TITLE: Curvature of a curve and projections on the unit normal
QUESTION [0 upvotes]: I must admit that the differences between arc-length and a general parameterization of a curve are confusing.
Consider a curve $\gamma$ parameterized not by arc-length. Let $n$ be the unit normal to the curve. Is it true that the curvature is $\ddot \gamma \cdot n$? This is not homework, I'm just trying to understand this concept.
If we take $\gamma (t)=\gamma (s(t))$ then we get $\ddot \gamma \cdot n=\frac {\gamma''(s(t))\cdot n}{dt/ds}$. Is that the curvature?
I'm trying to get a definition of a curvature using the projection on the unit normal to the curve that will be true for all parameterizations. Is that even possible? I know that it is true for the normal curvature (which involves a surface). Is it also true in this case?

REPLY [2 votes]: Curvature is defined by $\kappa=\langle\mathbf t',\mathbf n\rangle$, where $\bf t$ is the unit tangent vector. This definition is independent of the parametrization. Curvature is a geoemtric property.
For arc length parametrized curves holds $\bf t'=\gamma''$ , hence the curvature can be expressed as $\kappa=\langle \mathbf t',\mathbf n\rangle=\langle\gamma'',\mathbf n\rangle$.
But since $\gamma''$ is not a geometric property and might change with the used parametrization, $\langle\gamma'',\bf n\rangle$ can not give the curvature in general. The curvature must be independent of the parametrization.


TITLE: Metric with Harmonic Coefficient and Stress-Energy Tensor in General Relativity
QUESTION [2 upvotes]: I have two question:

Is there any possible implies or interest to use in general relativity a metric whose coefficients are harmonic functions?
What is the meaning (physical) if the stress-energy tensor ($T_{ij}$) has the components that are harmonic functions?

REPLY [2 votes]: I can answer to the first part of your question. A metric with harmonic coefficients is for example the FLRW metric for an universe with positive curvature. In this case the metric takes the form ($c=1$):
$$ds^2 = dt^2 - a^2(t) \left(dr^2 + \frac{1}{\sqrt{k}} \sin(r\sqrt{k}) d\Omega^2 \right)$$


TITLE: Suppose $G$ is cyclic group and $\sigma, \tau \in G$ are such that order$(\sigma)=12$ and order$(\tau)=21$
QUESTION [0 upvotes]: Suppose $G$ is cyclic group and $\sigma, \tau \in G$ are such that order$(\sigma)=12$ and order$(\tau)=21$. Then the order of the smallest group constaing $\sigma$ and $\tau$ is ____.

I think the solution is LCM$(21,12)=84$ because a cyclic group obeys the converse of langrange's theorem.

REPLY [4 votes]: Your intuition is correct. Let $H = \langle \sigma, \tau \rangle \subseteq G$. $H$ is therefore a cyclic group containing an element of order $12$ and of order $21$. In general, a cyclic group contains an element of a given order if and only if that order divides the order of the group - this is the converse to Lagrange's theorem I'm guessing you're talking about. Hence, the smallest possible order of $H$ is $lcm(12, 21) = 84$. As $\sigma, \tau \in G$, $12 \mid |G|$ and $21 \mid |G|$ so $84 \mid |G|$ and so this minimal possible order is attainable.


TITLE: Would I be able to feel a single-molecule wide string?
QUESTION [2 upvotes]: Imagine there is a material that is so strong it can form several meters string while being only a single molecule wide. It is securely attached and can withstand any human pressure. What happens if I slide my finger across it? Would it cut it or would I not notice anything?

REPLY [1 votes]: the idea of a single-molecule "fiber" of macroscopic length (called a "sinclair molecule chain") appeared in the science-fiction stories of Larry Niven some 50 years ago, an abbrievated analysis follows.
If we imagine a string consisting of a single linear molecule, that string would possess no dislocations, point defects, or grain boundaries, and its breaking strength would not be limited by the (relatively weak) forces that cause adjacent polymer chain molecules to adhere to each other in things like spider web silk.
As such, its breaking strength would be equal to the load required to actually break the chemical bond between any two atoms comprising the string's backbone, which would be many orders of magnitude greater than the load required to cause two of those chains (of shorter length) to slip past one another, or for dislocation travel (in a macroscopic fiber) to promote plastic deformation, grain boundary rotation, or the merging of point defects.
Niven invented the single-molecule chain to furnish an easily-concealed tool in the form of a long molecular chain with the cross-sectional area of one molecule and knobs on its ends which a person could unfurl and then use to cut through any solid with little effort. However, he didn't do the math- which of course had no effect on the entertainment value of the sci-fi story, but a reader who happened to be materials scientist would reject the premise.
A "real" polymer rope of finite cross-section might possess in cross-section somewhere around 10^20 polymer chains and have a breaking strength (based on slippage between adjacent chains) of order ~tens of thousands of pounds. Niven's premise was that the tensile strength of one molecule was of order ~10^20 greater than the strength of a collection of those molecules tangled up against one another in a solid, so that a single molecule string would possess the interatomic breaking strength of  ~10^20 tangled molecules held together with Van der Waals forces.
But what he failed to recognize was that a load of several hundred pounds applied to the ends of that single molecule string would produce a stress of ~ (load)/(cross-sectional area) and the area involved would be that of one molecule or more properly the cross-sectional area of the electron orbital overlap which furnished the interatomic bond.
With such a tiny number in the denominator of that fraction, the stress would be almost unimaginably gigantic, and pulling on it with your hands would easily break it.


TITLE: Why is the number of orders of $6$ sandwiches using $3$ types of sandwiches a combinations with repetitions problem?
QUESTION [1 upvotes]: There are $3$ types of sandwiches, namely chicken (C), fish (F) and ham (H), available in a restaurant. A boy wishes to place an order of $6$ sandwiches. Assuming that there is no limit in the supply of sandwiches of each type, how many such orders can the boy place?
My Attempt:
I know that this is a stars and bars problem and the solution is $8 \choose 2$. But why can't it simply be $3^6$ where each of the $6$ sandwich places has $3$ choices to fill as repetition is allowed. I know I am missing something trivial.

REPLY [1 votes]: Let us take Chicken and Ham sandwich and total order of 3. You are wondering why it should not be $2^3$ and why it is ${(3+2-1)\choose (2-1)}$
The $2^3=8$ choices that you mention are

AAA - 3 Chicken sandwich
AAB - 2 Chicken and 1 Ham
ABA
BAA
ABB - 2 Ham and 1 Chicken
BAB
BBA
BBB - 3 Ham

The total is $2^3$ but the chocies are merely ${(3+2-1)\choose (2-1)}$ = ${4\choose1}=4$


TITLE: CCD's and the photoelectric effect
QUESTION [1 upvotes]: Do charge coupled devices as found in telescopes use the photoelectric effect if not what eles librates the electons. Also what is charge intergration in reation to CCD's.

REPLY [2 votes]: First question:  depends on your definition of photoelectric effect.  In a CCD a photon is absorbed, and the energy given to an electron which is then promoted out of the Fermi sea, just as happens in the traditional experiment of a metal in a vacuum.  However, in a CCD, the electron has only enough energy to get to the conduction band (generates an electron-hole pair), not enough to leave the crystal. This process is sometimes called the photoelectric effect, or internal photoelectric effect.  But one could argue that it is not really the photoelectric effect because no current was generated outside of the crystal.
Second question:  What happens to those photo-generated electrons?  The microscopic geometry of the CCD is such that there are internal potential barriers that confine them to locations very close to where they were generated (that is, within a pixel). They stay there until the electronics apply voltages to sweep them out.  Before they are swept out, the pixel continues to collect additional photo-generated electrons as long as light is present on the pixel.  So the total number of electrons in a pixel for one exposure is the sum of all the electrons photo-generated during the exposure.  It has integrated the charge.


TITLE: Convergence rate of an iterative scheme
QUESTION [0 upvotes]: Given an iteration scheme of the form : $x_{k+1}=x_k+t_kd_k$. It is also known that there are $\delta,C>0,D\geq 0$ such that if $x_k \in B_{\delta}(x_*)$ holds we have the following estimate:
$ ||x_{k+1}-x_*||\leq C(||x_k-x_*||+D)||x_k-x_*||$.
What is the convergence rate of the iteration scheme if $C=D$ ?
I've tried to plug in $C=D$ inside this expression and tried to expand and see what the RHS looks like. I thought I might be able to factor out some term to make it look nice but was unable to do so. Any suggestions as to how to determine the rate of convergence of this scheme?

REPLY [0 votes]: Let $d_k = ||x_k-x_*||$.
We are given
$d_{k+1} \le C d_k(D+d_k)
=
CD d_k + C d_k^2
$.
This becomes
$\begin{align}
d_{k+1}
&\le C d_k(d_k+D) \\
&\le CD d_k+C d_k^2\\
&\le C D(CD d_{k-1}+C d_{k-1}^2)+C (CD d_{k-1}+C d_{k-1}^2)^2\\
&= C^2D^2 d_{k-1}+C^2 D d_{k-1}^2+C^3d_{k-1}^2 ( D+ d_{k-1})^2\\
&= C^2 D^2d_{k-1}+C^2 Dd_{k-1}^2+C^3d_{k-1}^2 ( D^2+ 2D d_{k-1}+d_{k-1}^2)\\
&= C^2 D^2d_{k-1}+C^2 Dd_{k-1}^2+C^3D^2d_{k-1}^2 + 2C^3D d_{k-1}^3+C^3d_{k-1}^4\\
&= C^2 D^2d_{k-1}+C^2D(1+CD) d_{k-1}^2+2C^3Dd_{k-1}^3+C^3d_{k-1}^4\\
\end{align}
$
If $D=C$,
this becomes
$d_{k+1}
\le C^4d_{k-1}+C^3(1+C^2) d_{k-1}^2+2C^4d_{k-1}^3+C^3d_{k-1}^4
$.
It looks like
$d_{k+1} = C^{m+1}D^{m+1}d_{k-m} + O(d_{k-m}^2)$
for integer $m \ge 0$.
Looking at this,
there doesn't seem to me
to be any obvious reason
why $D=C$ would be special.
$D=0$ would certainly be.


TITLE: Calculate limit using Maclaurin series
QUESTION [2 upvotes]: $$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$
So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$
What I did next was do Maclaurin expansion for three functions up until the third order:
$f(x) = 1 - \frac{x^2}{2} + o(x^3)$
$g(x) = x - \frac{x^3}{3} + o(x^3)$
$h(x) = x^3 + o(x^3)$
So when I plug in these values I get:
$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$
So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).
Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?

REPLY [1 votes]: Let us make the limit more "friendly" by a change of variable, $t:=\sin x$. We now have
$$\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{\arcsin^3t}=\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{t^3}.$$
Then by the generalized binomial theorem,
$$\lim_{t\to0}\frac{1-1+\dfrac t2t^2-\dfrac t2\left(\dfrac t2-1\right)\dfrac{t^4}2+\cdots}{t^3}$$ and higher order terms. You can conclude.


TITLE: Pentagon with two right angles (aka Van Aubel's Theorem)
QUESTION [0 upvotes]: My problem is the following: given that $ABCDE$ is a convex pentagon such that $AB=BC$, $CD=DE$, $M$ is middle point of side $EA$ and the angles $\widehat{ABC}=\widehat{CDE}=90°$, find the measure of the angle $\widehat{BMD}$.
I think that $\widehat{BMD}=90°$, too, and it is possible to prove it through the cosine theorem applied to $ACE$ and the Pythagorean theorem applied to $BMD$. However, this approach is hard and tedious, I wonder if an easier solution exists.

REPLY [0 votes]: Let $C={\bf 0}$, put  $A=2{\bf a}$, $E=2{\bf e}$, and denote by $*{\bf x}$ the vector ${\bf x}$ turned $90^\circ$ counterclockwise. Then
$$M={\bf a}+{\bf e}, \quad D={\bf e}-*{\bf e}, \quad B={\bf a}+*{\bf a}\ .$$

It follows that
$$\overrightarrow{BM}=({\bf a}+{\bf e})-({\bf a}+*{\bf a})={\bf e}-*{\bf a}$$
and then
$$\overrightarrow{DM}=({\bf a}+{\bf e})-({\bf e}-*{\bf e})={\bf a}+*{\bf e}=*({\bf e}-*{\bf a})=*\overrightarrow{BM}\ ,$$
which shows that $|BM|=|DM|$ as well.


TITLE: What is the function for the magnitude of air resistance/drag?
QUESTION [0 upvotes]: In John Taylor's Classical Mechanics, he gives a basic treatment on air resistance/drag. In order to do this, he models air resistance as $\mathbf{f}=-f(v)\mathbf{\hat{v}}$, and then Taylor expands $f(v)$, using only the linear and quadratic terms. Of the function itself, all he says is, "The function $f(v)$ that gives the magnitude of the air resistance varies with $v$ in
a complicated way, especially as the object's speed approaches the speed of sound." He gives no further resources as to what this actually is.
Does anybody know this proper formula for air resistance?

