text:
\begin{document}

\onehalfspace

\title{Induced Matchings in Graphs of Bounded Maximum Degree}

\author{Felix Joos\thanks{Institut f\"{u}r Optimierung und Operations Research,
Universit\"{a}t Ulm, Ulm, Germany,
e-mail: \texttt{felix.joos@uni-ulm.de}}
}

\date{}

\maketitle

\vspace{-0.5cm}

\begin{abstract}
For a graph $G$, let $\nu_s(G)$ be the induced matching number of $G$.
We  prove that
$\nu_s(G) \geq \frac{n(G)}{(\lceil\frac{\Delta}{2}\rceil+1) (\lfloor\frac{\Delta}{2}\rfloor+1)}$
for every graph of sufficiently large maximum degree $\Delta$	 and without isolated vertices.
This bound is sharp.
Moreover, there is polynomial-time algorithm which computes induced matchings of size as stated above.
\end{abstract}

{\small \textbf{Keywords:}  induced matching; strong matching; strong chromatic index}\\
\indent {\small \textbf{AMS subject classification:}
05C70,
05C15
}

\section{Introduction}

For a graph $G$, a set $M$ of edges is an \emph{induced matching} of $G$ if
no two edges in $M$ have a common endvertex and no edge of $G$ joins two edges of $M$.

The maximum number of edges that form an induced matching in $G$ is the {\it strong matching number $\nu_s(G)$ of $G$}.
We denote by $\Delta(G)$ the maximum degree of graph $G$
and let $n(G)=|V(G)|$ and $m(G)=|E(G)|$.

In contrast to the well known matching number $\nu(G)$, which can be computed in polynomial time \cite{ed},
it is NP-hard to determine the strong matching number even in bipartite subcubic graphs \cite{ca1,lo,stva}.

In fact, the strong matching number is even hard to approximate in restricted graphs classes as for example regular bipartite graphs \cite{dadelo}.

To the best of my knowledge,
the only known bound in terms of the order and the maximum degree
for $\nu_s(G)$ is obtained by the following simple observation \cite{zito}.
Let $G$ be a graph without isolated vertices.
There are at most $2\Delta(G)^2-2\Delta(G)+1$ many edges in distance at most $1$ from $e$ including $e$
and $m(G)\geq \frac{1}{2} n(G)$.
Thus a simple greedy algorithm implies
\begin{align*}
	\nu_s(G)\geq \frac{ n(G)}{2(2\Delta(G)^2-2\Delta(G)+1)},
\end{align*}
which is far away from being sharp if $G\not=K_2$.

It seems that the different behavior of $\nu(G)$ and $\nu_s(G)$ transfers to the corresponding partitioning problems.
The chromatic index $\chi'$ seems much simpler than the strong chromatic index $\chi_s'$,
defined as the minimum number of induced matchings one needs to partition the edge set.
While for $\chi'(G)$ Vizing's Theorem always gives $\chi'(G)\in\{\Delta(G),\Delta(G)+1\}$ \cite{vi},
no comparable result holds for the strong chromatic index.

A trivial greedy algorithm ensures $\chi_s'(G)\leq 2\Delta(G)^2-2\Delta(G)+1$.
Erd{\H{o}}s and Ne{\v{s}}et{\v{r}}il \cite{fashgytu2} conjectured $\chi_s'(G)\leq \frac{5}{4}\Delta(G)^2$,
which would  be best possible for even $\Delta$
because
equality holds for the graph
obtained from the $5$-cycle by replacing every vertex by an independent set of order $\frac{\Delta}{2}$.
The best general result in this direction is due to Molloy and Reed,
who proved that $\chi_s'(G)\leq 1.998\Delta(G)^2$ for sufficiently large maximum degree \cite{more}.
Thus Erd{\H{o}}s and Ne{\v{s}}et{\v{r}}il's conjecture is widely open and it is even unknown which technique is suitable to improve Molloy and Reed's result substantially.

In this paper I provide more insight concerning the behavior of induced matchings by improving the known lower bounds on $\nu_s(G)$
to a sharp lower bound
provided that the maximum degree is sufficiently large.

\begin{theorem}\label{result max degree}
There is an integer $\Delta_0$ such that for every graph $G$ of maximum degree $\Delta$ at least $\Delta_0$ and without isolated vertices,
\begin{align*}
\nu_s(G) \geq \frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}
\end{align*}
holds.
\end{theorem}
\noindent
The following construction shows that the bound in Theorem \ref{result max degree} is sharp.
Let $\Delta$ be an integer at least $3$ and
let the graph $H_1$ arise from the complete graph on $\left\lceil \frac{\Delta}{2} \right \rceil+1$ vertices by
attaching at each vertex $\left \lfloor \frac{\Delta}{2} \right\rfloor$ pendant vertices.
Let $H_2$ arise from the complete graph on $\left\lfloor \frac{\Delta}{2} \right \rfloor+1$ vertices by
attaching at each vertex $\left \lceil \frac{\Delta}{2} \right\rceil$ pendant vertices.
It follows that $\nu_s(H_i)=	1$ and $n(H_i)= \left(\left\lceil \frac{\Delta}{2} \right \rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)$; that is, the bound of Theorem \ref{result max degree} is sharp.
Note that $H_1=H_2$ if $\Delta$ is even.

For the sake of simplicity I do not try to optimize the constant $\Delta_0$ intensively.
We show  Theorem  \ref{result max degree} for $\Delta_0=1000$ but
with some more effort one can lower the bound down to $200$.

In \cite{jorasa} the same bound as in Theorem \ref{result max degree} is already shown by a simple inductive argument for graphs of girth at least $6$.
Hence one might ask whether the bound in Theorem \ref{result max degree} can be improved for graphs of large girth to $\frac{n(G)}{\Delta^c}$ for some $c<2$.
However, this is not the case.
By a result of Bollob{\'a}s \cite{bo},
for every $g \geq 3$ and $\Delta \geq 6$, there is a graph $H'$
of maximum degree $\lfloor\frac{\Delta}{2}\rfloor$, girth at least $g$, and independence number at most $\frac{4 \log \Delta}{\Delta}n(H')$.
Let $H$ arise from $H'$ by attaching to each vertex $\lceil\frac{\Delta}{2}\rceil$ many pendant vertices.
Note that $\nu_s(H)\leq \frac{4 \log \Delta}{\Delta}n(H')$ and $n(H)= \lceil\frac{\Delta}{2}\rceil n(H')$.
Thus $\nu_s(H) \leq \frac{8 \log \Delta}{\Delta^2}n(H)$ and the bound of Theorem \ref{result max degree} can only be improved by a $O(\log \Delta)$-factor.

Since the proof of Theorem \ref{result max degree} is constructive,
it is easy to derive a polynomial-time algorithm,
which computes an induced matching of size as guaranteed in Theorem \ref{result max degree}.

We use standard notation and terminology.
For a graph $G$, let $V(G)$ and $E(G)$ denote its vertex set and edge set, respectively.
For a vertex $v$, let $d_G(v)$ be its degree, let $N_G(v)$ be the set of neighbors of $v$, and
let $N_G[v]=N_G(v)\cup \{v\}$.
If the corresponding graph is clear from the context, we only write $d(v)$, $N(v)$ and $N[v]$, respectively.
A set $I$ of vertices of $G$ is independent if there is no edge joining two vertices in $I$.

\section{Proof of Theorem \ref{result max degree}}
We prove the theorem for $\Delta_0=1000$.
Let $G$ be a graph with maximum degree $\Delta$ at least $\Delta_0$ and without isolated vertices.
For a contradiction, we assume that $G$ is a counterexample such that
\begin{enumerate}[(1)]
	\item $\nu_s(G)$ is minimum and
	\item subject to (1), the order of $G$ is maximum.
\end{enumerate}
Since $\nu_s(G)\geq \frac{n(G)}{2\Delta^2}$, the graph $G$ is well-defined.

The choice of $G$ implies that if $v$ is a vertex of $G$ that is adjacent to a vertex of degree~$1$,
then $d(v)=\Delta$ because adding new vertices to $G$ and joining them to $v$ does not increase $\nu_s(G)$ but the order of $G$.

For some calculations it might help to know that $\frac{\Delta^2}{4}+\Delta+\frac{3}{4}
\leq \left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)$.

\begin{claim}\label{d1}
For every edge $uv$ of $G$, we have $d(u)+d(v) > \frac{\Delta}{4}$.
\end{claim}
\begin{proof}[Proof of Claim \ref{d1}]
For a contradiction, we assume that there is an edge $uv$ such that
$d(u)+d(v) \leq \frac{\Delta}{4}$.
Let $S=N[u]\cup N[v]$ and
let $I$ be the set of all isolated vertices of $G-S$.
Let $G'=G-S-I$.
Since $\nu_s(G)\geq \nu_s(G')+1$,
the choice of $G$ implies $\nu_s(G') \geq \frac{n(G')}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$.

By using the assumption $d(u)+d(v) \leq \frac{\Delta}{4}$,
we conclude $|S|+|I|\leq \left(\frac{\Delta}{4}-2\right)\Delta+2< \left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)$.
Therefore, $uv$ together with a maximum induced matching of $G'$ is an induced matching of $G$
of size at least $\frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$,
which contradicts the choice of $G$.
\end{proof}

\begin{claim}\label{d2}
Every vertex $v$ of $G$ is adjacent to at most $\frac{3}{4}\Delta$ many vertices of degree at most $9$.
\end{claim}
\begin{proof}[Proof of Claim \ref{d2}]
Choose $v$ such that the number of neighbors of degree at most $9$ is maximal.
Say $v$ has $\alpha\Delta$ many such neighbors.
For a contradiction, we assume that $\alpha>\frac{3}{4}\Delta$.
Let $u\in N(v)$ be of degree at most $9$.
As above, let $S=N[u]\cup N[v]$ and

let $I$ be the set of all isolated vertices of $G-S$.
Let $G'=G-S-I$.

