text:
\begin{document}

 \title[Vanishing of cohomology over {C}ohen--{M}acaulay
 rings]{Vanishing of cohomology\\ over {C}ohen--{M}acaulay rings}

 \author{Lars Winther Christensen}

 \address{Department of Math.\ and Stat., Texas Tech University,
   Lubbock, TX 79409, U.S.A.}

 \email{lars.w.christensen@ttu.edu}

 \urladdr{http://www.math.ttu.edu/\urltilda lchriste}

 \author{Henrik Holm}

 \address{Department of Basic Sciences and Environment, University of
   Copenhagen, Thorvaldsensvej 40, DK-1871 Frederiksberg C, Denmark}

 \email{hholm@life.ku.dk}

 \urladdr{http://www.matdat.life.ku.dk/\urltilda hholm}

 \date{23 February 2012}

 \keywords{AC ring, AB ring, Auslander's condition, Ext-index,
   c.i.~local homomorphism}

 \subjclass[2010]{13D07, 13H10}

\begin{abstract}
  A 2003 counterexample to a conjecture of Auslander brought attention
  to a family of rings---colloquially called AC rings---that satisfy a
  natural condition on vanishing of cohomology. Several results attest
  to the remarkable homological properties of AC rings, but their
  definition is barely operational, and it remains unknown if they
  form a class that is closed under typical constructions in ring
  theory. In this paper, we study transfer of the AC property along
  local homomorphisms of Cohen--Macaulay rings. In particular, we show
  that the AC property is preserved by standard procedures in local
  algebra. Our results also yield new examples of Cohen--Macaulay AC
  rings.
\end{abstract}

\maketitle

\section*{Introduction}

\noindent
Vanishing of Ext groups and functors play a crucial role in the study
of rings and their modules. For a fixed module $M$, vanishing of
$\Ext[]{n}{M}{N}$ for all modules $N$ and integers $n \gg 0$ says that
$M$ has finite projective dimension, in which case the functors
$\Ext[]{n}{M}{-}$ vanish for all $n > \pd[]{(M)}$. Finiteness of
$\pd{(M)}$ for all finitely generated $R$-modules is a lot to ask from
a ring---in the noetherian case it means that $R$ is regular.
Auslander conjectured that for every finitely generated module $M$,
also those of infinite projective dimension, there would be a sort of
an upper bound for non-vanishing of the groups $\Ext[]{n}{M}{N}$; see
\cite[ch.~V]{mas1}. The exact conjecture is that every Artin algebra
$R$ satisfies the following:

\begin{condition}
  \ac\ For every finitely generated $R$-module $M$ there
  exists an integer $\ab{M} \ge 0$ such that for every finitely
  generated $R$-module $N$ one has: $\Ext{n}{M}{N}=0$ for $n \gg 0$
  implies $\Ext{n}{M}{N}=0$ for $n > \ab{M}$.
\end{condition}

\noindent An integer $\ab{M}$ with this property is called an
\emph{Auslander bound} for $M$; the least such bound may be perceived
as a ``latent projective dimension'' of $M$.

An example, discovered by Jorgensen and \c{S}ega \cite{DAJLMS04},
disproved Auslander's conjecture. However, several classes of rings do
satisfy Auslander's condition \ac, and remarkable homological
properties that flow from \ac\ have been uncovered, for example, in
work of Huneke and Jorgensen \cite{CHnDAJ03}; see also \cite{LWCHHlc}.

The counterexample in \cite{DAJLMS04} is a commutative Gorenstein
local ring, which is a finite dimensional algebra, and even
Koszul. Nevertheless, the condition \ac\ is satisfied by all
commutative local rings that are complete intersection or Golod, and
that is striking: with regard to homological characteristics these two
classes of rings usually belong on opposite sides of the spectrum.

Among commutative noetherian local rings, those that satisfy \ac\ are
emerging as a family with intriguing homological properties, albeit
one whose position relative to the traditional classes of rings is not
easily described. In this paper, we give new examples---explicit and
abstract---of Cohen--Macaulay local rings that satisfy \ac.

\begin{center}
  $*\:*\:*$
\end{center}

From this point on, all rings considered are commutative, noetherian,
and local. In particular, $(R,\m)$ is such a ring, and $\Rhat$ denotes
its $\m$-adic completion.

Most Cohen--Macaulay rings of finite CM type are Golod and, therefore,
known from \prpcite[1.4]{DAJLMS04} to satisfy even a uniform version
of Auslander's condition:
\begin{condition}
  \uac\ There exists an integer $b \ge 0$ such that for all
  finitely generated $R$-modules $M$ and $N$ one has:
  $\Ext{n}{M}{N}=0$ for $n\gg 0$ implies $\Ext{n}{M}{N}=0$ for~$n >
  b$.
\end{condition}
\noindent In \secref{cm} we show by a direct argument that every
Cohen--Macaulay ring of finite CM type satisfies \uac.

An explicit collection of new examples of rings that satisfy \uac\ is
constructed via results about preservation of the \ac\ and \uac\
properties under standard operations in local algebra---such as
completion and reduction modulo regular elements. These results were
advertised in \cite[rmk.~5.7]{LWCHHlc}; they are proved in
\secref{stab}, and the examples follow in \secref{example}.

The canonical maps \mbox{$R \into \Rhat$} and \mbox{$R\onto R/(x)$},
where $x$ is a regular element, are archetypes of local homomorphisms
of finite flat dimension; they are even c.i.\ homomorphisms in the
sense of Avramov~\cite{LLA99}. A classical chapter of local algebra
studies transfer of ring theoretic properties along local
homomorphisms. In \secref{homos} we add to it as we prove,
essentially, that the \ac\ and \uac\ properties descend along
homomorphisms of finite flat dimension and ascend along
c.i. homomorphisms.

\section{Maximal Cohen--Macaulay modules}
\label{sec:cm}

For a finitely generated $R$-module $M$, the $n$th syzygy in its
minimal free resolution is denoted $\syz{n}{M}$. For $i > n \ge 0$
there are isomorphisms:
\begin{equation}
  \label{eq:is1}
  \Ext{i-n}{\syz{n}{M}}{N} \is \Ext{i}{M}{N}.
\end{equation}

Let $R$ be Cohen--Macaulay of dimension $d$. A finitely generated
$R$-module $M$ is called \emph{maximal Cohen--Macaulay}, abbreviated
MCM, if the equality $\dpt{M}=d$ holds. Every $d$th syzygy of a
finitely generated $R$-module is either $0$ or an MCM module, and
every syzygy of an MCM module is an MCM module; see \prpcite[(1.16)
and (1.3)]{yos}. These facts together with the theory of MCM
approximations, which is due to Auslander and Buchweitz
\cite{MAsROB89}, are central to this section.

\begin{rmk}
  \label{rmk:mcmab}
  Let $R$ be Cohen--Macaulay of dimension $d$. Dimension shifting
  \eqref{is1} shows that $R$ satisfies \ac\ if and only if every MCM
  $R$-module has an Auslander bound. Every MCM module is a (finite)
  direct sum of indecomposable MCM modules, so $R$ satisfies \ac\ if
  and only if every indecomposable MCM $R$-module has an Auslander
  bound. In particular, $R$ satisfies \uac\ if and only if the
  indecomposable MCM $R$-modules have a common Auslander bound.
\end{rmk}

A Cohen--Macaulay ring $R$ is said to be of \emph{finite CM type} if
there are only finitely many indecomposable MCM $R$-modules, up to
isomorphism.

\begin{thm}
  \label{thm:cm}
  A Cohen--Macaulay local ring of finite CM type satisfies \uac.
\end{thm}

\begin{proof}
  Let $R$ be Cohen--Macaulay of finite CM type.  By \rmkref{mcmab}, it
  suffices to prove that each of the finitely many indecomposable MCM
  $R$-modules has an Auslander bound. For an MCM $R$-module $X$, let
  $n(X)$ be the number of distinct, up to isomorphism, indecomposable
  MCM $R$-modules that occur as direct summands of any one of the
  syzygies $\syz{i}{X}$, for $i \ge 0$.