REPLY [1 votes]: There is no such function in general: air resistance depends not only on the velocity of the body through the medium, but also on the shape of the body and its rigidity (at least). I presume the above assumes a rigid sphere as the body, but even in that case, things are complicated.


TITLE: Banach fixed point theorem and inverse function
QUESTION [3 upvotes]: Let $U$ and $V$ be the open subsets in $\mathbb{R}^n$, $x\in U$ and $f:U\rightarrow V$ is a smooth function. There is an inverse function theorem which states that if the Jacobian determinant at $x$ is nonzero then there exists an open subset $U'\subset U$ such that $f|_{U'}$ is a diffeomorphism on its image.
I read some long and rather difficult proofs of this theorem. Also, I heard recently that there is an easy proof using Banach fixed point theorem. Could you tell me where I can read this? Unfortunately, I didn't manage to prove it myself.

REPLY [0 votes]: This seems to be what you're looking for: http://www.math.jhu.edu/~jmb/note/invfnthm.pdf


text:
\begin{document}

\title{Bijections from weighted Dyck paths to Schr\"oder paths}
\author{Dan Drake}
\address{Department of Mathematical Sciences\\Korea Advanced Institute
   of Science and Technology\\Daejeon, Korea}
\email{ddrake@member.ams.org}
\urladdr{http://mathsci.kaist.ac.kr/~drake}
\subjclass[2010]{Primary: 05A19; Secondary: 05A15, 05A05}
\keywords{lattice paths, \Sch numbers, matchings, $231$-avoiding permutations}
\date{\today}

\begin{abstract}
  Kim and Drake used generating functions to prove that the number of
  $2$-distant noncrossing matchings, which are in bijection with little
  \Sch paths, is the same as the weight of Dyck paths in which downsteps
  from even height have weight $2$. This work presents bijections from
  those Dyck paths to little \Sch paths, and from a similar set of Dyck
  paths to big \Sch paths. We show the effect of these bijections on the
  corresponding matchings, find generating functions for two new classes
  of lattice paths, and demonstrate a relationship with $231$-avoiding
  permutations.
\end{abstract}

\maketitle

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

This work begins with the work of Kim and the present author
~\cite{drake.kim:k-distant} in which they studied, among other things,
$2$-distant noncrossing matchings.
Such matchings---which will be defined shortly---are naturally
enumerated by little \Sch paths.
In the process of describing connections between $k$-distant noncrossing
matchings and orthogonal polynomials, Drake and Kim used generating
functions to show that little \Sch paths are equinumerous with a certain
set of labeled Dyck paths.
We present here a bijective proof of that fact; the bijection has a
number of interesting properties and is a consequence of a bijection
between big \Sch paths and a similar set of labeled Dyck paths.

We begin with definitions of the combinatorial objects mentioned above.
The notation $[n]$ refers to the set of positive integers from $1$ to
$n$.
A \textsl{matching of $[n]$} is a set of vertex-disjoint edges in the
complete graph on $n$ vertices so that every vertex is adjacent to
exactly one edge.
For our purposes, a matching can also be viewed as a permutation whose
cycles all have length $2$, or a set partition whose blocks all have
size $2$.
We will draw matchings by arranging the vertices horizontally and
drawing arcs, as in \autoref{fig:matching-example}.

\begin{figure}[h]
  \centering
  \inputtikz{matching-example}
  \caption{A matching of $[12]$.}
  \label{fig:matching-example}
\end{figure}

Drake and Kim~\cite{drake.kim:k-distant} define a \textsl{$k$-distant
crossing} as a pair of arcs $(i_{1}, j_{1})$ and $(i_{2}, j_{2})$, with
$i_{1} < i_{2} < j_{1} < j_{2}$ and $j_{1} - i_{2} \ge k$.
The arcs $(6, 11)$ and $(8,12)$ of the matching in
\autoref{fig:matching-example} form a $3$-distant crossing; the arcs
$(1,4)$ and $(3,7)$ form a $1$-distant crossing.
A \textsl{$k$-distant noncrossing matching} is simply a matching with no
$k$-distant crossing.
The matching in \autoref{fig:matching-example} is $4$-distant
noncrossing.
(This notion of $k$-distant crossing is different from the $k$-crossings
of matchings studied by, for example, Chen et al.\
~\cite{chen.deng.ea:crossings}; their work concerns sets of $k$ mutually
crossing edges, and ignores the distance between vertices.)

In a $2$-distant noncrossing matching, crossing edges are allowed as
long as the right vertex of the left edge is adjacent to the left
vertex of the right edge.
This fact allows us to describe a bijection from $2$-distant
noncrossing matchings to a certain class of lattice paths.
A lattice path of length $n$ is a sequence $(p_{0}, p_{1},\dots,p_{n})$
of points in $\N \times \N$; the $k$th step of the path is the pair
$(p_{k-1}, p_{k})$.
A step is called an \textsl{upstep} if the component-wise difference of
$p_{k} - p_{k-1}$ is $(1,1)$, and a \textsl{downstep} if the difference is
$(1, -1)$.
In this work, we will use paths with \textsl{double horizontal steps},
which is a pair of adjacent steps whose component-wise differences are
both $(1,0)$.
By a minor abuse of terminology, a double horizontal step will usually
be called a horizontal step.
A \textsl{little \Sch path} is a lattice path consisting of upsteps,
downsteps, and horizontal steps, such that no horizontal step occurs at
height zero.
See \autoref{fig:2dnc-matching-little-sch} for an example of such a path.

It is not difficult to describe a bijection from $2$-distant noncrossing
matchings to little \Sch paths: convert every vertex at the left end of
an arc to an upstep and every vertex at the right end of an arc to a
downstep---except for adjacent vertices involved in a crossing: convert
those vertices into a horizontal step.
This operation is a bijection because, given a little \Sch path, one can
recover the matching by drawing an opening half edge at every upstep,
two crossing half edges at every horizontal step, and a closing half
edge at every downstep.
Then connect every closing half-edge to the nearest opening half-edge to
create a matching.
See \autoref{fig:2dnc-matching-little-sch} for an example of this
correspondence.

\begin{figure}[h]
  \centering
  \inputtikz{2dncr-matching-little-sch}
  \caption{An example of the bijection between $2$-distant noncrossing
    matchings and little \Sch paths (left), and the correspondence
    between the edges incident to vertices of the matching and steps in
    the little \Sch path (right). }
  \label{fig:2dnc-matching-little-sch}
\end{figure}

The \textsl{little \Sch numbers} $s_{n}$ (sequence A1003 in the OEIS
~\cite{sloane:on-line}) count $2$-distant noncrossing matchings of
$[2n]$ and also little \Sch paths of length~$2n$.
If horizontal steps on the $x$-axis are allowed, one has a \textsl{big
\Sch path}; the number of such paths of length $2n$ is $S_{n}$, the
\textsl{big \Sch number} (sequence A6318) and it is well known that
$S_{n} = 2s_{n}$ for $n > 0$; see the next section and also Deutsch's
bijective proof~\cite{deutsch:bijective}.

We need several more definitions related to lattice paths.
The first is the step that \textsl{matches} a step.
For an upstep $u$, the matching step is the rightmost downstep to the
left of $u$ that leaves from the same height at which the upstep ends;
the definition for a downstep is similar.
For a horizontal step $h$ not on the $x$-axis, the matching step is the
leftmost downstep to the right of $h$ that leaves from the same height
as $h$; in the corresponding $2$-distant noncrossing matching, the
matching downstep corresponds to the rightmost vertex involved in the
two crossing edges.
For example, the step matching the first horizontal step in
\autoref{fig:2dnc-matching-little-sch} is the last downstep.
We will write paths using ``U'' for upsteps, ``HH'' for horizontal
steps, ``D'' for regular downsteps, and ``d'' for special downsteps,
which will be defined in \autoref{sec:descr-biject}.
The path in \autoref{fig:2dnc-matching-little-sch} is UUDHHUHHDUUDDD.

\subsection{Orthogonal polynomials and weighted Motzkin paths}
\label{sec:orth-polyn-weight}

When one has a sequence of positive numbers, in many cases it is
possible to describe that sequence as the moments of a sequence of
orthogonal polynomials.
In other words, given $\{\mu_{n}\}_{n \ge 0}$, define a measure (or a
linear functional on polynomials; the two are equivalent here) by
$\int x^{n} \,\mathrm{d} \mu = \mu_{n}$ and find polynomials $\{P_{n}(x)\}_{n
\ge 0}$ so that the integral
\begin{displaymath}
  \int P_{n}(x) P_{m}(x) \,\mathrm{d} \mu = 0
\end{displaymath}
when $n \ne m$ and is nonzero when $n = m$.
Many classical combinatorial sequences produce sequences of orthogonal
polynomials: the Catalan numbers produce Chebyshev polynomials of the
second kind, matching numbers produce Hermite polynomials, factorials
produce Laguerre polynomials, and so on.

Viennot described a completely combinatorial theory of orthogonal
polynomials ~\cite{viennot:theorie,viennot:combinatorial} in which the
moments of a sequence of orthogonal polynomials are expressed as
weighted Motzkin paths.
A Motzkin path is a lattice path that consists of upsteps, downsteps
and single horizontal steps (steps that move $(1,0)$); a weighted
Motzkin path has a weight $\lambda_ {n}$ associated with every downstep
leaving from height $n$, a weight $b_{n}$ for every horizontal step at
height $n$, and weight $1$ for all upsteps.
For many orthogonal polynomial moment sequences, the weights $b_{n}$
are zero, which means the corresponding moments may be described by
weighted Dyck paths; a \textsl{Dyck path} is just a Motzkin path with
no horizontal steps.

Drake and Kim~\cite{drake.kim:k-distant} showed that the number of
$2$-distant noncrossing matchings of $[2n]$---little \Sch numbers---is
the same as the total weight of weighted Dyck paths of length $2n$ in which
downsteps leaving from odd height have weight $1$, and downsteps leaving
from even height have weight $2$.
They proved this equality using equation (2) of Kim and Zeng
~\cite{kim.zeng:combinatorics} (or equation (1) of Vauchassade de
Chaumont and Viennot ~\cite{chaumont.viennot:polynomes}), which in the
present context is
\begin{equation}
  s_{n} = \sum_{k \ge 0} \frac{1}{n} \binom{n}{k} \binom{n}{k+1} 2^{k};
  \label{eq:little-sch-sum-of-narayana}
\end{equation}
in both works, the authors demonstrate that the sum above represents the
generating function for the weighted Dyck paths described above.
However, the sum also counts little \Sch paths, since
$\binom{n}{k}\binom{n}{k+1}/n$ is a \textsl{Narayana number} (sequence
A1263), which counts Dyck paths of length $2n$ with $k+1$ peaks and
$k$ ravines.
A peak is an upstep immediately followed by a downstep, and a ravine
is a downstep immediately followed by an upstep.
Between two consecutive peaks, there must be exactly one ravine, so
having $k+1$ peaks is equivalent to having $k$ ravines.
Any ravine can clearly be ``filled in'' and replaced with a horizontal
step, so a Dyck path with $k$ ravines corresponds to $2^{k}$ little
\Sch paths, which explains equation~\eqref{eq:little-sch-sum-of-narayana}.
On the other hand, any peak can be ``flattened'' into a horizontal
step, so we also have
\begin{equation}
  S_{n} = \sum_{k \ge 0} \frac{1}{n} \binom{n}{k} \binom{n}{k+1} 2^{k+1}
  \label{eq:big-sch-sum-of-narayana}
\end{equation}
because peaks can occur on the $x$-axis.
This provides one explanation for why there are twice as many big \Sch
paths as little ones.

\subsection{Plan of the paper}
\label{sec:plan-paper}

The aim of this work is to demonstrate a bijection from weighted Dyck
paths whose downsteps at even height have weight ~$2$ to little \Sch
paths.
That bijection will be a minor modification of a bijection from big \Sch
paths to a similar class of Dyck paths; both bijections will in turn be
consequences of more refined bijections between classes of little and
big hybrid paths, which are described in \autoref{sec:descr-biject}.
In \autoref{sec:hybrid-paths-matchings} we show the effect of those
bijections on the corresponding matchings, and then find generating
functions for little and big hybrid paths in
\autoref{sec:enum-hybr-paths}.
We finish by showing that our bijections are closely related to
$231$-avoiding permutations in \autoref{sec:hybrid-paths-231}.

