TITLE: Finding the polar form of a complex number
QUESTION [0 upvotes]: I have the following complex numbers : -3,18 +4,19i
I can calculate $r=\sqrt{a^2+b^2}$
Which gives r=5,26
now I know that cos $\theta = \frac{a}{r}$
gives $\theta=127,20$ degrees
But when I do the same with : $sin \theta = \frac{b}{r}$
it gives me $\theta= 52,80$ degrees
Why does it do this ? I know that it must give 127,20 degrees for both of them...

REPLY [2 votes]: If you draw your number in the complex plane you see that $\theta = 127.2$ could be correct but that $ \theta = 52.8$ lies is in a different qudarant, so that's definitely not the right answer.
I expect that you solved the equation $\sin \theta = \frac{b}{r}$ using the inverse sine function. When working with the inverse trig functions one has to remember that their range is limited (otherwise they wouldn't meet the definition of a function (each input has a single output)). Specifically:

Inverse sine $ = \sin^{-1}$ has domain $[-1, 1]$ and range
$[-\frac{1}{2}\pi, \frac{1}{2}\pi]$
Inverse cosine $ = \cos^{-1}$ has domain $[-1, 1]$ and range
$[0,\pi]$
Inverse tangent $= tan^{-1} $ has domain $(-\infty, \infty)$ and
range $(-\frac{1}{2}\pi, \frac{1}{2}\pi)$.


text:
\section{Notation and Preliminaries}\label{sec:notations}
Throughout the paper, unless otherwise specified, $\vecgal$ denotes a finite or infinite dimensional real vector space and $\cvx\subseteq \vecgal$ is a convex set. Given two distinct points $x$ and $y$ in $\vecgal$, we let $\oi{x}{y}=\enscond{tx+(1-t)y}{0<t<1}$ and $[x,y]=\{tx+(1-t)y : $ $0\leq t\leq 1\}$ denote the open and closed segments joining $x$ to $y$. We recall the following definitions, and we refer to~\cite{dubins1962extreme,klee_extremal_1957} for more details.

\paragraph{Lines, rays, and linearly closed sets}
A \textit{line} is an affine subspace of $\vecgal$ with dimension $1$. An open half-line, \ie{} a set of the form $\rho=\{ p+tv : $ $t>0 \}$, where $p,v\in \vecgal$, $v\neq 0$, is called a \emph{ray} (through $p$).
We say that the set $\cvx$ is \emph{linearly closed} (resp. linearly bounded) if the intersection of $\cvx$ and a line of $\vecgal$ is closed (resp. bounded) for the natural topology of the line. If $\vecgal$ is a topological vector space and $\cvx$ is closed for the corresponding topology, then $\cvx$ is linearly closed.

If $\cvx$ is linearly closed and contains some ray $\rho=p+\RR_+^*v$, it also contains the endpoint $p$ as well as the rays $q+\R_+v$ for all $q\in \cvx$. Therefore, if $\cvx$ contains a ray (resp. line), it recesses in the corresponding direction.

\paragraph{Recession cone and lineality space} The set of all $v\in\vecgal$ such that $\cvx+\RR_+^*v\subseteq \cvx$ is a convex cone called the \emph{recession cone of $\cvx$}, which we denote by $\rec{\cvx}$. If $\cvx$ is linearly closed then so is $\rec{\cvx}$, and  $\rec{\cvx}$ is the union of $0$ and all the vectors $v$ which direct the rays of $\cvx$. In particular, $\cvx$ contains a line if and only the vector space
\begin{equation}
\label{eq::lineality space}
\lin(\cvx)\eqdef \rec{\cvx}\cap (-\rec{\cvx})
\end{equation}
is non trivial. The vector space $\lin(\cvx)$
 is called the \emph{lineality space} of $\cvx$. It corresponds to the largest vector space of invariant directions for $\cvx$.

If $\vecgal$ is finite dimensional and $\cvx$ is closed, the recession cone coincides with the \emph{asymptotic cone}.

\paragraph{Extreme points, extremal rays, faces}
An \textit{extreme point} of $\cvx$ is a point $p\in \cvx$ such that $\cvx\setminus \{p\}$ is convex.
An \textit{extremal ray} of $\cvx$ is a ray $\rho \in \cvx$ such that if $x,y\in \cvx$ and $\oi{x}{y}$ intersects $\rho$, then $\oi{x}{y}\subset \rho$. If $\cvx$ contains the endpoint~$p$ of $\rho$ (\eg{} if $\cvx$ is linearly closed), this is equivalent to $p$ being an extreme point of $\cvx$ and $\cvx\setminus \rho$ being convex.

  Following~\cite{dubins1962extreme,klee_theorem_1963}, if $p\in \cvx$, the smallest face of $\cvx$  which contains $p$ is the union of $\{p\}$ and all the open segments in $\cvx$ which have $p$ as an inner point. We denote it by $\face{p}{\cvx}$. The (co)dimension of $\face{p}{\cvx}$ is defined as the (co)dimension of its affine hull. The collection of all elementary faces, $\{\face{p}{\cvx}\}_{p\in\cvx}$, is a partition of $\cvx$.
Extreme points correspond to the zero-dimensional faces of $\cvx$, while extreme rays are (generally a strict subcollection of the) one-dimensional faces.

\paragraph{Quotient by lines}
As noted above, if $\cvx$ is linearly closed, it contains a line if and only if the vector space $\lin(\cvx)$ defined in \eqref{eq::lineality space}

is nontrivial.  In that case, letting $W$ denote some linear supplement to $\lin(\cvx)$
, we may write
\begin{equation}
\label{eq::quotientbyline}
\cvx= \qcvx+\lin(\cvx), \textrm{ with } \qcvx\eqdef \cvx\cap W
\end{equation}
and the corresponding decomposition is unique (\ie{} any element of $\cvx$ can be decomposed in a unique way as the sum of an element of $\qcvx$ and $\lin(\cvx)$). The convex set $\qcvx$ (isomorphic to the projection of $\cvx$ onto the quotient space $\vecgal/\lin(\cvx)$) is then linearly closed, and the decomposition of $\cvx$ in elementary faces is exactly given by the partition $\{\face{p}{\qcvx}+\lin(\cvx)\}_{p\in\qcvx}$, where $\face{p}{\qcvx}$ is the smallest face of $p$ in $\qcvx$.

One may check that $\qcvx$ contains no line, as its recession cone $\rec{\qcvx}$, the projection of $\rec{\cvx}$ onto $W$ parallel to $\lin(\cvx)$, is a \emph{salient} convex cone.

config: arxiv


TITLE: Why is the electric field along a uniform line of charge some variation of $\int_{}^{}\frac{1}{\sqrt{x^{2} + a^{2}}}$?
QUESTION [1 upvotes]: Or, "why is it possible to use trig sub to find the electric field along a uniformly charged rod?"
When I was taking integral calculus, the professor explained that one of the motivations for trigonometric substitution was to calculate the electric field at a point along the axis of a uniform line of charge. Such a calculation could be expressed by some variant of:
$$\int_{}^{}\frac{1}{\sqrt{x^{2} + a^{2}}}dx$$
which could be solved by trig substitution.
Why is this the case? All I know is that the classic trig sub integral comes from the equation of an ellipse:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
But I don't know the link between the electric field along a line of charge and the ellipse equation, nor do I know why calculating the field results in a sum.

REPLY [1 votes]: Okay, I can give you an answer, but I don't know that you'll like it.  We'll start with Coulomb's law:
$$\vec E = \frac{q\hat r}{4\pi\epsilon_0r^2}$$
(This itself is a consequence of Gauss's law of electromagnetism, $\iint \vec E\cdot d\vec s = \iiint\frac{q}{\epsilon_0}dV$, considering a sphere with a point charge at the center and using the rotational symmetry to simplify.)
So in the case of the rod, you need to integrate,
$$\vec E = \int_{0}^{L}\frac{\lambda\hat r}{4\pi\epsilon_0r^2} dt = \int_{0}^{L}\frac{\lambda\vec r}{4\pi\epsilon_0r^3} dt$$
Note the parameters of integration; we're integrating now along the rod (which we're assuming, as the formula you gave seems to, goes along the x-axis with one end at the origin).  So now sub in $\vec r = (a - t)\hat x + b\hat y$ to get
$$\vec E = \int_{0}^{L}\frac{\lambda((a-t)\hat x+b\hat y)}{4\pi\epsilon_0((a-t)^2 + b^2)^{\frac{3}{2}}}dt$$
Use substitution to get
$$\vec E = \int_{-a}^{L-a}\frac{\lambda(-t\hat x+b\hat y)}{4\pi\epsilon_0(t^2 + b^2)^{\frac{3}{2}}}dt$$
Separate the components and use another substitution to get
$$\hat y\int_{-a}^{L-a}\frac{\lambda b}{4\pi\epsilon_0(t^2 + b^2)^{\frac{3}{2}}}dt - \hat x\int_{-a}^{L-a}\frac{\lambda t}{4\pi\epsilon_0(t^2 + b^2)^{\frac{3}{2}}}dt =  \hat y\int_{-a}^{L-a}\frac{\lambda b}{4\pi\epsilon_0(t^2 + b^2)^{\frac{3}{2}}}dt - \hat x\int_{a^2 + b^2}^{(L-a)^2 + b^2}\frac{\lambda}{8\pi\epsilon_0t^{\frac{3}{2}}}dt = \hat y\int_{-a}^{L-a}\frac{\lambda b}{4\pi\epsilon_0(t^2 + b^2)^{\frac{3}{2}}}dt + \hat x\frac{\lambda}{4\pi\epsilon_0}\left(\frac{1}{\sqrt{(L-a)^2 + b^2}} - \frac{1}{\sqrt{a^2 + b^2}}\right)$$
I don't think I'm wrong about the x-component, so I can only think that the formula you have there assumes that $a >> L$.  But that's how you get there.
As for the resemblance to elliptic integrals, to the extent that question has an answer, it's because they both involve stretching Euclidean distances - the arc lengths of a circle in one case, the position of a point charge in the other.


TITLE: No identity for Convolution of Lebesgue Integrable Functions
QUESTION [0 upvotes]: I want to show there's no identity for the convolution of two functions. I've found posts that answer this for Riemann integrable functions, but I'm working with the convolution given by $f\ast g = \int_{\mathbb{R}^d} f(x-y)g(y) dy$, where $f, g$ are Lebesgue integrable. So, I want to show there does not exist a $\delta\in{L^1(\mathbb{R}^d)}$ such that $f\ast\delta = f$.
My attempt:
Assume towards a contradiction that there is such a function $\delta$. Then we have $f\ast\delta = f$
and hence,
$$ \|\ f\ast\delta \|\ = \|\ f \|\ $$
but $$\|\ f \|\ \|\delta\|\ \ge \|\ f\ast\delta \|\ = \|\ f \|\ $$
which gives
$$ \|\ f\ast\delta \|\ = \|\ f \|\ $$
$$ \|\delta\|\ \ge \|\ f\ast\delta \|\ = 1 $$
With $ \|\ \delta \|\ \ge 1$, I believe I can derive a contradiction. I've tried integrating both sides, but that lead me nowhere. I'm not quite sure where to go from here. Any advice or tips?

REPLY [1 votes]: Suppose there exists a function $\delta \in L^1_{loc}(\mathbb{R}^d)$ such that $\delta \ast f = f$ for all $f\in C_c^\infty(\mathbb{R}^d)$. Then let $\phi_k$ be a sequence of smooth functions with $0\leq \phi_k \leq 1$, $\phi_k(0)=1$ and each with support contained in
$$
\left[-\frac{1}{k},\frac{1}{k}\right].
$$
The the dominated convergence theorem shows that
$$
\int_{\mathbb{R}^d} \delta(x)\phi_k(x)\,\mathrm{d}x \to 0.
$$
On the other hand,
$$
1 = \phi(0) = \delta\ast\phi(0) = \int_{\mathbb{R}^d} \delta(x)\phi_k(x)\,\mathrm{d}x
$$
which is clearly a contradiction.
The above gives a slightly stronger result then what you asked for.
Alternatively, suppose for the sake of contradiction that there exists a function $\delta \in L^1(\mathbb{R}^d)$ such that $f\ast \delta = f$ for all $f\in L^1(\mathbb{R}^d)$.
Then, consider the simple $L^1$-function $\rho = \mathbb{1}_{S}$ where
$$
S= \left[-\frac{1}{2},\frac{1}{2}\right]^d
$$
For almost every $x\in S$, we must have
$$
1 = \rho\ast\delta(x) = \int_{\mathbb{R}^d}\rho(x-y) \delta(y)\,\mathrm{d}y
=\int_{S+\{x\}} \delta(y)\,\mathrm{d}y.
$$
An application of the dominated convergence theorem therefore shows that
$$
\int_{S} \delta(y)\,\mathrm{d}y = 1.
$$
Similarly, for every $x_0\in \mathbb{Z}^d$ we can derive
$$
\int_{S_{x_0}} \delta(y)\,\mathrm{d}y = 1
$$
where
$$
S_{x_0}= \left[x_0-\frac{1}{2},x_0+\frac{1}{2}\right]^d.
$$
Finally, we conclude that
$$
\int_{\mathbb{R}^d} \delta(y)\,\mathrm{d}y
=\sum_{x_0\in\mathbb{Z}^d}\int_{S_{x_0}} \delta(y)\,\mathrm{d}y = \infty
$$
which is clearly a contradiction.