  Let $M$ be an indecomposable MCM $R$-module. If $n(M)$ is $1$, then
  $M$ is either free of rank $1$ or a direct summand of every syzygy
  module $\syz{i}{M}$ for $i \ge 0$, whence it has Auslander bound
  $0$; cf.~\eqref{is1}. Let $t \ge 1$ and assume that every MCM
  $R$-module $X$ with $n(X) \le t$ has an Auslander bound. Assume
  $n(M) = t+1$. If $M$ occurs as a direct summand of one of its
  syzygies, say, $\syz{l}{M}$ for some $l \ge 1$, then---by uniqueness
  of minimal free resolutions---$M$ occurs as a summand of infinitely
  many syzygies, namely $\syz{lm}{M}$ for all $m\in\NN$; whence $M$
  has Auslander bound $0$. If $M$ does not occur as a direct summand
  of $\syz{i}{M}$ for any $i \ge 1$, then one has $n(\syz{1}{M}) \le
  t$. By the induction hypothesis, $\syz{1}{M}$ has an Auslander
  bound, whence $M$ has one, again by dimension shifting \eqref{is1}.
\end{proof}

The \emph{embedding dimension} of $R$, $\edim{R}$, is the minimal
number of generators of its maximal ideal, and the \emph{codimension}
is the difference $\codim{R} = \edim{R} - \dimR$. The multiplicity
$\mult{R}$ of a Cohen--Macaulay ring is at least $\codim{R} +1 $; if
equality holds, then $R$ is said to have \emph{minimal multiplicity.}

\begin{rmk}
  Eisenbud and Herzog \cite{DEsJHr88} raise the question whether every
  complete Cohen--Macaulay ring of finite CM type and dimension at
  least $2$ has minimal multiplicity. Cohen--Macaulay rings of minimal
  multiplicity are Golod, see Avramov~\prpcite[5.2.4]{ifr}, and
  satisfy \uac\ for a strong reason, see \prpcite[1.4]{DAJLMS04}.
\end{rmk}

\begin{obs}
  \label{obs:Syz}
  Let $R$ be Cohen--Macaulay and assume that it has a dualizing
  module.  By \thmcite[A]{MAsROB89}, every finitely generated
  $R$-module $N$ has a maximal Cohen--Macaulay approximation: an exact
  sequence of finitely generated $R$-modules,
  \begin{equation*}
    0 \lra I \lra Y \lra N \lra 0,
  \end{equation*}
  where $Y$ is MCM, and $I$ is of finite injective dimension.  If $X$ is
  an MCM module, then one has $\Ext{n}{X}{I} =0$ for $n\ge 1$; see
  Yoshino \corcite[(1.13)]{yos} and \corcite[6.4]{MAsROB89}. Thus, for
  all $n \ge 1$ there are isomorphisms
  \begin{equation*}
    \Ext{n}{X}{Y} \is \Ext{n}{X}{N}.
  \end{equation*}
  From \eqref{is1} and the displays above, it is not hard to see that
  it suffices to verify the conditions \ac\ and \uac\ on MCM
  modules. That is, $R$ satisfies \ac\ if and only if it satisfies:
  \begin{condition}
    For every MCM $R$-module $X$ there exists $\ab{X} \ge 0$ such that
    for every MCM $R$-module $Y$ one has: $\Ext{n}{X}{Y}=0$ for $n\gg
    0$ implies $\Ext{n}{X}{Y}=0$ for~$n> \ab{X}$.
  \end{condition}
  And $R$ satisfies \uac\ if and only if it satisfies:
  \begin{condition}
    There exists an integer $b \ge 0$ such that for all MCM
    $R$-modules $X$ and $Y$ one has: $\Ext{n}{X}{Y}=0$ for $n\gg 0$
    implies $\Ext{n}{X}{Y}=0$ for $n> b$.
  \end{condition}
\end{obs}

\section{Completion and reduction modulo regular sequences}
\label{sec:stab}

Let $R$ be Cohen--Macaulay; we can now show that $R$ and $\Rhat$
satisfy \ac\ simultaneously and---in a generalization of
\prpcite[3.3.(1)]{CHnDAJ03}---that $R$ and $R/(x)$ satisfy \ac\
simultaneously, if $x$ is a regular element.

\begin{lem}
  \label{lem:regelt}
  Let $x$ be a regular element in $R$. If $R$ satisfies \acuac, then
  $R/(x)$ satisfies \acuac.
\end{lem}

\begin{proof}
  Let $M$ and $N$ be finitely generated $R/(x)$-modules. The change of
  rings spectral sequence \cite[XVI.\S$5.(2)_3$]{careil}
  \begin{equation*}
    \mathrm{E}_2^{p,q} = \Ext[R/(x)]{q}{\Tor{p}{R/(x)}{M}}{N} \Rightarrow
    \Ext{p+q}{M}{N}
  \end{equation*}
  has zero differentials on the second and all subsequent pages. For
  each $n\ge 0$ it follows from \prpcite[5.5]{careil} that there is an
  exact sequence
  \begin{equation*}
    0 \to \Ext[R/(x)]{n-1}{M}{N} \to \Ext{n}{M}{N} \to
    \Ext[R/(x)]{n}{M}{N}  \to 0.
  \end{equation*}
  If $b$ is an Auslander bound for $M$ as an $R$-module, then it
  follows that $\max\{0,b-1\}$ is an Auslander bound for $M$ as an
  $R/(x)$-module.
\end{proof}
\noindent
A slightly different argument for the result above appeared in the
proof of \prpcite[3.2.(1)]{CHnDAJ03}; the converse for Cohen--Macaulay
local rings is established in \corref[]{cmreg}.

The remaining proofs in this section use computations in the
derived category over $R$; first we recall some notation.  A complex
of $R$-modules is graded cohomologically,
\begin{equation*}
  M \deq \cdots \lra M^{n-1} \dra{n-1}{M} M^{n} \dra{n}{M}
  M^{n+1} \lra \cdots.
\end{equation*}
The suspension of $M$ is the complex $\Shift{M}$ with $\Shiftp{M}^n =
M^{n+1}$ and $\dif{\Shift{M}} = - \dif{M}$. The $n$th cohomology
module of $M$ is denoted \mbox{$\HH[n]{\mspace{-2mu}M}$}. Isomorphisms
in the derived category over $R$ are marked by the symbol~$\qis$; they
induce isomorphisms at the level of homology.  We use the standard
notation, $\RHom{-}{-}$ and $\Ltp{-}{-}$, for the right derived Hom
functor and the left derived tensor product functor. For all
$R$-modules $M$ and $N$ and all integers $n$ there are isomorphisms
\begin{equation*}
  \Ext{n}{M}{N} \is \HH[n]{\RHom[A]{M}{N}} \qand \Tor{n}{M}{N} \is
  \HHp[-n]{\Ltp{M}{N}}.
\end{equation*}

In the next work-hose lemma, the ring $S$ need not be local.

\begin{lem}
  \label{lem:descent}
  Let $\Rm$ be Cohen--Macaulay and assume that it has a dualizing
  module.  Let $S$ be an $R$-algebra of finite flat dimension and with
  $S/\m S \ne 0$. If $S$ satisfies \acuac, then $R$ satisfies
  \acuac.
\end{lem}

\begin{proof}
  By \obsref{Syz} it suffices to consider cohomology of MCM
  $R$-modules. For such modules $X$ and $Y$ there are isomorphisms in
  the derived category over $R$,
  \begin{align*}
    \Ltp{\RHom{X}{Y}}{S}
    & \qis \RHom[S]{\Ltp{X}{S}}{\Ltp{Y}{S}} \\
    & \qis \RHom[S]{\tp{X}{S}}{\tp{Y}{S}}.
  \end{align*}
  The first isomorphism is tensor evaluation, see Avramov and
  Foxby~\lemcite[4.4.(F)]{LLAHBF91}, and adjointness of Hom and
  tensor. The second isomorphism follows as one has $\Tor{n}{X}{S} =0=
  \Tor{n}{Y}{S}$ for $n\ge 1$; see \thmcite[(3.4)]{CFF-02}.  These
  isomorphisms and the assumption $S/\m S \ne 0$ yield:
  \begin{align*}
    \sup\setof{n}{\Ext{n}{X}{Y} \ne 0} =
    \sup\setof{n}{\Ext[S]{n}{\tp{X}{S}}{\tp{Y}{S}}\ne 0};
  \end{align*}
  see Foxby~\lemcite[2.1.(2)]{HBF77b}. Thus, if $b$ is an Auslander
  bound for the $S$-module $\tp{X}{S}$, then $b$ is also an Auslander
  bound for the $R$-module $X$.
\end{proof}

\begin{thm}
  \label{thm:completion}
  Let $\Rm$ be Cohen--Macaulay. Any one of the local rings
  \begin{equation*}
    R, \ \, \Rhat, \ \, \pows[R]{X}, \ \,\text{and} \ \, \poly[R]{X}_{(\m,X)}
  \end{equation*}
  satisfies \acuac\ if and only if they all satisfy \acuac.
\end{thm}

\begin{proof}
  The rings $R$, $\Rhat$, $\pows[R]{X}$, and $\poly[R]{X}_{(\m,X)}$
  are all local and Cohen--Macaulay.

  We first argue that $R$ and $\Rhat$ satisfy \acuac\
  simultaneously. If $\Rhat$ satisfies \acuac, then so does $R$ by
  \prpcite[5.5]{LWCHHlc}, as $\Rhat$ is a faithfully flat $R$-algebra.
  For the converse, assume that $R$ satisfies \acuac, and let
  $\x=x_1,\dots,x_d$ be a maximal $R$-regular sequence; its image in
  $\Rhat$ is a maximal $\Rhat$-regular sequence. There is an
  isomorphism of rings $R/(\x) \is \Rhat/(\x)$, and this ring is
  \acuac\ by \lemref{regelt}.  As $\Rhat$ has a dualizing module, it
  satisfies \acuac\ by \lemref{descent}.

