TITLE: Could I prove this coset theorem using just homomorphism like following?
QUESTION [1 upvotes]: In Dummit's Abstract Algebra, there is this Proposition 3.4 in page 80:

If there is some homomorphism from G to H as following:

Could I prove it by using only that homomorphism as following?
That homomorphism would bring $u$, $v$, $n$, and $m$ to exactly only one element in the codomain H: $U$, $V$, $N$, and $N$ respectively (otherwise it's not a function). So $UN = VN$ is only when $U = V$. As homomorphism only brings any values from domain to exactly only one value in codomain, uN = vN is only when u and v are in the same fiber of homomorphism. I said in the same fiber because also note that in my proof the set of elements of fiber above N is even not necessarily a subgroup of $G$. Where am I wrong?

REPLY [2 votes]: Let's write $u\sim v$ iff $u\in vN$.
There indeed exists a canonical function $\varphi:G\to G/\sim$ with fibers exactly the left cosets, but

In order to establish the quotient set $G/\sim$ we should already know that $\sim$ is an equivalence relation, which is basically the claim of this proposition.
We will get a group structure on $G/\sim$ iff $N$ is a normal subgroup, which means that the left and right cosets coincide. (Observe e.g. that taking elementwise inverses, we obtain $(gN)^{-1}=N^{-1}g^{-1}=Ng^{-1}$.)
In this case, $\varphi$ will be indeed a homomorphism.
But this is not required for proving that $\sim$ is an equivalence relation.

Note also, that for any subset $N$ of $G$, we have the following facts about the relation $\sim$ defined above:

$\sim$ is reflexive iff $1\in N$ (where $1$ denotes the identity element of $G$).
$\sim$ is symmetric iff $N$ is closed under inverses ($x\in N\implies x^{-1}\in N$).
$\sim$ is transitive iff $N$ is closed under multiplication ($x, y\in N\implies xy\in N$).


text:
\section{Stability of Dynamic Modified-CS and CS-residual}
In Theorem \ref{modcsbnd} (Theorem \ref{csresbnd}), we bounded the modified-CS (CS-residual) error as a function of $|\Delta|$ and $|\Delta_e|$. The bound is small as long as $|\Delta|$ and $|\Delta_e|$ are small. The important question is: under what conditions on the measurement model,  will $|\tDelta_e|$, $|\tDelta|$ (and hence $|\Delta_e|$, $|\Delta|$) remain stable, i.e. remain bounded by a small value compared to the signal support size, at all times?

When measurements are noiseless and we use noiseless-modified-CS, the answer to the above is trivial since in that case $\Delta$ and $\Delta_e$ contain the additions and removals from the true support respectively. Thus, as long as the maximum number of support additions and removals satisfy the exact reconstruction condition, the  reconstruction error will be stable at zero at all times. But when measurements are noisy, $\Delta_t$ also contains the undetected part of the previous support, besides the new additions. Similarly for $\Delta_e$. Thus, there is a need to explicitly find conditions under which $|\Delta|$ and $|\Delta_e|$ remain bounded by a time-invariant small value at all times. This, in turn, will imply that the signal estimation error is small compared to the signal power at all times.

\subsection{Signal model for studying stability}
For studying stability, we need to assume a signal model. We assume a simple deterministic model that (a) allows equal and nonzero number of additions/removals from the support at every time, (b) allows a new coefficient magnitude to gradually increase from zero, at a rate $M/d$, for a duration, $d$, and finally reach a constant value, $M$, (c) allows coefficients to gradually decrease and become zero (get removed from support) at the same rate, and (d) has constant signal power and support size.

\begin{sigmodel} Assume the following.
\ben

\item At $t=0$, the support size is $S_0$. The support contains $2S_a$ elements each with magnitude $M/d,2M/d, \dots (d-1)M/d$, and $S_0-(2d-2)S_a$ elements with magnitude $M$.

\item At each $t>0$, $S_a$ coefficients get added to the support at magnitude $M/d$. Denote this set by $\Aset_t$.

\item At each $t>0$, $S_a$ coefficients which had magnitude $M/d$ at $t-1$ get removed from the support (magnitude becomes zero). Denote this set by $\Rset_t$.

\item At each $t>0$, the magnitude of $S_a$ coefficients which had magnitude $(j-1) M/d$ at $t-1$ increases to $jM/d$. This occurs for all $1 \le j \le d-1$.

\item At each $t>0$, the magnitude of $S_a$ coefficients which had magnitude $(j+1)M/d$ at $t-1$ decreases to $jM/d$. This occurs for all $1 \le j \le (d-1)$.

\een
\label{sigmod2}
\end{sigmodel}

From the above model, the size and composition of the support at any time is the same as that at $t=0$. The size is $|N_t|=S_0$ and $N_t$ consists of $(d-1)S_a$ increasing coefficients and $(d-1)S_a$ decreasing coefficients, and $S_0- (2d-2)S_a$ constant coefficients with magnitude $M$.

To understand the model better, let us define the following.
\bd
Define the following
\ben
\item $\Dset_{t}(j): = \{i: |x_{t,i}| = jM/d, \ |x_{t-1,i}| = (j+1)M/d \}$.
\item $\Iset_{t}(j): = \{i: |x_{t,i}| = jM/d, \ |x_{t-1,i}| = (j-1)M/d \}$.
\item Small coeff. set, $\Sset_t(j):= \{i:  0 < |x_{t,i}| < j M/d \}$.
\item Newly added set, $\Aset_t:= \Iset_t(1)$.
\item Newly removed set, $\Rset_t:= \Dset_t(0)$.
\een
\ed
Consider a $d_0 \le d$. From the signal model, it is clear that at any time, $t$, $S_a$ elements enter the small coefficients' set, $\Sset_t(d_0)$, from the bottom (set $\Aset_t$) and $S_a$ enter from the top (set $\Dset_{t}(d_0-1)$). Similarly $S_a$ elements leave  $\Sset_t(d_0)$ from the bottom  (set $\Rset_t$) and $S_a$ from the top (set $\Iset_{t}(d_0)$). Thus,
\bea
\Sset_t(d_0) = \Sset_{t-1}(d_0)  \cup (\Aset_t \cup \Dset_{t}(d_0-1)) \setminus (\Rset_t \cup \Iset_{t}(d_0)) \ \ \
\label{sseteq}
\eea

\bd
Define
\ben
\item Increasing set, $\Inc_t := \Iset_t(1) \cup \Iset_t(2) \dots \cup  \Iset_t(d-1)$
\item Decreasing set, $\Dec_t := \Dset_t(1) \cup \Dset_t(2) \dots \cup  \Dset_t(d-1)$
\item Constant set, $\Con_t:= \{i : |x_{t,i}| = M \}$.
\een
Clearly $\Con_t = N_t \setminus (\Inc_t \cup \Dec_t)$.
\ed

With the above definitions, a generative model for Signal Model \ref{sigmod2} can be specified as follows. At each $t > 0$,
\ben

\item Select $\Dset_t(d-1) \subseteq \Con_{t-1}$ of size $S_a$ and select the newly added set, $\Iset_t(1) \subseteq N_{t-1}^c$ of size $S_a$.

\item Update the sets:
\bea
\Iset_t(j) \se \Iset_{t-1}(j-1), \ \forall \ j=2, \dots (d-1) \nn \\
\Dset_t(j) \se \Dset_{t-1}(j+1), \ \forall \ j=1, \dots (d-2) \nn \\
\Inc_t \se \Inc_{t-1} \cup \Iset_t(1) \setminus \Iset_t(d) \nn \\
\Dec_t \se \Dec_{t-1} \cup \Dset_t(d-1) \setminus \Dset_t(0) \nn \\
\Con_t \se \Con_{t-1} \cup  \Iset_t(d) \setminus \Dset_t(d-1) \nn \\
N_t \se \Inc_t \cup \Dec_t \cup \Con_t
\eea

\item Update the coefficient magnitudes:
\bea
|(x_{t})_{\Inc_{t}}| \se |(x_{t-1})_{\Inc_{t}}| + M/d \nn \\
|(x_{t})_{\Dec_{t}}| \se |(x_{t-1})_{\Dec_{t}}| - M/d \nn \\
|(x_{t})_{\Con_{t}}| \se |(x_{t-1})_{\Con_{t}}|
\eea

\item  The model does not assume anything about signs. But a reasonable generative model will keep the signs the same except when the coefficient first gets added (at that time, can set the sign to $\pm 1$ with equal probability).
\een

config: arxiv


text:
\begin{document}

\title{A Lower Bound of the First Dirichlet Eigenvalue of a Compact Manifold
with Positive Ricci Curvature
\thanks{2000 Mathematics Subject Classification Primary 58J50, 35P15; Secondary 53C21}}

\author{Jun LING}
\date{}

\maketitle

\begin{abstract}
We give a new estimate on the lower bound for the first Dirichlet
eigenvalue for a compact manifold with positive Ricci curvature in
terms of the in-diameter and the lower bound of the Ricci
curvature. The result improves the previous estimates.
\end{abstract}

\section{Introduction}\label{sec-intro}
If $(M,g)$ is an n-dimensional compact Riemannian manifold whose
Ricci curvature has a positive lower bound $(n-1)K$ for some
constant $K>0$ and whose non-empty boundary $\partial M$ has
nonnegative mean curvature with respect to the outward normal,
Reilly \cite{reilly} gave the following lower bound of the first
Dirichlet eigenvalue $\lambda$ of the Laplacian on $M$
\begin{equation}\label{nK-bound}
\lambda\geq nK.
\end{equation}
This estimate gives no information when the above constant $K$
vanishes. In such case, Li-Yau \cite{liy1} and Zhong-Yang
\cite{zy} provided another lower bound for the first non-zero
eigenvalue of a closed manifold
\[
\lambda\geq \frac{\pi^2}{d^2}.
\]

It is an interesting problem to find a unified lower bound of the
first Dirichlet eigenvalue $\lambda$ in terms of the lower bound
$(n-1)K$ of the Ricci curvature and the diameter $d$, in-diameter
$\tilde{d}$ and other geometric quantities, which do not vanish as
$K$ vanishes, of the manifold with positive Ricci curvature. D.
Yang \cite{yang} proved that
\begin{equation}\label{yang-bound}
\lambda \geq \frac14(n-1)K + \frac{\pi^2}{(\tilde{d})^2},
\end{equation}
where $\tilde{d}$ is the diameter of the largest interior ball in
$M$.

In this paper we give a new estimate on the lower bound of the
first Dirichlet eigenvalue $\lambda$. We have the following
result.

\begin{theorem} \label{main-thm}
If $(M,g)$ is an $n$-dimensional compact Riemannian manifold with
boundary. Suppose that Ricci curvature \textup{Ric}$(M)$ of $M$ is
bounded below by $(n-1)K$ for some constant $K>0$
\begin{equation}\label{ricci-bound}
\textup{Ric}(M)\geq (n-1)K
\end{equation}
and that the mean curvature of the boundary $\partial M$ with
respect to the outward normal is nonnegative, then the first
Dirichlet eigenvalue $\lambda$ of the Laplacian $\Delta$ of $M$
has the following lower bound
\begin{equation}                    \label{main-bound-in}
\lambda \geq  \frac12(n-1)K+\frac{\pi^2}{(\tilde{d})^2},
\end{equation}
where $\tilde{d}$ is the diameter of the largest interior ball in
$M$, that is, $\tilde{d}=2\sup_{x\in M}\{\textup{dist}(x,\partial
M)\}$.
\end{theorem}

Our result improves Yang' bound (\ref{yang-bound}) by doubling the
coefficient before $(n-1)K$. In the proof,  we use a function
$\xi$ that the author constructed in \cite{ling-d} for the
construction of the suitable test function instead of using the
Zhong-Yang's canonical function. That provides a new way to
sharpen the bound. In the next section, we derive some preliminary
estimates and conditions for test functions  first and we
construct the needed test function and prove the main result in
the last section.