TITLE: Determine a relation $R$ on the set of humans on this planet by putting $(x,y) \in R$ if condition
QUESTION [0 upvotes]: So i was given this question. Determine a relation $R$ on the set of humans on this planet by putting $(x,y) \in R $ if $x$ and $y$ are citizens of the same country. Is $R$ an equivalence relation?
I know the answer is no since transitivity fails but how do you test all three conditions and show that transitivity fails?

REPLY [0 votes]: Transitivity fails if you allow people to be citizens of multiple countries.  We could have Alice a citizen of A, Bob a citizen of A and B, and Charlie a citizen of B.  Then Alice R Bob because of A and Bob R Charlie because of B, but we do not have Alice R Charlie.


TITLE: Is the following set operation true?
QUESTION [1 upvotes]: Prove the following or else find a counter example:
For all sets $A$, $B$, and $C$, $$((A \cup B) − C) \cup (A \cap B) = ((A − B) \cup (B − A)) − C$$
For the life of me, I can't figure out if its true or not. I have tried drawing the Venn diagrams but I'm not too good with them and can't quite figure out if diagrams representing either side of above equation look alike.Help appreciated.

REPLY [1 votes]: It might help to let $C = \emptyset$, since it seems that's allowed. I'll assume here that $C = \emptyset$.
Just thinking about the right side, the set $(A - B) \cup (B - A)$ has a special name: it's the symmetric difference of $A$ and $B$. It's the set of everything that's in exactly one of $A$ or $B$. In particular, it doesn't contain anything in $A \cap B$.
But if you notice, the left side certainly contains everything in $A \cap B$, doesn't it? So, can you cook up a situation that takes advantage of this fact?

REPLY [0 votes]: Notice that in the Right set $((A-B)\cup (B-A)) - C$ nothing in $A\cap B$ is included. However such elements are obviously included in the left set, since they are one part of the union.
Thus any choice of sets A, B and C such that $A\cap B\neq \emptyset$ is a good answer.
Ex: $A={1,2}, B={2,3} C={4}$ or maybe if you want something simple $A=B=\{1\}$.

REPLY [0 votes]: Let $A=\{1,2,3\}$,
$B=\{2,3,4\}$ and
$C=\{0\}$
Then $$[(A\cup B)-C]\cup (A\cap B)=[\{1,2,3,4\}-\{0\}]\cup \{2,3\}$$
$$=\{1,2,3,4\}\cup \{2,3\}$$
$$=\{1,2,3,4\}$$
But $$[(A-B)\cup(B-A)]-C=[\{1\}\cup\{4\}]-\{0\}$$
$$=\{1,4\}-\{0\}$$
$$=\{1,4\}$$
So there's a counterexample.


TITLE: thermodynamic equilibrium under self-gravitation
QUESTION [2 upvotes]: Suppose you have a point mass M in space that is surrounded by a gas of particles mass m at temperature T. Let the gas distribution be spherically symmetric about M. What is the radial dependence of the equilibrium gas distribution? Wish to include the case were M=0.
This may be a trick question because I think the gas may evaporate over time and slowly cool. If this is so, then I want the answer to the question above for short time periods over which the evaporation is negligible.
Many thanks!

REPLY [1 votes]: Although detailed answers for the specific case of gas of state equation $P = \rho k_B T/m$ was given, I'll try to give a short but generic answer.
The problem can be divided into two sub-problems :

Finding the equation describing the equilibrium between the gravitational force (which depends on $\rho$ via the mass) and the pressure $P$
Finding the state equation of the considered gas, which relates $\rho$ and $P$.

The 1st problem leads to either the Newtonian hydrostatic equilibrium equation, or the Tolman–Oppenheimer–Volkoff equation in the relativistic case.
The 2nd problem depends on the physics, for stars the state equation is usually "polytropic".


TITLE: For $f \in l_1^*$ to be true
QUESTION [1 upvotes]: Do we have to check whether $f$ is linear and bounded ? As in $f \in L(X,Y)$ ?
To show that $f \in L(X,Y)$, it is enough to prove that $f$ is linear and for boundedness that
$\exists M > 0$ such that $||f(x)|| \leq M||x||$, where $x \in X$ and $ f(x) \in Y$
Does the same apply for dual space?

REPLY [1 votes]: Yeah, for $f$ being in $(l^1)^{*}$ you have to show the following:

$f$ is linear.
$f$ is bounded i.e., $|f(x)| \leq \|x\|_1$


text:
\begin{document}
\DeclareGraphicsExtensions{.eps}

\title{Optimal Power Allocation for GSVD-Based Beamforming in the MIMO Wiretap Channel\thanks{The authors are with the Dept. of Electrical Engineering and Computer Science, University of California, Irvine, CA 92697-2625, USA. e-mail:\{afakoori, swindle\}@uci.edu}}

\author{\normalsize S. Ali. A. Fakoorian, {\it Student Member, IEEE} and A. Lee Swindlehurst, {\it Fellow, IEEE}}

\maketitle

\begin{abstract}

This paper considers a multiple-input multiple-output (MIMO) Gaussian
wiretap channel model, where there exists a transmitter, a legitimate
receiver and an eavesdropper, each equipped with multiple
antennas. Perfect secrecy is achieved when the transmitter and the
legitimate receiver can communicate at some positive rate, while
ensuring that the eavesdropper gets zero bits of information. In this
paper, the perfect secrecy capacity of the multiple antenna MIMO
wiretap channel is found for aribtrary numbers of antennas under the
assumption that the transmitter performs beamforming based on the
generalized singular value decomposition (GSVD).  More precisely, the
optimal allocation of power for the GSVD-based precoder that achieves
the secrecy capacity is derived.  This solution is shown to have
several advantages over prior work that considered secrecy capacity
for the general MIMO Gaussian wiretap channel under a high SNR
assumption. Numerical results are presented to illustrate the proposed
theoretical findings.

\end{abstract}

\begin{keywords}
MIMO Wiretap Channel, Secrecy Capacity, Physical Layer Security,
Generalized Singular Value Decomposition
\end{keywords}

\section{Introduction}

The broadcast nature of a wireless medium makes it very
susceptible to eavesdropping, where the transmitted message is decoded by
unintended receiver(s). Recent information-theoretic research on
secure communication has focused on enhancing security at the physical
layer. The wiretap channel, first introduced and studied by Wyner [1],
is the most basic physical layer model that captures the problem of
communication security. Wyner showed that when an eavesdropper's
channel is a degraded version of the main channel, the source and
destination can achieve a positive secrecy rate, while ensuring that
the eavesdropper gets zero bits of information. The maximum secrecy
rate from the source to the destination is defined as the secrecy
capacity. The Gaussian wiretap channel, in which the outputs at the
legitimate receiver and at the eavesdropper are corrupted by additive
white Gaussian noise (AWGN), was studied in [2].

Determining the secrecy capacity of a Gaussian wiretap channel is in
general a difficult non-convex optimization problem, and has been
addressed independently in [3-7].  Oggier and Hassibi [3] and Khisti
and Wornell [4, 5] followed an indirect approach using a Sato-like
argument and matrix analysis tools. They considered the problem of
finding the secrecy capacity of the Gaussian MIMO wiretap channel
subject to a constraint on the total average power, and a closed-form
expression for the secrecy capacity in the high signal-to-noise-ratio
(SNR) regime was obtained in [5]. However, the optimal input
covariance matrix that achieves the secrecy capacity at high SNR is
not fully characterized, especially for the case where there is
non-trivial nullspace for the channel between the transmitter and
eavesdropper. When there is such a nullspace, [5] uses a complicated
beamforming matrix to transmit two groups of information-bearing
symbols into two different subspaces, one that lies in the nullspace
of the channel matrix at the unintended receiver, and the other
orthogonal to it.  As indicated in [5], most of the transmission power
is allocated to the first subspace, and only a small fraction of the
power is allocated to the other.  Furthermore, the available power is
distributed uniformly over the dimensions of the two subspaces.
In addition to the complexity of the beamforming matrix, the other
drawback of [5] is that the precise allocation of power between the
two subspaces is not clear, nor is the sensitivity of the secrecy
capacity to this power fraction quantified.

In [6], Liu and Shamai propose a more information-theoretic approach
using the enhancement concept, originally presented by Weingarten et
al. [8], as a tool for the characterization of the MIMO Gaussian
broadcast channel capacity. Liu and Shamai have shown that an enhanced
degraded version of the channel attains the same secrecy
capacity as does a Gaussian input distribution. From the mathematical
solution in [6] it was evident that such an enhanced channel exists;
however it was not clear how to construct such a channel until the
work of [7], which provided a closed-form expression for the secrecy
capacity under a \emph{covariance matrix} power constraint.  While
this result is interesting since the expression for the secrecy
capacity is valid for all SNR scenarios, there still exists no
computable secrecy capacity expression for the MIMO Gaussian wiretap
channel under an average total power constraint.  To date, a
solution has only been obtained for the so-called MISOME case in [4],
where the transmitter and eavesdropper may have multiple antennas,
but the desired receiver has only one.

In this paper, we investigate the non-convex optimization of the
secrecy rate for the general Gaussian multiple-input multiple-output
(MIMO) wiretap channel under a constraint on the total average power,
where the number of antennas is arbitrary for both the transmitter and
the two receivers.  We focus on the case where the transmitter uses
beamforming (linear precoding) based on the generalized singular value
decomposition (GSVD), as in [5].  In particular, we obtain the optimal power
allocation that achieves the secrecy capacity for the GSVD scheme.
The resulting power allocation is significantly different in nature
than the standard water-filling solution for achieving capacity in
MIMO links without secrecy considerations.

Compared with [5], our beamforming matrix is much simpler to
compute, and more importantly, the input covariance matrix that
achieves the secrecy capacity is completely characterized in terms of
the beamforming and power allocation matrices. We note that the analysis in this paper
characterizes the secrecy capacity of the Gaussian MIMO wiretap channel with
GSVD-based beamforming for \emph{any SNR} conditions, while [5] gives
the secrecy capacity of a \emph{general} Gaussian MIMO wiretap channel
(no restriction to GSVD beamforming), but only for the \emph{high SNR}
case.

The rest of this paper is organized as follows. In the next section,
we describe the assumed mathematical model.  We then present the GSVD
method and derive the optimal power allocation that achieves the
secrecy capacity in Section~III. Finally, we demonstrate our results
by means of several numerical examples in Section~IV.

\section{System Model}  \label{sec:ach1}

Consider a multiple-antenna wiretap channel with $n_t$ transmit
antennas and $n_r$ and $n_e$ receive antennas at the legitimate
recipient and the eavesdropper, respectively:
\begin{equation}\label{eq:r1}
\by_r=\textbf{H}_r \bx+ \bz_r
\end{equation}
\begin{equation}\label{eq:r2}
\by_e=\textbf{H}_e \bx+ \bz_e
\end{equation}
where $\bx$ is a zero-mean $n_t \times 1$ transmitted signal vector,
$\bz_r\in\mathbb{C}^{n_r\times1}$ and
$\bz_e\in\mathbb{C}^{n_e\times1}$ are the additive white Gaussian
noise (AWGN) vectors at the receiver and eavesdropper, respectively,
with i.i.d. entries distributed as $\mathcal{CN}(0, 1)$.  The matrices
$\textbf{H}_r\in\mathbb{C}^{n_r\times{n_t}}$ and
$\textbf{H}_e\in\mathbb{C}^{n_e\times n_t}$ represent the channels
associated with the receiver and the eavesdropper, respectively, and
are assumed to be quasi-static flat Rayleigh fading and independent of
each other, with i.i.d. entries distributed as $\mathcal{CN}(0,
\sigma_{r}^2)$ and $\mathcal{CN}(0, \sigma_{e}^2)$.  Similar to other
papers considering the secrecy capacity of the wiretap channel, we
assume that the transmitter and both receivers are aware of the
channel state information (CSI) for both links.

In a wiretap channel, the transmitter intends to send a confidential
message $W$ to the intended receiver while keeping it as secret as
possible from the eavesdropper. The corresponding
information-theoretic secrecy constraint is given by [1, 9]:
\begin{equation}\label{eq:3}
\lim_{N\to\infty} \frac{1}{N}  I (W ; Y_e^N)=  0
\end{equation}
where $N$ is the number of channel uses for sending the message $W$,
and $I(W ; Y_e^N)$ represents the mutual information between $W$ and
$Y_e^N$. Consequently, the secrecy capacity is defined as the maximum
number of bits that can be correctly transmitted to the intended
receiver while keeping the eavesdropper uninformed. Using the
single-letter characterization of the secrecy capacity of the wiretap
channel provided by Csiszar and Korner in [9], the secrecy capacity is
the solution of the following maximization problem:
\begin{equation}\label{eq:r4}
C_{sec}=\max_{U,X} [I(U; Y_r)- I(U; Y_e)]
\end{equation}
where $X$, $Y_r$ and $Y_e$ are random variable counterparts to the
specific realizations $\bx$, $\by_r$ and $\by_e$, respectively. $U$ is
an auxiliary variable, and the maximization is over all jointly
distributed $(U,X)$ such that $U\rightarrow X \rightarrow (Y_r, Y_e)$
forms a Markov chain, while the channel input $\bx$ satisfies an
average total power constraint
\begin{equation}\label{eq:r5}
\mathrm{Tr}(E\{\bx \bx^H\})\leq p
\end{equation}
where $(.)^H$ denotes the Hermitian (i.e., conjugate) transpose, $E$
is the expectation operator, and Tr(.) is the matrix trace.

