text:
\begin{document}
\maketitle

\begin{abstract}
We introduce algebraic structures known as \textit{psybrackets} and use them
to define invariants of pseudoknots and singular knots and links. Psybrackets
are Niebrzydowski tribrackets with additional structure inspired by the
Reidemeister moves for pseudoknots and singular knots. Examples and computations
are provided.
\end{abstract}

\parbox{5.5in} {\textsc{Keywords:} Pseudoknots, Singular knots, Psybrackets,
Niebrzydowski tribrackets, ternary quasigroups

\smallskip

\textsc{2020 MSC:} 57K12}

\section{\large\textbf{Introduction}}\label{I}

In \cite{MN,MN2}
algebraic structures known as \textit{knot-theoretic ternary quasigroups}
were introduced and investigated. With a notational change, these have
been studied by the third listed author and collaborators as
\textit{Niebrzydowski tribrackets} in papers such as \cite{GNT,NOO,NP,NP2}
and used to define invariants of classical knots and links, virtual links
and handlebody-links. Related objects known as \textit{biquasiles} have
been investigated by the first and third authors in \cite{KN} and by the
third author and collaborators in \cite{CNS,DSN2}.

\textit{Pseudoknots} arose in biology as a way of dealing with knotted
objects with only partial information about the crossings; see e.g.
\cite{CDDK,EP,LR,QR} etc. The mathematical formulation in \cite{H,HHJJMR,HJ}
defines pseudoknots and pseudolinks combinatorially as equivalence classes of
\textit{pseudoknot diagrams}, i.e., knot diagrams with ordinary classical
crossings together with \textit{precrossings} in which it is unknown which
strand is on top, under the equivalence relation determined by the
\textit{pseudoknot Reidemeister moves}.

\textit{Singular knots} are rigid vertex isotopy classes of 4-valent spatial
graphs. We can think of singular knots and links as knots and links in which
some strands are fused together at vertices known as \textit{singular
crossings}. In particular, the cyclic ordering of the edges around each
singular crossing is fixed.

Identifying singular crossings with precrossings, the singular
Reidemeister moves form a subset of the pseudoknot Reidemeister moves;
combinatorially, the two classes of objects differ only by a single move.

In \cite{NOR}, together with two collaborators the third listed author
exploited the similarity of the Reidemeister moves for pseudoknots
and singular knots to
introduce \textit{psyquandles}, algebraic coloring structures for
pseudoknots and singular knots extending the notion of \textit{biquandle
colorings} from the world of classical knots and links. Finite biquandles
give rise to integer-valued \textit{counting invariants}, which can be
enhanced in various ways to define new stronger invariants.

In this paper we apply the idea of Niebrzydowski tribrackets to the case of
pseudoknots and singular knots, defining \textit{psybrackets} analogously
to the way psyquandles extend biquandles. The paper is organized as follows.
In Section \ref{P} we review the basics of pseudoknots and singular knots.
In Section \ref{PSY} we define psybrackets and provide some examples. In
Section \ref{PCI} we define the psybracket counting invariant and provide some
computational examples to explore the power of the new invariants. We
conclude in Section \ref{Q} with some questions for future research.

\section{\large\textbf{Pseudoknots and Singular Knots}}\label{P}

In this section we review the basics of \textit{pseudoknots} and
\textit{singular knots}; the remainder of the paper will concern
algebraic structures from which we will derive invariants of
pseudoknots and singular knots.

\begin{definition}
An \textit{oriented pseudoknot diagram} has \textit{positive and negative
classical crossings} but also \textit{precrossings}.
\[\includegraphics{jk-sj-sn-12.pdf}\]
Replacing a precrossing with a classical crossing is known as
\textit{resolution}.
\end{definition}

Precrossings represent classical crossings for which it is unknown
which strand passes over and which passes under. We may regard a precrossing
as a linear combination of both crossings with a scalar coefficient
of $\frac{1}{2}$ for each; extending linearly, we may regard a pseudoknot
diagram as a linear combination of its resolutions. Interpreting the
scalar weights as probabilities, we obtain from a pseudoknot its
\textit{wereset} or \textit{weighted resolution set}, a
discrete probability distribution whose events are the classical knots
obtained by resolving all precrossings, with probabilities
given by the scalar coefficients.

\begin{example}
The pseudolink
\raisebox{-0.4in}{\includegraphics{jk-sj-sn-13.pdf}} has wereset
\[\left\{
\frac{1}{4}\raisebox{-0.4in}{\includegraphics{jk-sj-sn-14.pdf}},
\frac{1}{4}\raisebox{-0.4in}{\includegraphics{jk-sj-sn-15.pdf}},
\frac{1}{2}\raisebox{-0.4in}{\includegraphics{jk-sj-sn-16.pdf}}
\right\}.\]
\end{example}

Singular knots and links are 4-valent spatial graphs considered up to
\textit{rigid vertex isotopy}, where the cyclic ordering of the edges
entering a vertex is fixed. Such a rigid vertex is called a \textit{singular
crossing}; we can imagine singular crossings as points where the knot becomes
stuck to itself (transversely, not tangentially).

Identifying precrossings with singular crossings, the Reidemeister moves for
pseudolinks and singular links are the same except for one move: precrossings
can be introduced or removed via a Reidemeister I type move, while singular
crossings cannot. More precisely, a \textit{pseudolink} is an equivalence
class of pseudolink diagrams under the equivalence relation generated by
planar isotopy moves, the classical Reidemeister moves I, II and III,
\[\includegraphics{jk-sj-sn-17.pdf}\]
and the moves PI, PI$'$, PII, PIII and PIII$'$
\[\includegraphics{jk-sj-sn-4.pdf}.\]
A \textit{singular link} is an equivalence class of singular link diagrams under the equivalence relation generated by planar isotopy, the usual Reidemeister moves RI, RII and RIII, and the moves PII, PIII and PIII$'$.
Note that we depict here only an example of each move; a complete set of moves
includes mirror images and rotations. See \cite{BEHY} for a generating set of
singular knot moves and \cite{H,HHJJMR} for more about pseudoknot Reidemeister moves.

\begin{remark}
Singular knots and links may be regarded as the ``pseudo-framed case'' of
pseudoknots and pseudolinks, where the ``pseudo-writhe'' or number of
precrossings is preserved.
\end{remark}

As we will see, the algebraic conditions on our psybracket structure coming
from moves PI and PI$'$ are already implied by moves PII and PII$'$ so the
invariants we define will be valid for both pseudolinks and singular links;
on the other hand, they will not be able to distinguish pseudolinks which
differ only by PI and PI$'$ moves.

\section{\large\textbf{Psybrackets}}\label{PSY}

We begin with a definition.

\begin{definition}
Let $X$ be a set. A \textit{psybracket} structure on $X$ consists of
two maps $\langle,,\rangle_c,\langle,,\rangle_p:X\times X\times X\to X$
such that
\begin{itemize}
\item[(i)] For all $a,b,c\in X$ there exist unique $x,y,z,u,v\in X$
such that
\[\begin{array}{rcll}
\langle a,b,x\rangle_c & = & c & (i.i)\\
\langle a,y,b\rangle_c & = & c & (i.ii)\\
\langle z,a,b\rangle_c & = & c & (i.iii)\\
\langle u,b,c\rangle_p & = & b & (i.iv) \\
\langle a,b,v\rangle_p & = & b & (i.v),
\end{array}\]
\item[(ii)]
For all $a,b,c\in X$ we have
\[
\langle a,\langle a,b,c\rangle_c,c\rangle_p =\langle a,\langle a,b,c\rangle_p,c\rangle_c
\]
and
\item[(iii)]
For all $a,b,c,d\in X$ we have
\[\begin{array}{rcll}
\langle \langle a,b,c\rangle_c,c,d\rangle_c  & = &
\langle \langle  a,b,\langle b,c,d\rangle_c \rangle_c, \langle b,c,d\rangle_c, d\rangle_c & (iii.i)\\

& = &
\langle \langle  a,b,\langle b,c,d\rangle_p \rangle_c, \langle b,c,d\rangle_p, d\rangle_c &(iii.ii)\\
\langle \langle a,b,c\rangle_p,c,d\rangle_c  & = &
\langle \langle  a,b,\langle b,c,d\rangle_c \rangle_c, \langle b,c,d\rangle_c, d\rangle_p &(iii.iii)\\
\langle a,b,\langle b,c,d\rangle_c\rangle_c & = &
\langle a,\langle a,b,c\rangle_c,\langle \langle a,b,c\rangle_c,c,d\rangle_c\rangle_c &(iii.iv)\\

& = &
\langle a,\langle a,b,c\rangle_p,\langle \langle a,b,c\rangle_p,c,d\rangle_c\rangle_c &(iii.v)\\
\langle a,b,\langle b,c,d\rangle_p\rangle_c & = &
\langle a,\langle a,b,c\rangle_c,\langle \langle a,b,c\rangle_c,c,d\rangle_c\rangle_p &(iii.vi)
\end{array}\]
\end{itemize}
The set $X$ with the map $\langle,,\rangle_c$ forms a \textit{vertical
tribracket} as described in \cite{NOO}.
\end{definition}

The psybracket axioms are motivated by the following region coloring rules
for pseudoknot diagrams:

\[\includegraphics{jk-sj-sn-1.pdf}\]

Axioms (i.i)-(i.iii) are required by the Reidemeister II moves,
with the Reidemeister I moves following as special cases.
\[\includegraphics{jk-sj-sn-9.pdf}\]
\[\includegraphics{jk-sj-sn-8.pdf}\]
\[\includegraphics{jk-sj-sn-10.pdf}\]

Axioms (i.iv) and (i.v) are motivated by moves PI and PI$'$:
\[\includegraphics{jk-sj-sn-7.pdf}\]

Axiom (ii) is motivated by move PII:
\[\includegraphics{jk-sj-sn-6.pdf}\]

Axioms (iii.i)-(iii.iv) are motivated by Reidemeister III,
PIII and PIII$'$:
\[\includegraphics{jk-sj-sn-2.pdf}\]
\[\includegraphics{jk-sj-sn-3.pdf}\]
\[\includegraphics{jk-sj-sn-5.pdf}\]

\begin{definition}
Let $X$ and $Y$ be psybrackets. A map $f:X\to Y$ is a \textit{psybracket
homomorphism} if for all $a,b,c\in X$ we have
\[\langle f(a),f(b),f(c)\rangle_c=f(\langle a,b,c\rangle_c) \ \mathrm{and}\
\langle f(a),f(b),f(c)\rangle_p=f(\langle a,b,c\rangle_p).
\]
A bijective psybracket homomorphism is an \textit{isomorphism}.
\end{definition}

\begin{example}\label{ex:pb1}
A Niebrzydowski tribracket can be given the structure of a psybracket
by setting \[\langle a,b,c\rangle_p=\langle a,b,c\rangle_c.\] To see that
this definition satisfies the axioms, we need only note that replacing
the precrossings in the moves PI, PIII and PIII$'$ results in
valid classical Reidemeister moves and in the same diagram on both sides
of move PII. Similarly, setting
\[\langle a,b,c\rangle_p=d\]
where \[\langle a,d,c\rangle_c=b\]
yields a psybracket, as we can see by resolving the precrossings as
negative classical crossings.
\end{example}

\begin{example}
Let $G$ be a group. Then $G$ is a Niebrzydowski tribracket under the operation
\[\langle a,b,c\rangle =  ab^{-1}c\]
known as a \textit{Dehn tribracket}. Then the two psybracket structures
in Example \ref{ex:pb1} are
\[\langle a,b,c\rangle_c = \langle a,b,c\rangle_p= ab^{-1}c\]
and
\[\langle a,b,c\rangle_c = ab^{-1}c,\quad  \langle a,b,c\rangle_p= cb^{-1}a\]
respectively; we call these the \textit{positive} and \textit{negative
Dehn psybracket} structures on $G$.
\end{example}

We can specify a psybracket structure on a finite set $X=\{1,2,\dots, n\}$
with a pair of \textit{operation 3-tensors},
i.e. lists of $n$ square matrices of size $n\times n$ such that the
entry in matrix $a$ row $b$ column $c$ is $\langle a,b,c\rangle_c$ or
$\langle a,b,c\rangle_p$.

\begin{example}
The Dehn tribracket structure on $\mathbb{Z}_4=\{1,2,3,4\}$ (where we use
$4$ for the class of zero) is
\[\left[
\left[\begin{array}{rrrr}
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
\end{array}\right],
\left[\begin{array}{rrrr}
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
\end{array}\right],
\left[\begin{array}{rrrr}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
\end{array}\right],
\left[\begin{array}{rrrr}
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3 \\
1 & 2 & 3 & 4 \\
\end{array}\right]
\right].\]
The positive and negative Dehn psybracket structures on $\mathbb{Z}_4$
both have this same operation 3-tensor for
$\langle,,\rangle_p$ and for $\langle,,\rangle_c$.
\end{example}

\begin{example}
Let $X=\{1,2,3\}$. Using \textit{python} code, we compute that there are six
isomorphism classes of psybracket structures on $X$. Representatives
of each class are specified by the operation 3-tensors below:

\[\left[\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_p\right]
\]
\[\left[\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\2 & 1 & 3 \\3 & 2 & 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\3 & 2 & 1 \\1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\1 & 3 & 2 \\2 & 1 & 3
\end{array}\right]\right]_p\right]
\]
\[\left[\left[\left[\begin{array}{rrr}
2& 3& 1 \\3& 1& 2 \\1& 2& 3
\end{array}\right],\left[\begin{array}{rrr}
1& 2& 3\\2& 3& 1 \\3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3& 1& 2 \\1& 2& 3 \\2& 3& 1
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
2& 3& 1 \\3& 1& 2 \\1& 2& 3
\end{array}\right],\left[\begin{array}{rrr}
1& 2& 3 \\2& 3& 1 \\3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3& 1& 2 \\1& 2& 3 \\2& 3& 1
\end{array}\right]\right]_p\right]
\]
\[\left[\left[\left[\begin{array}{rrr}
2 & 3 & 1\\ 3 & 1 & 2\\ 1 & 2 & 3
\end{array}\right],\left[\begin{array}{rrr}
1 & 2 & 3\\ 2 & 3 & 1\\ 3 & 1 & 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1 & 2\\ 1 & 2 & 3\\ 2 & 3 & 1
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
3 & 2 & 1\\ 1 & 3 & 2\\ 2 & 1 & 3
\end{array}\right],\left[\begin{array}{rrr}
1 & 3 & 2\\ 2 & 1 & 3\\ 3 & 2 & 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3\\ 3 & 2 & 1\\ 1 & 3 & 2
\end{array}\right]\right]_p\right]
\]
\[\left[\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\3 & 2 & 1 \\2 & 1 & 3
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\2 & 1 & 3 \\1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\1 & 3 & 2 \\3 & 2 & 1
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 2 & 3 \\1 & 2 & 3 \\1 & 2 & 3
\end{array}\right],\left[\begin{array}{rrr}
2 & 3 & 1 \\2 & 3 & 1 \\2 & 3 & 1
\end{array}\right],\left[\begin{array}{rrr}
3 & 1 & 2 \\3 & 1 & 2 \\3 & 1 & 2
\end{array}\right]\right]_p\right]
\]
\[\left[\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\ 1 & 3 & 2 \\ 3 & 2 & 1
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\ 1 & 3 & 2 \\ 3 & 2 & 1
\end{array}\right]\right]_p\right]
\]
\end{example}