\section{Preliminary Estimates}\label{sec-pre-es}
The first basic estimate is of Lichnerowicz-type. Recall that the
classic Lichnerowicz Theorem \cite{lich} states that if $M$ is an
$n$-dimensional closed manifold whose Ricci curvature satisfies
(\ref{ricci-bound}) then the first non-zero eigenvalue has a lower
bound (\ref{nK-bound}). Reilly \cite{reilly} proved that this
Lichnerowicz-type estimate remains true for the first Dirichlet
eigenvalue $\lambda$ as well if the manifold has the same lower
bound for the Ricci curvature and has non-empty boundary whose
mean curvature with respect to the outward normal is nonnegative.
For the completeness and consistency, we use gradient estimate in
\cite{li}-\cite{liy1} and \cite{sy} to derive the
Lichnerowicz-type estimate.
\begin{lemma}               \label{nK-lemma}
Under the conditions in Theorem \ref{main-thm}, the estimate
(\ref{nK-bound}) holds.
\end{lemma}
\begin{proof}\quad
Let $v$ be a normalized eigenfunction of the first Dirichlet
eigenvalue such that
\begin{equation}                                \label{v-d-con}
\sup_{M}v=1, \quad \inf_{M}v=0.
\end{equation}
The function $v$ satisfies the following
\begin{equation}                                 \label{v-d-eq}
\Delta v=-\lambda v\quad \textrm{in }M
\end{equation}
and
\begin{equation}                                \label{v-d-boundary}
v=0\qquad \textrm{ on }\partial M.
\end{equation}

Take an orthonormal frame $\{e_1, \dots, e_n\}$ of $M$ about
$x_0\in M$. At $x_0$ we have
\[
\nabla_{e_j}(|\nabla v|^2)(x_0)= \sum_{i=1}^n2v_iv_{ij}
\]
and
\begin{eqnarray}
\Delta (|\nabla v|^2)(x_0)&=&2\sum_{i,j=1}^nv_{ij}v_{ij}+2\sum_{i,j=1}^nv_{i}v_{ijj}\nonumber\\
      &=&2\sum_{i,j=1}^nv_{ij}v_{ij}+2\sum_{i,j=1}^nv_{i}v_{jji}+2\sum_{i,j=1}^n\textrm{R}_{ij}v_iv_j\nonumber\\
      &=&2\sum_{i,j=1}^nv_{ij}v_{ij}+2\nabla v \nabla (\Delta v) + 2\textrm{Ric}(\nabla v, \nabla v)\nonumber\\
      &\geq&2\sum_{i=1}^nv_{ii}^2+ 2\nabla v \nabla (\Delta v) + 2(n-1)K|\nabla
      v|^2\nonumber\\
      &\geq&\frac{2}{n}(\Delta v)^2-2\lambda|\nabla v |^2 + 2(n-1)K|\nabla v|^2.\nonumber
\end{eqnarray}
Thus at all point $x\in M$,
\begin{equation}                        \label{basic2.002-d}
\frac12\Delta (|\nabla v|^2)\geq \frac1n \lambda^2 v^2 +
[(n-1)K-\lambda]|\nabla v|^2.
\end{equation}
On the other hand, after multiplying (\ref{v-d-eq}) by $v$ and
integrating both sides over $M$ and using (\ref{v-d-boundary}), we
have
\[
\int_{M}\lambda v^2\,dx=-\int_Mv\Delta v\, dx
\]
\[=-\int_{\partial M}v\frac{\partial}{\partial
\nu}v\,ds+\int_{M}|\nabla v|^2\, dx=\int_{M}|\nabla v|^2\, dx,
\]
where and below $\nu$ is the outward normal of $\partial M$. That
the integral on the boundary vanishes is due to
(\ref{v-d-boundary}). Integrating (\ref{basic2.002-d}) over $M$
and using the above equality, we get
\begin{equation}                                        \label{basic2.003-d}
\frac12\int_{\partial M}\frac{\partial}{\partial \nu}(|\nabla
v|^2)\,dx \geq\int_M(nK-\lambda)\frac{n-1}{n}\lambda v^2\,dx.
\end{equation}

We need show that $\frac{\partial}{\partial \nu}(|\nabla v|^2)\leq
0$ on $\partial M$. Take any $x_0\in \partial M$. If $\nabla
v(x_0)=0$, then it is done. Assume now that $\nabla v(x_0)\not=
0$. Choose a local orthonormal frame $\{e_1, e_2,\cdots, e_{n}\}$
of $M$ about $x_0$ so that $e_n$ is the unit outward normal vector
field near $x_0\in \partial M$ and $\{e_1, e_2,\cdots,
e_{n-1}\}|_{\partial M}$ is a local frame of $\partial M$ about
$x_0$. The existence of such local frame can be justified as the
following. Let $e_n$ be the local unit outward normal vector field
of $\partial M$ about $x_0\in \partial M$ and $\{e_1, \cdots,
e_{n-1}\}$ the local orthonormal frame of $\partial M$ about
$x_0$. By parallel translation along the geodesic
$\gamma(t)=\exp_{x_0}te_n$, we may extend $e_1$, $\cdots$,
$e_{n-1}$ to local vector fields of $M$. Then the extended frame
$\{e_1, e_2,\cdots, e_{n}\}$ is what we need. Note that
$\nabla_{e_n}e_i=0$ for $i\leq n-1$. Since $v|_{\partial M}=0$, we
have $v_i(x_0)=0$ for $i\leq n-1$. Using (\ref{v-d-con})-(\ref{v-d-boundary})
in the following arguments, then we have that at $x_0$,
\begin{eqnarray}\label{gra-v-leq-0}
& &{ }\frac{\partial}{\partial \nu}(|\nabla
v|^2)(x_0)=\sum_{i=1}^n 2v_iv_{in}=2v_nv_{nn}\nonumber\\
& &=2v_n(\Delta^M v -\sum_{i=1}^{n-1}v_{ii}) =2v_n(-\lambda
v-\sum_{i=1}^{n-1}v_{ii})\nonumber\\
&
&=-2v_n\sum_{i=1}^{n-1}v_{ii}=-2v_n\sum_{i=1}^{n-1}(e_ie_iv-\nabla^M_{e_i}e_i
v)\nonumber\\
& &=2v_n\sum_{i=1}^{n-1}\nabla^M_{e_i}e_iv
=2v_n\sum_{i=1}^{n-1}\sum_{j=1}^{n}g(\nabla^M_{e_i}e_i
,e_j)v_j\nonumber\\
& &=2v_n^2\sum_{i=1}^{n-1}g(\nabla^M_{e_i}e_i ,e_n)
=-2v_n^2\sum_{i=1}^{n-1}g(\nabla^M_{e_i}e_n ,e_i)\nonumber\\
& &=-2v_n^2\sum_{i=1}^{n-1}h_{ii}=-2v_n^2(x_0)m(x_0)\nonumber\\
& &\leq 0 \quad\textrm{by the non-negativity of }m,
\end{eqnarray}
where $g(,)$ is the Riemann metric of $M$, $(h_{ij})$ is the
second fundamental form of $\partial M$ with respect to the
outward normal $\nu$ and $m$ is the mean curvature of $\partial M$
with respect to $\nu$. Therefore (\ref{nK-bound}) holds.
\end{proof}

\begin{lemma}       \label{pre-es-d}
Let $v$ be, as the above, the normalized eigenfunction for the
first Dirichlet eigenvalue $\lambda$. Then $v$ satisfies the
following
\begin{equation}                        \label{basic3-d}
\frac{\left |\nabla v\right |^2}{b^2-v^2} \leq\lambda,
\end{equation}
where $b>1$ is an arbitrary constant.
\end{lemma}

\begin{proof}\quad
Consider the function
\begin{equation}                \label{p-of-x-def}
P(x)=|\nabla v|^2+Av^2,
\end{equation}
where $A=\lambda (1 +\epsilon)$ for small $\epsilon>0$. Function
$P$ must achieve its maximum at some point $x_0\in M$. We claim
that
\begin{equation}\label{gra-p-eq-0}
\nabla P(x_0)=0.
\end{equation}
If $x_0\in M\backslash \partial M$, (\ref{gra-p-eq-0}) is
obviously true. Suppose that $x_0\in\partial M$. Take the same
local orthonormal frame $\{e_1, e_2,\cdots, e_{n}\}$ of $M$ about
$x_0$ as in the proof of Lemma \ref{nK-lemma}, where $e_n$ is the
unit outward normal vector field near $x_0\in\partial M$, $\{e_1,
e_2,\cdots, e_{n-1}\}|_{\partial M}$ is a local frame of $\partial
M$ about $x_0$ and $\nabla_{e_n}e_i=0$ for $i\leq n-1$. Since
$v|_{\partial M}=0$, we have $v_i(x_0)=0$ for $i\leq n-1$.
$P(x_0)$ is the maximum implies that
\begin{equation}\label{p-i-eq-0}
P_i(x_0)= 0\qquad \textrm{for }i\leq n-1
\end{equation}
and
\begin{equation}\label{p-n-geq-0}
P_n(x_0)\geq 0.
\end{equation}
Using argument in proving (\ref{gra-v-leq-0}) and the non-negativity
of the mean curvature $m$ of $\partial M$ with respect to the outward
normal, we get
\[
\nabla_{e_n}(|\nabla v|^2)(x_0)\leq 0.
\]
Noticing that $v|_{\partial M}=0$, we have
\begin{equation}\label{p-n-leq-0}
P_n(x_0)=\nabla_{e_n}(|\nabla v|^2)(x_0)+2Av(x_0)v_n(x_0) )\leq 0.
\end{equation}
Now (\ref{p-i-eq-0}), (\ref{p-n-geq-0}) and (\ref{p-n-leq-0})
imply that $P_n(x_0)=0$ and $\nabla P(x_0)=0$.

Thus (\ref{gra-p-eq-0}) holds, no matter $x_0\not\in\partial M$ or
$x_0\in\partial M$. By (\ref{gra-p-eq-0}) and the Maximum
Principle, we have
\begin{equation}\label{max-prin}
\nabla P(x_0)=0 \qquad \textrm{and}\qquad \Delta P(x_0)\leq 0.
\end{equation}

We are going to show further that $\nabla v(x_0)=0$. If on the
contrary, $\nabla v (x_0)\not=0$, then we rotate the local
orthonormal frame about $x_0$ such that
\[
|v_1(x_0)|=|\nabla v(x_0)|\not=0\qquad \textrm{and}\qquad
v_i(x_0)=0,\quad i\geq2.
\]
From (\ref{max-prin}) we have at $x_0$,
\[
0=\frac12\nabla_{i} P=\sum_{j=1}^{n}v_jv_{ji}+Avv_i,
\]
\begin{equation}                \label{d1}
v_{11}=-Av \qquad \textrm{and}\qquad v_{1i}=0\quad i\geq 2,
\end{equation}
and
\begin{eqnarray}
&0 &\geq\frac12 \Delta P(x_0) =\sum_{i, j=1}^{n}\left(v_{ji}v_{ji}+v_{j}v_{jii}+Av_{i}v_{i}+Avv_{ii}\right)\nonumber\\
&{} &=\sum_{i, j=1}^{n}\left(v_{ji}^2+v_j(v_{ii})_j +\textrm{R}_{ji}v_{j}v_{i}+Av_{ii}^2 +Av v_{ii}\right)\nonumber\\
&{} &=\sum_{i, j=1}^{n}v_{ji}^2+\nabla v\nabla(\Delta v) +\textrm{Ric}(\nabla v,\nabla v)+A|\nabla v|^2 +Av\Delta v\nonumber\\
&{} &\geq v_{11}^2+\nabla v\nabla(\Delta v) + (n-1)K|\nabla v|^2+A|\nabla v|^2 +Av\Delta v\nonumber\\
&{} &=(-Av)^2-\lambda |\nabla v|^2+ (n-1)K|\nabla v|^2+A|\nabla v|^2 -\lambda Av^2\nonumber\\
&{} &=(A-\lambda + (n-1)K)|\nabla v|^2+Av^2(A-\lambda),\nonumber
\end{eqnarray}
where we have used (\ref{d1}) and (\ref{ricci-bound}). Therefore
at $x_0$,
\begin{equation}
0\geq (A-\lambda)|\nabla v|^2+A(A-\lambda)v^2, \label{d2}
\end{equation}
that is,
\[
|\nabla v(x_0)|^2+ \lambda(1+\epsilon)v(x_0)^2\leq 0.
\]
Thus $\nabla v(x_0) = 0$. This contradicts $\nabla v(x_0)\not= 0$.

Therefore in any case, if $P$ achieves its maximum at a point
$x_0$, then $\nabla v(x_0)=0$. Thus at $x_0$
\[
P(x_0)=|\nabla v(x_0)|^2 +Av(x_0)^2=Av(x_0)^2\leq A.
\]
and at all $x\in M$
\[
|\nabla v(x)|^2+Av(x)^2=P(x)\leq P(x_0)\leq A.
\]
Letting $\epsilon \rightarrow 0$ in the above inequality, the
estimate (\ref{basic3-d}) follows.
\end{proof}

We want to improve the upper bound in (\ref{basic3-d}) further and
proceed in the following way.