The auxiliary variable $U$ represents a precoding signal. In [6], Liu
and Shamai studied the optimization problem of (4) and showed that a
Gaussian $U = X$ is an optimal choice. In other words, Gaussian random binning
without prefix coding
is an optimal coding strategy for the MIMO Gaussian wiretap channel
[10]. Hence, a matrix characterization of the secrecy capacity is
given by
\begin{equation}\label{eq:r6}
C_{sec}=\max_{\bQ_x\succeq0} [I(X; Y_r)- I(X; Y_e)]
\end{equation}
where $\bQ_x=E\{\bx \bx^H\}$ is the input covariance matrix. The
non-convex maximization problem in (6) is considered under the power
constraint (5).

\section{Optimal Power Allocation for the GSVD-Based Gaussian MIMO Wiretap Channel}  \label{sec:ach1}

We consider the non-convex maximization problem in (6) for the case
that the transmitter performs beamforming by applying the generalized
singular value decomposition (GSVD) to the channel matrices
$\textbf{H}_r$ and $\textbf{H}_e$. Application of the GSVD technique
was first used by Khisti and Wornell (see e.g. [5]) who obtained a
closed-form expression for the secrecy capacity in the high SNR
regime. In this section, we first describe the GSVD beamforming method and next we derive
the optimal power allocation matrix that achieves the secrecy capacity
for any SNR, and we describe some important advantages of this scheme
over what is proposed in [5].$\\$
\textbf{Definition 1} \emph{(GSVD
Transform)}: Given two matrices \textbf{H$_r$} $\in$
$\mathbb{C}^{n_r\times{n_t}}$ and \textbf{H$_e$} $\in$
$\mathbb{C}^{n_e\times n_t}$, $gsvd($\textbf{H$_r$}, \textbf{H$_e$}$)$
returns unitary matrices $\bPsi_r$ $\in$ $\mathbb{C}^{n_r\times n_r}$
and $\bPsi_e$ $\in$ $\mathbb{C}^{n_e\times n_e}$, non-negative
diagonal matrices $\bC$ and $\bD$, and a matrix $\bA$ $\in$
$\mathbb{C}^{n_t\times q}$ with $q$=min$(n_t , n_e+n_r)$, such that
\begin{equation}\label{eq:r7}
\textbf{H$_r$}\bA = \bPsi_r\bC
\end{equation}
\begin{equation}\label{eq:r8}
\textbf{H$_e$}\bA = \bPsi_e\bD
\end{equation}
The nonzero elements of $\bC$ are in ascending order while the nonzero
elements of $\bD$ are in descending order. Moreover,
$\bC^T\bC+\bD^T\bD=\textbf{I}$.

The transmitted signal vector $\bx$ is constructed as
\begin{equation}\label{eq:r9}
\bx= \bA X_S ,\quad X_{S}\sim \mathcal{CN}(\textbf{0}, \bP)
\end{equation}
where $\bA$ is obtained from $gsvd($\textbf{H$_{r}$},
\textbf{H$_{e}$}$)$ as above, and each element of the vector $X_{S}$ represents
an independently encoded Gaussian codebook symbol that is beamformed
with the corresponding column of the $\bA$ matrix. $\bP$ is a positive
semi-definite diagonal matrix representing the power allocated by the
transmitter to the data symbols. In the following, we derive an
optimal source power allocation which achieves the secrecy capacity of
the GSVD-based MIMO Gaussian wiretap channel.  Substituting (9)
into the channel model (1)-(2) and using (7)-(8) yields
\begin{equation}\label{eq:r10}
\by_r= \bPsi_{r}\bC X_S + \bz_r
\end{equation}
\begin{equation}\label{eq:r11}
\by_e= \bPsi_{e}\bD X_S+ \bz_e
\end{equation}
Considering the above equations, the maximization problem in (6) is represented by
$$C_{sec}=\max_{\bQ_x} [I(X;Y_r)- I(X;Y_e)]=$$
$$\max_{\bP\succeq\textbf{0}, diagonal} \log|\textbf{I}+\bPsi_{r}\bC\bP\bC^{T}\bPsi_{r}^H|-\log|\textbf{I}+\bPsi_{e}\bD\bP\bD^{T}\bPsi_{e}^H|$$
\begin{equation}\label{eq:r12}
\text{subject to} \quad \text{Tr}(\bA^H \bA \bP)\leq p
\end{equation}
\textbf{Theorem 1:} Assuming that the transmitter applies the proposed
beamforming matrix $\bA$, the optimal source power allocation $\bP^*$
that achieves the secrecy capacity in problem (12) is given by
\begin{equation}\label{eq:r13}
p^*_i= \begin{cases} \max(0,\frac{-1+\sqrt{1-4c_i d_i+4(c_i-d_i)c_i d_i/(\mu a_i)}}{2c_i d_i}), & \text{if}\quad c_i > d_i\\ 0, & \mbox{otherwise}
\end{cases}
\end{equation}
where $p^*_i$, $c_i$, $d_i$ and $a_i$ are the $i$th diagonal elements
of the matrices $\bP^*$, $\bC^T\bC$, $\bD^T\bD$ and
diag$(\bA^H\bA)$, respectively. The Lagrange parameter $\mu>0$ is
chosen to satisfy the power constraint (5).$\\$
\textbf{Proof:} The optimization problem is non-convex. Our proof
technique involves applying the Karush-Kuhn-Tucker (KKT) conditions
(as necessary conditions), which help express the Lagrangian in the
form of an integral. This specific structure of the problem is then
exploited to obtain a closed-form solution for the optimal power
allocation strategy. Details can be found in the Appendix.

Eqs. (7) and (8) show that applying the GSVD transform to
\textbf{H$_r$} and \textbf{H$_e$} simultaneously diagonalizes
them. Thus, the GSVD transform creates a set of parallel independent
subchannels between the sender and the receivers, and it suffices to
use independent Gaussian codebooks across these subchannels. More
precisely, as (13) indicates, it is optimal to use only those
subchannels for which the output at the eavesdropper is a degraded
version of the output at the destination node. These subchannels
correspond to the condition $c_i> d_i$, or as shown in [5],
generalized singular values of $gsvd($\textbf{H$_r$},
\textbf{H$_e$}$)$ which are larger than 1. Clearly, if there are no
such subchannels, the achievable secrecy rate using this transmission
scheme would be zero [11].

It is interesting to note that the optimal source power allocation
(13) is different from the water-filling allocation that achieves
capacity for fading channels without the secrecy constraint. Moreover,
as we will observe in Section IV, (13) has an important role in
achieving the secrecy capacity even for moderately high SNRs. We have the
following result. $\\$ \textbf{Corollary 1:} The secrecy capacity of
the GSVD-based Gaussian MIMO wiretap channel is
\begin{equation}\label{eq:r14}
C_{sec}=\log|\textbf{I}+\bP^*\bC^T\bC|-\log|\textbf{I}+\bP^*\bD^T\bD|
\end{equation}
\textbf{Proof:} Follows directly from substituting (13) in to (12) and
by considering the fact that $\bPsi_{r}$ and $\bPsi_{e}$ are unitary
matrices.

It was shown in [5] that GSVD beamforming with uniform power
allocation is sufficient to attain the secrecy capacity of a MIMO
Gaussian wiretap channel in the high SNR regime.  However, for
the case where there is a non-trivial nullspace in the channel between the transmitter and
eavesdropper, the optimal input covariance matrix that achieves the
secrecy capacity is not fully
characterized in [5]. When there is such a nullspace, [5] uses a complicated
beamforming matrix to transmit two groups of information bearing
symbols into two different subspaces. The associated beamforming matrices are
obtained by performing an LQ decomposition on $\bA$ as well as
additional calculations which lead to Eq. (59) in [5].  The
aforementioned subspaces are identified as follows [5, Eq. (58)]:
$$\mathcal{S}_1=\text{Null}(\bH_e)\cap\text{Null}(\bH_r)^\perp$$
\begin{equation}\label{eq:r15}
\mathcal{S}_2=\text{Null}(\bH_e)^\perp\cap\text{Null}(\bH_r)^\perp
\end{equation}
where Null(.) denotes the nullspace of its matrix argument, while
$\perp$ denotes the orthogonal complement. It is important to note that
our transmission scheme in (9) and consequently our solution in Theorem 1 does not require
such calculations, and in fact yields the secrecy capacity of the Gaussian MIMO wiretap channel
with GSVD-based beamforming for any SNR under any assumptions regarding the nullspace of the
transmitter-to-eavesdropper channel.

As indicated in [5], most of the transmission power is assumed to be
allocated to the subspace $\mathcal{S}_1$ and only a small fraction
for $\mathcal{S}_2$, and the available power is distributed uniformly
over the dimensions of each of these two subspaces. The exact
allocation of power between these two subspaces is not specified in
[5], nor is the sensitivity of the secrecy capacity to the power
allocation studied. In the next section, we show that the secrecy
capacity of a MIMO Gaussian wiretap channel is in fact quite sensitive
to how power is allocated between these two subspaces, which
illustrates that using the proposed optimal power allocation is
essential.

\section{Numerical Results}  \label{sec:ach4}

In this section, we present numerical results to illustrate our
theoretical findings. In all of the following figures, the channel
matrices and background noise are modeled in the same way that we
described in Section II. In each figure, the values of $n_t$, $n_r$
and $n_e$, as well as $\sigma_r^2$, $\sigma_e^2$ and $p$, will be
depicted. In the simulations we compare the secrecy capacity obtained
by the optimal power allocation with the secrecy rate achieved by
uniform power allocation for a MIMO Gaussian wiretap channel with
GSVD-based beamforming. This comparison is performed for various
transmit powers (SNR), channel conditions and also different numbers of
antennas for the transmitter and receivers.  All displayed results are
calculated based on an average of at least 100 independent channel
realizations. In each trial, the secrecy capacity is obtained by
evaluating (14), while the secrecy rate achieved by the uniform power
allocation is obtained by using analytical results presented in
[5].

First we consider the case where the nullspace of the eavesdropper's
channel is non-trivial.  Figs. 1 and 2 investigate the allocation of
power between subspaces $\mathcal{S}_1$ and $\mathcal{S}_2$ as defined
in (15), for a case where $n_t=n_r=5, n_e=4$ and the transmit power is
$p=100$, or SNR=20 dB. The solid horizontal line illustrates the
secrecy capacity of a Gaussian MIMO wiretap channel with GSVD-based
beamforming and optimal power allocation, while the dashed curve
represents the secrecy rate achieved by uniform power allocation
versus the fraction of power used in subspace $\mathcal{S}_2$. Note
that, for this high SNR scenario, the peak of this curve is the
secrecy capacity of the \emph{general} MIMO Gaussian wiretap channel
[5].  Note also that, contrary to claims made in [5], the secrecy
rate is quite sensitive to the fraction of power allocated to the
two subspaces, and optimal performance requires a non-trivial
allocation of power to $\mathcal{S}_2$ (over 20\%).  The advantage
of the optimal GSVD-based power allocation approach is that this
imprecise distribution of power to the two subspaces is eliminated.

Fig. 3 compares the secrecy capacity achieved by the optimal power
allocation and the secrecy rate achieved by the uniform power
allocation for different transmit powers (SNRs). In this example,
there is no non-trivial null space between the transmitter and
eavesdropper. The figure shows that the difference between the optimal
and uniform power allocation is important, even at moderately high SNRs.  This is
especially true for the case where the desired receiver's channel is
of equal or better quality than the eavesdropper's channel ($\sigma_r^2\ge
\sigma_e^2$).

As the figure shows, the performance difference between
the optimal and uniform power allocation curves slowly decreases as the SNR
is increased. This is due to this fact that, as derived in [5] for the high SNR regime,
uniform power allocation is sufficient to achieve the secrecy capacity.

\section{Conclusions}

We have established the secrecy capacity for the Gaussian MIMO wiretap
channel assuming the transmitter uses GSVD-based beamforming. This
non-convex optimization problem is solved subject to an average
transmit power constraint. In particular, we have derived the optimal
power allocation for the GSVD-based beamformers that achieves the secrecy
capacity. Our numerical results demonstrate that the optimal power
allocation is necessary even for relatively high SNRs.

