ such that they all have the same energy but $$U |g_\phi\rangle = |g_\xi\rangle \text{ for } \phi\ne\xi.$$
Now, such a system could be in a state $\frac1N \sum_\phi |g_\phi\rangle$ or $\int \mathrm d\phi~f(\phi)~|g_\phi\rangle$ depending on whether the index is discrete or continuous. This state is also an eigenstate of the Hamiltonian with the ground energy. In such a case the symmetry is being broken by measurement. The standard reasoning applies: there must be some other system which is weakly coupled to $U + U^\dagger$ or so, such that the two get correlated and some sort of information about $\phi$ is getting leaked to an experimental apparatus. Then the system needs to fall into one of those eigenstates in particular and you have a nonunitary collapse to one of the ground states. That's the first thing that happens.
The second thing is dissipation. Of course you might have higher-energy excited states that are not $U$-sensitive, $U|e\rangle = |e\rangle.$ This is the sort of thing you are thinking about with your particle right on the tip of the Mexican hat. As long as the system is never coupled to an external environment, this sort of thing can remain in a strange state: it might not remain precisely over the center; the wavefunction is more likely to expand and contract and then expand again and contract again, eventually reforming the singularity. But once it is being coupled to that larger environment, one starts to imagine noise creeping in and decoherence happening and the system decaying down to a lower energy state.
There is no general theorem that I know of which says that this always happens, if you are not counting, say, the second law of thermodynamics or so. You do need to have some sort of dissipation and measurement and all of that other stuff to really make sense of it, and therefore you have to decide on a form under which those terms enter your Hamiltonian, and this prevents the derivation from being truly universal.
What usually happens in these non-general derivations on the condensed matter side is something like the following:

You assume very weak coupling to an infinite number of fermionic or bosonic creation and annihilation operators.
You use the state-matrix formulation to put all of those modes into a thermal state.
Then you look typically for second-order effects on the overall state matrix, usually using the interaction picture.
These second-order effects typically take the form of some sort of integral or double-integral, so you argue that the integral can be simplified with a Born-Markov approximation.
If you're lucky, some terms come out at the end having that beautiful characteristic Lindblad form, $A \rho A^\dagger - \frac12 A^\dagger A \rho - \frac12 \rho A^\dagger A,$ often with some sort of really strange limit needed on the coupling (like you have N modes with a coupling constant that decays like $1/\sqrt{N}$ or something) which you just brush past and say “assume this is some constant.”
That Lindblad form then describes a nonunitary dynamics and you can look there for things like dissipation, and you can try to convert it to a stochastic equation and do numerical simulations. But in the spontaneous-symmetry-breaking side, that can create a flow from the center of the Mexican hat out towards the ground states.

I have also seen a bunch of other approaches to quantum mathematics but I haven't worked in the field in years so I don't remember in detail whether they apply here. So for example I can't tell you whether the folks working with random matrices and pulling out spectrums of eigenvalues have some super-easy way to do these sorts of manipulations.


TITLE: Justification for a real number to a complex power
QUESTION [4 upvotes]: I searched extensively online and found that everyone would agree with the following:
$2^i = e^{\log{2^i}} = e^{i*\log{2}}$. About the second equality (taking the $i$ out of the $\log$), it seems natural. But I cannot justify this myself.
I understand that, $\log(z1*z2) = \log(z1)+\log(z2)$, and therefore, if I have something like $\log(i^2)$, then it's just $\log(i*i) = \log(i) + \log(i) = 2\log(i)$, but this is because we have a rational exponent. How can I do the similar thing for $2^i$? Or is there some other way to justify the step pulling the $i$ out of the $log$?
Thanks very much!

REPLY [1 votes]: By definition, if $a>0$, then $a^z=e^{z\times\log a}$. In particular, $2^i=e^{i\times\log2}$.
Of course, here $\log a$ is the only real logarithm of $a$. In general, a non-zero complex number has infinitely many logarithms, and therefore you shouldn't mention $\log z$ unless you state clearly which logarithm you have in mind. In particular, the equality $\log(z_1z_2)=\log(z_1)+\log(z_2)$ does't make sense in general.


TITLE: Fiber bundle on Stiefel manifold
QUESTION [5 upvotes]: Let $V_{n}(\mathbb{C}^{k})$ the Stiefel manifold of $n$-frame in $\mathbb{C}^{k}$. We can see $V_{n}(\mathbb{C}^{k})$ as a subset of $n$ copies of the cartesian product $S^{2k-1} \times \cdots \times S^{2K-1}$. So we have a bundle
$ V_{n-1}(\mathbb{C}^{k-1}) \rightarrow V_{n}(\mathbb{C}^{k}) \rightarrow S^{2k-1} $.
How can I find local trivializations?