\section{\large\textbf{Psybracket Counting Invariants}}\label{PCI}

As with previous knot coloring structures, by construction we have the
following result:

\begin{theorem}
Let $X$ be a finite psybracket, $K$ an oriented pseudoknot or singular knot
diagram, and $\mathcal{C}(K,X)$ the set of $X$-colorings of $K$. Then the
number of $X$-colorings of $K$,
\[\Phi_X^{\mathbb{Z}}(K)=|\mathcal{C}(K,X)|\]
is an integer-valued invariant of pseudoknots and singular knots we will
call the \textit{psybracket counting invariant} and denote by
$\Phi^{\mathbb{Z}}_X(K)$.
\end{theorem}

\begin{example}
Consider the pseudoknot $3_1.3$ below.
\[\includegraphics{jk-sj-sn-11.pdf}\]
It has two positive crossings and one
precrossing; resolving the precrossing one way yields a trefoil and the other
way yields an unknot. We can distinguish this pseudoknot from both the
trefoil and the unknot using psybracket counting invariants. Specifically,
the psybracket
\[X_1=\left[\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_p\right]
\]
gives us counting invariant values
$\Phi_{X_1}^{\mathbb{Z}}(3_1.3)=27$ and $\Phi_{X_1}^{\mathbb{Z}}(\mathrm{Unknot})=9$
while the psybracket
\[X_2=\left[\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\2 & 1 & 3 \\3 & 2 & 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\3 & 2 & 1 \\1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\1 & 3 & 2 \\2 & 1 & 3
\end{array}\right]\right]_p\right]
\]
gives us counting invariant values
$\Phi_{X_2}^{\mathbb{Z}}(3_1.3)=9$ and $\Phi_{X_2}^{\mathbb{Z}}(3_1)=27$.
\end{example}

\begin{example}
	We computed the coloring invariant $\Phi_{X}^{\mathbb{Z}}$ of a choice of orientation for the 2-bouquet graphs in \cite{O} using the psybracket $X$ with the operation matrix

	\[\left[\left[\left[\begin{array}{rrr}
	1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
	\end{array}\right],\left[\begin{array}{rrr}
	3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
	\end{array}\right],\left[\begin{array}{rrr}
	2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
	\end{array}\right]\right]_c,
	\left[\left[\begin{array}{rrr}
	1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
	\end{array}\right],\left[\begin{array}{rrr}
	3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
	\end{array}\right],\left[\begin{array}{rrr}
	2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
	\end{array}\right]\right]_p\right].
	\]

	The results are collected in the table.
	\begin{center}

		\begin{tabular}{ c|c }
			$L$& $\Phi_{X}^{\mathbb{Z}}(L)$ \\
			\hline
$0_1^k, 5^k_8,6^k_4, 6^k_5,1^l_1,6^l_{10}$ &$ 9$\\ \hline
$6^k_6, 6^k_7,5^l_1,6^l_7,6^l_{12}$  &$ 27 $\\ \hline
$2^k_1,3^k_1,5^k_6, 5^k_7, 6^k_{16},6^k_{17},6^k_{19},4^l_1,5^l_3,6^l_1, 6^l_3$ &$ 81 $\\ \hline
$5^k_2,5^k_4,5^k_5,6^k_{15},6^k_{19},3^l_1,6^l_2,6^l_4, 6^l_5,6^l_8$ &$ 243 $\\ \hline
$4^k_1, 4^k_2,4^k_3, 5^k_1, 5^k_3,6^k_{11},6^k_{13}, 6^k_{14},6^k_{18},6^l_6,6^l_9$ & $729 $\\ \hline
$6^k_{12},5^l_2, 6^l_{11}$ & $2187 $\\ \hline
$6^k_1,6^k_2, 6^k_3,6^k_8, 6^k_9,6^k_{10}$  & $6561 $
		\end{tabular}
	\end{center}
\end{example}

\begin{example}
	We computed the coloring invariant $\Phi_{X}^{\mathbb{Z}}$ of a choice of orientation for the pseudoknots up to 5 crossings in the pseudoknot tables in \cite{HHJJMR} using the psybracket $X$ with the operation matrix

	\[\left[\left[\left[\begin{array}{rrr}
	1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
	\end{array}\right],\left[\begin{array}{rrr}
	3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
	\end{array}\right],\left[\begin{array}{rrr}
	2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
	\end{array}\right]\right]_c,
	\left[\left[\begin{array}{rrr}
	1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
	\end{array}\right],\left[\begin{array}{rrr}
	3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
	\end{array}\right],\left[\begin{array}{rrr}
	2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
	\end{array}\right]\right]_p\right].
	\]

	The results are collected in the table.
	\begin{center}

		\begin{tabular}{ c|c }
			$L$& $\Phi_{X}^{\mathbb{Z}}(L)$ \\
			\hline
			$5_1.1, 5_1.2, 5_1.3,5_1.4,5_1.5,5_2.1$ & 9 \\
			\hline
			$3_1.1,3_1.2,3_1.3,4_1.1,4_1.2,4_1.3,4_1.4,4_1.5,5_2.2$ &27 \\
			\hline
			$5_2.3,5_2.4,5_2.6$ & 81\\
			\hline
			$5_2.5,5_2.7,5_2.9$ & 243\\
			\hline
			$5_2.8,5_2.10$ & 729\\
		\end{tabular}
	\end{center}
\end{example}

\begin{example}

We computed the coloring invariant $\Phi_{X}^{\mathbb{Z}}$ of a choice of orientation for the pseudoknots up to 5 crossings in the pseudoknot tables using the psybracket $X$ with the operation matrix

\[\left[\left[\left[\begin{array}{rrr}
1 & 2& 3\\ 2& 3& 1\\ 3& 1& 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 1& 2\\ 1& 2& 3\\ 2& 3& 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 3& 1\\ 3& 1& 2\\ 1& 2& 3
\end{array}\right]\right]_c,
\left[\left[\begin{array}{rrr}
1 & 3 & 2 \\2 & 1 & 3 \\3 & 2 & 1
\end{array}\right],\left[\begin{array}{rrr}
2 & 1 & 3 \\3 & 2 & 1 \\1 & 3 & 2
\end{array}\right],\left[\begin{array}{rrr}
3 & 2 & 1 \\1 & 3 & 2 \\2 & 1 & 3
\end{array}\right]\right]_p\right].
\]

The results are collected in the table.
\begin{center}

	\begin{tabular}{ c|c }
		$L$& $\Phi_{X}^{\mathbb{Z}}(L)$ \\
		\hline
		$3_1.2,3_1.3,4_1.2,4_1.3,5_1.1,5_1.3,5_1.4,5_2.1$ & 9 \\
		\hline
		$3_1.1,4_1.1,4_1.4,4_1.5,5_1.2,5_1.5,5_2.2$ &27 \\
		\hline
		$5_2.3,5_2.4,5_2.6$ & 81\\
		\hline
		$5_2.5,5_2.7,5_2.9$ & 243\\
		\hline
		$5_2.8,5_2.10$ & 729\\
	\end{tabular}
\end{center}

\end{example}

\section{\large\textbf{Questions}}\label{Q}

In this paper we have only initiated the study of the new topic of
psybrackets and their pseudo/singular knot and link invariants. There
are many interesting questions to be explored in this area; we suggest
a few of them here.

\begin{itemize}
\item As with counting invariants arising from other structures, many types
of \textit{enhancements} are possible. Applying a historically successful
strategy, we ask what invariants of psybracket-colored pseudoknots are possible.
Ideas might include cocycle enhancements analogous to those in \cite{NOO},
skein enhancements like those in \cite{ANR}, module enhancements like those in
\cite{NNS} and many more.
\item What is the structure of psybrackets? What kinds of products,
decompositions, functors to and from other algebraic categories are possible?
\item What generalizations are possible to the cases of pseudo/singular
trivalent graphs and handlebody knots or to the virtual and twisted virtual cases?
\end{itemize}

\bibliography{jk-sj-sn}{}
\bibliographystyle{abbrv}

\bigskip

\noindent
\textsc{Department of Mathematics, \\
Pusan National University, \\
Busan 46241, Republic of Korea
}

\medskip

\noindent
\textsc{Department of Mathematical Sciences \\
Claremont McKenna College \\
850 Columbia Ave. \\
Claremont, CA 91711}

\end{document}


TITLE: Why is the k=0 phonon neglected when calculating Debye-Waller factor?
QUESTION [5 upvotes]: When calculating Debye-Waller factor one gets the form:
$e^{-2W} = exp\left(-v\int\frac{d^dk}{2\pi^d}\sum_{s}\frac{\hbar}{2M\omega_s(\mathbf{k})}(\mathbf{q}\cdot\epsilon_s(\mathbf{k}))^2coth(\frac{1}{2}\beta\hbar\omega_s(\mathbf{k}))\right)$
where $d$ is the dimension of the solid.
For small $k$, assuming acoustic branches, $\omega_s(\mathbf{k}) \propto k$ is small, and the coth is proportional to $1/k$, so the integrand is proportional to $1/k^2$.
In 3 dimensions, we get $d^3k \propto k^2dk$ so the integral converges and Debye-Waller factor is finite.
In 2 dimensions and 1 dimensions though, the integral goes like $1/k$ and $1/k^2$ respectively, and does not converge.
Now, I've read that for a finite crystal, even though the integral is over the entire first Brillouin zone, the integral has a lower bound $k_{min} = \frac{2\pi}{L}$ where $L$ is the length of the crystal, which renders it proportional to $ln(L)$ in 2 dimensions or $L$ in 1 dimension. This will mean that for a finite crystal in 1/2D, $W$ will be finite (although very large possibly).
The question is why is the $k=0$ phonon discarded? Because it seems that $k=0$ is a valid solution to Hamilton equation in a harmonic crystal.

REPLY [4 votes]: The $k=0$ (acoustic) phonon is exactly zero, not approximately, and simply means the entire crystal is translated by a constant. This is a fully coherent motion, as all the atoms move in the same way, and thus has no impact on the scattering. One way to think about it is that the scattering shouldn't change if the entire crystal was moved from a lab in New York, to a lab in Los Angeles, but this is exactly what a $k=0$ acoustic phonon does, it just translates the whole system.


TITLE: is there any more odd near-perfect number?
QUESTION [1 upvotes]: we know the first odd near-perfect number is $3^4*7^2*11^2*19^2$(of course the number must be square).and from one paper ,which is studying the odd perfect number and giving some estimate on it, we can also use the inequality to describe the odd near-perfect number,such as
$n=\prod_{i=1}^{r}p_{i}^{n_{i}}$is an odd near-perfect number,then we have $n>\frac{1}{(2^\frac{1}{r}-1)^r},\sum_{i=1}^{r}\frac{1}{p_i}<1$.is there any good estimate about the odd near-perfect number?and is there any more odd near-perfect number?

REPLY [3 votes]: Indeed, so far the only known odd near-perfect number (a number $n$ with redundant divisor $d$ such that the sum of the proper divisors of $n$ is $n+d$) is the number found by Donovan Johnson, $173369889=3^4 \cdot 7^2 \cdot 11^2 \cdot 19^2$, see On perfect and near-perfect numbers by Paul Pollack and Vladimir Shevelev.
According to OEIS, it remains the only odd near-perfect number up to $1.4\cdot 10^{19}$


TITLE: If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ has coordinates $f^1 \ldots f^m$, and each $f^i$ is differentiable at $0$...
QUESTION [5 upvotes]: ...then does it follow that $f$ is differentiable at 0?
My motivation for asking this is as follows: in Spivak's Calculus on Manifolds, in theorem 2.9, he uses this with the additional condition that each $f^i$ is continuously differentiable in a nbh of 0, to conclude that $f$ is differentiable and I don't think is necessary.
Namely, if each $f^i$ has derivative $Df^i$, I claim the matrix with $i^{th}$ row $Df^i$ will serve as $Df$. Indeed, let $v_j$ be a sequence tending to 0 in $\mathbb{R}^n$,  we have (by the triangle inequality, if you wish)$$\frac{| f(v_j) - f(0) - \sum_i Df^i(v) |}{|v_j|} \leqslant \frac{\sum_i |f^i(v_j) - f^i(0) - Df^i(v_j)|}{|v_j|}$$Taking the limit as $j \rightarrow \infty$, each summand goes to 0 by the differentiability of $Df^i$ (and there's only $m$ of them), hence the limit is 0.
Is this wrong? Thanks in advance!
EDIT: btw, conditional on the above proof being right and/or the claim being right, does anyone know maybe what Spivak was going for?

REPLY [2 votes]: In case this is useful to anyone else, let me record the comments of Georges Elencwajg and Dylan Moreland-the answer is yes it's true, and the condition in Spivak is superfluous.
The proof is the one liner I wrote above, albeit with better notation: If $\lambda^i$ are the derivatives of the $f^i$ at 0, I claim the matrix with $i^{th}$ row $\lambda^i$ will serve as $Df(0)$. Indeed we have $$\frac{|f(v) - f(0) - \lambda^i(v)e_i|}{|v|} \leqslant \sum_i \frac{|f^i(v) - f^i(0) - \lambda^i(v)|}{|v|}$$Taking any $v_i \rightarrow 0$, applying the above inequality, and using the differentiability of each $f^i$ at 0 gives the result.
Thanks all for the assistance!


TITLE: Whistle Physics
QUESTION [20 upvotes]: I'm looking for a simple explanation of how a whistle operates.  I know that forcing air over a sharp lip can set up a wave in a resonating cavity, but how?  "Most whistles operate due to a feedback mechanism between flow instability and acoustics"--yes, but what does that feedback mechanism look like?
I was surprised to be unable to find a basic diagram online demonstrating how a whistle operates.  I did find lots of images like this:

. . . but such images are unhelpful since they don't show exactly what's producing the oscillation!