\appendix

We are interested in obtaining the power distribution that
achieves the secrecy capacity of the Gaussian MIMO wiretap channel in
problem (12) for the GSVD-based beamforming scheme presented in
Section III. This non-convex optimization problem is to be solved with the
average power constraint
$$\text{Tr}(E\{\bx \bx^H\})= \text{Tr}(\bA E\{X_{S} X_{S}^H\}\bA^H)$$
\begin{equation}\label{eq:r16}
=\text{Tr}(\bA \bP\bA^H)= \text{Tr}(\bA^H \bA \bP)\leq p
\end{equation}
By considering the fact that $\bPsi_{r}$ and $\bPsi_{e}$ are unitary
matrices and $\det(\textbf{I}+\bE\bF)=\det(\textbf{I}+\bF\bE)$, the
Lagrangian function $\mathcal{L}$ associated with this problem is given
by
\begin{equation}\label{eq:r17}
\mathcal{L}=\log|\textbf{I}+\bP\bC^T\bC|-\log|\textbf{I}+\bP\bD^T\bD|-\mu \text{Tr}(\bA^H\bA\bP)
\end{equation}
where $\mu>0$ is the Lagrange multiplier. Since $\bP$, $\bC^T\bC$ and
$\bD^T\bD$ are diagonal matrices, $\mathcal{L}$ can be written as
\begin{equation}\label{eq:r18}
\mathcal{L}=\sum_{i} [\log(1+p_ic_i) - \log(1+p_id_i)] -\mu \sum_{i} a_i p_i
\end{equation}
where $p_i$, $c_i$, $d_i$ and $a_i$ are the $i$th diagonal elements of
the matrices $\bP$, $\bC^T\bC$, $\bD^T\bD$ and diag$(\bA^H\bA)$,
respectively. Clearly, $p_i$, $c_i$, $d_i$ and $a_i$ all are real
non-negative numbers. Using a technique similar to that proposed in [12], the
optimal $p_i^*$ must maximize
\begin{equation}\label{eq:r19}
\mathcal{L}_i=\log(1+p_ic_i) - \log(1+p_id_i) -\mu a_i p_i=\int_0^{p_i} f_i(x)dx
\end{equation}
where $f_i(x)$ is defined as
\begin{equation}\label{eq:r20}
f_i(x)=\frac{1}{\ln 2} \left(\frac{c_i}{1+xc_i}-\frac{d_i}{1+xd_i}\right)-\mu a_i
\end{equation}
To obtain the optimal $p_i^*$ that maximizes $\mathcal{L}_i$ in (19),
we consider two cases based on the relationship between $c_i$ and
$d_i$.
\begin{enumerate}
\item $c_i\leq d_i$: In this case, (20) is always non-positive,
i.e., $f_i(x)\leq 0$. Hence, the maximum of $\mathcal{L}_i$ is
achieved by $p_i^*=0$.
\item $c_i > d_i$: In this case, since $f_i(x)$ is a decreasing function
for $x\geq 0$, the optimal $p_i^*$ that maximizes $\mathcal{L}_i$
depends on the value of the largest root of $f_i(x)=0$. Let $x_{Li}$
denote the largest root of $f_i(x)=0$, i.e.,
$$x_{Li}=\frac{-1+\sqrt{1-4c_i d_i+4(c_i-d_i)c_i d_i/(\mu a_i)}}{2c_i
d_i}$$ where we have used the fact that
$\bC^T\bC+\bD^T\bD=\textbf{I}$, or equivalently
$c_i+d_i=1$.\footnote{It is easy to verify that $\mu>0$ guarantees
$1-4c_i d_i+4(c_i-d_i)c_i d_i/(\mu a_i)\ge 0$ for the case $c_i >
d_i$. To do so, it is sufficient to prove that $1-4c_i d_i\ge 0$. We
have:$$c_i+d_i=1\rightarrow1-4c_i d_i= 1-4
(1-d_i)d_i=1-4d_i+4d_i^2=(1-2d_i)^2\ge 0$$ } If $x_{Li}$ is positive
then the maximum of $\mathcal{L}_i$ is achieved by $p_i^*=x_{Li}$,
otherwise $p_i^*=0$.
\end{enumerate}
Combining cases 1 and 2, we obtain the desired result as
\begin{equation}\label{eq:r21}
p_i^*= \begin{cases} \max(0,\frac{-1+\sqrt{1-4c_i d_i+4(c_i-d_i)c_i d_i/(\mu a_i)}}{2c_i d_i}), & \text{if}\quad c_i > d_i\\ 0, & \mbox{otherwise}
\end{cases}
\end{equation}
Finally, the Lagrange parameter $\mu>0$ is chosen to satisfy the power constraint (16).


text:
\begin{document}

\begin{abstract}
We study the Hausdorff dimension of $R$-analytic subgroups in an $R$-analytic profinite group, where $R$ is a pro-$p$ ring whose associated graded ring is an integral domain.
In particular, we prove that the set of such Hausdorff dimensions is a finite subset of the rational numbers.
\end{abstract}

\maketitle

\section{Introduction}

The study of Hausdorff dimension in profinite groups was initiated by Abercrombie in \cite{Abercrombie}, and has attracted special attention in recent times; see, for example, \cite{Abert,BK,YA,Ershov,FZ,Siegenthaler}.
If $G$ is a countably based infinite profinite group, a \emph{(normal) filtration\/} of $G$ is a descending series
$\{G_n\}_{n\in\mathbb{N}}$ of (normal) open subgroups of $G$ which is a base of neighbourhoods of $1$.
We can define a metric on $G$ by
\[
d(x,y)
=
\inf \, \{|G:G_n|^{-1}\ |\ xy^{-1}\in G_n\},
\]
and the topology induced by $d$ coincides with the original topology in $G$.
The metric $d$ defines a Hausdorff dimension function on all subsets of $G$, which we denote by  $\hDim_{\{G_n\}}$.
As shown in \cite[Theorem 2.4]{YA}, if the filtration $\{G_n\}_{n\in\mathbb{N}}$ is normal and $H$ is a closed subgroup of $G$, then
\[
\hDim_{\{G_n\}}(H)
=
\liminf_{n\rightarrow\infty} \, \frac{\log |H:H\cap G_n|}{\log |G:G_n|}.
\]

In \cite{YA}, Barnea and Shalev studied the Hausdorff dimension in $p$-adic analytic pro-$p$ groups with respect to the filtration $\{G^{p^n}\}_{n\in\N}$,
and proved that
\[
\hDim_{\{G^{p^n}\}}(H)=\frac{\dim(H)}{\dim(G)}
\]
for every closed subgroup $H$ of $G$.
Here, if $G$ is an analytic group, $\dim(G)$ denotes the dimension of $G$ as an analytic manifold.

In this paper we study Hausdorff dimension in $R$-analytic profinite groups, where $R$ is a pro-$p$ ring whose associated graded ring is an integral domain.
This is the natural setting for working with analytic groups over pro-$p$ rings, see \cite[Chapter 13]{Dixon}.
If $G$ is an $R$-analytic profinite group, then $G$ is virtually a pro-$p$ group, but the subgroups $G^{p^n}$ need not to form a filtration, since they need not to be open in $G$.
We consider instead the natural filtration induced by a standard open subgroup of $G$.
As we will see, the Hausdorff dimension of a closed subgroup $H$ of $G$ is the same with respect to the natural filtrations induced by all open standard subgroups of $G$.
This allows us to define the concept of \textit{standard Hausdorff dimension} of $H$, denoted by $\hDim_{\mathrm{Std}}(H)$.
We get the following theorem, in the spirit of the result of Barnea and Shalev.

\begin{main}
Let $G$ be an $R$-analytic profinite group and let $H$ be an $R$-analytic subgroup of
$G$.
Then
\begin{equation}
\label{T1}
\hDim_{\Std}(H)=\frac{\dim(H)}{\dim(G)}.
\end{equation}
\end{main}

An $R$-analytic subgroup $H$ of $G$ is a subgroup that is also an $R$-analytic submanifold of $G$; then $H$ is closed in $G$.
A crucial remark is that while the converse is true in a $p$-adic analytic group, i.e.\
every closed subgroup is analytic, it need not hold for an arbitrary pro-$p$ ring $R$.
For example, $\F_p[[t^d]]$ is a closed subgroup of $\F_p[[t]]$ for every positive integer $d$, but it is not analytic for $d>1$, even if it is an
$\F_p[[t]]$-analytic group in its own right.

Observe that (\ref{T1}) implies that the \textit{$R$-analytic spectrum\/}  of $G$, defined by
\[
\Spec_R(G)
=
\{ \hDim_{\Std}(H) \mid \text{$H$ is an $R$-analytic subgroup of $G$} \},
\]
is finite and consists of rational values.
On the other hand, the spectrum of $\F_p[[t]]$ corresponding to all closed subgroups is the full interval $[0,1]$, as shown in \cite[Lemma 4.1]{YA}.
Thus our main theorem is pointing to the fact that most closed subgroups of $\F_p[[t]]$ are non-analytic.

\vspace{10pt}

\noindent
\textit{Notation.\/}
We write $X^{(d)}$ to denote the cartesian product of $d$ copies of a set $X$.
The symbol $\subseteq_o$ indicates that a subset of a topological space is open.

\section{Preliminaries}

Throughout this paper $R$ is a pro-$p$ ring whose associated graded ring is an integral domain.
Hence $R$ is an integral domain, and we write $K$ for its field of fractions.
If $\mathfrak{m}$ is the maximal ideal of $R$, then $R/\mathfrak{m}$ is a finite field of characteristic $p$.
Set $H(n)=\dim_{R/\mathfrak{m}} (\mathfrak{m}^n/\mathfrak{m}^{n+1})$ for all $n\in\N$.
For large enough $n$, $H(n)$ coincides with a polynomial $P(n)$, called the \emph{Hilbert polynomial\/} of $R$ \cite[Chapter 8, Theorem C]{Eisenbud}.

Let $G$ be an $R$-analytic group.
Without loss of generality, we assume that the manifold structure is given by a full atlas.
A subgroup $H$ of $G$ is an \emph{analytic subgroup\/} if it is also an analytic submanifold of $G$.
We adopt Serre's definition of submanifold \cite[Section 3.11]{Serre}.
Every analytic subgroup is closed in $G$.
The converse is not true in general, but every open subgroup $H$ of $G$ (actually, any open subset) is a submanifold when considered with the restrictions of the charts of $G$, and $\dim H=\dim G$.

An $R$-analytic group $S$ with a global chart given by a homeomorphism $\phi:S\rightarrow (\mathfrak{m}^N)^{(d)}$ is called an \emph{$R$-standard group\/} of level $N$ if $\phi(1)=0$ and, for all $j\in\{1,\ldots,d\}$, there exist $F_j(X,Y)\in R[[X,Y]]$ (where $X$ and $Y$ are $d$-tuples of indeterminates), without constant term, such that
\begin{equation}
\label{formal group law}
\phi(xy)=F(\phi(x),\phi(y)), \quad \text{for every $x,y\in S$.}
\end{equation}
Here, $F=(F_1,\ldots,F_d)$ is called the \textit{formal group law\/} associated to $S$.
In these circumstances, we also say that $(S,\phi)$ is a standard group.

By using (\ref{formal group law}), $F$ defines a new group structure on $(\mathfrak{m}^N)^{(d)}$, other than its natural additive structure, and then
$\phi$ is an isomorphism between $S$ and $((\mathfrak{m}^N)^{(d)},F)$.
Every $R$-analytic group contains an open (and so analytic) $R$-standard subgroup $S$ \cite[Theorem 13.20]{Dixon}.
As a consequence, profinite $R$-analytic groups are countably based.

The formal group law of a standard group satisfies the following result \cite[pp.\ 331-334]{Dixon}.

\begin{lemma}
\label{fgl}
Let $F$ be a formal group law of dimension $d$ associated to a standard group $S$.
Then
\[
F(X,Y)=X+Y+G(X,Y),
\]
where every monomial involved in $G$ has total degree at least $2$, and contains a non-zero power of $X_r$ and of $Y_s$ for some $r,s\in\{1,\ldots,d\}$.
Moreover,
\[
F(X,I(X))=0=F(I(X),X)
\]
for some $I(X)=-X+H(X)\in R[[X]]$, and every monomial involved in $H$ has total degree at least $2$.
\end{lemma}

It follows that $(\mathfrak{m}^{N+n})^{(d)}$ is a subgroup of $((\mathfrak{m}^N)^{(d)},F)$ for all $n\ge 0$.
This allows us to introduce a special type of filtrations in an $R$-analytic group.

\begin{definition}
Let $G$ be an $R$-analytic group and let $S$ be a standard open subgroup of $G$, with global chart $(S,\phi)$.
Then for every positive integer $n\ge 0$, and for every subset $A\subseteq S$, we define
\[
\mathfrak{m}^n_\phi A = \phi^{-1}(\mathfrak{m}^n\phi(A)).
\]
We say that $\{\mathfrak{m}^n_\phi S\}_{n\in\mathbb{N}}$ is the \emph{natural filtration\/} of $G$ induced by $S$.
\end{definition}

Observe that
\[
\mathfrak{m}^n_\phi S=\phi^{-1}((\mathfrak{m}^{N+n})^{(d)}),
\]
which implies that $\mathfrak{m}^n_\phi S$ is an open subgroup of $G$, and also that
$\{\mathfrak{m}^n_\phi S\}_{n\in\mathbb{N}}$ is a filtration of $G$.
Actually, we have $\mathfrak{m}^n_\phi S\trianglelefteq S$
(see \cite[Proposition 13.22]{Dixon}).

\begin{lemma}
\label{iso}
Let $(S,\phi)$ be an $R$-standard group.
Then:
\begin{enumerate}
\item
For all $x,y\in S$ and $n\ge N$, we have $\phi(xy^{-1}) \in (\mathfrak{m}^n)^{(d)}$ if and only if
$\phi(x)-\phi(y) \in (\mathfrak{m}^n)^{(d)}$.
\item
For every $n\ge 0$, we have $|S:\mathfrak{m}^n_\phi S|=|(\mathfrak{m}^N)^{(d)}:(\mathfrak{m}^{N+n})^{(d)}|=q^{d\hspace{.4pt}f(n)}$,
where $f(n)=\sum_{i=N}^{N+n-1} \, H(i)$.
\item
$\phi$ is an isometry between the group $S$ with the metric induced by the filtration $\{\mathfrak{m}^n_\phi S\}_{n\in\mathbb{N}}$ and the group
$((\mathfrak{m}^N)^{(d)},+)$ with the metric induced by the filtration $\{(\mathfrak{m}^{N+n})^{(d)}\}_{n\in\mathbb{N}}$.
\end{enumerate}
\end{lemma}

\begin{proof}
(i)
Since $\phi(1)=0$, we may assume that $x\ne y$.
Let $k\in\N$ be such that
$\phi(xy^{-1})\in  (\mathfrak{m}^k)^{(d)} \smallsetminus (\mathfrak{m}^{k+1})^{(d)}$.
Then since
\[
\phi(x)
=
\phi(xy^{-1}y)
=
F(\phi(xy^{-1}),\phi(y))
=
\phi(xy^{-1}) + \phi(y) + G(\phi(xy^{-1}),\phi(y)),
\]
by Lemma \ref{fgl}, we have
\[
\phi(x) - \phi(y)
\equiv
\phi(xy^{-1})
\pmod{(\mathfrak{m}^{k+1})^{(d)}}.
\]
Hence also
$\phi(x)-\phi(y)\in (\mathfrak{m}^k)^{(d)} \smallsetminus (\mathfrak{m}^{k+1})^{(d)}$,
and (i) follows.