By Claim \ref{d1}, every vertex in $I$ that is adjacent to a vertex of degree at most $9$, has degree at least $10$.
Thus there are at most $(1-\alpha)\Delta+8$ many vertices in $S$ that are adjacent to vertices in $I$ of degree at most $9$.
Hence there are at most $\alpha(1-\alpha)\Delta^2+8\Delta$ many vertices in $I$ of degree at most $9$.
Furthermore, at most $8\alpha \Delta$ edges join vertices in $I$ and vertices in $N(v)\setminus \{u\}$ such that the vertices in $N(v)\setminus \{u\}$ have degree at most $9$.
Since $\alpha(1-\alpha)+ \frac{1}{10}(1-\alpha)^2< 0.22$, this implies
\begin{align*}
	|I|&\leq \alpha(1-\alpha)\Delta^2+8\Delta + \frac{1}{10}\left((1-\alpha)^2\Delta^2 +8\alpha\Delta \right)\\
	&< 0.22\Delta^2+9\Delta\\
	&\leq \frac{\Delta^2}{4}-9.
\end{align*}
Since $|S|\leq \Delta + 9$,
we obtain
$$|I|+|S|< \left(\left\lceil\frac{\Delta}{2}\right\rceil+1\right) \left(\left\lfloor\frac{\Delta}{2}\right\rfloor+1\right). $$
Again, the edge $uv$ together with a maximum induced matching of $G'$ is an induced matching of $G$
of size at least $\frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$,
which contradicts the choice of $G$.
\end{proof}

\noindent
Let $f: V(G)\rightarrow \mathbb{R}$ be such that
\begin{align*}
	f(v)= \sum_{w\in N(v):\ d(w)\not= \Delta}\frac{1}{d(w)}.
\end{align*}

\begin{claim}\label{d3}
If a vertex $v$ of $G$ is not adjacent to a vertex of degree $1$,
then $f(v)\leq \frac{2}{5}\Delta$.
\end{claim}
\begin{proof}[Proof of Claim \ref{d3}]
Let $v$ be a vertex that is not adjacent to a vertex of degree $1$.
By Claim \ref{d2}, the vertex $v$ has at most $\frac{3}{4}\Delta$ neighbors of degree at most $9$, which contribute to $f(v)$ at most $\frac{1}{2}$ each;
all remaining neighbors contribute at most $\frac{1}{10}$ each.
Thus $f(v)\leq \frac{1}{2}\cdot\frac{3}{4}\Delta+ \frac{1}{10}\cdot\frac{1}{4}\Delta = \frac{2}{5}\Delta$.
\end{proof}
\noindent
For the rest of the proof,
let $v\in V(G)$ be chosen such that $f(v)$ is maximal.

\begin{case}
$v$ is adjacent to a vertex of degree $1$.
\end{case}
Recall that this implies $d(v)=\Delta$.
Let $u\in N(v)$ be a vertex of degree~$1$.
As before, we want to combine $uv$ with a maximum induced matching of $G'=G-(N[v]\cup I)$ to obtain a contradiction, where $I$ are the isolated vertices of $G-N[v]$.

If $z\in I$ has degree $d<\Delta$, then $z$ contributes exactly $d$ times exactly $\frac{1}{d}$ to $f(w)$ for some $w\in N(v)$;
that is, the total contribution to $\sum_{w\in N(v)}f(w)$ is $1$.
Since no vertex in $I$ is adjacent to $u$,
there is no vertex $z\in I$ such that $d(z)=\Delta$.
This implies that
\begin{align}\label{sum f}
	|I|\leq \sum_{w\in N(v)}f(w).
\end{align}

Let $N_1$ and $N_{\Delta}$ be the set of vertices in $N(v)$ of degree
$1$ and $\Delta$, respectively.
Let $N_s$ be the set of vertices in $N(v)\setminus (N_1 \cup N_\Delta)$ of small degree, say such that their degree is between $2$ and $\frac{\Delta}{8}$.
Let $N_\ell=N(v)\setminus(N_1\cup N_s \cup N_\Delta)$, and let $n_1=|N_1|$, $n_s=|N_s|$, $n_\ell=|N_\ell|$, and $n_\Delta=|N_\Delta|$.

Since all vertices in $N_s\cup N_\ell$ do not have degree $\Delta$ and by the choice of $G$,
they are not adjacent to a vertex of degree $1$.
If $w \in N_1$, then $f(w)=0$ and $w$ contributes $1$ to $f(v)$.
If $w \in N_s$, then by Claim \ref{d1}, we conclude $f(w)\leq 1$, and the contribution of $w$ to $f(v)$ is at most $\frac{1}{2}$.
If $w \in N_\ell$, then by Claim \ref{d3} and the choice of $v$, we obtain $f(w)\leq \min\left\{\frac{2}{5}\Delta,f(v)\right\}$ and the contribution of $w$ to $f(v)$ is at most $\frac{8}{\Delta}$.
If $w \in N_\Delta$, then $f(w)\leq f(v)$ and $w$ contributes nothing to $f(v)$.
These observations imply both
\begin{align*}
	f(v) \leq \frac{8}{\Delta}n_\ell + \frac{1}{2}n_s + n_1
\end{align*}
and, by using (\ref{sum f}),
\begin{align*}
	|I|\leq f(v) n_\Delta + \min\left\{\frac{2}{5}\Delta,f(v)\right\}n_\ell+ n_s.
\end{align*}
In order to prove that $|I|\leq \lceil \frac{\Delta}{2}\rceil \lfloor \frac{\Delta}{2} \rfloor$,
we show that
\begin{align}\label{eq I}
	f' n_\Delta + \min\left\{\frac{2}{5}\Delta,f'\right\}n_\ell+ n_s \leq \left\lceil \frac{\Delta}{2}\right\rceil \left\lfloor \frac{\Delta}{2}\right \rfloor,
\end{align}
under the condition that $n_1,n_s,n_\ell,n_\Delta$ are non-negative integers and $n_1+n_s+n_\ell+n_\Delta=\Delta$
where
\begin{align}\label{eq fv}
	f'=\frac{8}{\Delta}n_\ell + \frac{1}{2}n_s + n_1.
\end{align}
Let $i(n_1,n_s,n_\ell,n_\Delta)=f' n_\Delta + \min\left\{\frac{2}{5}\Delta,f'\right\}n_\ell+ n_s$.
Obviously, $|I|\leq i(n_1,n_s,n_\ell,n_\Delta)$.

Inequality (\ref{eq fv}) implies $n_s+n_1\geq f'-8$.
Thus $n_\ell+n_\Delta = \Delta- n_1-n_s \leq \Delta-f'+8$ and
hence, by (\ref{eq I}), we obtain
\begin{align*}
	i(n_1,n_s,n_\ell,n_\Delta)\leq f'(\Delta-f'+8) +\Delta.
\end{align*}
If $f'\leq \frac{2}{5}\Delta+8$, then this implies that
$i(n_1,n_s,n_\ell,n_\Delta)\leq \frac{6}{25}\Delta^2+ \frac{24}{5}\Delta \leq
\frac{\Delta^2}{4}-1$, which implies the desired result.

Thus we may assume that $f'\geq \frac{2}{5}\Delta+8$.
Suppose $n_\ell \geq 1$ and hence $n_\Delta \leq \Delta-1$.
This implies that
\begin{align*}
	i(n_1,n_s,n_\ell-1,n_\Delta+1)-i(n_1,n_s,n_\ell,n_\Delta)
	&\geq - \frac{8}{\Delta}n_\Delta- \frac{2}{5}\Delta + \left(f'- \frac{8}{\Delta}\right)\cdot 1\\
	&\geq - \frac{8}{\Delta}(\Delta-1)- \frac{2}{5}\Delta + \frac{2}{5}\Delta+8- \frac{8}{\Delta}\\
	&= 0.
\end{align*}
Hence, we may assume that $n_\ell=0$.

Furthermore, we may assume that $n_\Delta \geq 2$; otherwise, by using $f',n_s\leq \Delta$, we conclude $i(n_1,n_s,n_\ell,n_\Delta) \leq 2\Delta$.
Suppose $n_s\geq 1$. Thus
\begin{align*}
	i(n_1+1,n_s-1,n_\ell,n_\Delta)-i(n_1,n_s,n_\ell,n_\Delta)\geq \frac{1}{2} \cdot 2 - 1\geq 0.
\end{align*}
Therefore, we may assume that $n_s=0$.
Thus $n_1=\Delta-n_\Delta$ and (\ref{eq fv}) implies that $f'=n_1$.
By using (\ref{eq I}), we conclude
$$i(n_1,n_s,n_\ell,n_\Delta)=n_\Delta(\Delta-n_\Delta)\leq \left\lceil \frac{\Delta}{2}\right\rceil \left\lfloor \frac{\Delta}{2}\right \rfloor.$$

Therefore, $|N[v]|+|I|\leq \left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)$
and the edge $uv$ together with a maximum induced matching of $G'$ yields
$\nu_s(G)\geq \frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$,
which is a contradiction to our choice of $G$.

\begin{case}
$v$ is not adjacent to a vertex of degree $1$.
\end{case}
Let $u\in N(v)$ such that $d(u)$ is minimal.
Let $S=N[u]\cup N[v]$ and $G'=G-S-I$ where $I$ is the set of isolated vertices of $G-S$.

By double counting the edges between $S$ and $I$, it is straightforward to see that $I$ contains at most $2\Delta$ vertices of degree $\Delta$.
Thus similarly as in (\ref{sum f}), we conclude that
\begin{align}\label{eq 4}
	|I|\leq \sum_{w\in S\setminus\{u,v\}}f(w) + 2\Delta.
\end{align}

If $d(u)\geq 10$,
then $f(v) \leq \frac{\Delta}{10}$.
Thus $|I|\leq \frac{\Delta^2}{5}+2\Delta$ and hence $|S|+|I|\leq \frac{\Delta^2}{4}$.
Therefore, $uv$ together with a maximum induced matching of $G'$ yields
$\nu_s(G)> \frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$,
which is a contradiction to our choice of $G$.

Thus we may assume that $d(u)\leq 9$ and hence trivially $\sum_{w\in N(u)\setminus \{v\}}f(w)\leq 8\Delta$ and $|S|\leq \Delta+9$.

Let $N_s$ be the set of neighbors of $v$ of degree at most $\frac{\Delta}{8}$, let $N_\ell=N(v)\setminus N_s$,
and let $\alpha= \frac{|N_s|}{\Delta}$ and hence $N_\ell \leq (1-\alpha)\Delta$.

The contribution of the vertices in $N_s$ to $f(v)$ is at most $\frac{\alpha\Delta}{2}$.
Using Claim \ref{d1}, we conclude that $f(w)\leq 1$ for $w\in N_s$.
The contribution of the vertices in $N_\ell$ to $f(v)$ is at most $8$
and $f(w)\leq f(v)$ for $w\in N_\ell$ by the choice of $v$.
This implies that $f(v)\leq \frac{\alpha\Delta}{2}+8$.
Note that $(1-\alpha)\frac{\alpha}{2}\leq \frac{1}{8}$.
Moreover, by (\ref{eq 4}), we obtain
\begin{align*}
	|I|& \leq  \sum_{w\in N(v)\setminus \{u\}}f(w)+\sum_{w\in N(u)\setminus \{v\}}f(w)+2\Delta\\
	& \leq \sum_{w\in N(v)\setminus \{u\}: w\in N_\ell}f(w)+\sum_{w\in N(v)\setminus \{u\}: w\in N_s}f(w)+8\Delta+2\Delta\\
	& \leq  (1-\alpha)\Delta f(v) + \alpha\Delta + 10\Delta\\
		&\leq (1-\alpha)\Delta \left(\frac{\alpha\Delta}{2}+8\right) + 11\Delta\\
	&\leq  \frac{\Delta^2}{4}-2\Delta.
\end{align*}
Thus $|I|+|S|\leq  \frac{\Delta^2}{4}$.
Therefore, $uv$ together with a maximum induced matching of $G'$ yields
$\nu_s(G)> \frac{n(G)}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}$,
which is the final contradiction.
\hfill $\Box$

\section{Graphs with Small Maximum Degree
}
Let $C_5^2$ be the graph obtained from the $5$-cycle by replacing every vertex by an independent set of order $2$ and
let $K_{3,3}^+$ be the graph obtained from the $5$-cycle by replacing the vertices by independent sets of orders $1,1,1,2$, and $2$, respectively.
Note that the graph $K_{3,3}^+$ can also be obtained from a $K_{3,3}$ by subdividing one edge once.
The graphs $C_5^2$ and $K_{3,3}^+$ show that Theorem \ref{result max degree} is not true for graphs of maximum degree $3$ or $4$.
However, I conjecture that these graphs are the only exceptions.