Define a function $F$ by
\[
Z(t)=\max_{x\in M,t=\sin^{-1} \left(v\left(x\right)/b\right)}
\frac{\left |\nabla v\right |^2}{b^2-v^2}/\lambda.
\]
The estimate in (\ref{basic3-d}) becomes
\begin{equation}                                \label{basic5-d}
Z(t)\leq 1\qquad \textrm{on } [0, \sin^{-1}(1/b)]
\end{equation}
For convenience, in this paper we let
\begin{equation}\label{alpha-delta}
\alpha =
\frac12(n-1)K\qquad\textrm{and}\qquad\delta=\alpha/\lambda.
\end{equation}
By (\ref{nK-bound}) we have
\begin{equation}\label{delta-bound}
\delta\leq \frac{n-1}{2n}.
\end{equation}
We have the following conditions for the test function $Z$.

\begin{theorem}                                                     \label{thm-barrier-d}
If the function $z:\ [0,\,\sin^{-1} (1/b)]\mapsto \mathbf{R}^1$
satisfies the following
\begin{enumerate}
 \item $z(t)\geq Z(t) \qquad t\in [0,  \sin^{-1}(1/b)]$,
 \item there exists some $x_0\in M$
       such that at point $t_0=\sin^{-1} (v(x_0)/b)$ \linebreak
       $z(t_0)=Z(t_0)$,
 \item $z(t_0)>0$,
 \item $z$ extends to a smooth even function, and
 \item $z'(t_0)\sin t_0\geq 0$,
\end{enumerate}
then we have the following
\begin{equation}                            \label{barrier-eq-d}
0\leq\frac12z''(t_0)\cos^2t_0 -z'(t_0)\cos t_0\sin t_0 -z(t_0)+
1-2\delta \cos^2t_0.
\end{equation}
\end{theorem}

\begin{proof}\quad
Define
\[ J(x)=\left\{ \frac{\left |\nabla v\right |^2}{b^2-v^2}
-\lambda z \right\}\cos^2t,
\]
where $t=\sin^{-1}(v(x)/b)$. Then
\[ J(x)\leq 0\quad\textrm{for } x\in M
\qquad \textrm{and} \qquad J(x_0)=0.
\]
So $J(x_0)$ is the maximum of $J$ on $M$.
If $\nabla v(x_0)=0$, then
\[ 0=J(x_0)=-\lambda z\cos^2 t.
\]
This contradicts the Condition 3 in the theorem. Therefore
\[ \nabla v(x_0)\not=0.
\]

We claim that
\begin{equation}\label{gra-j-eq-0}
\nabla J(x_0)=0.
\end{equation}
If $x_0\in M\backslash \partial M$, (\ref{gra-j-eq-0}) is
obviously true. Suppose that $x_0\in\partial M$. Take the same
local orthonormal frame $\{e_1, e_2,\cdots, e_{n}\}$ of $M$ about
$x_0$ as in the proof of Lemma \ref{nK-lemma}, where $e_n$ is the
unit outward normal vector field near $x_0\in\partial M$, $\{e_1,
e_2,\cdots, e_{n-1}\}|_{\partial M}$ is a local frame of $\partial
M$ about $x_0$ and $\nabla_{e_n}e_i=0$ for $i\leq n-1$. Since
$v|_{\partial M}=0$, we have $v_i(x_0)=0$ for $i\leq n-1$.
$J(x_0)$ is the maximum implies that
\begin{equation}\label{j-i-eq-0}
J_i(x_0)= 0\qquad \textrm{for }i\leq n-1
\end{equation}
and
\begin{equation}\label{j-n-geq-0}
J_n(x_0)\geq 0.
\end{equation}
Using argument in proving (\ref{gra-v-leq-0}) and the non-negativity
of the mean curvature $m$ of $\partial M$ with respect to the outward
normal, we get
\[
\left(|\nabla v|^2\right)_n\Big|_{x_0}\leq 0.
\]
The Dirichlet condition $v(x_0)=0$ implies that
$t(x_0)=0$ and $z'(t(x_0))=z'(0)=0$, since by the Condition 4 in the theorem
$z$ extends to a smooth even function. Therefore
\begin{equation}\label{j-n-leq-0}
J_n(x_0)=\frac{1}{b^2}\left(|\nabla
v|^2\right)_n-\lambda\cos t[z' \cos t -2z\sin t]t_n\Big|_{x_0}
 =\frac{1}{b^2}\left(|\nabla v|^2\right)_n\Big|_{x_0}
 \leq 0.
\end{equation}

Now (\ref{j-i-eq-0}), (\ref{j-n-geq-0}) and (\ref{j-n-leq-0})
imply (\ref{gra-j-eq-0}).

Thus (\ref{gra-j-eq-0}) holds, no matter $x_0\not\in\partial M$ or
$x_0\in\partial M$. By (\ref{gra-j-eq-0}) and the Maximum
Principle, we have
\begin{equation}                                                \label{es1}
\nabla J(x_0)=0\qquad \textrm{and}\qquad \Delta J(x_0)\leq 0.
\end{equation}
$J(x)$ can be rewritten as
\[  J(x)=\frac{1}{b^2}|\nabla v|^2-\lambda z\cos^2t.
\]
Thus (\ref{es1}) is equivalent to
\begin{equation}                                              \label{es2}
\frac{2}{b^2}\sum_{i}v_iv_{ij}\Big|_{x_0}=\lambda\cos t[z' \cos t
-2z\sin t]t_j\Big|_{x_0}
\end{equation}
and
\begin{eqnarray}                                              \label{es3}
0&\geq&\frac{2}{b^2}\sum_{i,j}v_{ij}^2+\frac{2}{b^2}\sum_{i,j}v_iv_{ijj}
 -\lambda (z''|\nabla t|^2+z'\Delta t)\cos^2t \\
 & &+4\lambda z'\cos t\sin t |\nabla t|^2 -
\lambda z\Delta\cos^2t\Big|_{x_0}.\nonumber
\end{eqnarray}
Rotate the frame so that
\[
|v_1(x_0)|=|\nabla v(x_0)|\not=0\qquad \textrm{and}\qquad
v_i(x_0)=0,\quad i\geq2.
\]
Then (\ref{es2}) implies
\begin{equation}                                             \label{es4}
v_{11}\Big|_{x_0}=\frac{\lambda b}{2}(z'\cos t-2z\sin t)
\Big|_{x_0}\quad\text{and}\quad v_{1i} \Big|_{x_0}=0\ \text{for }
i\geq2.
\end{equation}
Now we have
\begin{eqnarray}
|\nabla v|^2
\Big|_{x_0}&=&\lambda b^2z\cos^2t\Big|_{x_0},\nonumber\\
 |\nabla t|^2
\Big|_{x_0}&=&\frac{|\nabla v|^2}{b^2-v^2}=\lambda z
\Big|_{x_0},\nonumber\\
\frac{\Delta v}{b}\Big|_{x_0} &=&\Delta \sin t =\cos t\Delta
t-\sin t |\nabla t|^2
\Big|_{x_0},\nonumber\\
\Delta t\Big|_{x_0}&=&\frac{1}{\cos t}(\sin t|\nabla
t|^2+\frac{\Delta v}{b})
\nonumber\\
 &=&\frac{1}{\cos t} [ \lambda z\sin t-\frac{\lambda}{b}v] \Big|_{x_0}, \quad\textrm{and}
\nonumber\\
\Delta\cos^2t\Big|_{x_0}&=&\Delta \left(1-\frac{v^2}{b^2}\right)
 =-\frac{2}{b^2}|\nabla v|^2-\frac{2}{b^2}v\Delta v
\nonumber\\
  &=&-2\lambda z\cos^2t+\frac{2}{b^2}\lambda v^2\Big|_{x_0}. \nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
& {}&
\frac{2}{b^2}\sum_{i,j}v_{ij}^2\Big|_{x_0}\geq\frac{2}{b^2}v_{11}^2
\nonumber\\
& {}& =\frac{\lambda ^2}{2}(z')^2\cos^2t-2\lambda ^2zz'\cos t\sin
t
      +2\lambda ^2z^2\sin^2 t\Big|_{x_0}\nonumber,
\end{eqnarray}
\begin{eqnarray}
\frac{2}{b^2}\sum_{i,j}v_iv_{ijj}\Big|_{x_0}
&=&\frac{2}{b^2}\left(\nabla v\,\nabla
       (\Delta v)+\textrm{Ric}(\nabla v,\nabla v)\right)\nonumber\\
&\geq& \frac{2}{b^2}(\nabla v\,\nabla (\Delta v)+(n-1)K|\nabla v|^2)\nonumber\\
 &=&-2\lambda^2z\cos^2t+4\alpha \lambda z\cos^2t\Big|_{x_0},\nonumber
\end{eqnarray}
\begin{eqnarray}
&{}&  -\lambda (z''|\nabla t|^2+ z'\Delta t)\cos^2t\Big|_{x_0}\nonumber\\
&{}&=-\lambda^2 zz''\cos^2t-
\lambda^2zz'\cos t\sin t\nonumber\\
&{}&{ }+\frac{1}{b}\lambda^2z'v\cos t \Big|_{x_0},\nonumber
\end{eqnarray}
and
\begin{eqnarray}
&{}&4\lambda z'\cos t\sin t|\nabla t|^2-\lambda z\Delta
\cos^2t\Big|_{x_0}
\nonumber\\
&{}&=4\lambda^2zz'\cos t\sin
t+2\lambda^2z^2\cos^2t-\frac{2}{b}\lambda^2zv\sin t
\Big|_{x_0}.\nonumber
\end{eqnarray}
Putting these results into (\ref{es3}) we get
\begin{eqnarray}                                                   \label{es5}
0&\geq&-\lambda^2zz''\cos^2t+ \frac{\lambda^2}{2}(z')^2\cos^2t
+\lambda^2z'\cos t\left(z\sin t +\sin t\right)
  \nonumber\\
 & & {}+2\lambda^2z^2-2\lambda^2z +4\alpha \lambda z\cos^2t
 \Big|_{x_0},
\end{eqnarray}
where we used (\ref{es4}). Now
\begin{equation}                                                    \label{es6}
z(t_0)>0,
\end{equation}
by the Condition 3 in the theorem. Dividing two sides of (\ref{es5})
by $2\lambda^2z\Big|_{x_0}$, we have
\begin{eqnarray}
0&\geq&-\frac12z''(t_0)\cos^2t_0 +\frac12z'(t_0)\cos t_0\left(\sin t_0+\frac{\sin t_0}{z(t_0)}\right) +z(t_0) \nonumber\\
 & & {}  -1 +2\delta \cos^2t_0+\frac{1}{4z(t_0)}(z'(t_0))^2\cos^2t_0.\nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray}
0&\geq&-\frac12z''(t_0)\cos^2t_0 + z'(t_0)\cos t_0\sin t_0+z(t_0) -1 +2\delta \cos^2t_0\nonumber\\
 & & {}+\frac{1}{4z(t_0)}(z'(t_0))^2\cos^2t_0+\frac12z'(t_0)\sin t_0\cos
 t_0[\frac{1}{z(t_0)}-1].                               \label{es6.1}
\end{eqnarray}
Conditions 1, 2 and 5 in the theorem imply that
$0<z(t_0)=Z(t_0)\leq 1$ and $z'(t_0)\sin t_0\geq 0$. Therefore the last
two terms in (\ref{es6.1}) are nonnegative and
(\ref{barrier-eq-d}) follows.
\end{proof}

\section{Proof of Main Result}\label{sec-proof}

\begin{proof}[Proof of Theorem \ref{main-thm}]\quad
Let
\begin{equation}                                \label{z-def}
z(t)=1+\delta\xi(t),
\end{equation}
where $\xi$ is the functions defined by (\ref{xi-def}) in Lemma
\ref{xi-lemma}. We claim that
\begin{equation}                        \label{4.1-d}
Z(t)\leq z(t)\qquad \textrm{for }t\in  [0, \sin^{-1}(1/b)].
\end{equation}
Lemma \ref{xi-lemma} implies that for $t\in [0, \sin^{-1}(1/b)] $,
we have the following
\begin{eqnarray}
& &{}\frac{1}{2}z''\cos ^2t-z'\cos t\sin t-z
    =-1+ 2\delta\cos^2t,               \label{z-eq}\\
& &{}z'(t)\sin t\geq 0,\qquad \label{z'-geq0}\\
& &{}z\textrm{ is a smooth even function},\\
& &{}0<1-(\frac{\pi^2}{4}-1)\frac{n-1}{2n}\leq
1-(\frac{\pi^2}{4}-1)\delta=z(0)\leq z(t),\quad\textrm{and}\label{z-min}\\
& &{}z(t) \leq z(\frac{\pi}{2})=1. \label{z-max}
\end{eqnarray}