REPLY [2 votes]: Air in a chamber resonates like a spring: It has mass and springs back when compressed or decompressed. When air is blown across the opening, and slightly down, it will push the air inside down & cause a disturbance which will cause the air inside to start to resonate. As it resonates down, it deflects the air stream down into the opening, pushing the air down more. As the air inside springs back, it deflects the air stream upward, so it's no longer pushing down. It also drags air away from inside by the Bernoulli effect, helping to pull the air inside up. Any vortexes formed are incidental.


text:
\begin{document}

\begin{abstract}
We construct several families of manifolds that have pairs of genus $g$ Heegaard splittings that must be stabilized roughly $g$ times to become equivalent. We also show that when two unstabilized, boundary-unstabilized Heegaard splittings are amalgamated by a ``sufficiently complicated" map, the resulting splitting is unstabilized. As a corollary, we produce a manifold that has distance one Heegaard splittings of arbitrarily high genus. Finally, we show
that in a 3-manifold formed by a sufficiently complicated gluing, a low genus, unstabilized Heegaard splitting can be expressed in a unique way as an amalgamation over the gluing surface.
\end{abstract}

\maketitle

\noindent
Keywords: Heegaard Splitting, Stabilization, Normal Surface, Minimal Surface.

\markright{SPLITTINGS OF SUFFICIENTLY COMPLICATED 3-MANIFOLDS}

\section{Introduction.}

The purpose of this report is to outline the main results of a series of forthcoming papers by the author. The first three of these \cite{dividing}, \cite{polyhedra}, \cite{NormalIndexN} lay the groundwork for ``topological index $n$" surfaces. The final paper in the series \cite{results} will replace the present one, and uses this technology to establish the results described below.

Given a Heegaard surface $H$ in a 3-manifold, $M$, one can {\it stabilize} to obtain a splitting of higher genus by taking the connected sum of $H$ with the genus one splitting of $S^3$. Suppose $H_1$ and $H_2$ are Heegaard splittings of $M$, where $\rm{genus}(H_1) \ge \rm{genus}(H_2)$. It is a classical result of Reidemeister \cite{reidemeister} and Singer \cite{singer} from 1933 that as long as $H_1$ and $H_2$ induce the same partition of the components of $\partial M$, stabilizing $H_1$ some number of times produces a stabilization of $H_2$. Just one stabilization was proved to always be sufficient in large classes of 3-manifolds, including Seifert fibered spaces \cite{schultens:96}, genus two 3-manifolds \cite{rs:99}, and most graph manifolds \cite{Derby-Talbot2006} (see also \cite{sedgwick:97}). The lack of examples to the contrary has led to ``The Stabilization Conjecture": Any pair of Heegaard splittings requires at most one stabilization to become equivalent. (See Conjecture 7.4 in \cite{ScharlemannSurvey}.)

In this announcement we outline our program to construct families of counter-examples to the Stabilization Conjecture. This work, however, is broader than just the Stabilization Conjecture. The techniques developed to produce these examples also lead to the resolution of several other important questions in 3-manifold topology, as detailed below.

This work was announced in December of 2007 at a Workshop on {\it Triangulations, Heegaard Splittings, and Hyperbolic Geometry}, at the American Institute of Mathematics. At the same conference another family of counter-examples to the Stabilization Conjecture was announced by Hass, Thompson, and Thurston, and their preprint has since appeared on the arxiv \cite{HTT}. Their proof uses mainly geometric techniques. Several months later Johnson posted a preprint on the arxiv \cite{johnson} that claims to be a combinatorial version of their paper.

Our main family of counter-examples is described by the following theorem:

\begin{thm}
\label{t:MainFamily}
For each $n>5$ there is a closed 3-manifold which has a pair of inequivalent (unoriented) Heegaard splittings of genus $n$ which must be stabilized at least $n-3$ times to become equivalent.
\end{thm}

The pairs of splittings obtained by Hass, Thompson, and Thurston are isotopic, as unoriented surfaces. When the manifold has boundary oriented and unoriented isotopy classes are the same. Johnson uses this fact to produce unoriented counter-examples. In fact, when there is boundary present our construction is easier as well, and is outlined in Section \ref{s:Program} below. This sketch is representative of the techniques used to establish all of the results claimed here.

\section{Higher Genus Gordon Conjecture}

Given an analogous result about connected sums (see \cite{GordonConjecture}), it is natural to conjecture that amalgamations of unstabilized Heegaard splittings are unstabilized. That is, if $M_1$ and $M_2$ are glued along a surface $F$ of non-zero genus, and $H_i$ is an unstabilized Heegaard surface in $M_i$, then $H_1$ and $H_2$ can be amalgamated in $M_1 \cup M_2$ to an unstabilized Heegaard surface. Unfortunately, even if the genus of $F$ is one, then Schultens and Weidmann have shown this to be false \cite{SchultensWeidmann}. It is perhaps surprising, then, that we are able to establish the following:

\begin{thm}
\label{t:HighGenusGordon}
Let $M_1$ and $M_2$ denote irreducible, orientable, anannular 3-manifolds with incompressible, homeomorphic boundary. Let $M$ be the 3-manifold obtained from $M_1$ and $M_2$ by gluing their boundaries by some ``sufficiently complicated" homeomorphism. Let $H_i$ be an unstabilized, boundary-unstabilized Heegaard surface in $M_i$ of low genus. Then the amalgamation of $H_1$ and $H_2$ in $M$ is unstabilized.
\end{thm}

Just how low ``low genus" is depends on how complicated the gluing map between $M_1$ and $M_2$ is, unless $\bdy M_1 \cong T^2$. In this latter case the words ``low genus" may be removed from the statement of the theorem.

As a corollary to this result we construct an example whose existence has been conjectured by Yoav Moriah: a non-minimal genus Heegaard splitting which has Hempel distance \cite{hempel:01} exactly one.  Moriah has called the search for such examples the ``nemesis of Heegaard splittings" \cite{moriah}. In fact, we go further and produce a single manifold that has such splittings of arbitrarily high genus:

\begin{cor}
There exists a closed 3-manifold that contains non-minimal genus, unstabilized Heegaard splittings which are not strongly irreducible, of arbitrarily high genus.
\end{cor}

\begin{proof}
Let $M$ denote a 3-manifold with torus boundary, and strongly irreducible Heegaard splittings of arbitrarily high genus. Such an example has been constructed by Casson and Gordon. (See \cite{sedgwick:97}.)

Now let $M_1$ and $M_2$ be two copies of $M$, and let $H_g^i$ denote a genus $g$ strongly irreducible splitting in $M_i$. As $H_g^i$ is strongly irreducible, it is neither stabilized nor boundary-stabilized. Hence, if $M_1$ is glued to $M_2$ by a sufficiently complicated homeomorphism, it follows from Theorem \ref{t:HighGenusGordon} that the amalgamation of $H_g^1$ and $H_g^2$ is unstabilized. Finally, note that every amalgamation is weakly reducible.
\end{proof}

Finally, we prove the following theorem, which is also analogous to a result about connected sums established by the author in \cite{GordonConjecture}.

\begin{thm}
\label{t:HighGenusGordonIsotopy}
Let $M_1$ and $M_2$ denote irreducible, orientable, anannular 3-manifolds with incompressible, homeomorphic boundary.  Let $M$ be the 3-manifold obtained from $M_1$ and $M_2$ by gluing their boundaries by some ``sufficiently complicated" homeomorphism. Then every ``low genus" Heegaard splitting of $M$ has a \underline{unique} expression as the amalgamation of splittings of $M_1$ and $M_2$.
\end{thm}

Again, the words ``low genus" in the statement of this theorem can be removed if $M_1$ and $M_2$ are being glued along a torus.

Theorems \ref{t:HighGenusGordon} and \ref{t:HighGenusGordonIsotopy} together complete our picture of the ``low genus" Heegaard splittings of a 3-manifold constructed by gluing two manifolds $M_1$ and $M_2$ together by a ``sufficiently complicated" map.  Every unstabilized, low genus Heegard splitting of $M_1 \cup _\phi M_2$  is an amalgamation of ``component" unstabilized splittings of $M_1$ and $M_2$. Two such splittings are isotopic if and only if  their component splittings are isotopic. Hence, the unstabilized splittings of $M_1 \cup _\phi M_2$ are completely determined by the unstabilized splittings of $M_1$ and $M_2$.

\section{Normal and almost normal surfaces.}

A necessary step in the proofs of the above results is to put various classes of surfaces in some kind of ``normal form" with respect to a triangulation. We believe these results are of independent interest.

Two common classes of surfaces that have proved useful in 3-manifold topology are the {\it incompressible} and {\it strongly irreducible} \cite{cg:87} ones. In \cite{crit} the author introduced a third class, called {\it critical surfaces}. These were used by the author in \cite{GordonConjecture} to prove a conjecture of Gordon. Each of these categories is defined by a combinatorial condition on the set of compressing disks for the surface.

Following the work of Kneser \cite{kneser:29}, Haken proved that incompressible surfaces can be put into a nice form with respect to any triangulation \cite{haken:68}. Such surfaces were called {\it normal}, and intersect each tetrahedron in a collection of triangles and quadrilaterals. Rubinstein pushed this work further by proving that strongly irreducible Heegaard splittings can be isotoped to be {\it almost normal} (see also \cite{stocking:96}). Such surfaces look normal everywhere, with the exception of one piece in a single tetrahedron. This exceptional piece is either an octagon, or two normal disks connected by an unknotted tube. We call this piece the {\it almost normal exception}.

In \cite{bachman:98} the author extended Rubinstein's result to strongly irreducible surfaces that have non-empty boundary (see also \cite{bachman:98erratum}). For this work, a third type of almost normal exception was introduced, which looks like two normal disks connected by a band along $\bdy M$. Similar results have been obtained by Wilson \cite{wilson} and Coward \cite{coward}.

To prove the theorems listed above it was necessary to find a similar normal form for critical surfaces (possibly with non-empty boundary). The precise result is:

\begin{thm}
\label{t:CriticalNormal}
Given a fixed triangulation, any critical surface (with or without boundary) is isotopic to one which looks normal everywhere, except for exactly one 12-gon or two almost normal exceptions.
\end{thm}

This theorem, as well as new proofs of the normalization results mentioned before it, will appear in \cite{NormalIndexN}.

\section{Topological Barrier Surfaces}

A technical result that will be of interest to the experts is that when two boundary components of a (possibly disconnected) 3-manifold are glued by a ``sufficiently complicated" map the gluing surface in the resulting 3-manifold acts as a barrier to a wide variety of other surfaces. A precise statement is:

\begin{thm}
\label{t:barrier}
 \cite{NormalIndexN}
Let $M$ be a (possibly disconnected) 3-manifold with homeomorphic boundary components $F_1, F_2$. Let $M_\phi$ be the manifold obtained from $M$ by the gluing map $\phi:F_1 \to F_2$.  If $\phi$ is ``sufficiently complicated" then any incompressible, strongly irreducible, or critical surface in $M_\phi$ that has relatively low genus can be isotoped to lie entirely in $M$.
\end{thm}

If $F_1 \cong T^2$ then the words ``relatively low genus" can be dropped from the statement of the above theorem. In all other cases the more complicated the map $\phi$ is, the higher the genus of the surface can be that we can guarantee will lie entirely in $M$. Theorem \ref{t:barrier} is the heart of the techniques discussed here. This is the main technical result used to obtain proofs of Theorems \ref{t:MainFamily} through \ref{t:HighGenusGordonIsotopy}.

Theorem \ref{t:barrier} is a generalization of several previous results. Proofs of Theorem \ref{t:barrier} in the incompressible case have been given using both minimal surfaces and normal surfaces. Proofs in the strongly irreducible case have been given when $M$ is disconnected by Lackenby using (unstable) minimal surfaces \cite{lackenby:04} and Li using almost normal surfaces \cite{li}. There should certainly be a proof using minimal surfaces in the critical case, but it has only been conjectured that such surfaces can be isotoped to be minimal \cite{kyoto}.

In lieu of this we use techniques from normal surface theory to establish the result. We prove in \cite{NormalIndexN} that if Theorem \ref{t:barrier} is false, then there are normal-like surfaces in $M$ whose boundaries on $F_1$ and $F_2$ are a bounded distance apart in the curve complex after gluing via $\phi$.  Theorem \ref{t:barrier} then follows from \cite{js:98} in the case where $F_1 \cong T^2$ and $M$ is disconnected, from \cite{self} when $F_1 \cong T^2$ and $M$ is connected,  and from \cite{jr:Half} when $F_1$ has higher genus.

\section{The Program}
\label{s:Program}

In this section we illustrate the techniques used to obtain Theorems \ref{t:MainFamily} through \ref{t:HighGenusGordonIsotopy}. Here we sketch our proof that there is a family of counter-examples to the Stabilization Conjecture that are splittings of a manifold with torus boundary.

To begin, let $X$ be a compact, orientable, irreducible, annannular 3-manifold such that:
\begin{enumerate}
	\item One component of $\partial X$ is a torus $T$ which is incompressible in $X$.
	\item The other component of $\partial X$ is a high genus surface $F$, which is incompressible in $X$.
	\item There is a minimal genus Heegaard splitting $H_X$ of $X$ which separates $F$ from $T$, and has genus equal to $\gen(F)+1$.
\end{enumerate}

Such a manifold is not difficult to construct. We can obtain a new Heegaard splitting $G_X$ of $X$ by tubing $H_X$ to a copy of $T$ (pushed slightly into $X$). The resulting splitting does not separate the boundary components of $X$. It follows that $H_X$ and $G_X$ are not equivalent after {\it any} number of stabilizations in $X$. This observation is the key to the proof that the Heegaard splittings constructed presently give counterexamples to the Stabilization Conjecture.

Now let $Y$ be a 3-manifold with incompressible boundary, such that $\partial Y \cong F$. Let $H_Y$ denote a minimal genus Heegaard splitting of $Y$. Such a manifold can be constructed where $\gen(H_Y)=\gen(H_X)$. For each homeomorphism $\phi: \partial Y \to F$ let $M_{\phi}=X \cup _{\phi} Y$ denote the manifold obtained by gluing $X$ to $Y$ via $\phi$.

It is well known that $H_X$ and $H_Y$ give rise to a Heegaard splitting $H_\phi$ of $M_{\phi}$, called their {\it amalgamation} \cite{schultens:93}. Similarly, there is an amalgamation of $G_X$ and $H_Y$, which we call $G_\phi$.

\begin{thm}
\label{t:TorusBoundaryCounterExample}
If $\phi$ is ``sufficiently complicated" then one must stabilize $G_\phi$ at least $\gen(G_\phi)-4$ times to obtain a stabilization of $H_\phi$.
\end{thm}

This theorem is proved by using critical surfaces. To give a sketch of the proof, we must first describe how such surfaces arise.

