text:
\begin{document}

\title[Numbers with almost all convergents in a Cantor set]
  {Construction of numbers\\ with almost all convergents in a Cantor set}

\author{Damien Roy \and Johannes Schleischitz}

\begin{abstract}
In 1984, K. Mahler
asked how well elements in the Cantor middle third set
can be approximated by rational numbers from that set,
and by rational numbers outside of that set. We consider
more general missing digit sets $C$ and construct
numbers in $C$ that
are arbitrarily well approximable
by rationals in $C$, but badly approximable
by rationals outside of $C$.
More precisely, we construct them so that all but finitely many of
their convergents lie in $C$.
\end{abstract}

\keywords{Cantor set, continued fractions,
Diophantine approximation, parametric geometry of numbers}
\subjclass[2010]{Primary 11A55; Secondary 11J25, 11J82}

\maketitle

\section{Introduction and statement of the main result}
\label{intro}

Let $b\geq 3$ be an integer and $D$ be a proper subset of
$\{0,1,\ldots,b-1\}$ with at least two elements.
We consider the Cantor set $C$ which consists of all real numbers in the
interval $[0,1]$ whose base $b$ expansion involves only the
digits from the set $D$.  This is a compact subset of $\bR$ of
measure zero. It is called the \emph{middle third Cantor set} when $b=3$
and $D=\{0,2\}$.  In 1984, K.~Mahler \cite{Ma1984} proposed a problem
about this set, which also applies to any Cantor set.  He asked how
well irrational elements of $C$ can be approximated by rational
numbers from $C$, and how well they can be approximated by rational
numbers outside of $C$.  A construction of Y.~Bugeaud \cite{Bu2008}
(see also \cite[\S 2.2]{Sc2016})
generalizing earlier work \cite{Sh1979, PS1992} of J.~Shallit and
A.~J.~van der Poorten provides an interesting answer.  For any
monotone decreasing function $\psi\colon\bN\to(0,\infty)$ with
$\lim_{q\to\infty}q^2\psi(q)=0$, defined on the set $\bN$ of positive
integers, it yields an irrational element $\xi$ of $C$ and a
constant $c=c(b)>0$ such that
\[
 \begin{array}{ll}
  |\xi-p/q| \le \psi(q)
   &\text{for infinitely many $p/q \in \bQ\cap C$,}\\[2pt]
  |\xi-p/q| \ge c\,\psi(q)
   &\text{for all $p/q\in \bQ$,}
 \end{array}
\]
with $p/q$ in reduced form.  However, because the
construction is based on the folding lemma, such a number
$\xi$ possesses many good rational approximations $p/q$ besides
those for which $|\xi-p/q| \le \psi(q)$.  As we don't
know whether they belong to $C$ or not, we lack information
about approximation to $\xi$ by rational numbers outside
of $C$.  Our main result below is more precise in this aspect
and, at the same time, answers a question of L.~Fishman and D.~Simmons
in \cite[\S 2.1]{FS2015} by providing irrational elements
of $C$ with all but finitely many convergents inside $C$
(see \cite[Chapter I]{ScLNM} for the notion of convergents of
a real number, and the theory of continued fractions).

\begin{theorem}
 \label{mainres}
Let $C$ be as above.  Then, there is a constant $c_1$,
depending only on $b$ and $D$, with the following property.
For any $\epsilon>0$ and any function
$\Psi\colon \bN\to (0,1]$, there exists $\xi\in C$
whose convergents $p/q\in\bQ$ (in reduced form) with
denominator $q\ge c_1$ all lie in $C$ and satisfy
\begin{equation}
 \label{eq:mainres}
 \min\{\Psi(q), q^{-q}\}
  > \left| \xi - \frac{p}{q} \right|
  > c_2q^{-(1+\epsilon)q}\Psi(q)
\end{equation}
for a constant $c_2=c_2(b,\epsilon)>0$.
If $D$ contains $\{0,1\}$, we may take $c_1=1$, meaning that
all convergents of $\xi$, starting with $0/1$, belong to $C$.
\end{theorem}

In particular, the numbers $\xi$ of $C$ that we construct are
Liouville numbers that are $\psi$-approximable by rational
numbers inside $C$ and badly approximable by rational numbers
outside of $C$.  Indeed, if a fraction $p/q\in \bQ\setminus C$,
in reduced form, has denominator $q\ge c_1$, then $p/q$ is not
a convergent of $\xi$ and so $|\xi-p/q|\ge 1/(2q^2)$.  On the
other hand, a result of L.~Fishman and D.~Simmons
\cite[Corollary 1.2]{FS2015} shows the existence of a
constant $c_3=c_3(b)>0$ such that the inequality
$|\xi-p/q|\le c_3/q^2$ has infinitely many solutions
$p/q \in \bQ\setminus C$.  Thus the approximation to $\xi$
by rational numbers outside of $C$ is under control as well.
As the proof will show, we even obtain explicit
base $b$ expansions for the convergents of $\xi$ with large enough
denominators.

Note that, for a general Cantor set $C$, there may exist no element
of $C$ with all its convergents in $C$.  For example, if $b$ is a
large Fibonacci number and if $D=\{d,d+1\}$ where $d$ is the preceding
Fibonacci number, then $C\subseteq [d/(b-1),(d+1)/(b-1)]$ and all elements
of $C$ have the same initial convergents $0/1,\,1/1,\,1/2,\,2/3,\,3/5,\dots$ none
of which belong to $C$.

The original motivation for this paper was to determine
whether or not Schmidt and Summerer parametric geometry of
numbers \cite{Ro2015, SS2013} extends without
qualitative change when restricting to points of the form
$(1,\xi_1,\dots,\xi_n)$ with $\xi_1,\dots,\xi_n\in C$ instead
of the general points with $\xi_1,\dots,\xi_n\in \bR$.
For $n=1$, the question amounts to determine whether or not
for any irrational $\xi\in\bR$ there exists $\xi'\in C$ and a constant
$c>1$ such that, for any convergent $p/q$ of $\xi$ (resp.\ $p'/q'$
of $\xi'$), there exists a convergent $p'/q'$ of $\xi'$
(resp.\ $p/q$ of $\xi$) with $q\le cq'$ and $q'\le cq$.
We do not know the answer but we observed that if the denominators
of the convergents of $\xi$ grow very fast then $\xi'$ must have
essentially all its convergents in $C$ and the search for such
numbers $\xi'$ led us to the construction that we describe below.

\section{Proof of the theorem}
\label{sec:proof}

We will assume, without loss of generality, that $D$ consists of only
two digits $d_1,d_2$ with $0\le d_1 <d_2\le b-1$.  Let $D^*$
denote the monoid of finite words on the alphabet $D$ with the
product given by concatenation, and let $|w|$ denote the length of
a word $w\in D^*$.  Then, each rational number in $C$, except possibly $1$,
has an ultimately periodic base $b$ expansion of the form
\[
 (0.v\,\overline{w})_b = \frac{(vw)_b-(v)_b}{b^m(b^N-1)}
 \quad
 \text{with}
 \quad
 m=|v|
 \et
 N=|w|>0,
\]
where $v\in D^*$ is a possibly empty pre-period, and $w\in D^*$ is a non-empty
period.   The numerator in the right hand-side of the formula is the difference
of two integers $(vw)_b$ and $(v)_b$, written in base $b$.

For each non-empty word $w\in D^*$, let $w'$ be the word obtained
from $w$ by replacing its last letter or digit by the other element
of the set $D$, so that $w$ and $w'$ differ only in their last digits.
Our construction depends uniquely on the choice of a
strictly increasing sequence of non-negative integers $(m_i)_{i\ge 1}$.
We define a word $v$ and a sequence of words $(w_i)_{i\ge 1}$ in $D^*$ by
\[
 v=d_1^{m_1},
 \quad
 w_1=d_2d_1^{m_1}
 \et
 w_{i+1}=(w_i)^{m_{i+1}}w_i'
 \quad
 \text{for each $i\ge 1$.}
\]
Then the sequence of rational numbers, in reduced form,
\[
 \frac{p_i}{q_i}=(0.v\,\overline{w_i})_b
 \quad (i\ge 1)
\]
is contained in $C$ and converges to an element $\xi$ of $C$.  We claim that,
for an appropriate choice of $(m_i)_{i\ge 1}$, they are consecutive
convergents of $\xi$.  The simplest case is when $D=\{0,1\}$.  As we will
see, we can then choose $m_1=0$ so that $v$ is the empty word and
all fractions $p_i/q_i$ have purely periodic base
$b$ expansion.  The reader who wants to concentrate on this case may
skip the technical lemma \ref{lemma:choice_m1}.  We start with a simple
computation.

\begin{lemma}
\label{lemma:diff_p/q}
For each $i\ge 1$, we have
\begin{equation}
\label{eq:lemma:diff_p/q}
 \frac{p_{i+1}}{q_{i+1}}
 = \frac{p_i}{q_i} + \frac{(-1)^{i+1}u}{b^{m_1}(b^{N_{i+1}}-1)}
\end{equation}
where $u=d_2-d_1$ and $N_{i+1}=|w_{i+1}|$.
\end{lemma}

\begin{proof}
Since $w_i$ ends in $d_1$ for odd indices $i$ and in $d_2$ for even ones, we find
\[
 (0.v\,\overline{w_{i+1}})_b
 = (0.v\,\overline{w_i})_b + (-1)^{i+1}u(0.0^{m_1}\overline{\epsilon})_b
\]
where $\epsilon$ consists of $N_{i+1}-1$ zeros followed by a one. The result follows.
\end{proof}

\begin{lemma}
\label{lemma:criterion}
Suppose that the sequence $(q_i)_{i\ge 1}$ is strictly increasing.
Then  $(p_i/q_i)_{i\ge 1}$ consists of all convergents to $\xi$ with denominator
at least $q_1$ if and only if, for each $i\ge 1$, we have
$b^{m_1}(b^{N_{i+1}}-1)=uq_iq_{i+1}$.
\end{lemma}

\begin{proof}
The formula \eqref{eq:lemma:diff_p/q} from
Lemma \ref{lemma:diff_p/q} can be rewritten as
\[
 \det\begin{pmatrix} p_{i+1} &q_{i+1}\\ p_i &q_i \end{pmatrix}
 = (-1)^{i+1}\frac{uq_iq_{i+1}}{b^{m_1}(b^{N_{i+1}}-1)}
 \quad (i\ge 1).
\]
If $p_i/q_i$ and $p_{i+1}/q_{i+1}$ are consecutive convergents of $\xi$
then the above determinant is $\pm 1$ and so $uq_iq_{i+1}=b^{m_1}(b^{N_{i+1}}-1)$.
Conversely, suppose that the latter equality holds for each $i\ge 1$.
Then we have
\begin{equation*}
\label{eq1:lemma:criterion}
 \det\begin{pmatrix} p_{i+1} &q_{i+1}\\ p_i &q_i \end{pmatrix}
 = (-1)^{i+1}
 \quad (i\ge 1),
\end{equation*}
and, since $(q_i)_{i\ge 1}$ is increasing, we conclude that
the sequence $(p_i/q_i)_{i\ge 1}$ consists of all convergents of its
limit $\xi$, with denominator at least $q_1$.  We leave the verification
of this fact as an interesting exercise about continued fractions
(we do not have a precise reference to propose).
\end{proof}