\begin{conjecture}\label{conj general}
If connected graph $G\notin\{C_5^2,K_{3,3}^+\}$ with maximum degree $\Delta\geq 3$,
then
\begin{align*}
\nu_s(G) \geq \frac{1}{\left(\lceil\frac{\Delta}{2}\rceil+1\right) \left(\lfloor\frac{\Delta}{2}\rfloor+1\right)}n(G).
\end{align*}
\end{conjecture}
\noindent
Note that for $\Delta=3$, a result in \cite{jorasa}, and
for $\Delta \geq 1000$, Theorem \ref{result max degree} implies Conjecture~\ref{conj general}.

config: arxiv


TITLE: Missing premise in uniqueness portion of theorem
QUESTION [0 upvotes]: Problem statement, as written:
Let $f\colon A \rightarrow C$ and $g\colon A \rightarrow B$ be functions. Prove that there exists a function $h\colon B \rightarrow C$ such that $f = h \circ g$ if and only if $\forall x,y \in A$, $g(x) = g(y) \rightarrow f(x) = f(y)$. Prove that $h$ is unique.
The problem is... as written, $h$ isn’t unique. The only way I’ve been able to reconcile this theorem is by adding the premise, “let $g$ be surjective.” Can anyone confirm this? The following problem in the exercise set is here, and, as you can see, includes the phrase “suppose $g$ is bijective” in an analogous place to where “suppose $g$ is surjecrive” would make this problem work. Perhaps a typo or omission?
I can provide a link to my proof upon request- it just seems when my posts get too long you guys run away

REPLY [1 votes]: If $y \in B$ is not in the range of $g$,
it obviously doesn't matter what you take for $h(y)$.
So, assuming of course $C$ is not a singleton, what you need for $h$ to be unique is that $g$ is surjective.


TITLE: Formation of Six Letter Strings that Conform to Specific Criterias - Combination Theory
QUESTION [2 upvotes]: Problem:

How many strings of 6 characters can you create that starts and ends with the same letter (eg: "ANDREA"), reads equally forward as well as backward (eg: Consider that you are working with the capital letters of the English alphabet).

From what I understand, for these conditions to be met, the first three and last three letters will have to be the same and the permutations of the first three letters will be equal to that of the last three.
Did I miss anything? But I am not sure how to determine the combination. Can someone help me to better understand and resolve this problem?

REPLY [3 votes]: A palindrome is a word  (such as ABAABA) that reads the same way forwards as backwards.  You wish to count palindromes of length $6$ that can be formed with the capital letters of the English alphabet.  As you observed, choosing the first three letters completely determines the palindrome.  Since repetition of letters must be permitted, we have $26$ choices for each of those letters.  Hence, there are $26^3$ such words.


TITLE: For which $p \in \mathbb{R}_{>0}$ does the integral $\int_{[0,1]^n} \frac{\mathrm dx}{(x_1^p+2x_2^p + ... + nx_n^p)^{1/3}}$ converge?
QUESTION [11 upvotes]: I want to find out for which $p \in \mathbb{R}_{>0}$ the integral
$$\int_{[0,1]^n} \frac{\mathrm d x}{(x_1^p+2x_2^p + ... + nx_n^p)^{1/3}}$$
converges. To be honest, I have no idea or whatsoever how to do this. I computed the integral for $n=1$ and $n=2$ and found them to be existing for $p < 3n$ so far. But Mathematica takes forever to compute the integral for $n=3$, so I think I have to find a smart way to verify whether my guess is correct.
Can anyone give me a little hint, so that I can proceed? The problem is that we've never done such things before during classes or in exercises, so I really don't know where to begin.
Thanks in advance.

REPLY [12 votes]: This is to advocate an easy method leading to the answer without lingering on irrelevant details of the question. First, replacing each $x_i^p$ by $|x_i|^p$, one sees that it is equivalent to ask for the behaviour of the integral over $[-1,1]^n$, that is, for the behaviour of the integral around the point $x=0$. Second, since all the norms on $\mathbb{R}^n$ are equivalent, one can replace the denominator by $\|x\|^{p/3}$ for every norm $\|\ \|$, without changing the convergence or divergence. Choose the Euclidean norm and write $\|x\|=r$. By any spherical change of variables, $\mathrm{d}x$ is $r^{n-1}\mathrm{d}r$ times an infinitesimal element which involves the spherical coordinates of $x/\|x\|$. This reduces the question to the behaviour at $r=0^+$ of the integral
$$
\int_0 \frac{r^{n-1}\mathrm{d}r}{r^{p/3}}.
$$
Hence, there is convergence if and only if $n-1-p/3>-1$, that is, if and only if $p<3n$ (as the OP noticed when $n=1$ and $n=2$).


TITLE: Why when Y=-X I get negative p.d.f?
QUESTION [0 upvotes]: Am I doing something wrong here?
Suppose we know the pdf of the random variable X  fx(x) and we wish to find the pdf of the random variable Y fy(y), where Y=-X
Then fy(y)=-fx(-y). But that means the p.d.f of Y is negative since p.d.f of X is positive.
But we can't have a negative p.d.f
What's going on here?
Thanks in advance

REPLY [3 votes]: Let us restrict to the case where the CDF of $X$ is differentiable.
Then its derivative serves as PDF ($F_X'=f_X$) and:
$$F_{Y}\left(y\right)=P\left\{ Y\leq y\right\} =P\left\{ X\geq-y\right\} =1-F_{X}\left(-y\right)$$
so that: $$f_{Y}\left(y\right)=F_{Y}'\left(y\right)=f_{X}\left(-y\right)$$
Equality $f_{Y}\left(y\right)=f_{X}\left(-y\right)$ holds also if $F_X$ is not differentiable.


TITLE: $f_n \to 0$ $ a.e.$ and $\lim \int f_n d\mu =0$ but $\sup_n f_n$ is not in $L^1$
QUESTION [1 upvotes]: Give an example of a finite measure space $(X,M,\mu)$ and a sequence of functions $f_n:X \to[0, \infty)$ such that $f_n \to 0$ $a.e.$ and $\lim \int f_n d\mu=0$  but $\sup_n f_n$ is not in $L^1$
I have trouble coming up with such an example and prove it. Can someone help me with this problem?

REPLY [1 votes]: Take $X=[0,1]$, $f_n(x) = {1 \over x} \cdot 1_{({1 \over n+1}, {1 \over n}]}(x)$.
Then $f_n(x) \to 0$, $\int f_n = \ln(1+ {1 \over n}) \to 0$, for $x \neq 0$, $\sup_n f_n(x) = {1 \over x}$, hence $\sup_n f_n$ is not integrable.


TITLE: I have discovered a way to calculate the absolute value (area,volume, etc) of a n-dimentional shape, using it's coordinates only, how do I publish it?
QUESTION [0 upvotes]: Firstly, I want to preface by saying that I am no experience with the maths community at all, however I did take Maths and Further Maths for my A-Levels. What I have discovered is a way of using matrices and coordinates to calculate the absolute value of a n-dimentional shape in a systematic way, which means that it removes the decision making process that comes with working out the area or volume of a polygon or polyhedron, my method makes it trivial to do so and it works for n-dimensions.I think it will will also have uses in multi-dimentional statistics too.
Currently I have the method fully developed and I have conceptualised the proof, but needs writing (I have no experience in this). I have searched the internet to verify that my idea is novel, with faith in my google-fu I can say that it is. What I need is a bit of guidance to publish this method.
I am not looking for any monetary gain, but would like some recognition, which will help me when applying for university in four months.
Thanks for reading, all questions are welcome.

REPLY [1 votes]: How do I publish it ?

Anonymously, on a free content site.

I would like some recognition, which will help me when applying for university in $4$ months.

There's no point in getting recognition for $($re$)$discovering the multidimensional generalization of a well-known formula. Chances are it is already known $($to mankind$)$, albeit perhaps not as popular as its more earthy counterpart. Throughout my life, I have also $($re$)$discovered many things, such as binary powers, the $\Gamma$ and beta functions, Newton's binomial series, Theodore's spiral, etc., so I would kindly urge you to be more cautious in your assumptions about what is already known.


text:
\begin{document}

\title[Lattice isomorphisms]{Lattice isomorphisms between projection lattices of von Neumann algebras}

\author[M. Mori]{Michiya Mori}

\address{Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba, Tokyo, 153-8914, Japan.}
\email{mmori@ms.u-tokyo.ac.jp}

\subjclass[2010]{Primary 46L10, Secondary 16E50, 47B49, 51D25.}

\keywords{projection lattice; lattice isomorphism; ring isomorphism; locally measurable operator; von Neumann algebra}

\date{}

\begin{abstract}
Generalizing von Neumann's result on type II$_1$ von Neumann algebras, we characterize lattice isomorphisms between projection lattices of arbitrary von Neumann algebras by means of ring isomorphisms between the algebras of locally measurable operators.
Moreover, we give a complete description of ring isomorphisms of locally measurable operator algebras when the von Neumann algebras are without type II direct summands.
\end{abstract}

\maketitle
\thispagestyle{empty}

\section{Introduction}
Since the very first work \cite{MN} by Murray and von Neumann more than 80 years ago, the geometry of projections has played the central role in understanding the structure of von Neumann algebras (rings of operators).
For a von Neumann algebra $M$, let $\P(M)$ denote the projection lattice of $M$, that is, $\P(M):=\{p\in M\mid p=p^*=p^2\}$.
In this paper, we would like to consider the following question:
What is the general form of lattice isomorphisms between projection lattices of von Neumann algebras?

There are several important results related to this question.
Let us first think about finite dimensional factors.
The case $M=N=\M_n(\C)$ for $n=1, 2$ is not interesting at all.
Indeed, if $n=1$, then $\P(\M_n(\C))$ is $\{0, 1\}$, and a lattice automorphism of it is the identity mapping.
If $n=2$, then a bijection $\Phi$ on $\P(\M_n(\C))$ is a lattice automorphism if and only if $\Phi(0)=0$ and $\Phi(1)=1$.
If $M=N= \M_n(\C)$ for $3\leq n<\infty$, then the fundamental theorem of projective geometry gives an answer to our question.
Recall that a function $f\colon X\to Y$ between complex vector spaces is said to be \emph{semilinear} if it is additive and there exists a ring homomorphism $\sigma\colon \C\to \C$ satisfying $f(cx) = \sigma(c)f(x)$ for all $c\in \C$ and $x\in X$.
\begin{theorem}[Fundamental theorem of projective geometry]
Let $3\leq n<\infty$.
Suppose that $\Phi\colon \P(\M_n(\C))\to \P(\M_n(\C))$ is a lattice isomorphism.
Then there exists a semilinear bijection $f\colon \C^n\to \C^n$ such that $\Phi(p_{\xi}) = p_{f(\xi)}$ for every $\xi\in \C^n$, where $p_{\xi}$ denotes the projection from $\C^n$ onto $\C \xi$ for a vector $\xi\in \C^n$.
\end{theorem}

In the case of type I$_{\infty}$ factors, we can make use of a result below by Fillmore and Longstaff in 1984.
Recall that a projection $p\in \P(B(H))$ can be identified with its range $pH$, which is a closed subspace of $H$.
\begin{theorem}[{\cite[Theorem 1]{FL}}]
Let $X$ and $Y$ be infinite dimensional complex normed spaces.
Let $\cC(X)$ (resp.\ $\cC(Y)$) denote the lattice of all closed subspaces of $X$ (resp.\ $Y$), ordered by inclusion.
Suppose that $\Phi\colon \cC(X)\to \cC(Y)$ is a lattice isomorphism.
Then there exists a bicontinuous linear or conjugate-linear bijection $f\colon X\to Y$ such that $\Phi(C)=f(C)$ for any $C\in \cC(X)$.
\end{theorem}

See also the classical result \cite[Theorem 1]{KM}, in which Kakutani and Mackey studied orthocomplementation on the lattice $\P(B(H))$.