Let $P\in\mathbf{R}^1$ and $t_0\in [0,\sin^{-1}(1/b)]$ such that
\[ P=\max_{t\in [0,\sin^{-1}(1/b)]}\left(Z(t)-z(t)\right)=Z(t_0)-z(t_0).
\]
Thus
\begin{equation}\label{4.2}
Z(t)\leq z(t)+P\quad \textrm{for }t\in
[0,\sin^{-1}(1/b)]\qquad\textrm{and}\qquad Z(t_0)=z(t_0)+P.
\end{equation}
Suppose that $P>0$. Then $z+P$ satisfies the conditions in
Theorem \ref{thm-barrier-d} and therefore satisfies (\ref{barrier-eq-d}). So we have
\begin{eqnarray}
&{}&z(t_0)+P=Z(t_0)\nonumber\\
&{}&\leq  \frac12(z+P)''(t_0)\cos^2 t_0-(z+P)'(t_0)\cos t_0
 \sin t_0+1-2\delta \cos^2 t_0\nonumber\\
&{}&=\frac12z''(t_0)\cos^2t_0-z'(t_0)\cos t_0\sin
t_0+1-2\delta \cos^2 t_0\nonumber\\
&{}&=z(t_0).\nonumber
\end{eqnarray}
This contradicts the assumption $P>0$. Thus $P\leq 0$ and
(\ref{4.1-d}) must hold. That means
\begin{equation}\label{4.3-d}
\sqrt{\lambda}\geq \frac{|\nabla t|}{\sqrt{z(t)}}.
\end{equation}

Take $q_1$ on $M$ such that $v(q_1)=1 =\sup_M v$ and and $q_2\in
\partial M$ such that distance $d(q_1, q_2) = \textrm{ distance }d(q_1, \partial M)$.
Let $L$ be the minimum geodesic segment between $q_1$ and $q_2$.
We integrate both sides of (\ref{4.3-d}) along $L$ and change
variable and let $b\rightarrow 1$. Let
$\tilde{d}$ be the diameter of the largest interior ball in $M$.
Then
\begin{equation}\label{4.4-d}
\sqrt{\lambda}\,\frac{\tilde{d}}{2}\geq\int_{L}\,\frac{|\nabla
t|}{\sqrt{z(t)}}\,dl =\int_0^{\frac{\pi}{2}}
\frac{1}{\sqrt{z(t)}}\,dt \geq \frac{\left(\int_0^{\pi/2}\
\,dt\right)^\frac32}{(\int_0^{\pi/2}\ z(t)\,dt)^{\frac12}} \geq
\left( \frac{(\frac{\pi}{2})^3}{\int_0^{\pi/2}\  z(t)\,dt}
\right)^{\frac12}.
\end{equation}
Square the two sides. Then
\[
\lambda \geq \frac{\pi^3}{2(\tilde{d})^2\int_0^{\pi/2} \
z(t)\,dt}.
\]
Now
\[
\int_0^{\frac{\pi}{2}}\ z(t)\,dt=\int_0^{\frac{\pi}{2}}\ [1+
\delta \xi(t)]\,dt=\frac{\pi}{2}(1-\delta),
\]
by (\ref{xi-int}) in Lemma \ref{xi-lemma}. That is,
\[
\lambda \geq
\frac{\pi^2}{(1-\delta)(\tilde{d})^2}\quad\textrm{and}\quad
\lambda \geq \frac12(n-1)K + \frac{\pi^2}{(\tilde{d})^2}.
\]
\end{proof}

We now present a lemma that is used in the proof of Theorem
\ref{main-thm}.

\begin{lemma}                           \label{xi-lemma}
Let
\begin{equation}                        \label{xi-def}
\xi(t)=\frac{\cos^2t+2t\sin t\cos t +t^2-\frac{\pi^2}{4}}{\cos^2t}
\qquad \textrm{on}\quad [-\frac{\pi}{2},\frac{\pi}{2}\,].
\end{equation}
Then the function $\xi$  satisfies the following
\begin{eqnarray}
& &{}\frac{1}{2}\xi''\cos ^2t-\xi'\cos t\sin t-\xi
    =2\cos^2t\quad \textrm{in }(-\frac{\pi}{2},\frac{\pi}{2}\,),          \label{xi-eq}\\
& &{}\xi'\cos t -2\xi\sin t =4t\cos t                      \label{xi-eq2}\\
& &{}\int_0^{\frac{\pi}{2}}\xi(t)\, dt= -\frac{\pi}{2}          \label{xi-int}\\
& &{}1-\frac{\pi^2}{4}=\xi(0)\leq\xi(t)\leq\xi(\pm
\frac{\pi}{2})=0\quad
\textrm{on }[-\frac{\pi}{2},\frac{\pi}{2}\,],                                \nonumber\\
& &{} \xi' \textrm{ is increasing on }
[-\frac{\pi}{2},\frac{\pi}{2}\,] \textrm{ and }
\xi'(\pm \frac{\pi}{2}) =\pm \frac{2\pi}{3},                     \nonumber\\
& &{}\xi'(t)< 0 \textrm{ on }(-\frac{\pi}{2},0)\textrm{ and \ }
\xi'(t)>0 \textrm{ on }(0,\frac{\pi}{2}\,), \nonumber \\
& &{}\xi''(\pm\frac{\pi}{2})=2, \ \xi''(0)=2(3-\frac{\pi^2}{4})
\textrm{ and \ } \xi''(t)> 0 \textrm{ on }
[-\frac{\pi}{2},\frac{\pi}{2}\,], \nonumber\\
& &{}(\frac{\xi'(t)}{t})'>0 \textrm{ on } (0,\pi/2\,)\textrm{ and
\ } 2(3-\frac{\pi^2}{4})\leq \frac{\xi'(t)}{t}\leq \frac43
\textrm{ on } [-\frac{\pi}{2},\frac{\pi}{2}\,],\nonumber \\
& &{}\xi'''(\frac{\pi}{2})=\frac{8\pi}{15}, \xi'''(t)< 0 \textrm{
on }(-\frac{\pi}{2},0) \textrm{ and \ }  \xi'''(t)>0 \textrm{ on
}(0,\frac{\pi}{2}\,). \nonumber
\end{eqnarray}
\end{lemma}
\begin{proof}\quad
For convenience, let $q(t)= \xi'(t)$, i.e.,
\begin{equation}                                       \label{q-def}
q(t) = \xi'(t) = \frac{2(2t\cos t +t^2\sin t +\cos^2 t \sin t
-\frac{\pi^2}{4}\sin t)}{\cos^3 t}.
\end{equation}
Equation (\ref{xi-eq}) and the values $\xi(\pm \frac{\pi}{2})=0$,
$\xi(0)=1-\frac{\pi^2}{4}$ and $\xi'(\pm \frac{\pi}{2}) =\pm
\frac{2\pi}{3}$ can be verified directly from (\ref{xi-def}) and
(\ref{q-def}) .  The values of $\xi''$ at $0$ and $\pm
\frac{\pi}{2}$ can be computed via (\ref{xi-eq}). By
(\ref{xi-eq2}), $(\xi(t)\cos^2 t)' =4t\cos^2 t$. Therefore
\newline $\xi(t)\cos^2 t=\int_{\frac{\pi}{2}}^t \ 4s\cos^2 s\,ds$,
and
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\
\xi(t)\,dt=2\int_0^{\frac{\pi}{2}}\
\xi(t)\,dt=-8\int_0^{\frac{\pi}{2}}\left( \frac{1}{\cos^2(t)}
\int_t^{\frac{\pi}{2}}\ s\cos^2s\,ds\right)\,dt
\]
\[
=-8\int_0^{\frac{\pi}{2}}\left(\int_0^s\
\frac{1}{\cos^2(t)}\,dt\right)\ s\cos^2s\,ds
=-8\int_0^{\frac{\pi}{2}}\ s\cos s\sin s\,ds=-\pi.
\]
It is easy to see that $q$ and $q'$ satisfy the following
equations
\begin{equation}                                         \label{q-eq}
\frac12 q''\cos t -2q'\sin t -2q\cos t = -4 \sin t,
\end{equation}
and
\begin{equation}                                       \label{q'-eq}
\frac{\cos^2 t}{2(1+\cos^2 t)}(q')''-\frac{2\cos t\sin t}{1+\cos^2
t}(q')'-2(q')=-\frac{4}{1+\cos^2 t}.
\end{equation}
The last equation implies $q'=\xi''$ cannot achieve its
non-positive local minimum at a point in $(-\frac{\pi}{2},
\frac{\pi}{2})$. On the other hand, $\xi''(\pm\frac{\pi}{2})=2$,
by equation (\ref{xi-eq}), $\xi(\pm \frac{\pi}{2})=0$ and
$\xi'(\pm \frac{\pi}{2})=\pm \frac{2\pi}{3}$. Therefore
$\xi''(t)>0$ on $[-\frac{\pi}{2},\frac{\pi}{2}]$ and $\xi'$ is
increasing. Since $\xi'(t)=0$, we have $\xi'(t)< 0$ on
$(-\frac{\pi}{2},0)$ and $\xi'(t)>0$ on $(0,\frac{\pi}{2})$.

Similarly, from the equation
\begin{eqnarray}                                           \label{q''-eq}
&\frac{\cos^2 t}{2(1+\cos^2 t)}(q'')'' -\frac{\cos t\sin t
(3+2\cos^2 t)}{(1+\cos^2 t)^2}(q'')' -\frac{2(5\cos^2 t+\cos^4 t)}{(1+\cos^2 t)^2}(q'') \nonumber\\
&=-\frac{8\cos t\sin t}{(1+\cos^2 t)^2}
\end{eqnarray}
we get the results in the last line of the lemma.

Set $h(t)=\xi''(t)t-\xi'(t)$. Then $h(0)=0$  and $h'(t)=
\xi'''(t)t>0$ in $(0,\frac{\pi}{2})$. Therefore
$(\frac{\xi'(t)}{t})'=\frac{h(t)}{t^2}>0$ in $(0,\frac{\pi}{2})$.
Note that $\frac{\xi'(-t)}{-t}= \frac{\xi'(t)}{t}$,
$\frac{\xi'(t)}{t}|_{t=0}=\xi''(0)=2(3-\frac{\pi^2}{4})$ and
$\frac{\xi'(t)}{t}|_{t=\pi/2}=\frac43$. This completes the proof
of the lemma.
\end{proof}


TITLE: Is $G_i$ normal in $G_k$?
QUESTION [1 upvotes]: A group G is called solvable if it has a subnormal series whose factor groups (quotient groups) are all abelian, that is, if there are subgroups {1} = $G_0 < G_1 < ⋅⋅⋅ < G_k = G$ such that $G_{j − 1}$ is normal in $G_j$, and $G_j/G_{j − 1}$ is an abelian group, for $j = 1, 2, …, k$.
Question : Is $G_i$ normal in $G_k$ for $i \ge 2$?

REPLY [2 votes]: Not necessarily. Consider $$\{e\}\trianglelefteq\{e,(12)(34)\}\trianglelefteq \{e,(12)(34),(13)(24),(14)(23)\}\trianglelefteq A_4$$ $G_1$ is not normal in $A_4$. You can extend this sequence on the left by taking the direct product with an appropriate abelian group $H$: namely, $$\{0\}\times G_0\trianglelefteq 3\Bbb Z/6\Bbb Z\times G_0\trianglelefteq \Bbb Z/6\Bbb Z\times G_0\trianglelefteq\Bbb Z/6\Bbb Z\times G_1 \trianglelefteq\ \Bbb Z/6\Bbb Z\times G_2\trianglelefteq \Bbb Z/6\Bbb Z\times A_4$$ In the new sequence, $\widetilde G_3\not\trianglelefteq \widetilde G_5$.