(ii)
Observe that (i) implies that
\begin{equation}
\label{product vs sum 2}
xy^{-1} \in \mathfrak{m}^n_\phi S
\
\Longleftrightarrow
\
\phi(x)-\phi(y) \in (\mathfrak{m}^{N+n})^{(d)},
\end{equation}
or what is the same,
\[
x \cdot \mathfrak{m}^n_\phi S = y \cdot \mathfrak{m}^n_\phi S
\
\Longleftrightarrow
\
\phi(x)+(\mathfrak{m}^{N+n})^{(d)} = \phi(y)+(\mathfrak{m}^{N+n})^{(d)}.
\]
This proves (ii), since $\phi$ is a bijection.

(iii)
According to the definition of the metric associated to a fitration, it is clear that (ii) and (\ref{product vs sum 2}) together imply that $\phi$ is an isometry.
\end{proof}

Hausdorff dimension can be defined for any subset of a metric space, see \cite[Chapter 2]{Falconer} for its definition and main properties.
We need the following two lemmas about Hausdorff dimension in countably based profinite groups.

\begin{lemma}
\label{leqo}
Let $G$ be a countably based profinite group with filtration $\{G_n\}_{n\in\mathbb{N}}$.
Let $H$ be a closed subgroup of $G$, and let $U$ be a non-empty open subset of $H$.
Then
\[
\hDim_{\{G_n\}}(H) = \hDim_{\{G_n\}}(U).
\]
\end{lemma}

\begin{proof}
The proof is a straightforward consequence of \cite[Section 2.2]{Falconer}.
\end{proof}

Let $G$ be a countably based profinite group and let $S$ be an open subgroup of $G$.
If $\{S_n\}_{n\in\N}$ is a filtration of $S$, we can calculate the Hausdorff dimension of $X\subseteq S$ with respect to the metric induced by
$\{S_n\}_{n\in\N}$ in $S$ or in $G$, which we denote by $\hDim_{\{S_n\}}^S (X)$ and $\hDim_{\{S_n\}}^G (X)$, respectively.
Our next lemma shows that there is no need to make this distinction in the notation.

\begin{lemma}
\label{ino}
Let $G$ be a countably based profinite group and let $S$ be an open subgroup of $G$.
If $\{S_n\}_{n\in\mathbb{N}}$ is a filtration of $S$, then
\[
\hDim_{\{S_n\}}^G (X) = \hDim_{\{S_n\}}^S (X)
\]
for every $X\subseteq S$.
\end{lemma}

\begin{proof}
Let $d_S$ and $d_G$ be the metrics induced by $\{S_n\}_{n\in\N}$ in $S$ and $G$.
Then $d_S(x,y)=|G:S| \, d_G(x,y)$ for all $x,y\in S$, and the identity map from $(S,d_S)$ to $(S,d_G)$ is bi-Lipschitz.
Now the result follows from \cite[Corollary 2.4]{Falconer}.
\end{proof}

\section{Proof of the main theorem}

In this section we first prove that the Hausdorff dimension of a closed subgroup in an analytic profinite group with respect to a natural filtration is independent of the standard subgroup.
Then we prove the main theorem of our paper about Hausdorff dimension of analytic subgroups.

\begin{theorem}
\label{std}
Let $G$ be an $R$-analytic profinite group, and let $(S,\phi)$ and $(T,\psi)$ be two open standard subgroups of $G$.
Then
\[
\hDim_{\{\mathfrak{m}^n_\phi S\}}(H) = \hDim_{\{\mathfrak{m}^n_\psi T\}}(H)
\]
for every closed subgroup $H$ of $G$.
\end{theorem}

\begin{proof}
Let us write $S_n$ and $T_n$ for $\mathfrak{m}^n_\phi S$ and $\mathfrak{m}_\psi^n T$, and $N(S)$ and $N(T)$ for the levels of $S$ and $T$.
We first show that there exist non-negative integers $a$ and $b$ such that
\begin{equation}
\label{standard sandwich}
S_{n+a}\le T_n\le S_{n-b}
\end{equation}
for every $n\ge b$.
Since the charts $\phi$ and $\psi$ belong to the full atlas of $G$, the two functions $\psi\circ\phi^{-1}|_{\phi(S\cap T)}$ and $\phi\circ\psi^{-1}|_{\psi(S\cap T)}$ are analytic.
Since $\phi(S\cap T)$ is open in $\phi(S)$, it follows that $\psi\circ\phi^{-1}$ can be evaluated in $(\mathfrak{m}^{\ell})^{(d)}$ for some $\ell$.
By \cite[Lemma 6.45]{Dixon}, and observing that $\psi\circ\phi^{-1}(0)=0$, there exists a natural number $k$ such that
$$
\psi\circ\phi^{-1}((\mathfrak{m}^{n+k})^{(d)})\subseteq (\mathfrak{m}^n)^{(d)}
$$
for every $n\ge 0$.
This implies that $S_{n+a}\le T_n$ for $n\ge 0$, by choosing $a=\max\{k-N(S)+N(T),0\}$.
Arguing similarly with $\phi\circ\psi^{-1}$, we get (\ref{standard sandwich}).

Now, by \cite[Theorem 2.4]{YA}, we have
\begin{align*}
\hDim_{\{T_n\}}(H)
&=
\liminf_{n\rightarrow\infty} \, \frac{\log |H:H\cap T_n|}{\log |G:T_n|}
\\[5pt]
&\le
\liminf_{n\rightarrow\infty} \, \frac{\log |H:H\cap S_{n+a}|}{\log |G:S_{n+a}|-\log |T_n:S_{n+a}|}
\\[5pt]
&\le
\liminf_{n\rightarrow\infty} \, \frac{\log |H:H\cap S_{n+a}|}{\log |G:S_{n+a}|}\cdot \Bigg(1-\frac{\log |S_{n-b}:S_{n+a}|}{\log |G:S_{n+a}|} \Bigg)^{-1}
\\[5pt]
&=
\hDim_{\{S_n\}}(H).
\end{align*}
The last equality holds because, by (ii) of Lemma \ref{iso}, for $n$ large enough we have
$$
\log_q |S_{n-b}:S_{n+a}|=\sum_{i=N+n-b}^{N+n+a-1} \, P(i),
$$
which is polynomial in $n$ of degree $\deg P$; on the other hand, $\log_q |G:S_{n+a}|$ is asymptotically equivalent to
$$
\log_q |G:S| + \sum_{i=N}^{N+n+a-1} \, P(i)
$$
and, by the Euler-MacLaurin formula, this sum is a polynomial in $n$ of degree $\deg P+1$.
By swapping $S$ and $T$, the result follows.
\end{proof}

\begin{definition}
Let $G$ be an $R$-analytic profinite group and let $H$ be a closed subgroup of $G$.
Then the \emph{standard Hausdorff dimension\/} of $H$, $\hDim_{\Std}(H)$, is the Hausdorff dimension of $H$ calculated with respect to the natural filtration induced by any given standard open subgroup of $G$.
\end{definition}

We need the following lemma before proving our main theorem.

\begin{lemma}
\label{R-modules}
Let $R$ be a pro-$p$ ring, and let $E$ be a vector subspace of dimension $e$ of $K^{(d)}$.
Then, for every $N\in\N$, the Hausdorff dimension of $(\mathfrak{m}^N)^{(d)}\cap E$ in $R^{(d)}$ with respect to the filtration $\{(\mathfrak{m}^n)^{(d)}\}_{n\in\N}$ is $e/d$.
\end{lemma}

\begin{proof}
Let $\mathcal{B}=\{w_1,\ldots,w_e\}$ be a basis of $E$, and let $A$ be the $e\times d$ matrix over $K$ whose rows are the vectors in $\mathcal{B}$.
We may assume that $A$ is in reduced echelon form.
Set $W=\langle w_1,\ldots,w_e \rangle_R$.
Let $k$ be such that the product of all denominators of the entries of $A$ belongs to $\mathfrak{m}^k$.
One readily checks that
\begin{equation}
\label{sandwich W}
\mathfrak{m}^{n+k} \, W \subseteq (\mathfrak{m}^n)^{(d)}\cap E\subseteq \mathfrak{m}^n \, W.
\end{equation}
The required Hausdorff dimension is the limit inferior of the sequence
\[
c_n
=
\frac{\log |(\mathfrak{m}^N)^{(d)}\cap E:(\mathfrak{m}^n)^{(d)}\cap E|}{\log |R^{(d)}:(\mathfrak{m}^n)^{(d)}|},
\]
where $n\ge N$.
Now, from (\ref{sandwich W}), for $n\ge N+k$, we have
\[
|\mathfrak{m}^{N+k} \, W:\mathfrak{m}^n \, W|
\le
|(\mathfrak{m}^N)^{(d)}\cap E:(\mathfrak{m}^n)^{(d)}\cap E|
\le
|\mathfrak{m}^N \, W:\mathfrak{m}^{n+k} \, W|.
\]
Since $W$ is a free $R$-module of rank $e$, we have
$\dim_{R/\mathfrak{m}} (\mathfrak{m}^n\, W/\mathfrak{m}^{n+1}\, W)=e H(n)$, and consequently
\[
\frac{e (H(N+k)+\cdots+H(n-1))}{d (H(1)+\cdots+H(n-1))}
\le c_n
\le
\frac{e (H(N)+\cdots+H(n+k-1))}{d (H(1)+\cdots+H(n-1))}.
\]
Now, since $H(n)=P(n)$ for large enough $n$, we have
\[
\lim_{n\to\infty} \frac{H(N+k)+\cdots+H(n-1)}{H(1)+\cdots+H(n-1)}
=
\lim_{n\to\infty} \frac{P(N+k)+\cdots+P(n-1)}{P(1)+\cdots+P(n-1)}
=
1,
\]
since both sums in the last limit are polynomials in $n$ of degree $\deg P+1$ and with the same leading coefficient.
Arguing similarly with the upper bound for $c_n$ given above, we conclude that $\lim_{n\to\infty} \, c_n=e/d$, as desired.
\end{proof}

We are now ready to prove our main theorem.

\begin{proof}[Proof of the Main Theorem]
Let $\dim (G)=d$ and $\dim (H)=e$.
Since $H$ is an $R$-analytic submanifold of $G$, there exist an open subset $U$ of $H$ containing $1$, and a chart $(V,\phi)$ of $G$ such that
$U\subseteq V$ and $\phi(U)=E\cap \phi(V)$, for some vector subspace $E$ of $K^{(d)}$ of dimension $e$.

 From the proof of \cite[Theorem 13.20]{Dixon}, there exist a natural number $N$ and an open subgroup $S$ of $G$ such that $S\subseteq V$ and $(S,\phi)$ is a standard subgroup of $G$ of level $N$.
Since $U\cap S\subseteq_o U\subseteq_o H$, we have
\[
\hDim_{\Std}(H)=\hDim_{\{\mathfrak{m}^n_{\phi} S\}}(U\cap S),
\]
by Lemma \ref{leqo}.
Since $\phi$ is an isometry between $S$ and $((\mathfrak{m}^N)^{(d)},+)$ by Lemma \ref{iso}, we get
\[
\hDim_{\{\mathfrak{m}^n_{\phi} S\}}(U\cap S)=\hDim_{\{(\mathfrak{m}^n)^{(d)}\}}(\phi(U\cap S)).
\]
Now since $\phi (U\cap S)= E\cap (\mathfrak{m}^N)^{(d)}$, Lemma \ref{R-modules} yields that
\[
\hDim_{\{(\mathfrak{m}^n)^{(d)}\}}(\phi(U\cap S))=\frac{e}{d},
\]
which completes the proof.
\end{proof}

\begin{corollary}\label{c}
Let $G$ be an $R$-analytic group of dimension $d$.
Then
\[
\Spec_R(G) \subseteq \Big\{ 0,\frac{1}{d},\ldots,\frac{d-1}{d},1 \Big\}.
\]
In particular, the $R$-analytic spectrum of $G$ is finite and consists of rational numbers.
\end{corollary}

\noindent
\textit{Acknowledgement\/}.
We thank the anonymous referee for helpful comments.


TITLE: Traceless matrices and commutators
QUESTION [4 upvotes]: Any traceless $n\times n$ matrix with coefficients in a field of caracteristic $0$ is a commutator (or Lie bracket) of two matrices. What happens when the field has positive caracteristic?
When trying to reproduce the proof I have in the caracteristic $0$ case for the positive caractersitic case, I run into two problems:

multiples of the identity may have trace $=0$.
a matrix may have a spectrum equal to the whole field.

Are all traceless matrices commutators? If not, for which $n\in\mathbb{N}\setminus\lbrace 0  \rbrace$ and fields $k$ does it still hold? EDIT given a traceless matrix $M$, I want to know wether there are two matrices $A,B$ with $M=AB-BA$.