\section{Description of the bijection}
\label{sec:descr-biject}

Instead of working with Dyck paths in which downsteps from even height
have weight $2$, we will work with Dyck paths in which such downsteps
may or may not be labeled ``special''; the two ideas are clearly
equivalent.
Such paths will be called \textsl{even-special Dyck paths} and abbreviated
``ESDPs''; \textsl{odd-special Dyck paths} (ODSPs) are defined similarly.

We will first describe a bijection $E_{\infty}$ from odd-special Dyck
paths to big \Sch paths---our desired bijection from even-special Dyck
paths to little \Sch paths will follow from a minor modification of
that bijection.
The bijection $E_{\infty}$ will be a consequence of a more refined
bijection $E$ between two classes of what we will call big hybrid
paths.
\textsl{Big hybrid paths} include odd-special Dyck paths and any path
obtained by applying $E$ to a big hybrid path.
To understand this recursive definition, we must define the map $E$.

\begin{definition}
  \label{def:e-bijection}
  Given a hybrid path, the map $E$ does nothing to the path if the path
  contains no special steps.
  Otherwise, given a hybrid path with $k$ horizontal steps, $E$ yields a
  hybrid path with $k+1$ horizontal steps by the following procedure.
  Find the leftmost special step in the hybrid path.
  If that step is preceded by an upstep, \textsl{flatten} the upstep and
  special downstep by replacing them with a horizontal step.
  If the special step is preceded by a downstep $d$, find the upstep $u$
  that matches $d$ and let $P$ be the (possibly empty) subpath between
  $u$ and $d$.
  Replace $u$ with a horizontal step, delete $d$, \textsl{slide} $P$ so
  that it follows the horizontal step, and make the original special step
  an ordinary downstep.
\end{definition}

\begin{figure}
  \centering
  \subfloat[When the special step is preceded by an upstep, $E$ and $e$
  simply ``flatten'' the two steps.]{
  \label{fig:e-bijection-flatten}

  \parbox{8.25cm}{\centering \inputtikz{e-bijection-flatten}}
}

  \subfloat[When the special step is preceded by a downstep, $E$ and $e$
  find the matching up step and ``slide'' $P$.]{
  \label{fig:e-bijection-slide}
  \inputtikz{e-bijection-slide}
}

\caption{The action of bijections $E$ and $e$ on the leftmost special
  down step in a hybrid path.}
  \label{fig:e-bijection}
\end{figure}

\autoref{fig:e-bijection} demonstrates the flatten and slide operations.
All the paths in \autoref{fig:E-forward-backward} are big hybrid paths.

The map $E$ clearly preserves the total number of special and horizontal
steps and, for paths with at least one special step, reduces the number
of special steps by one.
It is also a bijection:

\begin{theorem}
  The map $E$ is a bijection from the set of odd-special Dyck paths of
  length $n$ with no special steps to the set of big \Sch paths of
  length $n$ with no horizontal steps.
  It is also a bijection from the set of big hybrid paths of length
  $n$ with $j$ special steps and $k$ horizontal steps to the set of
  big hybrid paths with $j-1$ special steps and $k+1$ horizontal steps.
  \label{thm:E-is-a-bijection}
\end{theorem}

We will show that $E$ is a bijection by describing a procedure for
finding the horizontal step that was added last; the operation described
in \autoref{def:e-bijection} and \autoref{fig:e-bijection} is obviously
reversible if we know which horizontal step was added last.
Before giving the proof, let's see why this identification is not as
simple as it may sound.
The problem is that sometimes $E$ ``moves forward'' and sometimes $E$
``moves backward''.
\autoref{fig:E-forward-backward} shows what we mean by this.
A horizontal step may be created by $E$ to the left, to the right, or in
the middle of the existing horizontal steps, so the left- or rightmost
horizontal step need not be the last one added.

\begin{figure}[h]
  \centering
  \subfloat[The horizontal steps added by $E$ move ``backwards'' when
  doing repeated slide operations.]{
  \label{fig:E-slide-backward-ex}
  \inputtikz{E-slide-ex}}

  \subfloat[The horizontal steps added by $E$ move ``forwards'' when
  doing repeated flatten operations.]{
  \label{fig:E-flatten-forward-ex}
  \inputtikz{E-flatten-ex}}

  \subfloat[Horizontal steps can also be added between existing horizontal
  steps.]{
  \label{fig:E-slide-middle-ex}
  \inputtikz{E-slide-middle-ex}}

  \caption{The horizontal steps created by $E$ are not necessarily added
    left-to-right. The three paths on the far right look similar, but
    their horizontal steps were added in different orders.}
  \label{fig:E-forward-backward}
\end{figure}

One may think that, since horizontal steps from slides are always
created at odd height and horizontal steps from flattenings are created
at even height, it might be possible to use that information to identify
the last-added step, but since slides change the height of parts of the
path by one, a simple examination of odd and even heights will not
suffice.

\begin{proof}[Proof of \autoref{thm:E-is-a-bijection}]
  The first statement of the theorem is trivial, as it is saying that
  $E$ acts as the identity on the set of Dyck paths.
  For the second statement, we must show that it is possible to identify
  which horizontal step was added last.
  This can be done with the following procedure.

  Partition the path into subpaths that consist of either a sequence
  of non-horizontal steps, or a horizontal step, its matching step, and
  all steps in between.
  The only part of a path altered by $E$ when adding a horizontal step
  is between the horizontal step and its matching downstep, so if a
  horizontal step~$b$ is to the right of the downstep matching a
  horizontal step~$a$, then $b$ must have been added after ~$a$.
  (This is a special case of \nonthmref{thm:21-order-adding-sp-steps}.)
  This fact tells us that the last-added horizontal step must be in the
  rightmost such subpath that contains a horizontal step.
  Call that subpath the \textsl{first active subpath}.
  \autoref{fig:partition-by-hh} illustrates this partitioning process.

  \begin{figure}[ht]
    \centering
    \inputtikz{partition-by-hh}
    \caption{The partitioning process to find the first active subpath,
      which is the rightmost subpath with a horizontal step.}
    \label{fig:partition-by-hh}
  \end{figure}

  If the first active subpath starts with a horizontal step on the
  $x$-axis, then, because horizontal steps on the $x$-axis can only be
  created with a flatten operation, the rightmost horizontal step in the
  subpath must be the last-added step.

  Otherwise, we may assume the first active subpath starts with a
  horizontal step at some positive height.
  If that step is at odd height, that step is the last-added
  horizontal step, because the slide operation of $E$ creates horizontal
  steps at odd height, and, as seen in
  \autoref{fig:E-slide-backward-ex}, as one moves forward along a
  sequence of downsteps, some of which are special steps, $E$ creates
  horizontal steps at the beginning of the first active subpath.

  If the step at the beginning of the first active subpath is at even
  height, we must partition the path again.
  Now partition the first active subpath into sequences of horizontal
  steps at the same height as the original horizontal step and subpaths
  that begin with an upstep and end at the downstep matching the upstep.
  Call these two kinds of sequences \textsl{valleys} and \textsl{hills},
  respectively.
  Using the same reasoning as before, the last-added horizontal step
  must be in the rightmost hill or valley that contains a horizontal step.
  Call that hill or valley the \textsl{second active subpath}.
  In the subpath of \autoref{fig:hills-valleys}, the final hill is the
  second active subpath.

  \begin{figure}[h]
    \centering
    \inputtikz{partition-by-hills-valleys}
    \caption{Partitioning the first active subpath into hills and
      valleys. The second active subpath is the rightmost hill or valley
      with a horizontal step.}
    \label{fig:hills-valleys}
  \end{figure}

  If the second active subpath is a valley, the rightmost step in the
  valley is the most recently added horizontal step because steps in a
  valley must come from the flattening operation.

  If the second active subpath is a hill, we recursively use the
  procedure described here to identify the last-added step within that
  hill.
  Since the hill begins at even height, the path is of the same form as
  the hybrid paths we began with.

  Since the paths have finite length and the recursion step uses a
  shorter path than it started with, this procedure always finishes, and
  since the ``exit points'' always identify what must be the most
  recently added step, the procedure as a whole will identify the
  last-added horizontal step of a hybrid path.
\end{proof}

For example, with the second active subpath in
\autoref{fig:hills-valleys}, we would recursively use the procedure on
the final hill.
The procedure in the proof above, given that hill (UUHHDd) as a single
path would identify HHD as the first active subpath then, since the
horizontal step is at even height, partition again and identify the
valley HH as the second active subpath, and finally declare that single
horizontal step as the most recently added horizontal step.
In \autoref{fig:e-bijection-ex}, step $11$ is the last-added
horizontal step.

\subsection{Consequences of the bijection}
\label{sec:consequences-of-E}

If we start with an odd-special Dyck path, we can use $E$ to iteratively
``evolve'' the path into a big \Sch path.
(In fact, we use $E$ to suggest the word ``evolve''.)
Let $E_{\infty}$ be the resulting map from odd-special Dyck paths to big
\Sch paths.
Since $E$ is a bijection and preserves the total number of special
steps and horizontal steps, we have the following corollary of
\autoref{thm:E-is-a-bijection}.

\begin{corollary} The map $E_{\infty}$ is a bijection from odd-special
Dyck paths of length $n$ with $k$ special steps to big \Sch paths of
length $n$ with $k$ horizontal steps.
  \label{thm:E-infinity-is-a-bijection}
\end{corollary}

The operation described in \autoref{def:e-bijection} and
\autoref{fig:e-bijection} does not refer to the parity of the heights
of the special steps, so we may use it with even-special Dyck paths.
Define the map $e$ the same way as $E$, but starting with even-special
Dyck paths.
\textsl{Little hybrid paths} are defined analogously to big hybrid
paths.
By simply switching ``odd'' and ``even'' in the proof of
\autoref{thm:E-is-a-bijection} and ignoring the possibility of
sequences of horizontal steps on the $x$-axis, we see that $e$ is also
a bijection:

\begin{corollary}
  The map $e$ is a bijection from the set of even-special Dyck paths of
  length $n$ with no special steps to the set of little \Sch paths of
  length $n$ with no horizontal steps.
  It is also a bijection from the set of little hybrid paths of length $n$ with
  $j$ special steps and $k$ horizontal steps to the set of hybrid paths
  with $j-1$ special steps and $k+1$ horizontal steps.
  \label{thm:e-is-a-bijection}
\end{corollary}

By defining $e_{\infty}$ analogously to $E_{\infty}$, we accomplish
our goal of showing bijectively that the little \Sch numbers enumerate
even-special Dyck paths of length $n$:

\begin{corollary}
  The map $e_{\infty}$ is a bijection from even-special Dyck paths of
  length $n$ with $k$ special steps to little \Sch paths of length $n$
  with $k$ horizontal steps.
  \label{thm:e-infinity-is-a-bijection}
\end{corollary}

\autoref{fig:e-bijection-ex} shows an example of $e_{\infty}$.

\begin{figure}[h]
  \centering
  \inputtikz{e-bijection-ex}
  \caption{An example of the bijection $e_{\infty}$.}
  \label{fig:e-bijection-ex}
\end{figure}

Using the reasoning behind equations
\eqref{eq:little-sch-sum-of-narayana} and
\eqref{eq:big-sch-sum-of-narayana}, which counted \Sch paths by changing
peaks or ravines in Dyck paths into horizontal steps, the bijections
above imply that the number of even-special Dyck paths of length $2n$
with $j$ special steps is, for positive $n$,
\begin{equation}
  \sum_{k \ge 0} N(n,k) \binom{k}{j}
  =
  N(n, j)\,\twofone{j-n}{j-n+1}{j+2}
  =
  \frac{1}{n}\binom{n}{j}\binom{2n-j}{n+1},
  \label{eq:esdps-with-j-sp}
\end{equation}
where $N(n, k)$ is again a Narayana number and the ${}_{2} F_{1}$
notation is a hypergeometric function evaluated at one, which we can sum
with the Chu-Vandermonde identity.
The above triangle of numbers is sequence A126216.
Similarly, the number of odd-special Dyck paths of length $2n$ with
$j$ special steps is, for positive $n$,
\begin{equation}
  \sum_{k \ge 0} N(n,k) \binom{k+1}{j}
  =
  N(n, j-1)\,\twofone{j-n}{j-n-1}{j}
  =
  \frac{1}{n-j+1}\binom{n}{j}\binom{2n-j}{n};
  \label{eq:osdps-with-j-sp}
\end{equation}
the middle expression is not defined when $j=0$, but in that case, the
sum on the left is just the sum of the Narayana numbers---a Catalan
number---so the rightmost expression is correct for all nonnegative $j$.
The triangle in equation \eqref{eq:osdps-with-j-sp} is sequence A60693.