TITLE: Differences between "cold" and "collisionless" plasma
QUESTION [4 upvotes]: I'm just stepping for the first time into plasma physics and I'm studying on Chen's "Introduction to plasma physics" and on the notes of professor Richard Fitzpatrick.
I just seem to not have clear the difference between the definitions of "cold plasma approximation" and "collisionless plasma" for what regards the two-fluid plasma modeling.
I used to think of cold plasma as basically low-enthalpy plasma, and see that as a consequence of collisionlessness. But professor Fitzpatrick derives fluid equations for a cold collisional plasma, so I must have missed the point.
Searching on google is just making things more complicated, since I'm reading all different points of view on the subject.
Could someone help me with a clarification on what exactly makes a plasma fall in the "cold approximation", and what's the difference between the fluid equations of a cold plasma and a collisionless plasma?

REPLY [0 votes]: My reference is J.A. Bittencourt's book Fundamentals of Plasma Physics.
Cold plasma
One understands under a cold plasma a particular choice of closure for the infinite series of moments of the Boltzmann equation.
In the momentum equation appears the pressure dyad (energy density). In a cold plasma this term is taken to be zero (no limit implied) and hence the series is truncated. As a result such a plasma fluid model only consists of mass and momentum equations.
This means however, that there is no thermal motion of particles (e.g. particle flux due to temperature gradient) or forces due to the divergence of the kinetic pressure tensor. But there are still collisions for momentum transfer.
In fact it follows that the plasma temperature is zero and the velocity distribution function is a Dirac $\delta$-function centered at the macroscopic
flow velocity $u(r,t)$ $$f_\alpha(r,v,t)=\delta(v-u(r,t))$$
Warm plasma
As a side note the next easier closure is the heat flux vector in the energy density equation to be taken zero. This is also called the adiabatic approximation.
Collisionless plasma
Now, the term collisionless plasma is studied in relation to propagation of electromagnetic waves in the plasma. In this context the collision frequency $\nu$ appears in the dielectric constant of the medium. Here, collisionless usually means that the collision frequency is much less than the wave frequency $\omega$. $$\nu \ll\omega$$. If this holds, the resulting dispersion relations can be significantly simplified.
However, to properly study the propagation of electromagnetic waves in the plasma, one has to have a model for the medium first, meaning one has to pick a closure relation for the momemt equations.
E.g., as another side note, the theory for wave propagation in a cold plasma (as defined above) is known as the magnetoionic theory.
As a result one has dispersion relations for a cold plasma with and without collisions as well as for a warm plasma and so on.


text:
\subsection{Sensing vertical (or horizontal) lines in MRI}
\label{sec:application_MRI}

We extend the analysis of Section~\ref{sec:appli} from the one-dimensional to the two-dimensional framework, leading a more realistic model for the MRI problem.

Let $n = 2^{2(J+1)}$ for some $J \in \Nbb$ and denote the one-dimensional Fourier-Haar transform introduced in Section~\ref{sec:applications_mri_iso} as $\phi = {\mathcal{F}}H^* \in \Cbb^{\sqrt{n} \times \sqrt{n}}$. Then, we define the two-dimensional Fourier-Haar transform as
$$
A_0 = \phi \otimes \phi \in\Cbb^{n \times n},
$$
where $\otimes$ denotes the Kronecker product. The splitting of the wavelet multi-index space $\{1,\ldots,\sqrt{n}\}^2$ and of the Fourier frequency space $\{-\sqrt{n}/2 + 1, \ldots, \sqrt{n}/2\}^2$ into subbands is carried out by tensorizing the one-dimensional splitting presented in Section~\ref{sec:applications_mri_iso}. Namely, we consider subbands defined, for $j,j' = 0,\ldots,J$, as
$$
\Omega_{j,j'}:= \Omega_{j} \times \Omega_{j'}
\quad \text{and} \quad
W_{j,j'}:= W_j \times W_{j'}.
$$
Moreover, we will represent the image wavelet coefficients as a vector $x \in \Cbb^n$ or, equivalently, as a matrix $X = \Cbb^{\sqrt{n} \times \sqrt{n}}$, such that $x = \vect(X)$, where $\vect: \Cbb^{\sqrt{n} \times \sqrt{n}} \to \Cbb^n$ is the vectorization operator that stacks the columns of a matrix atop, i.e., $\vect(x_1|\cdots|x_{\sqrt{n}}) = (x_1^*,\ldots,x_{\sqrt{n}}^*)^*$. Within this multi-level framework, we consider the following generalization of sparsity in levels.
\begin{definition}
\label{def:sparsity_in_levels_2D}
Let $x \in \Cbb^n$ and $X \in \Cbb^{\sqrt{n} \times \sqrt{n}}$ be such that $x = \vect(X)$. Then, given $S = \text{supp}(X)$, we define the following sparsities in levels for the vector $x$:
\begin{align}
\label{eq:scol}
s^{r}_{j} &:= \max_{0\leq \ell \leq J} \max_{ k \in {\Omega}_{\ell}  } \left| S \cap \Omega_{\ell,j} \cap R_k\right| ,

\end{align}
where $R_k {:=\{k\}\times \{1,\ldots,\sqrt{n}\}}$ represents the set corresponding to the $k$-th horizontal line or row of the wavelet multi-index space $\{1,\ldots,\sqrt{n}\}^2$  (see Figure \ref{fig:illusEns}).

\end{definition}

 \begin{figure}[t]
  \begin{center}

\begin{tikzpicture}

\draw (0,0) rectangle (1,1) node[midway,color=Cerulean] {\large$\Omega_{0,0}$};
\draw (1,0) rectangle (2,1) node[midway,color=Cerulean] {\large$\Omega_{0,1}$};
\draw (2,0) rectangle (4,1) node[midway,color=Cerulean] {\large ...};
\draw (4,0) rectangle (8,1) node[midway,color=Cerulean] {\large$\Omega_{0,J}$};

\draw (0,0) rectangle (1,-1) node[midway,color=Cerulean] {\large $\Omega_{1,0}$};
\draw (0,-1) rectangle (1,-3) node[midway,color=Cerulean] {\large $\vdots$};
\draw (0,-3) rectangle (1,-7) node[midway,color=Cerulean] {\large $\Omega_{J,0}$};

\draw (1,0) rectangle (2,-1) ;
\draw (2,0) rectangle (4,-1) node[midway] {};
\draw (4,0) rectangle (8,-1) ;

\draw (1,-1) rectangle (2,-3) ;
\draw (1,-3) rectangle (2,-7) ;

\draw (2,-1) rectangle (4,-3) node[midway,color=Cerulean] {\large $\ddots$};
\draw (2,-3) rectangle (4,-7) ;

\draw (4,-1) rectangle (8,-3) ;

\draw (4,-3) rectangle (8,-7) node[midway,color=Cerulean] {\large $\Omega_{J,J}$};

\draw (8.2,-3.7) node[right, color=Emerald]{$R_k$} ;
\fill [Emerald,opacity=0.6] (0,-3.5) rectangle (8,-3.8);

\draw[->,>=latex, color=Violet] (8.2,-4.5) to[bend left] (6.5,-3.8);
\draw (8.2,-4.5) node[right, color=Violet]{$R_k\cap \Omega_{J,J}$} ;

\fill [Violet,opacity=0.5] (4,-3.5) rectangle (8,-3.8);

\draw (0,2) -- (8,2);
\draw (0,2) node {$\bullet$} ;
\draw (1,2) node {$\bullet$} ;
\draw (2,2) node {$\bullet$} ;
\draw (4,2) node {$\bullet$} ;
\draw (8,2) node {$\bullet$} ;

\draw (0,2.2) node[above] {$1$} ;
\draw (8,2.2) node[above] {$\sqrt{n}$} ;

\draw (-1,-3.65) node {$\blacksquare$} ;
\draw (-1.2,-3.65) node[left] {$k$} ;

\draw[decoration={brace,raise=5pt},decorate]
  (0,2.7) -- node[above=6pt] {$\Omega_0$} (1,2.7);
\draw[decoration={brace,raise=5pt},decorate]
  (1,2.7) -- node[above=6pt] {$\Omega_1$} (2,2.7);
 \draw[decoration={brace,raise=5pt},decorate]
  (4,2.7) -- node[above=6pt] {$\Omega_J$} (8,2.7);

\draw (-1,1) -- (-1,-7);
\draw (-1,1) node {$\bullet$} ;
\draw (-1,0) node {$\bullet$} ;
\draw (-1,-1) node {$\bullet$} ;
\draw (-1,-3) node {$\bullet$} ;
\draw (-1,-7) node {$\bullet$} ;

\draw (-1.2,1) node[left] {$1$} ;
\draw (-1.2,-7) node[left] {$\sqrt{n}$} ;

\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,1) -- node[left=6pt] {$\Omega_0$} (-1.7,0);
\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,0) -- node[left=6pt] {$\Omega_1$} (-1.7,-1);
\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,-3) -- node[left=6pt] {$\Omega_J$} (-1.7,-7);

\end{tikzpicture}

 \caption{\label{fig:illusEns}2D view of the signal $x\in \Cbb^n$ to reconstruct. The vector $x$ can be reshaped into a $\sqrt{n}\times \sqrt{n}$ matrix. The \replace{indexes}{indices} can be partitioned according to the sets $\Omega_{j,j'} = \Omega_j \times \Omega_{j'} $ for $0\leq j,j'\leq J$. ${R}_k$ represents the coefficient \replace{indexes}{indices} corresponding to the $k$-th horizontal row. }
  \end{center}
 \end{figure}

The sparsities in levels $(s_j^r)_{0\leq j\leq J}$ are intrinsically anisotropic. According to this definition, the wavelet coefficients restricted to the vertical subband $\{1,\ldots,\sqrt{n}\}\times\Omega_j$ and to a generic row can be at most $s_j^r$-sparse.

 \begin{figure}[t]
  \begin{center}

\begin{tikzpicture}

\draw (0,0) rectangle (1,-1);
\draw (1,0) rectangle (2,-1);
\draw (2,0) rectangle (4,-1);
\draw (4,0) rectangle (8,-1);

\draw (0,-1) rectangle (1,-2);
\draw (1,-1) rectangle (2,-2);
\draw (2,-1) rectangle (4,-2);
\draw (4,-1) rectangle (8,-2);

\draw (0,-2) rectangle (1,-4);
\draw (1,-2) rectangle (2,-4);
\draw (2,-2) rectangle (4,-4);
\draw (4,-2) rectangle (8,-4);

\draw (0,-4) rectangle (1,-8);
\draw (1,-4) rectangle (2,-8);
\draw (2,-4) rectangle (4,-8);
\draw (4,-4) rectangle (8,-8);

\draw (0,1) -- (8,1);
\draw (0,1) node {$\bullet$} ;
\draw (1,1) node {$\bullet$} ;
\draw (2,1) node {$\bullet$} ;
\draw (4,1) node {$\bullet$} ;
\draw (8,1) node {$\bullet$} ;

\draw (0,1.2) node[above] {$1$} ;
\draw (8,1.2) node[above] {$\sqrt{n}$} ;

\draw[decoration={brace,raise=5pt},decorate]
  (0,1.7) -- node[above=6pt] {$\Omega_0$} (1,1.7);
\draw[decoration={brace,raise=5pt},decorate]
  (1,1.7) -- node[above=6pt] {$\Omega_1$} (2,1.7);
 \draw[decoration={brace,raise=5pt},decorate]
  (4,1.7) -- node[above=6pt] {$\Omega_J$} (8,1.7);

\draw (-1,0) -- (-1,-8);
\draw (-1,0) node {$\bullet$} ;
\draw (-1,-1) node {$\bullet$} ;
\draw (-1,-2) node {$\bullet$} ;
\draw (-1,-4) node {$\bullet$} ;
\draw (-1,-8) node {$\bullet$} ;

\draw (-1.2,0) node[left] {$1$} ;
\draw (-1.2,-8) node[left] {$\sqrt{n}$} ;

\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,0) -- node[left=6pt] {$\Omega_0$} (-1.7,-1);
\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,-1) -- node[left=6pt] {$\Omega_1$} (-1.7,-2);
\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-1.7,-4) -- node[left=6pt] {$\Omega_J$} (-1.7,-8);

\draw (0.25,-0.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-0.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.25,-0.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-0.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (0.25,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (0.75,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (0.25,-0.75) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (0.75,-0.75) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;

\draw (1.25,-0.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.75,-0.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.25,-0.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (1.25,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (1.75,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;

\draw (2.25,-0.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (2.75,-0.25) node[color=Gray] {\large$\blacksquare$} ;
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\draw (2.25,-0.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (2.25,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (2.75,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (3.75,-0.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;

\draw (0.25,-1.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-1.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-1.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (0.25,-1.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (0.75,-1.25) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;