REPLY [1 votes]: Not all traceless matrix are commutators given a finite field with order $p^n$. Note that $M^{p^n} - M$ will have trace zero. Use the fact that the trace of $M$ is the sum of its eigenvalues and that $M^k$ has trace as sum of the eigenvalues of $M$ raised to $k$.


TITLE: Show that MLE of $\lambda = \frac{n-T_n}{S_n+cT_n}$
QUESTION [3 upvotes]: $X_i$ are i.i.d exponential, mean $\lambda^{-1}$ for $1 \leq i \leq n$ and, the values are measured such that $X_i = c$ if $X_i \geq c$ and $X_i$ otherwise. Show that MLE of $\lambda = \frac{n-T_n}{S_n+cT_n}$ where $S_n= \sum_{j=1}^n X_jI(X_j < c)$ and $T_n = \sum_{j=1}^n I(X_j \geq c)$

Attempt:
The likelihood function is given by $L(\lambda | x_i, x_2, \ldots x_n) = \lambda^n e^{-\lambda \sum x_i} = \lambda^n e^{-\lambda(S_n+cT_n)}$ $\implies$ $log L = n log \lambda -\lambda(S_n+cT_n) $
and hence $\hat\lambda=\frac{n}{S_n+cT_n}$ which is different from the expected answer.

REPLY [2 votes]: Indicator variables to the rescue.
Few observations:
$P(X_i < x)  = 1 - \lambda e^{-x}$ if $ 0 \leq x < c$
$P(X_i = c) = P(X_i \geq c) = e^{-\lambda c}$ (This one looks counter intuitive at first look, but that is what the $X_i=c$ if $X_i \geq c$ returns)
Now,
$\begin{align}L( \lambda| x_i) &= \prod_{i=1}^{n}  (\lambda e^{-\lambda x_i}) I(0 \leq x_i < c) \times (e^{-\lambda c})I( x_i \geq c) \\&= \lambda^{\sum_{i=1}^n I(0 \leq x_i < c)}e^{-\lambda(\sum_{i=1}^n x_iI(0 \leq x_i < c)} \times e^{-\lambda c \sum_{i=1}^n I(x_i \geq c)} \\&= \lambda^{n-T_n}e^{-\lambda S_n} \times e^{-\lambda c T_n}\end{align}$
$\log L= (n-T_n) \log \lambda -\lambda(S_n+cT_n)$
$\frac{\partial \log L}{\partial \lambda} = \frac{n-T_n}{\lambda}-(S_n+cT_n)$
which gives $\hat \lambda = \frac{n-T_n}{S_n+cT_n}$


TITLE: What does it mean by a density argument?
QUESTION [7 upvotes]: When I read some books, it states "a density argument", what does it mean? When applying to symmetric function, what does it mean?

REPLY [6 votes]: For the sake of an answer (thanks to the comment above):

It is difficult to answer without a precise context, but usually "by a density argument" means that you had proved a statement for all elements $x$ in a dense subset $A$ of a topological set $X$, and you extend the statement to the whole $X$ using density. For example, if you prove that a continuous function $f$ is constant on a dense subset $A$ of $X$, then $f$ is constant on $X$ by density.


TITLE: Sign problem with integration by parts
QUESTION [2 upvotes]: I have a problem with a minus sign in the integration formula, but I don't know where it comes from.
Consider the domain $Q=\{(x,y)\in \mathbb R^2|x^2+y^2<1,y<0\}$, ie the lower half disk. Consider as well a $C^\infty$ function $g$ such that $g\equiv 0$ inside a neighborhood of the circumference $S=\{(x,y)|x^2+y^2=1\}$.
I want to compute the integral $\int_Q g_{y}$. With the usual orientation of $\mathbb R^2$ and using Fubini we have:
$\int_Q g_{y}=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^0g_{y}(x,y)dydx=\int_{-1}^1g(x,0)dx$
However we can also compute this integral integrating by parts and we get:
 $\int_Q g_{y}=-\int_Q 0\cdot g+\int_{\partial Q} 1 g\nu_2$
where $\nu_2$ is an outer normal to the boundary. On the boundary, $g$ is zero outside the axis so we only integrate on it, there the outer normal is $\nu=(0,1)$ so $\nu_2=1$ and the orientation of the border is from right to left since it has the induced orientation. So we get:
$\int_Q g_{y}=\int_{1}^{-1}g(x,0)dx$
Which has clearly the opposite sign. Where is the problem?

REPLY [1 votes]: The mistake you are making is in your your assumption that
$$
\int_{\partial Q} g=\int_{1}^{-1}g(x,0)dx.
$$
That formula is wrong. Just writing $\int_1^{-1}$ does not make sense. To calculate the line integral, you have to parametrize the segment; the integral on the right should be
$$
\int_{-1}^1g(t,0)dt
$$
The easiest way for me to see this is to notice that the integration by parts formula, in your case, is nothing but Green's Theorem.
Let $F(x,y)=(-g(x,y),0)$. Then $\nabla\cdot F\cdot\check k=0-(-g_y)=g_y$ (this is the integrand in Green).
So
$$
\int_Qg_y=\int_{\partial Q}F\cdot d\vec s
$$
As you said, $F$ is non-zero only on $y=0$. The parametrization of the segment $[-1,1]$ in counterclockwise way is $(-t,0)$ for $t\in[-1,1]$. So
$$
\int_{Q}g_y=\int_{\partial Q}F\cdot d\vec s=\int_{-1}^1(-g(-t,0),0)\cdot(-1,0)\,dt=\int_{-1}^1g(-t,0)dt=\int_{-1}^1g(x,0)dx
$$


TITLE: Confusing derivative problem
QUESTION [1 upvotes]: I'm having a lot of problem with this question. If $f(x) = 3xe^x \cdot g(x)$ and $g(0) = 10$ and $g'(0) = 3$, find $f'(0)$. Can someone please help me

REPLY [2 votes]: Using Product Rule, $\displaystyle f'(x)=3\frac{dx}{dx}e^xg(x)+3xe^x\frac{d \ g(x)}{dx}+3x\frac{d\ e^x}{dx}g(x)$
$\displaystyle\implies f'(x)=3e^xg(x)+3xe^xg'(x)+3xe^xg(x)$
$\displaystyle\implies f'(c)=3[e^cg(c)+c\cdot e^cg'(c)+c\cdot e^cg(c)]=\cdots$
Here $c=0$
Can you identify the usage of $g'(0),$ here?


TITLE: Deriving the commutation relations of the so(n) Lie algebra
QUESTION [1 upvotes]: The generators $(A_{ab})_{st}$ of the $so(n)$ Lie algebra are given by:
$$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$$,
where $a,b$ label the number of the generator, and $s,t$ label the matrix element.
Now, I need to prove the following commutation relation using the definition above:
$$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$$.
Here's my attempt.
$([A_{ij},A_{mn}])_{st}$
$ = (A_{ij})_{sp}(A_{mn})_{pt}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j]p}\delta_{p[m}\delta_{n]t}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$
Could you please suggest the next couple of steps? Should I expand all the antisymmetrised Kronecker delta's, or is there some sneaky shortcut to get to the answer?

REPLY [1 votes]: Notice that $-i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$ = $-i(A_{jm}\delta_{ni}-A_{jn}\delta_{mi}-A_{im}\delta_{nj}+A_{in}\delta_{mj})_{st}$.
Now,
$-\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$
$=-\delta_{si}\delta_{j[m}\delta_{n]t}+\delta_{sj}\delta_{i[m}\delta_{n]t}+\delta_{sm}\delta_{n[i}\delta_{j]t}-\delta_{sn}\delta_{m[i}\delta_{j]t}$
$=-\delta_{si}\delta_{jm}\delta_{nt}+\delta_{si}\delta_{jn}\delta_{mt}+\delta_{sj}\delta_{im}\delta_{nt}-\delta_{sj}\delta_{in}\delta_{mt}+\delta_{sm}\delta_{ni}\delta_{jt}-\delta_{sm}\delta_{nj}\delta_{it}-\delta_{sn}\delta_{mi}\delta_{jt}+\delta_{sn}\delta_{mj}\delta_{it}$
$=(-i)(-i)(\delta_{si}\delta_{jm}\delta_{nt}-\delta_{si}\delta_{jn}\delta_{mt}-\delta_{sj}\delta_{im}\delta_{nt}+\delta_{sj}\delta_{in}\delta_{mt}-\delta_{sm}\delta_{ni}\delta_{jt}+\delta_{sm}\delta_{nj}\delta_{it}+\delta_{sn}\delta_{mi}\delta_{jt}-\delta_{sn}\delta_{mj}\delta_{it})$
$=(-i)(-i)(\delta_{sj}\delta_{mt}\delta_{ni}-\delta_{sm}\delta_{jt}\delta_{ni}-\delta_{sj}\delta_{nt}\delta_{mi}+\delta_{sn}\delta_{jt}\delta_{mi}-\delta_{si}\delta_{mt}\delta_{nj}+\delta_{sm}\delta_{it}\delta_{nj}+\delta_{si}\delta_{nt}\delta_{mj}-\delta_{sn}\delta_{it}\delta_{mj})$
$=(-i)(-i)(\delta_{s[j}\delta_{m]t}\delta_{ni}-\delta_{s[j}\delta_{n]t}\delta_{mi}-\delta_{s[i}\delta_{m]t}\delta_{nj}+\delta_{s[i}\delta_{n]t}\delta_{mj})$
$=-i((A_{jm})_{st}\delta_{ni}-(A_{jn})_{st}\delta_{mi}-(A_{im})_{st}\delta_{nj}+(A_{in})_{st}\delta_{mj})$
Thus, the proof is complete.


TITLE: Constructing finite fields as quotients of a domain by its ideals
QUESTION [2 upvotes]: All of the finite fields of prime order can be constructed as quotients of $\mathbb{Z}$ by one of its ideals.
Is there a domain such that every finite field can be constructed as a quotient of that domain by one of its ideals?
What if "finite" is replaced with "of cardinality strictly smaller than a given cardinal"?

REPLY [2 votes]: Let $\kappa$ be the given (infinite) cardinal and let $R_{\kappa} = \Bbb{Z}[x_{\alpha} \mid \alpha < \kappa]$ be the ring of multivariate polynomials over a set of variables $x_{\alpha}$ of cardinality $\kappa$. Then any ring $S$ of cardinality at most $\kappa$ is a homomorphic image of $R_{\kappa}$ (under the homomorphism induced by any surjection carrying the $x_{\alpha}$ onto $S$). So $R_{\kappa}$ is an integral domain meeting your requirements.


text:
\begin{document}
\maketitle

\centerline{\scshape Michael Ruzhansky and Jens Wirth}
\medskip
{\footnotesize

 \centerline{Department of Mathematics, Imperial College London}
   \centerline{180 Queen's Gate}
   \centerline{London, SW7 2AZ, UK}
}

\bigskip

 \centerline{(Communicated by XXX)}

\begin{abstract}
In this note we provide dispersive estimates for Fourier integrals with parameter-dependent phase functions in terms of geometric quantities of associated families of Fresnel surfaces. The results are based on a multi-dimensional van der Corput lemma due to the first author.

Applications to dispersive estimates for hyperbolic systems and scalar higher order hyperbolic equations are also discussed.
\end{abstract}

\section{Introduction}
Dispersive estimates for solutions to linear evolution equations are vitally important for the study of a wide range of problems, including questions of stability of solutions to corresponding non-linear equations, descriptions of large-time asymptotic profiles of solutions or scattering results based on the time-dependent approach.

For the particular situation of hyperbolic evolution equations with a hyperbolic time-asymptotic profile, such estimates can be based on corresponding decay estimates for Fourier integral operators with real phases and certain well-controled amplitude functions. The aim of this note is to provide some robust estimates for such Fourier integrals in terms of geometric properties of associated Fresnel surfaces.

\section{Fourier integral operators, notation and main results}
We are interested in $L^1$--$L^\infty$ estimates for a family of Fourier integral operators depending on a parameter $t\in\mathbb R_+$ (thought of as time later on)
\begin{equation}\label{eq:Tdef}
   T_t u (x) = \int_{\mathbb R^n} \mathrm e^{\mathrm i ( x\cdot\xi + t\phi(t,x,\xi) )}a(t,x,\xi)\, \widehat u(\xi) \mathrm d\xi,
\end{equation}
where $a(t,x,\xi)$ is uniformly in $t$ a (global) symbol of order zero,
\begin{equation}\label{eq:symbEstXi}
  |\partial_\xi^\alpha a(t,x,\xi) | \le C_\alpha \langle\xi\rangle^{-|\alpha|},
\end{equation}
with $\langle\xi\rangle=\sqrt{1+|\xi|^2}$, and $\phi(t,x,\xi)$ is a homogeneous real-valued function satisfying
\begin{equation}
  C^{-1} |\xi| \le \phi(t,x,\xi) \le C |\xi|,\qquad |\partial_\xi^\alpha \phi(t,x,\xi)| \le C_\alpha' |\xi|^{1-|\alpha|}
\end{equation}
uniform in $t$ and $x$. For simplicity, we always assume $u\in\mathcal S(\mathbb R^n)$, estimates extend by continuity arguments to larger spaces. If we are only interested in estimates of $T_tu$ in the $L^\infty$-norm, we can treat $t$ and $x$ both as parameters and derive estimates for the corresponding stationary model operator
\begin{equation}\label{eq:Model}
  T : u \mapsto \int_{\mathbb R^n} \mathrm e^{\mathrm i(x\cdot\xi + \phi(\xi)) } a(\xi) \widehat u(\xi) \mathrm d\xi.
\end{equation}
 A precise understanding of the appearing constants in the estimate translates to estimates of the $t$-behaviour of the full model \eqref{eq:Tdef}.

