text:
\begin{document}

\title[Solvability of Poisson algebras]{Solvability of Poisson algebras}
\author{\textsc{Salvatore Siciliano}}
\address{Dipartimento di Matematica e Fisica ``Ennio De Giorgi", Universit\`{a} del Salento,
Via Provinciale Lecce--Arnesano, 73100--Lecce, Italy}
\email{salvatore.siciliano@unisalento.it}

\author{\textsc{Hamid Usefi}}
\address{Department of Mathematics and Statistics,
Memorial University of Newfoundland,
St. John's, NL,
Canada,
A1C 5S7}
\email{usefi@mun.ca}

\begin{abstract}  Let $P$ be a Poisson algebra with a Lie bracket $\{, \}$ over a field $\F$ of characteristic $p\geq 0$. In this paper, the Lie structure of $P$ is investigated. In particular, if $P$ is solvable with respect to its Lie bracket, then we prove that  the Poisson ideal $\mathcal{J}$  of $P$ generated by all elements $\{\{\{x_1, x_2\}, \{x_3, x_4\}\}, x_5\}$ with $x_1,\ldots ,x_5 \in P$ is associative nilpotent of index bounded by a function of the derived length of $P$. We use this result to further prove that if $P$ is solvable  and $p\neq 2$, then the Poisson ideal $\{P,P\}P$ is nil.
\end{abstract}
\subjclass[2010]{17B63, 16R10, 17B30,   17B50, 17B01.}
\keywords{Poisson algebra; Lie identity; solvable Lie algebra; nilpotent Lie algebra; Poisson ideal; nilpotent ideal; symmetric Poisson algebra.}
\date{\today}

  \maketitle

\section{Introduction}

Since the theory of polynomial identities (PI) has proven to be  extremely useful in the variety of associative, Lie, or Jordan algebras, in the past decade there has been great interest to develop analogues theories for Poisson algebras.
Recall that a Poisson algebra $P$ is a commutative associative algebra with a unity equipped with a Lie bracket $\{, \}$ that satisfies the Leibniz rule:
\begin{align*}
\{a \cdot b, c\} = a \cdot \{b, c\} + b \cdot \{a, c\}, \quad a, b, c \in P.
\end{align*}
Poisson algebras  naturally arise in different areas of algebra, topology and mathematical physics and have received a considerable attention over the years.
One of the main initiatives in the domain of PI Poisson algebras was taken by Farkas \cite{F1, F2} who introduced the analogues of standard polynomial identities. This work was further picked up in \cite{MPR} to study codimension growth in characteristic zero and prove that the tensor product of PI Poisson algebras is again a PI-algebra. Moreover, in \cite{GP} Giambruno and Petrogradsky established when the symmetric Poisson algebra $S(L)$ or the  truncated symmetric Poisson algebra $\s(L)$ of a restricted Lie algebra $L$ satisfies a nontrivial multilinear Poisson identical relation. More recently,  in \cite{MP}, Monteiro Alves and Petrogradsky focused on Lie identities of $S(L)$ and $\s(L)$, establishing in particular when these Poisson algebras are Lie nilpotent or solvable in odd characteristic. Further developments of these topics have been recently carried out by the first author in \cite{S}.

In this paper, we investigate the Lie identities of an arbitrary Poisson algebra $P$ and study the relationship between the Lie structure  and the associative structure of $P$. Motivation for these problems also arises from analogous problems for rings.
An important result for  Lie solvable varieties of associative algebras was proved independently by
Sharma and Srivastava \cite{SS} and Smirnov and Zalesskii \cite{SZ}.
Let $R$ be an associative algebra and consider the Lie product on $R$ given by $[x,y]=xy-yx$ for all $x,y\in R$. Denote by $\mathcal{J}$ the ideal of $R$ generated by all elements
$[[[x_1, x_2], [x_3, x_4]], x_5]$ with $x_1,\ldots ,x_5 \in R$. It is proved in \cite{SS} and \cite{SZ} that if $R$ is Lie solvable, then $\mathcal{J}$ is associative nilpotent. This result has been used numerously and  proven to be very useful (see e.g. \cite{AS, R, SU1, SU2}). The primary goal of this paper is to prove a similar statement for Poisson algebras. In Theorem \ref{S-Z}, we prove that if $P$ is a solvable Poisson algebra of derived length $n$, then the Poisson ideal generated by all elements $\{\{\{x_1,x_2\},\{x_3,x_4\}\},x_5\}$ is associative nilpotent of index bounded by a function of $n$. We also mention that an application of this result in a forthcoming paper allows to settle the solvability problem of $S(L)$ and $\s(L)$ in characteristic 2 posed in \cite[\S 5.3]{MP}.

In our second main result, we investigate the extent to which the ideal  $\{P, P\}P$ is  nil  (of bounded index) in case $P$ is solvable.
Jennings \cite{J} proved that if $R$ is a finitely generated Lie nilpotent ring, then the ideal $[R, R]R$  is nilpotent. In \cite{R}, for an associative algebra $A$ over a field of characteristic $p>0$, Riley showed that $[A, A]A$ is nil of bounded index whenever $A$ is Lie nilpotent, or Lie solvable and $p>2$. Now, in characteristic zero, it follows from \cite[Theorem 7.2]{MPR} that if a Poisson algebra $P$ is solvable, then $\{P,P\}P$ is nil. In Theorem \ref{solvablenil}, we strengthen this result by proving that if $P$ is solvable over a field of characteristic $p\neq 2$, then the Poisson ideal $\{P,P\}P$ is nil. Furthermore, if  $P$ is solvable and  $p\geq  3$ or $P$ is Lie nilpotent and  $p\geq  2$, then $\{P,P\}P$ is nil of bounded index. This generalizes the main result in \cite{MP} about the Lie structure of $S(L)$   to all reduced Poisson algebras.

\section{Definitions and notation}
  Let $P$ be a Poisson algebra over a field $\F$.
 The Poisson brackets are left-normed: $\{x_1,\ldots,x_n\}=\{\{x_1,\ldots,x_{n-1}\},x_n\}$, $n\geq 1$.
We will denote by $\langle S \rangle_\F$ the subspace spanned by a subset $S$ of $P$. We  use the symbol  $Z_P(P)$ for the Poisson center of $P$ (that is, the center of $P$ as a Lie algebra). When one deals with ideals of Poisson algebras or their properties, it is important
to distinguish which operation is considered. Our convention is
that the term \emph{ideal} refers to the associative multiplication, \emph{Lie ideal} refers to the Lie bracket, and \emph{Poisson ideal} refers to
both.

The terms of the derived series of a Lie ideal $L$ of $P$ are defined by $\delta_0(L) = L$ and $\delta_{n}(L) = \{\delta_{n-1}(L), \delta_{n-1}(L)\}$ for $n>0$. Moreover, we denote by $\gamma_n(L)$ ($n\geq 1$) the terms of the descending central series of $L$. In particular, $\gamma_2(L)=\delta_1(L)$  is the derived subalgebra of $L$. We say that $P$ is \emph{ Lie nilpotent} if $P$ is nilpotent as a Lie algebra. The upper Lie power series of $P$ is the chain of Poisson ideals of $P$ defined by $P^{(1)}=P$ and $P^{(n)}=\{P^{(n-1)},P^{(n-1)}\}P$ for every $n>1$. The  Poisson algebra $P$ is said to be \emph{strongly Lie nilpotent} of class $c$ if $P^{(c+1)}=0$ and $P^{(c)}\neq 0$.
  One says that the Poisson algebra $P$ is \emph{solvable} if $P$ is solvable as a Lie algebra. In this case, the minimal $n$ such that   $\delta_{n}(P)=0$ is called the \emph{derived length} of $P$.
	The upper derived series of $P$ is defined by setting $\ud_0(P)=P$ and $\ud_n(P)=\{\ud_{n-1}(P),\ud_{n-1}(P)\} P$ for every $n>0$. Note that $\ud_n(P)$ is a Poisson ideal of $P$ for every $n$. The Poisson algebra $P$ is said to be \emph{strongly solvable} if $\ud_n(P)=0$, for some $n$. In this case, the minimal $n$ with such a property is called the \emph{strong derived length} of $P$.  It is clear that strong solvability implies solvability, but the converse is in general not true.

We say that a Poisson ideal $I$ of $P$ is \emph{nil} if all elements of $I$ are nilpotent. In particular, $I$ is said to be \emph{nil of bounded index} if there exists a positive integer $r$ such that $x^r=0$ for every $x\in I$. Furthermore, $I$ is said to be \emph{ associative nilpotent} if $I$ is a nilpotent ideal of $P$ as an associative algebra.

\section{Lie properties of Poisson algebras}
 We start with the following result about strong solvability and strong Lie nilpotence, which represents the Poisson analogue of the corresponding results of Jennings for rings \cite[Theorems 5.5 and 6.5]{J}.
\begin{prop}\label{stronglynilp} Let $P$ be a  Poisson algebra over a field $\F$. Then the following statements hold.
\begin{enumerate}
\item If $P$ is strongly solvable, then $\ud_1(P)$  is associative
nilpotent of index at most $2^n-1$, where $n$ is the strong derived length of $P$.
\item $P$ is strongly Lie nilpotent if and only if $P$ is Lie nilpotent and strongly solvable.
\end{enumerate}
\end{prop}
\Proof
(1) It will be enough to show that, for every positive integer $k$, one has
\begin{equation}\label{sLieder}
\tilde{\delta}_1(P)^{2^k-1}\subseteq \tilde{\delta}_k(P).
\end{equation}
We proceed by induction on $k$. Let $k>1$, the claim being trivial for $k=1$. For all $x_1,x_2\in \ud_1(P)^{2^{k-1}-1}$ and $y_1,y_2\in P$ we have
 $$
x_1 x_2\{y_1,y_2\}=\{x_1 y_1,x_2 y_2\}-\{x_1,x_2 y_2\}y_1-\{x_1 y_1,x_2\}y_2+\{x_1,x_2\}y_1 y_2.
$$
It follows that
\begin{align*}
 \ud_1(P)^{2^{k}-1}&=\ud_1(P)^{2^{k-1}-1} \cdot \ud_1(P)^{2^{k-1}-1} \cdot \ud_1(P)\\
&\subseteq \{\ud_{k-1}(P), \ud_{k-1}(P)\}P= \tilde{\delta}_k(P),
\end{align*}
as claimed.

(2) Necessity is clear. Conversely, suppose that $P$ is both Lie nilpotent and strongly solvable. We proceed by induction on  the Lie nilpotence class $c$ of $P$. The claim is trivial for $c=1$.  Suppose then $c>1$. Then the Poisson algebra
$\bar{P}=P/\gamma_c(P)P$ is Lie nilpotent of class at most $c-1$ and so,  by the induction hypothesis,
$\bar{P}$ is strongly nilpotent. Hence we have $P^{(r)}\subseteq \gamma_c(P)P$    for some positive integer $r$. As $\gamma_c(P) \subseteq Z_P(P)$, it follows that

$$P^{(2r-1)}\subseteq \gamma_c(P)P^{(r)}\subseteq \gamma_c^2(P)P.$$
We can then continue in this fashion to show that
\begin{equation}\label{gammac}
P^{(k(r-1)+1)}\subseteq \gamma_c^{k}(P)P
\end{equation}
for every nonnegative integer $k$.
 Now, by the first part of the theorem we know that $\ud_1(P)=\gamma_2(P)P$ is associative nilpotent. As $\gamma_c(P)\subseteq \ud_1(P)$, by Equation (\ref{gammac}) we conclude that $P^{(2k(r-1)+1)}=0$ for some $k$ sufficiently large, which proves the assertion. \qed

We now deal with solvability. Our aim is prove the Poisson version of a result obtained by Sharma and Srivastava \cite{SS} and, independently, by Smirnov and Zalesski \cite{SZ}.
We will make use of the following result. Its proof can be found in \cite[Lemma 8.7]{MP} for $I=P$, by the same arguments actually work for any Lie ideal $I$ of the Poisson algebra $P$.

\begin{lem}\label{MP} Let $P$ be a Poisson algebra and $I$ a Lie ideal of $P$. Then
$$
\gamma_m(I)\gamma_n(I)\subseteq \gamma_{m+n-2}(I)P
$$
for every $m,n\geq 2$.
\end{lem}

Our first main result is the following:

\begin{thm}\label{S-Z} Let $P$ a solvable Poisson algebra of derived length $n$. Then the
Poisson ideal generated by all elements $\{\{x_1,x_2\},\{x_3,x_4\},x_5\}$ is associative
nilpotent of index bounded by a function of $n$.
\end{thm}

\Proof
We can suppose $n\geq 2$, the claim being trivial when $P$ is abelian.  We have divided the proof into a sequence of steps.  Throughout the proof, $I$ is a Lie ideal of $P$.