  If $\pows[R]{X}$ or $\poly[R]{X}_{(\m,X)}$ satisfies \acuac, then
  so does $R$ by \lemref{regelt}. For the converse, assume that $R$
  and, therefore, $\Rhat$ satisfies \acuac. The local ring
  $\pows[\Rhat]{X}$, which is Cohen--Macaulay and has a dualizing
  module, satisfies \acuac\ by \lemref{descent} applied to the
  surjection $\pows[\Rhat]{X} \onto \Rhat$. The isomorphisms
  \begin{equation*}
    \widehat{\poly[R]{X}_{(\m,X)}} \is \pows[\Rhat]{X} \is \widehat{\pows[R]{X}}
  \end{equation*}
  and the assertion established in the first part of the proof now
  show that $\pows[R]{X}$ and $\poly[R]{X}_{(\m,X)}$ satisfy \acuac.
\end{proof}

\begin{cor}
  \label{cor:cmreg}
  Let $R$ be Cohen--Macaulay and $\x=x_1,\dots,x_n$ be an $R$-regular
  sequence. The ring $R/(\x)$ satisfies \acuac\ if and only if $R$
  satisfies \acuac.
\end{cor}

\begin{proof}
  The ``if'' part is immediate from \lemref{regelt}. For the converse,
  assume that $R/(\x)$ satisfies \acuac. It is a Cohen--Macaulay
  ring, so by \thmref{completion} the completion
  \mbox{$\smash{\widehat{R/(\x)} \is \Rhat/(\x)}$} satisfies
  \acuac. As the image of $\x$ in $\Rhat$ is a regular sequence, the
  ring $\Rhat$ satisfies \acuac\ by \lemref{descent}, and then
  another application of \thmref[]{completion} shows that $R$
  satisfies \acuac.
\end{proof}

A Cohen--Macaulay ring reduces modulo a regular sequence to an
artinian ring, and by \corref{cmreg} the two rings satisfy \acuac\
simultaneously. In a clear analogy, the
conditions \ac\ and \uac\ can be verified on modules of finite length:

\begin{prp}
  Let $R$ be Cohen--Macaulay. It satisfies \ac\ if and only if it
  satisfies the condition:
  \begin{condition}
    For every $R$-module $K$ of finite length there exists an integer
    $\ab{K} \ge 0$ such that for every $R$-module $L$ of finite length
    one has: $\Ext{n}{K}{L}=0$ for $n\gg 0$ implies $\Ext{n}{K}{L}=0$
    for $n> \ab{K}$.
  \end{condition}
  Furthermore, $R$ satisfies \uac\ if and only if it satisfies the
  condition:
  \begin{condition}
    There exists an integer $b \ge 0$ such that for all $R$-modules
    $K$ and $L$ of finite length one has: $\Ext{n}{K}{L}=0$ for $n\gg
    0$ implies $\Ext{n}{K}{L}=0$ for $n> b$.
  \end{condition}
\end{prp}

\begin{proof}
  Assume that $R$ satisfies \acuac\ for modules of finite
  length. For $\Rhat$-modules $K$ and $L$ of finite length and
  integers $n \ge 0$ there are isomorphisms
  \begin{equation*}
    \Ext[\Rhat]{n}{K}{L} \is \Ext[\Rhat]{n}{\tp{\Rhat}{K}}{L} \is \Ext{n}{K}{L}.
  \end{equation*}
  As $R$-modules, $K$ and $L$ also have finite length; hence it
  follows that $\Rhat$ satisfies \acuac\ for modules of finite
  length. In view of \thmref{completion} we can now assume that $R$ is
  complete and, in particular, that it has a dualizing module.

  By \obsref{Syz} it is sufficient to prove that $R$ satisfies
  \acuac\ for MCM modules. Let $X$ and $Y$ be MCM $R$-modules, and
  let $\x=x_1,\dots,x_d$ be a maximal $R$-regular sequence. As $\x$ is
  regular on $X$ and $Y$, there are isomorphisms
  \begin{equation}
    \label{eq:aa}
    \tag{1}
    X/\x X \qis \Ltp{X}{R/(\x)} \qand Y/\x Y \qis \Ltp{Y}{R/(\x)}
  \end{equation}
  in the derived category over $R$. Nakayama's lemma,
  see~\lemcite[2.1.(2)]{HBF77b}, yields the first equality below,
  while the second follows from \lemcite[4.4.(F)]{LLAHBF91} and
  \eqref{aa}.
  \begin{equation}
    \label{eq:ab}
    \tag{2}
    \begin{split}
      \sup\setof{n}{\HH[n]{&\RHom{X}{Y} \ne 0}}\\
      &= \sup\setof{n}{\HHp[n]{\Ltp{\RHom{X}{Y}}{R/(\x)}}\ne 0}\\
      &= \sup\setof{n}{\HH[n]{\RHom{X}{Y/\x Y} \ne 0}}.
    \end{split}
  \end{equation}
  From \eqref{aa} and adjointness of Hom and tensor one gets
  \begin{equation}
    \label{eq:ac}
    \tag{3}
    \RHom{X/\x X}{Y/\x Y} \qis \RHom{X}{\RHom{R/(\x)}{Y/\x Y}}.
  \end{equation}
  The Koszul complex on $\x$ is a free resolution of $R/(\x)$, so
  there is an isomorphism
  \begin{equation}
    \label{eq:ad}
    \tag{4}
    \Shift[d]{\RHom{R/(\x)}{-}} \qis \Ltp{-}{R/(\x)}.
  \end{equation}
  The next equalities follow from \eqref{ab} and Nakayama's lemma,
  from \lemcite[4.4.(F)]{LLAHBF91} and \eqref{ad}, and from
  \eqref{ac}, respectively.
  \begin{align*}
    \sup\setof{n}{\HH[n]{&\RHom{X}{Y}}\ne 0}\\
    &= \sup\setof{n}{\HHp[n]{\Ltp{\RHom{X}{Y/\x Y}}{R/(\x)}}\ne 0}\\
    &= \sup\setof{n}{\HH[n]{\RHom{X}{\Shift[d]{\RHom{R/(\x)}{Y/\x
              Y}}}}\ne 0 }\\
    &= \sup\setof{n}{\HH[n]{\RHom{X/\x X}{Y/\x Y}}\ne 0 } -d.
  \end{align*}
  That is, one has
  \begin{align*}
    \sup\setof{n}{\Ext{n}{X}{Y}\ne 0}= \sup\setof{n}{\Ext{n}{X/\x
        X}{Y/\x Y}\ne 0 } -d.
  \end{align*}
  As the $R$-modules $X/\x X$ and $Y/\x Y$ have finite length, a
  straightforward argument finishes the proof, cf.~proof of
  \lemref{descent}.
\end{proof}

\section{An example}
\label{sec:example}

One can construct new examples of Cohen--Macaulay local rings that
satisfy \acuac\ from existing ones through the process of adjoining
power series variables and killing regular elements. To get examples
not covered in the literature, see the list \cite[A.1]{LWCHHlc}, the
new ring $R$ should not be Golod or complete intersection (c.i.); and
if $R$ is Gorenstein, then its codimension, $\codim{R}$, should be at
least $5$, and its multiplicity, $\mult{R}$, should be at least
$\codim{R}+3$. Furthermore, $R$ should not be of finite CM type,
cf.~\thmref{cm}.

\enlargethispage{\baselineskip}
\begin{prp}
  \label{prp:heitmann}
  Let $Q$ be a Cohen--Macaulay local ring and assume that it satisfies
  \acuac. For integers $d,n \ge 1$ and $s_i \ge 2$, the local ring
  \begin{equation*}
    R = \pows[Q]{X_1,\dots,X_d,Y_1,\dots,Y_n}/(Y_1^{s_1},\dots,Y_n^{s_n})
  \end{equation*}
  is Cohen--Macaulay with $\codim{R} = \codim{Q} + n$, and it
  satisfies \acuac. Moreover, the following hold:
  \begin{prt}
  \item $R$ is not of finite CM type.
  \item If $Q$ is not c.i., then $R$ is not c.i.\ and $R$ is not
    Golod.
  \item If $Q$ has infinite residue field and one has $s_i \ge 4$ for
    some $i\in\set{1,\ldots,n}$, then the inequality $\mult{R} \ge
    \codim{R} + 3$ holds.
  \end{prt}
\end{prp}

\begin{proof}
  Clearly, $R$ is Cohen--Macaulay with $\codim{R} = \codim{Q} + n$,
  and it is immediate from \thmref{completion} and \corref{cmreg} that
  $R$ satisfies \acuac.

  (a): Let $\q$ be the maximal ideal in $Q$. The image of $Y_1$ in
  $R_{(\q,Y_1,\dots,Y_n)}$ is a zero-divisor, so $R$ is not an
  isolated singularity; in particular, $R$ is not of finite CM type;
  see Huneke and Leuschke~\corcite[2]{CHnGJL02}.

  (b): Assume that $Q$ is not c.i. By \cite[(5.10)]{LLA99} the ring
  $R$ is not c.i. As $Q$ is singular, one has $\codim{R} =\codim{Q} + n
  \ge 2$. The $R$-module $R/(Y_1)$ has constant Betti numbers equal to
  $1$. It follows from~\thmcite[5.3.3.(5)]{ifr} that $R$ is not Golod.

  (c): Let $\mathfrak{a}$ be a minimal reduction of the maximal ideal $\q$ in
  $Q$. It is elementary to verify that $\mathfrak{b} =
  (\mathfrak{a},X_1,\ldots,X_d)$ is a reduction of $\m =
  (\q,X_1,\ldots,X_d,Y_1,\ldots,Y_n)$, the maximal ideal in
  $R$. Assume that $Q/\q$ is infinite. It follows that $\mathfrak{a}$ is
  minimally generated by $\dim[]{Q}$ elements and, therefore,
  $\mathfrak{b}$ is generated by $\dim[]{Q}+d=\dimR$ elements. Hence,
  $\mathfrak{b}$ is a minimal reduction of $\m$; see Swanson and
  Huneke~\prpcite[8.3.7 and cor.~8.3.5.(1)]{icirm}.  As $R$ is
  Cohen--Macaulay, \prpcite[11.2.2]{icirm} yields $\mult{R} =
  \lgt[]{R/\mathfrak{b}}$. If one has $s_i \ge 4$ for some $i$, then
  $(\m/\mathfrak{b})^3$ is non-zero, whence
  \begin{equation*}
    \lgt[]{R/\mathfrak{b}} \ge \edim{R/\mathfrak{b}} + 3 \ge
    \edim{R} - \dimR + 3 = \codim{R} + 3. \qedhere
  \end{equation*}
\end{proof}

We thank Louiza Fouli for clarifying for us the basics of multiplicities
that are used in the proof of part (c) above.