\subsection{Fresnel surfaces and contact indices} We turn to the analysis of \eqref{eq:Model} and define the geometric quantities which determine its $L^\infty$ behaviour. The consideration follows essentially Sugimoto \cite{sugi94}, \cite{Sug96}, for the non-degenerate situations see the earlier works of Strichartz \cite{Str70} or Brenner \cite{Bre75}.

We assume that the phase $\phi(\xi)$ is non-negative, 1-homogeneous and smooth, and define the associated Fresnel surface (also called slowness surface)
\begin{equation}
  \Sigma = \{ \xi\in\mathbb R^n : \phi(\xi) = 1 \}.
\end{equation}
We introduce two indices
for such a Fresnel surface $\Sigma$, assuming that
it is of class $C^k$ with $k$ being sufficiently large.
For $p\in\Sigma$ we denote by $\mathbb T_p$ the
tangent hyperplane to $\Sigma$ at $p$. Then for any plane $H$
of dimension 2 which contains $p$ and the normal of
$\Sigma$ at $p$ we denote by $\gamma(\Sigma,p,H)$
the order of contact between the curve
$\Sigma\cap H$ and its tangent $H\cap \mathbb T_p$. Furthermore, we set
\begin{equation}
 \gamma(\Sigma) = \sup_{p} \sup_{H} \gamma(\Sigma,p,H),\qquad
 \gamma_0 (\Sigma) = \sup_{p} \inf_H \gamma(\Sigma, p,H).
\end{equation}
The definition implies directly that
$2\le\gamma_0(\Sigma)\le\gamma(\Sigma)$.
Furthermore, for isotropic problems $\Sigma$ is a dilation of the sphere $\mathbb S^{n-1}$ and
$\gamma(\mathbb S^{n-1})=\gamma_0(\mathbb S^{n-1})=2$.
Moreover, if the Gaussian curvature of $\Sigma$ never
vanishes $\Sigma$ is convex and we have $\gamma(\Sigma)=2$.

These indices are directly related to mapping properties of \eqref{eq:Model}. The following theorem is a combination of results from \cite{sugi94}, \cite{Sug96} in the analytic setting,
improved by the author \cite{Ru09}, \cite{RS10} to the smooth
setting as well as to the limited regularity cases.
Moreover, the corresponding results in
\cite{Ru09}, \cite{RS10} do not require the positive homogeneity
of the phase in $\xi$ and allow the dependence on parameters.

As usual, $B^r_{p,q}(\mathbb R^n)$ denotes the Besov space of regularity $r$ modelled over $L^p$.
\begin{theorem}
  Assume that $\phi$ is smooth on $\mathbb R^{1+n}\times(\mathbb R^n\setminus\{0\})$, positive and $1$-homogeneous in $\xi$, $\Sigma$ smooth  and $a\in S^0_{1,0}(\mathbb R^n)$. Then the operator $T$ defined by
  \eqref{eq:Model} maps
  \begin{equation}
  B^r_{1,2}(\mathbb R^n) \to L^\infty(\mathbb R^n)
  \end{equation}
  where $r = n-\frac{n-1}{\gamma(\Sigma)}$ if $\Sigma$ is convex and
  $r = n-\frac{1}{\gamma_0(\Sigma)}$ if $\Sigma$ is arbitrary.
\end{theorem}
The operator norm in this estimate depends on finitely many symbol seminorms of $a$, the estimates of $\phi$ and its derivatives together with some quantitative measures of the contact between $\Sigma$ and tangent lines.

To control the order of contact of $\Sigma$ by tangent lines quantitatively, we will introduce another quantity $\varkappa(\Sigma)$. First, assume that $\Sigma$ is convex and that it is of class $C^{\gamma(\Sigma)+1}$. For $p\in\Sigma$, rotating $\Sigma$ if
necessary, we may assume that it is parameterised
by points $\{(y,h(y)), y\in \Omega\}$ near $p$ for an open set
$\Omega\subset\mathbb R^{n-1}$. For $p=(y,h(y))$, let us define
\begin{equation}
\varkappa(\Sigma,p)=\inf_{|\omega|=1} \sum_{j=2}^{\gamma(\Sigma)}
\left| \frac{\partial^j}{\partial\rho^j} h(y+\rho\omega)|_{\rho=0}
\right|.
\end{equation}
From the definition of $\gamma(\Sigma)$
it follows that $\varkappa(\Sigma,p)>0$ for all $p\in\Omega$.
Indeed, from the definition of $\gamma(\Sigma,p,H)$ it follows
that if $\omega$ is such that $y+\rho\omega\in H$, then
\begin{equation}\label{EQ:ass-kappa1}
\varkappa(\Sigma,p,H)=\left|\frac{\partial^{\gamma(\Sigma,p,H)}}
{\partial\rho^{\gamma(\Sigma,p,H)}} h(y+\rho\omega)|_{\rho=0}
\right|>0.
\end{equation}
Now, we clearly have
$\sum_{j=2}^{\gamma(\Sigma)}
\left| \frac{\partial^j}{\partial\rho^j} h(y+\rho\omega)|_{\rho=0}
\right|\geq \varkappa(\Sigma,p,H)$,
and hence we have $\varkappa(\Sigma,p)>0$ since the set
$|\omega|=1$ is compact.
Noticing that $\varkappa(\Sigma,p)$
is a continuous function of $p$, by compactness of
$\Sigma$ it
follows that if we define
\begin{equation}
\varkappa(\Sigma)=\min_{p\in\Sigma} \varkappa(\Sigma,p),
\end{equation}
then $\varkappa(\Sigma)>0$.
If $\Sigma$ is not convex, we define similarly
\begin{equation}
\varkappa_0(\Sigma)=\min_{p\in\Sigma}
\sup_{|\omega|=1} \sum_{j=2}^{\gamma_0(\Sigma)}
\left| \frac{\partial^j}{\partial\rho^j} h(y+\rho\omega)|_{\rho=0}
\right|.
\end{equation}
Again, we have $\varkappa_0(\Sigma)>0$.

Considering again the example of spheres $\Sigma=r\mathbb S^{n-1}$ of radius $r$, we see that $\gamma(\Sigma)=\gamma_0(\Sigma)=2$ and that $\varkappa(\Sigma)=\varkappa_0(\Sigma)\sim \frac 1r$ is a measure of scalar curvature.

\subsection{Main results}\label{sec:main}
Estimates of \eqref{eq:Tdef} depend on indices of a family of Fresnel surfaces defined in terms
$\phi(t,x,\xi)$,
\begin{equation}\label{eq:Sigma-Def}
   \Sigma_{t,x}  =\{ \xi\in\mathbb R^n : \phi(t,x,\xi) = 1\},\qquad t\ge t_0, x\in\mathbb R^n.
\end{equation}
For a first result we assume that $\Sigma_{t,x}$ is convex for all $t\ge t_0$ and all $x\in\mathbb R^n$.

\begin{theorem}\label{thm:2.3}
Assume that $\phi(t,x,\xi)$ is real-valued, continuous in $t$ and $x$ and smooth in $\xi\in\mathbb R^n\setminus\{0\}$, homogeneous of order one in $\xi$ and such that for some $t_0>0$ and $C>0$ we have
\begin{equation}
    C^{-1} |\xi| \le \phi(t,x,\xi) \le C|\xi| \quad\text{and}\quad |\partial_\xi^\alpha \phi(t,x,\xi)|\le C|\xi|^{1-|\alpha|}
\end{equation}
for all $t\ge t_0$, all $x$, $\xi\ne0$ and all multi-indices $\alpha$ with $|\alpha|\le \max\{ \gamma+1, \lfloor (n-1)/\gamma\rfloor +2\}$. Assume further that all the sets $\Sigma_{t,x}$ defined by \eqref{eq:Sigma-Def}
are convex for all $t\ge t_0$ and assume that
\begin{equation}
 \limsup_{t\to\infty}\sup_x \gamma(\Sigma_{t,x}) \le\gamma
 \quad\text{together with}\quad
 \liminf_{t\to\infty} \inf_x \varkappa(\Sigma_{t,x})>0.
\end{equation}
Suppose further that the amplitude $a(t,x,\xi)$ satisfies
\begin{equation}
   |\partial_\xi^\alpha a(t,x,\xi)| \le C_\alpha \langle\xi\rangle^{-|\alpha|}\quad\text{for all}\quad |\alpha|\le \lfloor (n-1)/\gamma\rfloor +1.
\end{equation}
Then the operator family $T_t$ defined by \eqref{eq:Tdef} satisfies for all $t\ge t_0$ the estimate
\begin{equation}
   \| T_t u\|_{L^\infty(\mathbb R^n)} \le C t^{-\frac{n-1}\gamma} \|u\|_{B^r_{1,2} (\mathbb R^n)}
\end{equation}
with $r=n-\frac{n-1}\gamma$.
\end{theorem}

In the non-convex situation decay rates can be much lower. Results can be improved by using a more detailed analysis of the geometric situation near points of high contact order.

\begin{theorem}\label{thm:2.4}
Assume that $\phi(t,x,\xi)$ is real-valued, continuous in $t$ and $x$ and smooth in $\xi\in\mathbb R^n\setminus\{0\}$, homogeneous of order one in $\xi$ and such that for some $t_0>0$ and $C>0$ we have
\begin{equation}
    C^{-1} |\xi| \le \phi(t,x,\xi) \le C|\xi| \quad\text{and}\quad |\partial_\xi^\alpha \phi(t,x,\xi)|\le C|\xi|^{1-|\alpha|}
\end{equation}
for all $t\ge t_0$, all $x$, $\xi\ne0$ and all multi-indices $\alpha$ with $|\alpha|\le\gamma_0+1$. Assume further that all the sets $\Sigma_{t,x}$ defined by \eqref{eq:Sigma-Def} satisfy
\begin{equation}
 \limsup_{t\to\infty}\sup_x \gamma_0(\Sigma_{t,x}) \le\gamma_0
 \quad\text{together with}\quad
 \liminf_{t\to\infty} \inf_x \varkappa_0(\Sigma_{t,x})>0.
\end{equation}
 Suppose further that the amplitude $a(t,x,\xi)$ satisfies
\begin{equation}
   |\partial_\xi^\alpha a(t,x,\xi)| \le C_\alpha
   \langle\xi\rangle^{-|\alpha|}\quad\text{for all}\quad
   |\alpha|\le 1.
\end{equation}
Then the operator family $T_t$ defined by
\eqref{eq:Tdef} satisfies for all $t\ge t_0$ the estimate
\begin{equation}
   \| T_t u\|_{L^\infty(\mathbb R^n)} \le C
   t^{-\frac{1}{\gamma_0}} \|u\|_{B^r_{1,2} (\mathbb R^n)}
\end{equation}
with $r=n-\frac{1}{\gamma_0}$.
\end{theorem}

The proofs of both theorems are similar to the ones in \cite{RW10} and full details are omitted here. The dependence on $x$ is parametric, and the assumption assure that the resulting estimates are uniform in $x$.

If we make further regularity assumptions with respect to the $x$-variable it follows that the operators $T_t$ are uniformly $L^2$-bounded such that interpolation implies the usual $L^p$--$L^q$ decay estimates. For completeness, we include a corresponding statement. The assumptions are chosen such that after substituting $x=ty$ and
$t\xi=\eta$ Theorem~2.5 of \cite{RS06} applies:

\begin{theorem}[\cite{RS06}]\label{Th2.5}
Let $\Gamma_y,\Gamma_\xi\subset\mathbb R^n$ be open cones.
Let operator $T$ be defined by
\begin{equation}\label{T2}
 Tu(x)
 =\int_{\mathbb R^n}\int_{\mathbb R^n}
 e^{i(x\cdot\xi+\varphi(y,\xi))}a(y,\xi)u(y)\ \mathrm dy
 \ \mathrm d\xi,
\end{equation}
where $a(y,\xi)\in C^\infty(\mathbb R^n_y\times\mathbb R^n_\xi)$,
${\rm supp}\ a\subset \Gamma_y\times\Gamma_\xi$, and
$\varphi(y,\xi)\in C^\infty(\Gamma_y\times\Gamma_\xi)$ is
a real-valued function.
Assume that
\[
|\partial_y^\alpha\partial_\xi^\beta a(y,\xi)|
\leq C_{\alpha\beta},
\]
for $|\alpha|, |\beta|\leq 2n+1$.
Also assume that
\[
|\det\partial_y\partial_\xi\varphi(y,\xi)|\geq C>0
\quad\text{on}\quad\Gamma_y\times\Gamma_\xi,
\]
that
$$
|\partial_\xi\varphi(x,\xi)-\partial_\xi
\varphi(y,\xi)|\geq C|x-y|\quad
\text{for} \quad
x,y\in\Gamma_y,\; \xi\in\Gamma_\xi,
$$
and that
\[
|\partial_y^\alpha\partial_\xi \varphi(y,\xi)\leq C_{\alpha}|,
\qquad
|\partial_y\partial_\xi^\beta \varphi(y,\xi)|\leq C_{\beta}
\quad\text{on supp}\; a
\]
for $1\leq |\alpha|,|\beta|\leq 2n+2$.
Then the operator $T$ is $L^2(\mathbb R^n)$-bounded, and satisfies
\[
\|T\|_{L^2\to L^2}\leq C
\;\text{\rm sup}_{|\alpha|,|\beta|\leq 2n+1}
 \|\partial_y^\alpha\partial_\xi^\beta a(y,\xi)\|
  _{L^\infty(\mathbb R^n_y\times\mathbb R^n_\xi)}.
\]
\end{theorem}