The choice of $m_1$ is the most delicate part of the argument.
It depends on the factorisation of $u=d_2-d_1$ in the form
\[
 u = u_1u_2,
\]
where $u_1,u_2$ are positive integers with the prime factors of $u_1$
dividing $b$ and those of $u_2$ not dividing $b$.  In the
statement below, $\varphi$ denotes Euler's totient function.

\begin{lemma}
\label{lemma:choice_m1}
Suppose that $m_1=N-1$ where $N=\varphi(u_2^2(b-1)^2)$.
Then, we have $u_1|b^{m_1}$ and the reduced fraction
$p_1/q_1=(0.v\,\overline{w_1})_b$ satisfies $u_2q_1=b^{N}-1$.
\end{lemma}

\begin{proof}
Since $N\ge u_2(b-1)\ge b-1$, we have $m_1\ge b-2\ge
u_1-1\ge v_p(u_1)$ for any prime divisor $p$ of $u_1$,
where $v_p$ denotes the valuation at $p$. Since any such
prime $p$ divides $b$, it follows that $u_1|b^{m_1}$.

For the second assertion, set $S=1+b+\cdots+b^{N-1}$,
so that $b^N-1=(b-1)S$.  We find
\[
 (0.v\,\overline{w_1})_b
 = (0.d_1^{m_1}\overline{d_2d_1^{m_1}})_b
 = (0.\overline{d_1^{m_1}d_2})_b
 = \frac{(d_1^{m_1}d_2)_b}{b^N-1}
 = \frac{ d_1S+u}{b^N-1}.
\]
Thus, we simply need to show that $\gcd(d_1S+u, b^N-1)=u_2$ or,
equivalently, that
\[
 \min\{v_p(d_1S+u), v_p(b^N-1)\} = v_p(u_2)
\]
for every prime factor $p$ of $b^N-1$.  Fix such a prime number $p$.

Since $b$ is coprime to $u_2(b-1)$, we have $b^N\equiv 1 \mod u_2^2(b-1)^2$.
Thus, $u_2^2(b-1)$ divides $S$.  If $p$ divides $u_2(b-1)$, this implies
that
\[
 v_p(S)\ge v_p(u_2^2(b-1))>v_p(u_2)=v_p(u),
\]
so $v_p(d_1S+u)=v_p(u_2)<v_p(S)\le v_p(b^N-1)$ and we are done.
Otherwise, $p$ divides $S$ but not $u$, so it does not divide $d_1S+u$
and we are done again.
\end{proof}

\begin{lemma}
\label{lemma:recursion}
Let $u_1$ and $u_2$ be as above.  Suppose that
\begin{itemize}
  \item[(i)]
  $u_1|b^{m_1}$ \et $u_2q_1=b^{m_1+1}-1$,
  \smallskip
  \item[(ii)]
  $q_0p_1(m_2+1)\equiv -1 \mod q_1$ \quad \text{where}\quad $q_0=b^{m_1}/u_1$,\
  \smallskip
  \item[(iii)]
  $q_i|m_{i+1}$ \quad for each $i\ge 2$.
\end{itemize}
Then, for each $i\ge 1$, we have
\begin{equation}
 \label{eq:lemma:recursion}
  b^{m_1}(b^{N_i}-1)=uq_{i-1}q_i
  \quad
  \text{where}
  \quad
  N_i=|w_i|.
\end{equation}

\end{lemma}

\begin{proof}
We proceed by induction on $i$.  If $i=1$, we find
$uq_0q_1=b^{m_1}u_2q_1=b^{m_1}(b^{N_1}-1)$ since
$N_1=m_1+1$.  Suppose now that \eqref{eq:lemma:recursion}
holds for some integer $i\ge 1$.  Since $N_{i+1}=(m_{i+1}+1)N_i$,
we have
\[
 b^{N_{i+1}}-1 = (b^{N_i}-1)S_{i+1}
 \quad\text{where}\quad
 S_{i+1}=1+b^{N_i}+\cdots+b^{m_{i+1}N_i},
\]
and so Lemma \ref{lemma:diff_p/q} yields
\[
 \frac{p_{i+1}}{q_{i+1}}
 = \frac{p_i}{q_i}
   + \frac{(-1)^{i+1}u}{uq_{i-1}q_iS_{i+1}}
 = \frac{R_{i+1}}{q_{i-1}q_iS_{i+1}}
 \quad\text{where}\quad
 R_{i+1}=q_{i-1}p_iS_{i+1}+(-1)^{i+1}.
\]
To complete the induction step, we simply need to show
that $q_i$ divides $R_{i+1}$ because, since
$q_{i-1}S_{i+1}$ is coprime to $R_{i+1}$, this implies
that $q_{i+1}=q_{i-1}S_{i+1}$ and so
\[
 b^{m_1}(b^{N_{i+1}}-1)
  =b^{m_1}(b^{N_i}-1)S_{i+1}
  =uq_{i-1}q_iS_{i+1}
  =uq_iq_{i+1}.
\]
When $i=1$, we use the fact that
$b^{N_1}\equiv 1 \mod q_1$ by Condition (i).
This implies that $S_2\equiv m_2+1 \mod q_1$ and thus,
using Condition (ii), we obtain, as needed,
\[
 R_2\equiv q_0p_1(m_2+1)+1 \equiv 0 \mod q_1.
\]
Now suppose that $i>1$.  Then, \eqref{eq:lemma:recursion}
has the following two consequences.  On the one hand, in
combination with Lemma \ref{lemma:diff_p/q}, it yields
\[
 p_iq_{i-1}-q_ip_{i-1}
  =q_{i-1}q_i\left(\frac{p_i}{q_i}-\frac{p_{i-1}}{q_{i-1}}\right)
  = (-1)^{i},
\]
so $p_iq_{i-1}\equiv(-1)^{i}\mod q_i$, and thus $R_{i+1}\equiv
(-1)^i(S_{i+1}-1) \mod q_i$.  On the other hand, it shows that $q_i$
divides $b^{m_1}(b^{N_i}-1)$, so
$q_i$ divides $b^{m_1}q_i^*$ where $q_i^*=\gcd(q_i,b^{N_i}-1)$.
Modulo $q_i^*$, we have $b^{N_i}\equiv 1$ and $m_{i+1}\equiv 0$
by Condition (iii), thus $S_{i+1}\equiv m_{i+1}+1\equiv 1$.
Since $u_1|b^{m_1}$ and $N_i> N_1>m_1$, we also have
$S_{i+1}\equiv 1 \mod b^{m_1}$.  As $b^{m_1}$ and $q_i^*$ are
coprime, this implies that $b^{m_1}q_i^*$ divides $S_{i+1}-1$
and so $R_{i+1}\equiv (-1)^i(S_{i+1}-1)\equiv 0 \mod q_i$.
\end{proof}

\begin{proof}[\textbf{Proof of Theorem \ref{mainres}}]
Fix a choice of $\epsilon>0$ and of a function $\psi\colon\bN\to(0,1]$.
If $d_2\neq 1$, we take $m_1$ as in Lemma \ref{lemma:choice_m1}
so that Condition (i) of Lemma \ref{lemma:recursion} holds.
Otherwise, we have $d_1=0$ and $u=d_2=1$, and we set $m_1=0$.
This yields $p_1/q_1=(0.\overline{1})_b=1/(b-1)$, and so
Condition (i) still holds.  Moreover, in both cases, the product $q_0p_1
=p_1b^{m_1}/u_1$ is coprime to $q_1$.  Thus, the integers $m_2$
satisfying Condition (ii) of Lemma \ref{lemma:criterion} form
a congruence class modulo $q_1$.  We choose $m_2$ to be the smallest
positive element of that class with $m_2\ge q_1$ for which the
corresponding fraction $p_2/q_2=(0.v\,\overline{w_2})_b$ satisfies
$1/(q_1q_2)<\psi(q_1)$.
More generally, once $m_i$ and $p_i/q_i$ are constructed for
some index $i\ge 2$, we choose $m_{i+1}$ to be the smallest
positive multiple of $q_i$ such that $p_{i+1}/q_{i+1}
=(0.v\,\overline{w_{i+1}})_b$ satisfies
\begin{equation}
 \label{eq1:proof}
 \frac{1}{q_iq_{i+1}}<\psi(q_i).
\end{equation}
This is possible at each step $i\ge 1$ because $N_{i+1}=
|w_{i+1}|=(m_{i+1}+1)N_i$ tends to infinity with $m_{i+1}$
and so, according to Formula \eqref{eq:lemma:recursion}
in Lemma \ref{lemma:recursion}, the ratio
\begin{equation}
 \label{eq2:proof}
 \frac{1}{q_iq_{i+1}}=\frac{u}{b^{m_1}(b^{N_{i+1}}-1)}
\end{equation}
tends to $0$.