For type I factors, we may observe a correspondence between lattices and rings.
Let $H$ be a Hilbert space with $\dim H\geq 3$.
For any lattice automorphism $\Phi\colon \P(B(H))\to \P(B(H))$,
take a mapping $f\colon H\to H$ as above.
It is a semilinear bijection if $\dim H<\infty$; a linear or conjugate-linear bounded bijection if $\dim H=\infty$.
Hence we may construct a ring automorphism $\Psi\colon B(H)\to B(H)$ such that $\Phi(l(x))= l(\Psi(x))$ for every $x\in B(H)$ (namely, $\Psi(x) := f\circ x\circ f^{-1}$), where $l(x)$ denotes the left support projection of $x$.
It is easy to see that the converse also holds.
That is, any ring automorphism $\Psi\colon B(H)\to B(H)$ determines a lattice automorphism $\Phi$ of $\P(B(H))$ such that $\Phi(l(x))= l(\Psi(x))$ for every $x\in B(H)$.

We next consider finite von Neumann algebras.
In 1930's, motivated by the geometry of projection lattices of type II$_1$ factors,
von Neumann produced the beautiful theory on the correspondence between complemented modular lattices and regular rings.
One of his achievements \cite[Part II, Theorem 4.2]{N}, applied to the case of arbitrary type II$_1$ von Neumann algebras, reads as follows.
\begin{theorem}[von Neumann]\label{vN}
Let $M$ and $N$ be von Neumann algebras of type II$_1$.
Suppose that $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism.
Then there exists a unique ring isomorphism $\Psi\colon S(M)\to S(N)$ between the algebras of measurable operators such that $\Phi(l(x))=l(\Psi(x))$ for any $x\in S(M)$.
\end{theorem}
See the next section for the definition of undefined terms and see also Section \ref{question} for further details about von Neumann's theory.

In the general setting of von Neumann algebras, with an additional assumption, Dye obtained the following result in 1955.
\begin{theorem}[{\cite[Corollary of Theorem 1]{Dy}}, see also {\cite[Theorem 1]{Fe}}]\label{dye}
Let $M$ and $N$ be von Neumann algebras without type I$_2$ direct summands.
Suppose that $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism with
\[
pq=0 \iff \Phi(p)\Phi(q)=0
\]
for any $p, q\in \P(M)$.
Then there exists a real $^*$-isomorphism $\Psi\colon M\to N$ that extends $\Phi$.
\end{theorem}

Each of the above results implies that lattice isomorphisms between projection lattices are closely related to ring isomorphisms.
See also McAsey's survey \cite{Mc} which discusses projection lattice isomorphisms in various settings.
It is natural to imagine that we can give a similar result for arbitrary lattice isomorphisms in the general setting of von Neumann algebras.
The main theorem of this paper realizes it.
\begin{thm}
Let $M$ and $N$ be two von Neumann algebras.
Suppose that $M$ does not admit type I$_1$ nor I$_2$ direct summands, and that $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism.
Then there exists a unique ring isomorphism $\Psi\colon LS(M)\to LS(N)$ such that $\Phi(l(x)) = l(\Psi(x))$ for all $x\in LS(M)$.
\end{thm}
Here, $LS(M)$ and $LS(N)$ mean the algebras of locally measurable operators of $M$ and $N$, respectively (see Section \ref{LSdef}).
We remark that the converse of Theorem A can be verified without difficulty.
Namely, any ring isomorphism $\Psi\colon LS(M)\to LS(N)$ determines a unique lattice isomorphism $\Phi\colon \P(M)\to \P(N)$ such that $\Phi(l(x)) = l(\Psi(x))$ for all $x\in LS(M)$ (Proposition \ref{converse}).
Therefore, Theorem A naturally gives rise to the following
\begin{q}
Let $M$, $N$ be von Neumann algebras.
What is the general form of ring isomorphisms from $LS(M)$ onto $LS(N)$?
\end{q}
We may answer this question for type I von Neumann algebras using ring isomorphisms of their centers (Proposition \ref{I}).
Moreover, we obtain
\begin{thmb}
Let $M, N$ be von Neumann algebras of type I$_{\infty}$ or III.
If $\Psi\colon LS(M)\to LS(N)$ is a ring isomorphism, then there exist a real $^*$-isomorphism $\psi\colon M\to N$ (which extends to a real $^*$-isomorphism from $LS(M)$ onto $LS(N)$) and an invertible element $y\in LS(N)$ such that  $\Psi(x)=y\psi(x)y^{-1}$, $x\in LS(M)$.
\end{thmb}
We leave the case of type II von Neumann algebras as an open question.

In Section \ref{preliminaries}, we introduce some tools we use later.
Section \ref{main} is devoted to the proof of Theorem A.
The proof is based on the combination of von Neumann's strategy in \cite[Part II, Chapter IV]{N} and a binary relation on the projection lattice which we call LS-orthogonality.
After that we give a proof of Dye's theorem as an application of Theorem A.
We consider Question in Section \ref{extra}, and prove Theorem B.
This paper ends with comparison of our result with von Neumann's theory and several suggestions of further research directions (Section \ref{question}).

\section{Preliminaries}\label{preliminaries}
Let $M\subset B(H)$ be a von Neumann algebra.
We use the symbols $\sim$ to mean the Murray--von Neumann equivalence relation on $\P(M)$.
That is, for $p, q\in \P(M)$, $p\sim q$ means that there exists a partial isometry $v\in M$ such that $p= vv^*$ and $q= v^*v$.
As usual, for $p, q\in \P(M)$, $p\perp q$ means that $p$ and $q$ are orthogonal. That is, $pq=qp=0$, or equivalently, $pH\perp qH$ in the Hilbert space $H$.
We use the symbol $p^{\perp}:= 1-p$ for $p\in \P(M)$.
The symbol $\Z(M)=\{x\in M\mid xy=yx\text{ for all }y\in M\}$ means the center of $M$.

For $n\in \N=\{1, 2, \ldots\}$, we say that $M$ has \emph{order $n$} if there exists a collection $p_1, \ldots, p_n$ of mutually orthogonal projections in $M$ such that $p_1\sim p_2\sim\cdots\sim p_n$ and $\sum_{k=1}^n p_k= 1$.
It is well known that every von Neumann algebra without finite type I direct summands has order $n$ for any $n\in \N$ \cite[Lemma 6.5.6]{KR}.
In particular, such an algebra has order $3$.
It follows that every von Neumann algebra $M$ without type I$_1$ and I$_2$ direct summands can be decomposed into the ($\ell^{\infty}$-)direct sum of von Neumann algebras $M_n$, $3\leq n<\infty$, such that $M_n$ has order $n$ for every $n$.
If $M$ has order $n\in \N$, then $M$ can be identified with the algebra $\M_n(\hat{M})$ of $n\times n$ matrices with entries in some von Neumann algebra $\hat{M}$.

\subsection{Various isomorphisms of von Neumann algebras}
For $^*$-algebras $A$ and $B$, a (not necessarily linear) bijection $\psi\colon A\to B$ is called
\begin{itemize}
\item a \emph{semigroup isomorphism} if it is multiplicative,
\item a \emph{ring isomorphism} if it is additive and multiplicative,
\item a \emph{real algebra isomorphism} if it is a real-linear ring isomorphism,
\item an \emph{algebra isomorphism} if it is a complex-linear ring isomorphism,
\item a \emph{real $^*$-isomorphism} if it is a real algebra isomorphism and satisfies $\psi(x^*)=\psi(x)^*$ for any $x\in A$,
\item a \emph{$^*$-isomorphism} if it is a complex-linear real $^*$-isomorphism, and
\item a \emph{conjugate-linear $^*$-isomorphism} if it is a conjugate-linear real $^*$-isomorphism.
\end{itemize}

\begin{lemma}\label{isoms}
Let $M$ and $N$ be von Neumann algebras.
Suppose that $\psi\colon M\to N$ is a bijection.
\begin{enumerate}
\item If $M$ is without type I$_1$ direct summands and $\psi$ is a semigroup isomorphism, then $\psi$ is a ring isomorphism.
\item If $M$ does not admit a finite dimensional ideal and $\psi$ is a ring isomorphism, then $\psi$ is a real algebra isomorphism.
\item If $\psi$ is a real algebra isomorphism, then there exist a real $^*$-isomorphism $\psi_0\colon M\to N$ and an invertible element $y\in N$ such that $\psi(x) =y\psi_0(x)y^{-1}$ for any $x\in M$.
\item If $\psi$ is a real $^*$-isomorphism, then there exist central projections $p\in M$ $q\in N$, a $^*$-isomorphism $\psi_1\colon Mp\to Nq$, and a conjugate-linear $^*$-isomorphism $\psi_2\colon Mp^{\perp}\to Nq^{\perp}$ such that $\psi(x) = \psi_1(xp) + \psi_2(xp^{\perp})$ for any $x\in M$.
\end{enumerate}
\end{lemma}
\begin{proof}
Each item is easily obtained by known results.

(1) We may take a projection $p\in \P(M)$ such that both of the central supports of $p$ and $1-p$ are equal to $1$.
It is easy to see that the following hold:
(a) If $x\in M$ satisfies $xM=\{0\}$, then $x=0$;
(b) If $x\in M$ satisfies $pMx=\{0\}$, then $x=0$;
(c) If $x\in M$ satisfies $pxpMp^{\perp}=\{0\}$, then $pxp=0$.
Hence we may apply Martindale's theorem \cite[Theorem]{Ma} to obtain the desired conclusion.

The item (2) is a consequence of Kaplansky's result \cite[Theorem]{Ka}.

We prove (3) and (4) at the same time.
Let $\psi\colon M\to N$ be a real algebra isomorphism.
We know that $\psi(i)^2 =\psi(i^2)=\psi(-1)= -1$ and that $\psi(i)$ is central in $N$.
It follows that $\psi(i) = qi-q^{\perp}i$ for some central projection $q$ of $N$.
Put $p:= \psi^{-1}(q)$, which is a central projection of $M$.
If $\psi$ is a real $^*$-isomorphism, then $\psi$ restricted to $Mp$ is a $^*$-isomorphism from $Mp$ onto $Nq$, and $\psi$ restricted to $Mp^{\perp}$ is a conjugate-linear $^*$-isomorphism from $Mp^{\perp}$ onto $Nq^{\perp}$, hence the proof of (4) is complete.
If $\psi$ is merely a real algebra isomorphism, then $\psi$ restricted to $Mp$ is an algebra isomorphism from $Mp$ onto $Nq$, and $\psi$ restricted to $Mp^{\perp}$ determines an algebra isomorphism from $Mp^{\perp}$ onto $\overline{Nq^{\perp}}$, where $\overline{Nq^{\perp}}$ means the complex conjugation of the von Neumann algebra $Nq^{\perp}$.
See e.g.\ \cite[Section 2.3]{P} for the definition of complex conjugation of von Neumann algebras.
Lastly, we may use the result on the general form of algebra isomorphisms between von Neumann algebras (\cite[Theorem I]{O}, see also \cite{Ga} and \cite[Section 4.1]{Sa}) to obtain the desired conclusion.
\end{proof}

\subsection{The algebra of locally measurable operators}\label{LSdef}
Let $M\subset B(H)$ be a von Neumann algebra.
In this paper, the algebra $LS(M)$ of locally measurable operators with respect to $M$, which we briefly describe below, plays a crucial role.