Loosely speaking, a {\it Generalized Heegaard Splitting} GHS $H$ of a 3-manifold $M$ is a pair of sets of orientable, connected surfaces, $\thick{H}$ and $\thin{H}$, such that each component $M'$ of the complement of $\thin{H}$ in $M$ contains exactly one element $H_+$ of $\thick{H}$, and $H_+$ is a Heegaard surface for $M'$. From the point of view of  \cite{st:94}, GHSs arise naturally from handle structures. Conversely, given a GHS one can always find an associated handle structure.

There are two natural ways to take a GHS and produce a ``simpler" one. These can both be described by the following procedure:
	\begin{enumerate}
		\item Beginning with a GHS, form an associated handle structure.
		\item Either (a) swap the order of attachment of a 2-handle that had followed a 1-handle, or (b) cancel a 1-handle and a 2-handle.
		\item Form the associated GHS of the new handle structure.
	\end{enumerate}
In Case (a) we say the new GHS is obtained from $H$ by a {\it weak reduction}. In Case (b) we say it was obtained by {\it destabilization}. There is a complexity such that in either case the new GHS is ``smaller" than the original. If there are no weak reductions or destabilizations for a GHS, then each of its thick levels are strongly irreducible. Scharlemann and Thompson showed that this implies that its thin levels must then be incompressible.

A {\it Sequence Of GHSs} (SOG) is defined to be any sequence where each element is obtained from its predecessor or successor by a weak reduction or destabilization. If an element $H$ of a SOG $\bf H$ is greater than both its predecessor and its successor then we say it is {\it maximal} in $\bf H$.

Just as there are two ways to take a GHS and produce a smaller one, there are ways one can ``reduce" a SOG. If a SOG can not be reduced, then each thick level of each maximal GHS is either strongly irreducible or critical and each thin level of each maximal GHS is incompressible.

\begin{figure}
\psfrag{H}{$H_X$}
\psfrag{h}{$G_X$}
\psfrag{G}{$H_Y$}
\psfrag{F}{$F$}
\psfrag{T}{$T$}
\psfrag{1}{$H _\phi$}
\psfrag{2}{$G_\phi$}
\psfrag{*}{$H_*$}
\[\includegraphics[width=5 in]{NSFInitialSOG.eps}\]
\caption{The initial SOG $\bf H$. Note that $H_X$ separates $F$ and $T$ in $X$, while $G_X$ does not.}
\label{f:NSFInitialSOG}
\end{figure}

We are now prepared to sketch the proof of Theorem \ref{t:TorusBoundaryCounterExample}. First, observe that the basic set-up of the hypotheses gives us the SOG $\Gamma$ of $X \cup _\phi Y$ pictured in Figure \ref{f:NSFInitialSOG}. If there is no way to reduce the SOG $\Gamma$ then it would follow that the minimal genus common stabilization $H_*$ of $H_\phi$ and $G_\phi$ is critical. Since no Heegaard surface in $X \cup _\phi Y$ can be isotoped to be disjoint from $F$, it now follows from Theorem \ref{t:barrier} that the genus of $H_*$ must be very high. Hence, if Theorem \ref{t:TorusBoundaryCounterExample} is false then there must be some way to reduce the SOG $\Gamma$.

We now assume $\Lambda$ is obtained from $\Gamma$ by a maximal sequence of complexity reducing moves. It follows from Theorem \ref{t:barrier} that the surface $F$ must be an element of $\thin{\Lambda^i}$ for each GHS $\Lambda^i \in \Lambda$.

\begin{figure}
\psfrag{F}{$F$}
\psfrag{T}{$T$}
\[\includegraphics[width=5 in]{BreakF.eps}\]
\caption{An SOG with $F$ as a unique thin level of each GHS can be ``broken" at $F$ to obtain SOGs of $X$ and $Y$.}
\label{f:BreakF}
\end{figure}

If $F$ is a unique element of $\thin{\Lambda^i}$, for each element $\Lambda^i \in \Lambda$, then the SOG $\Lambda$ can be ``broken" at $F$ to obtain separate SOGs of $X$ and $Y$. See Figure \ref{f:BreakF}. The first GHS of the SOG of $X$ thus obtained comes from weak reductions and destabilizations of the Heegaard surface $H_X$. As such it separates the surfaces $F$ and $T$. The final GHS of this SOG is descended from $G_X$, and therefore does not separate $F$ and $T$. We now have a contradiction, as there can be no SOG which interpolates between GHSs that partition the boundary of $X$ differently.

We conclude that there must be an element $\Lambda^*$ of $\Lambda$ for which the surface $F$ appears as two elements of $\thin{\Lambda^*}$. Hence, there is an element of $\thick{\Lambda^*}$ which lies between two copies of $F$, and is thus a Heegaard surface in a submanifold homeomorphic to $F \times I$. A classification Theorem of Scharlemann and Thompson \cite{st:93} tells us that this surface must be two copies of $F$ connected by a tube. (The classification includes one more possibility which is ruled out.)

Each GHS has a well-defined {\it genus}, which is the sum of the genera of its thick surfaces minus the sum of the genera of its thin surfaces. The above argument leads us to the inequality
\[genus(\Lambda^*) \ge genus(X)+genus(Y)\]
But, an analysis of the complexity reducing moves used to obtain $\Lambda$ now shows that there must be a GHS of $\Gamma$ whose genus is at least that of $\Lambda^*$. Finally, the genus of the surface $H_*$ is the largest genus of all the GHSs in $\Gamma$. Putting these inequalities together gives us $\gen(H_*) \ge \gen(X)+\gen(Y)$. But the genus of $G_\phi$ is $1+\gen(X)+\gen(Y)-\gen(F)$. The difference between these is the number of times we must stabilize $G_\phi$ in order to obtain a stabilization of $H_\phi$. This number is $\gen(F)-1$, which is also equal to $\gen(G_\phi)-4$.

config: arxiv


TITLE: If $\gamma : J \mapsto M$ is a smooth curve in a smooth manifold M, then $\gamma'(t) \neq 0$ $\forall t \in J$ iff $d\gamma$ is injective.
QUESTION [0 upvotes]: If $\gamma : J \mapsto M$ is a smooth curve in a smooth manifold M, then $\gamma'(t) \neq 0$ $\forall t \in J$ iff $d\gamma$ is injective.
Here $J$ is just an open interval of $\mathbb{R}$
I'm just reading through Lee's book on smooth manifolds and this question pops up on page 78. Any help would be much appreciated!
Note : I don't understand how this map can be injective since each tangent space $T_{t}J$ has a zero element which the liner map $d_{j}$ will therefore send to zero. So $dj$ will send many elementsof $TJ$ to zero?

REPLY [0 votes]: "$d\gamma$ is injective" just means here that pointwise, it is an injective (i.e. nonzero) linear map $T_t J \approx \mathbb{R} \rightarrow T_{\gamma(t)} M.$
The statement is then obvious (since $\gamma'(t) = d\gamma_{|t}(1)$ where $1 = \partial/\partial_t$ under the previous identification)


TITLE: If $p^k m^2$ is an odd perfect number, then what is the optimal constant $C$ such that $\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$?
QUESTION [2 upvotes]: The topic of odd perfect numbers likely needs no introduction.
The question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then what is the optimal constant $C$ such that $$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}?$$

(Note that the special prime $p$ satisfies $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.)
MY OWN PROOF FOR $C = 1$
It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$.  This implies that
$$\frac{\sigma(m^2)}{p^k} \neq m^2 - p^k$$
since $\sigma(m^2)/p^k$ is always odd.
Now, suppose to the contrary that
$$\sigma(m^2) > p^k m^2 - p^{2k}.$$
$$p^{2k} + \sigma(m^2) > p^k m^2$$
$$2p^{2k} + 2\sigma(m^2) > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 2\bigg)$$
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 2} \leq \frac{2p^{2k}}{p^k - 1}$$
where we have used the lower bound $\sigma(p^k) \geq p^k + 1$.
But from the paper Dris (2012), we have the lower bound
$$3p^k \leq \sigma(m^2)$$
while we also have the upper bound
$$\frac{2p^{2k}}{p^k - 1} = 2(p^k + 1) + \frac{2}{p^k - 1}.$$
This implies that
$$3p^k < 2(p^k + 1) + \frac{2}{p^k - 1}$$
$$p^k - 2 < \frac{2}{p^k - 1}$$
$$(p^k - 1)(p^k - 2) < 2.$$
This last inequality is a contradiction, as $p^k \geq 5$.
We therefore conclude that the inequality
$$\frac{\sigma(m^2)}{p^k} < \frac{m^2 - p^k}{C}$$
holds when $C=1$.

Does the inequality still hold when the constant $C > 1$?  If so, then what is the optimal value for $C$?

REPLY [1 votes]: (Last updated on March 2, 2021 - 5:04 PM Manila time)
Let $p^k m^2$ be an odd perfect number with special prime $p$.
Suppose to the contrary that
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{C}$$
and that $C=3$.
Following in the footsteps of this answer, we have
$$3\sigma(m^2) > p^k m^2 - p^{2k}$$
$$6\sigma(m^2) + 2p^{2k} > 2p^k m^2 = \sigma(p^k)\sigma(m^2)$$
$$2p^{2k} > \sigma(m^2)\bigg(\sigma(p^k) - 6\bigg)$$
Assuming $\sigma(p^k) \neq 6$ (that is, assuming it is not the case that $p=5$ and $k=1$ holds), then
$$\sigma(m^2) < \frac{2p^{2k}}{\sigma(p^k) - 6} \leq \frac{2p^{2k}}{p^k - 5} = \frac{2p^{2k} - 50}{p^k - 5} + \frac{50}{p^k - 5} = 2(p^k + 5) + \frac{50}{p^k - 5}.$$
But according to Broughan et al. (2013), we have the lower bound $315p^k \leq \sigma(m^2)$.
This implies that
$$315p^k < 2(p^k + 5) + \frac{50}{p^k - 5}$$
$$(313p^k - 10)(p^k - 5) < 50$$
which is clearly a contradiction if $p \neq 5$ or $k \neq 1$.
We therefore have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $\sigma(p^k) \neq 6$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{3}.$$

SECOND ATTEMPT
Using mathlove's approach in his/her answer below, together with Broughan et al. (2013)'s result, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 315$.
As in mathlove's answer, let the quantity $F$ be given by
$$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{315} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{315} := f_1(k).$$
Since $f_1(k)$ is increasing, we get
$$F \geq f_1(k) \geq f_1(1) = \frac{p+1}{2} - \frac{p}{315} \geq \frac{188}{63} \approx 2.98412698412698$$
whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{188}{63}}.$$

THIRD ATTEMPT
Using mathlove's approach in his/her answer below, together with mathlove's answer to a closely related question, we have the lower bound $\frac{\sigma(m^2)}{p^k} \geq 3^3 \times 5^3 = 3375$.
As in mathlove's answer, let the quantity $F$ be given by
$$F = \frac{\sigma(p^k)}{2} - \frac{p^{2k}}{\sigma(m^2)} \geq \frac{\sigma(p^k)}{2} - \frac{p^k}{3375} = \frac{p^{k+1} - 1}{2(p-1)} - \frac{p^k}{3375} := f_2(k).$$
Since $f_2(k)$ is increasing, we get
$$F \geq f_2(k) \geq f_2(1) = \frac{p+1}{2} - \frac{p}{3375} \geq \frac{2024}{675} = 2.99\overline{851}$$
whence we have the following proposition:

If $p^k m^2$ is an odd perfect number with special prime $p$, then
$$\frac{\sigma(m^2)}{p^k} \leq \frac{m^2 - p^k}{\frac{2024}{675}}.$$


text:
\begin{document}

\title[On the stability of the Schwartz class]{On the stability of the
  Schwartz class \\ under the magnetic Schr\"odinger flow}
\author[G. Boil]{G. Boil} \email{gregory.boil@univ-rennes1.fr}
\author[N. Raymond]{N. Raymond} \email{nicolas.raymond@univ-rennes1.fr
} \author[S. V\~u Ng\d{o}c]{S. V\~u Ng\d{o}c}
\email{san.vu-ngoc@univ-rennes1.fr}

\address[]{IRMAR, Universit\'e de Rennes 1, Campus de Beaulieu,
  F-35042 Rennes cedex, France}

\begin{abstract}
  We prove that the Schwartz class is stable under the magnetic
  Schr\"odinger flow when the magnetic $2$-form is non-degenerate and
  does not oscillate too much at infinity.
\end{abstract}

\maketitle

\section{Introduction}

\subsection{Motivation and context}
This paper is devoted to describing the solutions to the magnetic
Schr\"odinger equation. Let $\mathbf{B}$ be a smooth and closed
$2$-form on $\R^d$. Let $\mathbf{A} : \R^d\to\R^d$ be a $1$-form
(identified with a vector field) such that
$\dd \mathbf{A}=\mathbf{B}$. The magnetic Schr\"odinger operator is
the essentially self-adjoint differential operator
\[\mathscr{L}_{h}=(-ih\nabla-\mathbf{A})^2=\sum_{j=1}^d L^2_{j}\,,\]
where $h>0$ and, for all $j\in\{1,\ldots, j\}$,
$L_{j}=-ih\partial_{j}-A_{j}$. Its domain is given by
\[
\begin{split}
  \mathrm{Dom}(\mathscr{L}_{h}) &= \{\psi\in L^2(\R^d) :
  (-ih\nabla-\mathbf{A})\psi\in L^2(\R^d)\,,
  (-ih\nabla-\mathbf{A})^2\psi\in L^2(\R^d)\}\\
  &=\{\psi\in L^2(\R^d) : (-ih\nabla-\mathbf{A})^2\psi\in
  L^2(\R^d)\}\,.
\end{split}
\]
The time dependent magnetic Schr\"odinger equation is given by
\begin{equation}\label{eq.SE}
  -ih\partial_{t}\psi=\mathscr{L}_{h}\psi\,,\qquad
  \psi(0)=\psi_{0}\in\mathrm{Dom}(\mathscr{L}_{h})\,.
\end{equation}
By Stone's theorem, this Cauchy problem admits a unique solution,
evolving in the domain of $\mathscr{L}_{h}$, and it is given by
\[
\forall t\in\R\,,\quad
\psi(t)=e^{\frac{it\mathscr{L}_{h}}{h}}\psi_{0}\,.
\]
By unitarity of the flow, we have
\[
\forall t\in\R\,,\quad \|\psi(t)\|=\|\psi_{0}\|\,, \quad
\|\mathscr{L}_{h}\psi(t)\|=\|\mathscr{L}_{h}\psi_{0}\|\,,
\]
where $\|\cdot\|$ denotes the usual norm on $L^2(\R^d)$. This norm
controls the rough phase space localization of the quantum state
$\psi(t)$; a natural question is to know to which extent a strong
phase space localization of $\psi_0$ is preserved by the flow. More
precisely, this paper was inspired by the following rather naive
question. Is it true that
\begin{equation}\label{eq.question}
  \psi_{0}\in\mathscr{S}(\R^d)\quad\Longrightarrow\quad \forall t\in\R\,,\quad\psi(t)\in\mathscr{S}(\R^d) \quad?
\end{equation}
If so, what kind of explicit control do we have in terms of the
Schwartz semi-norms?