\draw (1.25,-1.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.75,-1.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (2.25,-1.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.25,-1.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (2.75,-1.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.75,-1.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.25,-1.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.75,-1.75) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (3.25,-1.75) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;
\draw (2.75,-1.75) node[color=RubineRed,opacity=0.2] {\large$\blacksquare$} ;

\draw (0.25,-2.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-2.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.25,-3.25) node[color=Gray] {\large$\blacksquare$} ;

\draw (0.25,-4.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.75,-6.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (0.25,-7.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (1.75,-2.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.25,-2.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.75,-3.75) node[color=Gray] {\large$\blacksquare$} ;

\draw (1.25,-4.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.75,-5.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (1.25,-6.25) node[color=Gray] {\large$\blacksquare$} ;

\draw (2.25,-2.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.75,-3.25) node[color=Gray] {\large$\blacksquare$} ;
\draw (2.75,-3.25) node[color=Gray] {\large$\blacksquare$} ;

\draw (2.75,-4.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.75,-5.75) node[color=Gray] {\large$\blacksquare$} ;
\draw (3.25,-7.25) node[color=Gray] {\large$\blacksquare$} ;

\draw (6,-8.7) node[color=RubineRed] {$\dots$};
\draw (0.5,-8.9) node[rotate=45,color=RubineRed]{$s_0^r=2$};
\draw (1.5,-8.9) node[rotate=45,color=RubineRed]{$s_1^r=2$};
\draw (3,-8.9) node[rotate=45,color=RubineRed]{$s_2^r=3$};

\end{tikzpicture}
 \caption{\label{fig:struct_sparsity_ex}Example of a signal $x\in \Cbb^n$ with support denoted by squares in grey for the first vertical subbands. The quantities $(s_j^r)_j$  are the maximal degrees of sparsity restricted to rows and respectively to vertical subbands $\cup_{j'} \Omega_{j',j}$. In the Figure, they are reached for support locations highlighted in red.
 }
  \end{center}
 \end{figure}

{Equipped with this notion of structured sparsity, we consider measurements corresponding to vertical lines in the frequency space. This sensing strategy models MRI acquisition in a more realistic way than isolated Fourier measurements and corresponds to the setting \ref{block_finite_setting} in Section~\ref{subsec:sampling}. Specifically, we partition the set of $n$ rows of the matrix $A_0$ into $\sqrt{n}$ blocks of size $\sqrt{n} \times n$ defined for $k = 1,\ldots,\sqrt{n}$ as
$$
D_{k} := \phi_{k,:} \otimes \phi = \begin{pmatrix} \phi_{k,1} \phi | \cdots | \phi_{k,\sqrt{n}} \phi\end{pmatrix} \in \mathbb{C}^{\sqrt{n} \times n},
$$
where $\phi_{k,:}$ denotes the $k$-th row of $\phi$. Drawing the block $D_k$ corresponds to sampling along the vertical line $\{1,\ldots,\sqrt{n}\}\times\{k\}$ in the frequency space.

In this setting, the analysis based on the quantity $\Theta(S,\pi)$ leads to the following result.}

\begin{corollary}
\label{corol:linesHaarMRI}
Let $A_0\in \Cbb^{n \times n}$ be the two-dimensional Fourier-Haar transform and consider the splitting of the Haar and of the Fourier spaces into subbands as defined above.
Fix $\varepsilon \in (0,1)$.
Suppose that $x \in \Cbb^{n}$ is an $S$-sparse vector with random signs and associated structured sparsities {in levels} $(s_j^{r})_{0\leq j \leq J}$.
Then, {the same} recovery guarantees of Theorem \ref{thm:noisy} hold with probability at least $1-\varepsilon$ by

drawing vertical lines \replace{with probability}{according to the probability distribution} $\left( \pi_k \right)_{1\leq k \leq \sqrt{n}}$ defined as
$$ {\pi}_k = \frac{2^{-j(k)} \sum_{j=0}^J 2^{-|j(k)-j|/2} s^{r}_{j}}{
\sum_{\ell=1}^{\sqrt{n}}
2^{-j(\ell)} \sum_{j=0}^J
2^{-|j(\ell) -j|/2} s^{r}_{j}  },
$$
with a number $m$ of drawn vertical lines of the order
\begin{equation}
\label{eq:horLinesMRI}
m \gtrsim  \sum_{j=0}^J \left(  s^{r}_{j}  + \sum_{j'=0 \atop j'\neq j}^J
2^{-|j -j'|/2} s^{r}_{j'}  \right) \cdot\ln^2 \left( \frac{6 n }{\varepsilon}\right).
\end{equation}
{In particular, the probability \replace{}{distribution} $\left( \pi_k \right)_{1\leq k \leq \sqrt{n}}$ is constant on each frequency subband $W_j$.}

\end{corollary}

The proof of this results is given in Appendix~\ref{proof:corollinesHaarMRI_lambda}.
Corollary \ref{corol:linesHaarMRI} meets the results of \cite[Corollary 4.10]{boyer2017compressed}: the bounds on the number of blocks of measurements are of the same order. Nevertheless, note that Corollary \ref{corol:linesHaarMRI} requires an additional assumption on the sign randomness. Moreover, recall that Theorem~\ref{thm:noisy} also ensures stable and robust recovery.

We can improve this result by analyzing the quantity $\Lambda(S,\pi)$.

\begin{corollary}
\label{corol:linesHaarMRI_lambda}
Let $A_0 \in \Cbb^{n \times n}$ be the two-dimensional Fourier-Haar transform and consider the splitting of the Haar and of the Fourier spaces into subbands as defined above. Fix $\varepsilon \in (0,1)$.
Suppose that $x \in \Cbb^{n}$ is an $S$-sparse vector with random signs and associated structured sparsities $(s_j^{r})_{0\leq j \leq J}$.
 If
$$
\min_{0\leq j \leq J}    \left(  s_{j}^{r} +  \sum_{j' = 0 \atop j'\neq j}^J   2^{-|j-j'|} s_{j'}^{r} \right) \cdot \log_2(\sqrt{n})  \gtrsim \ln(3 n/\varepsilon),
$$
then, the same recovery guarantees of Theorem \ref{thm:killthetalog} hold with probability at least $1-\varepsilon$ by

drawing vertical lines \replace{with probability}{according to the probability distribution} $\left( \pi_k \right)_{1\leq k \leq \sqrt{n}}$ defined as
$$ {\pi}_k
=
{\frac{2^{-j(k)} \sum_{j = 0}^{J} 2^{-|j-j(k)|} s_{j}^r}{\sum_{\ell = 1}^{\sqrt{n}} 2^{-j(\ell)} \sum_{j = 0}^{J} 2^{-|j(\ell)-j|} s_{j}^r}},
$$
with a number $m$ of drawn vertical lines of the order
\begin{equation}
\label{eq:horLinesMRI_lambda}
m \gtrsim  \left( \sum_{j=0}^J  s_{j}^{r}  +\sum_{j'=0 \atop j'\neq j}^J   2^{-|j-j'|} s_{j'}^{r} \right)\cdot \log_2(\sqrt{n})  \cdot \ln \left( \frac{6 n }{\varepsilon}\right).
\end{equation}
In particular, the probability \replace{}{distribution} $\left( \pi_k \right)_{1\leq k \leq \sqrt{n}}$ is constant on each frequency subband $W_j$.
\end{corollary}

The proof of Corollary~\ref{corol:linesHaarMRI_lambda} is given in Appendix~\ref{proof:corollinesHaarMRI_lambda}.

Note that the bound \eqref{eq:horLinesMRI_lambda} improves Corollary~\ref{corol:linesHaarMRI} by  attenuating the interference between different subbands sparsities.

\begin{remark}
Results analogous to those presented in this Section hold when sampling horizontal lines in the Fourier space, up to replacing the structured sparsities $s_j^r$ in Definition~\ref{def:sparsity_in_levels_2D} with
$$
s_j^c := \max_{0 \leq \ell \leq J} \max_{k \in \Omega_\ell} |S \cap \Omega_{j,\ell} \cap C_k|,
$$
where $C_k = \{1,\ldots,\sqrt{n}\}\times \{k\}$ is the $k$-th vertical line of the wavelet multi-index space. In this case, sampling along the $k$-th horizontal line corresponds to draw a block $D_k = \phi \otimes \phi_{k,:}$. The proofs in Appendix~\ref{proof:corollinesHaarMRI_lambda} can be easily adapted to this setting by interchanging the role of rows and columns in the Haar and in the Fourier multi-index spaces, respectively (or, equivalently, by switching the order of tensorization).
\end{remark}
\begin{remark}
There is a mismatch between the notation adopted here and  in \cite{boyer2017compressed}, where the role of rows and columns in the Haar and Fourier spaces is switched. As a result, horizontal Fourier sampling and structured sparsities $s_j^c$ in \cite{boyer2017compressed} correspond to vertical Fourier sampling and structured sparsities $s_j^r$, respectively, in our framework. Although choosing one of these two conventions might be considered just a matter of taste, we have changed notation in order to adhere to the way the two-dimensional Fourier transform \replace{in}{is} performed by the \textsc{Matlab}$^{\tiny\textregistered}$ command \texttt{fft2}. In our setting, fully-sampled measurements $y = A_0 x = (\mathcal{F}H^* \otimes \mathcal{F}H^*)x$ correspond to $Y = \mathcal{F}H^* X (\mathcal{F}H^*)^T$, where $x = \vect(X)$ and $y = \vect(Y)$. In particular, the block measurement $D_k x = ((\mathcal{F}H^*)_{k,:}\otimes \mathcal{F}H^*)x$ corresponds to the $k$-th column (or vertical line) of the matrix $Y$.
\end{remark}


TITLE: What is this question asking for exactly?
QUESTION [1 upvotes]: So I'm given a question that asks:
"A coffee filter is cone shaped with radius = 4 and height of 8. Suppose filter is filled with water up to a height of level h. Find an expression for the volume of water in the filter in terms of h (recall the volume of cone is $\frac{1}{3}\pi r^2 h$). Hint: look for similar figures and make a proportion"
at first glance I'm thinking I just have to rewrite the volume formula for h but the hint makes it seem like it's not that simple.

REPLY [3 votes]: Consider the image representing your situation:
By similar triangles it follows that
$$ \frac{4}{r}=\frac{8}{h}~\Leftrightarrow4h = 8r \Leftrightarrow r = \frac{h}{2} $$
As the volume of the cone is $V=\frac{1}{3}\pi r^2h$
Plugging the radius in terms of height in we get
$$ V = \frac{\pi h^3}{12}$$


TITLE: Questions about existence of injections between infinite sets and the sets of all infinite topologies on them
QUESTION [5 upvotes]: 1) If $X$ is an infinite set and $T_X$ the set of all infinite topologies on $X$ is it in general true that there is no injection $f_T:T_X \to X$?
2) What conditions on $X$ assure an injection (if that´s ever possible)?
By infinite topologies, I mean topologies with an infinite number of sets (subsets of $X$) as its elements.