REPLY [2 votes]: As I mentioned in the comments, given compact groups $H\subseteq K\subseteq G$, there is always a fiber bundle of the form $$K/H\rightarrow G/H \rightarrow G/K$$ where the projection map $\pi$ maps $gH$ to $gK$ and which has structure group $K$.
I don't know of any particularly easy or obvious proof of this fact.  The only reference I have for it is a bunch of hand written notes of my advisor, but there are probably other references out there.  The key lemma is that one always has a slice and that the isotropy representation in such a case must be trivial.
The most general form of it I know is: if $K$ is a compact Lie group acting smoothly on a manifold $M$ with all isotropy groups conjugate (say, to a subgroup $H$), then the projection $M\rightarrow M/K$ is a fiber bundle map with structure group $K$ and fiber $K/H$.
Anyway, using this on the inclusions $U(k-n)\subseteq U(k-1)\subseteq U(k)$ (via the usual block embeddings) gives you the fibration you're looking for.
For your particular example, if $G$ acts transitively on a manifold $M$ with isotropy subgroup $H$, then there is a $G$-equivariant diffeomorphism $G/H\cong M$. This is proven in Lee's Introduction to Smooth Manifolds Theorem 9.24.
So, in order to show, for example, $U(k)/U(k-1)\cong S^{2k-1}$, it suffices to find a transitive $U(k)$-action on $S^{2k-1}$ with isotropy subgroup $U(k-1)$.  Well, $U(k)$ naturally acts on $\mathbb{C}^k$ and preserves lengths of vectors, so it also acts on $S^{2k-1}\subseteq\mathbb{C}^k$. It's not too hard to see this action is transitive. The set of elements in $U(k)$ which fix the North pole is easily seen to be $U(k-1)$ block embedded.
For the Stiefel manfolds, we have $U(k)/U(k-n)\cong V_n(\mathbb{C}^k)$. Note that $U(k)$ naturally acts on $\mathbb{C}^k$, and hence on the collections of ordered tuples of vectors in $\mathbb{C}^k$. Given any $n$ orthonormal vectors $\{v_1,..., v_n\}$, complete them to an orthonormal basis $\{v_1,..., v_n, w_{n+1}, ... ,w_k\}$. The matrix whose columns are the $v$'s and $w$'s is an element of $U(k)$ which maps the first $n$ standard basis vectors to $\{v_1,...,v_n\}$.
This proves that $U(k)$ acts transitively on $V_n(\mathbb{C}^k)$. I'll leave it to you to check that the matrices in $U(k)$ which fix the first $n$ standard basis vectors are precisely the matrices of the form $\operatorname{diag}(I,B)$ with $B\in U(k-n)$.


TITLE: How is solving Proca equation equivalent to scalar field equation?
QUESTION [2 upvotes]: My prof. told me that using differential forms proca equation reduces to solving for scalar field equation. How is that? I can’t see how does one relate to Scalar equation using differential forms.
Proca equation: $$\mathcal{L} = \frac{-1}{16}F^{\mu v}F_{\mu v} + \frac{1}{8\pi}m^2A_\mu A^\mu.$$
Equation of motion for Proca: $$\partial_\mu F^{\mu v} + m^2 A^v = 0.$$

REPLY [1 votes]: The Lagrangian that you use for the Proca equation looks a bit unusual, I will factor out $\frac{1}{8}$ and change the factor  ($\frac{m}{\sqrt{\pi}} \rightarrow \frac{m}{\sqrt{2}}$): Then written out in differential forms it looks like:
$$L = 8{\cal{L}} =-\frac{1}{2} dA \wedge \star dA + \frac{1}{2} m^2   A\wedge \star A$$
We will derive the L-E equations via variation (actually I did most of it already in another post, so I will shorten the derivation a little bit, the details can be looked up in the post https://physics.stackexchange.com/a/432941/30506 ):
$$\delta S =\int\delta L = -\frac{1}{2} \int (d\delta A \wedge \star dA + dA \wedge \star d\delta A) +  \frac{1}{2} m^2  \int (\delta A \wedge \star A + A\wedge \star \delta A) =   -\frac{1}{2} \int 2 d\delta A \wedge \star dA + m^2 \int \delta A \wedge \star A $$
In the next manipulation we will use the product rule:
$$d(\delta A\wedge \star d A) = d\delta A \wedge\star dA - \delta A \wedge d\star dA $$ therefore we have:  $$-d\delta A\wedge \star dA = - d(\delta A\wedge \star dA ) -\delta A\wedge d\star dA $$.
We will this substitute in the first term of the varied action:
$$\delta S =\int  - d(\delta A\wedge \star dA )  -\int  \delta A\wedge d\star dA  + m^2  \int \delta A \wedge \star A $$
Finally the first term is an integral over an absolute derivative which be transformed into a surface integral on whose surface the variation $\delta A=0$.   So we get finally:
$$0=\delta S  = - \int \delta A \wedge  (d\star dA  - m^2  \star A )$$
As the last expression has to be zero for all variations  $\delta A$, the result of the variation is:
$$d\star dA  - m^2   \star A =0 $$  or a bit more nicely written (actually at the moment I don't know if $\star \star =1$ or $\star \star =-1$, but I will check that up).
$$ \star d\star dA = m^2 A$$
In some books $ \delta: =  \star d\star$ (this $\delta$, however, has nothing to do with the variation) and with $F=dA$ we get:
$$\delta F =m^2 A $$
The definition of the hodge operator can be looked up in the other post  https://physics.stackexchange.com/a/432941/30506  I mentioned at the beginning.