A densely defined closed operator $x$ on $H$ is said to be \emph{affiliated with $M$} (and we write $x\eta M$) if
$yx\subset xy$ for any $y\in M'$, where $M':=\{y\in B(H)\mid ay=ya\text{ for any }a\in M\}$ denotes the commutant of $M$.
An operator $x\eta M$ is said to be \emph{measurable} with respect to $M$ if the spectral projection $\chi_{(c, \infty)}(\lvert x\rvert)\in \P(M)$ is a finite projection in $M$ for some real number $c>0$.
An operator $x\eta M$ is said to be \emph{locally measurable} with respect to $M$ if there exists an increasing sequence $\{p_n\}_{n\geq 1}$ of central projections in $M$ such that $p_n\nearrow 1$ and $xp_n$ is measurable with respect to $M$ for any $n$.
We write $S(M)$ (resp.\ $LS(M)$) to mean the collection of all measurable (resp.\ locally measurable) operators with respect to $M$.
If $x, y\in S(M)$ (resp. $LS(M)$), then $x^*$ and the closures of $xy$, $x+y$ are in $S(M)$ (resp. $LS(M)$).
Using this fact, we can consider $S(M)$ and $LS(M)$ as $^*$-algebras that contain $M$.
In what follows, we abbreviate the symbol of the closure of an unbounded operator unless it is confusing.
We remark that $LS(M)=M$ holds if and only if $M$ is the direct sum of finite number of type I and III factors.
We also remark that if $M$ is finite then $LS(M)=S(M)$ is the collection of all affiliated operators.
See \cite{Y} and \cite{Se} for more details of (locally) measurable operators.

In \cite[Lemma 2.2]{Mor2}, the author obtained the following result.

\begin{lemma}\label{inverse}
Let $M$ be a von Neumann algebra and $a\in M_+$.
Then the following two conditions are equivalent:
\begin{enumerate}
\item The element $a$ is invertible in the algebra $LS(M)$.
\item For any $b\in M_+\setminus \{0\}$, there exists an $x\in M_+\setminus \{0\}$ such that $x\leq a$ and $x\leq b$.
\end{enumerate}
\end{lemma}

For $x\in LS(M)$, let $l(x)\in \P(M)$ denote the left support of $x$.
That is, $l(x) :=\bigwedge \{p\in \P(M)\mid px = x\}$.
Similarly, we write $r(x) :=\bigwedge \{p\in \P(M)\mid x = xp\}$.
Then $l(x)= \chi_{(0, \infty)}(\lvert x^*\rvert)$ and $r(x)= \chi_{(0, \infty)}(\lvert x\rvert)$ hold.
We remark that, for $x, y\in LS(M)$, we have $xy=0$ if and only if $r(x)l(y)=0$.
Indeed, if $r(x)l(y)=0$, then $xy=xr(x)l(y)y=0$.
If $xy=0$, then we have $\lvert x\rvert \lvert y^*\rvert = 0$,
which implies $\chi_{(\varepsilon, \infty)}(\lvert x\rvert)\chi_{(\varepsilon, \infty)}(\lvert y^*\rvert)=0$ for every $\varepsilon>0$.
Take the limit $\varepsilon\to 0$ in the strong operator topology to obtain   $r(x)l(y)=0$.

\subsection{Halmos's two projection theorem}\label{halmos}
In order to play with projection lattices, it is useful to look at the relative position of a pair of projections.
For it, we make use of Halmos's two projection theorem \cite{H} from the viewpoint of von Neumann algebra theory.
Here we recapitulate the argument in \cite[Lemma 2.2]{Mor1}.

Let $M\subset B(H)$ be a von Neumann algebra and $p, q\in \P(M)$.
Put
\[
e_1= p- p\wedge q - p\wedge q^{\perp},\quad e_2=p^{\perp} - p^{\perp}\wedge q - p^{\perp}\wedge q^{\perp},
\]
and $x:= e_1(q-p\wedge q-p^{\perp}\wedge q)e_2$.
By an elementary calculation, we see that $l(x)=e_1$ and $r(x)=e_2$.
By polar decomposition, we may take a partial isometry $v=v_{p, q}\in M$ such that $x= v\lvert x \rvert = \lvert x^* \rvert v$, $vv^*= e_1$ and $v^*v= e_2$.

We can identify each $y\in (e_1+e_2)M(e_1+e_2)$ with
$\begin{pmatrix}
e_1ye_1 & e_1yv^* \\
vye_1 & vyv^*
\end{pmatrix} \in \M_2(e_1Me_1)$.
Then $q-p\wedge q-p^{\perp}\wedge q\, (\leq e_1+e_2)$ is identified with
\[
\begin{split}
&\begin{pmatrix}
e_1(q-p\wedge q-p^{\perp}\wedge q)e_1 & e_1(q-p\wedge q-p^{\perp}\wedge q)v^* \\
v(q-p\wedge q-p^{\perp}\wedge q)e_1 & v(q-p\wedge q-p^{\perp}\wedge q)v^*
\end{pmatrix} \\
=& \begin{pmatrix}
e_1(q-p\wedge q-p^{\perp}\wedge q)e_1 & \lvert x^* \rvert \\
\lvert x^* \rvert & v(q-p\wedge q-p^{\perp}\wedge q)v^*
\end{pmatrix}
\in \M_2(e_1Me_1).
\end{split}
\]
Put $a:= (e_1(q-p\wedge q-p^{\perp}\wedge q)e_1)^{1/2}$ and $b:= (v(q-p\wedge q-p^{\perp}\wedge q)v^*)^{1/2}$, which are positive injective operators in $M_{p, q}:= e_1Me_1$.
Since $\begin{pmatrix}
a^2 & \lvert x^* \rvert \\
\lvert x^* \rvert & b^2
\end{pmatrix}$ is a projection, some calculations show that $a, b$ and $\lvert x^* \rvert$ commute, $a^2+b^2= e_1$ and $\lvert x^* \rvert = ab$.
Thus $q-p\wedge q-p^{\perp}\wedge q$ corresponds to $\begin{pmatrix}
a^2 & ab \\
ba & b^2
\end{pmatrix}$.

Therefore, we may decompose $p$ and $q$ in the following manner:
\[
p= 1\oplus 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
1&0\\
0&0
\end{pmatrix},\quad
q= 0\oplus 1\oplus 1\oplus 0\oplus
\begin{pmatrix}
a^2&ab \\
ab&b^2
\end{pmatrix},
\]
where $H$ is decomposed as $H = (p\wedge q^{\perp}) H \oplus (p^{\perp} \wedge q) H \oplus (p\wedge q) H \oplus (p^{\perp} \wedge q^{\perp}) H \oplus (e_1+e_2)H$,
and $a$ and $b$ are positive injective operators in $M_{p, q}\,\, (= e_1Me_1)$ such that $a^2+b^2=1_{M_{p, q}}$.

\subsection{Center-valued norm}\label{cvn}
Let $M$ be a von Neumann algebra of type I or III and $x\in LS(M)$.
Then there exists a unique minimal element $\vertiii{x} \in LS(\Z(M))_+\,\,(\subset LS(M))$ with $\lvert x\rvert\leq \vertiii{x}$.
The mapping $\vertiii{\cdot}\colon LS(M)\to LS(\Z(M))_+$ is called the \emph{center-valued norm}.
Remark that if $M$ is a factor, then $\Z(M)$ can be identified with $\C$ and we have $\vertiii{x} = \lVert x \rVert\in \R$ for every $x\in M$.
Be cautious of the fact that we cannot take such a mapping for a type II von Neumann algebra.
That's why we will need to exclude type II cases in the proof of Theorem B.

As is expected, the center-valued norm possesses e.g.\ the following properties:
For any $x, y\in LS(M)$ and $a\in LS(\Z(M))$, we have (i) $\vertiii{x}=0\Longrightarrow x=0$. (ii) $\vertiii{x+y} \leq \vertiii{x}+\vertiii{y}$. (iii) $\vertiii{a} = \lvert a\rvert$. (iv) $\vertiii{ax} = \lvert a\rvert\vertiii{x}$. (v) $\vertiii{xy}\leq \vertiii{x}\vertiii{y}$.
See for example \cite[Section 2]{AAKD} and references therein for further information about the center-valued norm.

\section{Lattice isomorphisms of projection lattices}\label{main}
Part of this section heavily depends on von Neumann's argument in \cite[Part II, Chapter IV]{N}.
The aim of this section is to give a proof of
\begin{thm}
Let $M$ and $N$ be two von Neumann algebras.
Suppose that $M$ does not admit type I$_1$ nor I$_2$ direct summands, and that $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism.
Then there exists a unique ring isomorphism $\Psi\colon LS(M)\to LS(N)$ such that $\Phi(l(x)) = l(\Psi(x))$ for all $x\in LS(M)$.
\end{thm}

Before its proof, we consider the converse of Theorem A.
\begin{proposition}\label{converse}
Let $M$ and $N$ be von Neumann algebras.
Suppose that $\Psi\colon LS(M)\to LS(N)$ is a ring isomorphism.
Then there exists a unique lattice isomorphism $\Phi\colon \P(M)\to \P(N)$ such that $\Phi(l(x)) = l(\Psi(x))$ for any $x\in LS(M)$.
\end{proposition}
\begin{proof}
It is easy to see that $\Psi(0)=0$.
Let $x, y\in LS(M)$ satisfy $l(x)\leq l(y)$.
Then we have $\{z\in LS(M)\mid zx\neq 0\}\subset \{z\in LS(M)\mid zy \neq 0\}$ and hence $\{z\in LS(N)\mid z\Psi(x)\neq 0\}\subset \{z\in LS(N)\mid z\Psi(y) \neq 0\}$, which in turn leads to $l(\Psi(x))\leq l(\Psi(y))$.
We obtain $l(x)\leq l(y)\Longleftrightarrow l(\Psi(x))\leq l(\Psi(y))$ for any $x, y\in LS(M)$.
Therefore, the mapping $\Phi\colon \P(M)\to \P(N)$ defined by $\Phi(p) = l(\Psi(p))$, $p\in \P(M)$, satisfies the desired condition.
\end{proof}

\begin{remark}
The same proof is valid even if we replace a ring isomorphism with a semigroup isomorphism.
However, Martindale's result \cite{Ma} implies that a semigroup isomorphism $\Psi\colon LS(M)\to LS(N)$ is automatically a ring isomorphism if $M$ is without type I$_1$ direct summands.
\end{remark}

We begin the proof of Theorem A.
Let us first check the uniqueness of $\Psi$.