\section{Hybrid paths as matchings}
\label{sec:hybrid-paths-matchings}

This work began with an investigation of certain matchings, and since
little hybrid paths were developed to describe our bijection, it is
fitting that we examine the connection between little hybrid paths and
matchings.
We already know that little \Sch paths correspond to $2$-distant
noncrossing matchings, so first we will describe an interpretation
of even-special Dyck paths.
Using the bijection from \Sch paths to matchings, a Dyck path with no
special steps corresponds to a noncrossing matching, so it is reasonable
to interpret special steps in the path as special edges in the matching.
For example, the path UUdUUUDDdD corresponds to the noncrossing
matching $\{(1,10), \spthing{(2,3)}, \spthing{(4,9)}, (5,8), (6,7)\}$
in which the edges between $2$ and $3$ and between $4$ and $9$ are
special.

To interpret paths with both special steps and horizontal steps and
understand the action of $e$ in terms of paths, we need to define
nesting.
An edge $(a,b)$ in a matching \textsl{nests} the edge $(c,d)$ if $a <
c < d < b$.
An edge $a$ in a matching \textsl{immediately nests} edge $b$ if $a$
nests $b$, and any other edge that nests $b$ also nests $a$.

Before describing the action of the flatten and slide operations on
``hybrid matchings'', we need one observation.

\begin{lemma}
  Let $h$ be a double horizontal step in a little hybrid path.
  Let $d$ be the downstep matching $h$ and $u$ the upstep matching $d$.
  The step in the path corresponding to the rightmost (respectively,
  leftmost) vertex involved in the $1$-distant crossing at $h$ is either
  $d$ (resp., $u$) or the leftmost (resp., rightmost) horizontal step to
  the right (resp., left) of $h$ which is at the same height as $h$,
  whichever is closer to $h$.
  \label{thm:horiz-to-crossing-matching-steps}
\end{lemma}

\begin{proof}
  A key idea in this proof is that a sequence of steps in a hybrid
  path that begins with an upstep and ending with the matching
  downstep---what we called a hill in the proof of
  \autoref{thm:E-is-a-bijection}--- corresponds to a set of vertices
  in the matching that form a ``submatching''.
  Let $h$ be a double horizontal step in a little hybrid path, and
  partition the path as we did to find the second active subpath in the
  proof of \autoref{thm:E-is-a-bijection} (but ignore the height of
  $h$).
  The first step in $h$ corresponds to an opening half edge $x$.
  To what vertex will $x$ be connected?
  Any hill to the right of $h$ corresponds to a group of vertices that
  form a submatching, and hence $x$ will not be connected to any of
  them.
  If there is a double horizontal step between $h$ and $d$, then $x$
  will be connected to the vertex corresponding to the second step of the
  leftmost such double horizontal step; otherwise, $x$ will be connected
  to the vertex corresponding to $d$.
  The proof for the ``respectively'' part of the statement is similar.
\end{proof}

For example, in \autoref{fig:2dnc-matching-little-sch}, the horizontal
step at positions $4$ and $5$ corresponds to the $1$-distant crossing in
the matching in the same position; step $14$ in the \Sch path is the
downstep that matches the horizontal edge, so the right vertex of the
edge incident to vertex $4$ is vertex $14$; and step $1$ in the path is
the upstep that matches step $14$, so vertex $1$ is the left vertex of
the edge incident to vertex $5$.
Another example is UHHUDHHD; vertex $2$ in the corresponding matching
(which is $\{(1, 3), (2, 7), (4, 5), (6, 8)\}$) connects to vertex $7$
because the second horizontal step is at the same height as the first,
and is between the first horizontal step and its matching downstep.

Now we can describe the corresponding action of the bijection $e$ on
hybrid matchings.

\begin{figure}[ht]
  \centering
  \subfloat[The flatten operation when applied to a matching just ``swaps
  tails''.]{
  \label{fig:e-bijection-matching-flatten}
  \inputtikz{e-bijection-matching-flatten}}

  \subfloat[The slide operation when applied to a matching. Note that the
  edge from $c$ to $b-1$ may actually be several edges, as in UUUHHDdD,
  where $b-1$ is $6$ and $c$ is $3$.]{
  \label{fig:e-bijection-matching-slide}
  \inputtikz{e-bijection-matching-slide}}

  \caption{The effect of $e$ on matchings. This is the matchings version
  of \autoref{fig:e-bijection}.}
  \label{fig:e-bijection-matchings}
\end{figure}

\begin{theorem}
  The analogue of the flatten operation for hybrid paths works as
  follows on matchings: given a special edge connecting vertices $c$ and
  $c+1$, find the edge that immediately nests that special edge; say it
  connects $a$ and $b$.
  Then ``swap the tails'': replace the edges $(c,c+1)$ and $(a,b)$ with
  edges $(c,b)$ and $(a,c+1)$.
  \label{thm:e-matchings-flatten}
\end{theorem}

\begin{proof}
  The special downstep in the path (which is immediately preceded by an
  upstep, since we are doing a flatten operation) becomes a horizontal
  edge.
  So there will be $1$-distant crossing at vertices $c$ and $c+1$.
  We only need to find the other two vertices involved; because of the
  way matchings are constructed from paths and
  \nonthmref{thm:horiz-to-crossing-matching-steps}, those other two
  vertices are $a$ and $b$.
\end{proof}

To define the analogue of the slide operation, we need to define the
\textsl{transitive left endpoint} of an edge.
Given an edge $e$, the transitive left endpoint of that edge is simply
the left endpoint of $e$---unless the edge is the right edge in a
$1$-distant crossing; then the transitive left endpoint of $e$ is the
transitive left endpoint of the left edge in the crossing.
For example, the transitive left endpoint of the edge $(6,8)$ in the
matching $\{(1,5), (2,3), (4,7), (6,8)\}$ is $1$. In little hybrid
paths, the transitive left endpoint corresponds to finding the upstep
that matches a downstep; because of
\nonthmref{thm:horiz-to-crossing-matching-steps}, a matching
upstep-downstep pair may not correspond to the left and right vertices
of a single edge.

\begin{theorem}

  The analogue of the slide operation for paths works as follows on
  matchings: given a special edge $(a,b)$, let $c$ be the transitive left
  vertex of the edge incident to vertex $b-1$.
  The slide operation on hybrid paths corresponds to replacing $(a,b)$
  and $(c,b-1)$ with edges $(a,c+1)$ and $(c,b)$ and sliding all half edges
  incident to vertices from $c+1$ to $b-2$ to the right by one
  vertex.
  \label{thm:e-matchings-slide}
\end{theorem}

\begin{proof}
  Since we are doing a slide operation, the special step in the path
  must be preceded by a ordinary downstep, which means $b-1$ must be the
  right vertex of an ordinary edge nested by the special edge.
  The new $1$-distant crossing created by the slide operation will be at
  the upstep that matches the downstep at $b-1$, which as we saw above
  is the transitive left endpoint of the edge incident to $b-1$ in the
  matching. We create the new crossing at vertices $c$ and $c+1$; the
  slide operation on the path moves all steps from $c+1$ to $b-2$ to
  $c+2$ to $b-1$, so all half edges incident to vertices from $c+1$ to
  $b-2$ are moved to the right one vertex.
\end{proof}

\autoref{fig:e-bijection-matchings} demonstrates these two operations
for matchings.

\section{Enumeration of hybrid paths}
\label{sec:enum-hybr-paths}

Having defined and used hybrid paths it is natural to wonder just many
of them there are.
All big \Sch paths and odd-special Dyck paths are big hybrid paths, and
Dyck paths, which are counted by the Catalan number $C_{n}$, are both
big \Sch paths and OSDPs, so there are certainly at least $2 S_{n} -
C_{n}$ big hybrid paths, but there are paths such as HHUd which are
neither \Sch paths nor odd-special Dyck paths.
\autoref{tab:hybrid-paths} shows the number of all hybrid paths,
little and big, for some small values of $n$.

\begin{table}[ht]
  \centering
  \begin{tabular}{lrrrrrrrrrrrr}
   $n$:  & 0   &  2  &   4  &   6  &    8  &    10  &    12  &     14  &
   16  &      18  &       20  & 22\\
  \hline
   little:  & 1 &  1  &   4  &  18  &   87  &   439  &  2278  &  12052  &
   64669  &  350733  &  1918152  &  10560678\\
   big:  & 1   &  3  &  11  &  47  &  219  &  1075  &  5459  &  28383  &
   150131  &  804515  &  4355163  & 23768079\\
  \end{tabular}

  \caption{The number of little and big hybrid paths.}
  \label{tab:hybrid-paths}
\end{table}

One way to count hybrid paths is to begin with even- and odd-special
Dyck paths with $j$ special steps, which are counted in equations
\eqref{eq:esdps-with-j-sp} and \eqref{eq:osdps-with-j-sp}; repeatedly
applying $e$ or $E$ to such a Dyck path will produce $j$ hybrid paths.
Multiplying those equations by $j+1$ and summing over $j$ yields, for
little hybrid paths,
\begin{equation}
  C_{n}\,\threeftwo{-n}{-n+1}{2}{-2n}{1}{-1},
  \label{eq:little-hyb-3f2}
\end{equation}
where $C_{n}$ is a Catalan number and the hypergeometric function is now
evaluated at $-1$. Similarly, the number of big hybrid paths is
\begin{equation}
  C_{n}\,\threeftwo{-n}{-n-1}{2}{-2n}{1}{-1}.
  \label{eq:big-hyb-3f2}
\end{equation}

Another way to enumerate these paths is to find their generating
functions.
Let $L(x)$ and $B(x)$ be the generating functions for little and big
hybrid paths, respectively.
We will use the following generating functions: $E(x)$ and $O(x)$ for
even- and odd-special Dyck paths, and $s(x)$ and $S(x)$ for little and
big \Sch paths.
Of course, we already know that $E(x) = s(x)$, $O(x) = S(x)$, $S(x) = 2
s(x) - 1$, and
\begin{displaymath}
  s(x) = \frac{2}{1 + x + \sqrt{x^2 - 6x + 1}},
\end{displaymath}
but it will be helpful to use different names to keep different types of
paths separate.
In all the generating functions considered here, paths of length $2n$
are weighted by $x^{n}$.

\begin{theorem}
Let $R = \sqrt{x^{2} - 6x + 1}$. The ordinary generating function
for little hybrid paths is
\begin{equation}
  L(x) = \frac{R+1-x}{2} \cdot
         \frac{2(R+x)}{R (R+x+1)} \cdot
         \frac{2}{R+x+1}
  \label{eq:little-hybrid-gf-R}
\end{equation}
and the ordinary generating function for big hybrid paths is
\begin{equation}
  B(x) = \left(\frac{R+1-x}{2} - 1 + x -
               \frac{R(R+x)}{2} + \frac{3}{2} - \frac{3x}{2} \right)
         \frac{2(R+x)}{R (R+x+1)} \cdot
         \frac{2}{R+x+1}.
\label{eq:big-hybrid-gf-R}
\end{equation}
More explicitly, we have
\begin{equation}
  L(x) = \frac{1 -5x + \sqrt{x^{2} -6x +1}}
{\left(
(x-1)^{2} + (x+1)\sqrt{x^{2} -6x+1}
\right) \sqrt{x^{2}-6x+1}}
\label{eq:little-hybrid-gf}
\end{equation}
and
\begin{equation}
  B(x) = \frac{7x - 2x^{2} - 1 + \sqrt{x^{2} - 6x + 1}}{2x \sqrt{x^{2}
      - 6x + 1}} - 1.
\label{eq:big-hybrid-gf}
\end{equation}
\end{theorem}

\begin{proof}
  We will decompose little and big hybrid paths to express $L$ and $B$
  in terms of each other and then solve the system.
  A key idea is that raising a big hybrid path up one unit and
  sandwiching it between and upstep and downstep yields a valid little
  hybrid path, since all the step height parities have effectively been
  reversed.
  Doing the same thing to a little hybrid path yields a big hybrid path,
  although the resulting path will never have a horizontal step at height
  ~$1$.

  Every nonempty little hybrid path may be decomposed into an upstep, a
  big hybrid path $P$, a downstep, then a little hybrid path $Q$, as shown
  in \autoref{fig:little-hybrid-decomp}.
  Any pair $P$ and $Q$ is allowed, unless $P$ has a special step and $Q$
  has a horizontal step.

  \begin{figure}[h]
    \centering
    \inputtikz{little-hybrid-decomp}
    \caption{A decomposition of a nonempty little hybrid path.}
    \label{fig:little-hybrid-decomp}
  \end{figure}

  Assume that $P$ has a special step, so that $Q$ has no horizontal
  step.
  Since $S(x)$ counts big hybrid paths with only upsteps, downsteps, and
  horizontal steps---in other words, with no special steps---the
  generating function for big hybrid paths with a special step is $B(x) -
  S(x)$; similar reasoning shows that the generating function for little
  hybrid paths with no horizontal steps is simply $E(x)$.
  Thus the generating function for little hybrid paths with a special step
  in their first components is $x (B(x) - S(x)) E(x)$.