We can now formulate the corresponding $L^2$-result:

\begin{theorem}\label{THM:L2}
Let
\begin{equation}\label{eq:Tdef11}
   T_t u (x) = \int_{\mathbb R^n} \mathrm e^{\mathrm i ( x\cdot\xi + t\phi(t,x,\xi) )}a(t,x,\xi)\, \widehat u(\xi) \mathrm d\xi,
\end{equation}
Assume $\phi(t,x,\xi)$ is real-valued, $1$-homogeneous in $\xi$ and satisfies
\begin{equation}\label{eq:phase-asss}
   |\det(\mathrm I + t \partial_x\partial_\xi \phi(t,x,\xi))| \ge C_0 >0,\qquad |t^{|\alpha|}\partial_x^\alpha\partial_\xi^\beta \phi(t,x,\xi)| \le C_{\alpha,\beta}
\end{equation}
uniform in $t\ge t_0$, $x$, $\xi\ne0$ and for $1\le |\alpha|,|\beta|\le  2n+2$. Assume further that $a(t,x,\xi)$ is supported in
$t |\xi|\ge C$ for some constant $C$ and that
\begin{equation}
   |t^{|\alpha|}\partial_x^\alpha\partial_\xi^\beta a(t,x,\xi)| \le C_{\alpha,\beta}
\end{equation}
for all $|\alpha|, |\beta|\le 2n+2$ and $t\ge t_0$. Then the family $T_t$ in \eqref{eq:Tdef11}, $t\ge t_0$, is uniformly $L^2$-bounded,
\begin{equation}
   \|T_t u\|_{L^2(\mathbb R^n)} \le C \| u\|_{L^2(\mathbb R^n)}.
\end{equation}
\end{theorem}
Let us sketch how Theorem \ref{Th2.5} implies
Theorem \ref{THM:L2}. Using homogeneity, we can rewrite
\eqref{eq:Tdef11} as
\begin{equation}\label{eq:Tdef2}
   T_t u (x) = t^{-n}\int_{\mathbb R^n} \mathrm e^{\mathrm i (
   z\cdot\eta + \phi_t(t,z,\eta) )}a_t(t,z,\eta)\,
   \widehat u(\eta/t)
\mathrm d\eta,
\end{equation}
with $x=tz, t\xi=\eta$, and
$$
\phi_t(t,z,\eta)=\phi(t,tz,\eta),\quad
a_t(t,z,\eta)=a(t,tz,\eta/t).
$$
By rescaling, the powers of $t$ in the $L^2$-estimate
cancel, and it can be shown that the adjoint to the operator in
\eqref{eq:Tdef2} satisfies assumptions on Theorem
\ref{Th2.5} uniformly for $t\geq t_0$.

We also note that the assumptions on the mixed derivatives of
$\phi$ as in \eqref{eq:phase-asss} can be relaxed by looking only at
$|\alpha|=1$ or $\beta=1$, as in Theorem \ref{Th2.5}.

\subsection{Appendix: Fourier transform of surface carried measures} Let $\Sigma\subseteq\mathbb R^n$ be a closed hypersurface. Then the indices $\gamma(\Sigma)$ and $\gamma_0(\Sigma)$ are related to decay estimates for Fourier transforms of surface carried measures. Consider for $f\in C^\infty(\Sigma)$ the integral
\begin{equation}
  u(x) = \int_\Sigma \mathrm e^{\mathrm i x\cdot\xi} f(\xi) \mathrm d\xi,
\end{equation}
where $\mathrm d\xi$ denotes the $(n-1)$-dimensional surface measure on $\Sigma$. Then the following theorem is valid.
\begin{theorem}
\begin{enumerate}
\item Assume $\Sigma$ is a smooth hypersurface and convex. Then
\begin{equation}
   |u(x)| \le C \langle x\rangle^{-\frac{n-1}{\gamma(\Sigma)}} \|f\|_{C^r(\Sigma)}
\end{equation}
with $r\ge (n-1)/\gamma(\Sigma)+1$.
\item For general (non-convex) smooth hypersurface $\Sigma$ it follows that
\begin{equation}
   |u(x)| \le C \langle x\rangle^{-\frac{1}{\gamma_0(\Sigma)}} \|f\|_{C^1(\Sigma)}.
\end{equation}
\end{enumerate}
\end{theorem}
The proof of the first statement in \cite{sugi94} uses real analyticity of the surface, the stronger results are due to methods of \cite{Ru09} and \cite{RS10}. Smoothness of the surface can be further relaxed to the class $C^{\gamma}$ with $\gamma=\gamma(\Sigma)$ (or $\gamma_0(\Sigma)$, respectively).

\section{Applications to hyperbolic evolution equations}
The estimates discussed so far have applications for dispersive estimates of hyperbolic evolution equations. Without aiming for completeness we will discuss some possible applications.

\subsection{Wave models} As a first example we consider wave models with time-dependent coefficients,
\begin{equation}\label{eq:wave}
   \partial_t^2 u - \sum_{i,j=1}^n a_{ij}(t) \partial_{x_i}\partial_{x_j} u
    = 0
\end{equation}
in $\mathbb R\times\mathbb R^n$ and with Cauchy data $u(0,\cdot)=u_0$, $u_t(0,\cdot)=u_1$, both assumed to be Schwartz for simplicity. Denoting
\begin{equation}
  \mathcal T\{\ell\} = \{ a\in C^\infty(\mathbb R) : |\partial_t^k a(t) | \le c_k \langle t\rangle^{\ell-k} \},
\end{equation}
we assume  $a_{ij} \in \mathcal T\{0\}$, $a_{ij}(t)=\overline{a_{ji}(t)}$, together with the strict hyperbolicity assumption
\begin{equation}
a(t,\xi) =  \sum_{i,j=1}^n a_{ij}(t) \xi_i\xi_j \ge C |\xi|^2,
\end{equation}
for some $C>0$ uniform in $t$. Within the zone $\{(1+t)|\xi|\gtrsim 1\}$ solutions to \eqref{eq:wave} are given by Fourier integrals of the form \eqref{eq:Tdef} with phase functions
\begin{equation}
  \phi_\pm(t,\xi) =\pm  \frac 1t \int_0^t \sqrt{a(\theta,\xi)}\mathrm d\theta.
\end{equation}
In this situation the Fresnel surfaces are given by
\begin{equation}
  \Sigma_t = \left\{ \xi\in\mathbb R^n :  t = \int_0^t \sqrt{a(\theta,\xi)}\mathrm d\theta \right\}
\end{equation}
and satisfy $\gamma(\Sigma_t)=2$ together with uniform upper and lower bounds for $\varkappa(\Sigma_t)$.
This follows directly from \cite[Lemma~3.1]{R04}. Thus Theorem~\ref{thm:2.3} is applicable and
in combination with a uniform bound on the $L^2$-energy one concludes that solutions to \eqref{eq:wave} satisfy
\begin{equation}
   \|\partial_t u(t,\cdot)\|_q + \| \nabla u(t,\cdot)\|_q \le C_{pq} (1+t)^{-\frac{n-1}2\left(\frac1p-\frac1q\right)} \left(
   \|u_1\|_{W^{p,r_p+1}} + \|u_2\|_{W^{p,r_p}}\right)
\end{equation}
with $1<p\le 2\le q<\infty$, $pq=p+q$ and $r_p = n(1/p-1/q)$.

For a detailed consideration of such models we refer to Reissig \cite{R04}, \cite{RS05a}. It is possible to include further lower order terms $b(t)\partial_t u$, $\sum_j c_j(t) \partial_{x_j} u$ and $d(t)u$ with $b, c_j\in \mathcal T\{-1\}$ and $d\in\mathcal T\{-2\}$ (together with further assumptions to control the low-frequency behaviour). We refer to \cite{Wir06} for an example of such a treatment. If lower order terms belong to different classes, the asymptotic behaviour of solutions might change its type, see, e.g., \cite{Wir07} a discussion about dissipative wave equations and effective dissipation.

The assumptions on coefficients can be weakened to classes
\begin{equation}
  \mathcal T_\nu\{\ell\} = \{ a\in C^\infty(\mathbb R) : |\partial_t^k a(t) | \le c_k \langle t \log (e+t) ^\nu\rangle^{\ell-k} \}
\end{equation}
with parameter $\nu\in[0,1]$ where $\nu\in(0,1)$ implies a small loss of decay and $\nu=1$ yields a polynomial loss. It can be shown that without further assumptions $\nu>1$ leads generically to supra-polynomial growth of the energy. Additional assumptions to compensate this behaviour for wave equations with variable speed of propagation were considered by Hirosawa \cite{Hiro07}, see also \cite{Wir10b}, but the estimates of Theorems~\ref{thm:2.3} and \ref{thm:2.4} need to be refined in order to obtain dispersive estimates in this situation.

\subsection{Higher order scalar equations}\label{sec:3.2}
In a similar way we can treat homogeneous higher order equations of the form
\begin{align}
   &\mathrm D_t^m u + \sum_{k=0}^{m-1} \sum_{|\alpha|=m-k} a_{k,\alpha} (t) \mathrm D_t^k \mathrm D_x^\alpha u = 0,\\
   & \mathrm D_t^j u(0,\cdot) = u_j,\quad j=0,1,\ldots, m-1,
\end{align}
where $\mathrm D = -\mathrm i\partial$, $a_{k,\alpha}\in\mathcal T\{0\}$ and the characteristic equation
\begin{equation}
  \tau^m + \sum_{k=0}^{m-1}\sum_{|\alpha|=m-k} a_{k,\alpha} (t) \tau^k \xi^\alpha = 0
\end{equation}
has $m$ uniformly distinct real solutions $\tau_1(t,\xi)$,\ldots, $\tau_m(t,\xi)$, $|\tau_i(t,\xi) - \tau_j(t,\xi) | \ge C|\xi|$ for $i\ne j$ and some $C>0$. Under the stronger assumption that $a_{k,\alpha}'(t)\in L^1(\mathbb R_+)$, this model was treated by Matsuyama and the first author, \cite{MR10}, the full form follows from \cite{RW10}. If we assume
\begin{equation}\label{EQ:hh-int}

    \bigg| \sum_{k\ne j} \int_0^t \frac{\partial_t \tau_j(\theta,\xi)}{\tau_j(\theta,\xi)-\tau_k(\theta,\xi)}\mathrm d \theta  \bigg|<C
\end{equation}
uniform in $j$, $t$ and $\xi\ne0$, solutions are given as sums of Fourier integrals of the form \eqref{eq:Tdef} with phases
\begin{equation}\label{eq:phi-j}
   \phi_j(t,\xi) = \frac1t \int_0^t \tau_j(\theta,\xi)\mathrm d\theta
\end{equation}
and uniformly bounded amplitudes. Again, the results from Section~\ref{sec:main} provide dispersive estimates for solutions to such problems. We remark that for higher order equations the associated Fresnel surfaces $\Sigma_t$ need not be convex and therefore the full generality of Theorems~\ref{thm:2.3} and \ref{thm:2.4} is of importance here.

We note that the analysis in \cite{MR10} is based on the
asymptotic integration method for constructing solutions
for equations with homogeneous symbols,
and it requires only $C^1$-regularity of coefficients,
i.e. only that $a_{k,\alpha}\in C^1$ in terms of the regularity
in time, which is crucial in applications to the Kirchhoff
equations, see e.g. \cite{MR09}.

\subsection{Hyperbolic systems}\label{sec:3.3}
In \cite{RW08}, the authors announced
results on dispersive estimates
for a general class of pseudo-differentiable
hyperbolic systems with time-dependent coefficients
and arbitrary $\gamma$.
D'Abbico, Lucente and Taglialatela, \cite{dAbbico:2009},
considered differential
hyperbolic systems with $\gamma=2$ , while in \cite{RW10}
the authors gave a rigorous
treatment for general pseudo-differentiable hyperbolic
systems with time-dependent coefficients and
arbitrary $\gamma$. For the sake of simplicity we
formulate the results for the homogeneous situation. We consider
\begin{equation}\label{eq:system}
   \mathrm D_t U = A(t,\mathrm D_x) U,\qquad U(0,\cdot)=U_0
\end{equation}
for a $m\times m$ matrix $A(t,\mathrm D_x)$ of $t$-dependent Fourier multipliers. We denote
\begin{equation}
  \mathcal S\{m_1,m_2\} = \{ a(t,\xi) \in C^\infty(\mathbb R\times (\mathbb R^n\setminus\{0\}) ) :
   |\xi|^{|\alpha|-m_1} \mathrm D_\xi^\alpha a(t,\xi) \in \mathcal T\{m_1\} \},
\end{equation}
the condition being uniform in $\xi$.
We assume that the symbol $A(t,\xi)\in\mathcal S\{1,0\}$ is $1$-homogeneous in $\xi$, a self-adjoint matrix and its eigenvalues
$\tau_1(t,\xi)$,\ldots, $\tau_m(t,\xi)$ are uniformly distinct, $|\tau_i(t,\xi) - \tau_j(t,\xi) | \ge C|\xi|$ for $i\ne j$ and some $C>0$. Without further assumptions it follows that the solutions to \eqref{eq:system} can be represented in the form
\eqref{eq:Tdef} with phases given by \eqref{eq:phi-j}.

Assumptions can be relaxed, without assuming self-adjointness we have to require real eigenvalues together with a uniform symmetrisability of the matrix $A(t,\xi)$ together with further assumptions to control the large-time influence of lower order terms. We refer to \cite{RW10} for details.