\begin{lemma}
Let $M$ be a von Neumann algebra without type I$_1$ direct summands.
For any $x\in M$, there exists a subset $F\subset M$ with $\# F\leq 9$, $\sum_{y\in F} y = x$, and the following property:
For any $y\in F$, there exists a pair $p, q\in \P(M)$ of mutually orthogonal projections such that $p\sim q$ and either $pyp=y$ or $pyq=y$.
\end{lemma}
\begin{proof}
It suffices to consider the case where $M$ has fixed order $2\leq n<\infty$.
Then we may identify $M$ with $\M_n(\hat{M})$ for some von Neumann algebra $\hat{M}$.
We may write $x\in M$ as $x = (x_{ij})_{1\leq i, j\leq n} \in \M_n(\hat{M})$.
It is easy to see that we can take integers $n_0:=0\leq n_1\leq n_2\leq n=:n_3$ such that $n_1, n_2-n_1, n_3-n_2\leq n/2$.
For $1\leq k, l\leq 3$, define $x^{kl} = (x^{kl}_{ij})_{1\leq i, j\leq n}\in \M_n(\hat{M})$ by
$x^{kl}_{ij} = x_{ij}$ if $n_{k-1}+1\leq i\leq n_k$ and $n_{l-1}+1\leq j\leq n_l$, and $x^{kl}_{ij} = 0$ otherwise.
(Here, we are decomposing $x$ into $3\times 3$ blocks.)
Then the nine operators $x^{kl}$, $1\leq k, l\leq 3$, (some of which may be $0$) satisfy the desired condition.
\end{proof}

\begin{lemma}\label{id}
Let $M$ be a von Neumann algebra without type I$_1$ direct summands.
Suppose that $\Psi\colon LS(M)\to LS(M)$ is a ring isomorphism with $l(\Psi(x))=l(x)$ for all $x\in LS(M)$.
Then $\Psi$ is the identity mapping on $LS(M)$.
\end{lemma}
\begin{proof}
Let $p\in \P(M)$.
We prove $\Psi(p)=p$.
Since $pp^{\perp} =0$, we have $\Psi(p)\Psi(p^{\perp})=0$, which implies $0=r(\Psi(p))l(\Psi(p^{\perp}))=r(\Psi(p))p^{\perp}$.
We obtain $r(\Psi(p))\leq p$.
We also have the equation $\Psi(p)^2=\Psi(p^2)=\Psi(p)$.
Hence we obtain $(p-\Psi(p))\Psi(p)=0$, which implies $0=(p-\Psi(p))l(\Psi(p))=(p-\Psi(p))p$ and $p-\Psi(p)=0$.

In what follows, let $p, q\in \P(M)$ be mutually orthogonal mutually Murray--von Neumann equivalent projections.
We next prove that $\Psi(x)=x$ if $x\in M\,\,(\subset LS(M))$ satisfies $pxq=x$.
By additivity, we may assume $\lVert x\rVert\leq 1/2$.
Then there exists a projection $e\in \P(M)$ such that $e\leq p+q$, $peq=x$.
Indeed, let $x=v\lvert x\rvert=\lvert x^*\rvert v$ be the polar decomposition.
Take an operator $a\in(pMp)_+$ such that $\lVert a\rVert\leq \pi/4$ and $\lvert x^*\rvert=\sin a\cos a=(\sin 2a)/2$.
Then
\[
e:= \cos^2 a + v^*(\sin a\cos a) + (\sin a\cos a)v + v^*(\sin^2 a)v
\]
satisfies this property.
We obtain $\Psi(x)=\Psi(peq)=\Psi(p)\Psi(e)\Psi(q)=peq=x$.

Suppose that $x\in M$ satisfies $pxp=x$.
Take a partial isometry $v\in M$ such that $vv^*=p$ and $v^*v=q$.
Then we have $p(xv)q = xv$ and $qv^*p=v^*$.
Hence $\Psi(x)=\Psi(xvv^*)=\Psi(xv)\Psi(v^*)= xvv^* = x$.

By the additivity of $\Psi$ and the preceding lemma, we see that $\Psi$ fixes every element in $M$.
Let $x\in LS(M)$ and let $x=v\lvert x\rvert$ be its polar decomposition.
It is clear that $\Psi(1)=1$.
Since $v, (\lvert x\rvert + 1)^{-1}\in M$, we obtain
\[
\begin{split}
\Psi(x) = \Psi(v\lvert x\rvert) &= \Psi(v)\Psi(\lvert x\rvert)\\
&=v(\Psi(\lvert x\rvert + 1)-1)= v(\Psi((\lvert x\rvert + 1)^{-1})^{-1} -1)\\
&= v((\lvert x\rvert + 1)-1) = v\lvert x\rvert=x
\end{split}
\]
\end{proof}

Hence we obtain the uniqueness of $\Psi$ in Theorem A.
Indeed, if two ring isomorphisms $\Psi, \Psi'\colon LS(M)\to LS(N)$ satisfies $l(\Psi(x))=l(\Psi'(x))$ for all $x\in LS(M)$, then we have $l(\Psi^{-1}\circ\Psi'(x))=l(x)$ for all $x\in LS(M)$, hence the preceding proposition implies $\Psi^{-1}\circ\Psi'(x)=x$ for all $x\in LS(M)$.
\medskip

We introduce a binary relation on $\P(M)$, which is a key to the proof of Theorem A.
Let $p, q\in \P(M)$ be two projections with $p\wedge q = 0$.
By Subsection \ref{halmos}, we decompose $p$ and $q$:
\begin{equation}\label{eq:halmos}
p= 1\oplus 0\oplus 0\oplus
\begin{pmatrix}
1&0\\
0&0
\end{pmatrix},\quad
q= 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
a^2&ab \\
ab&b^2
\end{pmatrix}.
\end{equation}
We say that $p$ is \emph{LS-orthogonal} to $q$ if the operator $b\in M_{p, q}$ is  invertible in $LS(M_{p, q})$.

\begin{lemma}\label{orthogonalize}
Let $M$ be a von Neumann algebra and $p, q\in \P(M)$.
Suppose that $p$ is LS-orthogonal to $q$.
Then there exists an invertible element $S= S_{p, q}\in LS(M)$ such that $S(p\vee q)^{\perp}=(p\vee q)^{\perp}S= (p\vee q)^{\perp}$, $Sp= p$ and $l(SqS^{-1})=p\vee q - p$.
\end{lemma}
\begin{proof}
Put $S:= 1\oplus1\oplus1\oplus
\begin{pmatrix}
1&-ab^{-1} \\
0&b^{-1}
\end{pmatrix}$ with respect to the decomposition as above.
Then $S$ is an element in $LS(M)$ with inverse $S^{-1}= 1\oplus1\oplus1\oplus
\begin{pmatrix}
1&a \\
0&b
\end{pmatrix}$.
It is easy to see that
\[
S(p\vee q)^{\perp}=(p\vee q)^{\perp}S= (p\vee q)^{\perp} = 0\oplus0\oplus1\oplus
\begin{pmatrix}
0&0 \\
0&0
\end{pmatrix}.
\]
We also have
\[
Sp = 1\oplus 0\oplus 0\oplus \begin{pmatrix}
1&0 \\
0&0
\end{pmatrix} = p
\]
and
\[
l(SqS^{-1}) = l\left(0\oplus 1\oplus 0\oplus \begin{pmatrix}
0&0 \\
a&1
\end{pmatrix}\right) = 0\oplus 1\oplus 0\oplus \begin{pmatrix}
0&0 \\
0&1
\end{pmatrix} = p\vee q- p.
\]
\end{proof}

\begin{lemma}\label{LS}
Let $M$ be a von Neumann algebra and $p, q\in \P(M)$ be two projections with $p\wedge q = 0$.
Then the following are equivalent.
\begin{enumerate}
\item The projection $p$ is LS-orthogonal to $q$.
\item There exists a lattice automorphism $\Phi$ of $\P(M)$ such that $\Phi(p)\perp\Phi(q)$.
\item If a projection $p_0\in \P(M)$ satisfies $p_0\leq p$ and $p_0\vee q = p\vee q$, then $p_0=p$.
\item The projection $q$ is LS-orthogonal to $p$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)\Rightarrow (2)$ Take $S\in LS(M)$ as in the preceding lemma and let $\Phi$ be the unique lattice isomorphism such that $\Phi(l(x))= l(SxS^{-1})$, $x\in LS(X)$.

$(2)\Rightarrow (3)$ Clear.

$(3)\Rightarrow (1)$
We use the decomposition \eqref{eq:halmos}.
By Lemma \ref{inverse}, if $(1)$ does not hold, then there exists an element $d\in M_{p, q +}\setminus \{0\}$ such that $\{x\in M_{p, q +}\mid x\leq b,\,\, x\leq d\} = \{0\}$.
Take the nonzero spectral projection $p_1:= \chi_{(\lVert d\rVert/2, \lVert d\rVert]}(d)\in \P(M_{p, q})$.
It follows that
\begin{equation}\label{Mpq}
\{x\in M_{p, q +}\mid x\leq b,\,\, x\leq p_1\} = \{0\}.
\end{equation}
Indeed, if $x\in M_{p, q +}$ satisfies $x\leq b$ and $x\leq p_1$, take a positive real number $c$ with $c\leq 1$ and $c\leq \lVert d\rVert/2$, then $cx\leq cb\leq b$ and $cx\leq cp_1\leq d$, hence $cx=0$ and we obtain $x=0$.
Put $p_0 := 1\oplus 0\oplus 0\oplus \begin{pmatrix}
1-p_1&0 \\
0&0
\end{pmatrix}\in \P(M)$.
Then $p_0\leq p$ and $p_0\neq p$.
We prove that $p_0\vee q = p\vee q$, or equivalently, $\begin{pmatrix}
1-p_1&0 \\
0&0
\end{pmatrix} \vee \begin{pmatrix}
a^2&ab \\
ab&b^2
\end{pmatrix} = 1_{\mathbb{M}_2(M_{p, q})}$, which is in turn equivalent to
\begin{equation}\label{0}
\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix} = 0_{\mathbb{M}_2(M_{p, q})}.
\end{equation}
We have
\[
\begin{split}
&\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\left(\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\right)\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\\
&\leq \begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix}\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix} = \begin{pmatrix}
p_1&0 \\
0&0
\end{pmatrix}
\end{split}
\]
and
\[
\begin{split}
&\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\left(\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\right)\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\\
&\leq \begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix} = \begin{pmatrix}
b^2&0 \\
0&0
\end{pmatrix}.
\end{split}
\]
Since the square root mapping preserves the order of positive operators, \eqref{Mpq} implies that the square root of the operator
\[
\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\left(\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\right)\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}
\]
is equal to $0$. Hence
\[
\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}\left(\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\right)\begin{pmatrix}
1&0 \\
0&0
\end{pmatrix}=0,
\]
or equivalently,
$\begin{pmatrix}
p_1&0 \\
0&1
\end{pmatrix} \wedge \begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}\leq \begin{pmatrix}
0&0 \\
0&1
\end{pmatrix}$ holds. However, we know $\begin{pmatrix}
0&0 \\
0&1
\end{pmatrix}\wedge\begin{pmatrix}
b^2&-ab \\
-ab&a^2
\end{pmatrix}  = 0$, so we finally obtain \eqref{0}.