  On the other hand, if $P$ doesn't have a special step, then it is a big
  \Sch path and $Q$ can be any little hybrid path.
  The generating function for little hybrid paths with no special step
  in their first components is therefore $x S(x) L(x)$.

  Every nonempty little hybrid path can be uniquely decomposed in this way
  and falls into exactly one of the above categories, so adding in the
  empty path we have
  \begin{displaymath}
    L(x) = 1 + x (B(x) - S(x)) E(x) + x S(x) L(x),
  \end{displaymath}
  or, solving for $L$,
  \begin{equation}
    L(x) = \frac{1 + x ( B(x) - S(x)) E(x)}{1 - x S(x)}.
    \label{eq:L-gf-in-terms-of-B}
  \end{equation}

  The decomposition for big hybrid paths is slightly more involved.
  Given a big hybrid path with an upstep, let $s$ be the first downstep
  to return to the $x$-axis and decompose the path as in
  \autoref{fig:big-hybrid-decomp}.
  In any such big hybrid path, either there is or is not a special
  step before $Q$.

  \begin{figure}[h]
      \centering
      \inputtikz{big-hybrid-decomp}
      \caption{A decomposition of a big hybrid path with an upstep. The
        path begins with a possibly empty sequence of horizontal steps.
        The step $s$ is the first downstep to return to the $x$-axis;
        it may or may not be a special step.}
      \label{fig:big-hybrid-decomp}
  \end{figure}

  Assume that there is a special step before $Q$, and that $s$ is special.
  In that case, $P$ can be any little hybrid path, because if $s$ is
  special, we know that no horizontal steps can appear in $P$ at height
  $1$---such a step has $s$ as its matching downstep and can only
  be created with a slide operation that would make $s$ an ordinary
  step.
  Since there are no horizontal steps at height $1$, $P$ can be any
  little hybrid path, and since there are special steps preceding $Q$, it
  cannot have any horizontal steps, and hence is an odd-special Dyck path.
  The generating function for the possibly empty sequence of horizontal
  steps at the beginning is $1/(1-x)$, so the generating function for
  paths of this type is $x L(x) O(x) / (1 - x)$.

  If a special step appears before $Q$ and $s$ is ordinary, then $P$
  must have a special step and cannot have a horizontal step at height
  ~$1$.
  Assume there is such a horizontal step ~$h$.
  The step $h$ is at odd height and its matching step is $s$, so the
  only way $h$ could be created is by a slide operation that converts $s$
  from a special to an ordinary step, but since we process special steps
  left to right, $s$ would be converted from special to ordinary only if
  there were no special steps in $P$, which is a contradiction.
  This means that $P$ can be any little hybrid path with a special step;
  the generating function for such paths is $L(x) - s(x)$.
  The subpath $Q$ can therefore be any odd-special Dyck path, so the
  generating function for all such big hybrid paths is $x (L(x) - s(x))
  O(x) / (1-x)$.

  Finally, if there is no special step preceding $Q$, then $P$ can be any
  big \Sch path, and $Q$ can be any big hybrid path.
  The generating function for such big hybrid paths is $x S(x) B(x) / (1
  -x)$.

  Every big hybrid path with an upstep falls into exactly one of the
  categories above, so, including paths that consist only of a sequence
  of horizontal steps on the axis, we have
  \begin{displaymath}
    B(x) = \frac{1}{1-x} + \frac{x L(x) O(x)}{1-x} + \frac{x (L(x) - s(x))
      O(x)}{1-x} + \frac{x S(x) B(x)}{1 -x},
  \end{displaymath}
  or, solving for $B$,
  \begin{equation}
   B(x) = \frac{1 + x L(x) O(x) + x (L(x) - s(x))}{1 - x - x S(x)}.
  \label{eq:B-gf-in-terms-of-L}
  \end{equation}

  Solving the system of equations \eqref{eq:L-gf-in-terms-of-B} and
  \eqref{eq:B-gf-in-terms-of-L} and using the fact that
  \begin{gather*}
    s(x) = E(x) = \frac{2}{1+x+R} = \frac{1+x-R}{4x} \quad \text{and}\\
    S(x) = O(x) = \frac{4}{1+x+R} - 1 = \frac{1+x-R}{2x} - 1,
  \end{gather*}
  we obtain the desired expressions for $L(x)$ and $B(x)$.
\end{proof}

It may seem that the generating functions $L$ and $B$ were described in
equations \eqref{eq:little-hybrid-gf-R} and \eqref{eq:big-hybrid-gf-R}
in an unusual way, but the expressions show that $L$ and $B$ are in some
sense built out of familiar generating functions for paths:
\begin{displaymath}
  \frac{2(R+x)}{R (R+x+1)} = 1 + 2x + 7x^{2} + 30x^{3} + 141x^{4}
  + \cdots
\end{displaymath}
is the generating function for sequence A116363, which counts dot
products of rows of Pascal's and Catalan's triangle, and of course
$2/(R+x+1)$ is the generating function for the little \Sch numbers.
Also appearing in both $L$ and $B$ is
\begin{displaymath}
  \frac{R+1-x}{2} = 1 - x - x S(x),
\end{displaymath}
a minor modification of the generating function for the big \Sch
numbers.
In $B$, we see that we have exactly $-x S(x)$; the remaining terms in
$B$ are
\begin{displaymath}
  -\frac{R(R+x)}{2} + \frac{3}{2} - \frac{3x}{2}
\end{displaymath}
which is $1 + x + x^{2} S(x)$.

\section{The bijections $E$ and $e$ and 231-avoiding permutations}
\label{sec:hybrid-paths-231}

While using the bijections $E$ or $e$, one can keep track of the order
in which horizontal steps are added and thereby associate a permutation
to an even- or odd-special Dyck path.
For example, the paths in \autoref{fig:E-forward-backward} correspond to
the permutations $321$, $123$, and $132$; the path in
\autoref{fig:e-bijection-ex} corresponds to $12387465$.
In this section we will see that every permutation so obtained must avoid
the pattern $231$. This is very interesting, since $231$-avoiding
permutations are counted by the Catalan numbers and hence are in
bijection with Dyck paths; see Mansour et al. \cite[\S
3.1]{mansour.deng.ea:dyck}.

A permutation $\pi = \pi_{1}\pi_{2}\cdots \pi_{n}$ written in one-line
notation \textsl{contains} a pattern $\sigma$ (another permutation) if
there is some subset of the $\pi_{i}$'s that are order-isomorphic to
$\sigma$.
A permutation \textsl{avoids} a pattern if it does not contain it.
The permutation $12584367$ contains the pattern $231$ because the subset
$583$ is order-isomorphic to $231$, and avoids the pattern $3124$.
The notation $S_{n}(231)$ refers to the set of $231$-avoiding
permutations of $[n]$.

We start with a lemma that tells us exactly when the horizontal steps
created by two special steps are added out of order---that is, when
two special steps create the pattern $21$.
Given a special step $s$, let $h(s)$ refer to the horizontal step
created when $s$ is turned into an ordinary step.

\begin{lemma}
  \label{thm:21-order-adding-sp-steps}
  Given two special steps $a$ and $b$ in a hybrid path with $a$ to the
  left of $b$, $h(b)$ is created to the left of $h(a)$ if and only if
  $b$ is preceded by a downstep $d$ and the upstep matching $d$ is to
  the left of $a$.
\end{lemma}

\begin{proof}
  We only need to examine three possibilities: $b$ is preceded by an
  upstep, $b$ is preceded by a downstep whose matching upstep is to
  the right of $a$, and $b$ is preceded by a downstep whose matching
  upstep is to the left of $a$.
  To work through those three cases, we need to use the fact that for
  any special step $s$, $h(s)$ is created to the left of $s$ and to
  the right of the downstep matching $s$.
  (When doing a flatten, ``to the left'' and ``to the right'' are weak
  inequalities, since the horizontal step will be created in the same
  position as those steps.)

  In the first case, if $b$ is preceded by an upstep, then $h(b)$ will
  clearly be to the left of $h(a)$, since $h(b)$ will be created at
  the position of $b$, which is to the right of $a$.

  If $b$ is preceded by a downstep whose matching upstep is to the
  right of $a$, let $u$ be that matching upstep.
  Since $h(b)$ will be created at $u$ and the following step, $h(b)$
  is to the right of $a$ and hence to the right of $h(a)$.

  Finally, if $b$ is preceded by a downstep whose matching upstep is
  to the \emph{left} of $a$, let $u$ be that matching upstep.
  See \autoref{fig:e-bijection-slide}; $b$ would be the special step
  pictured in that figure, $u$ would be the upstep, and $a$ would be
  somewhere in the subpath $P$, and since the upstep matching $a$ is
  also in that subpath, $h(b)$ will be created to the left of $h(a)$.
\end{proof}

With that result, we can easily prove the following theorem.

\begin{theorem}
  \label{thm:231-avoiding}
  The permutation corresponding to the order in which horizontal steps
  are added while transforming an even- or odd-special Dyck path into a
  small or large \Sch path avoids the pattern~$231$.
\end{theorem}

\begin{proof}
  Consider any three special steps $a$, $b$, and $c$ in a hybrid path,
  appearing in that order left to right.
  If these three steps cause the corresponding permutation to contain
  $231$, then we must have $h(b)$, $h(c)$, and $h(a)$ in that order.
  The bijections $E$ and $e$ process special steps left to right, so we
  first create $h(a)$, and then create $h(b)$ to the left of that.
  Now we must have $h(c)$ created to the right of $h(b)$, which means
  by \nonthmref{thm:21-order-adding-sp-steps} either $c$ is preceded by an
  upstep, or is preceded by a downstep whose matching upstep is to the
  right of $b$, but both of those possibilities cause $h(c)$ to be to
  the right of $h(a)$, which is a contradiction.
\end{proof}

The permutations produced are therefore a subset of $231$-avoiding
permutations; next we will see that every such permutation can be
obtained from some even- or odd-special Dyck path.

\begin{theorem}
  \label{thm:hybrid-paths-all-231-avoiding-perms-possible}
  Every $231$-avoiding permutation can be obtained from some
  odd-special Dyck path.
\end{theorem}

\begin{proof}
  Given a $231$-avoiding permutation $\pi = \pi_{1}\cdots \pi_{n}$, we
  use the following recursive procedure to construct an odd-special Dyck
  path that, when using $E_{\infty}$, will create horizontal steps in the
  order specified by $\pi$.

  To begin with, the empty permutation corresponds to the empty path.
  Given a nonempty permutation $\pi$ of length $n$, if $\pi_{n} = n$,
  find the path corresponding to $\pi_{1}\cdots\pi_{n-1}$ and
  append the steps Ud to that path.
  (Recall that U refers to an upstep, and D and d refer to ordinary
  and special downsteps, respectively.)
  We will call this an \textsl{append} operation; it corresponds to the
  flatten operation.
  If a path $P$ corresponds to the permutation $\pi_{1}\pi_{2}\dots
  \pi_{n-1}$, then the path obtained by appending Ud to $P$ will
  correspond to $\pi_{1}\pi_{2}\dots \pi_{n-1} n$.

  If the permutation does not end with $n$, we will need the
  \textsl{lift} operation, which is defined as follows: given an
  odd-special Dyck path of the kind produced by this procedure, find
  an upstep leaving from the $x$-axis and let $Q$ be the subpath
  consisting of that upstep and everything following it.
  The lift operation, illustrated in \autoref{fig:lift}, replaces $Q$
  with a path consisting of two upsteps, then $Q$, then an ordinary
  downstep and a special downstep.
  The lift operation corresponds to the slide operation of $E$ and
  $e$.

  \begin{figure}[h]
      \centering
      \inputtikz{lift}
      \caption{The lift operation. If the resulting path has $n$ special
        steps, in the corresponding permutation all numbers from $P$
        will precede $n$, and all numbers from $Q$ will follow $n$.}
      \label{fig:lift}
  \end{figure}

  \newcommand{\corrto}{\leftrightarrow}

  For ease of description, define \textsl{good insertion} to be the
  operation of inserting $n$ into a $231$-avoiding permutation of
  $[n-1]$ anywhere except at the end so that the resulting
  permutation is also $231$-avoiding.
  We write $P \corrto \pi$ if a path $P$, constructed using the append
  and lift operations, corresponds to the permutation $\pi$.

  Assume $P \corrto \pi$, where $\pi \in S_{n}(231)$.
  Say that we obtain $\pi'$ by good insertion of $n+1$ after the $k$th
  entry of $\pi$.
  We need to show first that we can do a lift operation following the
  $k$th special step of $P$, and second that the corresponding path $P'$
  corresponds to $\pi'$.