These questions are motivated by the recent investigation of the
propagation of coherent states by the magnetic Hamiltonian flow in two
dimensions (see the Ph.~D. thesis of the first
author~\cite{theseBoil}). The present paper gives a positive answer to
\eqref{eq.question}. Our explicit estimates of the Schwartz semi-norms
(in terms of the semiclassical parameter $h$), combined with the use
of the Birkhoff normal form from \cite{RVN15}, turn out to be the key
ingredients in the study by~\cite{BVN18} of the propagation of
coherent states up to times of order $h^{-N}$, for all
$N\in\mathbb{N}$. This gives a quantum analog to the low energy (say
of order $\varepsilon$) classical propagation for times of order
$\varepsilon^{-\infty}$ (see \cite[Theorem 1.2]{RVN15}). Taking into
account the analysis of \cite{HKRVN16}, one can even hope to extend
these results to three dimensions where the classical dynamics has a
more complex behavior.

Independently of this motivation, the answer to \eqref{eq.question}
has an interest of its own, especially because it lives at the
confluence of two closely related domains: hypoellipticity and
semiclassical analysis with magnetic fields. On these vast subjects,
the literature is enormous, and we only refer to \cite{Hormander67,
  HN, HM88, HN92, T02, T04, FH10, R17}. In this paper, we will use
many classical ideas from these two contexts, and provide an
elementary and self-contained presentation.

\subsection{Main results}
Let us now describe our assumptions and results.

Let $\mathscr{P}$ be the class defined by
\[\mathscr{P}=\{\psi\in\mathscr{C}^\infty(\R^d) :
\forall\alpha\in\mathbb{N}^d\,,\exists
(C,m)\in\R_{+}\times\mathbb{R}_{+}\,,\forall x\in\R^d\,,
|\partial^\alpha\psi|\leq C\langle x\rangle^{m} \}\,.\]

The following assumption will hold throughout the paper, where we
identify $\mathbf{B}$ with its antisymmetric matrix obtained in the
usual basis $(\dd x_j\wedge \dd x_k; j<k)$.
\begin{assumption}\label{hyp.0}
  We assume that
  \begin{enumerate}[\rm i.]
  \item $\mathbf{A}$ belongs to $\mathscr{P}$ (in particular $\mathbf{B}\in\mathscr{P}$),
  \item there exists $b_{0}>0$ such that, for all $x\in\R^d$,
    \[\mathrm{Tr}^+\mathbf{B}(x)\geq b_{0}\,,\]
    where $\mathrm{Tr}^+\mathbf{B}(x)$ denotes the sum of the moduli
    of the eigenvalues with positive imaginary part of the matrix
    $\mathbf{B}(x)$,
  \item for all $\alpha\in\mathbb{N}^d$, there exists $C>0$ such that,
    for all $x\in\R^d$,
    $\|\partial^\alpha\mathbf{B}(x)\|\leq C\|\mathbf{B}(x)\|$, where
    $\|\cdot\|$ denotes a norm on the space of matrices.
  \end{enumerate}
\end{assumption}
Assumption \ref{hyp.0} is stronger than really necessary as we can see
in our proofs. In this context, we will use the following lemma (see
\cite[Theorem 2.2]{HM96}).
\begin{lemma}\label{lem.minmag}
  We have
  \[\inf\mathrm{sp}(\mathscr{L}_{h})=h\inf_{x\in\R^d}\mathrm{Tr}^+\mathbf{B}(x)+o(h)\,.\]
  In particular, there exist $C>0$ and $h_{0}>0$, such that, for all
  $h\in(0,h_{0})$, $\mathscr{L}_{h}$ is invertible and
  \[\|\mathscr{L}_{h}^{-1}\|\leq Ch^{-1}\,.\]
\end{lemma}
In the following, we will always assume that $h$ is small enough and
such that $\mathscr{L}_{h}$ is invertible.

\begin{definition}
  For all $n\in\mathbb{N}$, we let
  \[\mathrm{Dom}(\mathscr{L}^n_{h})=\{\psi\in L^2(\R^d) :
  \forall\ell\in\{1,\ldots, n\} :
  \mathscr{L}_{h}^{\ell-1}\psi\in\mathrm{Dom}(\mathscr{L}_{h})\}\,.\]
  The operator $\mathscr{L}^n_{h}$ is defined by induction by
  \[\forall \psi\in\mathrm{Dom}(\mathscr{L}^n_{h})\,,\quad
  \mathscr{L}^n_{h}\psi=\mathscr{L}_{h}(\mathscr{L}^{n-1}_{h}\psi)\,.\]
\end{definition}
The operator $(\mathrm{Dom}(\mathscr{L}^n_{h}),\mathscr{L}^n_{h})$ is
self-adjoint and invertible. The following theorem proves some
magnetic elliptic estimates, showing that iterations of the magnetic
laplacian $\mathscr{L}_h$ control iterations of the magnetic
derivatives $(L_j)_{1\leq j \leq d}$. This will be an important tool
on the proof of the main result of the paper.
\begin{theorem}\label{thm.elliptic}
  Let Assumption~\ref{hyp.0} hold. Let $n\in\mathbb{N}$. There
  exist $h_{0}>0$ and $C>0$ such that, for all $h\in(0,h_{0})$, and
  all $\psi\in\mathrm{Dom}(\mathscr{L}^n_{h})$,
  \begin{equation}\label{eq.elliptic}
    \sum_{\sigma\in\mathfrak{A}(2n)} \|L_{\sigma}\psi\|\leq Ch^{-3n/2}\|\mathscr{L}_{h}^n\psi\|\,,
  \end{equation}
  where, for $k\in\N$,
  $\mathfrak{A}(k)=\cup_{p=0} ^k \{1,\ldots,d\}^{\{1,\ldots,p\}}$, and
  for $p\in\N$, for $\sigma\in\{1,\ldots,d\}^{\{1,\ldots,p\}}$,
  $L_{\sigma} := L_{\sigma(1)}\ldots L_{\sigma(p)}$, with the convention $L_{\emptyset}=\mathrm{Id}$.
\end{theorem}
In the case where $\mathbf{A}$ is
bounded, Theorem \ref{eq.elliptic} is closely related to \cite[Theorem
3]{T02}, which deals with the context of general G\aa rding
inequalities.

\begin{definition}
  For all $k\in\mathbb{N}$ and all $\psi\in\mathscr{S}(\R^d)$, we let
  \[p_{k}(\psi)=\underset{\underset{ |\alpha|+|\beta|\leq
      k}{(\alpha,\beta)\in\mathbb{N}^{2d}}}{\max}\|x^\alpha \partial^\beta
  \psi\|_{\infty}\,.\]
\end{definition}

We can now state the main result of this paper.
\begin{theorem}\label{thm.evol}
  Let Assumption~\ref{hyp.0} hold.  For all $t\in\R$, we have
  \[e^{\frac{it\mathscr{L}_{h}}{h}}\mathscr{S}(\R^d)\subset
  \mathscr{S}(\R^d)\,.\]
  More precisely, for all $M\in\mathbb{N}^*$, for all
  $k\in\mathbb{N}$, there exist $h_{0}>0$, $C>0$, $N\in\mathbb{N}^*$
  and $K\in\mathbb{N}$, such that, for all $h\in(0,h_{0})$, and for
  all $\psi_{0}\in \mathscr{S}(\R^d)$, and all $t\in[0,h^{-M}]$,
  \[p_{k}(e^{\frac{it\mathscr{L}_{h}}{h}}\psi_{0})\leq
  Ch^{-N}p_{K}(\psi_{0})\,.\]
\end{theorem}
Theorem \ref{thm.evol} is related to the (pseudo-differential)
analysis in \cite[Section 7]{T04}. In this work, under the assumption
that the derivatives of order two or higher of the symbol of the
propagator should be bounded, a parametrix of the evolution operator
was constructed. Closely related is also the paper~\cite[Corollary
2.11]{Rob07}, based on the analysis of coherent states, where the
derivatives of order three or higher of the symbol have to be
bounded. Our approach here is more directly related to the structure
of the magnetic Laplacian, and is reminiscent of the analysis
in~\cite{Hormander67} of the ellipticity of certain algebras of
non-commuting vector fields.

\subsection{Organisation of the proofs}
In Section \ref{sec.elliptic}, we prove Theorem \ref{thm.elliptic} by
using a regularization argument involving exponentially weighted
estimates and commutator estimates. In Section \ref{sec.evol}, we
apply our magnetic elliptic estimates to prove Theorem \ref{thm.evol}.

\newpage
\section{Magnetic elliptic estimates}\label{sec.elliptic}
This section is devoted to the proof of Theorem \ref{thm.elliptic}.

\subsection{Density argument}\label{sec.density}
Let us explain here why it is sufficient to prove Theorem
\ref{thm.elliptic} for $\psi\in\mathscr{S}(\R^d)$.

Let us consider $f\in L^2(\R^d)$. There is a unique
$\psi\in\mathrm{Dom}(\mathscr{L}^n_{h})$ such that
$\mathscr{L}^n_{h}\psi=f$. Consider
$f_{k}\in\mathscr{C}^\infty_{0}(\R^d)$ converging to $f$ in
$L^2(\R^d)$ and consider the unique
$\psi_{k}\in\mathrm{Dom}(\mathscr{L}^n_{h})$ such that
$\mathscr{L}^n_{h}\psi_{k}=f_{k}$. Note that, by continuity of
$(\mathscr{L}^n_{h})^{-1}$, $\psi_{k}$ converges to $\psi$ in
$L^2(\R^d)$.

\begin{lemma}\label{lem.agmon}
  We let
  \[H^\infty_{\exp}(\R^d)=\{\psi\in L^2(\R^d) : \forall
  \alpha\in\mathbb{N}^d\,,\exists \beta>0 : e^{\beta\langle
    x\rangle}\partial^\alpha\psi\in L^2(\R^d)\}\,.\]
  Consider $f\in H_{\exp}^\infty(\R^d)$. Then, the unique solution
  $u\in\mathrm{Dom}(\mathscr{L}_{h})$ to $\mathscr{L}_{h}u=f$
  satisfies $u\in H^\infty_{\exp}(\R^d)$.
\end{lemma}

\begin{proof}
  The proof follows from the classical Agmon estimates (see
  \cite{Agmon85, H88}). Consider $\beta>0$ such that
  $e^{\beta\langle x\rangle}f\in L^2(\R^d)$. Let $\varepsilon>0$ and
  $\Phi_\varepsilon=\beta\min(\langle x\rangle,\varepsilon^{-1})$.  We
  have the Agmon formula (see \cite[Section 4.2]{R17}):
  \[\langle\mathscr{L}_{h}u,
  e^{2\Phi_\varepsilon}u\rangle=\int_{\R^d}|(-ih\nabla-\mathbf{A})e^{\Phi_\varepsilon}u|^2
  \dd x-h^2\|\nabla\Phi_\varepsilon e^{\Phi_\varepsilon}u\|^2\,.\]
  In particular,
  \begin{equation}\label{eq.agmonineg}
    \int_{\R^d}|(-ih\nabla-\mathbf{A})e^{\Phi_\varepsilon}u|^2\dd x-\beta^2h^2\|e^{\Phi_\varepsilon}u\|^2\leq \|e^{\Phi_\varepsilon}f\|\|e^{\Phi_\varepsilon}u\|\,.
  \end{equation}
  On the other hand, by Lemma \ref{lem.minmag}, we get, for some
  $c>0$,
  \[(hc-\beta^2h^2)\|e^{\Phi_\varepsilon}u\|^2\leq\int_{\R^d}|(-ih\nabla-\mathbf{A})e^{\Phi_\varepsilon}u|^2\dd
  x-\beta^2h^2\|e^{\Phi_\varepsilon}u\|^2\,.\]
  Choosing $\beta$ small enough, we get, for some $C(h)>0$ independent
  of $\varepsilon$,
  \[\|e^{\Phi_\varepsilon}u\|\leq C(h)\|e^{\Phi_\varepsilon}f\|\,.\]
  Then, we take the limit $\varepsilon\to 0$ and apply Fatou's Lemma
  to find
  \begin{equation}\label{eq.AgmonL2}
  \|e^{\beta\langle x\rangle}u\|\leq C(h)\|e^{\beta\langle
    x\rangle}f\|\,.
 \end{equation}
  Coming back to \eqref{eq.agmonineg}, and replacing $\beta$ by
  $\tilde\beta<\beta$, we get
  \[\int_{\R^d}|e^{\Phi_\varepsilon}(-ih\nabla-\mathbf{A}-ih\nabla\Phi_{\varepsilon})u|^2\dd x-\tilde\beta^2h^2\|e^{\Phi_\varepsilon}u\|^2\leq \|e^{\Phi_\varepsilon}f\|\|e^{\Phi_\varepsilon}u\|
\,.\]
Thus,
  \[\begin{split}
  \frac{h^2}{2}\int_{\R^d}|e^{\Phi_\varepsilon}\nabla u|^2\dd x &\leq \|e^{\Phi_\varepsilon}f\|\|e^{\Phi_\varepsilon}u\|+\tilde\beta^2h^2\|e^{\Phi_\varepsilon}u\|^2+2\|(ih\nabla\Phi_{\varepsilon}+\mathbf{A})e^{\Phi_{\varepsilon}}u\|^2\\
&\leq  \|e^{\Phi_\varepsilon}f\|\|e^{\Phi_\varepsilon}u\|+\tilde\beta^2h^2\|e^{\Phi_\varepsilon}u\|^2+4\|\nabla\Phi_{\varepsilon}e^{\Phi_{\varepsilon}}u\|^2+4\|\mathbf{A}e^{\Phi_{\varepsilon}}u\|^2\,.
\end{split}
 \]
 Using that $\mathbf{A}\in\mathscr{P}$, \eqref{eq.AgmonL2}, and Fatou's Lemma, we get $e^{\tilde\beta\langle x\rangle}\nabla u\in L^2(\R^d)$.