TITLE: One Line, Broken into 2 Equal Length Segments, Intersecting 3 Vectors
QUESTION [1 upvotes]: 3 vectors intersected by line segment
Given three vectors that start from the origin, there is a line segment that intersects these three vectors such that the portion of the segment between Vector 1 and Vector 2 is equal in length to the portion of the segment between Vector 2 and Vector 3 (length of Segment Portion A equals length of Segment Portion B). I'm fairly certain that there are infinite such line segments, but that they will all have the same slope. I need a way to calculate the slope of this line segment given the slopes of the three vectors. I have been working on calculating it by simulation but there has to be an easier way.

REPLY [0 votes]: angle and point between with geometric solution
There is a euclidean construction to find the line segments. Consider the drawing. It shows 2 rays emanating from A, they are AJ and AI. Between them is the point D. We wish to find the line through D intersecting the 2 rays at F and G such that FD = DG.
Construct ray AD from the common point A of the 2 rays and the point D. Next construct the perpendicular line to ray AD through AD intersecting the rays at B and C.  Next construct the circle c2 with radius DC and center D. The circle intersect line BC at E. Construct a line through E parallel to ray AJ it intersects ray AI at F. The line from F through D to G on AJ is the needed line. That is FD = DG
The proof
We can see that this is true since by construction ED = DC also by construction angle DEF = angle DCG. Also angle FDE = angle GDC, since they are vertical angles of an intersection. This means that triangle DFE is congruent to triangle DGC so FD = DG.


TITLE: Show that the first child can not win
QUESTION [0 upvotes]: Three children have 10 pieces numbered from 0 to 9 on both sides. They play the following game:
-The first child chooses a piece, so a number, preserves it and passes the number on a sheet
-The second one chooses a piece, preserves it and passes on the sheet the sum of the number chosen by him and the number written on the sheet of the first child
-The third child chooses a piece, preserves it and passes on the sheet the amount of the second child's written number and the number he chooses
Then the game resumes with the first child. When the pieces are finished, they are re-inserted into the box and the game is resumed from the child that he was following. The game is won by the first child who obtain 145 on sheet.
Show that the first child can not win this game.
I think that this problem has something wrong because I denote by $A, B, C$ the children. They move in this way:$ A, B,C,A,B,C,A,B,C, A$. At this step $B$ will write on the sheet $0+1+2+...+9=45$. After that  we have the sequence $B, C,A,B,C,A,B,C,A,B$. At this step $A $ will write 90 on the sheet. And the last sequence $C, A, B, C, A, B, C,A,B,C$. At the last step $C$ will write 135. If $A$ chooses 0,$ B $ chooses 1, $C $ chooses 2 and $A$ chooses 7 then $A$ wins.

REPLY [3 votes]: If the players not perfect players, then $A$ can indeed win, just as you show.
However, I am sure that what the problem statement is missing is that the three players are perfect players. That is: they will all try to win, and they will all make the best possible move in order for them to win. Thus, with your example, $C$ in the end would not pick $2$, but pick $9$, and win.
The good news is that you already did the hard work for this puzzle: No matter how the first $10$ rounds play out, it will always be true after 30 moves, the number $135$ is written on the paper, and it is now $A$'s turn ... can you see how, if $B$ and $C$ are perfect players, $A$ will never win?
Interesting follow-up question: Suppose the highest priority of all the players is to win, but that if they know they can't win, they will try to prevent others from winning in the hopes of not losing either. Can you see how, if all players are perfect players, $B$ and $C$ cannot win either? (indeed, this is not a game I would like to play ... I have other things to do ...)


TITLE: Explanation of this proof of $d\theta$ not being an exact form
QUESTION [1 upvotes]: I have been reading Spivak's Introduction to Differential Geometry and I cannot get this. I have already seen other proofs of the following statement, and I get those, I just do not follow this one.