We claim that, upon putting $N_0=1$, we have
\begin{equation}
 \label{eq3:proof}
 b^{m_iN_{i-1}} \le q_i < b^{(m_i+1)N_{i-1}}=b^{N_i}
\end{equation}
for each $i\ge 1$.  For $i=1$, this follows from
\[
 b^{m_1}
   \le \frac{(b-1)b^{m_1}}{u_2}
   \le \frac{b^{m_1+1}-1}{u_2}=q_1
   \le b^{m_1+1}.
\]
If $i>1$ and if we assume that \eqref{eq3:proof} holds
for all smaller values of $i$, then we have $q_{i-1}\ge q_0$
and, since $u\le b^{m_1}\le uq_0$, we find
\[
 b^{N_i-N_{i-1}}
   \le \frac{b^{N_i}}{q_{i-1}+1}
   < \frac{b^{m_1}(b^{N_i}-1)}{uq_{i-1}}
   = q_i
   < \frac{b^{m_1+N_i}}{uq_0}
   \le b^{N_i}.
\]
So, by induction, \eqref{eq3:proof} holds for all $i\ge 1$.

In particular, the sequence $(q_i)_{i\ge 1}$ is strictly increasing
and thanks to \eqref{eq:lemma:recursion}, Lemma \ref{lemma:criterion}
shows that $(p_i/q_i)_{i\ge 1}$ is a sequence of consecutive convergents
to its limit $\xi\in C$. Fix an index $i\ge 1$.  By the theory of
continued fractions, we have
\[
 \frac{1}{2q_iq_{i+1}}
  < \left|\xi-\frac{p_i}{q_i}\right|
  < \frac{1}{q_iq_{i+1}}.
\]
According to \eqref{eq1:proof}, this implies
that $|\xi-p_i/q_i|<\psi(q_i)$.  By \eqref{eq3:proof}, we also
have $q_{i+1} \ge b^{m_{i+1}N_i}\ge q_i^{q_i}$ because $m_{i+1}\ge q_i$,
and this further yields $|\xi-p_i/q_i|\le 1/q_{i+1}<q_i^{-q_i}$.

To get a lower bound for $|\xi-p_i/q_i|$ when $i\ge 2$, we note that,
if $m_{i+1}>q_i$, then, by virtue of \eqref{eq2:proof}
and of the choice of $m_{i+1}$, we have
\[
 \psi(q_i)
 \le \frac{u}{b^{m_1}(b^{N_{i+1}-q_iN_i}-1)}
  \le \frac{1}{b^{N_{i+1}-q_iN_i}}.
\]
This is again true if $m_{i+1}=q_i$ because, by hypothesis,
$\psi(q_i)\le 1$.  Thus we obtain
\[
 \left|\xi-\frac{p_i}{q_i}\right|
  > \frac{1}{2q_iq_{i+1}}
  = \frac{u}{2b^{m_1}(b^{N_{i+1}}-1)}
  \ge \frac{1}{2b^{m_1+N_{i+1}}}
  \ge \frac{\psi(q_i)}{2b^{m_1+q_iN_i}}
  \ge \frac{\psi(q_i)}{2b^{m_1}q_i^{q_i(1+m_{i})/m_{i}}},
\]
where the last inequality uses \eqref{eq3:proof}. As $m_i$ tends to
infinity with $i$ (because $m_i\ge q_{i-1}$ for $i\ge 2$), this means
that $|\xi-p_i/q_i|>\psi(q_i)q_i^{-(1+\epsilon)q_i}$ for each sufficiently
large index $i$.

This proves the theorem with $c_1=q_1$ as $q_1$
depends only on $b$ and $u$.  If $d_2=1$, we have
$D=\{0,1\}$ and we can even take
$c_1=1$ because $0=0/1$ is the only missing
convergent of $\xi$ among $(p_i/q_i)_{i\ge 1}$, and
$0=(0.\overline{0})_b$ belongs to $C$.
\end{proof}

\section*{Acknowledgments}
The research of Johannes Schleischitz was supported by the Schr\"odinger Scholarship J 3824 of the Austrian Science Fund (FWF), while that of D.~Roy was partly supported by an NSERC discovery grant.


TITLE: Finding an orthogonal matrix $P$ of the quadric $3x^2+3z^2-4xy-8xz+4yz=1$
QUESTION [0 upvotes]: I am trying to find an orthogonal matrix $P$ of the quadric $3x^2+3z^2-4xy-8xz+4yz=1$.

I have found that the eigenvalues are $-1$ (with algebraic multiplicity 2) and $8$ (with algebraic multiplicity 1) with corresponding eigenvectors {<1,0,1>,<-0.5,-2,0.5>} and {<2,-1,-2>} respectively. I know, once normalised, that these eigenvectors form the columns of $P$, but must the $det(P)=1$? If so, my matrix $P$ is incorrect as I can't manipulate the eigenvectors to adhere to this property. This is a question from a past exam paper and help on this problem would be greatly appreciated

REPLY [1 votes]: We just need to normalize the vectors by $$u_i=\frac{v_i}{|v_i|}$$
in order to obtain $P$ such that $P^TP=I$.


TITLE: One-way wave equation IBVP
QUESTION [4 upvotes]: Plese help me to find the solution of te following equation.

For values of $x$ in the interval $[-2,3]$ and $t>0$ we consider the one way wave equation $$u_t+u_x=0$$
  with initial data \begin{align*}
u(0,x)= \left\{\begin{array}{ll}1-|x|\;\; \text{if}\; 0\le |x|\le 1\\0\;\;\;\text{otherwise} \end{array}
\right.\end{align*}
  and boundary data $u(t,-2)=0.$

I tried to use the method of characteristics in the following way:
$\frac{dx}{ds}=1$ and $\frac{dt}{ds}=1$
but i didn't manage to find the solution.
Please help me to do so. Thanks

REPLY [2 votes]: I guess that a graphical answer is more in the spirit of this exercise. By using the method of characteristics, we find the set of curves $x = t + x_0$ along which $u$ is constant. These curves are represented in the $x$-$t$ plane below:

The boundary is represented by the thick black line. All curves starting on the boundary carry the value $u=0$, besides the curves in the red area defined by $-1\leq x-t \leq 1$ which carry the value $1 - |x-t|$. Therefore the solution for $x \in [-2,3]$, $t>0$ reads
$$
u(x,t) = \left\lbrace
\begin{aligned}
&1 - |x-t| && \text{if}\quad {-1}+t\leq x \leq 1+t,\quad (0\leq t\leq 4) \\
&0 && \text{otherwise}
\end{aligned}\right.
$$


text:
\begin{document}

\maketitle

\begin{abstract}
    We study  homological properties of a locally complete intersection ring by importing facts from homological algebra over  exterior algebras. One  application is showing that  the thick subcategories of the bounded derived category of a locally complete intersection ring are self-dual under Grothendieck duality. This was proved by Stevenson when the ring is a quotient of a regular ring modulo a regular sequence; we offer two independent proofs in the more general setting. Second, we use these techniques to supply new proofs that  complete intersections possess symmetry of complexity.
\end{abstract}

\section*{Introduction}

Homological algebra over complete intersections is profoundly linked to the homological algebra over exterior algebras.
This was clarified in  \cite{AI3} where Avramov and Iyengar established a process to obtain homological information over complete intersections from the corresponding results over graded Hopf algebras.  Their techniques provided new, easier proofs of many known  results over  complete intersections.

For example, the complexity of a module measures the polynomial rate of growth of its Betti numbers while the injective complexity measures  the polynomial rate of growth of the module's Bass numbers. Using the process described above, Avramov and Iyengar easily deduced that over  complete intersections the complexity of a module agrees with its injective complexity, and both of these values are bounded above by the complexity of the residue field; they also employ their methods to show the latter   is exactly  the codimension of the complete intersection.

In this article, we adopt techniques from  \cite{AI3} to acquire further information about complete intersections. For the rest of the introduction $R$ is a commutative noetherian ring. The first main result is framed in terms of the derived category of $R$, denoted $\D(R)$.

We let  $\D^f(R)$ denote the full subcategory   of $\D(R)$ consisting of those complexes of $R$-modules whose total homology is  finitely generated. It inherits the structure of a triangulated category from $\D(R)$.  Recently, there has been much interest in understanding  the structure of thick subcategories of $\D^f(R)$, see for example  \cite{ABIM,CI,DGI,Letz,Pol,St,Tak}. Our first main result is the following:
\begin{Thm}\label{t1}If $R$ is locally complete intersection, then
each thick subcategory of $\D^f(R)$ is self-dual under Grothendieck duality. That is, for any thick subcategory $\T$ of $\D^f(R)$ and object $M$ in $\T$,  $\RHom_R(M,R)$ is in $\T$, as well.
\end{Thm}
 Stevenson in \cite[4.11]{St} proved the result under the additional assumption that $R$ is a quotient of a regular ring modulo a regular sequence. One can also deduce Theorem \ref{t1} from \cite[4.11]{St} in conjunction with recent results of  Letz \cite[3.12 \& 4.5]{Letz}; details are provided in Remark \ref{rs}.

In this article, we present two proofs of Theorem \ref{t1} both of which rely  on a  local-to-global principle of Benson, Iyengar and Krause (see \ref{lg}) and the structure of thick subcategories in the derived category of an exterior algebra over a field (cf. \cite{CI}). The first proof uses the theory of cohomological support discussed in Section \ref{sCI}. Namely, we show that the containment of thick subcategories is encoded in the  support varieties of Avramov and Buchweitz defined in \cite{AB} (see Theorem \ref{Th} for a precise statement).

The second proof makes direct use of the graded Hopf algebra structure of the exterior algebra to show that thick subcategories  over an exterior algebra on generators of homological degree one are fixed by Grothendieck duality (see Theorem \ref{closed}).
 Furthermore, it is worth noting that  the both of the proofs of Theorem \ref{cor1}  do not rely on the full classification of thick subcategories in \cite[8.8]{St2}; making the proofs here simpler even in the case that $R$ is a quotient of a regular ring modulo a regular sequence.