{\bf Step 1}: $\gamma_3(I)\gamma_4(I)\subseteq \delta_2(I)P$. Let $x_1,x_2,x_3,x_4,x_5,x_6,x_7 \in I$.  We clearly have

$$
\{\{x_1,x_2\},\{x_3\{x_4,x_5\},x_6,x_7\} \}\in \{\gamma_2(I),\gamma_3(I)\}\subseteq \delta_2(L).
$$
On the other hand one has
\begin{align*}
&\{\{x_1,x_2\},\{x_3\{x_4,x_5\},x_6,x_7\} \}=\{\{x_1,x_2\},\{x_3,x_6,x_7\}\{x_4,x_5\}+\\
&\{x_3,x_6\}\{x_4,x_5,x_7\}+\{x_3,x_7\}\{x_4,x_5,x_6\}+x_3\{x_4,x_5,x_6,x_7\} \}\\
 &\equiv \{\{x_1,x_2,x_3\} \cdot \{x_4,x_5,x_6,x_7\} \}\quad (\mathrm{mod}\, \delta_2(I)P),
\end{align*}
and the claim follows at once.

{\bf Step 2}: $\gamma_2(I)\{\gamma_2(I),P\}\subseteq \gamma_3(I)P$. Let $x_1,x_2,x_3,x_4\in I$ and $x_5\in P$. We have
\begin{align*}
\{x_3,x_4, \{x_1,x_2 x_5\}\}&=\{x_3,x_4, \{x_1,x_2\}x_5+x_2\{x_1,x_5\}\}\\
&\equiv \{ x_1,x_2\}\{x_3,x_4,x_5 \}\quad (\mathrm{mod}\, \gamma_3(I)P).
\end{align*}
Since $\{x_3,x_4, \{x_1,x_2 x_5\}\}\in \gamma_3(I)P$, this yields the claim.

{\bf Step 3}: $\{I^2,P\}\subseteq I$. Let $x_1,x_2\in I$ and $x_3\in P$. We have
$$
\{x_1x_2,x_3\}=\{x_2,x_1x_3\}-\{x_2x_3,x_1\}\in I,
$$
yielding the claim.

{\bf Step 4}: $\gamma_2^2(I)\gamma_3(P)\subseteq I+\{\gamma_2(I),P\}P$. Let $x_1,x_2,x_3,x_4\in I$ and $x_5,x_6,x_7\in P$. Consider the element $x=\{x_1,x_2x_6,\{x_3x_5,x_4\},x_7\}\in \{\gamma_2(I),P\}$.  Then
$$
x=\{\{x_1,x_2\}x_6+x_2\{x_1,x_6\},\{x_3,x_4\}x_5+x_3\{x_5,x_4\},x_7\}.
$$
Since $x_2\{x_1,x_6\}$ and $x_3\{x_5,x_4\}$ are in $I^2$,  by Step 3 we have
$$
\{\{x_1,x_2\}x_6,\{x_3,x_4\}x_5,x_7\}\in I+\{\gamma_2(I),P\}P.
$$
Moreover
$$
\{\{x_1,x_2, \{x_3,x_4\}x_5\}x_6,x_7\} \in \{\gamma_2(I),P\}P
$$
and so, by the Jacobi identity, we infer that
$$
\{\{x_1,x_2\}\{x_6,\{x_3,x_4\}x_5\},x_7\} \in I+\{\gamma_2(I),P\}P.
$$
In turn, this implies
$$
\{\{x_1,x_2\}\{x_3,x_4\}\{x_6,x_5\},x_7\} \in I+\{\gamma_2(I),P\}P.
$$
It follows that
$$
\{x_1,x_2\}\{x_3,x_4\}\{x_5,x_6,x_7\} \in I+\{\gamma_2(I),P\}P,
$$
and the desired conclusion follows at once.

{\bf Step 5}: $\big(\gamma_2(I)^{5}\{\gamma_3(P),\gamma_3(P),P\} \big)^3\subseteq \delta_2(I)P$.
From Step 4 it follows that
$$
\{\gamma_2(I)^2\gamma_3(P), \gamma_2(I)^2\gamma_3(P)\}\subseteq \gamma_2(I)+\{\gamma_2(I),P\}P.
$$
This forces
$$
\gamma_2(I)^4\{\gamma_3(P), \gamma_3(P)\}\subseteq \gamma_2(I)+\{\gamma_2(I),P\}P
$$
and then
\begin{equation}\label{ref6}
\gamma_2(I)^4\{\gamma_3(P), \gamma_3(P),P\}\subseteq \{\gamma_2(I),P\}P.
\end{equation}
Now, by Step 1 and Lemma \ref{MP} we obtain $\gamma_3(I)^3\subseteq \delta_2(I)P$ and so, by Step 2, we have  $\gamma_2(I)^3\{\gamma_2(I),P\}^3\subseteq \delta_2(I)P$. Therefore, in view of relation (\ref{ref6}) we get
$$
\gamma_2(I)^3\big(\gamma_2(I)^4\{\gamma_3(P),\gamma_3(P),P\} \big)^3\subseteq \delta_2(I)P,
$$
which is equivalent to the claim.

{\bf Step 6}: $\{\gamma_3(P),\gamma_2(P),P\}\{\gamma_2(P),\gamma_3(P),\gamma_3(P)\}\subseteq \{P,\gamma_3(P),\gamma_3(P)\}P$. Let $J,K$ be Lie ideals of $P$. Suppose that $K\subseteq L$, where $L$ is the Lie ideal consisting of all elements $x\in P$ such that $\{P,x\}\subseteq J$. We will show that
\begin{equation}\label{ref8}
\{L,J,P\}\{P,P,K,K\}\subseteq \{J,K,P\}P.
\end{equation}
From this, the desired conclusion will follow by setting  $J=K=\gamma_3(P)$ and $L=\gamma_2(P)$.
Let $x_1\in L$, $x_2\in J$, $x_3,x_4,x_5\in P$, $x_6,x_7\in K$. We have
\begin{align*}
&\{x_4x_6,x_7,x_2,x_3\}=\{\{x_4,x_7\}x_6+x_4\{x_6,x_7\},x_2,x_3\}\\
&=\{\{x_4,x_7,x_2\}x_6+\{x_4,x_7\}\{x_6,x_2\}+\{x_4,x_2\}\{x_6,x_7\}+x_4\{x_6,x_7,x_2\},x_3\}\\
&\equiv \{\{x_4,x_7,x_2\}\{x_6,x_3\}+\{x_4,x_7,x_3\}\{x_6,x_2\}+\{x_4,x_2,x_3\}\{x_6,x_7\}\\
&+\{x_4,x_2\}\{x_6,x_7,x_3\}+\{x_4,x_3\}\{x_6,x_7,x_2\}
\quad (\mathrm{mod}\,\{J,K,P\}P ).
\end{align*}
As $\{x_4x_6,x_7,x_2,x_3\}\in \{J,K,P\}$,  replacing $x_6$ with $\{x_5,x_6\}$ in the previous relation yields
\begin{align}\label{ref8a} \nonumber
\{x_4,x_7,x_2\}\{x_5,x_6,x_3\}&+\{x_4,x_7,x_3\}\{x_5,x_6,x_2\}\\
&+\{x_4,x_2,x_3\}\{x_5,x_6,x_7\}\in \{J,K,P\}P.
\end{align}
Now, by switching $x_6$ with $x_7$ and $x_4$ with $x_5$ in (\ref{ref8a}) we get
\begin{align*}
\{x_4,x_7,x_2\}\{x_5,x_6,x_3\}&+\{x_4,x_7,x_3\}\{x_5,x_6,x_2\}\\
&+\{x_5,x_2,x_3\}\{x_4,x_7,x_6\}\in \{J,K,P\}P
\end{align*}
and then
$$
\{x_4,x_2,x_3\}\{x_5,x_6,x_7\}\equiv \{x_5,x_2,x_3\}\{x_4,x_7,x_6\} \quad (\mathrm{mod}\,\{J,K,P\}P ),
$$
or equivalently
\begin{equation}\label{ref8b}
\{x_4,x_2,x_3\}\{x_6,x_5,x_7\}\equiv \{x_5,x_2,x_3\}\{x_7,x_4,x_6\} \quad (\mathrm{mod}\,\{J,K,P\}P ).
\end{equation}
Note that
\begin{equation}\label{8c}
\{x_5,x_6,x_4,x_7\}\equiv \{x_5,x_6,\{x_4,x_7\}\} \quad (\mathrm{mod}\,\{J,K,P\}P ).
\end{equation}
Therefore,  replacing $x_4$ by $x_1$ and $x_6$ by $\{x_6,x_4\}$  in (\ref{ref8b}) and using  (\ref{8c}) we get
\begin{align*}
&\{x_1,x_2,x_3\}\{x_6,x_4,x_5,x_7\}\equiv \{x_5,x_2,x_3\}\{x_7,x_1,\{x_6,x_4\}\}\\
&\equiv  \{x_5,x_2,x_3\}\{x_6,x_4,x_1,x_7\}\equiv  \{x_5,x_2,x_3\}\{x_6,x_1,x_4,x_7\}\\
&\equiv  \{x_5,x_2,x_3\}\{x_6,x_1,\{x_4,x_7\}\}\equiv  \{x_1,x_2,x_3\}\{x_4,x_7,x_5,x_6\}\\
&\equiv  \{x_1,x_2,x_3\}\{x_6,x_5,\{x_4,x_7\}\} \equiv  \{x_1,x_2,x_3\}\{x_6,x_5,x_4,x_7\} \quad (\mathrm{mod}\,\{J,K,P\}P ).
\end{align*}
Consequently, by the Jacobi identity we must have
$$
\{x_1,x_2,x_3\}\{x_6,\{x_4,x_5\},x_7\}\in \{J,K,P\}P
$$
and then
$$\{x_1,x_2,x_3\}\{x_4,x_5,x_6,x_7\}\in \{J,K,P\}P,$$
from which (\ref{ref8}) follows at once.

{\bf Step 7}: \emph{Let $A,B,C$ be Lie ideals of $P$ such that $\{P,B\}\subseteq A$. Then $\{P,C\}\{A,B\}\subseteq \{A,C\}P$}. Let $x_1\in P$, $x_2\in A$, $x_3\in B$, $x_4\in C$. We have
$$
\{\{x_1x_2,x_3\},x_4\} \equiv \{x_1,x_4\}\{x_2,x_3\} \quad (\mathrm{mod}\, \{A,C\}P).
$$
Since $\{\{x_1x_2,x_3\},x_4\}\subseteq \{A,C\}$, the claim follows.

{\bf Step 8}: $\{\delta_2(P),P\}^3\gamma_3(P)^4\subseteq \{\gamma_3(P),\gamma_3(P),P\}P$.  By applying Step 7 for $A=\gamma_3(P)$, $B=\gamma_2(P)$ and $C=\{\gamma_3(P),\gamma_2(P)\}$ we get

$$
\{\gamma_3(P),\gamma_2(P), P\}\{\gamma_3(P),\gamma_2(P)\}\subseteq \{\gamma_2(P),\gamma_3(P), \gamma_3(P)\}P.
$$
Thus, by Step 6 we obtain
\begin{equation}\label{ref11}
\{\gamma_3(P),\gamma_2(P), P\}^2\{\gamma_3(P),\gamma_2(P)\}\subseteq \{\gamma_3(P),\gamma_3(P),P\}P.
\end{equation}
We have
 \begin{align}\label{pezzo}\nonumber
\{\delta_2(P),P\}&=\{\gamma_2(P),\gamma_2(P),P\}\\
&\subseteq \{\gamma_2(P),P,\gamma_2(P)\}=\{\gamma_3(P),\gamma_2(P)\}.
\end{align}
Moreover
\begin{align}\label{ref11a}\nonumber
\{\gamma_2(P),\gamma_2(P),\gamma_2(P)\}&\subseteq \{\gamma_2(P),\gamma_2(P),P,P\}\\
&\subseteq \{\gamma_2(P),P,\gamma_2(P),P\}=\{\gamma_3(P),\gamma_2(P), P\}.
\end{align}
Let $x_1,x_2,x_3,x_4 \in \gamma_2(P)$ and $x_5\in P$. We have
$$
\{x_1,x_2,\{x_3,x_4x_5\}\}\equiv \{x_1,x_2,x_5\}\{x_3,x_4\} \quad (\mathrm{mod}\, \{\gamma_2(P),\gamma_2(P),\gamma_2(P)\}P).
$$
As $\{x_1,x_2,\{x_3,x_4x_5\}\}\in \{\gamma_2(P),\gamma_2(P),\gamma_2(P)\}$, we have
$$
\{\gamma_2(P),\gamma_2(P),P\}\{\gamma_2(P),\gamma_2(P)\}\subseteq \{\gamma_2(P),\gamma_2(P),\gamma_2(P)\}P
$$
and so, by (\ref{ref11a}), we conclude that
\begin{equation}\label{ref11b}
\{\delta_2(P),P\}\delta_2(P)P\subseteq \{\gamma_3(P),\gamma_2(P), P\}P.
\end{equation}
Furthermore, for every $x_1,x_2\in \gamma_2(P)$ and $x_3,x_4\in P$ one has
$$
\{x_1,\{x_2x_3,x_4\}\}\equiv \{x_1,x_3\}\{x_2,x_4\}\quad (\mathrm{mod}\, \delta_2(P)P).
$$
Since
$\{x_1,\{x_2x_3,x_4\}\}\in \delta_2(P)$, it follows that
\begin{align}\label{ref11c}
\gamma_3(P)^2\subseteq \delta_2(P)P.
\end{align}
At this stage, the combination of  (\ref{ref11}) with (\ref{pezzo}),  (\ref{ref11b}) and  (\ref{ref11c}) yields
 $$
\{\delta_2(P),P\}^3\gamma_3(P)^4\subseteq \{\gamma_3(P),\gamma_3(P),P\}P.
$$
{\bf Step 9}:  \emph{We have $\{\delta_2(P),P\}^t=0$, where
$t=15^{n-2}+\frac{3}{2}(15^{n-2}-1).$} By Step 5 and Step 8 we infer that
$$
\gamma_2(I)^{15} \gamma_3(P)^{12}\{\delta_2(P),P\}^9\subseteq \delta_2(I)P.
$$
In particular, for $I=\delta_{i}(P)$ where $i\geq 0$ the previous relation becomes
\begin{equation}\label{passogen}
\delta_{i+1}(P)^{15} \gamma_3(P)^{12}\{\delta_2(P),P\}^9\subseteq \delta_{i+2}(P)P.
\end{equation}
Now, it follows from (\ref{passogen})  that
\begin{equation}\label{primopasso}
\delta_{n-1}(P)^{15} \gamma_3(P)^{12}\{\delta_2(P),P\}^9=0.
\end{equation}
Furthermore, by combining  (\ref{passogen}) (applied for $i=n-3$) and (\ref{primopasso}) we obtain