Considering the equation $\mathscr{L}_{h}u=f$, we get, in the sense of tempered distributions,
\begin{equation}\label{eq.Deltau}
-h^2\Delta u=f-|\mathbf{A}|^2u-ih(\nabla\cdot\mathbf{A})u-2ih(\mathbf{A}\cdot\nabla) u\,.
\end{equation}

Noticing that we have just controlled the terms of order at most one, we deduce that, for
  some $\beta>0$,
  \[e^{\beta\langle x\rangle}\Delta u\in L^2(\R^d)\,,\]
and also
  \[
  \Delta \left(e^{\beta\langle x\rangle} u\right)\in L^2(\R^d)\,.
  \]
  By Fourier transform, we get
  \[e^{\beta\langle x\rangle}u\in H^2(\R^d)\,,\]
  which implies that, for all $\alpha\in\mathbb{N}^d$ with
  $|\alpha|\leq 2$,
  \[e^{\beta\langle x\rangle}\partial^\alpha u\in L^2(\R^d)\,.\]
  The higher order derivatives can be controlled by induction (taking
  successive derivatives of \eqref{eq.Deltau}).
\end{proof}

\begin{remark}
  By the Sobolev embeddings, we have
  $H^\infty_{\exp}(\R^d)\subset\mathscr{S}(\R^d)$.
\end{remark}

By Lemma \ref{lem.agmon}, we see that
$\psi_{k}\in\mathscr{S}(\R^d)$. Assume that \eqref{eq.elliptic} holds
for any $\psi \in \mathscr{S}(\R^d)$, with
$\psi=\psi_{k}-\psi_{\ell}$, for all $(k,\ell)\in\mathbb{N}^2$, this
shows that, for all $\sigma\in\mathfrak{A}(2n)$,
$(L_{\sigma}\psi_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in
$L^2(\R^d)$. Thus, in the sense of distributions,
$L_{\sigma}\psi\in L^2(\R^d)$. It remains to use again
\eqref{eq.elliptic} with $\psi=\psi_{k}$ and to take the limit
$k\to+\infty$.

\subsection{Preliminary lemmas}

\begin{lemma}\label{lem.prelim}
  For all $\psi\in\mathrm{Dom}(\mathscr{L}_{h})$, we have
  \[\|(-ih\nabla-\mathbf{A})\psi\|^2\leq\frac{1}{2}\|\psi\|^2_{\mathscr{L}_{h}}\]
  where
  $\|\psi\|^2_{\mathscr{L}_{h}}= \|\psi\|^2+ \|\mathscr{L}_h
  \psi\|^2$.
\end{lemma}
\begin{proof}
  We recall that, by definition of the domain, for all
  $\psi\in\mathrm{Dom}(\mathscr{L}_{h})$,
  \[\langle\mathscr{L}_{h}\psi,\psi\rangle=\|(-ih\nabla-\mathbf{A})\psi\|^2=\sum_{j=1}^d\|L_{j}\psi\|^2\,,\]
  so that
  \[\sum_{j=1}^d\|L_{j}\psi\|^2\leq
  \|\psi\|\|\mathscr{L}_{h}\psi\|\,.\]
\end{proof}

\begin{lemma}\label{lem.ell-B}
  There exist $C>0$ and $h_{0}>0$ such that, for all
  $\psi\in\mathscr{S}(\R^d)$ and all $h\in(0,h_{0})$,
  \begin{equation}\label{eq.B}
    \|\langle\mathbf{B}\rangle\psi\|+\|\langle\mathbf{B}\rangle^{\frac{1}{2}}(-ih\nabla-\mathbf{A})\psi\|\leq Ch^{-1}\|\psi\|_{\mathscr{L}_{h}}\,.
  \end{equation}
  Moreover, for all $\psi\in\mathrm{Dom}(\mathscr{L}_{h})$, we have
  \[\mathbf{B}\psi\in L^2(\R^d)\quad \mbox{ and }\quad
  |\mathbf{B}|^{\frac{1}{2}}(-ih\nabla-\mathbf{A})\psi\in
  L^2(\R^d)\,,\]
  and \eqref{eq.B} holds for $\psi\in\mathrm{Dom}(\mathscr{L}_{h})$.

\end{lemma}
\begin{proof}
  By integration by parts and using Assumption \ref{hyp.0},
  \begin{equation}\label{eq.1}
    \begin{split}
      \int_{\R^d}\langle\mathbf{B}\rangle|(-ih\nabla-\mathbf{A})\psi|^2\dd x&=\langle \langle\mathbf{B}\rangle(-ih\nabla-\mathbf{A})\psi,(-ih\nabla-\mathbf{A})\psi\rangle\\
      &\leq\langle \mathscr{L}_{h}\psi,\langle\mathbf{B}\rangle\psi\rangle+C\|\langle\mathbf{B}\rangle\psi\|\|(-ih\nabla-\mathbf{A})\psi\|\\
      &\leq
      C\|\langle\mathbf{B}\rangle\psi\|\|\psi\|_{\mathscr{L}_{h}}\,.
    \end{split}
  \end{equation}
  Then, we have
  \[\int_{\R^d}|h\mathbf{B}|^2|\psi|^2\dd
  x=\sum_{(k,\ell)\in\{1,\ldots,d\}^2}
  \int_{\R^d}|hB_{k,\ell}|^2|\psi|^2\dd x\] and we write
  \begin{multline*}
    \sum_{(k,\ell)\in\{1,\ldots,d\}^2}\int_{\R^d}|hB_{k,\ell}|^2|\psi|^2\dd x=\sum_{(k,\ell)\in\{1,\ldots,d\}^2}|\langle[L_{k},L_{\ell}]\psi, hB_{k,\ell}\psi\rangle|\\
    \leq
    Ch\|\langle\mathbf{B}\rangle\psi\|\|(-ih\nabla-\mathbf{A})\psi\|+Ch\int_{\R^d}\langle\mathbf{B}\rangle|(-ih\nabla-\mathbf{A})\psi|^2\dd
    x\,,
  \end{multline*}
  where we used an integration by parts and Assumption \ref{hyp.0}.

  By \eqref{eq.1}, it follows
  \[\int_{\R^d}|\mathbf{B}|^2|\psi|^2\dd x\leq
  Ch^{-1}\|\langle\mathbf{B}\rangle\psi\|\|\psi\|_{\mathscr{L}_{h}}\]
  and then
  \[\|\langle\mathbf{B}\rangle\psi\|\leq
  Ch^{-1}\|\psi\|_{\mathscr{L}_{h}}\,.\]
  Using again \eqref{eq.1}, the conclusion follows.
\end{proof}

\subsection{Case $n=1$}
The estimate of Theorem \ref{thm.elliptic} is obvious when $n=0$. Let us consider the case when $n=1$ to explain the principle producing these estimates.

\begin{lemma}\label{lem.ordre2sch}
  There exist $C>0$, $h_{0}>0$ such that, for all
  $\psi\in\mathscr{S}(\R^d)$ and all $h\in(0,h_{0})$,
  \[\|L^2_{1}\psi\|+\|L^2_{2}\psi\|+\|L_{1}L_{2}\psi\|+\|L_{2}L_{1}\psi\|\leq
  Ch^{-1}\|\mathscr{L}_{h}\psi\|\,.\]
\end{lemma}

\begin{proof}
  Let us consider $\psi\in\mathscr{S}(\R^d)$ and let
  \[\mathscr{L}_{h}\psi=f\,.\]
  Consider $j\in\{1,\ldots, d\}$. We have
  \[\mathscr{L}_{h}(L_{j}\psi)=L_{j}f+[\mathscr{L}_{h},
  L_{j}]\psi\,,\] and
  \[\|(-ih\nabla-\mathbf{A})(L_{j}\psi)\|^2=\langle L_{j}f,
  L_{j}\psi\rangle+\langle[\mathscr{L}_{h}, L_{j}]\psi,
  L_{j}\psi\rangle\,,\] We have
  \[[\mathscr{L}_{h}, L_{j}]=\sum_{k=1}^d
  [L^2_{k},L_{j}]=\sum_{k=1}^d\left( [L_{k},L_{j}]L_{k}+L_{k}[L_{k},
    L_{j}]\right)\,.\] Thus,
  \[|\langle[\mathscr{L}_{h}, L_{j}]\psi, L_{j}\psi\rangle|\leq
  C\|\mathbf{B}\psi\|\|(-ih\nabla-\mathbf{A})\psi\|+C\int_{\R^d}
  |\mathbf{B}||(-ih\nabla-\mathbf{A})\psi| ^2\dd x\,.\]
  We have, for all $\varepsilon\in(0,1)$,
  \[|\langle L_{j}f, L_{j}\psi\rangle|\leq
  \varepsilon\|L_{j}^2\psi\|^2+C\|f\|^2\,,\] and then
  \begin{multline*}
    \|(-ih\nabla-\mathbf{A})(L_{j}\psi)\|^2\\
    \leq
    C\|f\|^2+C\|\mathbf{B}\psi\|\|(-ih\nabla-\mathbf{A})\psi\|+C\int_{\R^d}
    |\mathbf{B}||(-ih\nabla-\mathbf{A})\psi| ^2\dd x\,.
  \end{multline*}
  With Lemma \ref{lem.prelim}, noting that
  $\|\psi\|_{\mathscr{L}_h} ^2 \leq C (1+ h^{-2})\|f\|^2$, we find
  \[\|(-ih\nabla-\mathbf{A})(L_{j}\psi)\|^2\leq Ch^{-3}\|f\|^2\,.\]
\end{proof}

\subsection{Induction}
Let $n\in\mathbb{N}^*$. Let us assume that, for all
$k\in\{1,\ldots, n\}$, the ellipticity property \eqref{eq.elliptic} is
true.  Let us consider $f,\psi\in\mathscr{S}(\R^d)$ such that
\[\mathscr{L}_{h}\psi=f\,.\]
Consider $\sigma\in\mathfrak{A}(2n)$. Since the functions are in the
Schwartz class, all the following computations are justified.

We have
\[\mathscr{L}_{h}L_{\sigma}\psi=L_{\sigma}f+[\mathscr{L}_{h},
L_{\sigma}]\psi\,,\] and then
\begin{equation}\label{eq.Lalpha}
  \langle\mathscr{L}_{h}L_{\sigma}\psi,L_{\sigma}\psi\rangle=\langle L_{\sigma}f, L_{\sigma}\psi\rangle+\langle[\mathscr{L}_{h}, L_{\sigma}]\psi, L_{\sigma}\psi\rangle\,.
\end{equation}
By using the Cauchy-Schwarz inequality and the induction assumption,
we have for all $\varepsilon\in(0,1)$,
\[|\langle L_{\sigma}f, L_{\sigma}\psi\rangle|\leq
Ch^{-3n}\|\mathscr{L}^n_{h}f\|\|\mathscr{L}^n_{h}\psi\|\leq
Ch^{-3n}\|\mathscr{L}^{n+1}_{h}\psi\|^2+Ch^{-3n}\|\mathscr{L}^{n}_{h}\psi\|^2\,,\]
so that, by Lemma \ref{lem.minmag},
\[|\langle L_{\sigma}f, L_{\sigma}\psi\rangle|\leq
Ch^{-(3n+2)}\|\mathscr{L}^{n+1}_{h}\psi\|^2\,.\]
Let us now deal with
$\langle[\mathscr{L}_{h}, L_{\sigma}]\psi, L_{\sigma}\psi\rangle$. The
commutator $[\mathscr{L}_{h}, L_{\sigma}]$ is the sum of various
terms. Each of them is the composition of at most $2n-2$ of the
$L_{j}$ and with exactly one of the $B_{k,\ell}$. By commuting the
$B_{k,\ell}$ to put it on the left, and using Assumption \ref{hyp.0},
we get
\[\|[\mathscr{L}_{h}, L_{\sigma}]\psi\|\leq
C\sum_{\tau\in\mathfrak{A}(2n-2)}\|\langle\mathbf{B}\rangle
L_{\tau}\psi\|\,.\]
By applying Lemma \ref{lem.ell-B}, and then the induction assumption,
we get
\[\|[\mathscr{L}_{h}, L_{\sigma}]\psi\|\leq
Ch^{-1}\sum_{\tau\in\mathfrak{A}(2n-2)}\|
L_{\tau}\psi\|_{\mathscr{L}_{h}}\leq
Ch^{-(3n-1)/2}\|\mathscr{L}^n_{h}\psi\|\,.\] Thus, we deduce
\[|\langle\mathscr{L}_{h}L_{\sigma}\psi,L_{\sigma}\psi\rangle|\leq
Ch^{-3n-1/2}\|\mathscr{L}^{n+1}_{h}\psi\|^2\,.\]
This shows that, for all $\gamma\in\mathfrak{A}(2n+1)$,
\begin{equation}\label{eq.2n+1}
  \|L_{\gamma}\psi\|\leq Ch^{-(3n/2+1)}\|\mathscr{L}^{n+1}_{h}\psi\|\,.
\end{equation}
Now, we want to get the control for $\gamma \in
\mathfrak{A}(2n+2)$.
Let $\sigma \in \mathfrak{A}(2n+1)$. We consider again
\eqref{eq.Lalpha}. By integration by parts, we can write
\[\langle L_{\sigma}f, L_{\sigma}\psi\rangle=\langle
L_{\check\sigma}f, L_{\hat\sigma}\psi\rangle\,,\]
with $\check\sigma \in \mathfrak{A}(2n)$ and
$\hat\sigma\in\mathfrak{A}(2n+2)$. Thus, by Cauchy-Schwarz, and the
induction assumption, for all $\varepsilon>0$, there exists $C>0$ such
that
\begin{equation}\label{eq.RHS1}
  |\langle L_{\sigma}f, L_{\sigma}\psi\rangle|\leq\varepsilon\|L_{\hat\sigma}\psi\|^2+Ch^{-3n/2}\|\mathscr{L}^{n+1}_{h}\psi\|^2\,.
\end{equation}
As previously, we have
\[\|[\mathscr{L}_{h}, L_{\sigma}]\psi\|\leq
C\sum_{\tau\in\mathfrak{A}(2n-1)}\|\langle\mathbf{B}\rangle
L_{\tau}\psi\|\,.\]
We use Lemma \ref{lem.ell-B} and \eqref{eq.2n+1} to find
\[\|[\mathscr{L}_{h}, L_{\sigma}]\psi\|\leq
Ch^{-1}\sum_{\tau\in\mathfrak{A}(2n-1)}\|
L_{\tau}\psi\|_{\mathscr{L}_{h}}\leq
Ch^{-(3n/2+2)}\|\mathscr{L}^{n+1}_{h}\psi\|\,.\]
Then, with Cauchy-Schwarz and \eqref{eq.2n+1}, we get
\begin{equation}\label{eq.RHS2}
  |\langle[\mathscr{L}_{h}, L_{\sigma}]\psi, L_{\sigma}\psi\rangle|\leq Ch^{-(3n+3)}\|\mathscr{L}^{n+1}_{h}\psi\|^2\,.
\end{equation}
From \eqref{eq.Lalpha}, \eqref{eq.RHS1}, and \eqref{eq.RHS2}, summing
over $\sigma \in \mathfrak{A}(2n+1)$, and choosing $\varepsilon$ small
enough, we get \eqref{eq.elliptic} with $n$ replaced by $n+1$.