  The path Ud corresponds to the permutation $1$, and both claims are
  true for that path-permutation pair.
  To prove the two claims in general, we need two propositions:

  \begin{proposition}
    If $\pi \in S_{n}(231)$, good insertion can be done after the $k$th
    entry of $\pi$ if and only if the first $k$ entries of $\pi$ form a
    permutation of $[k]$.
  \end{proposition}

  \begin{proposition}
    Assume that $P \corrto \pi$.
    The $k$th special step of $P$ ends on the $x$-axis if and only if
    the first $k$ entries of $\pi$ form a permutation of $[k]$.
  \end{proposition}

  The proof of the first is elementary and left to the reader.
  As for the second, let $s$ be the $k$th special step of $P$.
  Assume $s$ ends on the $x$-axis.
  Then by \nonthmref{thm:21-order-adding-sp-steps}, the horizontal step
  for every special step to the right of $s$ will be created to the right
  of $s$, which means the first $k$ entries of $\pi$ form a permutation of
  $[k]$.
  On the other hand, if $s$ does not end on the $x$-axis, then because
  of the definition of the lift operation, there must be a special step to
  the right of $s$ that is immediately preceded by a downstep $d$, with
  the upstep matching $d$ to the left of $s$.
  Therefore, by \nonthmref{thm:21-order-adding-sp-steps}, there will be a
  number bigger than $k$ among the first $k$ entries of $\pi$, so
  $\pi_{1}\cdots \pi_{k}$ will not form a permutation of $[k]$.

  The first claim above is now clear.
  The second claim is also easy to see: say $\pi = \pi_{1}\cdots
  \pi_{n}$ and $\pi'$ is obtained by good insertion after $\pi_{k}$.
  If one lifts $P$ after the $k$th special step, the resulting path $P'$
  will correspond to the permutation $\pi'$ because, following the lift
  operation, the special steps corresponding to $\pi_{1}\cdots\pi_{k}$
  and to $\pi_{k+1}\cdots \pi_{n}$ will be turned into horizontal steps
  in the same order (they are either unchanged or simply raised by $2$
  units), and the final step of $P'$ is a special downstep that will be
  turned into a horizontal step that follows every horizontal step
  created by the first $k$ special steps of $P'$ and precedes every
  horizontal step created by the $(k+1)$st to $n$th special steps.
\end{proof}

\autoref{fig:permutation-recursion} shows an example of this procedure.
One can also obtain every $231$-avoiding permutation with even-special
Dyck paths simply by taking a path produced by this procedure and
sandwiching it between an upstep and an ordinary downstep.
That produces an even-special Dyck path that clearly corresponds to
the same permutation.

\begin{figure}[h]
  \centering
  \inputtikz{permutation-recursion}
  \caption{An example of the recursive procedure to build an OSDP
    corresponding to a $231$-avoiding permutation. Here we see how the
    path for $15324$ is built up from the empty path. Below each path
    is the corresponding permutation.}
  \label{fig:permutation-recursion}
\end{figure}

We close with an interesting conjecture.
The paths produced for $231$-avoiding permutations of $[n]$ are not
all the same length; the lengths range from $2n$ for the path
corresponding to $123\cdots n$ to $4n-2$ for the path corresponding to
$n (n-1)\cdots 2 1$.
An obvious question to ask is: how are the lengths distributed?
In other words, find the coefficients of the polynomial
\begin{displaymath}
  \sum_{\pi \in S_{n}(231)} q^{\pathlen(\pi)},
\end{displaymath}
where $\pathlen(\pi)$ is the length of the path corresponding to $\pi$
using the construction above.
We can also sum that expression over all $n$, since for a given length
there can be only finitely many permutations that correspond to a path
of that length, and ask what generating function we get.

The lengths appear to have the Narayana distribution; see sequence
A1263 and Sulanke~\cite{sulanke:narayana}. The table below shows the
polynomials for some small values of $n$.

\begin{table}[h]
  \centering
  \begin{tabular}{ll|ll}
    $n$ & distribution of lengths & $n$ & distribution of lengths\\
    \hline
    $1$ & $q^{2}$  & 4 & $q^{14} + 6q^{12} + 6q^{10} + q^{8}$ \\
    $2$ & $q^{6} + q^{4}$ & 5 & $q^{18} + 10q^{16} + 20q^{14} + 10q^{12} + q^{10}$ \\
    $3$ & $q^{10} + 3 q^{8} + q^{6}$ & 6 &  $q^{22} + 15q^{20} + 50q^{18} + 50q^{16} + 15q^{14} + q^{12}$ \\
  \end{tabular}
    \caption{The lengths of the OSDPs corresponding to $231$-avoiding
      permutations of $[n]$ appear to be Narayana-distributed.}
  \label{tab:lengths-narayana-dist}
\end{table}

\begin{conjecture}
  The lengths of the paths corresponding to $231$-avoiding permutations
  of $[n]$ using the above construction have the Narayana distribution;
  that is,
  \begin{equation}
      \sum_{\pi \in S_{n}(231)} q^{\pathlen(\pi)} =
      q^{2n} \sum_{k \ge 0} N(n, k) q^{2k},
  \label{eq:conj-dist}
  \end{equation}
  where $N(n,k)$ is a Narayana number. We also have
  \begin{equation}
    \sum_{n \ge 0} \  \sum_{\pi \in S_{n}(231)} q^{\pathlen(\pi)} =
    \frac{1-q^{2}+q^{4}-\sqrt{1-2q^{2}-q^{4}-2q^{6}+q^{8}}}{2q^{4}}.
  \label{eq:conj-gf}
  \end{equation}
\end{conjecture}

The right-hand side of equation~\eqref{eq:conj-gf} is the generating
function for generalized Catalan numbers described by Stein and
Waterman~\cite{stein.waterman:new} (see the $m=1$ column of Table 1),
and by Vauchassade de Chaumont and Viennot
\cite{chaumont.viennot:polynomes}. Those numbers are sequence A4148,
and count secondary structures of RNA molecules according to the number
of bases.

\section{Included Sage code}
\label{sec:included-sage-code}

\newcommand{\myurl}[1]{\href{http://#1}{\texttt{#1}}}

This preprint includes Sage code
(see \myurl{sagemath.org}) for working with the various
paths, bijections, matchings, and permutations described here.
The code is included with the preprint source; visit
\myurl{arxiv.org/abs/1006.1959}, click ``Other formats'', then
``Download source'', and look for the file
\texttt{code\_for\_dyck\_schroeder.sage}.

\bibliographystyle{amsplainurl} \bibliography{dyck-schroeder}

\end{document}

config: arxiv


TITLE: Question about an exact sequence
QUESTION [1 upvotes]: Hello,
I would be glad, if someone could answer a question concerning the following:
http://www.math.uni-bonn.de/people/schroer/preprints/repdim.pdf
On page 5 they show (3)=>(1). The last step is not clear to me.
I wanted to ask, why (and under which general conditions) the Hom-functor can be omitted.
Thanks.

REPLY [4 votes]: For any abelian category $\mathcal{A}$, an object $M\in\mathcal{A}$ is called a generator if, given an injective nonsurjective morphism $U\to V$ in $\mathcal{A}$, there always exists a morphism $M\to V$ that cannot be factorized as $M\to U\to V$.
In particular, an object $M$ in the abelian category of left $A$-modules is a generator provided that the left $A$-module $A$ is a direct summand of a direct sum of copies of $M$.  The converse is not true, however (there are also generators of other types sometimes).
Given a generator $M$ of an abelian category $\mathcal{A}$ and a sequence $X\to Y\to Z$ in $\mathcal{A}$, exactness of the sequence of abelian groups $Hom_{\mathcal A}(M,X)\to Hom_{\mathcal A}(M,Y) \to Hom_{\mathcal{A}}(M,Z)$ at the middle term $Hom_{\mathcal A}(M,Y)$ implies exactness of the sequence $X\to Y\to Z$ at the middle term $Y$.
One can easily prove this claim by considering the injective morphism from the image $U$ of the morphism $X\to Y$ to the kernel $V$ of the morphism $Y\to Z$.

REPLY [2 votes]: This is a general property of generators in Grothendieck categories (see 1.2 in this paper).
In the category of $A$-modules one can be more explicit: Let $S = \text{End}_A(M)$. Then the Hom's can be eliminated by tensoring with $M \otimes_S -$, since $$M\otimes_S \text{Hom}_A(M,N) \cong N$$ is a natrual isomorphism for each $A$-module $N$ (see 2.4c) of the linked paper).


TITLE: Projections of multivariate normal distribution
QUESTION [3 upvotes]: Given a random vector X with the multivariate normal distribution F(X), we know that, for two vectors a and b, the projections $A=\sum_j a_j X_j $  and  $B=\sum_i b_i X_i $ are univariate normal.
I'm interested in the joint distribution of A and B. Is their joint distribution normal? Is the dependence between A and B described only by their correlation? (do they have only linear dependence?)
Thank you for any insight. References are highly appreciated as well.

REPLY [3 votes]: The vector (A,B) consisting of two scalars is a "two-dimensional projection" of the multivairate normal vector X. More generally, if you multiply X with any matrix such that you get a new vector with dimension <= than the original dimension of X, you again get a multivariate normal vector. That is, all the components of the new vector will be normal, and their dependence is determined only by their correlations.


TITLE: Does the Alcubierre drive use negative energy to compress space infront of it?
QUESTION [1 upvotes]: The contraction in most images looks like a gravity well.  Is it that you use negative energy to BOTH contract and expand space or positive and negative to contract (positive energy) and then expand with then negative energy?

REPLY [2 votes]: The geometry of the Alcubierre drive looks like this:

The exotic matter is arranged as a torus, and it compresses spacetime on one side and expands it on the other so the direction of motion is normal to the plane of the torus. No additional normal or exotic matter is required.
One of the lesser appreciated features of the drive should be immediately apparent from the drawing - the drive is symmetrical so how do we decide which way it moves? That's a good question, and the somewhat unsatisfactory answer is that the Alcubierre metric describes the system in motion and doesn't tell us how to get the thing moving in the first place. I've seen claims that by some manipulation of the metric you can show that it can be interpreted as multiplying the original velocity. So you choose the direction of motion by setting your spaceship in motion before you turn on the drive. There are more details on the NASA web site if you fancy tackling the maths.


text:
\chapter{Introduction}

The term \emph{Hamiltonian} is used in this thesis in two distinguished but not entirely disconnected respects. For one, \term{Hamiltonian quantisation} describes the operator formalism as the geometric field theory in continuum is transformed -- quantised -- to an algebraic theory of quantum operators.  The second meaning refers to the time-evolution of the physical systems, which can be written explicitly as a \term{Hamiltonian operator} acting on the relevant entities. Here we treat time as a simple parameter space characterising these dynamics; while acknowledging that this in all likelihood is not a fundamental aspect of physics, it nevertheless is useful simplification from a variety of perspectives.
  \footnote{See \cite{ROVELLI2011} for a take on how time could be an emergent property of a quantum system.}

The switch from field geometry to operator algebra is not the only change brought in by the quantisation: some effects are also seen in the symmetries of the system. The geometric field theory incorporates a simple but far-reaching idea of invariance, which means that some internal components of the theory can be transformed according to certain rules \emph{without affecting the observable outcome}. This is known as the gauge symmetry invariance, although on the outset it has little to do with our intuition of visual symmetry since these symmetric similarities lie in the underlying mathematical formulation rather than in geometric shapes as such. One of the problems -- or properties -- of quantisation is that it can break some of these symmetries as ones moves from the classical to the quantum. The parts of the theory reflecting this are called anomalous and the effect itself an anomaly.

Hamiltonian anomaly is then an anomalous quantisation effect that is seen on the level of quantum operator dynamics. There are several ways to portray this formally, and \emph{one of the aims of this thesis is to put together many of these perspectives}. To this end we also introduce a handful of fundamental mathematical ideas which form the basis for quantisation anomalies. A central argument is that the \emph{symmetry anomalies have a topological origin}, or that they can be best approached by using topological tools such as K-theory and group cohomology. Many of these methods are not particularly new; what could be consider more contemporary is the use of higher structures in the sense of category theory. There is an undergoing process in mathematical and theoretical physics in general to figure out how gauge field theories could look from such a perspective. To this we add a small contribution concerning mostly categorical groups, motivated by the fact that it is in the group theory where one still finds the most appropriate mathematical realisations for symmetry.

Indeed, underlying the gauge symmetry are what are called \term{current groups}. Formally these can be formed of smooth maps $\map{M,G}$ from a smooth manifold $M$ to a Lie group $G$, defining infinite-dimensional Lie groups.
  \footnote{Subject to appropriate conditions: in general, the proper context for spaces of Lie group -valued maps is the category of infinite-dimensional Fréchet Lie groups, see for instance \cite{NEEB2001}.}

  \marginnote{The first physical current algebra was constructed by Murray Gell-Mann (1929--2019) to describe hadrons.}
Their algebraic counterpart $\map{M,\lie{g}}$ consisting of continuous maps to the Lie algebra $\lie{g}$ of the symmetry group $G$ form the setting to what is called \term{current algebra}.
Historically, current algebras started as a formal means to understand the basics of nuclear interaction, leading to the theory of chromodynamics.