Exchanging the roles of $p$ and $q$, we also obtain $(2)\Leftrightarrow(4)$.
\end{proof}

Let us recall the setting of Theorem A:
Let $M$, $N$ be von Neumann algebras.
Suppose that $M$ is without type I$_1$ and I$_2$ direct summands and $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism.
By the preceding lemma, we see that $\Phi$ preserves LS-orthogonality in both directions, that is, for any $p, q\in \P(M)$, $p$ and $q$ are LS-orthogonal if and only if $\Phi(p)$ and $\Phi(q)$ are LS-orthogonal.

In what follows, we show the existence of $\Psi$ as in the statement of Theorem A in the case $M$ has order $3$.
Thus $M$ can be identified with $\M_3(\hat{M})$ for some von Neumann algebra $\hat{M}$.
Put
\[
e_1^M :=\begin{pmatrix}
1&0&0 \\
0&0&0 \\
0&0&0
\end{pmatrix},\,\,
e_2^M :=\begin{pmatrix}
0&0&0 \\
0&1&0 \\
0&0&0
\end{pmatrix},\,\,
e_3^M :=\begin{pmatrix}
0&0&0 \\
0&0&0 \\
0&0&1
\end{pmatrix} \in \P(\M_3(\hat{M})).
\]

Put $e_1:=\Phi(e_1^M)$, $e_2:=\Phi(e_2^M)$, $e_3:=\Phi(e_3^M)$.
We know that $e_1\vee e_2$ is LS-orthogonal to $e_3$, and $e_1$ is LS-orthogonal and $e_2$.
In addition, we know $e_1\vee e_2\vee e_3 = 1$.
Take $S_{e_1\vee e_2, e_3}$ and $S_{e_1, e_2}$ as in the statement of Lemma \ref{orthogonalize}.
Consider the lattice automorphism $\varphi\colon \P(N)\to \P(N)$ determined by the condition $\varphi(l(x))= l(S_{e_1, e_2}S_{e_1\vee e_2, e_3} x S_{e_1\vee e_2, e_3}^{-1}S_{e_1, e_2}^{-1})\,(=l(S_{e_1, e_2}S_{e_1\vee e_2, e_3} x))$, $x\in LS(N)$.
Then a moment's calculation shows that $\varphi(e_1), \varphi(e_2), \varphi(e_3)$ are mutually orthogonal and $\varphi(e_1)+ \varphi(e_2)+ \varphi(e_3) = 1_N$.

\begin{lemma}\label{sim}
We have $\varphi(e_1)\sim \varphi(e_2)\sim \varphi(e_3)$ in $N$.
\end{lemma}
\begin{proof}
Subsection \ref{halmos} implies:
For $p, q\in \P(N)$, if $p\vee q = 1$ and $p\wedge q = 0$, then $p^{\perp}\sim q$.
Since $\varphi\circ\Phi$ is a lattice isomorphism, we obtain
\[
\varphi(e_1) = \varphi\circ\Phi(e_1^M) \sim \left(\varphi\circ\Phi\left(\frac{1}{2}\begin{pmatrix}
1&1&0 \\
1&1&0 \\
0&0&2
\end{pmatrix}\right)\right)^{\perp}\sim \varphi\circ\Phi(e_2^M) = \varphi(e_2).
\]
Similarly, we obtain $\varphi(e_1)\sim \varphi(e_3)$.
\end{proof}

It suffices to consider $\varphi\circ \Phi$ instead of $\Phi$.
Hence we may identify $N$ with $\M_3(\hat{N})$ for some von Neumann algebra $\hat{N}$, and we may assume $\Phi(e_1^M)=e_1^N$, $\Phi(e_2^M)=e_2^N$ and $\Phi(e_3^M)=e_3^N$, where
\[
e_1^N :=
\begin{pmatrix}
1&0&0 \\
0&0&0 \\
0&0&0
\end{pmatrix},\,\,
e_2^N :=
\begin{pmatrix}
0&0&0 \\
0&1&0 \\
0&0&0
\end{pmatrix},\,\,
e_3^N :=
\begin{pmatrix}
0&0&0 \\
0&0&0 \\
0&0&1
\end{pmatrix} \in \P(\M_3(\hat{N})).
\]

Let $x\in LS(\hat{M})$.
Suppose that $\hat{M}\subset B(K)$.
Viewing $x$ as a closed operator, we see that the collection
\[
\left\{\begin{pmatrix}
\xi \\
x\xi \\
0
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x\right\}
\]
is a closed subspace in $K\oplus K\oplus K$.
Take the projection $P_{12}[x]\in \P(B(K\oplus K\oplus K))$ onto this subspace.
Then we have
\begin{equation}\label{eq:proj}
P_{12}[x] = \begin{pmatrix}
(1+x^*x)^{-1}&(1+x^*x)^{-1}x^*&0 \\
x(1+x^*x)^{-1}&x(1+x^*x)^{-1}x^*&0 \\
0&0&0
\end{pmatrix}
\end{equation}
and hence $P_{12}[x]\in \P(\M_3(\hat{M}))$.
Similarly, let $P_{13}[x], P_{23}[x]\in \P(\M_3(\hat{M}))$ denote the projections onto
\[
\left\{\begin{pmatrix}
\xi \\
0 \\
x\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x\right\},\quad
\left\{\begin{pmatrix}
0 \\
\xi \\
x\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x\right\},
\]
respectively.

\begin{lemma}\label{P_x}
Let $Q\in \P(\M_3(\hat{M}))$.
Then the following conditions are equivalent:
\begin{enumerate}
\item There exists an $x\in LS(\hat{M})$ such that $Q=P_{12}[x]$.
\item $Q\vee e_2^M = e_1^M\vee e_2^M$, and $Q$ is LS-orthogonal to $e_2^M$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(1)\Rightarrow (2)$
Let $Q=P_{12}[x]$.
Since $(1+x^*x)^{-1}$ is a positive injective operator, we have $Q\vee e_2^M = e_1^M\vee e_2^M$ by \eqref{eq:proj}.
Let $x=v\lvert x\rvert$ be the polar decomposition.
By \eqref{eq:proj}, we have
\[
Q=P_{12}[x] = \begin{pmatrix}
(1+\lvert x\rvert^2)^{-1}&(1+\lvert x\rvert^2)^{-1}\lvert x\rvert v^*&0 \\
v\lvert x\rvert(1+\lvert x\rvert^2)^{-1}&v\lvert x\rvert(1+\lvert x\rvert^2)^{-1}\lvert x\rvert v^*&0 \\
0&0&0
\end{pmatrix}.
\]
Hence we have
\[
Q\wedge e_2^M \leq \begin{pmatrix}
0&0&0 \\
0&v\lvert x\rvert(1+\lvert x\rvert^2)^{-1}\lvert x\rvert v^*&0 \\
0&0&0
\end{pmatrix}.
\]
Since $1-v\lvert x\rvert(1+\lvert x\rvert^2)^{-1}\lvert x\rvert v^*$ is a positive injective operator, we see that $Q\wedge e_2^M=0$.
As in \eqref{eq:halmos}, we may decompose $e_2^M$ and $Q$ in the following form:
\[
e_2^M= 1\oplus 0\oplus 0\oplus
\begin{pmatrix}
1&0\\
0&0
\end{pmatrix},\,\,
Q= 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
a^2&ab \\
ab&b^2
\end{pmatrix}.
\]
We also have
\[
e_1^M= 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
0&0\\
0&1
\end{pmatrix}
\]
with respect to the same decomposition.
Recall that $(1+x^*x)^{-1}$ is invertible in $LS(\hat{M})$, or equivalently,
$e_1^MQe_1^M$ is invertible in $LS(e_1^M M e_1^M)$.
This means that
\[
0\oplus 1\oplus 0\oplus
\begin{pmatrix}
0&0 \\
0&b^2
\end{pmatrix}
\]
is invertible in $LS(e_1^M M e_1^M)$, which in particular implies the invertibility of $b$ in $LS(M_{e_2^M, Q})$.\\
$(2)\Rightarrow (1)$
As in \eqref{eq:halmos}, we may decompose $e_2^M$ and $Q$ in the following form:
\[
e_2^M= 1\oplus 0\oplus 0\oplus
\begin{pmatrix}
1&0\\
0&0
\end{pmatrix},\,\,
Q= 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
a^2&ab \\
ab&b^2
\end{pmatrix}.
\]
Note that $b$ is invertible as a locally measurable operator.
Consider the partial isometry
\[
w= 0\oplus 1\oplus 0\oplus
\begin{pmatrix}
0& 0\\
a&b
\end{pmatrix}.
\]
We have $ww^*=e_1^M$ and $w^*w=Q$.
A moment's reflection shows that there exist $w_1, w_2\in \hat{M}$ such that $w_1\geq 0$, $w_1$ is invertible in $LS(\hat{M})$ and
$w=\begin{pmatrix}
w_1&w_2&0 \\
0&0&0 \\
0&0&0
\end{pmatrix}\in \M_3(\hat{M})$.
Put $x= w_2^*w_1^{-1}$.
Since $ww^*=e_1^M$, we obtain $w_1^2 + w_2w_2^*=1_{\hat{M}}$.
Hence
\[
1+x^*x= 1+ w_1^{-1}w_2w_2^*w_1^{-1} = 1+ w_1^{-1} (1-w_1^2) w_1^{-1} = w_1^{-2}.
\]
It follows by \eqref{eq:proj} that
\[
P_{12}[x] = \begin{pmatrix}
(1+x^*x)^{-1}&(1+x^*x)^{-1}x^*&0 \\
x(1+x^*x)^{-1}&x(1+x^*x)^{-1}x^*&0 \\
0&0&0
\end{pmatrix}
= \begin{pmatrix}
w_1^2&w_1w_2&0 \\
w_2^*w_1&w_2^*w_2&0 \\
0&0&0
\end{pmatrix}= w^*w=Q.
\]
\end{proof}

\begin{corollary}
Let $k\in \{12, 13, 23\}$.
There exists a bijection $\psi_{k}\colon LS(\hat{M})\to LS(\hat{N})$ such that $\Phi(P_k[x]) = P_k[\psi_k(x)]$.
Moreover, $x\in LS(\hat{M})$ is invertible in $LS(\hat{M})$ if and only if $\psi_{k}(x)$ is invertible in $LS(\hat{N})$
\end{corollary}
\begin{proof}
Since $\Phi$ is a lattice isomorphism with $\Phi(e_1^M)= e_1^N$ and  $\Phi(e_2^M)= e_2^N$, the first half of the case $k=12$ follows from the preceding lemma.
For $x\in LS(\hat{M})$, let $P_{21}[x]\in \P(\M_3(\hat{M}))$ denote the projection onto
\[
\left\{\begin{pmatrix}
x\xi \\
\xi \\
0
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x\right\},
\]
thus
\[
P_{21}[x] = \begin{pmatrix}
x(1+x^*x)^{-1}x^*&x(1+x^*x)^{-1}&0 \\
(1+x^*x)^{-1}x^*&(1+x^*x)^{-1}&0 \\
0&0&0
\end{pmatrix}.
\]
It is easy to  see that, for $x, y\in LS(\hat{M})$, the equation $P_{12}[x]=P_{21}[y]$ holds if and only if $x$ is invertible in $LS(\hat{M})$ and $y=x^{-1}$.
Therefore, Lemma \ref{P_x} implies that an operator $x\in LS(\hat{M})$ is invertible in $LS(\hat{M})$ if and only if $P_{12}[x]$ is LS-orthogonal to $e_1^M$ and $P_{12}[x]\vee e_1^M = e_1^M\vee e_2^M$.
Thus $\psi_{12}$ preserves invertibility.
The other cases can be shown similarly.
\end{proof}