 $$
\delta_{n-2}(P)^{15^2} \gamma_3(P)^{12(15+1)}\{\delta_2(P),P\}^{9(15+1)}=0.
$$
At this stage, for every $j\geq n$ we see  by induction that
$$
\delta_{n-j}(P)^{15^j} \gamma_3(P)^{12\cdot (\sum_{k=0}^{j-1}15^k)}\{\delta_2(P),P\}^{9\cdot (\sum_{k=0}^{j-1}15^k)}=0,
$$
thus
 \begin{equation}\label{passoj}
\delta_{n-j}(P)^{15^j} \gamma_3(P)^{\frac{6}{7}\cdot (15^j-1)}\{\delta_2(P),P\}^{\frac{9}{14}\cdot (15^j-1)}=0.
\end{equation}
Since $\{\delta_2(P),P\}\subseteq \delta_2(P) \subseteq \gamma_3(P)$, for $j=n-2$ the formula (\ref{passoj}) implies
 $$
\{\delta_2(P),P\}^{15^{n-2}+\frac{3}{2}(15^{n-2}-1)}=0,
$$
as claimed. This completes the proof. \qed

Let $P$ be a Poisson algebra over a field of characteristic zero. If $P$ is solvable, then it follows from \cite[Theorem 7.2]{MPR} that $\ud_1(P)$ is nil. Our next goal is to extend this result in  characteristic $p>2$. In this case, we will obtain the stronger conclusion that  $\ud_1(P)$ is nil of bounded index. Note that this is not true in characteristic zero.   Consider for instance the Poisson-Grassmann algebra ${\bf G}$ on a countable-dimensional vector space $V=\mathrm{span} \{e_1,e_2,\ldots\}$ over a field  of characteristic zero (cf. \cite[\S 6]{MPR}). Then  ${\bf G}$ is Lie nilpotent of class 2 (see \cite[Theorem 6.1(5)]{MPR}). However,
$$
\{{\bf G},{\bf G}\}=\{e_{i_1}e_{i_2}\cdots e_{i_{2n}}\vert\, i_1<i_2<\cdots<i_{2n}\}
$$
is not nil of bounded index since, for every positive integer $n$, the element $e_1e_2+e_3e_4+\cdots+e_{2n-1}e_{2n}$ has index of nilpotency $n+1$.

We first prove a couple of preliminary lemmas.

\begin{lem}\label{x,y}
	 Let $I$ be a Poisson ideal of a   Poisson algebra  $P$ over a field $\F$.  Then, for every  $a,b,c,x,y\in I$ and
	 positive integer $m$, we have
	 \begin{enumerate}
	 	\item $	(\{a, b, c\}-\{b, c, a\})\gamma_m(I)\su \gamma_{m+1}(I)P$.
	\item $\{x,y\}^2\gamma_m(I)\su \gamma_{m+1}(I)P$.
	 \end{enumerate}
\end{lem}
\Proof Let $d, e\in I$. Direct computations imply that
\begin{align*}
(\{a, b, c\}-\{b, c, a\})\{d, e\}=& \{ac, d, b, e\} -a\{c, d, b, e\} +\{c, d, be, a\}\\
& +b\{d, c, e, a\} +\{d, c, b,  a\}e + \{d,a,b,c\}e \\
&-\{a,d,b,e\}c+\{b,c , ae, d\}+ a\{c,b,e,d\}\\
&+\{c,b,a,d\}e+\{b,a,ec,d\}+e\{a,b,c,d\}\\
&+\{a,d,be,c\}+\{a,b,e,d\}c-b\{a, d, e, c\}.
\end{align*}
It is now enough to take  $d\in \gamma_{m-1}(I)$ to deduce Part (1).

By Part (1), for all $a,b\in I$, we have
\begin{equation}\label{xyy}
\{a,b,b\}\gamma_m(I)\su \gamma_{m+1}(I)P.
\end{equation}
Using Part (1) again, we get  $\{a, b, c\}z\equiv \{b, c, a\}z\; (\text{mod } \gamma_{m+1}(I)P)$, for every $a, b, c\in I$ and  $z\in \gamma_m(I)$.
It follows from the Jacobi identity that
\begin{align}\label{abc}
3\{a, b, c\}\gamma_m(I)\su \gamma_{m+1}(I)P.
\end{align}
Let $x,y,z\in I$. A direct expansion yields
\begin{align}\label{ab}
\{x,y\}\{y,z\}&=3\{xy,y,z\}-3\{x,y,z\}y\nonumber\\
&-\{xz,y,y\}+\{x,y,y\}z+x\{z,y,y\}.
\end{align}
The result now follows  from (\ref{xyy}),  \eqref{abc}, and \eqref{ab}.  \qed

We note that Equation \eqref{abc} immediately implies that if $P$ is Lie nilpotent and $\text{char}(\F)\neq 3$, then $\gamma_{3}(P)P$ is associative nilpotent. This also follows from  \cite[ Lemma 8.7]{MP}.

We also need the following  technical lemma after which we would be ready to prove the second main result of this paper.

\begin{lem}\label{delta2J}
	Let $P$ be a Poisson algebra and denote by $\mathcal{J}$  the Poisson ideal of $P$ generated by all elements $\{\{x_1, x_2\}, \{x_3, x_4\}, x_5\}$ with $x_1,\ldots ,x_5 \in P$. Then $4\{\{x_1, x_2\}, \{x_3, x_4\}\}^3\in   \mathcal{J}$,  for all $x_1,\ldots, x_4\in P$.
\end{lem}
\Proof Let $z_1=\{x_1, x_2\}$ and $z_2=\{x_3, x_4\}$. Note that
\begin{align*}
2\{z_1, z_2\}^2=& \{z_1^2, z_2, z_2\}- 2z_1\{z_1, z_2, z_2\}\equiv \{z_1^2, z_2, z_2\}\quad (\mathrm{mod}\, \mathcal{J}).
\end{align*}
Furthermore,
\begin{align*}
2\{z_1^2, z_2, z_2\} \{z_1, z_2\}=& \{z_2^2, z_1^2, z_2, z_1\}+2z_2\{z_1^2, z_2, z_2, z_1\} \equiv 0 \quad (\mathrm{mod}\, \mathcal{J}).
\end{align*}
We conclude that $4\{z_1,z_2\}^3\in   \mathcal{J}$, as required. \qed

\begin{thm}\label{solvablenil}  Let $P$ be   Poisson algebra  over a field $\F$ of characteristic $p\geq 0$.
	\begin{enumerate}
		\item If $P$ is solvable and $p\neq 2$, then $\{P, P\}P$ is a  nil ideal.
		\item  If $P$ is solvable and $p\geq 3$, then $\{P, P\}P$ is  nil of bounded index.
		\item   If $P$ is Lie nilpotent and $p>0$, then $\{P, P\}P$  is  nil of bounded index.
	\end{enumerate}
\end{thm}
\Proof   Let $x,y\in P$. Note that, by Lemma \ref{x,y},  $\{x,y\}^3\in \gamma_{3}(P)P$ and  $\{x,y\}^5\in \gamma_{4}(P)P$. As a consequence, by Step 1 of the proof of Theorem \ref{S-Z}, we deduce that  $\{x,y\}^8\in \delta_{2}(P)P$.
Now, a typical element $z$ of $\{P, P\}P$  is  of the form
$z=\sum_{i=1}^s \{x_i, y_i\}z_i,$ where $x_i,y_i,z_i\in P$ for every $1\leq i\leq s$.

To prove Part (1), we note that
 $z^{8s}\in \delta_{2}(P)P$. Therefore it suffices to show that $\delta_{2}(P)P$ is nil.
Now, every element $a$ of $\delta_2(P)P$ is  of the form
$a=\sum_{i=1}^r\{u_i,v_i\}x_i$, where $u_i, v_i\in \{P, P\}$ and $x_i\in P$, for every $1\leq i\leq r$.
Since $p\neq 2$, it follows from Lemma \ref{delta2J} that  each $\{u_i,v_i\}^3\in \mathcal{J}$. Thus  $a^{3r}\in \mathcal{J}$.   Theorem \ref{S-Z}  now implies that $a$ is nilpotent.

Suppose that $p\geq 3$ and let us prove Part (2). For what was showed above, for all $x, y\in P$ we have $\{x,y\}^8\in \delta_{2}(P)P$, so that $\{x,y\}^{p^2}\in \delta_{2}(P)P$. Hence $z^{p^2}=\sum_{i=1}^s \{x_i, y_i\}^{p^2}z_i^{p^2}\in \delta_{2}(P)P$.
It is now enough to observe that  $\delta_{2}(P)P$ is nil of bounded index. Indeed, every element $b$ of $\delta_2(P)P$ is of the form
$b=\sum_{i=1}^r\{u_i,v_i\}x_i$, where $u_i, v_i\in \{P, P\}$ and $x_i\in P$, for every $1\leq i\leq r$.
Since $p\neq 3$, it follows from Lemma \ref{delta2J} that  each $\{u_i,v_i\}^p\in \mathcal{J}$. Hence  $b^{p}\in \mathcal{J}$, and by  Theorem \ref{S-Z}   we obtain the desired conclusion.

To prove Part (3), it is enough  by Part (2)  to prove the statement for  $p=2$. By Lemma \ref{x,y} and  the Jacobi identity, we have
 \begin{align}\label{abc2}
 \{a, b, c\}\gamma_m(P)\su \gamma_{m+1}(P)P,
 \end{align}
for every  $a, b, c\in P$. Let $s$ be the nilpotence class of $P$ and $r$ the smallest integer such that $2^r\geq s+1$.
 We deduce from Equation \eqref{abc2} that $z^{2^r}=0$, for every $z\in  \gamma_{3}(P)$. We can now replace $P$ with $P/\gamma_{3}(P)P$ and assume that $\gamma_{3}(P)=0$. It then follows from Lemma \ref{x,y} that
 $\{x,y\}^4=0$, for every $x, y\in P$. Since every element $z\in \{P, P\}$ is a linear combination of
 the $\{x_i,y_i\}$'s, it then follows that $z^4=0$. This finishes the proof.
\qed

\begin{rem} \emph{The assumption on the characteristic of the ground field cannot be dropped in Theorem \ref{solvablenil}(1). In fact, the Hamiltonian Poisson algebra ${\bf H}_2$ in two indeterminates over a field $\F$ of characteristic 2 is solvable of derived length 3 (see \cite[Lemma 11.3]{MP}). However, as ${\bf H}_2$ is a domain, $\tilde{\delta}_1({\bf H}_2)$ is not nil.
}
\end{rem}

Let us now discuss some consequences of our previous results.
We say that a Poisson algebra $P$ is \emph{reduced} if $P$ is reduced as an associative algebra (that is, $P$ is free of nonzero nilpotent elements). By Theorem \ref{solvablenil} and Proposition \ref{stronglynilp} we have

\begin{cor}\label{reduced} Let $P$ be a reduced Poisson algebra over a field $\F$. Then the following statements hold.
\begin{enumerate}
\item $P$ is Lie nilpotent if and only if $P$ is abelian.
\item $P$ is strongly solvable if and only if $P$ is abelian.
\item If $\F$ has characteristic not 2, then $P$ is solvable if and only if $P$ is abelian.
\end{enumerate}
\end{cor}

Let $L$ be a Lie algebra $L$ over a field $\F$ and denote by $S(L)$ its symmetric algebra, which we identify with the polynomial ring $\F[x_1,x_2,\ldots]$ where $x_1,x_2,\ldots$ is an $\F$-basis of $L$ over $\F$. The Lie bracket $\{x,y\}$ of $L$ can be uniquely extended to a Poisson bracket of  $S(L)$ so that this commutative algebra becomes a Poisson algebra, called the \emph{symmetric Poisson algebra} of $L$.
The Lie identities of $S(L)$ have been investigated by Monteiro Alves and Petrogradsky in \cite{MP} (see also \cite{S} for further developments). In particular, by generalizing a result of Shestakov in \cite{Sh}, in \cite[Theorem 4.4]{MP} the authors proved the theorem quoted below:

\begin{thm}
Let $L$ be a Lie algebra over a field $\F$, and $S(L)$ its symmetric Poisson algebra. The following conditions are equivalent:
\begin{enumerate}
\item $L$ is abelian;
\item $S(L)$ is strongly Lie nilpotent;
\item $S(L)$ is Lie nilpotent;
\item  $S(L)$ is strongly solvable;
\item   $S(L)$  is solvable (here assume that $\charac (\F) \neq 2$).
\end{enumerate}
\end{thm}
 As the Poisson algebra $S(L)$ is reduced, the previous theorem is now a consequence of Corollary \ref{reduced}.