Using the current algebra one studies the commutator relations of physical currents. Corresponding to a gauge symmetry one can define current (densities) $j^a(x)$ as elements in the current algebra and then impose equal-time commutator relations akin to
  \footnote{Here the indices $a,b,c$ refer to the generators of the algebra $\lie{g}$, and the arguments $x,y$ are points on the manifold $M$ of dimension $2k+1$. The currents are expressed in a simplified form without space-time indices; we will discuss the more general case in the main text.}
\[
  \comm{j^a(x)}{j^b(y)}_{t=0} = i \lambda^{ab}_c j^c(x) \delta^{2k+1}(x-y) .
\]
It is in these commutators that one also finds the earliest manifestation of the Hamiltonian anomaly: additional terms may appear on the right side of the above equation~\cite{GI1955,SCHWINGER1959}. We show how the appearance of the anomaly in such commutators is derived from the topology of the Dirac Hamiltonians in the case of massless fermion fields. The topological argument also makes it clear how the commutator anomaly is a result of the impossibility of defining a proper vacuum state in the space of observables.

\section{Outline}

The bulk of this thesis is naturally a distillation of already-established research. In Chapter~\ref{chap:gauge} we briefly recap the central ideas in gauge theory and its quantisation, focusing on the fermionic field theory. This is done largely just to set down common notation and to define the geometric backdrop.

In Chapter~\ref{chap:higher} we discuss the use of higher structures in gauge theory. We focus mostly on constructing categorical groups as in Section~\ref{sec:cat_groups}, in which the treatment of $3$-loop group extensions is largely based on an earlier publication~\cite{MN2019}.
  \footnote{The joint publication \cite{MN2019} contains significant contributions by the author.}

Here we explain in more detail an alternative approach which previously was mentioned only in passing (\cite[Rem.~3]{MN2019}). The broader aim beyond these examples is to envision the concept of current groups in this higher framework. We will also give a brief introduction to bundle gerbes, which have been successfully used as a geometric handle to the cohomology classification of anomalies.

The topological nature of quantum Dirac operators is discussed in Chapter~\ref{chap:ktheory}, in which we explore the rudiments of K-theory and operator index theorems. This forms the operator-theoretical foundations for the further treatment of the Hamiltonian anomaly.

Finally in Chapter~\ref{chap:hanomaly}, we review various perspectives on the Hamiltonian anomaly and try to clarify the connections in-between. What can be considered as new is Section~\ref{sec:general_comm}, in which we link the bundle gerbe cohomology to general anomalous commutator terms and derive them explicitly in a simplified example over the spatial manifold $M=S^3$. In this we rely on the topological arguments presented earlier in Chapter~\ref{chap:ktheory}.

Stylistically we aim to strike a balance between the basic treatments of quantum physics and more recent mathematical research. For instance, in Chapter~\ref{chap:gauge} we introduce the groups of gauge connections and their transformations in the standard textbook fashion while noting that the proper mathematical entity involved is not a pair of groups but a groupoid, and later come back to it with more details in Chapter~\ref{chap:higher}. However, while giving space to introduce abstract mathematical topics, we also try to keep in mind the goal of concrete realisations useful to physics. For the brave intent on diving deep into the modern mathematical treatment of gauge theories we refer \cite{SCHREIBER2017}.


TITLE: Map an open interval to another one using a logistic curve
QUESTION [1 upvotes]: Given $a,b \in \mathbb{R}$, such that $a < b$ and $m,M \in \mathbb{R}$ such that $m < M$, I would like to map the interval $[a,b]$ to the interval $[m,M]$ using a continuous, monotonically increasing curve shaped like a logistic curve (https://en.wikipedia.org/wiki/Logistic_function). If $\varphi \, : \, [a,b] \, \longrightarrow \, [m,M]$ is the function I am looking for, I would like to set :
$$ \forall t \in [a,b] \, \varphi(t) = \begin{cases} m & \text{if} \, t=a \\[1mm] ??(t) & \text{if} \, a < t < b \\[1mm] M & \text{if} \, t = b \end{cases} $$
where $??$ is some function which would have the same shape as a logistic function. The function
$$ \ell \, : \, t \in \mathbb{R} \, \mapsto \, m + \frac{M-m}{1+e^{-t}} $$
maps $\mathbb{R}$ to $(m,M)$. So, in order to map $(a,b)$ to $(m,M)$, I could first map $(a,b)$ to $(0,1)$ (using an affine function) and then map $(0,1)$ to $\mathbb{R}$ (using $y \, \mapsto \, -\ln(\frac{1}{y} - 1)$, for example). But this does not give me the desired function (it actually gives me an affine function in the end).
How shall I address this problem ?

REPLY [2 votes]: Use a smoothstep such as $S(t) = 3t^2 - 2t^3$, which does a smooth $C^1$ transition from $0$ to $1$ over $[0,1]$.
Then, for $ a < t < b$, use
$$\varphi(t)=m+(M-m)S(\dfrac{t-a}{b-a})$$
This gives you a smooth $C^1$ transition from $m$ to $M$ over $[a,b]$.
For a smooth $C^2$ transition, use $S(t) = 6t^5 - 15t^4 + 10t^3$.


TITLE: Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0 ,y=mx$
QUESTION [3 upvotes]: Find the volume of the part of the cylinder $\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$ which lies between the plane $y=0, y=mx$
The limits of integration which I set went as follows:
$${\int \int}_{D} \int_{y=0}^{mx}\mathrm dy\mathrm dx\mathrm dz$$ where $$D=\big \lbrace \: (x,z) \mid \frac{x^2}{a^2}+\frac{z^2}{c^2}=1 \: \big \rbrace$$
Now let $x=au,z=bv$, then $D$ transforms into $\big \lbrace \: (u,v) \mid u^2 + v^2=1 \: \big \rbrace$
Then the integral becomes $$\int_{u=-1}^1 \int_{v=-\sqrt{1-u^2}}^{v=\sqrt{1-u^2}} ma^2cu\mathrm du\mathrm dv$$

REPLY [3 votes]: @Antimomy , your approach is absolutely OK. Now just solve the integration.
$$\int_{u=-1}^{u=1}\int_{v=-\sqrt{1-u^2}}^{v=-\sqrt{1-u^2}}ma^2cudvdu=a^2cm\int_{u=-1}^{u=1}\int_{v=-\sqrt{1-u^2}}^{v=-\sqrt{1-u^2}}udvdu$$
In this step, if you change it into the polar form then you will get the answer $0$. Because, see in the picture the same volume is in the negative side also.
First use the properties of definite integral then you can change it into polar. So,
$$V=4a^2cm\int_{u=0}^{u=1}\int_{v=0}^{v=-\sqrt{1-u^2}}udvdu$$
Now, put $u=rcos\theta$, $v=rsin\theta$. For positive $u$, $v$ the limits will be, $0\leq r\leq 1$ and $0\leq \theta \leq \frac{\pi}{2}$.
$$V=4a^2cm\int_0^{\frac{\pi}{2}}\int_0^1r^2cos\theta drd\theta=\frac{4}{3}a^2cm\int_0^\frac{\pi}{2}cos\theta d\theta=\frac{4}{3}a^2cm.$$


TITLE: Find a value $x$, such that if we had $a$ cents and $b$ cents stamps, we would make all values of money $\geq x$
QUESTION [2 upvotes]: Find a value $x$, such that if we had $a$ cents and $b$ cents stamps, we would make all values of money $\geq x$
This is the classic stamp question used for induction, when we have suppose, $4$ cent stamps and $7$ cent stamps, and we want to prove that any amount of postage $n\geq 18$ can be made using these stamps.
What I want to know is that if there is a formula/algorithm for figuring out the minimum amount of cent stamps that can be made? (in this case it is 18)
The way I usually do it is by writing down some numbers and then look for the starting pointer, whereon after, all postages can be made.
Does such an algorithm exist that would give us the value $n$ we are looking for?

REPLY [1 votes]: Yes! I'd recommend you try to guess then prove a formula for yourself, but if you get stuck, a solution is here.


TITLE: Question on Generic Freeness, ref. [Matsumura, page 185]
QUESTION [3 upvotes]: I am sure this must have been answered somewhere but I can't find them, so I shall try my luck here.

Let $A$ be a Noetherian integral domain and $M$ a finitely generated $A$-module. Then there exists $0 \neq f\in A$ such that $M_f$ is a free $A_f$-module.

I have tried reading several sources but it boils down to the proof that we may write $M$ as a composition series, say
$$0\subseteq M_1\subseteq\cdots\subseteq M_{n-1}\subseteq M$$
where each
$$M_i/M_{i-1}\cong A/\frak p_i$$
for some prime $\frak p_i$. According to the remark in [Matsumura, page 185], it suffices to choose $a$ in the intersection of all those $\frak p_i$ which are not zero and we get the localisation of $M_i/M_{i-1}$ at $a$ a free $A_a$ module. How is this possible?
If we make things very simple say $M=Am$ with $Ann(m)=\frak p$, why would choosing $p\in\frak p$ would allow $M_p$ to be a free $A_p$ module?
And second question (a bit different, but related nonetheless) is that returning to the special case where $M=Am$, why should the set
$$U=\{{\frak{q}} \in Spec A: M_{\frak{q}}\text{ is a free }A_{\frak{q}}\text{ module}\}$$
be non empty? I am reasoning out this way that given an exact sequence
$$0\rightarrow {\frak{p}} \rightarrow A\rightarrow Am\rightarrow 0$$
if we localise at the zero ideal, we get
$$0\rightarrow {\frak{p}}_{(0)} \rightarrow K(A)\rightarrow Am\otimes_{A} K(A)\rightarrow 0$$
so by dimension counting, ${\frak{p}}_{(0)}$ should be a zero vector space, and hence $U$ should be non empty by taking the zero ideal, but here I am contradicting myself in some sense that if $\frak p$ is not zero, then its localisation is probably not zero. Can someone point out the mistake to me?
Thanks!

REPLY [3 votes]: I don't know to what refers "How is this possible?", but $(A/\mathfrak p_i)_a\simeq A_a/\mathfrak p_iA_a$, and there are two cases: $\mathfrak p_i\ne 0$ and then $a\in \mathfrak p_i$, so $\mathfrak p_iA_a=A_a$ (which gives $(A/\mathfrak p_i)_a=0$), or $\mathfrak p_i=0$ and then we get $A_a$. (Now I hope that you know how to finish the proof. If not, then let me know.)
If $M=Am$ then $M\simeq A/\operatorname{Ann}(m)$ and as before $M_p\simeq(A/\mathfrak p)_p\simeq A_p/\mathfrak pA_p=0$. Moreover, $M_{\mathfrak q}\simeq A_{\mathfrak q}/\mathfrak pA_{\mathfrak q}$ and whether $\mathfrak p\nsubseteq\mathfrak q$ then $\mathfrak pA_{\mathfrak q}=A_{\mathfrak q}$. (For instance, if $\mathfrak p\ne 0$ you can take $\mathfrak q=0$. Otherwise, $\mathfrak p=0$ and thus $M\simeq A$.)


TITLE: Derangement problem but with functions instead of permutations, using inclusion-exclusion.
QUESTION [0 upvotes]: How many functions $f : [n] \to [n]$ are there with no fixed points, i.e. such that $f(i) \ne i$ for all $i$? Use inclusion-exclusion to solve it.