As a consequence of Theorem \ref{t1} we   obtain asmyptotic information over complete intersections. For example, we recover a result of Avramov and Buchweitz \cite[6.3]{AB} that says the eventual vanishing of Ext is equivalent to the eventual vanishing of Tor over locally complete intersections (cf. Corollary \ref{cor}). Furthermore, in the local case we can use Theorem \ref{t1} to show complexity is symmetric in $M$ and $N$;
recall   the complexity of a pair of objects $M$ and $N$ of $\D^f(R)$ is the polynomial rate of growth of the minimal number of generators of $\Ext_R^n(M,N)$ (see \ref{cx} for a precise definition).
\begin{Thm}\label{t2}
If $R$ is  complete intersection, then $\cx_R(M,N)=\cx_R(N,M)$ for each pair of objects $M$ and $N$ in $\D^f(R).$
\end{Thm}

 This was first proven by Avramov and Buchweitz \cite{AB}; an alternative proof was provided by the second author in \cite{Po}. In contrast, we give two new proofs of this result  in this paper; both of which use the homological properties over exterior algebras. The first proof deduces Theorem \ref{t2} from Theorem \ref{t1}, illustrating how the containment  of thick subcategories  and duality can provide asymptotic information.  The second proof directly links the complexity of a pair of objects $M$ and $N$  in $\D^f(R)$ with the complexity of a pair  of objects in the derived category of an exterior algebra over a field. This  proof fills in a missing piece of the work in \cite{AI3}; cf. \cite[6.9]{AI3} and the discussion in Remark \ref{rcx}.

\begin{ack}
Both authors are indebted to Srikanth Iyengar for  many helpful discussions, as well as suggesting the two authors collaborate because of their many common interests. We are also very happy to thank Benjamin Briggs for several useful comments on an earlier draft of this paper as well as numerous conversations that helped  clarify some  ideas in Section \ref{a1}. We also thank Janina Letz and Greg Stevenson for their comments on a preliminary draft of this work.
\end{ack}

\section{Background, Notation, and Terminology} Throughout this article $R$ will be a commutative noetherian ring.

 \subsection{(Locally) Complete Intersections}
 Suppose $(R,\m,k)$ is local. Recall the \emph{embedding dimension} of $R$ is $\dim_k\m/\m^2$, the minimal number of generators for $\m$, and the \emph{codimension of R} is
\[
\dim_k\m/\m^2-\dim R.
\]

A  local ring $(R,\m,k)$ is  \emph{complete intersection} if its $\m$-adic completion $\widehat{R}$ is isomorphic to $Q/I$ where $Q$ is a regular local ring and  $I$  is generated by  a $Q$-regular sequence. In fact, the presentation can be chosen so that $Q$ and $R$ have the same embedding dimension  and $I$ is generated by $c$ elements where $c$ is the codimension of $R$.

More generally,  $R$ is \emph{locally complete intersection}  provided that the local ring $R_\p$ is  complete intersection for each prime ideal $\p$ of $R$.

\subsection{Derived Category of a DG Algebra}

Let $A$ be a DG $R$-algebra. We briefly discuss the derived category of DG $A$-modules and set notation used throughout the rest of the article. See \cite[Section 3]{ABIM} or \cite[Chapter 6]{FHT} for more details.

Let  $\D(A)$ denote the derived category of (left) DG $A$-modules. Recall that $\D(A)$ is a triangulated category with $\shift$ being the suspension functor; for each $X$ in $\D(A)$, $\shift X$ is the DG $A$-module given by $\shift X_i=X_{i-1}$, $a\cdot (\shift x)=(-1)^{|a|}ax$ and $\del^{\shift X}=-\del^X$. We let $\D^f(A)$ denote the full subcategory of $\D(A)$ consisting of those objects $X$ of $\D(A)$ such that $\h(X)$ is a finitely generated graded $\h(A)$-module.

Each DG $A$-module $X$ admits a semiprojective resolution. That is, there exists a surjective quasi-isomorphism $P\to M$ such that $\Hom_A(P,-)$ preserves surjective quasi-isomorphisms.
 For any $Y$ in $\D(A)$, we set  \[\RHom_A(X,Y)\coloneqq\Hom_A(P,Y)\] where $P\to X$ is a semiprojective resolution and  \[\Ext_A(X,Y)\coloneqq\h\left(\RHom_A(X,Y)\right),\] which naturally inherits  a graded $\Ext_A(Y,Y)$-$\Ext_A(X,X)$-bimodule structure.

\subsection{Koszul Complexes}
\label{kos} Background on Koszul complexes can be found in \cite[Section 1.6]{BH}. We recall the necessary facts here.

For a list of elements $\x=x_1,\ldots, x_n$ in $R$, we set $\Kos^R(\x)$ to be the Koszul complex of $\x$ on $R$, which is regarded as a DG $R$-algebra in the usual way.

When $R$ is local with maximal ideal $\m$, set $K^R$ to be the Koszul complex on a minimal generating set for $\m$. It is well-defined up to an isomorphism of DG $R$-algebras.

Fix  a prime ideal  $\p$ of $R$ and  let $M$ be an object of $\D(R).$ We set
$$M(\p)\coloneqq M_\p\otimes_{R_\p} K^{R_\p}$$   which is a DG $K^{R_\p}$-module.
Restricting scalars along the morphism of DG algebras $R_\p \to K^{R_\p}$ we may regard $M(\p)$ as an object of $\D(R_\p)$.

 \subsection{Koszul Complexes over  Complete Intersections} \label{koszulci}
 Let $(R,\m,k)$ be  complete intersection of codimension $c$ and  $\Lambda$  be the  exterior algebra over $k$ on  generators $e_1,\ldots, e_c$ of homological degree $1$.   Let $\mathsf{t}\colon \D(R)\to \D(K^{R})$ be the functor  $-\otimes_R K^{R}$.

 By  \cite[6.4]{AI3} there is a quasi-isomorphism of DG algebras $K^R\simeq  \Lambda$ that
   induces an equivalence of triangulated categories $\mathsf{j}\colon\D(K^R)\to  \D(\Lambda)$;  this restricts to an equivalence $ \D^f(K^R)\xrightarrow{\equiv}\D^f(\Lambda)$ that is compatible with Grothendieck duality
(see \cite[3.6]{ABIM} or \cite[2.5]{AI3}). Hence, when $R$ is complete intersection we have the following composition
\[\mathsf{jt}:\D^f(R)\to \D^f(K^R)\xra{\equiv}\D^f(\Lambda);\] this is the main bridge for  importing results over graded exterior algebras to  complete intersections.
Throughout the rest of the paper, $\mathsf{j}$ and $\mathsf{t}$ will always denote the functors introduced here.

\subsection{Thick Subcategories}\label{t}
Let $A$ be a DG algebra and $\T$ be a full subcategory of $\D(A)$. We say $\T$
 is \emph{thick} if it is a triangulated subcategory that is closed under taking direct summands. For an object $M$ of $\D(A)$, we let $\thick_{\D(A)} (M)$ denote the smallest thick subcategory of $\D(A)$ containing $M$.  This can be realized as the intersection of all thick subcategories  of $\D(A)$ containing $M$; alternatively, this has an inductive construction discussed in \cite[2.2.4]{ABIM}.

\subsection{Local-to-Global Principle}\label{lg}

The main results in the present paper rely  on the following \emph{local-to-global principle} of Benson, Iyengar and Krause (see by \cite[5.10]{BIK2}).
Namely,  for objects $M$ and $N$ of $\D^f(R)$, $M$ is in $\thick_{\D(R)}(N)$ if and only if $M(\p)$ is in $\thick_{\D(R_\p)}\left(N_\p\right)$ for each prime ideal $\p$ of $R$.
As $N(\p)$ is an object of $\thick_{\D(R_\p)}\left( N_\p\right)$, we  restate the local-to-global principle as:
\[M\text{ is in }\thick_{\D(R)} (N) \iff   M(\p)\text{ is in }\thick_{\D(R_\p)}\left(N(\p)\right)\]
for each  prime ideal $\p$ of $R.$

\subsection{Homogeneous Support}
Let $\S$ be a commutative noetherian graded ring. We let $\Spec^* \S$ denote the homogeneous spectrum of $\S$. That is, $\Spec^*\S$ consists of the homogeneous prime ideals of $\S$. For a graded $\S$-module $X$ and $\p\in \Spec^*\S$, $X_\p$  denotes the homogeneous localization of $X$ at $\p$. The homogeneous support of $X$ is $$\supp_\S X=\{\p\in \Spec^*\S: X_\p\neq 0\}.$$

\section{Cohomological Support Varieties}
\label{sCI}

Throughout this section we fix the following notation.
Let $(R,\m,k)$ be  complete intersection with  codimension $c$ and embedding dimension $\nu.$
Let $\S$ denote the graded $k$-algebra $k[\chi_1,\ldots,\chi_c]$  where each $\chi_i$ has homological degree $-2$.
 We set  $\Lambda$ to be the  exterior algebra over $k$ on  generators $e_1,\ldots, e_c$ of homological degree $1$. Finally, let $\mathsf{j}$ and $\mathsf{t}$ be the functors from \ref{koszulci}.

\begin{chunk}\label{act}
In \cite[Theorem 5]{Sj}, Sj{\"o}din described the graded $k$-algebra structure of $\Ext_R(k,k)$. It contains $\S$ as a polynomial subalgebra in  such a way that
$$
\Ext_R(k,k)\cong \S\otimes_k \bigwedge \shift^{-1}k^{\nu}
$$ as graded $\S$-modules; see also \cite[10.2.3]{IFR} for more details. Thus,
 $\S$ acts on $\Ext_R(k,M)$  through the $\Ext_R(k,k)$-action for each  $M$  in $\D(R)$. We define  \emph{the cohomological support of $M$ over $R$} to be $$\V_R(M)\coloneqq\supp_\S \Ext_R(k,M).$$
\end{chunk}

\begin{chunk}\label{civar}
 By
 \cite[5.1]{AI3} (see also  \cite[7.4]{ABIM}), there is an isomorphism of graded $k$-algebras  $$
\Ext_\Lambda(k,k)\cong \S.$$
 For any $X$ in $\D(\Lambda)$, we define  the \emph{cohomological support of $X$  over $\Lambda$} to be $$\V_\Lambda(X)\coloneqq \supp_{\S}\Ext_\Lambda(k,X).$$
These varieties can detect  the containment of thick subcategories in $\D^f(\Lambda).$ Namely, in  \cite[4.4]{CI}, Carlson and Iyengar showed
 \[
 X\text{ is in } \thick_{\D(\Lambda)} (Y)\iff \V_\Lambda(X)\subseteq \V_\Lambda(Y)
 \]  for any pair of objects $X$ and $Y$ in $\D^f(\Lambda)$. This  essentially follows from  the celebrated theorem of Hopkins \cite[11]{H} and Neeman \cite[1.2]{N} (see  also  \cite[3.2]{CI} for the version needed) and a BGG correspondence (cf. \cite[7.4]{ABIM}).
\end{chunk}