When $\F$ has characteristic $p>0$, the Poisson bracket of $S(L)$ naturally induces a Poisson bracket on the factor algebra ${\bf s}(L)=S(L)/I$, where $I$ is the ideal generated by the elements $x^p$ with $x\in L$. The conditions under which the Poisson algebra ${\bf s}(L)$ is solvable in characteristic different from 2 or Lie nilpotent were also determined by  Monteiro Alves and Petrogradski in \cite{MP}. In characteristic 2, solvability of $S(L)$ and ${\bf s}(L)$ was left by the authors as an open problem and a related conjecture was proposed (see \cite[\S 5.3, Conjecture]{MP}). A counterexample for this conjecture was found in \cite{S}.  However, by means of Theorem \ref{S-Z}, a corrected version of  \cite[\S 5.3]{MP} can be proved, thereby completing the classification. This will be accomplished in a  forthcoming separate paper.


TITLE: Logical trap in R topology
QUESTION [1 upvotes]: We know that in metric spaces, Bolzano-Weierstrass (BW) (each infinite set owns a cluster point) and Borel-Lebesgue (BL) properties are equivalent, i.e. compactness and countably compactness are equivalent (and also sequential compactness).
In the set of real numbers $\mathbb{R}$, the BW theorem is often written : "Any bounded infinite set has a cluster point". So according to the equivalence, we can say that in $\mathbb{R}$, "any bouded set verifies (BL)", which is not true because we know that only bounded and closed sets do...
Can you help me to find where I am wrong among this...?
EDIT : to be more precise, here is the wrong reasoning:
"In $\mathbb{R}$, the collection of all infinite bounded sets verifies the (BW) property; since this property is equivalent to the (BL) property and since this one means compactness, all infinite bounded sets is compact."

REPLY [0 votes]: OK, thanks to Henning Makholm, I realized my mistake.
The definition of a cluster point refers to a subset $E$ of a topological space $X$, whereas the notion of compactness refers to the topology in itself, i.e. $E$ is a compact set ("of $X$") if and only if it is compact as a topological space for the topology of subspace. Hence, for compactness, the (BW) property must be studied from the viewpoint of the topological space, and not from the viewpoint of $E$ being a subset of $X$.
For example, in $X=\mathbb{R}$, it is true that any infinite subset of $E=]0,1]$ has a cluster point in $X$, even the infinite subset $A=\{\frac{1}{n} / n\in\mathbb{N^*}\}$ which has $0$ as cluster point in $X$. So $E$ verifies in $X$ the property that lies in (BW), but $E$ is not compact (or "is not a compact subset of $X$") because as a topological space, $E$ doesn't fulfil (BW) since the infinite set $A$ has no cluster point in $E$.


TITLE: Proof of Hilbert Projection Theorem
QUESTION [1 upvotes]: If M is a closed subspace of the Hilbert space H and $x \in H$, then:
There exists a unique element $\hat{x} \in M$ such that:
$\|x-\hat{x} \|=\inf_{y \in M}\|x-y \|$
To proof of the existence of such a point (the uniqueness is proven separately) is given as:
Proof of existence of $\hat{x} \in M$
If $d=\inf_{y \in M} \|x-y \|^2$ then there is a sequence $\{y_n \}$ of elements of  M such that $\|y_n-x \|^2\rightarrow d$. Apply the parallelogram law and using the fact that $(y_m +y_n)/2 \in M$ we can write
$0 \leq \|y_m-y_n \|^2=-4\|(y_m+y_n)/2-x \|^2 + 2(\|y_n-x \|^2+ \|y_m-x\|^2)$
$ \leq -4d + 2(\|y_n-x \|^2+ \|y_m-x\|^2)$  $\rightarrow 0$ as $m,n \rightarrow \infty$
The first term on the second line seems fairly clear:
If $y_n=y_m$ then you have an equality so that $=-4\|(y_m+y_n)/2-x \|^2= -4d$.
If $y_n \neq y_m$ then the negative sign means that the inequality holds i.e. $=-4\|(y_m+y_n)/2-x \|^2 < -4d$. As $y_m$ must be further from x than $y_n$.
The second term (which is unchanged) is causing me some confusion.
Again if $y_n=y_m$ then you have an equality so that $2(\|y_n-x \|^2+ \|y_m-x\|^2)=4d$. This then means that the expression sums to zero and so in this case there is no problem.
However
If $y_n \neq y_m$ then the second term $\|y_m-x\|^2> d$. Indeed it can be arbitrarily large?
I can see that the expression only works if $y_m=y_n$ but surely this is a proof
that the point is unique not that is exists?
Can someone please explain how this is a proof of the existence of such an element?
I mean the existence of such an element is assumed as we are told that there is a sequence $\{y_n \}$ of elements of  M such that $\|y_n-x \|^2\rightarrow d$. But I don't see how the expression above contradicts this? If $y_m \neq y_n$ then the expression above could be arbitrarily large whether or not $\|y_n-x \|^2\rightarrow d$? So there is no contradiction that I can see?

REPLY [1 votes]: Because $\|y_n -x\|^2 \to d$, for all $\epsilon$ there exists $N$ s.t.
$$\|y_n - x\|^2 \leq d + \epsilon$$
for all $n \geq N$.
This means the last equation satisfies
$$-4d + 2(\|y_n-x \|^2+ \|y_m-x\|^2) \leq -4d + 2\left((d + \epsilon) + (d + \epsilon)\right) = 4\epsilon$$
for all $m,n \geq N$.
So the limit does go to zero.


TITLE: Independent Transcendental Numbers
QUESTION [12 upvotes]: I've been thinking about numbers which have not yet been proven nor disproven to be transcendental, such as $e + \pi,\, \pi - e,\, \frac{\pi}{e},\, \gamma,\,\zeta(3),$ etc. Some of these numbers haven't even been proven to be irrational, so it naturally led to me questioning whether these numbers' transcendence could perhaps be independent of $\sf ZF$ or $\sf ZFC$. Do (or can) there exist numbers such that their transcendence or irrationality is independent of $\sf ZF$ or $\sf ZFC$?
I'm aware that this question may be unsolved, so relevent references would be appreciated as well!

Note that by number I mean one which is not defined conditionally using some other independent statement such as the continuum hypothesis or the axiom of choice. (Side question: what happens if we change this definition to demanding the number can be computed to arbitrary precision?)

REPLY [5 votes]: You can code $\Pi^0_1$ statements to define a real as follows: Suppose $R(n)$ is a recursive predicate. Define $x_R = \sum \{2^{-n!} : (\forall m < n)R(m)\}$. Then it is not hard to check that $x_R$ is transcendental iff $(\forall n)R(n)$. Notice that using a computer program for $R(n)$, you can estimate $x_R$ within arbitrary precision. Since there are recursive predicates $R(n)$ (e.g., "$n$ does not code a proof of $0=1$ in ZFC") for which $(\forall n)R(n)$ is undecidable in ZFC, you have the sort of examples you are looking for.


text:
(*
  Title: Ad-Hoc Duality for Orderings and Lattices
  Author: Georg Struth
  Maintainer: Georg Struth <g.struth@sheffield.ac.uk>
*)

section \<open>Ad-Hoc Duality for Orderings and Lattices\<close>

theory Order_Duality
  imports Sup_Lattice

begin

text \<open>This component presents an "explicit" formalisation of order and lattice duality. It augments the data type based one
 used by Wenzel in his lattice components \cite{Wenzel}, and complements the "implicit" formalisation given by locales. It uses a functor dual,
supplied within a type class, which is simply a bijection (isomorphism) between types, with the constraint that
the dual of a dual object is the original object. In Wenzel's formalisation, by contrast, dual is a bijection, but not
 idempotent or involutive. In the past, Preoteasa has used a similar approach with Isabelle~\cite{Preoteasa11b}.\<close>

text \<open>Duality is such a fundamental concept in order and lattice theory that it probably deserves to be included in the type classes
for these objects, as in this section.\<close>

class dual =
  fixes dual :: "'a \<Rightarrow> 'a" ("\<partial>")
  assumes inj_dual: "inj \<partial>"
  and invol_dual [simp]: "\<partial> \<circ> \<partial> = id"

text \<open>This type class allows one to define a type dual. It is actually a dependent type
for which dual can be instantiated.\<close>

typedef (overloaded) 'a dual = "range (dual::'a::dual \<Rightarrow> 'a)"
  by fastforce

setup_lifting type_definition_dual

text \<open>At the moment I have no use for this type.\<close>

context dual
begin

lemma invol_dual_var [simp]: "\<partial> (\<partial> x) = x"
  by (simp add: pointfree_idE)

lemma surj_dual: "surj \<partial>"
  unfolding surj_def by (metis invol_dual_var)

lemma bij_dual: "bij \<partial>"
  by (simp add: bij_def inj_dual surj_dual)

lemma inj_dual_iff: "(\<partial> x = \<partial> y) = (x = y)"
  by (meson inj_dual injD)

lemma dual_iff: "(\<partial> x = y) = (x = \<partial> y)"
  by auto

lemma the_inv_dual: "the_inv \<partial> = \<partial>"
  by (metis comp_apply id_def invol_dual_var inj_dual surj_dual surj_fun_eq the_inv_f_o_f_id)

end

text \<open>In boolean algebras, duality is of course De Morgan duality and can be expressed within the language.\<close>

sublocale boolean_algebra \<subseteq> ba_dual: dual "uminus"
  by (unfold_locales, simp_all add: inj_def)

definition map_dual:: "('a \<Rightarrow> 'b) \<Rightarrow> 'a::dual \<Rightarrow> 'b::dual" ("\<partial>\<^sub>F") where
  "\<partial>\<^sub>F f = \<partial> \<circ> f \<circ> \<partial>"

lemma map_dual_func1: "\<partial>\<^sub>F (f \<circ> g) = \<partial>\<^sub>F f \<circ> \<partial>\<^sub>F g"
  by (metis (no_types, lifting) comp_assoc comp_id invol_dual map_dual_def)

lemma map_dual_func2 [simp]: "\<partial>\<^sub>F id = id"
  by (simp add: map_dual_def)

lemma map_dual_nat_iso: "\<partial>\<^sub>F f \<circ> \<partial> = \<partial> \<circ> id f"
  by (simp add: comp_assoc map_dual_def)

lemma map_dual_invol [simp]: "\<partial>\<^sub>F \<circ> \<partial>\<^sub>F = id"
  unfolding map_dual_def comp_def fun_eq_iff by simp

text \<open>Thus map-dual is naturally isomorphic to the identify functor: The function dual is a natural transformation
between  map-dual and the identity functor, and, because it has a two-sided inverse --- itself, it is a natural isomorphism.\<close>

text \<open>The generic function set-dual provides another natural transformation (see below). Before introducing it, we introduce useful notation
for a widely used function.\<close>

abbreviation "\<eta> \<equiv> (\<lambda>x. {x})"

lemma eta_inj: "inj \<eta>"
  by simp

definition "set_dual = \<eta> \<circ> \<partial>"

lemma set_dual_prop: "set_dual (\<partial> x) = {x}"
  by (metis comp_apply dual_iff set_dual_def)

text \<open>The next four lemmas show that (functional) image and preimage are functors (on functions). This does not really belong here,
but it is useful for what follows. The interaction between duality and (pre)images is needed in applications.\<close>

lemma image_func1: "(`) (f \<circ> g) = (`) f \<circ> (`) g"
  unfolding fun_eq_iff by (simp add: image_comp)

lemma image_func2: "(`) id = id"
  by simp

lemma vimage_func1: "(-`) (f \<circ> g) = (-`) g \<circ> (-`) f"
  unfolding fun_eq_iff by (simp add: vimage_comp)