REPLY [0 votes]: For $k\in[n]$ let $A_k$ be the set of functions from $[n]$ to $[n]$ that have $k$ as a fixed point. Then the principle of inclusion-exclusion says that
$$\begin{align*}
\left|\bigcup_{k\in[n]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_k\right|\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}kn^{n-k}\,:
\end{align*}$$
there are $\binom{n}k$ subsets $I\subseteq[n]$ such that $|I|=k$, and there are $n^{n-k}$ functions from $[n]\setminus I$ to $[n]$ whose set of fixed points includes all of $I$. There are $n^n$ functions from $[n]$ to $[n]$, so there are
$$\begin{align*}
n^n-\sum_{k=1}^n(-1)^{k+1}\binom{n}kn^{n-k}&=n^n+\sum_{k=1}^n(-1)^k\binom{n}kn^{n-k}\\
&=\sum_{k=0}^n(-1)^k\binom{n}kn^{n-k}
\end{align*}$$
functions from $[n]$ to $[n]$ with no fixed points.
As a sanity check let $n=3$. If we list $f(1),f(2)$ and $f(3)$ in that order, the $8$ functions from $[3]$ to $[3]$ with no fixed points can be described by the triples $231$, $312$, $211$, $311$, $212$, $232$, $331$, and $332$, and indeed
$$\sum_{k=0}^3(-1)^k\binom3k3^{3-k}=27-27+9-1=8\,.$$


TITLE: Does this set of functions separate points of $\Omega$?
QUESTION [3 upvotes]: Let $\Omega$$\subset\mathbb{C}$ be a open connected set (domain). Let $z\in\Omega$ I want to show that if all bounded holomorphic functions on $\Omega$ are not just the constant functions (denoted by $H^\infty(\Omega)\not\equiv\mathbb{C}$), then, for every $0\neq\xi\in\mathbb{C}$, there exists a bounded holomorphic function $f$ on $\Omega$ such that $f’(z)\xi\neq 0$
The proof goes like this (see proof of Proposition 2.5.1 in this book).
Suppose $H^\infty(\Omega)\not\equiv\mathbb{C}$ and let $f_0\in H^\infty(\Omega), a_1,a_2\in\Omega$, such that $f_0(a_1)\neq f_0(a_2)$.
Define for $j=1,2$, $f_j$ as
\begin{equation*}
    f_j(z) = \begin{cases}
              \frac{f_0(z)-f_0(a_j)}{z-a_j}& \text{if } z\neq a_j,\\
               f_0’(a_j) & \text{if } z= a_j.
          \end{cases}
\end{equation*}
Then it is said that $f_0,f_1,f_2$ separates points of $\Omega$. And $rank(f_0’,f_1’,f_2’)=1$ on $\Omega$ and hence proved.
I do not understand how $f_0,f_1,f_2$ separates points of $\Omega$? And what is the meaning of $rank(f_0’,f_1’,f_2’)=1$?

REPLY [1 votes]: The statement about separating points means that for any distinct $z,w \in \Omega$ there exist $j \in \{0,1,2\}$ with $f_j(z) \ne f_j(w)$, and the statement about the rank of the derivatives just means that these three derivatives do not simultaneously vanish at any point in $\Omega$. And since you are working in one complex dimension here, you don't need $\xi$, just need to show that given $z_0 \in \Omega$ (which you omitted from your assumptions, but this is the way it is phrased in the referenced book) there exists a bounded holomorphic function $f$ in $\Omega$ with $f'(z_0) \ne 0$.
I think there is a more elementary proof than the one in the book. By assumption there exists a non-constant function $f$ in $\Omega$. If $f'(z_0) \ne 0$, then you are done. Otherwise, the Taylor expansion at $z_0$ looks like $f(z) = w_0 + a_n (z-z_0)^n + \ldots$ with $w_0 = f(z_0)$ and $a_n \ne 0$ for some $n \ge 2$. Then $g(z) = \frac{f(z)-w_0}{(z-z_0)^{n-1}}$ is another bounded function in $\Omega$ (which is not hard to show) with $g(z) = a_n (z-z_0) + \ldots$, so $g'(z_0) = a_n \ne 0$.


TITLE: Ricci theorem and 4-th divergence of Energy-momentum/ Einstein tensor
QUESTION [0 upvotes]: I know that Ricci theorem says that absolute differential of metric $g_{ij}$ :
$D(g_{ij}) = (\nabla_{k} g_{ij}) \text{d}x^{k} = 0$
So we can write : $\nabla_{k} g_{ij} = 0$
with $k$ that can be equal to $i$ or $j$.
On the other side, in General relativity, we have the 4-th divergence of $G_{\mu\nu}$ and $T_{\mu\nu}$ which is equal to 0 :
$\nabla_{\mu}T_{\mu\nu} = \nabla_{\mu}G_{\mu\nu} = 0$
Can we do the link between Ricci theorem for metric $g_{ij}$ and these 4-th divergences equal to zero ?
I mean, we cannot have (unlike with Ricci theorem) :
$\nabla_{k}T_{\mu\nu} = \nabla_{k}G_{\mu\nu} = 0$ for $k\neq \mu$ and $k\neq \nu$, isn't it ?
So, can we say, from a particular point of view, that Ricci theorem is more general, in the way that we have for any $k$ index (including $i$ or $j$) : $\nabla_{k} g_{ij} = 0$ ??
UPDATE 1:
You say that "The tensor equation $\nabla_{\rho} g_{\mu\nu} = 0$ ensures invariance of quantities such as $g_{\mu\nu}v^{\mu} w^{\nu}$ under parallel transportations : how to prove it ?
Maybe If I take (with $\tau$ curvilinear abscissa) : $\nabla_{\rho} g_{\mu\nu} \dfrac{\text{d}x^{\rho}}{\text{d}\tau} = 0\quad\quad(1)$,
which would imply :
$(\partial_{\rho} g_{\mu\nu} - g_{\mu\alpha}\Gamma^{\alpha}_{\rho\nu} - g_{\nu\alpha}\Gamma^{\alpha}_{\rho\mu}) \dfrac{\text{d}x^{\rho}}{\text{d}\tau} = 0$
but how to introduce $v^{\mu}$ and $w^{\nu}$ ?
I know that elementary length (like $\text{d}\tau$ above) $\text{d}s^2 = g_{\mu\nu} \text{d}x^{\mu} \text{d}x^{\nu}$ is invariant, so we could have :
$1 = g_{\mu\nu} \dfrac{\text{d}x^{\mu}}{\text{d}s} \dfrac{\text{d}x^{\nu}}{\text{d}s} = g_{\mu\nu} v^{\mu} w^{\nu} = \text{constant}\quad\quad(2)$
but how to connect (2) and (1) ? or get (2) from (1) ?
Any help is welcome, regards

REPLY [0 votes]: The ``divergence-free'' equations,
\begin{equation}
\nabla_{\mu}G^{\mu\nu} = \nabla_{\mu}T^{\mu\nu}=0
\end{equation}
are just conservation laws (remember that the repeated index $\mu$ is summed so the above equations are continuity equations that suggest local conservation).
The fact that $\nabla_{\rho}g_{\mu\nu}=0$ is a completely different statement. To make it more obvious, remember that the metric tensor is essentially nothing more than the inner product of your axes unit vectors; it is a way of saying how ``much perpendicular your coordinates are''. The tensor equation $\nabla_{\rho}g_{\mu\nu}=0$ ensures invariance of quantities such as $g_{\mu\nu}w^{\mu}v^{\nu}$ under parallel transportations and it basically comes from the strong equivalence principle and the fact that the metric tensor is a ... tensor! (in flat spacetime $\eta_{\mu\nu}$, i.e. local inertial frames, the equation $\partial_{\rho}\eta_{\mu\nu}$ holds so it must hold for all frames of reference).
To sum it up, $\nabla_{\rho}g_{\mu\nu}=0$ is a unique tensor equation holding for the metric tensor itself, but $\nabla_{\mu}G^{\mu\nu} = \nabla_{\mu}T^{\mu\nu}=0$ are just manifestations of conservation laws in curved spacetimes.
UPDATE 1
The invariance of a quantity $(\dots)$ under parallel transportation along a direction with tangent 4-vector $t^{\rho}$ is expressed by the condition $t^{\rho}\nabla_{\rho}(\dots)=0$. For the invariant quantity $g_{\mu\nu}w^{\mu}v^{\nu}$ this means,
\begin{equation}\begin{aligned}
t^{\rho}\nabla_{\rho} (g_{\mu\nu}w^{\mu}v^{\nu}) &= t^{\rho}\nabla_{\rho}g_{\mu\nu}w^{\mu}v^{\nu} + t^{\rho}g_{\mu\nu}\nabla_{\rho}w^{\mu}v^{\nu} + t^{\rho}g_{\mu\nu}w^{\mu}\nabla_{\rho}v^{\nu} \\
&= t^{\rho}\nabla_{\rho}g_{\mu\nu}w^{\mu}v^{\nu} = 0 \\
^{\text{To hold for any }t^{\rho}, w^{\mu}, v^{\nu}}\Rightarrow \nabla_{\rho}g_{\mu\nu} = 0
\end{aligned}\end{equation}
where the fact that parallel transportation of 4-vectors also means $t^{\rho}\nabla_{\rho}w^{\mu} = t^{\rho}\nabla_{\rho}v^{\nu} = 0$ was used.


TITLE: Help with set theory question about binary relations
QUESTION [0 upvotes]: On my assignment I was asked the question:
Determine, with reason, if the binary relation is reflexive, symmetric, antisymmetric, or transitive.
Let X be any set containing at least three distinct elements a, b, c ∈ X. Let S be the
relation on P(X) such that (A, B) ∈ S when A ∩ B = {a}.
I was wondering if anyone can point me in the right direction here as I don't even know where to begin. Where do the A and B come from? If S is the relation on P(X) then does it mean that (A,B) is only in the relation if its intersection returns only the element {a}? I can't seem to wrap my head around this question, any help would be appreciated. Thank you all for taking the time to read this.

REPLY [1 votes]: That $S$ is a relation on $P(X)$ mean that $A$ and $B$ are subsets of $X$. So for any two sets $A,B\subseteq X$ we say the relation $S(A,B)$ hold, or in other words $(A,B)\in S$ if $A\cap B=\{a\}$. Some hints to look at the properties:

Reflexivity: R is reflexive if for each $A\subseteq X$ it hold that $(A,A)\in S$ i.e. $A\cap A = \{a\}$, does this hold?
Symetry: R is symetric if for each $A,B\subseteq X$, such that $(A,B)\in S$ it hold that $(B,A)\in S$. If you translate $(A,B)\in S$ to the definition of S, then does it hold?
Anti-symetry: If $A,B \subseteq X$ and both $(A,B),(B,A)\in S$ then $A=B$. If you translate  $(A,B),(B,A)\in S$ by the definition, then does this hold?
Transitivity: If $A,B,C\subseteq X$ and $(A,B),(B,C)\in S$ then $(A,C)\in S$. Does this hold?


TITLE: How much power is needed to keep ideal gas in a fix volume?
QUESTION [1 upvotes]: I was playing with the idea of entropy and an interesting thought came out. Imagine a physical cube with side $a$. Inside the cube is $N$ particles of an ideal gas at a temperature $T$. One particle of the gas has the mass of $m$.

It can be seen that each particle would be moving at a velocity of v, bouncing off the wall of the cube. Let's take a particle that starts moving at the center of a face of the cube and bouncing off the centers of other faces of the cube. Now if we replace the cube with a series of machine guns surrounding the what is now an imaginary cube, the guns are programmed so that whenever it sees a particle about to pass the "cube", it will fire an identical particle at speed v so that it will hit the particle in the "cube" at the exact moment and angle so that the inside particle is steered back inside the "cube" just like when there was a physical cube.

If this happens, the frequency of the gun getting shot will be the frequency of a particle going to the side of the imaginary cube. This is:
$$f = \frac{v\sqrt{2}}{a}$$
where $v$ is the velocity of a particle and $a$ is the side length of the cube
For each time a gun fires, in order to work, it must give its "bullet" a total of kinetic energy is:
$$KE = \frac{1}{2} mv^2$$
So the power of the guns system would be:
$$P = KE.f = \frac{v\sqrt{2}}{a} \frac{1}{2} mv^2 = \frac{mv^3}{a\sqrt{2}}$$
Since $$v = \sqrt{\frac{3kT}{m}}$$
so:
$$P = \frac{m}{a\sqrt{2}} \biggl(\frac{3kT}{m}\biggl)^\frac{3}{2}$$
$$P = \frac{(3kT)^\frac{3}{2}}{a\sqrt{2m}}$$
If we imagine that all particles move like this (I think this will happen if we take the average of where each particles hit the "cube") and write the equation in term of $V$:
$$P = \frac{N(3kT)^\frac{3}{2}}{V^\frac{1}{3}\sqrt{2m}}$$
So I concluded that this is the minimum power required to keep a collection of $N$ particles of ideal gas at temperature $T$ in a volume $V$ (without sqreading all over the place).
However, this seems really wrong, because the power required is incredibly large. There is not a lot of energy required if we can reuse the kinetic energy of bounced of "bullets", but the amount of power of the gun system is still very large.
For example, in order to keep 1 mol of Hydrogen gas in a 1m^3 space at a temperature of 273K, the amount of power needed is $1.2516 * 10^7 W$
Can someone tell me whether this is correct or not and can we really conclude that everything is constantly emitting energy and get them back constantly?
Thank you so much for your help.

REPLY [0 votes]: You don't need to keep firing new particles at it. Two identical particles could both be trapped in identical boxes, colliding with each other and confining each other, thus requiring no energy input. When you keep firing more particles at it, the same particles are reflected back with the same speed so the energy entering the system equals the energy leaving it. This means no energy input is required to confine a gas. This makes sense because you wouldn't expect a container to lose energy ad infinitum.