There is a way to relate the supports defined over $R$ and $\Lambda.$ This was first noticed  in the case that $R$ is artinain \cite[5.11]{CI}; however, the same proof works without any restriction on the Krull-dimension of $R$ and so we sketch it for the convenience of the reader in the following remark and lemma.
\begin{chunk}\label{injection} First, there is a canonical injective map  $\eta\colon \Ext_{\Lambda}(k,k)\to \Ext_R(k,k)$ of graded  $k$-algebras that can be factored as $$\Ext_{\Lambda}(k,k)\xra{\cong} \Ext_{K^R}(k,k)\hookrightarrow \Ext_R(k,k)$$ where the isomorphism is induced by the inverse of the equivalence $\mathsf{j}$ from \ref{koszulci}.   Moreover, the  image of $\eta$ is exactly the polynomial subalgebra $\S$ of $\Ext_R(k,k)$ mentioned in \ref{act}. Therefore, the   cohomological supports over $R$ and those over $\Lambda$ can naturally be thought of as subsets of the same  $\Spec^*\S$. Moreover, we have the following connection.
\end{chunk}

\begin{lemma}\label{supp2}
For each  $M$ in  $\D(R)$,
$\V_R(M)=\V_R(\mathsf{t} M)=\V_\Lambda(\mathsf{jt} M).$
\end{lemma}
\begin{proof} First, consider the isomorphisms of graded $\S$-modules
\begin{align*}
\Ext_R(k,\mathsf{t}M)&\cong \Ext_{K^R}( \mathsf{t}k, \mathsf{t} M) \\
&\cong  \Ext_\Lambda(\mathsf{jt} k, \mathsf{jt}M) \\
&\cong \bigoplus_{i=0}^{\nu}\shift^{-i}\Ext_\Lambda(k,\mathsf{jt}M)^{{\nu \choose i}}
\end{align*}
where the third isomorphism holds because $\mathsf{t} k\cong\bigoplus_{i=0}^{\nu}\shift^{-i}k^{{\nu \choose i}} $ and $\mathsf {j}(k)\simeq k$, see \cite[3.9]{ABIM} for the latter.
Also, we have the isomorphism of graded $\S$-modules
$$\Ext_R(k,\mathsf{t}M)\cong \bigoplus_{i=0}^{\nu}\shift^i\Ext_R(k,M)^{{\nu \choose i}}.$$ Therefore, the isomorphisms of $\S$-modules show $$\V_R(M)=\V_R( \mathsf{t} M)=\V_\Lambda(\mathsf{jt}M).\qedhere$$
\end{proof}

\begin{proposition}\label{prop}
For  $M$ and $N$ in $\D^f(R)$,
\[\mathsf{t} M\text{ is in }\thick_{\D(R)}(\mathsf{t} N)\iff \V_R(M)\subseteq \V_R(N).\]
\end{proposition}
\begin{proof} The forward direction is trivial from the first equality in Lemma \ref{supp2}.

Conversely, assume $\V_R(M)\subseteq \V_R(N)$.
Using Lemma \ref{supp2}, this reads as  \[\V_\Lambda(\mathsf{jt} M)\subseteq \V_\Lambda(\mathsf{jt} N).\] Thus,  \ref{civar} implies that  $\mathsf{jt} M$ is an object of   $\thick_{\D(\Lambda)}  ( \mathsf{jt} N).$ As $\mathsf{j}$ is an equivalence we conclude that $\mathsf{t} M$ is an object of $\thick_{\D(K^R)}(\mathsf{t} N)$. The result follows by restricting scalars along the morphism of DG $R$-algebras $R\to K^R$.
\end{proof}

We end this section with the following technical lemma which will be put to use in Section \ref{a1}. Note that since  $\Lambda$ has trivial differential, for each DG $\Lambda$-module $X$ we can negate differential of $X$ to obtain a DG $\Lambda$-module. Namely, let $X'$ denote the DG $\Lambda$-module whose underlying graded $\Lambda$-module is $X$ and its differential is $\del^{X'}\coloneqq -\del^X.$ When $\Lambda$ is concentrated in even degrees, $X\cong X'$ as DG $\Lambda$-modules (cf. \ref{twist}). However, as the generators of $\Lambda$ have degree 1 we do not know whether these are isomorphic. Instead, we show they have the same cohomological support, and hence, generate the same thick subcategory.
\begin{lemma}\label{l:neg}
If $X$ is in $\D^f(\Lambda)$, then $\V_\Lambda(X)=\V_\Lambda(X').$ Moreover, we have the following equality of thick subcategories:
\[
\thick_{\D(\Lambda)}(X)=\thick_{\D(\Lambda)}(X').
\]
\end{lemma}
\begin{proof}
By \cite[4.2]{CI}, there exists a  semiprojective resolution $F\xra{\simeq} k$ over $\Lambda$ such that $F$ admits a DG $\S$-module structure compatible with the $\S$-action on $\Ext_R(k,Y)$ for any $Y$ in $\D(\Lambda).$ As $k$ has trivial differential, the same is true of the  semiprojective DG $\Lambda$-resolution equipped with a DG $\S$-module structure $F'\xra{\simeq} k$.

Define $\Phi: \Hom_\Lambda(F,X)\to \Hom_\Lambda(F',X')$ given by \[
\alpha\mapsto (-1)^{|\alpha|}\alpha.
\] As $\S$ is concentrated in even degrees this is an isomorphism of DG $\S$-modules. Therefore, $\h(\Phi)$ establishes the following isomorphism of graded $\S$-modules
\[
\Ext_\Lambda(k,X)\cong \Ext_\Lambda(k,X');
\]  so $X$ and $X'$ have the same cohomological support. The equality of thick subcategories now follows from \ref{civar}.
\end{proof}

\section{Duality of Thick Subcategories  via Support}
\label{sd2}

In this section we give the first proof of our result on the duality of thick subcategories over  locally complete intersections (see Theorem \ref{cor1}). The  main idea behind it is that the theory of cohomological supports, discussed in  Section \ref{sCI}, both detects  containment of thick subcategories and is unaffected by duality. The first theorem addresses the former point while \ref{l2}  the latter.
\begin{theorem}\label{Th}
Let  $R$ be locally complete intersection. For   $M, N$   in $\D^f(R)$,   \[M\text{ is in } \thick_{\D(R)}(N)\iff
    \V_{R_\p}(M_\p)\subseteq \V_{R_\p}(N_\p)\] for each prime ideal $\p$ of $R$.
\end{theorem}
\begin{proof}
First, assume $M$ is an object of $\thick_{\D(R)}N$. Hence,   $M_\p$ is an object of  $\thick_{\D(R_\p)}N_\p$ and so it follows easily that  $\V_{R_\p}(M_\p)\subseteq \V_{R_\p}(N_\p)$ for $\p\in \Spec R.$

Conversely, suppose  $\V_{R_\p}(M_\p)\subseteq \V_{R_\p}(N_\p)$ for each $\p\in \Spec R.$  By Proposition \ref{prop}, $M(\p)$ is in $\thick_{\D(R_\p)} \left(N(\p)\right)$ for each prime ideal $\p$ of $R$.
 Finally, we apply \ref{lg} to conclude that $M$ is in $\thick_{\D(R)}(N).$
 \end{proof}

\begin{chunk}\label{l2} It is well-known  that  cohomological  support over complete intersections  is closed under duality. That is,
  \[
\V_R(M)=\V_R(\RHom_R(M,R))\] for each $M$  in $\D^f(R)$,
This was shown for closed points of $\Spec^*\S$  in \cite[3.3]{AB}, and  the general setting was shown in  \cite[4.1.5]{Po}.  Alternatively, a new proof is  obtained in the present work by combining  Lemma \ref{supp2} and  Theorem \ref{closed}; this proof sticks with the theme of establishing results for complete intersections by passing to an exterior algebra.
\end{chunk}

\begin{theorem}\label{cor1}
If $R$ is locally complete intersection, then every thick subcategory of $\D^f(R)$ is closed under $\RHom_R(-,R)$. In particular, for each $M$ in $\D^f(R)$
\[
\thick_{\D(R)}(M)=\thick_{\D(R)}(\RHom_R(M,R)).
\]
\end{theorem}
\begin{proof}[First proof of Theorem \ref{cor1}]
As $M$ is an object of  $\D^f(R)$, there is a natural isomorphism
$$
\RHom_R(M,R)_\p\cong \RHom_{R_\p}(M_\p,R_\p)
$$
for each prime ideal $\p$ of $R$. So   \ref{l2} shows
$$
\V_{R_\p}(M_\p)=\V_{R_\p}(\RHom_{R_\p}(M_\p,R_\p))=\V_{R_\p}(\RHom_R(M,R)_\p)
$$ for each prime ideal $\p$ of $R$.
Now we obtain \[\thick_{\D(R)}(M)=\thick_{\D(R)}(\RHom_R(M,R))\]as an immediate consequence of Theorem \ref{Th}.
\end{proof}

\begin{remark}\label{rs}
Theorem \ref{cor1} can also be proved by combining results of Stevenson and  Letz (see \cite[4.11]{St} and \cite[3.12 \& 4.5]{Letz}, respectively).  Stevenson used his classification of  thick subcategories of the singularity category of a regular ring modulo a regular sequence in \cite[8.8]{St2}  to show Theorem \ref{cor1} holds  for such rings. Letz showed for $M$ and $N$ in $\D^f(R)$, \[
M\text{ is in }\thick_{\D(R)}(N)\iff M\otimes \widehat{R_\p}\text{ is in }\thick_{\D(\widehat{R_\p})} (N\otimes_R \widehat{R_\p})\] for each prime ideal $\p$ of $R$ where $\widehat{R_\p}$ is the  $\p R_\p$-adic completion of $R_\p.$  So their work, indeed, offers a different argument for Theorem \ref{cor1}.  \end{remark}