lemma vimage_func2: "(-`) id = id"
  by simp

lemma iso_image: "mono ((`) f)"
  by (simp add: image_mono monoI)

lemma iso_preimage: "mono ((-`) f)"
  by (simp add: monoI vimage_mono)

context dual
begin

lemma image_dual [simp]: "(`) \<partial> \<circ> (`) \<partial> = id"
  by (metis image_func1 image_func2 invol_dual)

lemma vimage_dual [simp]: "(-`) \<partial> \<circ> (-`) \<partial> = id"
  by (simp add: set.comp)

end

text \<open>The following natural transformation between the powerset functor (image) and the identity functor
is well known.\<close>

lemma power_set_func_nat_trans: "\<eta> \<circ> id f = (`) f \<circ> \<eta>"
  unfolding fun_eq_iff comp_def by simp

text \<open>As an instance, set-dual is a natural transformation with built-in type coercion.\<close>

lemma dual_singleton: "(`) \<partial> \<circ> \<eta> = \<eta> \<circ> \<partial>"
  by auto

lemma finite_dual [simp]: "finite \<circ> (`) \<partial> = finite"
  unfolding fun_eq_iff comp_def using inj_dual finite_vimageI inj_vimage_image_eq by fastforce

lemma finite_dual_var [simp]: "finite (\<partial> ` X) = finite X"
  by (metis comp_def finite_dual)

lemma subset_dual: "(X = \<partial> ` Y) = (\<partial> ` X = Y)"
  by (metis image_dual pointfree_idE)

lemma subset_dual1: "(X \<subseteq> Y) = (\<partial> ` X \<subseteq> \<partial> ` Y)"
  by (simp add: inj_dual inj_image_subset_iff)

lemma dual_empty [simp]: "\<partial> ` {} = {}"
  by simp

lemma dual_UNIV [simp]: "\<partial> ` UNIV = UNIV"
  by (simp add: surj_dual)

lemma fun_dual1: "(f = g \<circ> \<partial>) = (f \<circ> \<partial> = g)"
  by (metis comp_assoc comp_id invol_dual)

lemma fun_dual2: "(f = \<partial> \<circ> g) = (\<partial> \<circ> f = g)"
  by (metis comp_assoc fun.map_id invol_dual)

lemma fun_dual3: "(f = g \<circ> (`) \<partial>) = (f \<circ> (`) \<partial> = g)"
  by (metis comp_id image_dual o_assoc)

lemma fun_dual4: "(f = (`) \<partial> \<circ> g) = ((`) \<partial> \<circ> f = g)"
  by (metis comp_assoc id_comp image_dual)

lemma fun_dual5: "(f = \<partial> \<circ> g \<circ> \<partial>) = (\<partial> \<circ> f \<circ> \<partial> = g)"
  by (metis comp_assoc fun_dual1 fun_dual2)

lemma fun_dual6: "(f = (`) \<partial> \<circ> g \<circ> (`) \<partial>) = ((`) \<partial> \<circ> f \<circ> (`) \<partial> = g)"
  by (simp add: comp_assoc fun_dual3 fun_dual4)

lemma fun_dual7: "(f = \<partial> \<circ> g \<circ> (`) \<partial>) = (\<partial> \<circ> f \<circ> (`) \<partial> = g)"
  by (simp add: comp_assoc fun_dual2 fun_dual3)

lemma fun_dual8: "(f = (`) \<partial> \<circ> g \<circ> \<partial>) = ((`) \<partial> \<circ> f \<circ> \<partial> = g)"
  by (simp add: comp_assoc fun_dual1 fun_dual4)

lemma map_dual_dual: "(\<partial>\<^sub>F f = g) = (\<partial>\<^sub>F g = f)"
  by (metis map_dual_invol pointfree_idE)

text \<open>The next facts show incrementally that the dual of a complete lattice is a complete lattice.\<close>

class ord_with_dual = dual + ord +
  assumes ord_dual: "x \<le> y \<Longrightarrow> \<partial> y \<le> \<partial> x"

begin

lemma dual_dual_ord: "(\<partial> x \<le> \<partial> y) = (y \<le> x)"
  by (metis dual_iff ord_dual)

end

lemma ord_pres_dual:
  fixes f :: "'a::ord_with_dual \<Rightarrow> 'b::ord_with_dual"
  shows "ord_pres f \<Longrightarrow> ord_pres (\<partial>\<^sub>F f)"
  by (simp add: dual_dual_ord map_dual_def ord_pres_def)

lemma map_dual_anti: "(f::'a::ord_with_dual \<Rightarrow> 'b::ord_with_dual) \<le> g  \<Longrightarrow> \<partial>\<^sub>F g \<le> \<partial>\<^sub>F f"
  by (simp add: le_fun_def map_dual_def ord_dual)

class preorder_with_dual = ord_with_dual + preorder

begin

lemma less_dual_def_var: "(\<partial> y < \<partial> x) = (x < y)"
  by (simp add: dual_dual_ord less_le_not_le)

end

class order_with_dual = preorder_with_dual + order

lemma iso_map_dual:
  fixes f :: "'a::order_with_dual \<Rightarrow> 'b::order_with_dual"
  shows "mono f \<Longrightarrow> mono (\<partial>\<^sub>F f)"
  by (simp add: ord_pres_dual ord_pres_mono)

class lattice_with_dual = lattice + dual +
  assumes sup_dual_def: "\<partial> (x \<squnion> y) = \<partial> x \<sqinter> \<partial> y"

begin

subclass order_with_dual
  by (unfold_locales, metis inf.absorb_iff2 sup_absorb1 sup_commute sup_dual_def)

lemma inf_dual: "\<partial> (x \<sqinter> y) = \<partial> x \<squnion> \<partial> y"
  by (metis invol_dual_var sup_dual_def)

lemma inf_to_sup: "x \<sqinter> y = \<partial> (\<partial> x \<squnion> \<partial> y)"
  using inf_dual dual_iff by fastforce

lemma sup_to_inf: "x \<squnion> y = \<partial> (\<partial> x \<sqinter> \<partial> y)"
  by (simp add: inf_dual)

end

class bounded_lattice_with_dual = lattice_with_dual + bounded_lattice

begin

lemma bot_dual: "\<partial> \<bottom> = \<top>"
  by (metis dual_dual_ord dual_iff le_bot top_greatest)

lemma top_dual: "\<partial> \<top> = \<bottom>"
  using bot_dual dual_iff by force

end

class boolean_algebra_with_dual = lattice_with_dual + boolean_algebra

sublocale boolean_algebra \<subseteq> badual: boolean_algebra_with_dual _ _ _ _ _ _ _ _ uminus
  by unfold_locales simp_all

class Sup_lattice_with_dual = Sup_lattice + dual +
  assumes Sups_dual_def: "\<partial> \<circ> Sup = Infs \<circ> (`) \<partial>"

class Inf_lattice_with_dual = Inf_lattice + dual +
  assumes Sups_dual_def: "\<partial> \<circ> Supi = Inf \<circ> (`) \<partial>"

class complete_lattice_with_dual = complete_lattice + dual +
  assumes Sups_dual_def: "\<partial> \<circ> Sup = Inf \<circ> (`) \<partial>"

sublocale Sup_lattice_with_dual \<subseteq> sclatd: complete_lattice_with_dual Infs Sup infs "(\<le>)" le sups bots tops "\<partial>"
  by (unfold_locales, simp add: Sups_dual_def)

sublocale Inf_lattice_with_dual \<subseteq> iclatd: complete_lattice_with_dual Inf Supi infi "(\<le>)" le supi boti topi "\<partial>"
  by (unfold_locales, simp add: Sups_dual_def)

context complete_lattice_with_dual
begin

lemma Inf_dual: "\<partial> \<circ> Inf = Sup \<circ> (`) \<partial>"
  by (metis comp_assoc comp_id fun.map_id Sups_dual_def image_dual invol_dual)

lemma Inf_dual_var: "\<partial> (\<Sqinter>X) = \<Squnion>(\<partial> ` X)"
  using comp_eq_dest Inf_dual by fastforce

lemma Inf_to_Sup: "Inf = \<partial> \<circ> Sup \<circ> (`) \<partial>"
  by (auto simp add: Sups_dual_def image_comp)

lemma Inf_to_Sup_var: "\<Sqinter>X = \<partial> (\<Squnion>(\<partial> ` X))"
  using Inf_dual_var dual_iff by fastforce

lemma Sup_to_Inf: "Sup = \<partial> \<circ> Inf \<circ> (`) \<partial>"
  by (auto simp add: Inf_dual image_comp)

lemma Sup_to_Inf_var: "\<Squnion>X = \<partial> (\<Sqinter>(\<partial> ` X))"
  using Sup_to_Inf by force

lemma Sup_dual_def_var: "\<partial> (\<Squnion>X) = \<Sqinter> (\<partial> ` X)"
  using comp_eq_dest Sups_dual_def by fastforce

lemma bot_dual_def: "\<partial> \<top> = \<bottom>"
  by (smt Inf_UNIV Sup_UNIV Sups_dual_def surj_dual o_eq_dest)

lemma top_dual_def: "\<partial> \<bottom> = \<top>"
  using bot_dual_def dual_iff by blast

lemma inf_dual2: "\<partial> (x \<sqinter> y) = \<partial> x \<squnion> \<partial> y"
  by (smt comp_eq_elim Inf_dual Inf_empty Inf_insert SUP_insert inf_top.right_neutral)

lemma sup_dual: "\<partial> (x \<squnion> y) = \<partial> x \<sqinter> \<partial> y"
  by (metis inf_dual2 dual_iff)

subclass lattice_with_dual
  by (unfold_locales, auto simp: inf_dual sup_dual)

subclass bounded_lattice_with_dual..

end

end

subset_name: curated


text:
\section{Multifidelity model management strategies}
\label{sec:MM:strategies}
Model management in multifidelity methods defines how different models are employed during execution of the outer loop and how outputs from different models are combined. Models are managed such that low-fidelity models are leveraged for speedup, while judicious evaluations of the high-fidelity model establish accuracy and/or convergence of the outer-loop result. This section describes a categorization of model management methods into three types of strategies. The following sections then survey specific model management methods in the context of uncertainty propagation, statistical inference, and optimization.

As shown in Figure~\ref{fig:MMTypes}, we distinguish between three types of model management strategies: adaptation, fusion, and filtering.

\subsection{Adaptation} The first model management strategy uses adaptation to enhance the low-fidelity model with information from the high-fidelity model while the computation proceeds. One example of model management based on adaptation is global optimization with EGO, where a kriging model is adapted in each iteration of the optimization process \cite{jones_efficient_1998,venter_construction_1998}. Another example is the correction of low-fidelity model outputs via updates, which are derived from the high-fidelity model. It is common to use additive updates, which define the correction based on the difference between sampled high-fidelity and low-fidelity outputs, and/or multiplicative updates, which define the correction based on the ratio between sampled high-fidelity and low-fidelity outputs \cite{alexandrov_trust-region_1998,alexandrov_approximation_2001}. The correction model is then typically built using Taylor series expansion based on gradients, and possibly also on higher-order derivative information \cite{michael_eldred_second-order_2004}. In \cite{kennedy_predicting_2000}, low-fidelity models are corrected (calibrated) with Gaussian process models to best predict the output of the high-fidelity model.
Another multifidelity adaptation strategy is via adaptive model reduction, where projection-based reduced models are efficiently adapted as more data of the high-fidelity model become available during solution of the outer-loop application problem. Key to online adaptive model reduction is an efficient adaptation process. In \cite{Peherstorfer201521,aDEIM}, the basis and operators of projection-based reduced models are adapted with low-rank updates. In \cite{NME:NME4800}, an $h$-adaptive refinement of the basis vectors uses clustering algorithms to learn and adapt a reduced basis from high-fidelity model residuals. The work \cite{Amsallem2015} adapts localized reduced bases to smooth the transition from one localized reduced basis to another localized basis.

\subsection{Fusion} The second model management strategy is based on information fusion. Approaches based on fusion evaluate low- and high-fidelity models and then combine information from all outputs. An example from uncertainty propagation is the control variate framework \cite{Bratley,MCMethods,nelson_control_1987}, where the variance of Monte Carlo estimators is reduced by exploiting the correlation between high- and low-fidelity models. The control variate framework leverages a small number of high-fidelity model evaluations to obtain unbiased estimators of the statistics of interest, together with a large number of low-fidelity model evaluations to obtain an estimator with a low variance. Another example from uncertainty propagation is the fusion framework introduced in \cite{koutsourelakis_accurate_2009}, which is based on Bayesian regression.