All assumptions on $Q$ in \prpref{heitmann} are satisfied by the ring
in \exaref{Q}; thus for all choices of $d,n \ge 1$ and $s_i \ge 2$ the
resulting ring $R$ in \prpref[]{heitmann} is not among the rings on
the list \cite[A.1]{LWCHHlc} nor is it of finite CM type.

\begin{exa}
  \label{exa:Q}
  Let $k$ be a field of characteristic $0$ and consider the local
  $k$-algebra
  \begin{equation*}
    Q = \poly{U,V,W}/(U^2-V^2,V^2 - W^2,UV,VW).
  \end{equation*}
  It has radical cube zero and $\edim{Q}=3$, so it satisfies \uac\ by
  \prpcite[1.1.(3)]{DAJLMS04}. Moreover, $Q$ is not Gorenstein; in
  particular, it is not c.i.
\end{exa}

Through a result of Heitmann \cite{RCH93}, the construction in
\prpref{heitmann} also provides examples of unique factorization
domains that satisfy \acuac.

\begin{obs}
  Let $Q$ be a complete Cohen--Macaulay local ring that satisfies
  \acuac. Assume that $Q$ is not c.i.\ and that no integer is a zero
  divisor in~$Q$; the ring in \exaref[]{Q} serves as an example.

  Let $R$ be as in \prpref{heitmann}. If $Q$ is not artinian or $d$ is
  at least $2$, then $R$ has depth at least $2$. By
  \thmcite[8]{RCH93}, there is a local unique factorization domain,
  $D$, with $\widehat{D} \is R$. It follows from \thmref{completion}
  that $D$ satisfies \acuac.  Of course, $R$ is also the completion
  of the ring
  $\poly[Q]{X_1,\dots,X_d,Y_1,\dots,Y_n}/(Y_1^{s_1},\dots,Y_n^{s_n})$
  localized at the irrelevant maximal ideal, but that ring is not a
  domain.
\end{obs}

\section{Local homomorphisms}
\label{sec:homos}

Let $\Rm$ and $\Sn$ be local rings, and let $\mapdef{\f}{R}{S}$ be a
\emph{local} homomorphism, that is, a ring homomorphism with $\f(\m)
\subseteq \n$. The \emph{semicompletion} of~$\f$~is its composite with
the embedding $S \into \Shat$; it is written
$\mapdef{\grave{\f}}{R}{\Shat}$. The semicompletion of $\f$ admits a
Cohen factorization; that is, there exist local homomorphisms
\begin{equation*}
  R \xra{\dot{\f}} R' \xra{\f'} \Shat,
\end{equation*}
such that $\grave{\f}=\f'\dot{\f}$, where $\dot{\f}$ is flat with
regular closed fiber $R/\m R'$, the ring $R'$ is complete, and $\f'$
is surjective; see Avramov, Foxby, and
Herzog~\thmcite[(1.1)]{AFH-94}. If $\f$ is of finite flat
dimension---that is, $S$ has finite flat dimension as an $R$-module
via $\f$---then the surjection $\f'$ is of finite projective
dimension; see~\cite[(3.3)]{AFH-94}.

A Gorenstein local ring satisfies \ac\ if and only if it satisfies
\uac; following \cite{CHnDAJ03} such a ring is called an \emph{AB
  ring}. The \ac\ and \uac\ properties descend along local
homomorphisms of finite flat dimension:

\begin{thm}
  Let $\mapdef{\f}{R}{S}$ be a local homomorphism of finite flat
  dimension.
  \begin{prt}
  \item If $S$ is Cohen--Macaulay and satisfies \acuac, then $R$ is
    Cohen--Macaulay and satisfies \acuac.
  \item If $S$ is an AB ring, then $R$ is an AB ring.
  \end{prt}
\end{thm}

\begin{proof}
  Assume that $S$ is Cohen--Macaulay and satisfies \acuac, then
  $\Shat$ has the same properties; see \thmref{completion}.  Consider
  a Cohen factorization $R \to R' \to \Shat$ of the semicompletion
  $\grave{\f}$.  By \cite[(3.10)]{AFH-94} the rings $R'$ and $R$ are
  Cohen--Macaulay. The ring $R'$ is complete, hence it has a dualizing
  module, and by \lemref{descent} it satisfies \acuac. Finally, $R$
  satisfies \acuac\ by \prpcite[5.5]{LWCHHlc}; this proves part
  (a). For part (b) it suffices to note that if $S$ is Gorenstein,
  then $R$ is Gorenstein; see Avramov and Foxby
  \thmcite[(2.4)]{LLAHBF90}.
\end{proof}

The notation $\Rmk$ specifies that $k$ is the residue field of the local
ring $\Rm$.

\begin{lem}
  \label{lem:ascent}
  Let $\mapdef{\f}{\Rmk}{(S,\n,k)}$ be a local homomorphism of
  complete $k$-algebras. If $R$ is Cohen--Macaulay and satisfies
  \acuac, then $\f$ has a Cohen factorization $R \to R' \to S$,
  where $R'$ is Cohen--Macaulay and satisfies \acuac.
\end{lem}

\begin{proof}
  Let $y_1,\ldots,y_n$ be a generating set for $\n$. As the field $k$
  is a Cohen ring for $S$, in the sense of \cite[(1.0.2)]{AFH-94}, the
  inclusion $k \into S$ extends by the assignment $Y_i \mapsto y_i$ to
  a surjective homomorphism $\pi\colon \pows[k]{Y_1,\ldots,Y_n} \onto
  S$. Set $R' = \pows[R]{Y_1,\ldots,Y_n}$, then $R'$ is complete and a
  flat local extension of $R$ with regular closed fiber $R'/\m R' \is
  \pows[k]{Y_1,\ldots,Y_n}$. The homomorphism $\f$ extends by the
  assignment $Y_i \mapsto y_i$ to a homomorphism $R' \to S$; it is
  surjective because $\pi$ is surjective. Thus, $R \into R' \onto S$
  is a Cohen factorization of $\f$. If $R$ is Cohen--Macaulay and
  satisfies \acuac, then $R'$ has the same properties; see
  \thmref{completion}.
\end{proof}

Following \cite[3.1]{LLA99}, a local homomorphism $\mapdef{\f}{R}{S}$
is called \emph{c.i.}, for complete intersection, if there is a Cohen
factorization $R \to R' \to \Shat$ of the semicompletion $\grave{\f}$
in which the kernel of the surjection $R' \onto S$ is generated by an
$R'$-regular sequence.

Here is an ascent result for the \ac\ and \uac\ properties:

\begin{thm}
  Let $\mapdef{\f}{\Rmk}{(S,\n,k)}$ be a c.i.\ local homomorphism of
  $k$-algebras.
  \begin{prt}
  \item If $R$ is Cohen--Macaulay and satisfies \acuac, then $S$ is
    Cohen--Macaulay and satisfies \acuac.
  \item If $R$ is an AB ring, then $S$ is an AB ring.
  \end{prt}
\end{thm}

\begin{proof}
  Assume that $R$ is Cohen--Macaulay and satisfies \acuac, then
  $\Rhat$ has the same properties; see \thmref{completion}.  By
  \lemref{ascent}, the semicompletion $\grave{\f}$ has a Cohen
  factorization $R \to R' \to \Shat$, in which $R'$ is Cohen--Macaulay
  and satisfies \acuac.  As $\f$ is c.i., the kernel of the
  surjection $R' \onto \Shat$ is generated by an $R'$-regular
  sequence; see \cite[(3.3)]{LLA99}. It follows from \corref{cmreg}
  that $\Shat$ is Cohen--Macaulay and satisfies \acuac\ and,
  therefore, $S$ is Cohen--Macaulay and satisfies \acuac; see
  \thmref{completion}. This proves part (a).

  For part (b), recall that c.i.\ homomorphisms are Gorenstein by
  \thmcite[(2.4)]{LLAHBF90}. Thus, if $R$ is Gorenstein, it then
  follows from \cite[(3.11)]{AFH-94} that $S$ is Gorenstein.
\end{proof}

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TITLE: Proof of existence of $a$ in $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$
QUESTION [4 upvotes]: In deriving the derivative of function $f(x)=\exp(x)$, it is often pointed out that in the general case of $f_a(x)=a^x$ the following expression can be deduced from the definition of the derivative:
$$
\frac{d}{dx}a^x = a^x\cdot\left(\lim_{h\rightarrow0}\frac{a^h-1}{h}\right)
$$
with $e$ defined as the real number for which the above limit term is equal to 1.
However, this seems unsatisfactory to me as the existence of a real number $a$ satisfying the equation $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$ is not established.
Can somebody please point me to a proof of the existence of such a number or provide it if possible?
Thanks.