This achieves the proof of Theorem \ref{thm.elliptic} when
$\psi\in\mathscr{S}(\R^d)$ and it remains to use the discussion of
Section \ref{sec.density}.

\section{Application to the evolution problem}\label{sec.evol}
We can now prove Theorem \ref{thm.evol}. Let
$\psi_{0}\in\mathscr{S}(\R^d)$. We denote by $\psi(\cdot)$ the
solution to the Schr\"odinger equation \eqref{eq.SE}.

\begin{notation}
For $\kappa, \lambda \in \N$, we define $\Pi_{\kappa,\lambda}$
the set of the operators $P$ that are composition of operators taken
among $(L_j)_{1\leq j\leq d}$ and and $(x_k)_{1\leq k\leq d}$ with
$\kappa$ occurences of $x$ and $\lambda$ occurences of $L$. We also set $\displaystyle{\Pi=\bigcup_{(\kappa,\lambda)\in\N^2}\Pi_{\kappa,\lambda}}$.
\end{notation}

The aim of this
section is to prove the following proposition.
\begin{proposition}\label{prop.P}
  Let $(\kappa,\lambda)\in\mathbb{N}^2$ and consider $P\in\Pi_{\kappa,\lambda}$ . There exist
  $h_{0}>0$, $C\geq 0$ and $N\in\mathbb{N}$ such that, for all
  $t\geq 0$, and all $\psi_{0}\in\mathscr{S}(\R^d)$, for
  $\lambda \leq 2n \leq \lambda+1$,
  \[\|P\psi(t)\|\leq C h^{-N}\left(1+ t^\kappa\right) \sum_{|\alpha|
    \leq \kappa : |\alpha| + \nu \leq \kappa + n} \|\mathscr{L}_h ^\nu
  x^\alpha \psi_0\|\,.\]
\end{proposition}
This proposition implies the control of the Schwartz semi-norms and
achieves the proof of Theorem \ref{thm.evol}. Indeed, from Sobolev embeddings, for all $k\in\N$, there exists $K \in \N$ such that for all $ f \in \mathscr{S}(\R^d)$,
\[p_k(f) \leq \max_{|\alpha|, |\beta| \leq K}\ \|
x^\alpha \partial^\beta f\|\,.\]
Using now that $\partial_j = (-ih)^{-1} (L_j +A_j)$, and
$A \in \mathscr{P}$, there exists $C>0$ and $N\in\mathbb{N}$ such
that, for all $f\in L^2(\R^d)$, if $\norm{P f}<+\infty$ for all
$P\in \Pi$, then $x^\alpha\partial^\beta f\in L^2(\R^d)$ for all
$\alpha,\beta\in \N^d$, and
\[\|x^\alpha \partial^\beta f \| \leq Ch ^{-N} \sum_{P \in R} \|P f\|\,,\]
where $R$ is a finite part of $\Pi$. Then, Proposition~\ref{prop.P} implies, for $N$, $\kappa$ and $n$ large enough,
\[\|x^\alpha \partial^\beta \psi(t)\| \leq Ch^{-N} \sum_{P\in R} \left\|P \psi(t)\right\| \leq Ch^{-N} (1+t^\kappa) \sum_{|\gamma| \leq \kappa, |\nu|\leq n} \|\mathscr{L}_h ^\nu x^\gamma \psi_0\|\,.\]
Finally, we conclude by
\[\|\mathscr{L}_h ^\nu x^\alpha \psi_0\| \leq C \|(1+x^2)^m\mathscr{L}_h ^\nu x^\alpha \psi_0\|_\infty \leq C p_K(\psi_0)\,,\]
for $m$, $K$ large enough.

\subsection{Case when $\kappa=0$}
For all $t\geq 0$, we have, by definition,
\[\psi(t)=e^{\frac{it}{h}\mathscr{L}_{h}}\psi_{0}\,.\]
For all $\ell\in\mathbb{N}$, we get
\[\mathscr{L}_{h}^\ell\psi(t)=e^{\frac{it}{h}\mathscr{L}_{h}}\mathscr{L}^\ell_{h}\psi_{0}\,.\]
Applying Theorem \ref{thm.elliptic}, this establishes the estimate of
Proposition \ref{prop.P} when $\kappa=0$.

\subsection{Case when $\kappa=1$}
Before starting the induction procedure, let us understand first the
mechanism with only one occurence of $x$. Let $j\in\{1,\ldots,
d\}$. We have
\[-ih\partial_{t}(x_{j}\psi)=\mathscr{L}_{h}(x_{j}\psi)+[x_{j},\mathscr{L}_{h}]\psi\,.\]
Note that $[x_{j},\mathscr{L}_{h}]=2hL_{j}$. With the Duhamel formula,
we have, for all $t\geq 0$,
\begin{equation}\label{eq.duh}
  x_{j}\psi(t)=e^{\frac{it}{h}\mathscr{L}_{h}}(x_{j}\psi_{0})+\int_{0}^t e^{\frac{i(t-s)}{h}\mathscr{L}_{h}}2L_{j}\psi(s)\dd s\,.
\end{equation}
With Lemma \ref{lem.prelim}, we get
\[\|x_{j}\psi(t)\|\leq
\|x_{j}\psi_{0}\|+2\int_{0}^t\|L_{j}\psi(s)\|\dd s\leq
\|x_{j}\psi_{0}\|+\sqrt{2}\int_{0}^t\|\psi(s)\|_{\mathscr{L}_{h}}\dd s
\,.\] Since the evolution is unitary, we get, for all $t\geq 0$,
\[\|x_{j}\psi(t)\|\leq
\|x_{j}\psi_{0}\|+2\int_{0}^t\|L_{j}\psi(s)\|\dd s\leq
\|x_{j}\psi_{0}\|+t\sqrt{2}\|\psi_{0}\|_{\mathscr{L}_{h}} \,.\]
More generally, with \eqref{eq.duh}, we have, for all
$\ell\in\mathbb{N}$,
\[\mathscr{L}^\ell_{h}x_{j}\psi(t)=e^{\frac{it}{h}\mathscr{L}_{h}}\mathscr{L}^{\ell}_{h}(x_{j}\psi_{0})+\int_{0}^t
e^{\frac{i(t-s)}{h}\mathscr{L}_{h}}2\mathscr{L}^{\ell}_{h}L_{j}\psi(s)\dd
s\,.\] so that
\[\begin{split}
  \|\mathscr{L}^\ell_{h}x_{j}\psi(t)\|&\leq\|\mathscr{L}^{\ell}_{h}(x_{j}\psi_{0})\|+2\int_{0}^t \|\mathscr{L}^{\ell}_{h}L_{j}\psi(s)\|\dd s\\
  &\leq \|\mathscr{L}^{\ell}_{h}(x_{j}\psi_{0})\|+Ch^{-N}\int_{0}^t\|\mathscr{L}_{h}^{\ell+1}\psi(s)\|\dd s\\
  &\leq\|\mathscr{L}^{\ell}_{h}(x_{j}\psi_{0})\|+Cth^{-N}\|\mathscr{L}_{h}^{\ell+1}\psi_{0}\|\,.
\end{split}
\]
It remains to apply Theorem \ref{thm.elliptic} and to commute the
$x_{j}$ with the $L_{k}$.

\subsection{Induction}
Let us now end the proof of Proposition \ref{prop.P} by induction. We set the following two induction assumptions. For $\kappa\in\N$, let
\begin{multline*}
  \mathscr{Q}_\kappa : \forall n\in\N, \, \forall \alpha\in\N^d, \, |\alpha| =\kappa, \, \exists N \in \N \ \text{s.t.} \\
  \|\mathscr{L}_h ^n x^\alpha \psi(t)\| \leq \|\mathscr{L}_h ^n
  x^\alpha \psi_0\| +C \sum_{|\beta| \leq \kappa-1 : |\beta| + \nu
    \leq \kappa + n}\|\mathscr{L}_h ^\nu x^\beta \psi_0\| h^{-N}
  (1+t^\kappa)\,;
\end{multline*}
and
\begin{equation*}
  \mathscr{P}_\kappa : \forall P \in \Pi_{\kappa,\lambda}, \ \|P \psi(t)\| \leq C \sum_{|\alpha| \leq \kappa : |\alpha| + \nu \leq \kappa +n} \|\mathscr{L}_h ^\nu x^\alpha \psi_0\| h^{-N} (1+t^\kappa), \ \text{for} \, \lambda \leq 2n \leq \lambda +1\,.
\end{equation*}
We have proved propositions $\mathscr{P}_0$, $\mathscr{Q}_0$ and
$\mathscr{Q}_1$. We assume now that for a given $\kappa\in \N^*$, for
any $k\leq\kappa$, $\mathscr{Q}_k$ and $\mathscr{P}_k$ hold, and we
prove $\mathscr{P}_{\kappa+1}$ and $\mathscr{Q}_{\kappa +1}$. We begin
with $\mathscr{Q}_{\kappa+1}$. Let $\alpha\in\N^d$, with
$|\alpha| = \kappa +1$ and $n\in\N$. We have
\[-ih\partial_t \big(\mathscr{L}_h ^n x^\alpha \psi(t)\big) =
\mathscr{L}_h ^{n+1} x^\alpha \psi(t) + \mathscr{L}_h ^n [x^\alpha,
\mathscr{L}_h] \psi(t)\,.\] Then, noting that
\[[x^\alpha, \mathscr{L}_h] = h P_1\]
with $P_1$ a sum of elements in $\Pi_{\kappa,1}$, we get from the
Duhamel formula
\begin{equation*}
  \mathscr{L}_h ^n x^\alpha \psi(t) = e^{\frac{it}{h}\mathscr{L}_h} \mathscr{L}_h ^n x^\alpha \psi_0 + i \int_0 ^t e^{i\frac{t-s}{h} \mathscr{L}_h} \mathscr{L}_h ^n P_1 \psi(s) \mathrm{d}s\,.
\end{equation*}
As $\mathscr{L}_h ^n P_1$ is a sum of elements in
$\Pi_{\kappa,(2n+1)}$, we can apply $\mathscr{P}_\kappa$, and
integrating in time and using the unitariness of
$e^{i\frac{t}{h}\mathscr{L}_h}$, we find, for some integer $N \in \N$
\begin{equation*}
  \|\mathscr{L}_h ^n x^\alpha \psi(t)\| \leq \|\mathscr{L}_h ^n x^\alpha \psi_0\| + C h^{-N} \sum_{|\beta| \leq \kappa : |\beta| + \nu \leq \kappa + n+1} \|\mathscr{L}_h ^{\nu} x^\beta \psi_0\| (1+t^{\kappa+1})
\end{equation*}
that proves $\mathscr{Q}_{\kappa + 1}$. It remains to prove
$\mathscr{P}_{\kappa+1}$. We consider so some
$P \in \Pi_{\kappa+1,\lambda}$, for a given $\lambda \in \N$. Then,
because of the commutation relation of the $(x_k)_{1\leq k \leq d}$
with the $(L_j)_{1 \leq j \leq d}$, that is
$[x_k,L_j] = -ih\delta_{k,j}$, there is $\alpha \in \N^d$,
$|\alpha| = \kappa+1$ such that
\[P = P_2 + P_3 x^\alpha\]
with $P_2$ a sum of elements in $\Pi_{\kappa,\lambda-1}$ and
$P_3 \in \Pi_{0,\lambda}$. So
\[\|P \psi(t)\| \leq \|P_2 \psi(t)\| + \|P_3 \psi(t)\|\,.\]
Then, applying $\mathscr{P}_\kappa$, we get, for some integer $N\in\N$
\begin{equation}\label{eq.P2}
  \|P_2 \psi(t)\| \leq C h^{-N} (1+t^\kappa) \sum_{|\beta|\leq \kappa : |\beta|+ \nu \leq \kappa + n} \|\mathscr{L}_h ^\nu x^\beta \psi_0\|
\end{equation}
and applying Theorem~\ref{thm.elliptic} along with
$\mathscr{Q}_{\kappa+1}$, for some integer $N\in\N$, and for
$\lambda \leq 2n \leq \lambda+1$, we have
\begin{equation}\label{eq.P3}
  \begin{split}
    \|P_3 x^\alpha \psi(t)\| &\leq C h^{-n}\|\mathscr{L}^n x^\alpha \psi(t)\| \\
    &\leq C h^{-N} (1+t^{\kappa+1}) \left( \|\mathscr{L}^n x^\alpha
      \psi_0\| + \sum_{|\beta| \leq \kappa : |\beta| + \nu \leq \kappa
        + n +1} \|\mathscr{L}_h ^{\nu} x^\beta \psi_0\| \right).
  \end{split}
\end{equation}
Now, gathering \eqref{eq.P2} and \eqref{eq.P3}, we find, for some
integer $N$,
\[\|P \psi(t)\| \leq C h^{-N} (1+t^{\kappa+1}) \sum_{|\beta| \leq
  \kappa +1: |\beta| + \nu \leq \kappa+n+1} \|\mathscr{L}_h ^\nu
x^\beta \psi_0\|\]
that proves $\mathscr{P}_{\kappa+1}$ and achieves the proof of
Proposition~\ref{prop.P}.

\subsection*{Acknowledgments}
The authors are grateful to Karel Pravda-Starov for sharing useful
references.


text:
\begin{definition}[Definition:Property]
A '''property''' is a concept that specifies an aspect of an [[Definition:Object|object]].
In the phrase ''living thing'', ''living'' is a property that ''thing'' has. It creates a distinction between ''things'' that are ''living'' and ''things'' that are ''not living''.
\end{definition}

config: wiki


TITLE: Must every right-inverse of a linear transformation be a linear transformation?
QUESTION [4 upvotes]: Let T be a linear transformation $\mathbb{R}^2 \rightarrow \mathbb{R}^3$. Let S be the right-inverse of T. Does S have to be linear transformation?
Thanks in Advance.