Co-kriging is another example of a multifidelity method that uses model management based on fusion. Co-kriging derives a model from multiple information sources, e.g., a low- and a high-fidelity model \cite{annels_geostatistical_1991,myers_matrix_1982,Perdikaris20150018}. Co-kriging is often used in the context of optimization if gradient information of the high-fidelity model is available, see \cite{forrester_multi-fidelity_2007}. The work \cite{laurenceau_building_2008} compares kriging and co-kriging models on aerodynamic test functions. In \cite{yamazaki_derivative-enhanced_2012}, gradients are computed cheaply with the adjoint method and then used to derive a co-kriging model for design optimization in large design spaces. In \cite{Han2013177}, co-kriging with gradients and further developments of co-kriging are compared for approximating aerodynamic models of airfoils.

\subsection{Filtering} The third model management strategy is based on filtering, where the high-fidelity model is invoked following the evaluation of a low-fidelity filter. This might entail evaluating the high-fidelity model only if the low-fidelity model is deemed inaccurate, or it might entail evaluating the high-fidelity model only if the candidate point meets some criterion based on the low-fidelity evaluation.
One example of a multifidelity filtering strategy is a multi-stage MCMC algorithm. For example, in two-stage MCMC \cite{christen_markov_2005,Fox97samplingconductivity}, a candidate sample needs to be first accepted by the likelihood induced by the low-fidelity model before the high-fidelity model is evaluated at the candidate sample. As another example, in the multifidelity stochastic collocation approach in \cite{narayan_stochastic_2014}, the stochastic space is explored with the low-fidelity model to derive sampling points at which the high-fidelity model is then evaluated. A third example is multifidelity importance sampling, where the sampling of the high-fidelity model is guided by an importance sampling biasing distribution that is constructed with a low-fidelity model \cite{Peherstorfer2016490}.

config: arxiv


TITLE: Uniform Convergence of integrals
QUESTION [10 upvotes]: If a sequence of functions $f_n$ are uniformly convergent in a given interval $[a,b]$ to a function $f$, are all riemann integrable, then the integral $$\int ^b_af_ndx\rightarrow\int^b_afdx$$ and $f$ is riemann integrable. but is the convergence uniform?
More precisely, is it true that $\forall \epsilon >0, \exists N:\forall n>N$,
$$\left| \int ^b_af_ndx-\int^b_afdx\right|<\epsilon$$
How would I prove it? Or is it obvious?

REPLY [10 votes]: This is tautological. By definition, $\int_a^b f_n$ tends to $\int_a^bf$ means the $\epsilon$ condition you are trying to prove. So if you already know the convergence, there is nothing to do. But actually, I suggest you prove that $\int_a^b f_n$ tends to $\int_a^bf$. It will be a good exercise to clarify these notions.
Hint: the integral is linear and
$$
|\int_a^bg(t)dt|\leq \int_a^b|g(t)|dt\leq (b-a)\sup_{t\in[a,b]}|g(t)|=(b-a)\|g\|_\infty.
$$

REPLY [7 votes]: It doesn't make sense to speak of uniform convergence in this context. The integrals $$I_n := \int_a^b f_n(x) \, dx \qquad \qquad I:=\int_a^b f(x) \, dx$$ are real numbers which do not depend on another parameter (since $a,b$ are fixed). $$I_n = \int_a^b f_n(x) \, dx \to I=\int_a^b f(x) \, dx \qquad (n \to \infty)$$ is (by the definition of convergence of sequences) equivalent to the following statement: $$\forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: |I_n - I| \leq \varepsilon \\ \Leftrightarrow \forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: \left|\int_a^b f_n(x) \, dx - \int_a^b f(x) \, dx \right| \leq \varepsilon$$


TITLE: Orbifolds and singular points
QUESTION [1 upvotes]: If I understand correctly, we roughly define the singular locus $\Sigma_O$ of an orbifold $O$ to be the set of points where the orbifold fails to be a manifold. In particular, if $x \in O$ has a neighbourhood $U \subseteq X_O$ which is homeomorphic to $\bar{U}/\Gamma_i$ for $\bar{U} \subseteq \mathbb{R}^n$, then we can associate to $x$ the isotropy group $I_{\bar{x}}$ of any $\bar{x} \in \bar{U}$ projecting to $x$, which is well defined up to isomorphism. We then have
$$
\Sigma_0 = \{ x \in X_O : I_x \ne \emptyset \}.$$
Now, for two-dimensional orbifolds $O$, it turns out that every singular point $x \in \Sigma_O$ has a neighbourhood modelled on either $\mathbb{R}^2/Z_2$, $\mathbb{R}^2/Z_n$, or $\mathbb{R}^2/D_n$. We call such points mirror points, elliptic points, or corner reflectors, respectively.
My question is really one of visualisation: How should I visualise these different classes? I know for example that elliptic points are exemplified by taking the quotient of $\mathbb{R}^2$ by $C_n$ acting by rotations about the origin, giving you an infinite cone with an elliptic point at the apex. But what are some nice, easily visualised examples of corner reflectors and mirrors? Thanks.

REPLY [0 votes]: The figures are from P.Scott's  The Geometries of 3 manifolds
I think the section about the two dimensional orbifolds will be very helpful visualizing the singular points.


text:
\begin{document}
\maketitle

\begin{abstract}
It has been conjectured that the sequence $(3/2)^n$ modulo 1 is uniformly distributed. The distribution of this sequence is significant in relation to unsolved problems in number theory including the Collatz conjecture. In this paper, we describe an algorithm to compute $(3/2)^n$ modulo 1 to $n=10^8$. We then statistically analyze its distribution. Our results strongly agree with the hypothesis that $(3/2)^n$ modulo 1 is uniformly distributed.
\end{abstract}

\section{Introduction}
\label{Introduction}
The distribution of $(3/2)^n$ modulo 1 is an interesting topic because of its connections with Mahler's Z-numbers \cite{mahler1968unsolved}, Waring's problem on writing integers as sums of $n$th powers \cite{hilbert1909beweis}, and ergodic-theoretic aspects of the Collatz conjecture \cite{lagarias19853x+}. Pisot and Vijayaraghavan proved that this sequence has infinitely many accumulation points \cite{pisot1938repartition}\cite{vijayaraghavan1941fractional}. It is thought to be uniformly distributed, but it has never even been proven dense in the unit interval \cite{finch2003mathematical}. Thus, whether $(3/2)^n$ modulo 1 is uniformly distributed or not is still an unsolved problem.

One straightforward way to understand this sequence is to compute powers of $3/2$ and statistically evaluate the distribution of fractional parts. When generating these powers by multiplication, they quickly result in large numbers that a computer cannot fully represent in floating point. Multiplication is also a computationally expensive operation. To that end, for the problem of efficiently generating powers of $3/2$ modulo 1 we make the following contributions:
\begin{itemize}
  \item We describe a parallelizable, binary addition algorithm of complexity $\mathcal{O}(n^2)$ to compute the fractional part of the sequence $(3/2)^n$ (section \ref{Concept}, appendix \ref{Source}).
  \item We compute the results to $n=10^8$ and conduct statistical testing procedures to assess the strength of the evidence against the uniformity hypothesis for the distribution of the sequence in the unit interval (section \ref{Results}).
\end{itemize}
Section \ref{Background} presents background, section \ref{Concept} offers a description of the proposed parallelization algorithm, section \ref{Results} describes the testing statistical testing procedures employed, section \ref{Related} discusses related work, and section \ref{Conclusion} offers conclusions.

\subsection*{Acknowledgements}
\label{know} J.J.P. Veerman wishes to thank the Department of Mathematics at the University of Crete for its generous hospitality and in particular to Nikos Frantzikinakis for introducing him to this subject.

\section{Background}
\label{Background}
Any real number $x$ can be decomposed into the sum of its parts by letting $\floor*{x}$ denote the integral part and $\{x\} = x - \floor*{x}$ denote the fractional part.

\theoremstyle{definition}
\begin{definition}
A sequence $(x_n)_{n=1}^\infty$ is said to be \textit{uniformly distributed modulo }1 (abbreviated u.d. mod 1) if the proportion of terms of the sequence $\{x_n\}_{n=1}^\infty$ falling in a subinterval $[a,b]$ of $[0,1]$ is proportional to the length of the subinterval \cite{lerma1995distribution}.
That is, if $$\lim_{N\rightarrow \infty} \displaystyle{\frac{\#\{n \leq N: \{x_n\} \in [a,b]\}}{N}} = b-a $$
for every $0 \leq a < b \leq 1.$ Here, $\#(A)$ denotes the cardinality of a set $A$.
\end{definition}

\noindent The next theorem is due to Weyl \cite{weyl1916gleichverteilung}.
\theoremstyle{theorem}
\begin{theorem}
For almost all $x > 1$ and for all $n \in \mathbb{N}$, the sequence $x^n$ is u.d. mod 1.
\end{theorem}
\noindent The exceptional set for which this theorem does not hold has Lebesgue measure 0. A few examples of such sequences in the exceptional set are outlined below, as well as in Finch and Lerma \cite{finch2003mathematical}\cite{lerma1995distribution}:
\begin{enumerate}
  \item All integers $x > 1$, since $x^n \equiv 0$ (mod 1).
  \item Pisot-Vijayaraghavan (or P.V.) numbers, real algebraic integers $x > 1$ whose Galois conjugates lie inside the open unit disc $\{z \in \mathbb{C}: |z| < 1 \}$. Clearly, integers greater than 1 are P.V. numbers. If $x$ is a P.V number, then $\lim_{n\rightarrow \infty} x^n \equiv 0$ (mod 1) geometrically \cite{salem1963algebraic}. Examples are the silver ratio $\delta_s = 1 + \sqrt{2}$ and the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$.
\end{enumerate}
Remarkably, candidate sequences of the measure 1 set for which $x^n$ is u.d. mod 1 are easy to find but difficult to prove. One such candidate is constructed in Drmota \cite{drmota2006sequences}.

\section{Program Design}
\label{Concept}
\subsection{Binary Representation}
\label{Binary}
A simple relationship exists in binary between the sequences $3^n$ and $(3/2)^n$. The number 3/2 is represented in binary as 1.1. A standard binary addition algorithm to compute $(3/2)^2$ involves shifting the radix point one place to the left and then adding the original binary value to the shifted value. This gives $$1.10 + .11 = 10.01.$$

Similarly, the integer 3 is represented in binary digits as 11. To find $3^2 = 9$, a bit shift operation $(<<)$ can be applied to 11 to obtain 110. This term is then added to the original term, which gives $$110 + 11 = 1001.$$

By continuing this way we obtain ordered sequences of binary numbers representing powers of $(3/2)$ and $3$. Here are the first ten.

\begin{center}
\begin{tabular}{ r c r c }
 1.1 & $(3/2)^1 \quad \quad $ & 11 & $3^1$ \\
10.01 & $(3/2)^2 \quad \quad $ & 1001 & $3^2$ \\
11.011 & $(3/2)^3 \quad \quad $ & 11011 & $3^3$ \\
101.0001 & $(3/2)^4 \quad \quad $ & 1010001 & $3^4$ \\
111.10011 & $(3/2)^5 \quad \quad $ & 11110011 & $3^5$ \\
1011.011011 & $(3/2)^6 \quad \quad $ & 1011011011 & $3^6$ \\
10001.0001011 & $(3/2)^7 \quad \quad $ & 100010001011 & $3^7$ \\
11001.10100001 & $(3/2)^8 \quad \quad $ & 1100110100001 & $3^8$ \\
100110.011100011 & $(3/2)^9 \quad \quad $ & 100110011100011 & $3^9$ \\
111001.1010101001 & $(3/2)^{10} \quad \quad $ & 1110011010101001 & $3^{10}$
\end{tabular}
\end{center}

These sequences produce the same binary digits, the only difference being the inclusion of the radix point in the calculation of $(3/2)^j$ for $1 \leq j \leq n$. A shift of one position in the radix occurs for every iteration because the trailing digit of 1 is always added to 0. Thus, in order to determine $(3/2)^j$ from the calculation of $3^j$, the radix point should be located $j$ positions from the right.

For our purposes, the advantages of generating powers of $3$ instead of powers of $(3/2)$ are twofold. First, a hardware implementation of binary integer addition can be used rather than a software implementation of string addition in order to keep track of the location of the radix point. This improves the speed of program execution. Second, the program can run in parallel by providing powers of 3 as starting values. Large starting values of $3^j$ are calculated using an arbitrary precision package.
\subsection{Algorithm Description}
\label{Algorithm}
This algorithm is an arbitrary precision method specifically tailored to compute the sequence $\{(3/2)^n\}$. A technique called \textit{variable length quantity} is used. That is, we concatenate 63 bits of every 64 bit unsigned long integer to form each value of $3^j$, $1 \leq j \leq n$. The 64th bit serves as a marker to determine when a 64 bit unsigned long integer will overflow. This method is preferable to a standard arbitrary precision package for speed of execution.