$d\theta$ is closed but it is not exact

I need some explanation, the proof is simple, but I do not quite get why is there a contradiction, the proof goes as it follows:
Proof: Let $\omega=d\theta$, and suppose there exists a $C^1$ function $f:\mathbb{R}^2\setminus \{ 0\} \rightarrow \mathbb{R}$ such that $\omega=df$, clearly this implies that $\partial f/ \partial x=\partial \theta / \partial x$ and  $\partial f/ \partial y=\partial \theta / \partial y$, and hence $f=\theta + constant$ on $\mathbb{R}^2\setminus L$, which is impossible. $\square$
Note:
Here $\omega$ is defined (as always) as $$\omega = \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$.
I do not get why is it impossible, some explanation would really be appreciation, thanks in advance.

REPLY [5 votes]: This problem illustrates that $d\theta$ is a particularly misleading and hence poor notation for this $1$-form: It isn't the exterior derivative of anything!
(More generally the common notation $dV$ for a volume form on an oriented manifold suffers the same issue. For this reason I prefer a slightly more verbose notation like $\operatorname{vol}$ for a volume form.)
This notation also makes the provided proof possibly confusing. Here's a bit more detail:

If $\omega = df$ for some differentiable (and hence continuous) function $f$, then in particular that equality holds on the slit plane $$S := \Bbb R^2 \setminus L , \qquad L := \{(x, 0) \in \Bbb R^2 : x \geq 0\} ,$$
  where in particular we can define the usual coordinates $(r, \theta) : S \to (0, \infty) \times (0, 2 \pi)$, so that
  $$df \vert_S = \omega\vert_S = d\theta .$$
(The mentioned abuse of notation $d\theta$ makes a sidebar important here: In the preceding equation, $d\theta$ is the exterior derivative of the function $\theta$, and hence is defined only $S$; written there it is not the form $\omega$, but rather its restriction to $S$.)
So, on $S$ we have
  $$0 = df\vert_S - d\theta = d(f\vert_S - \theta) ,$$ and in particular (since $S$ is connected) $f\vert_S - \theta$ is some constant function $C$, that is, as claimed, $$f\vert_S = \theta + C .$$
On the other hand, $f\vert_S$ cannot be the restriction of a continuous function $f$ on $\Bbb R^2 - \{(0, 0)\}$: At any point $(x, 0)$, $x > 0$, in the slit, the limit of $f$ at that point as you approach from above the $x$-axis is $C$, but the limit from below is $C + 2 \pi$, so $\lim_{p \to (x, 0)} f$ does not exist, hence $f$ is not continuous there, a contradiction.

A much faster proof is available once you have access to integration: By Stokes' Theorem (in this case, the Fundamental Theorem of Line Integrals, essentially), if $\omega$ were an exact form, then we would have $\int_\gamma \omega = 0$ for any curve $\gamma$. But parameterizing a convenient closed anticlockwise curve that encloses the origin once---say, the unit circle---and computing directly gives $\int_\gamma \omega = 2 \pi$, a contradiction.


TITLE: Using $\bigvee$ and $\bigwedge$ instead $\exists$ and $\forall$
QUESTION [14 upvotes]: My professor of Algebra use some "strange" notation for me. He uses $\bigvee$ instead $\exists$ and $\bigwedge$ instead $\forall$. For example $$\displaystyle\bigwedge_{x\in \mathbb{Z}}\bigwedge_{m\in \mathbb{Z}\backslash\{0\}}\bigvee_{q,r\in \mathbb{Z}}(x=qm+r \wedge 0\leq r<|m|)$$ is same as $(\forall x \in \mathbb{Z})(\forall m \in \mathbb{Z}\backslash \{0\}) (\exists q,r \in \mathbb{Z}) (x=qm+r \wedge 0\leq r<|m|) $. If we know the set with which we are working, then we say $\displaystyle\bigwedge_{x}\bigvee_{y}(x+y=0)$ (without saying $x \in \text{Set}$). I asked him for this notation, and he said that I can see this in

K.Kuratowski, A.Mostowski, Set theory, PWN, Warszawa, 1976.

I found this book in library and it's really true.
Could someone say something more about this notation? Is this standard notation in mathematics? Did you see it anywhere else?

REPLY [20 votes]: I try to give you an argument why this notation makes sense.
Consider
$$\displaystyle\bigvee_x A(x)$$ as an infinite version of $\vee$.
For example, if $x$ comes from a countable set $\{x_1,x_2,x_3,\ldots\}$, then consider $\displaystyle\bigvee_x A(x)$ as $$A(x_1) \vee A(x_2) \vee A(x_3) \vee \ldots.$$
This expression is true as long as there is at least one $x_i$ such that $A(x_i)$ is true. So equivalently, there exists an $x_i$ such that $A(x_i)$ is true, which is $\displaystyle\exists x : A(x)$.
You can do the same for $\bigwedge$.