REPLY [2 votes]: For an infinite set $X$, there are $2^{2^{|X|}}$ topologies with infinitely many elements on $X$.
The upper bound is obvious, the lower bound is given by my comment (and the slight correction in YCor's): for any ultrafilter $U$ on $X$, define $T_U$ to have as opens the empty set and the elements of $U$. It is clear that this forms a topology, and that you can recover $U$ from it.
Moreover, clearly $T_U$ has infinitely many open sets.
So $U\mapsto T_U$ defines an injection $\beta X\to T_X$, and it's well known that $|\beta X| = 2^{2^{|X|}}$.
So there can be no injection $T_X\to X$ (in fact, no injection $T_X\to P(X)$)
YCor's comment below mine elaborates a bit on what one can ask of these topologies and how many we get (Hausdorff, metrizable)


text:
\begin{document}

\begin{frontmatter}

\title{Tuning the Branching Ratio in a Symmetric Potential Energy Surface with a Post-Transition State Bifurcation using External Time Dependence}

 \author[label1]{V. J. Garc\'ia-Garrido}
 \author[label2]{M. Katsanikas}
 \author[label2]{M. Agaoglou}
 \author[label2]{S. Wiggins\corref{mycorrespondingauthor}}
 \ead{S.Wiggins@bristol.ac.uk}

 \address[label1]{Departamento de F\'isica y Matem\'aticas, Universidad de Alcal\'a, \\ Alcal\'a de Henares, 28871, Spain.\\[.2cm]}

  \address[label2]{School of Mathematics, University of Bristol, \\ Fry Building, Woodland Road, Bristol, BS8 1UG, United Kingdom.}

 \cortext[mycorrespondingauthor]{Corresponding authors}

\begin{abstract}
Chemical selectivity, as quantified by a branching ratio,  is a phenomenon relevant for many organic chemical reactions. It may be exhibited on a  potential energy surface (PES) that
features a valley-ridge inflection point (VRI) in the region between two sequential index-1 saddles, with one saddle having higher energy than the other. Reaction occurs when a trajectory crosses the region of the higher energy saddle (the ``entrance channel'') and approaches the lower energy saddle. On both sides of the lower energy saddle, there are two wells and  the question we address is that given an initial ensemble of trajectories, what is the relative fraction of trajectories that enter each well.  For a symmetric PES this fraction is $1:1$. We consider a symmetric PES subject to a time-periodic forcing characterized by an amplitude, frequency, and phase. In this letter we analyse how the branching ratio depends on these three parameters.
\end{abstract}

\begin{keyword}
 Chemical selectivity \sep Branching ratio \sep Valley-ridge inflection point \sep Time-dependent forcing.
\MSC[2019] 34C37 \sep 70K44 \sep 34Cxx \sep 70Hxx
\end{keyword}

\end{frontmatter}

\section{Introduction}
\label{sec:intro}

Designing reactions with a specified outcome for the  products is a sought after goal in many areas of reaction dynamics. An approach to this problem in organic chemistry is to choose a reacting molecule whose description via a potential energy surface (PES) displays features that lead to multiple possibilities for the products
\cite{campos2019designing}. This ``selectivity'', i.e. the distribution of products as quantified by a branching ratio, occurs in many organic reactions and is the topic of numerous reviews \cite{carpenter1998dynamic, ess2008, rehbein2011, hare2017post}. Many studies of selectivity have highlighted the fact that the phenomenon is intimately coupled with the dynamics
\cite{black2012dynamics, yang2018dynamics, lourderaj2008classical, ma2017perspective, nyman2014computational, salem1971narcissistic, wang1973dyanmics, tantillo2010carbocation}.

The importance of dynamics naturally leads to a study of the behavior of trajectories on a potential energy surface having features that may reveal the mechanisms governing selectivity, and the design of such potential energy surfaces is a topic of current interest \cite{campos2019designing}. A model that has served as a paradigm for studying this topic is a potential energy surface (PES) having two sequential index-1 saddles with no intervening energy minimum. In particular, between the two index-1 saddles, one of higher energy than the other, there is a valley ridge inflection (VRI) point. Reaction occurs when a trajectory crosses the region of the higher e\-nergy saddle (the ``entrance channel'') and approaches the lower energy saddle. On both sides of the lower energy saddle, there are two wells. The question of interest is which well does the trajectory enter (``product selectivity'').

This problem was analyzed in \cite{AGKW2020a} where the two potential wells in the PES were symmetric. For the symmetric PES, the product distributions are known in advance: 50\% go to one well and 50\% go to the other. The main point of that analysis was to understand the phase space structures that enforced the $1:1$ branching ratio. It was expected that this knowledge could then be used to understand how symmetry breaking perturbations in phase space, or external time dependent forcing, could influence the phase space mechanisms that enforce the $1:1$ branching ratio in the symmetric case. This letter makes a contribution to this topic by considering the role of external time dependent forcing on the selectivity.

In order to highlight only the role of the forcing on selectivity, we consider the same symmetric PES as considered in \cite{AGKW2020a}, but subjected to a time-periodic external forcing. The role of the forcing is highlighted in the symmetric case since without the forcing we know a priori that the branching ratio is 1:1. The time-dependent term depends on three parameters: an amplitude ($A$), a frequency ($\omega$), and a phase  ($\phi$), and we consider the effect of each of these parameters on selectivity.

This letter is outlined as follows. In Section \ref{sec:model} we introduce the relevant landscape features of the symmetric PES that determines the two degree-of-freedom (DoF) Hamiltonian model used to understand the emergence of selectivity in these type of chemical systems and we describe the external time-periodic forcing that we apply to this Hamiltonian system. In Section \ref{sec:setup} we describe the design of the numerical calculation that allows us to compute  the branching ratio and  assess the effect of the parameters that define the time-periodic forcing term on selectivity. In Section \ref{sec:results} we discuss the results and in Section \ref{sec:conc} we summarize our findings and discuss further directions for research.

\section{The Time-Dependent Hamiltonian Model}
\label{sec:model}

We use the potential energy surface (PES) given in \cite{AGKW2020a}, which is symmetric with respect to the $x$-axis. For this symmetric PES we know that the expected product ratio from the branching of trajectories is $1:1$.

An equipotential of the PES is depicted in Fig. \ref{pes_conts}, which shows an exit/entrance channel that is characterized by an index-1 saddle (upper index-1) and an index-1 saddle (lower index-1) which is an energy barrier separating two potential wells. Moreover, the PES has a valley-ridge inflection point (VRI), which is located between both index-1 saddles. We have also indicated the location of the VRI point and the blue arrows indicate the possible fates of  trajectories that enter the system through the channel of the high energy index-1 saddle. Table \ref{tab:tab1} gives the configuration space coordinates and energies of  the critical points.

\begin{figure}[htbp]
	\begin{center}
		\includegraphics[scale=0.4]{figure1}
	\end{center}
	\caption{This figure is from \cite{AGKW2020a}, Equipotential contours of the potential energy surface described in Eq. \eqref{pes_modelVRI}. We denote index-1 saddles (upper and lower) as red diamonds, minima of the potential wells as red points and the valley-ridge inflection point as a blue point. The blue arrows indicate the possible paths for trajectories that enter through the upper channel of the system, illustrating the chemical selectivity mechanism.}
	\label{pes_conts}
\end{figure}

\begin{table}[htbp]
	\begin{tabular}{| l | c | c | c|}
		\hline
		Critical point \hspace{1cm} & \hspace{0.6cm} Location $(x,y)$ \hspace{0.6cm} & \text{Energy} $(V)$ & \hspace{.6cm} \text{Stability} \hspace{.6cm} \\
		\hline\hline
		Index-1 Saddle (Upper) & $(0,0)$ & 0 & saddle $\times$ center \\
		\hline
		Index-1 Saddle (Lower) & $(1,0)$ & -4/3 & saddle $\times$ center \\
		\hline
		Potential Well (Top) & $(1.107146,0.879883)$ & -1.94773 & center \\
		\hline
		Potential Well (Bottom) & $(1.107146,-0.879883)$ & -1.94773 & center  \\
		\hline
		\end{tabular}
		\caption{Location and energies of the critical points of the PES, together with their linear stability behavior when considered as equilibrium points of Hamilton's equations in the time-independent case, i.e. $A=0$.}
	\label{tab:tab1}
\end{table}

The 2 DoF Hamiltonian model that we study is the sum of kinetic energy,  potential energy, and a time-periodic forcing term:

\begin{equation}
H(x,y,p_x,p_y) = \dfrac{p_x^2}{2 m_x} + \dfrac{p_y^2}{2 m_y} + V(x,y) - y A \sin(\omega t + \phi) \;,
\label{hamiltonian}
\end{equation}

\noindent
where $A$, $\omega$, and $\phi$ are variable parameters, $m_{x}$, $m_{y}$ represent the masses of the $x$ and $y$ DoF, respectively.  and the PES has the form:

\begin{equation}
V(x,y) = \dfrac{8}{3}x^3 - 4x^2 + \dfrac{1}{2} y^2 + x y^2 \left(y^2 - 2\right) \;,
\label{pes_modelVRI}
\end{equation}

\noindent
For this study we choose $m_x = m_y = 1$, and thus Hamilton's equations of motion are the following:

\begin{equation}
\begin{cases}
\dot{x} = \dfrac{\partial H}{\partial p_x} = p_x \\[.4cm]
\dot{y} = \dfrac{\partial H}{\partial p_y} = p_y  \\[.4cm]
\dot{p}_x = -\dfrac{\partial H}{\partial x} = 8 x \left(1 - x\right) + y^2\left(2 - y^2 \right) \\[.4cm]
\dot{p}_y = -\dfrac{\partial H}{\partial y} = y \left[4 x \left(1 - y^2\right) - 1\right] + A \sin(\omega t + \phi)
\end{cases}
\;.
\label{ham_eqs}
\end{equation}

\section{Set-up for the Calculation}
\label{sec:setup}

In this section we describe the setup for our numerical calculations. We emphasize that for a time-dependent Hamiltonian total energy is not conserved. However, we will use energy surface information  for the time independent Hamiltonian system, i.e. $A=0$, with total energy $H = 0.1$ (the same as considered in \cite{AGKW2020a}).

In Fig. \ref{fig:experiment_setup} we describe the main components of phase space and configuration space that we use to initiate our numerical experiment, as well as to interpret the results of the numerical experiment. The red line $x = -0.005$ in configuration space shown in the figure  is used to construct the dividing surface on which we initialize the trajectories. One thousand points are equally spaced along this line. For each point we choose the momentum $(p_x, p_y) = (p_x>0, 0)$ such that the total energy of the time-independent Hamiltonian is $H = 0.1$. The condition $p_x > 0$ corresponds to trajectories that enter that react. These are the initial conditions that are then evolved under the time dependent dynamics \eqref{ham_eqs}. The resulting reacting trajectories can have three possible fates for the length of integration time that we consider. They can enter the top well, the bottom well, or they can reflect off the right hand wall of the potential energy surface without entering either well and exit the reacting region through a neighborhood of the high energy index-1 saddle.

In Fig. \ref{fig:experiment_setup} we have indicated the locations of the top and bottom wells with yellow dots and the red line represents the ensemble of initial conditions used in the simulation.  The blue lines in the figure  at $y = \pm 0.5$ establish the conditions used to determine that a given trajectory has entered the top or bottom well, respectively. When the $y$ coordinate of a trajectory exceeds $y=0.5$ we consider it to have entered to top well, and the integration of that trajectory is stopped, and  when the $y$ coordinate of a trajectory decreases below $y=-0.5$ we consider it to have entered to bottom well, and the integration of that trajectory is stopped.

\begin{figure}[htbp]
	\begin{center}
		\includegraphics[scale=0.35]{figure2}
	\end{center}
	\caption{Configuration space representation of the different elements used to setup the numerical experiment for the calculation  of the branching ratio. We have indicated the locations of the top and bottom wells with yellow dots,  and the red line represents the ensemble of initial conditions used in the simulation. The blue lines at $y = \pm 0.5$ establish the conditions used to determine that a given trajectory has entered the top or bottom well, respectively.}
	\label{fig:experiment_setup}
\end{figure}

\section{Results}
\label{sec:results}

In this section we discuss the results of our calculations. We consider first the behavior of the system when it is subjected to an external forcing, for which we use three different values of the angular frequency $\omega = \frac{\pi}{3},\frac{\pi}{2},\pi$. The simulations are carried out in the parameter space determined by the amplitude ($A$) and phase $(\phi)$ of the wave, and the values taken for these parameters move in the ranges $0 \le A \le 0.35$ and $0 \le \phi \le \pi$.

We begin by illustrating the main effect that time dependence can have on the branching ratio. In Fig. \ref{fig:top_bot_fraction} A we show the total fraction of trajectories that enter the top well  (blue curve) and the bottom  well (red curve) as a function of time for $A=0$, i.e.  for the symmetric, time independent PES.  The result is as we expect for the symmetric case. The two curves lie on top of each other. In Fig. \ref{fig:top_bot_fraction} B we consider the case $A=0.1$, $\phi =0$, and $\omega = \frac{\pi}{3}$. In this case we see that the red and blue curves split, and that the branching ratio is no longer $1:1$. We now want to explore the parameter dependence of this effect more completely.

Figures \ref{fig:paramSp_pi_div_3}, \ref{fig:paramSp_pi_div_2} and \ref{fig:paramSp_pi} correspond respectively to $\omega = \frac{\pi}{3}$, $\omega = \frac{\pi}{2}$ and $\omega = \pi$. For each figure, panel A gives the fraction of trajectories that enter the top well, panel B gives the fraction of trajectories that enter the bottom well, and panel C gives the total fraction of trajectories that do not interact with either well. Notice that Figs. \ref{fig:paramSp_pi_div_3} and \ref{fig:paramSp_pi_div_2} show similar trends in all three panels. For a range of phases (roughly, $0 \le \phi / \pi \le 0.6$)  the fraction of trajectories  in the top well increases as A increases and  for a range of phases (roughly, $0.6 \le \phi / \pi \le 1$)   the fraction of trajectories in the bottom well increases as $A$ increases. Note that essentially all reacting trajectories enter either the top or bottom well. The behaviour in Fig. \ref{fig:paramSp_pi} is somewhat different. For this larger frequency a significant fraction of the trajectories do not either  well before returning to cross the higher energy saddle.