REPLY [0 votes]: If $T:X\to Y$ is injective but not surjective, consider a left-inverse $S$, i.e. a map $Y\to X$ such that $$S(T(x)) = x\text{ for all }x\in X.$$ Note that this equation only says something about how $S$ acts on the range (image) of $T$. For the points of $Y$ that are outside the range of $T$, we are free to prescribe absolutely any behavior for $S$.
Similarly, if $T:X\to Y$ is instead surjective but not injective, we consider a right-inverse $S$, $$T(S(y)) = y\text{ for all }y\in Y.$$ This means that for a $y\in Y$, the map $S$ just picks one of the inverse images of $y$ under $T$. Every time we have a $y$ that is "hit" by more than one $x$ under $T$, we have the freedom to choose any of them as our value $S(y)$.
The above is true in general (in the category of arbitrary sets and maps). If we consider the case where $X$ and $Y$ are vector spaces (or more concretely $X=\mathbb{R}^n$ and $Y=\mathbb{R}^m$), and where $T$ is a linear map, it is clear (I think; left to the reader) that our freedom to mingle with $S$ concerns full subspaces (or complements of subspaces) and we can pick a completely "wild" behavior of $S$. In particular, we can ensure that $S$ is not a linear map.


TITLE: Find a closed form for the sum with harmonic number $\sum_{n=0}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}$
QUESTION [3 upvotes]: In this link gave an answer for a question which invoved the sum of generalized harmonic numbers with power $2n+1$.
And I thought about this sum:
$$S=\sum_{n=1}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}$$
I tried to search in our website but still can't find some similiar questions.
The "generalized" version can be found in Ali Shadhar's book.
$$S_1=\sum_{n=1}^\infty (-1)^{n}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=-\beta{(4)}+\frac{35\pi}{64}\zeta{(3)}-\frac{\pi^2}{48}G$$
Can someone give me a hint for this sum? Every help is welcomed, thank you very much.

REPLY [4 votes]: In the book you mentioned in page 68, we have
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}x^{n}=\operatorname{Li}_4(x)-2\operatorname{Li}_4(1-x)+2\ln(1-x)\operatorname{Li}_3(1-x)+\frac12\operatorname{Li}_2^2(x)$$
$$-\ln^2(1-x)\operatorname{Li}_2(1-x)-\frac13\ln(x)\ln^3(1-x)+2\zeta(4).
$$
Using
$$\sum_{n=0}^\infty (-1)^{n}a_{2n+1}=\mathfrak{J}\sum_{n=1}^\infty i^n a_n$$
we have
$$\sum_{n=0}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}=\mathfrak{J}\sum_{n=1}^\infty i^{n}\frac{H_n^{2}}{n^2}$$
$$=2\zeta(4)+\mathfrak{J}\{\operatorname{Li}_4(i)-2\operatorname{Li}_4(1-i)+2\ln(1-i)\operatorname{Li}_3(1-i)+\frac12\operatorname{Li}_2^2(i)$$
$$-\ln^2(1-i)\operatorname{Li}_2(1-i)-\frac13\ln(i)\ln^3(1-i)\}$$


text:
\section{Introduction}

We often encounter the following optimization problem in machine learning, signal processing, and stochastic control:
\begin{align}\label{eqn-obj-1}
\min_{X}f(X)\quad\textrm{subject to}\quad X\in \Omega,
\end{align}
where $f:\RR^d\rightarrow \RR$ is a loss function, $\Omega:\triangleq \{X\in \RR^d:g_i(X)= 0, i=1,2,...,m\}$ denotes a feasible set, $m$ is the number of constraints, and $g_i:\RR^{d}\rightarrow \RR$'s are the differentiable functions that impose constraints into model parameters. For notational simplicity, we define $\cG(X)=[g_1(X),...,g_m(X)]^\top$ and $\Omega= \{ X\in\RR^d: \cG(X)=0\}$. Principal component analysis (PCA), canonical correlation analysis (CCA), matrix factorization/sensing/completion, phase retrieval, and many other problems \citep{friedman2001elements,sun2016geometric,bhojanapalli2016global,li2016symmetry,ge2016matrix,chen2017online,zhu2017global} can be viewed as special examples of \eqref{eqn-obj-1}. Many algorithms have been proposed to solve \eqref{eqn-obj-1}. For the unconstrained ($\Omega=\RR^d$) or a simple constraint $\cG(X)$, e.g., the spherical constraint, $\cG(X):=\norm{X}_2-1$, we can apply simple first order algorithms such as the projected gradient descent algorithm \citep{luenberger1984linear}.

However, when $\cG(X)$ is complicated, the aforementioned algorithms are often not applicable or inefficient. This is because the projection to $\Omega$ does not admit a closed form expression and can be computationally expensive in each iteration. To address this issue, we convert \eqref{eqn-obj-1} to a min-max problem using the Lagrangian multiplier method. Specifically, instead of solving \eqref{eqn-obj-1}, we solve the following problem:
\begin{align}\label{eqn-Lag-1}
\min_{X\in \RR^d} \max_{Y \in \RR^m}~ \cL(X,Y):= f(X)+Y^\top \cG(X),
\end{align}
where $Y\in\RR^m$ is the Lagrangian multiplier. $\cL(X,Y)$ is often referred as the Lagrangian function in existing literature \citep{boyd2004convex}. The existing literature on optimization also refers to $X$ as the primal variable and $Y$ as the dual variable. Accordingly, \eqref{eqn-obj-1} is called the primal problem. From the perspective of game theory, they can be viewed as two players competing with each other and eventually achieving some equilibrium. When $f(X)$ is convex and $\Omega$ is convex or the boundary of a convex set, the optimization landscape of \eqref{eqn-Lag-1} is essentially convex-concave, that is, for any fixed $Y$, $\cL(X,Y)$ is convex in $X$, and for any fixed $X$, $\cL(X,Y)$ is concave in $Y$. Such a landscape further implies that the equilibrium of \eqref{eqn-Lag-1} is a saddle point, whose primal variable is equivalent to the global optimum of \eqref{eqn-obj-1} under strong duality conditions. To solve \eqref{eqn-Lag-1}, we resort to primal-dual algorithms, which iterate over both $X$ and $Y$ (usually in an alternating manner). The global convergence rates to the equilibrium are also established accordingly for these algorithms \citep{lan2011primal,chen2014optimal,iouditski2014primal}.

When $f(X)$ and $\Omega$ are nonconvex, both \eqref{eqn-obj-1} and \eqref{eqn-Lag-1} become much more computationally challenging, NP-Hard in general. Significant progress has been made toward solving the primal problem \eqref{eqn-obj-1}. For example, \cite{ge2015escaping} show that when certain tensor factorization satisfies the so-called strict saddle properties, one can apply some first order algorithms such as the projected gradient algorithm, and the global convergence in polynomial time can be guaranteed. Their results further motivate many follow-up works, proving that many problems can be formulated as strict saddle optimization problems, including PCA, multiview learning, phase retrieval, matrix factorization/sensing/completion, complete dictionary learning \citep{sun2016geometric,bhojanapalli2016global,li2016symmetry,ge2016matrix,chen2017online,zhu2017global}. Note that these strict saddle optimization problems are either unconstrained or just with a simple spherical constraint. However, for many other nonconvex optimization problems, $\Omega$ can be much more complicated. To the best of our knowledge, when $\Omega$ is not only nonconvex but also complicated, the applicable algorithms and convergence guarantees are still largely unknown in existing literature.

To handle the complicated $\Omega$, this paper proposes to investigate the min-max problem \eqref{eqn-Lag-1}. Specifically, we first define a special class of Lagrangian functions, where the landscape of  $\cL(X,Y)$ enjoys the following good properties:
\begin{itemize}[leftmargin=*]
\item {\it There exist only two types of equilibria -- stable and unstable equilibria. At an unstable equilibrium, $\cL(X,Y)$ has negative curvature with respect to the primal variable $X$. More details in Section~\ref{sec-equi}.}
\item {\it All stable equilibria correspond to the global optima of the primal problem \eqref{eqn-obj-1}}.
\end{itemize}
Both properties are intuitive. On the one hand, the negative curvature in the first property enables the primal variable to escape from the unstable equilibria along some decent direction.
On the other hand, the second property ensures that we do not get spurious local optima of \eqref{eqn-obj-1}, that is all local minima must also be global optima.

We then study a generalized eigenvalue (GEV) problem, which includes CCA, Fisher discriminant analysis (FDA, \cite{mika1999fisher}),  sufficient dimension reduction (SDR, \cite{cook2005sufficient}) as special examples. Specifically, GEV solves
 \begin{align}
	X^*=\argmin_{X \in \RR^{d\times r}}~f(X):={-\tr(X^\top A X)}~~\textrm{s.t.}~~X \in\cT_{B}:=  \{ X\in \RR^{d\times r} : X^\top B X=I_r \}, \label{eqn:original}
\end{align}
where $A,B \in \RR^{d \times d}$ are symmetric, $B$ is positive semidefinite. We rewrite \eqref{eqn:original} as a min-max problem,
\begin{align}
\min_X\max_Y\cL(X,Y)=-\tr( X^\top A X) + \langle Y,X^\top B X-I_r \rangle,\label{eqn:minimax}
\end{align}
 where $Y\in \RR^{r\times r}$ is the Lagrangian multiplier. Theoretically, we show that the Lagrangian function in \eqref{eqn:minimax} exactly belongs to our previously defined class. Motivated by our defined landscape structures, we then solve an online version of \eqref{eqn:minimax}, where we can only access independent unbiased stochastic approximations of $A,~B$ and directly accessing $A$ and $B$ is prohibited. Specifically, at the $k$-th iteration, we only obtain independent $A^{(k)}$ and $B^{(k)}$ satisfying
 \begin{align*}
 \EE A^{(k)} =A\quad\textrm{and}\quad\EE B^{(k)} = B.
 \end{align*}
Computationally, we propose a simple stochastic primal-dual algorithm, which is a stochastic variant of the generalized Hebbian algorithm (GHA, \cite{gorrell2006generalized}). Theoretically, we establish its asymptotic rate of convergence to stable equilibria for our stochastic GHA (SGHA) based on the diffusion approximations~\citep{kushner2003stochastic}. Specifically, we show that, asymptotically, the solution trajectory of SGHA weakly converges to the solutions of stochastic differential equations (SDEs). By studying the analytical solutions of these SDEs, we further establish the asymptotic sample/iteration complexity of SGHA under certain regularity conditions~\citep{harold1997stochastic,li2016near,chen2017online}. To the best of our knowledge, this is the first asymptotic sample/iteration complexity analysis of a stochastic optimization algorithm for solving the online version of GEV problem. Numerical experiments are presented to justify our theory.

Our work is closely related to several recent results on solving GEV problems. For example, \cite{ge2016efficient} propose a multistage semi-stochastic optimization algorithm for solving GEV problems with a finite sum structure. At each optimization stage, their algorithm needs to access the exact $B$ matrix, and compute the approximate inverse of $B$ by solving a quadratic program, which is not allowed in our setting. Similar matrix inversion approaches are also adopted by a few other recently proposed algorithms for solving GEV problem \citep{allen2016doubly,arora2017stochastic}. In contrast, our proposed SGHA is a fully stochastic algorithm, which does not require any matrix inversion.

Moreover, our work is also related to several more complicated min-max problems, such as Markov Decision Process with function approximation, Generative Adversarial Network, multistage stochastic programming and control~\citep{sutton2000policy,shapiro2009lectures,goodfellow2014generative}. Many primal-dual algorithms have been proposed to solve these problems. However, most of these algorithms are even not guaranteed to converge. As mentioned earlier, when the convex-concave structure is missing, the min-max problems go far beyond the existing theories. Moreover, both primal and dual iterations involve sophisticated stochastic approximations (equally or more difficult than our online version of GEV). This paper makes the attempt on understanding the optimization landscape of these challenging min-max problems. Taking our results as an initial start, we expect more sophisticated and stronger follow-up works that apply to these min-max problems.

\noindent \textbf{Notations}.  Given an integer $d$, we denote  $I_d$ as a $d \times d$ identity matrix, $[d] = \{1,2,\ldots,d \}$. Given an index set $\cI\subseteq [d]$ and
 a matrix $X \in \RR^{d \times r}$, we denote $\Ic = [d] \backslash \cI$ as the complement set of $\cI$, $X_{:,i}$ ($X_{i,:}$) as the $i$-th column (row) of $X$, $X_{i,j}$ as the $(i,j)$-th entry of $X$, and $X_{:,\cI}$ ($X_{\cI,:}$) as the column (row) submatrix of $X$ indexed by $\cI$, $\vect(X) \in \RR^{dr}$ as the vectorization of $X$, $\textrm{Col}(X)$ as the column space of $X$, and $\textrm{Null}(X)$ as the null space of $X$. Given a symmetric matrix $X \in \RR^{d \times d}$, we denote $\lambda_{\min\slash\max}(X)$ as its smallest\slash largest singular value, and denote the eigenvalue decomposition of $X$ as $X = O \Lambda O^{\top}$, where $\Lambda=\diag(\lambda_1,...\lambda_d)$ with $\lambda_1\geq\cdots\geq\lambda_{d}$,  denote $\norm{X}_2$ as the spectral norm of $X$. Given two matrices $X$ and $Y$, $X \otimes Y$ as the Kronecker product of $X$, $Y$.


TITLE: Extension, restriction, and coextension of scalars adjunctions in the case of noncommutative rings?
QUESTION [6 upvotes]: If we have a ring homomorphism $f: R \to S$, then we can put an $(R,S)$ or $(S, R)$ bimodule structure on $S$ and define the extension $f_!:= S \otimes_R (-) : R \text{Mod} \to S \text{Mod}$, restriction $f_*: S \text{Mod} \to R \text{Mod}$ and coextension $f_*:= \text{Hom}_{R}(S, -) : R \text{Mod} \to S \text{Mod}$ of scalars functors. These constructions do not depend at all on the commutativity of the rings $R$ and $S$, but in every source I've seen that mentions extension and coextension of scalars, it is assumed that the rings are commutative, such as in the nLab article.
Is there any reason that the rings are assumed to be commutative? Does the adjunction still hold in the noncommutative case? Are there any properties of these constructions that are "less nice" in the noncommutative case? Or is it simply for convenience?
Most of my algebra knowledge is self-taught, so I apologize if I'm missing something obvious.

REPLY [4 votes]: As mentioned by Qiaochu Yuan in the comments, one nice thing about the case where $f: C \to K$ is a morphism of commutative rings is that scalar extension $f_!$ is strong monoidal.
In particular, for $f: C \to K$ a morphism of commutative rings,
\begin{align}
f_!(V) \otimes_K f_!(W) =& (K \otimes_C V) \otimes_K (K \otimes_C W)\cong (V \otimes_C K) \otimes_K (K \otimes_C W) \\ \cong& V \otimes_C K \otimes_C W
\cong K \otimes_C V \otimes_C W = f_!(V \otimes_C W).
\end{align}
This is given as Proposition 3 in section II.5.1 of Bourbaki's Algebra I.
As a corollary, it follows that restriction of scalars is lax monoidal.