We begin by seeding an unsigned long vector called \texttt{value} $\in \mathbb{R}^1$ with the number 3. To generate powers of 3, two arithmetic operations are performed: a shift operation $(<<)$ and an add operation $(+)$. These operations are carried out as described in section \ref{Binary} subject to the following conditions: the shift operation sets a shifted bit when overflow of a 63 bit vector word occurs, and the add operation sets a carry bit when overflow of a 63 bit vector word occurs. A new word is added to the vector when the carry or shift bit, or both, indicates. After this occurs the first time, \texttt{value} becomes an element of $\mathbb{R}^2$. At every iteration, a counter is incremented to track the designated location of the radix point to convert the sequence $3^j$ to $(3/2)^j$.

Next, to determine the distribution of the fractional parts of the sequence $(3/2)^j$, the interval $[0,1)$ is subdivided into $2^r$ bins of equal length. The bin number that corresponds to the fractional part of \texttt{value} for each iteration is determined by selecting the $r$ most significant bits of the fractional part, i.e. the digits in positions $[j, j+r-1]$ from the right. The bin number for each iteration is equal to the decimal representation of the digits $[j, j+r-1]$. For each bin, a counter is incremented every time the bin is hit. Thus, the value of the counter gives the number of elements in the bin.

Below we show a portion of the results for the first 40 iterations using $r=2^{10}$ bins. The fractional part to be binned is underlined. Notice, on the last iteration \texttt{value} spans $\mathbb{R}^2$, but the most significant bits of the fractional part of \texttt{value} spans $\mathbb{R}^1$. The source code provides details for the case where the most significant bits of the fractional part spans $\mathbb{R}^2$ or greater (see appendix \ref{Source}).\\

\begin{scriptsize}
\texttt{\raggedleft
000000000000000000000000000000000000000000000000000000000000001\underline{1} \\
00000000000000000000000000000000000000000000000000000000000010\underline{01} \\
0000000000000000000000000000000000000000000000000000000000011\underline{011} \\
000000000000000000000000000000000000000000000000000000000101\underline{0001} \\
00000000000000000000000000000000000000000000000000000000111\underline{10011} \\
0000000000000000000000000000000000000000000000000000001011\underline{011001} \\
000000000000000000000000000000000000000000000000000010001\underline{0001011} \\
00000000000000000000000000000000000000000000000000011001\underline{10100001} \\
0000000000000000000000000000000000000000000000000100110\underline{011100011} \\
000000000000000000000000000000000000000000000000111001\underline{1010101001} \\
00000000000000000000000000000000000000000000001010110\underline{0111111101}1 \\
0000000000000000000000000000000000000000000010000001\underline{1011111100}01 \\
\textbf{\vdots} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\
0011100000111101100100010\underline{1110000101}11000010111111111100000001011 \\
...000000000000000000000000000000000001  \ 001010001011100010110100\underline{0101001000}101001000111111110100000100001 \\
\color{white}hey!}
\end{scriptsize}
\noindent The following table displays the bin numbers for this example using $2^{10}$ bins.

\begin{table}[H]
\centering
\label{table:1}
\captionsetup{justification=centering}
\caption{}
\begin{tabular}{ | c | c | c | }
\hline
Iteration & Most significant fractional digits & Corresponding bin number \\ \hline
1 & 1000000000& 512 \\ \hline
2 & 0100000000& 256 \\ \hline
3 & 0110000000& 384 \\ \hline
4 & 0001000000& 064 \\ \hline
5 & 1001100000& 608 \\ \hline
6 & 0110010000& 400 \\ \hline
7 & 0001011000& 088 \\ \hline
8 & 1010000100& 644 \\ \hline
9 & 0111000110& 454 \\ \hline
10 & 1010101001& 681 \\ \hline
11 & 0111111101& 509 \\ \hline
12 & 1011111100& 764 \\ \hline
\vdots & \vdots& \vdots \\ \hline
39 & 1010101001& 901 \\ \hline
40 & 0101001000& 328 \\ \hline
\end{tabular}
\end{table}

\section{Results}
\label{Results}
We now conduct a numerical experiment for large $n$, and implement a statistical testing procedure to assess the strength of the evidence for the uniformity of the distribution $(3/2)^n$. In particular, we are interested in assigning $n=10^8$ values in $[0,1]$ to $r=2^{25}$ bins that partition the interval, to determine whether the values are uniformly distributed across the different bins.  The number of ways in which $n$ objects can be assigned to $r$ bins ($r\leq n$) is given by the multinomial coefficient $n!/\lrp{x_1!x_2!\cdots x_r!}$.

Hence, assuming that all bins are equally probable (i.e., the numbers are uniformly distributed within the unit interval) and letting $X_i$ ($1\leq i \leq r$) denote the random variable that counts the number of values that fall in bin $i$, the random vector $\boldsymbol{X}=(X_1,\ldots,X_r)$, jointly follows a multinomial distribution with probability mass function (pmf) given by
\begin{eqnarray}
\Pr(X_1=x_1,\ldots,X_r=x_r \given  H_0)&=& \frac{n!}{x_1!x_2!\cdots x_r!} \left(\frac{1}{r}\right)^{x_1+\ldots+x_r}\nonumber\\
&=&{n\choose x_1,\ldots,x_r}\left(\frac{1}{r}\right)^n.\label{eq:H0}
\end{eqnarray}
The conditioning on $H_0$ is used to represent the distribution of $\boldsymbol{X}$ under the null hypothesis of  uniformity. In general , letting $\bpi=(\pi_1,\pi_2,\ldots,\pi_r)$ the multinomial pmf is given by
\begin{eqnarray}
f(\bx \given \bpi)\;=\; \Pr(X_1=x_1,\ldots,X_r=x_r \given \bpi)&=& \frac{n!}{x_1!x_2!\cdots x_r!} \pi_1^{x_1}\cdots\pi_r^{x_r}.\label{eq:H1}
\end{eqnarray}

Thus, testing for uniformity amounts to deciding if the vector $$\bpi=\lrp{\pi_1,\pi_2,\ldots,\pi_r}\text{ is equal to }\bpi_0=\lrp{1/r,1/r,\ldots,1/r},$$ based on the observed vector of counts for the $r$ bins. This statistical inference problem can be addressed from different viewpoints depending on the branch of statistics considered, each providing fundamentally different inferential statements.  From the frequentist perspective, the strength of the evidence \emph{against} $H_0: \bpi=\bpi_0$ can be assessed through a testing procedure that yields a falsifiability statement.  Conversely, the Bayesian inferential setting allows one to compare the evidence present in the data for both $H_0$ and $H_1: \bpi\not=\bpi_0$ through a probabilistic statement.  Throughout the remainder of the section we describe each approach and provide the results obtained for our data in each case.

\subsection{Frequentist Testing}
Defining $\hat{\bpi}=\lrb{X_1/n, X_2/n,\ldots, X_r/n}$, the test statistic $T(\hat{\bpi},\bpi_0)$ is a random variable that characterizes the divergence between the multinomial distributions characterized by the densities \eqref{eq:H0} and \eqref{eq:H1}.  Let the vector of observed counts for the $r$ bins be given by $\bx=(x_1, x_2, \ldots, x_r)$, and denote by $\hat{\bp}=(x_1/n, x_2/n, \ldots, x_r/n)$ the vector of estimated values for the multinomial probabilities.  Furthermore, let $\mcal{R}(\alpha)$ denote the $\alpha$-level rejection region for the test.  Then the testing procedure is characterized by the decision rule
\bean

\text{If }\tau\in\mcal{R}(\alpha) \hfill & \longrightarrow & \text{reject }H_0\\
\text{If }\tau\not\in\mcal{R}(\alpha) \hfill & \longrightarrow & \text{do not reject }H_0,

\eean
where $\tau=T(\hat{\bp},\bpi_0)$ (i.e., the value of the test statistic evaluated at $\hat{\bp}$).  The rejection region $\mcal{R}(\alpha)$ depends on the distribution of the test statistic $T(\hat{\bpi},\bpi_0)$ and the level $\alpha$, often chosen to be in the set $\lrb{0.01,0.05,0.1}$.

The traditional asymptotic results for the distribution of $T(\hat{\bpi},\bpi_0)$ are obtained for the case with $r/n\rightarrow 0$ as $n\rightarrow\infty$ \cite{read2012goodness,Bahadur08}.  However, our problem of interest includes an exceedingly large number of bins, hence, a more suitable assumption for the asymptotic relationship between $n$ and $r$ is $r/n\rightarrow \nu$ with $\nu\in(0,\infty)$. In particular, \cite{Holst1972} shows that some traditional test statistics for the $r/n\rightarrow 0$ case, such as the likelihood ratio test, break down if $r/n\rightarrow \nu$; however, \cite{Holst1972} demonstrates that the popular $\chi$-squared statistic, given by
\begin{eqnarray}
T(\hat{\bpi},\bpi_0)&=& \lrp{\frac{r}{n}\sum_{i=1}^r X_i^2}-n \;=\; rn\lrb{\lrp{\sum_{i=1}^r \lrp{\frac{X_i}{n}}^2}-\frac{1}{r}},
\end{eqnarray}
is robust to this type of asymptotic behavior.  Assuming that $H_0$ holds, $T(\hat{\bpi},\bpi_0)$ is asymptotically distributed chi-squared with $r-1$ degrees of freedom \cite{pearson1900x} (denoted by $\chi_{r-1}^2$) as $r/n\rightarrow\nu$ while $n$ and $r$ approach $\infty$. Hence, for this test statistic the $\alpha$-level rejection region is given by $\mcal{R}(\alpha)=[\chi_{r-1,1-\alpha},\infty)$, where $\chi_{r-1,1-\alpha}$ denotes the $\alpha$-level critical value for the $\chi_{r-1}^2$ distribution (e.g., see Chapter 8.2 in \cite{casella2002statistical}).

Hence, the test statistic evaluated at $\hat{\bp}$ corresponds to
\bea
\tau \;=\; T(\hat{\bp},\bpi_0) &=& \lrp{\frac{r}{n}\sum_{i=1}^r x_i^2}-n,\label{eq:tau}
\eea
implying that if $\hat{p}_i=\frac{x_i}{n}=\frac{1}{r}$ for all $i=1,\ldots,r$, then $\tau=0$, which explains why the rejection region only encompasses the upper tail of the distribution.

Figure \ref{fig:chisqtest} below depicts the probability density functions obtained for the test statistic under the null hypothesis $H_0$ for $r=2^{25}$ and $n=10^8$.  The region shaded in grey represents $P\lrp{T(\hat{\bpi},\bpi_0) > \tau}$.
As mentioned above, the $\alpha$-level hypothesis test conducted rejects $H_0$ whenever $T(\hat{\bpi},\bpi_0)\in\mcal{R}(\alpha)$, or equivalently whenever $P\lrp{T(\hat{\bpi},\bpi_0) > \tau} < \alpha$, which is clearly not the case for for $\alpha\in\lrb{0.01, 0.05,0.1}$ (see Figure \ref{fig:chisqtest}). Thus, from the frequentist standpoint there is no evidence in the data for the test to reject the null hypothesis in favor of $H_1$.

\begin{figure}[htbp]
\begin{center}
\includegraphics[scale=0.5]{FigAll.pdf}
\caption{Probability density function for $T(\hat{\bp},\bpi_0)$ under the null hypothesis of uniformity with $n=10^8$ and $r=2^{25}$.}
\label{fig:chisqtest}
\end{center}
\end{figure}

We tried analogous tests for the pairs $(n=(10^8/2^3), r=2^{22})$, $(n=(10^8/2^5), r=2^{20})$,  and $(n=(10^8/2^7), r=2^{18})$, with $P\lrp{T(\hat{\bpi},\bpi_0) > \tau}$ equal to 0.45, 0.94, and 0.01, respectively.  Note that the three tests with larger $n$ and $r$ values, lead to probabilities that are consistent with $H_0$.  While the test for $(n=(10^8/2^7), r=2^{18})$ has a small probability under $H_0$, this result should be interpreted in contrast to all other possible multinomial weights. In the next section we will revisit this result from the Bayesian standpoint, which allows us to test for the strength of $H_0$ among all other possibilities in the $r$-dimensional simplex, as opposed to simply providing an argument of the falsibiability of $H_0$.

\subsection{Bayesian Inference}

While the frequentist testing framework assumes that the vector of multinomial probabilities $\bpi$ is fixed but unknown, in the Bayesian context $\bpi$ is considered a random vector. The Bayesian testing machinery uses Bayes rule to calculate the conditional probability for the hypotheses of interest given the observed data, denoted by $P(H_0\given \bx)$ and $P(H_1\given \bx)$, respectively, and uses these probabilities as the measure of evidence for or against them.  These probabilities are referred to in the statistical literature as \emph{posterior probabilities}.  The difficulty in calculating these probabilities stems from choosing \emph{suitable} prior probability distributions over the $r$-dimensional simplex $\mcal{S}_r$ (for the vector of multinomial probabilities $\bpi$) and over the space of hypotheses $\lrb{H_0,H_1}$.   What is a \emph{suitable} prior probability distribution has been a topic of intense debate in the statistical literature, but the consensus is that in the absence of prior information, these statements should avoid biasing the outcome of the tests in any direction, and should thus be as uninformative as possible to maximize the ability of the information contained in the data to drive the conclusions.