 In Section \ref{a1}, we  give our second proof of Theorem \ref{cor1}. This requires an analysis of duality over a graded exterior algebra which is discussed there. We end this section with the following application  that recovers a theorem of Avramov and Buchweitz \cite[6.3]{AB}. In Section \ref{s:complexity}, we strengthen the equivalence of (1) and (2) in Corollary \ref{cor} when $R$ is further assumed to be local.

\begin{corollary}\label{cor}
Let $R$ be locally complete intersection. For $M$ and $N$ in $\D^f(R)$, the following are equivalent:
\begin{enumerate}
    \item $\Ext_R^i(M,N)=0$ for all $i\gg 0$;
    \item$\Ext_R^i(N,M)=0$ for all $i\gg 0$;
    \item$\Tor^R_i(M,N)=0$ for all $i\gg 0$.
\end{enumerate}
\end{corollary}
\begin{proof} It suffices to show the equivalence of (1) and (3) since Tor is symmetric in $M$ and $N$. Furthermore, the equivalence of (1) and (3) is the same as
\[
\RHom_R(M,N) \text{ is in }\D^f(R)\iff M\ot_RN \text{ is in }\D^f(R);
\]so we show the latter equivalence holds. For the rest of the proof let $(-)^\vee$ be $\RHom_R(-,R).$ Observe for any pair of objects $M$ and $N$ in $\D(R),$ we have the following adjunction isomorphism:
\begin{equation}
\label{eeiso}\tag{$\dagger$}
(M\ot_RN)^\vee\simeq \RHom_R(M,N^\vee).
\end{equation}

$\impliedby:$ As $M\ot_R N$ is in $\D^f(R)$ and Theorem \ref{cor1},  it follows that $(M\ot_RN)^\vee$ is in $\D^f(R).$ Hence, by (\ref{eeiso}),
$\RHom_R(M,N^\vee)$ is in $\D^f(R)$. By Theorem \ref{cor1}, $N$ is in $\thick_{\D(R)}(N^\vee)$, and so   $\RHom_R(M,N)$ is in $\D^f(R),$ as well.

$\implies:$
By Theorem \ref{cor1}, our assumption is equivalent to $\RHom_R(M,N^\vee)$ is in $\D^f(R). $ By (\ref{eeiso})
and  Theorem \ref{cor1} we conclude that $(M\ot_RN)^{\vee\vee}$ is in $\D^f(R)$, as well.
Finally, as $R$ is Gorenstein,
the natural map \[
M\to M^{\vee\vee}
\] is an isomorphism in $\D^f(R)$. Therefore, $M\ot_R N$ is in $\D^f(R),$ as needed.
\end{proof}

\section{Duality of Thick Subcategories via Exterior Algebras}\label{a1}

For a finite dimensional Hopf algebra $H$ over a field $k$ and finite dimensional $H$-module $M$, it is well known that $M$ is a direct summand of $M\otimes_k \Hom_k(M,k)\otimes_k M$ (see, for example, \cite[3.1.10]{benson});  the  $H$-module structure on the latter needs both the antipode and co-multiplication of $H$. In this section we make use of the DG version of this fact for a graded exterior algebra.

Throughout $k$ will be a field and  $\Lambda$ is the exterior algebra on  generators $e_1,\ldots,e_c$ of homological degree $1$ over $k$. It is easily checked $\Lambda$ is a graded Hopf algebra with co-multiplication and antipode  determined by
$$
\Delta(e_i)=e_i\otimes 1+1\otimes e_i \
\text{ and } \
 \sigma(a)=(-1)^{|a|}a,
 $$
 respectively (cf. \cite[5.3]{AI3}).

The main goal of the   section is  to prove the next theorem and as an application we give a second short proof of Theorem \ref{cor1}. The proof of Theorem \ref{closed} requires some additional setup  and can be found after \ref{split}.

\begin{theorem}\label{closed}
For an object  $M$ of $\D^f(\Lambda)$,
 \[\thick_{\D(\Lambda)} (M)=\thick_{\D(\Lambda)} (\RHom_\Lambda(M,\Lambda)).\]
\end{theorem}

\begin{chunk}
\label{dual}
Let $\rho\colon \Lambda\to \Lambda$ be an anti-endomorphism of graded $k$-algebras,
which is nothing more than an endomorphism  since $\Lambda$ is graded-commutative.  The map $\rho$ prescribes a natural left DG $\Lambda$-module structure on the graded $k$-space $M^*=\Hom_k(M,k)$;  define $M^*(\rho)$ to be the left DG $\Lambda$-module whose underlying graded $k$-space is $M^*$ and its differential and $\Lambda$-action are  given by \begin{align*} \del^{M^*(\rho)}(f)&\coloneqq-(-1)^{|f|}f\del^M=(-1)^{|f|+1}f\del^M\\
a\cdot f&\coloneqq(-1)^{|a||f|}f(\rho(a)-).\end{align*} We are particularly interested in the relationship between $M^*(\id)$ and $M^*(\sigma)$. Furthermore, as $\Hom_k(\Lambda,k)\cong \shift^{-c} \Lambda$ as DG $\Lambda$-modules a direct calculation yields
\begin{equation}\label{e:duals}
\RHom_{\Lambda}(M,\Lambda)\simeq \shift^c\Hom_k(M,k)=\shift^cM^*(\id).
\end{equation}
\end{chunk}

\begin{chunk}\label{twist}
Let $M$ be a DG $\Lambda$-module. Since $\Lambda$ has trivial differential we can twist the differential \text{and} $\Lambda$-action of $M$ to obtain an isomorphic DG $\Lambda$-module. We define $M_\tau$ to be the DG $\Lambda$-module whose underlying graded $k$-space is $M$ equipped with differential and $\Lambda$-action
\begin{align*}
    \del^{M_\tau}&\coloneqq -\del^M \\
    a\cdot m&\coloneqq (-1)^{|a|}am.
\end{align*}The map  $M\to M_\tau$ given by
\[
m\mapsto (-1)^{|m|}m
\] is easily checked to be an isomorphism of DG $\Lambda$-modules.
\end{chunk}

\begin{proposition}\label{dualiso}
For any $M$ in  $\D(\Lambda)$, $M^*(\sigma)$ and $M^*(\id)_\tau$ have the same underlying graded $\Lambda$-module while their differentials are negatives of one another.
\end{proposition}
\begin{proof}
This follows directly from the definitions in \ref{dual} and \ref{twist}.

\end{proof}

\begin{chunk}\label{structure}
Let $M$ and $ N$ be left DG $\Lambda$-modules, then $M\otimes_k N$ is a left DG $\Lambda\otimes_k \Lambda$-module.
We define $\Hom_k(M,N)$ to be a left DG $\Lambda\otimes_k\Lambda$-module
\begin{align*}
\del^{\Hom_k(M,N)}(f)\coloneqq\del^Nf-(-1)^{|f|}f\del^M\\
(a_1\otimes a_2)\cdot f\coloneqq(-1)^{|a_2||f|}a_1 f(\sigma(a_2)-).
\end{align*}
Hence, both $M\otimes_k N$ and $\Hom_k(M,N)$ inherit a DG $\Lambda$-module structure via $\Delta$.
There is a natural morphism of DG $\Lambda$-modules
\[
\vp_{M,N}\colon N\otimes_k M^*(\sigma)\to  \Hom_k(M,N)\]
which is an isomorphism in $\D(\Lambda)$ when $\h(M)$ is a finite rank $k$-space (see  \cite[4.8]{AI3}).
\end{chunk}

\begin{chunk}\label{split}
Let $M$ be a DG $\Lambda$-module. Consider the morphism of DG $\Lambda$-modules
$$
\pi_M\colon k\rightarrow \Hom_k(M,M)
$$
mapping $1$ to $\id_M$.
$$
M\cong k\otimes_k M\xrightarrow{\pi_M\otimes M} \Hom_k(M,M)\otimes_k M.
$$
For an object $M$ in $\D^f(\Lambda)$ the composition  \[ M\cong k\otimes_k M\xrightarrow{\pi_M\otimes M} \Hom_k(M,M)\otimes_k M\xra{\vp_{M,M}^{-1}\otimes M } M\otimes_k M^*(\sigma)\otimes_k M\] splits in $\D^f(\Lambda)$; this is
essentially the same argument from  the classical case  (cf. \cite[3.1.10]{benson}).

\end{chunk}

\begin{proof}[Proof of Theorem \ref{closed}]
As  $\thick_{\D(\Lambda)}(k)=\D^f(\Lambda)$,  it follows that  \[M\otimes_k M^*(\sigma)\otimes_k M\text{ is an object of } \thick_{\D(\Lambda)}(M^*(\sigma)).\] Hence, \ref{split} implies that  $M$ is in $\thick_{\D(\Lambda)}(M^*(\sigma))$, as well.
Since $(M^*(\sigma))^*(\sigma)\cong M$, then by symmetry we have  shown
\begin{equation}
    \thick_{\D(\Lambda)}(M)=\thick_{\D(\Lambda)} \left(M^*(\sigma)\right).
    \label{e:thick}
\end{equation}

Now we only need to observe that
\begin{align*}
    \thick_{\D(\Lambda)}(M)&=\thick_{\D(\Lambda)} (M^*(\sigma)) \\
    &= \thick_{\D(\Lambda)} (M^*(\sigma),-\del^{M^*(\sigma)}) \\
    &=\thick_{\D(\Lambda)}(M^*(\id)) \\
    &=\thick_{\D(\Lambda)}\left(\RHom_\Lambda(M,\Lambda)\right)
\end{align*}
where the first equality holds by (\ref{e:thick}), the second equality is Lemma \ref{l:neg}, the third equality
is from  \ref{twist} and Proposition \ref{dualiso}, and the last equality holds by \ref{dual}(\ref{e:duals}). Therefore, we have justified the theorem.