REPLY [0 votes]: One possible approach is to define $e$ as the limit of the sequence
$$(1+\frac{1}{n})^n$$
It takes a bit of work to show that the above sequence is increasing, and not hard to show that is bounded above by $3$.
With some more work one shows that
$$(1+\frac{1}{n})^{n+1}$$
is decreasing. They both have the same limit, which one defines to be the number $e$.
From here one checks that
$$e^{\frac{1}{n+1}}<1+\frac{1}{n}<e^{\frac{1}{n}}$$
and this implies
$$\lim_{x\to 0_{+}}\frac{e^x-1}{x}=1$$
and from this
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$


TITLE: Order statistics when variables have different distributions
QUESTION [0 upvotes]: Let a,b and c be random variables, where a~U[0,8] and b~U[3,8]. Let c=max{a,b}, What is the mean of c?
In general, let $(x_1, ..., x_n)$ be independently distributed in different supports. What is the distribution of its maximum, i.e. of its highest order statistic?
I have the solution for my first question (from simulating it in Matlab) but I am looking for the general expression. I encountered the Bapat-Beg Theorem but I am convinced there is something simpler I can use. Intuitive explanations are very welcomed.

REPLY [1 votes]: First, it is clear that the support of $c$ must be $(3,8).$
The maximum must take values in this interval.
Second, the cdf of $c$ must be
$$F_c(t) = P(c \le t) = P(a \le t, b \le t) = P(a \le t)\;P(b \le t) = \cdots ,$$
for $3 < t < 8.$ This is a very common way to find the
distribution of the maximum of two independent random variables.
Third, find the density function and integrate to find the mean.
This is a case in which simulation gives you a pretty good
idea of the answer as a check on your analysis.
 m = 10^6;  a = runif(m, 0, 8);  b = runif(m, 3, 8)
 c = pmax(a,b);  hist(c, prob=T);  mean(c)
 ## 6.022239

The histogram shows a trapezoidal PDF, and the simulation gives a reasonably good estimate of
the expectation (maybe one or two place accuracy).


TITLE: Express the following sentence using logical connectives, predicates and quantifiers:
QUESTION [1 upvotes]: There is someone who can fool exactly one person besides himself or herself

Predicate given: F(x, y): x can fool y.

My answer attempt:
ヨxヨy (F(x, y) ⋀ ((x = y) V ( (x ≠ y) ⋀ ∀k(F(x, k) -> (k = y))))

Way of thinking:

There exists a person x and there exists a person y for which x can fool y and if so, then either x IS y OR if x is not y, and if for all k, x can fool k, then k is indeed y.
 Please correct me if I am wrong.

REPLY [2 votes]: You're almost there. I would say:
$$\exists x (\exists y: F(x,y) \land (\forall z: (F(x,z) \land (z \neq x)\rightarrow z=y)$$
In words: there is some $x$ that can fool some $y$ and if we ever find someone that is fooled by $x$ and is not $x$ itself, it must have been $y$.


TITLE: Separating an orbit and a point by neighborhoods
QUESTION [2 upvotes]: Let $G$ be a group acting on the Hausdorff topological space $X$ discontinuously (meaning each orbit does not contain any limit point in $X$). Let $x,y\in X$ be two points that are not $G$ equivalent, meaning there is no $g\in G$ such that $g.x=y$. I wonder how to show the following:

There exists open neighborhoods $U$ of $x$  and $V$ of $y$ such that $gU\cap V=\varnothing$ for all $g\in G$.

It is easy to see that we can find an open neighborhood $V$ of $y$ such that $G.x\cap V=\varnothing$. But I don't know how to continue.
Source: Iwaniec topics in classical automorphic forms, page 29.

REPLY [1 votes]: I think the stuff from the picture is bad.
As I understood the definition, $\Gamma$ acts on $X$ transitively. This means that an orbit $\Gamma x=X$ for any point $x\in X$, see, for instance, Wikipedia. So each two points of $X$ are $\Gamma$-equivalent and all formal claims about not $\Gamma$-equivalent points of $X$ trivially true. :-)
The stability groups of points need not to be finite. For instance, let $X$ be any infinite discrete space and $\Gamma$ be a group of all bijections of the set $X$. Then $\Gamma$ acts on $X$ transitively and discontinuously, but all stability groups $\Gamma_x$, $x\in X$  are infinite.
If $X$ has no limit points in $X$ then each point of $X$ is isolated. This makes the claim “for any $x\in X$ there exists a small neighborhood $U\dots$” also trivial.
Trying to obtain non-trivial facts from the definitions, we can relax transitivity condition in the definition of a discontinuous action (as you did).

There exists open neighborhoods $U$ of $x$  and $V$ of $y$ such that $gU\cap V=\varnothing$ for all $g\in G$.

But this is not necessarily true. For a counterexample for the claim it suffices in example provided by Moishe Kohan put $x=(1,0)$ and $y=(0,1)$. Indeed, for each neighborhood $U$ of $x$ there exists $\varepsilon>0$ such that $A=\{1\}\times [0,\varepsilon)\subset U$. It is easy to see that $y\in\overline{GA}$.


TITLE: Theoretical Proof of Pauli's Exclusion Principle
QUESTION [8 upvotes]: "No theoretical proof of the Pauli's Exclusion Principle can be given as yet and for the present it must be regarded as something empirical added to and regulating the vector atom model."
I've found it in Atomic & Nuclear Physics by N. Subrahmanyam & Brij Lal.
My question is "What was the motivation behind Pauli's Exclusion Principle? Is that just an ad-hoc intuitive attempt of Pauli to explain the Zeeman effect or it was derived on the basis of mathematics? If mathematical then how? "

REPLY [6 votes]: The Pauli exclusion principle can be derived from relativistic quantum field theory. Even though the Pauli exclusion principle is still important in the non-relativistic approximation, the reason for it is deeply rooted in relativistic QFT.
The theorem is called the spin-statistics theorem. The inputs to the theorem include

Lorentz symmetry,
Something called the spectrum condition, which says that the total energy must have a finite lower bound.

The result is easiest to derive for the case of a free (non-interacting) spin-1/2 particles. This is explained in more detail in Name YYY's answer. Here's a words-only version that includes a little extra context. In QFT, particles are phenomena that the theory predicts, rather than ingredients used to construct the theory. The theory is constructed in terms of quantum fields, which are represented as operators acting on a Hilbert space. For example, all electrons (and antielectrons) are manifestations of the electron field, just like all photons are manifestations of the electromagnetic field. For a free spin-1/2 field, Lorentz symmetry requires that the equation of motion be linear in the space- and time-derivatives, rather than being quadratic like it would be for a scalar or vector field. Because of this, the field operators need to anticommute with each other at spacelike separation (rather than commuting with each other as they would for a scalar or vector field) in order for the energy operator to have an energy spectrum with a finite lower bound. This anticommutativity is the Pauli exclusion principle. This derivation, in the special case of a free spin-1/2 field, is included in many QFT textbooks, such as section 3.5 in Peskin and Schroeder's An Introduction to Quantum Field Theory.
For general derivations:

The classic reference is the book by Streater and Wightman, PCT, Spin and Statistics, and All That (1980). They prove the theorem starting from a system of axioms for relativistic QFT called the Wightman axioms. They also prove another general theorem that is usually called the CPT theorem (they called it "PCT" instead of "CPT" — same thing), which is often presented together with the spin-statistics theorem because they share the same inputs.
It can also be derived from a different system of axioms in the context of algebraic QFT. The relevant theorem in this case is called the Doplicher-Roberts reconstruction theorem, which actually proves much more: it helps explain why field operatores are useful in the first place, in addition to explaining the Pauli exclusion principle. This theorem is reviewed on page 92 in "Algebraic Quantum Field Theory", https://arxiv.org/abs/math-ph/0602036.

REPLY [2 votes]: From my research, the short answer to your question is Pauli came up with the principle to explain more on what was happening within an atom, but he himself could not explain where it comes from. The Pauli exclusion principle has no derivation. While Wolfgang Pauli was trying to answer the big quantum question at the time, why don't all the electrons go into the lowest energy state, he had come up with the Pauli exclusion principle. At the time many physicists were confused by this because they had no idea where it even came from, and why it even exists in the first place. In fact, Pauli himself was troubled that he couldn't explain his own principle from logic and that he couldn't derive it from any quantum equations. This is a common theme within quantum physics, where there are many laws and equations that have no proper explanation on why they happen, and scientists are still trying to find more answers.
My source: https://www.aps.org/publications/apsnews/200701/history.cfm


TITLE: limit with greatest integer function
QUESTION [2 upvotes]: How can I evaluate this limit?
$$\lim_{x\to-7} \frac{[x]^2+15[x]+56}{\sin(x+7)\sin(x+8)}$$ where $[x]$ denotes the greatest integer less than or equal to $x$
I could easily factor the numerator. But I cannot apply any of the standard limits due to $[x]$.

REPLY [2 votes]: The way the question is phrased makes me wonder if you've understood the question, either the nature of limits or the nature of the greatest-integer function.  Your final sentence makes me suspect the latter.
If $x<-7$ then $\lfloor x\rfloor=-8$.
If $x>-7$ then $\lfloor x\rfloor= -7$.
It you need that explained to you you then you haven't understood the greatest-integer function and that would be what you need to look at.
In one case the numerator is $(-8)^2+15(-8)+56=0$ and in the other case it is $(-7)^2 + 15(-7)+56 = 0$.  Either way the fraction is equal to $0$ in small neighboorhoods about $-7$ unless the denominator is $0$.  And in small neighborhoods about $-7$, the denominator is $0$ only if $x=-7$, so essentially your asking for $\lim\limits_{x\to-7} 0$.