We finish this letter by briefly illustrating the response that the system displays when it is subjected to a forcing characterized by a very high frequency. In particular we take $\omega = 8 \pi$ and perform a similar analysis as the one we carried out above in the amplitude-phase parameter space. The expected behavior of the system is that, since the oscillations are very fast,  the trajectories generated from the underlying dynamics do not have enough time to respond to such rapid oscillations and, consequently,  they will follow paths very similar to the scenario where the system is fully time independent, i.e. $A = 0$, where the branching ratio is $1:1$. This is confirmed from the output of our calculations, which is included in Fig. \ref{fig:paramSp_8pi}. In panel A we can clearly see how the fraction of trajectories that go to the top well, calculated from all those that enter either well, give values very close to $0.5$ for the range of parameter values considered. Moreover, it is interesting to note that the fraction of trajectories that escape the system in this regime of high frequency forcing is indeed very small. This is depicted in panel B, where we see that the escape fraction exhibits interesting wave-like patterns in its distribution. We complement this analysis by doing a parameter study, but in this case fixing an amplitude of the forcing, $A = 0.15$, and sweeping through the angular frequency $\pi / 4 \leq \omega \leq 10 \pi$ and phase $0 \leq \phi \leq \pi$. From Fig. \ref{fig:paramSp_ampl} we can conclude that the fraction of trajectories that go to the top well vary from $0.35$ to approximately $0.83$ in the range of frequencies $0 \leq \omega \leq 2\pi$, but that it stabilizes to a value close to $0.5$ for frequencies higher than $\omega = 2\pi$. Moreover, panel B illustrates the escape fraction, and we can see how this quantity varies from $6\%$ to $12\%$ for the  interval of small frequencies $\pi / 4 \leq \omega \leq 2\pi$, while it drops below $6\%$ when the forcing frequency is increased beyond this interval.

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.26]{figure3a}
		B)\includegraphics[scale=0.26]{figure3b}
	\end{center}
	\caption{Time evolution of the fraction of trajectories that enter the top or bottom wells, measured from the total number of trajectories that visit any of the wells. A) Time-independent system, $A=0$. B) Time-dependent system with a time dependence characterized by a forcing with amplitude $A = 0.1$, phase $\phi = 0$ and angular frequency $\omega = \pi / 3$.}
	\label{fig:top_bot_fraction}
\end{figure}

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.19]{figure4a}
		B)\includegraphics[scale=0.19]{figure4b}
		C)\includegraphics[scale=0.22]{figure4c}
	\end{center}
	\caption{Fraction of trajectories escaping the system or that enter either of the wells. The amplitude-phase parameter space has been calculated for a time dependent forcing characterized by an angular frequency $\omega = \pi / 3$. A) Trajectories that enter the top well from all those that enter any of the wells. B) The same but for the bottom well. In the panels A and B we calculated the fraction of the trajectories from all those that enter any of the wells. C) Trajectories that escape the system without entering either of the wells, calculated from the total number of trajectories of the initial ensemble.}
	\label{fig:paramSp_pi_div_3}
\end{figure}

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.18]{figure5a}
		B)\includegraphics[scale=0.18]{figure5b}
		C)\includegraphics[scale=0.22]{figure5c}
	\end{center}
	\caption{Fraction of trajectories escaping the system or that enter either of the wells. The amplitude-phase parameter space has been calculated for a time dependent forcing characterized by an angular frequency $\omega = \pi / 2$. A) Trajectories that enter the top well from all those that enter any of the wells. B) The same but for the bottom well. In the panels A and B we calculated the fraction of the trajectories from all those that enter any of the wells. C) Trajectories that escape the system without entering either of the wells, calculated from the total number of trajectories of the initial ensemble.}
	\label{fig:paramSp_pi_div_2}
\end{figure}

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.19]{figure6a}
		B)\includegraphics[scale=0.19]{figure6b}
		C)\includegraphics[scale=0.22]{figure6c}
	\end{center}
	\caption{Fraction of trajectories escaping the system or that enter either of the wells. The amplitude-phase parameter space has been calculated for a time dependent forcing characterized by an angular frequency $\omega = \pi$. A) Trajectories that enter the top well from all those that enter any of the wells. B) The same but for the bottom well. In the panels A and B we calculated the fraction of the trajectories from all those that enter any of the wells. C) Trajectories that escape the system without entering either of the wells, calculated from the total number of trajectories of the initial ensemble.}
	\label{fig:paramSp_pi}
\end{figure}

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.18]{figure7a}
		B)\includegraphics[scale=0.17]{figure7b}
	\end{center}
	\caption{Fraction of trajectories escaping the system or that enter either of the wells. The amplitude-phase parameter space has been calculated for a time dependent forcing characterized by an angular frequency $\omega = 8\pi$. A) Trajectories that enter the top well from all those that enter any of the wells. B) Fraction of trajectories that escape the system without entering either of the wells, calculated from the total number of trajectories of the initial ensemble.}
	\label{fig:paramSp_8pi}
\end{figure}

\begin{figure}[htbp]
	\begin{center}
		A)\includegraphics[scale=0.19]{figure8a}
		B)\includegraphics[scale=0.19]{figure8b}
	\end{center}
	\caption{Fraction of trajectories escaping the system or that enter either of the wells. The $(\omega,\phi)$-parameter space has been calculated for a time dependent forcing characterized by an amplitude $A = 0.15$. A) Trajectories that enter the top well from all those that enter any of the wells. B) Fraction of trajectories that escape the system without entering either of the wells, calculated from the total number of trajectories of the initial ensemble.}
	\label{fig:paramSp_ampl}
\end{figure}

\section{Summary and Outlook}
\label{sec:conc}

In this letter we have shown that chemical selectivity is exhibited on a symmetric  potential energy surface (PES) that features a valley-ridge inflection point (VRI) in the region between two sequential index-1 saddles, with one saddle having higher energy than the other when the system is subject to a time-periodic forcing. The forcing function depends on three parameters: amplitude, phase, and frequency. In the absence of forcing the branching ratio is $1:1$ as a consequence of the symmetry. However, we show that even though the PES is symmetric, the time-dependent forcing allows us to ``tune'' the branching ratio, essentially to any value that we desire by appropriate choice of parameters. We note the efforts that have been made in designing PESs that enforce a desired branching ratio \cite{campos2019designing}. Our results offer the possibility of a different approach to achieving a desired branching ratio.

Our analysis is based solely on trajectory calculations utilizing a numerical experiment guided by the geometry of the symmetric PES. In particular, we have not analysed the time-dependent phase space structure that governs selectivity in such a manner like it was analyzed for the symmetric, time-independent, PES in \cite{AGKW2020a}. This would be an intriguing analysis, and potentially very insightful, that we will leave for future investigations.

\section*{Acknowledgements}
The authors acknowledge the support of EPSRC Grant No. EP/P021123/1 and Office of Naval Research Grant No. N00014-01-1-0769.

\section*{Graphical Abstract}

\begin{figure}[htbp]
	\begin{center}
		\includegraphics[scale=0.19]{graphical_abstract}
	\end{center}
	\caption{Time-dependent forcing applied to the trajectories entering the potential energy surface through the upper channel governed by an index-1 saddle point. This interaction allows for the control of the selectivity mechanism in the chemical system.}
	\label{fig:paramSp_ampl}
\end{figure}

\bibliography{SNreac}

\end{document}


TITLE: Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$.
QUESTION [4 upvotes]: QUESTION: Let $p$ be a prime number and let $G$ be a finite $p\text{-group}$. Let $M$ be a maximal subgroup of $G$.   Show that $M$ is a normal subgroup of $G$ and that $| G: M | = p$.
THE HINT GIVEN IS: By strong induction on $n$, where $| G | = p ^{n}$. Let $ y \in Z(G) - \{1 \}$, a convenient $x$ power belonging to $Z(G) - \{1\}$ has order $p$. Consider $G / \langle x \rangle$.
ANSWER GIVEN:
By induction on $n$, where $| G | = p^ n$. According to the tip, consider $x \in Z (G)$ of order $ p $ and let $N = \langle x \rangle$. The group $ G / N $ has order $ p^{n-1} $, so we can apply induction. If $ N $ is a subgroup of $ M $ then $ M / N $ is normal for the $ p $ index in $ G $. Now suppose $ N $ is not a subgroup of $ M $. Being $ M $ maximal we get $ NM = G $. On the other hand being $ | N | = p $ prime we have $ N \cap M = \{1 \} $ logo $ p^{n-1} = | G / N | = | MN / N | = | M / M \cap N | = | M | $ e we deduct $| G: M | = p $. In addition $ M $ is a subgroup of $ N_G (M) $ and $ N $ is a subgroup of $ N_G (M) $ because $ N \leq Z (G) $. It follows that $ G = NM \leq N_G (M) $ so $ M $ is a normal subset of $ G $.
MY QUESTIONS: I didn't understand the following steps showed in this proof.

The induction used in its solution.
If $N$ is subgroup of $M$
then one is stated that $M/N$ is normal (why?)
If $N$ is NOT a
subgroup of $M$ then one is stated that $NM=G$ (again, why?)
Why
$N \cap M=\{1\}?$
Why $N \leq N_G(M)$ ?
Why $NM\leq N_G(M)$?

REPLY [3 votes]: The induction is on the exponent of the group. Every $p$ group has order $p^n$ for some $n$, so the induction is assuming that the theorem is true for every $p^m$ for $m<n$.
$N$ is part of the center so it commutes with everything, which is stronger than being normal.
If $N$ is not a subgroup of $M$ then some element of $NM$ must not be in $M$, which means $M$ is a subgroup of $NM$. $M$ is maximal, so that means $NM=G$.
$N$ is prime and cyclic so if any element of $N$ other than the identity is in $M$, then all of $N$ must be in $M$, which would make it a subgroup of $M$, which is assumed not to be true in this case.
Again, $N$ is part of the center, so it commutes with everything.
Of course $M$ normalizes itself, and $N$ commutes with everything in $M$ so it normalizes $M$. Therefore the product of these two groups normalizes $M$.


TITLE: Differentiate under expectation sign.
QUESTION [1 upvotes]: Let $f(x)$ be the probability density function of a $\text{Uniform}(0,1)$ distribution. Let $Z\sim\text{Normal}(0,1)$ and $g(x)=E[f(x+Z)]$. If one plots $g(x)$ in a computer, one observes like an asymetric bell shape.
We have that $f'=0$ exists except at two points, $0$ and $1$. But the probability that $x+Z$ belongs to $\{0,1\}$ is $0$. My question is why $g'(x)=E[f'(x+Z)]=0$ is not true.
I think that one can use the dominated convergence theorem: fixed $x\in\mathbb{R}$, $f(x+Z)$ is differentiable at $x$ almost surely and $|f'(x+Z)|\leq 1$ almost surely. So the expectation and the differentiation should commute. What is wrong with this reasoning?
Edit: I write the use of the dominated convergence theorem more in detail: by definition of derivative, $$\lim_{h\rightarrow0} \frac{f(x+h+Z)-f(x+Z)}{h}=f'(x+Z).$$ By the mean value theorem, $$ \left|\frac{f(x+h+Z)-f(x+Z)}{h}\right|=|f'(x+\xi_h+Z)|\leq 1, $$ where $|\xi_h|<|h|$ and $E[1]=1<\infty$ (thus integrable). Then one uses the dominated convergence theorem, by interchanging the limit and the expectation. What is wrong here?

REPLY [0 votes]: $$g'(x) = \lim_{h \to 0} \dfrac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \mathbb E\left[ \frac{f(x+h+Z)-f(x+Z)}{h} \right]$$
For $h > 0$, $$f(x+h+Z)-f(x+Z) = \cases{1 & if $-x-h < Z < -x$\cr
                                        -1 & if $1-x-h < Z < 1-x$\cr
                                         0 & otherwise}$$
so $g'(x) = \lim_{h \to 0} h^{-1} (\mathbb P(-x-h<Z<-x)-\mathbb P(1-x-h<Z<1-x)) = \varphi(-x)-\varphi(1-x)$
where $\varphi$ is the density of $Z$.