\end{proof}

As an application of the theory above (in particular, Theorem \ref{closed}) we now present a second proof of Theorem \ref{cor1}.

\begin{proof}[Second proof of Theorem \ref{cor1}]
First, if $R$ is complete intersection we show that for any $M$ in $\D^f(R)$, \begin{equation}\thick_{\D(R)}\left(M\otimes_R K^R\right)=\thick_{\D(R)}\left(\RHom_R(M,R)\otimes_R K^R\right).\label{eq}\end{equation}

Now observe that Theorem \ref{closed} shows
\[
\thick_{\D(\Lambda)}\left(\mathsf{j}\mathsf{t} M\right)=\thick_{\D(\Lambda)}\left(\RHom_\Lambda(\mathsf{j}\mathsf{t} M,\Lambda)\right)
\] where $\mathsf{j}$ and $\mathsf{t}$ are the  functors  introduced in \ref{koszulci}.
Therefore,
\[
\thick_{\D(K^R)} (\mathsf{t} M)=\thick_{\D(K^R)}\left(\RHom_{K^R}(\mathsf{t} M,K^R)\right)
\]
and since  \[
\RHom_{K^{R}}(\mathsf{t} M,K^{R})\simeq \mathsf{t} \RHom_R(M,R)
\]
it follows that
\begin{equation*}\label{eq1}
\thick_{\D(K^{R})}(\mathsf{t}M) =\thick_{\D(K^{R})}\left(\mathsf{t}\RHom_R(M,R)\right).
\end{equation*} So restricting scalars along $R\to K^R$ finishes the proof of (\ref{eq}) in the case that  $R$ is complete intersection.

Now we return to the general setting; namely, assume that $R$ is locally complete intersection. Let $\p$ be  a prime ideal of $R$, then by assumption $R_\p$ is complete intersection and
\begin{align*}
    \thick_{\D(R_\p)}\left(M(\p)\right)&= \thick_{\D(R_\p)}\left(\RHom_{R_\p}(M_\p,R_\p)\otimes_{R_\p}K^{R_\p}\right)\\
    &=\thick_{\D(R_\p)}\left(\RHom_{R}(M,R)(\p)\right);
\end{align*}
the first equality is from the already established equality (\ref{eq}) and the second equality
 is immediate from the isomorphism $$\RHom_{R_\p}(M_\p,R_\p)\otimes_{R_\p}K^{R_\p}\simeq \RHom_R(M,R)\otimes_R K^{R_\p}.$$ Since these equalities of thick subcategories hold for each prime ideal $\p$ of $R$, an application of the local-to-global principle in \ref{lg}   establishes the desired result.
\end{proof}

\section{Symmetry of Complexity}\label{s:complexity}

 In this section, we offer two proofs of Theorem \ref{symcx}; both methods indicate how the symmetry of complexity over  complete intersections follow from studying properties of exterior algebras. Theorem \ref{symcx} was originally shown in  \cite[5.7]{AB} using support varieties and the use of intermediate hypersurfaces. A second proof was given by the second author in \cite[4.3.1]{Po} by studying the cohomological support of certain DG modules over a  graded commutative ring of finite global dimension.

Throughout this section  $(R,\m,k)$ is a commutative noetherian local ring. Also, we let $(-)^\vee$ denote the functor $\RHom_R(-,R).$
\begin{chunk}\label{cx}Let $M$ and  $N$ be  in $\D^f(R)$. The \emph{complexity of the  pair $(M,N)$},  denoted $\cx_R(M,N)$, is the least non-negative integer $d\in \mathbb N$ such that \[\dim_k(\Ext^n_R(M,N)\otimes_R k)\leq an^{d-1}\] for all $n\gg 0$ and some $a\in\mathbb R$. That is, $\cx_R(M,N)$ measures the polynomial rate of growth of the minimal number of generators of $\Ext_R^n(M,N)$.
\end{chunk}

\begin{theorem}\label{symcx}
Let $R$ be  complete intersection. For  each pair of objects $M$ and $N$ in $\D^f(R)$,  $\cx_R(M,N)=\cx_R(N,M).$
\end{theorem}

\begin{remark}\label{r1}
The first proof of Theorem \ref{symcx} is a straightforward application  of Theorem \ref{cor1} and two standard facts:
\begin{enumerate}
    \item when $R$ is Gorenstein there is a natural isomorphism of graded $R$-modules
    \[\Ext_R(M,N)\cong \Ext_R(N^\vee,M^\vee)\] for each $M$ and $N$ in $\D^f(R)$;
    \item  when $R$ is complete intersection \[\cx_R(X,Y)\leq \cx_R(M,N)\] whenever $X$ is in $\thick_{\D(R)}(M)$ and $Y$ is in $\thick_{\D(R)}(N)$ (see \cite[5.6]{AB} or \cite[4.2.5 \& 4.2.9]{Po}).
\end{enumerate}
\end{remark}

\begin{proof}[First proof of Theorem \ref{symcx}]
From Remark \ref{r1}(1) it follows easily that
$$
\cx_R(M,N)=\cx_R(N^\vee,M^\vee).
$$
Using Theorem \ref{cor1}, we obtain $M^\vee$ is an object of $\thick_{\D(R)}(M)$ and $N^\vee$ is an object of   $\thick_{\D(R)}(N)$. So  Remark \ref{r1}(2) establishes the inequality
$$
\cx_R(M,N)=\cx_R(N^\vee,M^\vee)\leq \cx_R(N,M)
.$$
By symmetry the result follows.
\end{proof}

\begin{remark}\label{rcx}
As discussed in the introduction, the theme of \cite{AI3} is to deduce homological results over $R$, when  $R$ is complete intersection, using the bridge
\[\mathsf{jt}\colon\D^f(R)\to\D^f(K^R)\xra{\equiv} \D^f(\Lambda)\]from \ref{koszulci}. However, in  \cite[8.9]{AI3} it is remarked that the authors did not see how to deduce Theorem \ref{symcx} from studying this bridge. The second proof of Theorem \ref{symcx}, given below, shows that  one can in fact arrive at the symmetry of complexity over the complete intersection $R$ as a direct consequence of the symmetry of complexity over  $\Lambda.$
\end{remark}

\begin{proof}[Second proof of Theorem \ref{symcx}]
Assume $R$ is complete intersection.

Observe that
\[
\Ext_{R}(M,\mathsf{t}N)\cong \Ext_{K^R}(\mathsf{t} M, \mathsf{t} N )
\cong \Ext_\Lambda(\mathsf{jt}M, \mathsf{jt}N)
\] where the first isomorphism is adjunction and the second one uses that  $\mathsf{j}$ is an equivalence. Also, by \cite[4.2.7]{Pol}
\[\cx_R(M,N)=\cx_R(M,\mathsf{t}N),\]
and so combining  this with the isomorphisms above we have \begin{equation}\label{e5}\cx_R(M,N)=\cx_\Lambda(\mathsf{jt}M, \mathsf{jt}N).\end{equation}
Similarily, it follows that
\begin{equation}\label{e8}\cx_R(N,M)=\cx_\Lambda(\mathsf{jt}N, \mathsf{jt}M).\end{equation}
Note \cite[5.3]{AI3} established
\[
\cx_\Lambda(\mathsf{jt} M, \mathsf{jt}N)=\cx_\Lambda(\mathsf{jt}N, \mathsf{jt}M);
\]  this equality along with the ones in  (\ref{e5}) and (\ref{e8})  establish $\cx_R(M,N)=\cx_R(N,M)$, as claimed.
\end{proof}

\bibliographystyle{amsplain}
\bibliography{ref}

\end{document}

config: arxiv


TITLE: Definition of exterior algebra
QUESTION [6 upvotes]: I'm studying tensor algebra and exterior algebra and having problems with the definition of $\bigwedge(\mathbb{V})$.
Let $\mathbb{V}$ be any vectorial space over a field $K$. Define
$T^k\mathbb{V}:=\underbrace{\mathbb{V}\otimes \mathbb{V}\otimes\cdots\otimes\mathbb{V}}_{k-times}$, where $T^0\mathbb{V}=K$.
Define the tensorial algebra $T(\mathbb{V}):=\bigoplus_{k=0}^{\infty}T^k(\mathbb{V})=K\oplus\mathbb{V}\oplus(\mathbb{V}\otimes\mathbb{V})\oplus\cdots$.
Multiplication on $T(\mathbb{V})$ is determinated by the canonical isomorphism $T^k(\mathbb{V})\oplus T^j(\mathbb{V})\rightarrow T^{k+j}(\mathbb{V})$, so $T(\mathbb{V})$ is naturally a graded algebra.
Define exterior algebra (and here is my doubt) $\bigwedge(\mathbb{V})$ over a $K$-vector space $\mathbb{V}$ as the quotient algebra  of the tensorial algebra $T(\mathbb{V})$ by the two sided ideal $I$ spanned by the elements of the form $x\otimes x$ with $x\in\mathbb{V}$,
$\bigwedge(\mathbb{V}):=\dfrac{T(\mathbb{V})}{I}$
My question is, first of all how are specifically the elements of $I$? It is a subspace of $T(\mathbb{V})$.

REPLY [5 votes]: Multiplication on $T(\mathbb{V})$ is determinated by the canonical isomorphism $T^k(\mathbb{V})\oplus T^j(\mathbb{V})\rightarrow T^{k+j}(\mathbb{V})$, so $T(\mathbb{V})$ is naturally a graded algebra.

I think that direct sum should be a tensor.

Define exterior algebra (and here is my doubt) $\bigwedge(\mathbb{V})$ over a $K$-vector space $\mathbb{V}$ as the quotient algebra  of the tensorial algebra $T(\mathbb{V})$ by the two sided ideal $I$ spanned by the elements of the form $x\otimes x$ with $x\in\mathbb{V}$,
  $$\bigwedge(\mathbb{V}):=\dfrac{T(\mathbb{V})}{I}$$
  My question is, first of all how are specifically the elements of $I$? It is a subspace of $T(\mathbb{V})$.