TITLE: Show that $f \in L_\infty(\mathbb{R})$ as soon as $f'$ is a Radon measure of finite total variation.
QUESTION [1 upvotes]: Let $f$ be a generalized function in $\mathcal{D}'(\mathbb{R})$ such that its derivative $\mathrm{D} f = \mu \in \mathcal{M}(\mathbb{R})$, the space of signed Radon measure with finite total variation, endowed with the total variation norm
$ \| \mu \|_{\mathcal{M}} =  |\mu|(\mathbb{R}) < \infty.$
Is there any simple argument to say that $f \in L_\infty(\mathbb{R})$ is (a.e.) bounded.

REPLY [2 votes]: Suppose $f\in L^{loc}_1(\mathbb{R})$ and $\nu_f:=Df\in\mathcal{M}(\mathbb{R})$. Let $F(x)=\nu_f(-\infty,x]$. It follows from Lebesgue integration by parts that for any $\phi\in\mathcal{D}(\mathbb{R})$,
$$\int_{\mathbb{R}}\phi(s)\nu_f(ds)=-\int_{\mathbb{R}}F(s-)\phi'(s)\,ds=-\int_{\mathbb{R}} F(s)\,\phi'(s)\,ds$$
On the other hand,
$$\int_\mathbb{R} \phi(s)\nu_f(ds)=-\int_\mathbb{R} f(s)\phi'(s)\,ds$$
Hence $$\int_{\mathbb{R}}F(s)\,\phi'(s)\,ds=\int_\mathbb{R} f(s)\,\phi'(s)\,ds$$
for all $\phi\in\mathcal{D}(\mathbb{R})$. The problem reduces to

Problem 0: Prove that if $u\in\mathcal{D}'(\mathbb{R})$ is such that $Du(\phi)=-u(\phi')=0$ for all $\phi\in\mathcal{D}(\mathbb{R})$, then $u\equiv constant$, that is, for some $c\in\mathbb{R}$, $u(\phi)=c\int_\mathbb{R}\phi(s)\,ds$ for all $\phi\in\mathcal{D}(\mathbb{R})$.

This would imply that $f(t)=\nu_f(-\infty,t]+c$ (Lebesgue almost surely) for some $c$ and hence, $f\in L_\infty$.
Problem 0 has been discussed in MSE. See here for example.

Edit: This is to address the more general setting of the OP.
Suppose $u\in \mathcal{D}'(\mathbb{R})$ and $\mu:=Du\in\mathcal{M}(\mathbb{R})$. Define $f(x)=\mu(-\infty,x]$. $f$ is a bounded function of finite bounded variation. By Lebesgue's integration by parts, for any $\phi\in\mathcal{D}(\mathbb{R})$
$$Df(\phi):=-\int f(x)\phi'(x)\,dx=\int\phi(x)\,\mu(dx)=Du(\phi)$$
Hence $f$, as an element in $\mathcal{D}'(\mathbb{R})$, and $u$ have the same derivative. Hence,  they differ by a constant $c$. This means that $u$ admits a representation in $L^{loc}_1(\mathbb{R})$, namely
$$u(\phi)=\int  (f(x)+c)\phi(x)\,dx,\qquad \phi\in\mathcal{D}(\mathbb{R})$$
As a consequence, $u\in L_\infty$.
Thanks to @MaoWao for his observation.


TITLE: Using fast flowing rapid to pump water
QUESTION [0 upvotes]: I am here in Zambia. I have a fast flowing river at a safari camp and I want to use this to pump water to a header tank some 30ft above the river. I am looking to create a tapered box with one end 100cm x 50cm ish and the other end tapering to a 40mm pipe. The box would be about 2m long and have a fall within the river of about 20cm (river bed gradient). So the water flows into the wide end of my box and the pressure will push water up through my 40mm pipe and into my tank. And this will happen as long as the river flows which so far is several million years. My question therefore is will this work. Or am I wasting my time. The river flow is probably 1m per second.

REPLY [1 votes]: According to the Bernoulli equation, a fluid stream traveling at velocity v can support a column of water of height $\frac{v^2}{2g}$ (above the water surface).  For 1 m/sec, that translates into 0.05 m, or 5 cm.  That's quite a bit less than 30 ft.  And that would only be if the flow up the column is zero.  You would need a stream velocity in excess of 15 m/s to really make this work.


text:
\subsection{Proof of Proposition \ref{prop:uniform_convergence}}
\label{appen:uniformconv}
We first prove the pointwise convergence of $g_{p}(v)$ to $\E g_{p}(v)$: there exists $c>0$, s.t., for any $v\in[0,2]$ and $\veps>0$,
\begin{equation}
\P\left(\left|g_{p}(v)-\E g_{p}(v)\right|>\varepsilon\right)\leq ce^{-c^{-1}p\min\left\{ \tfrac{\veps^2}{\sampratio},\veps \right\} }.\label{eq:concentration_Gp_pointwise}
\end{equation}

Recall that $g_{p}(v)=\mathcal{G}_{p}\left( {\vacav}{v} \right)$, so it is equivalent to prove $|\mathcal{{G}}_{p}({\vacav}v)-\E\mathcal{{G}}_{p}({\vacav}v)|\to 0$. We first control the moment generating function of $\mathcal{{G}}_{p}({\vacav}v)-\E\mathcal{{G}}_{p}({\vacav}v)$. Let $\vb$ be an i.i.d. copy of $\va$. For all $|\lambda|\leq\frac{p}{2\sqrt{2}\pi}$, we can apply Theorem 2.2 of \cite[p.176]{pisier1986probabilistic} to get
\begin{align*}
\E[\exp\lambda(\mathcal{{G}}_{p}({\vacav}v)-\E\mathcal{{G}}_{p}({\vacav}v))] &\leq \E e^{ \frac{\pi\lambda}{2}(\vb v)^{\T}\nabla\mathcal{{G}}_{p}({\vacav}v) }\\
&\tleq{\text{(a)}}\E_{\dmtx,\noise}\E_{\vacav}e^{\frac{2\lambda^{2}\pi^{2}}{p}\|\vacav\|^{2}}\\
& =\exp\left[-\frac{n}{2}\log\left(1-\frac{4\lambda^{2}\pi^{2}}{p^{2}}\right)\right]\\
& \tleq{\text{(b)}}\exp\left(\frac{4\sampratio\lambda^{2}\pi^{2}}{p}\right).
\end{align*}
In step (a), we take expectation over $\vb$ and use $|v|\leq 2$ and $\|\nabla\mathcal{{G}}_{p}({\vacav}v)\| \leq {2\|{\vacav}\|}$, as implied by (\ref{eq:Gp_Lipschitz}); In step (b), we use the inequality $\log(1+x)\geq\frac{x}{1+x}$,
for $x>-1$ and the condition that $|\lambda|\leq\frac{p}{2\sqrt{2}\pi}$.
As a result, for any $\veps \geq 0$ and $\lambda\in\left[0,\frac{p}{2\sqrt{2}\pi}\right]$,
\begin{align}
\label{eq:Chernoff1}
\P\left( g_{p}(v)-\E g_{p}(v) > \varepsilon \right) & \leq e^{-\lambda\varepsilon+\frac{4\sampratio\lambda^{2}\pi^{2}}{p}}.
\end{align}
After minimizing the exponent on the RHS of (\ref{eq:Chernoff1})
over $\lambda\in\left[0,\frac{p}{2\sqrt{2}\pi}\right]$,

we can get for any $\veps\in[0,\sqrt{8}\pi\sampratio]$, $\P\left(g_{p}(v)-\E g_{p}(v)>\varepsilon\right)\leq e^{-\tfrac{p\varepsilon^{2}}{16\sampratio\pi^{2}}}$; for any $\veps > \sqrt{8}\pi\sampratio$, $\P\left(g_{p}(v)-\E g_{p}(v)>\varepsilon\right)\leq e^{-\tfrac{p\varepsilon}{4\sqrt{2}\pi}}$. The other direction also holds by the same reasoning. Thus,
\begin{align}
\label{eq:gpv_concentrate}

\P\left(\left|g_{p}(v)-\E g_{p}(v)\right|>\varepsilon\right)&\leq 2 e^{-\tfrac{p}{16\pi^2}\min\{\tfrac{\veps^2}{\sampratio }, \veps\}}.
\end{align}