For our particular problem selecting a prior distribution on the space of hypotheses is straightforward. This space solely contains two elements, therefore, only $\psi_0=P(H_0)$ (the prior probability that  $H_0$ holds) has to be specified.  By construction, conditional on $H_0$ the vectors $\bpi$ and $\bpi_0$ are equal with probability one. Furthermore, let $g$ denote a conditional  density function with respect to the Lebesgue measure on $\lrb{\bpi\not=\bpi_0}$, with $g(\bpi_0)=0$.  Using Bayes theorem and the definitions above, the posterior probability for $H_0$ is
\bea
P(H_0\given \bx)&=&\frac{\psi_0 \, f(\bx\given \bpi_0) }{\psi_0 \, f(\bx\given \bpi_0) + (1-\psi_0) \int_{\mcal{S}_r}f(\bx\given \bpi)g(\bpi)d\,\bpi}\nonumber\\
&=& \lrp{1+\frac{(1-\psi_0)}{\psi_0} \frac{1}{B_g(\bx)}}^{-1},\label{eq:post}
\eea
where $B_g(\bx)=f(\bx\given \bpi_0)\big/ \int_{\mcal{S}_r}f(\bx\given \bpi)g(\bpi)d\,\bpi$ is the Bayes factor, which corresponds to a likelihood ratio for $H_0$ relative to $H_1$ (where the likelihood under $H_1$ is averaged over all possible $\bpi\in\mcal{S}$ according to $g$).

In the absence of prior information the usual choice for $\psi_0$ is often $\psi=0.5$, since it sets the ratio in $(1-\psi_0)/\psi_0$ in \eqref{eq:post}  to 1, making
\bean
P(H_0\given \bx) &=& \frac{B_g(\bx)}{1+ B_g(\bx)},
\eean
which is a function driven by the observed data and the function $g$.  The choice of $g$ strongly affects the outcome of the testing problem, with its influence manifesting through $B_g(\bx)$.  Given that $P(H_0\given \bx)$ is a monotone nondecreasing function of $B_g(\bx)$, instead of choosing a particular function $g$, in \cite{Delampady1990} the authors propose to obtain a lower bound $\underline{B}_G(\bx)=\inf_{g\in G}B_g(\bx)$ over entire classes of densities $G$.  This in turns determines a conservative lower bound for $P(H_0\given \bx)$.  Here, we follow the guidance provided in \cite{Delampady1990}, and restrict to the Dirichlet distribution, characterized by the class of density functions of the form
$$G=\lrb{ g_{c}(\bpi)=\frac{\Gamma(c)}{\lrp{\Gamma( c/r)}^r} \prod_{i=1}^r\pi_{i}^{(c/r)-1}\text{, for some }c>0\text{ and }\bpi\in\mcal{S}_r},$$
whose elements $g_{c}(\bpi)$ are conjugate to the likelihood $f(\bx\given \bpi)$.  These density functions are centered around $\bpi_0$ (i.e., $E_{g_c}[\bpi]=\bpi_0$), and spread their mass around the mean in a way that prevents biases towards particular values, lending some measure of ``objectivity'' to the testing procedure. The lower bound for the Bayes factor in the class $G$ is given by
\bean
\underline{B}_G(\bx) &=&\inf_{c>0}\frac{\Gamma(c+n)\lrp{\Gamma(c/r)}^r \lrp{1/r}^{n}}{\Gamma(c)\prod_{i=1}^r\Gamma(x_i+c/r)},
\eean
which approaches $\underline{B}^{*}(\bx)=\inf_{0<\rho<1} \rho^{1-r}\exp{\lrb{-\frac{1}{2}(1-\rho^2)\tau}}$ as $n\rightarrow\infty$, with $\tau$ defined as in \eqref{eq:tau}.

Applying the results from \cite{Delampady1990} described above to the data generated with $n=10^8$ and $r=2^{25}$, the value of the asymptotic lower bound $\underline{B}^{*}(\bx)=1$.  A Bayes factor of 1 represents indifference between two competing hypotheses.  That being said, the fact that the Bayes factor is bounded below at 1 indicates that, at its worst, the evidence for $\bpi_0$ is as strong as the evidence in favor all other possible values of $\bpi\in \lrb{\mcal{S}_r \setminus \lrb{\bpi_0}}$ combined.   Thus, choosing $\psi_0=0.5$, we have that
\bean
P(H_0\given \bx) &\geq& \frac{B^*(\bx)}{1+ B^*(\bx)} \; =\; \frac{1}{2},
\eean
implying that the posterior probability for $\lrb{\bpi=\bpi_0}$ is at least as large as the posterior probabilities for all other possible values in $\mcal{S}_r$ combined; an overwhelming amount of evidence in favor of $H_0$.

Revisiting the alternative experiments considered, we have that, excluding the pair $n=(10^8/2^7)$ and $r=2^{18}$, all other $(n,r)$ combinations lead lower bound for $P(H_0\given \bx)$ of $1/2$, which again provides strong evidence in favor of $H_0$.  For the pair $n=(10^8/2^7)$ and $r=2^{18}$, we find a lower bound of 8\%.  Implying that although the evidence is not overwhelming in this case, there is some evidence for the null hypothesis, especially in light of the fact that the space of all other possibilities is uncountable.

\section{Related Work}
\label{Related}
To our knowledge no research has been published to evaluate numerically the distribution of the sequence $(3/2)^n$ modulo 1. However, the distribution of this sequence has been studied in relation to another topic: Mahler's Z-numbers. A Z-number is a real number $\xi$ such that $\{\xi(3/2)^n\} \leq 1/2$ for all $n \in \mathbb{N}$. Mahler conjectured in 1968 that no Z-numbers exist \cite{mahler1968unsolved}.

Recently, Akiyama proved that for coprime integers $p>q> 1$, where $p > q^2$, there exists $\xi > 0$ such that $\{\xi(p/q)^n\}$ stays in a Cantor set for all $n \in \mathbb{N}$ \cite{akiyama2008mahler}. Furthermore, Dubickas also proved that for coprime $p>q> 1$, there exists $\xi > 0$ such that $\{\xi(p/q)^n\}$ lies in a short interval of $[0,1]$. In particular, his work implies that $\Vert \{\xi (3/2)^{2n}\} \Vert < 14/45$ \cite{dubickas2010powers}.

Waring's problem is the conjecture that every natural number can be expressed as the sum of $k$ $n$th powers of non-negative integers \cite{finch2003mathematical}. Waring's problem can be solved if the inequality $$\{(3/2)^n\} \leq 1 - (3/4)^n$$ holds. Kubina and Wunderlich demonstrated computationally that this inequality is met for $2 \leq n \leq 471,600,000$ \cite{kubina1990extending}. No counterexample to this inequality has been found, and it is believed that this inequality can be extended to $$(3/4)^n < \{(3/2)^n\} < 1 - (3/4)^n$$ for $n > 7$, though this statement has never been proven \cite{finch2003mathematical}.

As Vijayaraghavan \cite{vijayaraghavan1941fractional} remarked in 1941, it is not known whether $$\limsup\limits_{n\rightarrow \infty} \{(3/2)^n\} - \liminf\limits_{n\rightarrow \infty} \{(3/2)^n\} > 1/2,$$ and this is still an open problem, but in 1995, Flatto, Lagarias, and Pollington \cite{flatto1995range} showed that $$\limsup\limits_{n\rightarrow \infty} \{(3/2)^n\} - \liminf\limits_{n\rightarrow \infty} \{(3/2)^n\} \geq 1/3.$$

\section{Conclusions and Further Work}
\label{Conclusion}
The statistical tests conducted for the numerical experiment strongly support the hypothesis that the sequence $(3/2)^n$ modulo 1 is uniformly distributed when larger sample sizes are considered, both from the frequentist and from the Bayesian standpoint.  For a smaller dataset tested, the evidence is not as conclusive, but even in this case the uniformity hypothesis is not unlikely when considering the outcome of the Bayesian testing procedure.

For the numerical experiment described in Section \ref{Results}, the number of iterations was taken to $n= 10^8$, but it could reasonably be taken to $n=10^9$ or $10^{10}$ with the aid of a supercomputer. The program was parallelized to run four instances total on two computers with Intel(R) Core(TM) i7-2600 CPU @ 3.40GHz, 3401 Mhz, 4 Core(s), 8 Logical Processor(s). The runtime of the program for these computers was $11$ days, with a more precise runtime given by $t = (2.378 \num{1e-10})n^2$, for $n$ denoting the number of iterations and $t$ denoting time (in seconds). Starting values for parallelization are noted in Figure \ref{speed} below.

\begin{figure}[H]
\centering
\includegraphics[width=\textwidth]{extrapolation1.pdf}
\caption{Program runtime (extrapolated)}
\label{speed}
\end{figure}

\bibliography{references}
\bibliographystyle{acm}

\appendix

\section{Source Code}
\label{Source}
All source code can be found at www.github.com/paulaneeley.

\end{document}


TITLE: Understanding Conservative and Curl
QUESTION [1 upvotes]: There are several things I need to clarify on Curl.
1)
Is the conservativeness of  a gradient field only applicable for a Closed curve?
If the field is gradient and if c (curve) is not closed then $\int F.dr$ may not be zero? So I can't say  a gradient field is always conservative.can I?
What it says is that if  $\overrightarrow F$ is a gradient Field then work done along any closed curve will be zero.
2) It says that,
 Let  Vector Field F =$ M \hat i+N \hat j$ where  $M= \partial f/\partial x =f_x $ and $ N=\partial f/\partial y =f_y $. Thus if $\overrightarrow F= \nabla f$ (F is a gradient field) then $M_y = N_x$.    ----(1)
By definition curl  $\overrightarrow F$ = $N_x - M_y$  . ---(2)
My question is By (1) and (2) does it imply that if  $\overrightarrow F$  is a gradient field then curl  $\overrightarrow F$ = 0.
But curl  $\overrightarrow F$ =0 desn't imply that field is conservative(gradient ) does it? Because to say that when curl  $\overrightarrow F$=0 that the field is conservative , the field should be defined everywhere in the region ,
3) Curl is used on vector fields not necessarily gradient fields right ?
Please tell me if I have made any wrong conclusions.

REPLY [1 votes]: please note:
1.if field is conservative, then $\int F.dr$ is path independent and depends only on end points, now this implies that if path is closed then obviously $\int F.dr$=0
2. if $\overrightarrow F$ is conservative then curl$\overrightarrow F$ = 0 and vice versa is also true
3. since curl is cross product , so curl is only defined for vector fields


TITLE: What is the constraint acceleration equation of the system?
QUESTION [1 upvotes]: In the question there is a hint given, $y_A$ is not constant. From this hint I began to look at the pulley that holds $m_1$ and $m_2$, and can see that it has the same tension as $m_3$. From this observation, I (think I) have a means to involve all the masses within a given equation.
My force equation for pulley A is $$T_B - (m_1 + m_2)g = (m_1 + m_2)a$$
and for $m_3$
$$T_B-m_3 g= m_3 a$$
Solving both equations for $T_B$ and setting equal to each other I get
$$m_1 a + m_2 a + (m_1 + m_2)g = m_3 g + m_3 a$$
and then solving for a I get
$$a= \frac{m_3 g - (m_1 + m_2)g}{m_1 + m_2 - m_3}$$
What I would like to know is if my assumption about pulley 2 is correct and my subsequent work.

REPLY [2 votes]: Your procedure is not correct. This looks like a homework problem, so I just give few hints:

write equations with respect to some inertial reference frame, such as ground
each mass has its own acceleration - this is very important!
since pulley is considered massless, tension in the cable from both sides has equal magnitude
write equations of motion for the three masses separately; this will give you 3 equations with 5 unknowns ($a_1$, $a_2$, $a_3$, $T_A$, and $T_B$)
to find unique solution you need 2 more equations: (i) equation of motion for the pulley A will give relationship between tensions $T_A$ and $T_B$, and (ii) displacements for the three masses will give relationship between accelerations $a_1$, $a_2$, and $a_3$.

Here are links to my answers to similar problems that can help you with setting up the equations:
What is positive rotation direction of a pulley in the Atwood machine?
How to find a condition for direction of motion in a system with hanging pulley?


TITLE: Calculus - Continuous Function Proof Help
QUESTION [2 upvotes]: Regarding this question:
Let $f$ be function that is continuous on $[0,1]$.
Suppose that $f(x)>0$ for every $x \in [0,1]$.
Then there exists an $\varepsilon > 0$ such that $f(x)>\varepsilon$ for every $x \in [0,1]$.
--
So I want to say: because $f$ is continuous on $[0,1]$, and $f(x)>0$ for every $x \in [0,1]$, so $f(0)>0$ and $f(1)>0$. So $f(0)f(1)>0$.
From the intermediate value theorem, there is NO such a point $c \in R$ that $0<c<1$ such that $f(c)=0$.
From here I need some hint to continue.
Thanks a lot!

REPLY [1 votes]: This is immediate using the extreme value theorem.