As it's mentioned in its name, the tensor algebra is an algebra, which means it's a vector space together with a product $\otimes: TV\times TV\to TV$. In particular it is a ring (under addition of vectors and tensor product) and that is where the notion of ideal comes from.
By definition, an ideal of a ring is a subset of a ring such that:

It's a subgroup under addition,
If you multiply any element of the ideal by ANY element of the ring, the result will be again an element of the ideal, for example, whenever you multiply an even integer by ANY integer, the result is again an even integer. Therefore, the even integers form an ideal of $\mathbb{Z}$.

Now, the question is how do I construct the ideal generated by a set? Let's do this for the example at hand. (I'll do the explicit construction, the less hands-on approach consists of intersecting all the ideals containing $S$.) Consider the set $S=\{x\otimes x|x\in V\}\subset V\otimes V$. The ideal generated by $S$ should satisfy the two properties we mentioned above:

It must be a subgroup under addition. (1) $0\in S$, (2) We must add inverses, that is, add all the elements of the form $-(x\otimes x)$ for all $x\in V$, (3) it must be closed under addition, so we must allow all possible finite sums of its elements.

With this in mind, what we have is that the ideal generated by $S$ must at least containg the set
$$
S\subset S'=\{\sum_{i}(\pm 1)x_i\otimes x_i|\text{the sum is finite and }x_i\in V\}.
$$

Tensoring any element of the ideal must be an element of the ideal again! The way to solve this is rather obvious, we just add all the possibilities to $S'$! The explicit formula of the ideal then is:
$$
\boxed{
I=\{\sum_i r_i\otimes (x_i\otimes x_i)\otimes s_i|\text{the sum is finite and }x_i\in V, r_i,s_i\in TV\}.
}$$

I hope I didn't end up making it look more complicated than it is, feel free to ask questions.
The essential idea behind this construction is that quotienting by this ideal is basically imposing the relation $x\otimes x=0$ (in the quotient) or equivalently (try to show it),
$$
x\otimes y + y\otimes x\sim0\iff x\otimes y \sim -y\otimes x,
$$
which turns out to be very useful in a lot of contexts, for example, notice that the determinant function somehow behaves like this: if you exchange two columns it changes signs.


TITLE: Find the length of the line r(t) for t ∈ [0, ∞)
QUESTION [0 upvotes]: When $σ(t)$ is the curve $σ(t) = cos(t) i + sin (t) j$ and set $r(t) =e^{-t} σ(t)$
I gave it a shot. I might be a little rusty at this:
$L[0, ∞)=\int_0^∞\sqrt{x_1'(t) + x_2'(t)+ ...+x_n'(t)} dt$
\begin{equation}L= [0, ∞)=\int_0^∞\sqrt{-e^{-2t}cos^2(t)-e^{-2t}sin^2(t)-e^{-2t}sin^2(t)+e^{-2t}cos^2(t)}  dt \end{equation}
And in a  few steps
\begin{equation}L= [0, ∞)=\int_0^∞\sqrt{e^{-2t}}\sqrt{-1+cos(2t)}  dt \end{equation}
Is this correct? Can someone help me with what to do next

REPLY [1 votes]: If a curve $\gamma$ is given in the so-called polar form $\phi\mapsto r(\phi)$, which is the same as the parametric representation
$$\gamma:\quad t\mapsto r(t)(\cos t,\sin t)\qquad(a\leq t\leq b)\ ,$$
then from $$x(t)=r(t)\cos t,\quad y(t)=r(t)\sin t$$ we get
$$x'(t)=r'(t)\cos t-r(t)\sin t,\quad y'(t)=r'(t)\sin t+r(t)\cos t\ .$$
In the case $r(t)=e^{-t}$ we therefore have
$$x'^2(t)+y'^2(t)=r'^2(t)+r^2(t)=2e^{-2t}\ .$$
It follows that
$$L(\gamma)=\int_0^\infty\sqrt{x'^2(t)+y'^2(t)}\>dt=\sqrt{2}\int_0^\infty e^{-t}\>dt=\sqrt{2}\ .$$


TITLE: What is the distribution of x (P(X=x))?
QUESTION [0 upvotes]: Consider drawing balls, one at a time, without replacement, from an urn containing M balls, K of which are defective. Let the random variable X(Y) denote the number of the draw on which the first defective (nondefective) ball is obtained. Let Z denote the number of the draw On which the rth defective ball is obtained.
My problem is that I don't know what this notation X(Y) means, does it mean that Y is contained in X?
because later it asks me for the joint f(x, y), so X and Y are two different variables

REPLY [0 votes]: Putting a second option in parentheses is another way of saying “respectively”. The second sentence of your question is equivalent to:

Let the random variables $X$ and $Y$ denote the numbers of the draws on which the first defective ball and the first nondefective ball, respectively, are obtained.

That is, $X$ is the number of the draw on which the first defective ball is obtained, and $Y$ is the number of the draw on which the first non-defective ball is obtained.


TITLE: solving triangles
QUESTION [0 upvotes]: I have $\displaystyle\frac{56.851}{\tan(42.0892^\circ)}$, how do I figure that out on a calculator?
I tried and I keep getting syntax error and I don't know how to make degrees.
Google was no help either.

REPLY [1 votes]: First make sure your calculator is set to degrees or do the conversion mentioned in the other posts.
You shouldn't need to put in the degrees symbol if the calculator is set to degrees (ie you can see either D or Deg somewhere on the display) if instead you can see R, Rad, G or Grad then you need to set it to degrees instead. If it is set to degrees then the number on  which a trig function is operating is already assumed to be in degrees. Your question gives the angle in decimal degrees so you don't need to worry about converting to degrees, minutes and seconds and back again (if you don't understand the last sentence don't worry you probably don't need to).
Older calculators required you to enter the value of the angle 42.0892 first then press the trigonometric button to get the result. Normally if you can see the calculation as you are typing it in then you can just do the calculation in the natural order otherwise you need to do the calculations bit by bit and build up to the final steps.
As a test (with the calculator set to degrees) type 45 and press the tan button, if you have the older style of calculator this will return a value of 1 (the answer to doing tan(45)).
If this is the case then I would work out your calculation using the following steps:

Make sure your calculator is set to degrees
Type in the angle 42.0892
Press tan
The answer is the tan of 42.0892, make it into a reciprocal (there's probably a $\frac{1}{x}$ button or $x^{-1}$ button somewhere on the calculator)
Now multiply the resulting value by the numerator (56.851)
Press equals that should be your answer.

If you can't see a reciprocal button, put the answer to tan(42.0892) into the calculator memory, enter 56.851 and divide it by the stored value.
If you do it right you'll get an answer of 63 rounded to to 2sf (I won't give you the full answer as you should write down the full display and then round it to a suitable degree of accuracy yourself).


TITLE: What does the (absolute) value of the Earth Mover Distance between two histograms mean?
QUESTION [0 upvotes]: When using the Earth Mover Distance as a metric between two 2D histograms which represent normalized probability distributions, what does the value from the resulting optimization actually mean ?
For example say that I call (in algorithmic notation where $P$ and $Q$ are the two histograms in question):
emd(P,Q) = 1.56e-06

What does this absolute value actually mean ?

REPLY [0 votes]: The Earth Mover's Distance is also known as the Wasserstein metric.
The interpretation of the resulting value is a metric that defines you how similar two probability distributions (represented by your histograms) are.
Note that this only makes sense if your histograms are actually normed to have a sum of 1, to satisfy the condition on being a probability distribution.
Edit: A graphical introduction to this distance metric is given here.


TITLE: Fourier transform of Bessel function
QUESTION [0 upvotes]: Please help me find the the representation in $k$-space (Fourier transform) of the following function:
$$
f(k_t, k_z, m) = J_m(k_t\rho)\exp(ik_zz)\exp(im\phi)/N
$$
The properties of $f$ given below might come in handy. The equations are from a paper by S.J. Van Enk and G. Nienhuis(2007) (https://doi.org/10.1080/09500349414550911) eqns.35 and 36 under the 4th subheading.
$$
(\hat{P}^2_x + \hat{P}^2_y)f = \hbar^2k_t^2f,\\ \hat{P}_zf = \hbar k_zf\\ \hat{L}_zf = \hbar m f
$$

REPLY [0 votes]: The $k_z$ is irrelevent, so this is just a two-dimensional problem.
I suspect that this is a homework problem, so I'll just give a hint:
Start by showing that if  you have a function of the form $f(r,\theta)=f(r)e^{-il\theta}$ then the integral definition of the Bessel function $J_l(kr)$ shows that the  two dimensional Fourier transform is
$$
\tilde f({\bf k})= 2\pi i^l e^{-il\theta_k} \int_0^\infty J_l(kr)f(r)r dr.
$$
Here $k$, $\theta_k$ are the polar coordinates of the vector ${\bf k}=(k_x,k_y)$.
Now you can use the Hankel transform formula
$$
\int_0^\infty J_l(kr) J_l(k'r) rdr = \frac 1 k \delta(k-k'),
$$
(which is just a disguised version of the Fourier inversion theorem) to evaluate your FT.


TITLE: Probability question about distinguishable and non distinguishable objects
QUESTION [2 upvotes]: so for part a I got the answer as m choose 1 times (1/m)^b
but for part B I am having different approaches and dont know which one is correct
approach 1:
m choose 2 times (2/m)^m
approach 2:
m choose 2 times (1/m)^m
approach 3:
(i was thinking since they are indistinguishable numbers in piles and empty sets)
(n+1-r) choose (r-7)

REPLY [1 votes]: Think of the balls, urns as being labelled. That does not affect the probabilities.
There are $m^n$ equally likely ways to distribute the $n$ objects among the $m$ urns.
Call a distribution good if at most $2$ urns contain balls. We count the good distributions.
These are of two types (i) all the balls end up in one urn and (ii) the balls end up in two urns.
It is clear that there are $\binom{m}{1}$ type (i) distributions.
For type (ii), there are $\binom{m}{2}$ ways to choose the two lucky urns. There are $2^n-2$ ways to distribute the balls among the $2$ urns so that neither urn is empty. So the number of distributions of type (ii) is $\binom{m}{2}(2^n-2)$.
Add to find the number of good distributions.