To show uniform convergence (\ref{eq:concentration_gp_uniform}), it suffices to prove the Lipschitz continuity of $g_{p}(v)$ and $\E g_{p}(v)$. From Lemma 1 of \cite{laurent2000adaptive}, we have for all $x>0$, $\P\Big(\|\vacav\|^2\geq \sampratio +\tfrac{2\sqrt{\sampratio x}}{\sqrt{p}}+\tfrac{2x}{p}\Big)\leq \exp(-x)$.
Let $x=n(\sqrt{y+1}-1)^2$, we have for $y\geq 2$, $\P\left(  \tfrac{\|\va\|^2}{\sampratio}-1\geq y\right)\leq\exp\left(-\tfrac{ny}{4}\right)$.
Therefore, by taking $y=K/\sampratio$, we get for any $K\geq2\sampratio$,
\begin{align}
\label{eq:K_concentrate}
\P(\|\vacav\|^{2}>K) \leq \P(\|\vacav\|^{2}-\sampratio>K/2) \leq \exp(-\tfrac{pK}{4}).
\end{align}
Combining it with (\ref{eq:gp_derivative_upperbd}), we know for $K\geq2 \sampratio$, $g_{p}(v)$ is $2K$-Lipschitz
with probability greater than $1-\exp(-\tfrac{pK}{4})$.
From (\ref{eq:gp_derivative_upperbd}), we can also get $\left|\frac{d\E g_{p}(v)}{dv}\right|\leq 2 \sampratio$, so $\E g_{p}(v)$
is $2 \sampratio$-Lipschitz continuous over $v\in[0,2]$. Combining the Lipschitz continuity
of $g_{p}(v)$ and $\E g_{p}(v)$ with (\ref{eq:concentration_Gp_pointwise}),
we can obtain (\ref{eq:concentration_gp_uniform}) by a standard epsilon-net argument as follows. We need to consider different values of $\veps$:
\begin{enumerate}
  \item If $\veps \geq \sampratio$, we construct an epsilon-net of $[0,2]$ formed by the following points: $v_k = \frac{k}{4}$, $k=1,2,\ldots,8$. For any $v\in[0,2]$, denote $v^{*}$ as the closest point to $v$ in the above epsilon-net. By construction, $|v-v^{*}|\leq \frac{1}{8}$. If $g_{p}(v)$ is $2K$-Lipschitz, then for any $v\in[0,2]$,
      \begin{align}
      |g_{p}(v)-\E g_{p}(v)| &\leq |g_{p}(v)-g_{p}(v^{*})| + |g_{p}(v^{*})-\E g_{p}(v^{*})| + |\E g_{p}(v^{*})-\E g_{p}(v)|\nonumber\\
      &\leq \frac{K}{4} + |g_{p}(v^{*})-\E g_{p}(v^{*})| + \frac{\veps}{2},
      \end{align}
      where we have used the Lipschitz continuity of $g_{p}(v)$ and $\E g_{p}(v)$, as well as $\veps \geq \sampratio$. Then $\sup_{v\in[0,2]}|g_{p}(v)-\E g_{p}(v)|\geq 2\veps$, only if at least one of following holds: (i) $K\geq 2\veps\geq 2\sampratio$, (ii) there exists a $k\in\{1,2,\ldots,8\}$, s.t., $|g_{p}(v_{k})-\E g_{p}(v_{k})| \geq \veps$. Combining (\ref{eq:gpv_concentrate}) and (\ref{eq:K_concentrate}) and applying the union bound, we get for $\veps\geq \sampratio$,
      \begin{equation}
      \label{eq:gpv_concentrate1}
       \P(\sup_{v\in[0,2]}|g_{p}(v)-\E g_{p}(v)|\geq 2\veps)\leq 18 e^{-\tfrac{p\veps}{16\pi^2} }.
      \end{equation}
  \item If $\veps< \sampratio$, we construct an epsilon-net of $[0,2]$ formed by the following points: $v_k = 2k/\lceil\frac{8\sampratio}{\veps}\rceil$, $k=1,2,\ldots,\lceil\frac{8\sampratio}{\veps}\rceil$. In this case, for any $v\in[0,2]$, we have $|v-v^{*}|\leq \frac{\veps}{8\sampratio}$. Then similar as previous argument, we have $\sup_{v\in[0,2]}|g_{p}(v)-\E g_{p}(v)|\geq 2\veps$, only if at least one of following holds: (i) $g_{p}(v)$ is not $4\sampratio$-Lipschitz, (ii) there exists a $k \leq \lceil\frac{8\sampratio}{\veps}\rceil$, s.t., $|g_{p}(v_{k})-\E g_{p}(v_{k})| \geq \veps$. Combining (\ref{eq:gpv_concentrate}) and (\ref{eq:K_concentrate}) and applying the union bound, we get:
      \begin{equation}
      \label{eq:gpv_concentrate2}
       \P(\sup_{v\in[0,2]}|g_{p}(v)-\E g_{p}(v)|\geq 2\veps)\leq \frac{16\sampratio}{\veps} e^{-\tfrac{p\veps^2}{16\pi^2\sampratio} }.
      \end{equation}

\end{enumerate}
Combining (\ref{eq:gpv_concentrate1}) and (\ref{eq:gpv_concentrate2}),

together with symmetry and the union bound, we directly get (\ref{eq:concentration_gp_uniform}).


TITLE: Uniqueness of ODE by checking Lipschitz condition
QUESTION [1 upvotes]: There are many questions on this site about this topic, but as far as I could find, all omitted the step I am not certain about. $I \subseteq \mathbb{R},u_0 \in \mathbb{R}$
The IVP is given as $$\begin{cases}
      u'(t)=2tu(t)=f(t,u(t)) & \text{for}\ t\in I \\
      u(0)=u_0
    \end{cases} $$
I want to first check for existence and uniqueness, so I want to know if it satisfies the Lipschitz Condition in the second variable. I know that if $$\frac{\partial f}{u(t)} $$ is bounded, then by the MVT the Lipschitz Condition is satisfied.
But how do I differentiate $\frac{\partial f}{u(t)} $?
I looked at questions like these: Derivative of a function with respect to another function., but I can't make that work for my example and in examples often times only the solution is given, which I can not follow.

REPLY [1 votes]: A first order IVP is of the form, $$y'=f(x,y(x))\;\;\;,y(x_0)=y_0$$
Here $x$ denotes the independent variable whereas, $y$ is the dependent one.
Computationally, $\displaystyle \frac{\partial f(x,y(x))}{\partial y(x)}$ and $\displaystyle \frac{\partial f}{\partial y}$ are one and the same as, $y$ in itself is a variable.
So for computational ease, we replace $y(x)$ by $y$ and proceed with usual partial differentiation.


TITLE: Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
QUESTION [3 upvotes]: Let $x'=f(t,x)$ be a differential equation with $f$ in the hypothesis of Picard's theorem. Let $\varphi$ be a solution such that its interval of definition contains $(t_0,+\infty)$ for some fixed $t_0\in \mathbb{R}$. Suppose $\lim_{t\to+\infty} \varphi(t)=a \in \mathbb{R}$. Must the constant function $t\mapsto a$ be a solution of the equation?
(This is a question I've asked myself while studying the logistic model $x'=ax(1-x)$ and wondering how to justify the solutions look the way they do.)
EDIT: By "the hypothesis of Picard's theorem", I mean $f$ continuous and locally Lipschitz with respect to $x$.
I'm interested also in particular cases (e.g. autonomous system) where this holds.

REPLY [2 votes]: This is an elaboration of my previuous comment, written at Bruno's request.
Consider the equation $x'=f(x)$ with $f\colon\mathbb{R}\to\mathbb{R}$ continuous and let $x\colon(t_0,\infty)\to\mathbb{R}$ be a solution such that $\lim_{t\to\infty}x(t)=a$. I claim that $f(a)=0$, and hence $z(t)\equiv a$ is a constant solution. The proof is by contradiction.
Suppose $f(a)\ne0$. Without loss of generality we may assume that $f(a)>0$. We have
$$
\lim_{t\to\infty}x'(t)=\lim_{t\to\infty}f(x(t))=f(a).
$$
Let $\delta=f(a)/2$. There exists $t_1\ge t_0$ such that $x'(t)\ge\delta$ if $t\ge t_1$. In particular,
$$
x(t)=x(t_1)+\int_{t_1}^tx'(t)\,dt\ge x(t_1)+(t-t_1)\delta\quad\forall t\ge t_1,
$$
which contradicts the fact that $\lim_{t\to\infty}x(t)=a$.
The proof carries over to autonomous systems of equations.


TITLE: Probability distribution from mean time to failure
QUESTION [0 upvotes]: A factory has different categories of machines which require frequent
  adjustments and repair. Each category of machine fails uniformly after
  continuous operation and the failure profile of the different
  categories of machines is given by its mean time to failure (MTTF).

With the above information I need to assume a probability distribution for their failure profiles. So what are the meaningful/safe assumptions that I should make about the distribution?What actually do we know about machines when we know their MTTF and that they fail uniformly after continuous operation?

REPLY [1 votes]: "Fail uniformly" probably means that they fail at a constant rate (with a different rate for each type of machine). Uniformly is not a good word in this context as it implies the machine failure time follows a uniform distribution (this would be roughly the same number of machines failing in every time period, which is not realistic).
Failing at a constant rate $\lambda$ per unit time happens when the number of fails X in the interval from 0 to t (so $\lambda$t fails on average) follows a Poisson distribution
$P(X = k) = \frac{{{(\lambda t)}^{k}}\exp (-\lambda t)}{k!}$
If roughly half of these machines have failed up to some time $t_0$ then half of the remaining will fail in the next interval up to $t_0$ and so on, decreasing all the time. If T is the time to failure of a machine, which is a random variable, then the probability that the machine is still working after time t is the same as the probability that X is 0 i.e. there has been no fail up to time t. Then
$P(T>t) = P(X = 0) = \exp (-\lambda t)$
This is the Reliability of the machine type, which is the proportion of machines which are still working after time t. The failure rate $\lambda$ is 1/MTTF for that type of machine. So for a given MTTF we can work out the proportion of machines which will fail up to time t using the reliability.
e.g If lightbulbs have a MTTF of 300hrs then the proportion which are still working after 200hrs is
$exp(-\frac{1}{300}*200) = 0.513$ or 51%
The proportion which have failed up to time t is 1 - Reliability. This can be written as an integral up to time t of the probability density for T. Differentiation will then give the density for the variable T, but the Reliability is a more meaningful thing to focus on for practical purposes.


TITLE: Is the topology on compact connected Lie groups metrizable?
QUESTION [1 upvotes]: The question is in the title, really.
Suppose $G$ is a compact connected Lie group. Is there a metric on $G$ which induces the underlying topology? (so in particular $G$ is compact and connected wrt this metric).
I am not really from this area (my differential geometry is slack, to say the least). I am purely interested in the existence of such a metric.
Thanks.

REPLY [0 votes]: The short answer is yes, I believe. The metric that you would be looking for would be the one induced by the Killing form. There is a really good discussion on the construction of the Killing form and its properties here:
http://scipp.ucsc.edu/~haber/ph251/KillingForm.pdf


text:
\section{REFERENCE ALGORITHMS}
To compare our new approach with the existing strategies we present the most important reference algorithms.

For our experiments, it is sufficient to consider the most restrictive case (node placement) and the most flexible approach (free placement). At the moment, we do not consider algorithms for infrastructural placement, since all algorithms can easily be adapted to generate appropriate solutions.

The algorithms in Table \ref{tab:algorithms} are suitable for certain placement constraints.

\begin{table} [htbp]
\begin{center}
\caption{Overview of the presented algorithms and their usability. }
\label{tab:algorithms}
\begin{tabular}{|c|c|}
\hline
Node placement  & Node and Free placement \\
\hline
2-Approx &  k-Means  \\

Greedy & Evolutionary Algorithm  \\

GRASP &   \\

\hline
\end{tabular}
\end{center}
\end{table}
\vspace*{-0.7cm}

\subsection{2-Approx}
The 2-Approx choses a random vertice at the beginning, which becomes the location of the first center node. After that, the algorithm calculates for every placed center node the distance to all vertices. It chooses the vertice with the largest distance to their closest center node as the new location of the next center node. This iteration runs until the specified amount of center nodes is reached.
This algorithm is 2-approximable.

Generally, $\beta$-approximable algorithm with factor $\beta$, guarantees a solution with cost $d$ where  $d \leq \beta  \cdot optimum$.

The algorithm guarantees that the maximum value of a distance from a vertice to its nearest center node is not larger than twice the maximum considering the optimal placement location of the center node \cite{Gonzalez85,Hochbaum85}. This bound is given without knowledge of the actual optimum.

\subsection{k-Means}
The following group of clustering algorithms can be adapted for the free placement and node placement constraints. The main idea is to define k center nodes, one for each cluster. In this case a cluster is a group of vertices. Already placed nodes obviously influence the location of following placed nodes. In contrast to the iterative approach, these algorithms place all center nodes at the same time. This makes backtracking unnecessary. Placement changes of a center node have a direct effect to other center nodes.

Various approaches for clustering exist for a fixed number of clusters, they differ mainly with regard to the initial placement of center points. The \citet{MacQueen67} algorithm is one of the less complex k-Means algorithms. It relies on randomly selected locations of the vertices, which are used as the initial locations of the center nodes. Compared to the algorithm from \citet{Lloyd82}, also known as Voronoi relaxation, it starts with completely randomly placed center nodes in the area. Another typical initialization is used in the k-Means++ algorithm \cite{Arthur07}. Here, the location of the first center node is chosen randomly at a vertice location. The other center nodes are also placed randomly at vertice locations, however, the probability is skewed to favor certain locations. The selection probability is increased proportionally with their squared distances to already selected locations.

After initialization, all vertices are assigned to their respective center nodes. For each group of vertices related to a center node an updated location is calculated. The new center node is the geometrical center of all vertices in a group. This process is repeated until center node locations do no longer change.

During every iteration step, the vertices are reassigned to the nearest center node.

The algorithms are run multiple times due to random initialization. To adapt these strategies for the three constraints, the locations of the center nodes are mapped to the nearest possible location either in every step or at the end of the optimization.

\subsection{Greedy and GRASP}
The Greedy strategy \citet{Jamin01} initially places the center nodes at the location of predefined vertices iteratively. During each iteration it tries all possible placement combinations for the next center node and ultimately selects the location that provides the biggest benefit with respect to the optimization criterion. The Greedy strategy repeats this process until all center nodes are placed.

To improve the quality further they include backtracking, to test whether already placed center nodes can be placed in a better way or may be removed completely.

The advanced approach of GRASP tries several starting locations as initialization for a greedy local optimization \cite{Pitsoulis2009
}. A weighted greedy randomized strategy is used for the initialization process. These starting locations are subsequently iteratively improved using local search. It updates the location of randomly chosen center nodes in a greedy manner.

\subsection{Evolutionary Algorithm}
Finally we use several evolutionary algorithms to optimize the center node locations. The SEREIN framework \cite{serein} is used to implement these algorithms. We employ the standard implementation of a genetic algorithm (GA) provided by SEREIN and use a population with 25 individuals evolving over 80 generations.

In addition, a Particle Swarm Optimization (PSO) and Simulated Annealing (SA) approach is implemented as well.
The parameters for the algorithms were determined experimentally using meta-optimization.


TITLE: Let $f(x)$ be a function whose third derivative exists and $f(x)=0$ has n distinct real roots in the interval $[a,b]$
QUESTION [0 upvotes]: State with proof whether the given statement is true or false,
Statement :
$$
\frac{d}{d x}\left(f(x) f^{\prime}(x) f^{\prime \prime}(x)\right)
$$ has $3n-4$ distinct real roots in $[a,b]$.
I have tried to solve this problem by Rolle's theorem therefore no of real distinct roots of $f'(x)$ and $f''(x)$ are $n-1$ and $n-2$ respectively, their sum with n gives $3n-3$, therefore the derivative of the function in the statement should have $3n-4$ roots, but the answer given is false.
Please rectify my error and provide a solution.

REPLY [2 votes]: As $f(x)=0$ has $n$ distinct zeros, $f'(x)=0$ will have $n-1$ distinct roots, $f''(x)=0$ will have $n-2$ distinct roots.
then $g(x)=f(x)f'(x)f''(x)$ will be zero $n+n-1+n-2=3n-3$ times. So by Rolle's Theorem $g'(x)=0$ will have $3n-4$ real distinct  roots.


TITLE: Converting $x=\frac{1}{2}\cos\theta\;;\;\; y=2\sin\theta $ to Cartesian form
QUESTION [3 upvotes]: How can we transform these parametric equations to Cartesian form?
$x=\frac{1}{2} \cos\theta, \quad y=2\sin\theta \quad\text{ for}\;\;0 \leq \theta \leq \pi$

REPLY [5 votes]: We can use the fact that $\sin^2(\theta) + \cos^2(\theta) = 1,$ and then express the given parametric equations in terms of $\cos\theta$ and $\sin\theta$, respectively.
$x=\frac{1}{2}\cos\theta \iff 2x = \cos\theta$
$y=2\sin\theta \iff \frac12y=\sin\theta$.
Now using the identity: $$\sin^2(\theta) + \cos^2(\theta) = 1,$$ you simply need to substitute into the identity the expressions above for $\sin\theta$ and $\cos\theta$.
Doing so will give you a Cartesian function in $x$ and $y$.


TITLE: Representation of base M digits in binary digits
QUESTION [0 upvotes]: In my textbook I came across the following sentence-
"If $M = 2^{m}$, with $m$ is a positive integer, then each base $M$
digit can be represented by a sub-sequence of $m$ binary digits"
Why is this true?
Let's say we are in base $M=16$. Then $M=2^{4}$ , i.e. $m=4$.
What I'm guessing the above is saying is that if I have some base 16 digits, e.g. $1, 16, 256, 4096$ then all of these digits can be represented by 4 binary digits each? Personally, I can't see how this works. Clearly $256$ cannot be represented by 4 binary digits only.
Thanks for the help!

REPLY [2 votes]: The numbers $1,16,256$ and $4096$ are not base $16$ digits, but rather they are powers of $16$, i.e. $16^0,16^1,16^2,16^3$. We need instead to consider digits.
Taking base $M=16=2^4$, then the digits are $0,1,2,\dots,E,F$, and these can be represented by equivalent binary strings $0000,0001,0010,\dots,1110,1111$, each of which contains $4$ binary digits.
So, given a base $16$ number, say $E2F_{16}$, we can immediately convert this to binary as $1110\,0010\,1111_2$.
As claimed, a similar conversion can be made in any base $M=2^m$ using $m$ binary digits for each base $M$ digit.


TITLE: understanding an argument involving derivative and integral of a dirac delta generalised function
QUESTION [0 upvotes]: An argument about moving the one-dim wave equation in vector space to position space involves the below equality.
Note: the primes on the variable x do not denote derivatives of any sort. They are merely placeholders for distinguishing different positions.
$\frac{\partial }{\partial x}\left (  \int_{a}^{b}\int_{a}^{b} \delta \left ( x-x' \right )\left ( \frac{\partial }{\partial x}\delta \left ( x'-x'' \right ) \right )f\left ( x'',t \right )dx'dx'' \right )$
leads to
$\frac{\partial }{\partial x}\int_{a}^{b}\left ( \frac{\partial }{\partial x} \delta \left ( x-x'' \right )\right )f\left ( x'',t \right )dx''$
I would like to understand the above argument.
Any illumination would be helpful.

REPLY [2 votes]: We have for any $g$
$$\int_a^b \delta(x-x') g(x'-x'') dx' = g(x-x'')$$
Now take $g(x'-x'') = \frac{\partial}{\partial x} \delta(x'-x'')$ and you have your result.