In particular, the operators $\psi_{12}(1), \psi_{13}(1)$ are invertible in $LS(\hat{N})$.
Consider the lattice automorphism $\phi$ of $\P(\M_3(\hat{N}))$ determined by
$\phi(l(x)) = l(SxS^{-1})$, where $S= \begin{pmatrix}
1&0&0 \\
0&\psi_{12}(1)^{-1}&0 \\
0&0&\psi_{13}(1)^{-1}
\end{pmatrix}$.
We see that $\phi(e_i^N) = e_i^N$, $i=1, 2, 3$, and
\[
\phi\circ \Phi(P_{12}[1_{\hat{M}}]) = P_{12}[1_{\hat{N}}],\,\, \phi\circ \Phi(P_{13}[1_{\hat{M}}]) = P_{13}[1_{\hat{N}}].
\]
Considering $\phi\circ \Phi$ instead of $\Phi$, we may assume
$\Phi(P_{12}[1_{\hat{M}}]) = P_{12}[1_{\hat{N}}]$ and $\Phi(P_{13}[1_{\hat{M}}]) = P_{13}[1_{\hat{N}}]$,
or equivalently, $\psi_{12}(1)=\psi_{13}(1)=1$.

\begin{lemma}\label{multiplicative}
For any $x, y\in LS(\hat{M})$, we have
\[
P_{13}[xy]=
(P_{23}[-x] \vee P_{12}[y])\wedge (e_1^M\vee e_3^M).
\]
\end{lemma}
\begin{proof}
Let $\hat{M}\subset B(K)$.
We know that the range of $P_{23}[-x] \vee P_{12}[y]$ is the closure of
\[
V:=\left\{\begin{pmatrix}
\eta \\
\xi+ y\eta \\
-x\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x,\,\, \eta\in \operatorname{dom}y\right\}.
\]
In particular, we have
$\begin{pmatrix}
\eta \\
0 \\
xy\eta
\end{pmatrix}\in V$ for any $\eta\in \operatorname{dom} y$ with $y\eta\in \operatorname{dom} x$.
Since the collection $\{\eta\in \operatorname{dom} y\mid y\eta\in \operatorname{dom} x\}$ is a core of the operator $xy\in LS(\hat{M})$, we have $P_{13}[xy]\leq (P_{23}[-x] \vee P_{12}[y])\wedge (e_1^M\vee e_3^M).$

We claim that the orthogonal complement $V^{\perp}$ of $V$ is
\[
\left\{\begin{pmatrix}
-y^*x^*\zeta  \\
x^*\zeta \\
\zeta
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \zeta\in \operatorname{dom}x^*,\,\, x^*\zeta\in \operatorname{dom}y^*\right\}.
\]
It is clear that any $\begin{pmatrix}
-y^*x^*\zeta  \\
x^*\zeta \\
\zeta
\end{pmatrix}$ as above is an element in $V^{\perp}$.
If $\begin{pmatrix}
\zeta_1  \\
\zeta_2 \\
\zeta_3
\end{pmatrix}\in V^{\perp}$, then
\[
0 = \left\langle \begin{pmatrix}
\zeta_1  \\
\zeta_2 \\
\zeta_3
\end{pmatrix}, \begin{pmatrix}
0 \\
\xi \\
-x\xi
\end{pmatrix}\right\rangle = \langle\zeta_2, \xi\rangle - \langle\zeta_3, x\xi\rangle
\]
for any $\xi\in \operatorname{dom}x$, and hence we obtain $\zeta_3\in \operatorname{dom}x^*, \zeta_2= x^*\zeta_3$.
By the equation
\[
0 = \left\langle \begin{pmatrix}
\zeta_1  \\
\zeta_2 \\
\zeta_3
\end{pmatrix}, \begin{pmatrix}
\eta \\
y\eta \\
0
\end{pmatrix}\right\rangle
\]
for $\eta\in \operatorname{dom}y$, we obtain the claim.
Let $\begin{pmatrix}
h_1  \\
h_2 \\
h_3
\end{pmatrix}$ belong to the range of $(P_{23}[-x] \vee P_{12}[y])\wedge (e_1^M\vee e_3^M)$, which is equal to the orthogonal complement of
$V^{\perp}\cup \left\{\begin{pmatrix}
0\\
k\\
0
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, k\in K\right\}$.
Then we have $h_2=0$ and
\[
0 = \left\langle \begin{pmatrix}
h_1  \\
0 \\
h_3
\end{pmatrix}, \begin{pmatrix}
-y^*x^*\zeta  \\
x^*\zeta \\
\zeta
\end{pmatrix}\right\rangle = -\langle h_1, y^*x^*\zeta\rangle + \langle h_3, \zeta\rangle
\]
For any $\zeta\in \operatorname{dom}x^*$ with $x^*\zeta\in \operatorname{dom}y^*$.
We know that $\{\zeta\in \operatorname{dom}x^*\mid x^*\zeta\in \operatorname{dom}y^*\}$ is a core of the operator $y^*x^*\in LS(\hat{M})$.
Thus we obtain $h_1\in \operatorname{dom}(y^*x^*)^* = \operatorname{dom}(xy)$  and $h_3 = (xy) h_1$ (here we view $xy$ as a closed operator in $LS(\hat{M})$).
\end{proof}

\begin{lemma}
We have $\psi_{12}=\psi_{13}=\psi_{23}=: \psi$.
Moreover, $\psi\colon LS(\hat{M}) \to LS(\hat{N})$ is multiplicative.
\end{lemma}
\begin{proof}
Let $x, y\in LS(\hat{M})$.
By the preceding lemma, we have
\[
P_{13}[xy]=
(P_{23}[-x] \vee P_{12}[y])\wedge (e_1^M\vee e_3^M)
\]
and hence
\[
\begin{split}
P_{13}[\psi_{13}(xy)]=\Phi(P_{13}[xy]) &=
\Phi\left((P_{23}[-x] \vee P_{12}[y])\wedge (e_1^M\vee e_3^M)\right)\\
&=
(\Phi(P_{23}[-x]) \vee \Phi(P_{12}[y]))\wedge (\Phi(e_1^M)\vee \Phi(e_3^M))\\
&=
(P_{23}[\psi_{23}(-x)] \vee P_{12}[\psi_{12}(y)])\wedge (e_1^N\vee e_3^N).
\end{split}
\]
It follows by the preceding lemma again (applied to $N$ instead of $M$) that
\[
(P_{23}[\psi_{23}(-x)] \vee P_{12}[\psi_{12}(y)])\wedge (e_1^N\vee e_3^N) = P_{13}[-\psi_{23}(-x)\psi_{12}(y)].
\]
Thus we obtain $P_{13}[-\psi_{23}(-x)\psi_{12}(y)]= P_{13}[\psi_{13}(xy)]$, which implies $-\psi_{23}(-x)\psi_{12}(y) = \psi_{13}(xy)$.

In particular,
putting $x=y=1$, we obtain $\psi_{23}(-1)= -1$.
Putting $x=1$, we obtain $-\psi_{23}(-1)\psi_{12}(y) = \psi_{13}(y)$, hence $\psi_{12}(y) = \psi_{13}(y)$.
Moreover, putting $y=1$, we obtain $-\psi_{23}(-x)\psi_{12}(1) = \psi_{13}(x)$, hence $-\psi_{23}(-x) =\psi_{13}(x)$.
Thus $\psi_{12}(x) \psi_{12}(y) = -\psi_{23}(-x)\psi_{12}(y) = \psi_{13}(xy) =\psi_{12}(xy)$.
Therefore, $\psi_{12}$ is multiplicative.
It follows that $\psi_{12}(-1)$ is central in $LS(\hat{N})$, $\psi_{12}(-1)^2=1$ and $\psi_{12}(-1)y\neq y$ for any $y\neq 0$, and hence we obtain $\psi_{12}(-1)=-1$.
We reach the equation $\psi_{13}=\psi_{12}=\psi_{23}$.
\end{proof}

\begin{lemma}
The mapping $\psi$ is additive.
\end{lemma}
\begin{proof}
Let $x, y\in LS(\hat{M})$.
Consider the projections
\[
f= (P_{12}[x] \vee e_3^M) \wedge (P_{13}[1] \vee e_2^M)\quad\text{and}\quad
g= (P_{12}[y] \vee P_{13}[1]) \wedge (e_2^M\vee e_3^M).
\]
By an argument similar to that in the proof of Lemma \ref{multiplicative}, we can check the following:
The range of $f$ is equal to
\[
\left\{\begin{pmatrix}
\xi \\
x\xi\\
\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x\right\}
\]
and the range of $g$ is equal to
\[
\left\{\begin{pmatrix}
0 \\
-y\eta\\
\eta
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \eta\in \operatorname{dom}y\right\},
\]
hence $(f\vee g) \wedge (e_1^M\vee e_2^M) = P_{12}[x+y]$.
Apply $\Phi$ to both sides to obtain the desired conclusion.
\end{proof}

We define a mapping $\Psi\colon LS(\M_3(\hat{M}))\to LS(\M_3(\hat{N}))$ by $\Psi((x_{ij})_{ij}) := (\psi(x_{ij}))_{ij}$, $x_{ij}\in LS(\hat{M})$, $i, j = 1, 2, 3$.
The preceding lemmas implies that $\Psi$ is a ring isomorphism from $LS(\M_3(\hat{M}))$ onto $LS(\M_3(\hat{N}))$.

\begin{lemma}
We have $\Phi(l(x)) = l(\Psi(x))$ for any $x\in LS(\M_3(\hat{M}))$.
\end{lemma}
\begin{proof}
We partly imitate Dye's argument in the proof of \cite[Lemma 7]{Dy}.
By Lemma \ref{id}, it suffices to show that the lattice isomorphism $\Phi'\colon \P(\M_3(\hat{M})) \to \P(\M_3(\hat{N}))$ determined by $l(\Psi(x)) = \Phi'(l(x))$, $x\in LS(\M_3(\hat{M}))$, satisfies $\Phi=\Phi'$.
For $x\in LS(\hat{M})$, we have
\[
\begin{split}
\Phi(P_{12}[x]) = P_{12}[\psi(x)] &= l\begin{pmatrix}
1&0&0 \\
\psi(x)&0&0 \\
0&0&0
\end{pmatrix}\\
&= l\left(\Psi \begin{pmatrix}
1&0&0 \\
x&0&0 \\
0&0&0
\end{pmatrix}\right)
= \Phi'\left(l\begin{pmatrix}
1&0&0 \\
x&0&0 \\
0&0&0
\end{pmatrix}\right)= \Phi'(P_{12}[x]).
\end{split}
\]
Similarly, we see that $\Phi(p)=\Phi'(p)$ for any $p\in \{P_k[x] \mid x\in LS(\hat{M}),\,\, k=12,\,23,\,13\}$.

Let $x_2, x_3\in LS(\hat{M})$.
We consider the projection $P_{x_2, x_3}\in \P(\M_3(\hat{M}))$ onto the closed  subspace
\[
\left\{\begin{pmatrix}
\xi \\
x_2\xi \\
x_3\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in \operatorname{dom}x_2\cap \operatorname{dom}x_3\right\}.
\]
It is not difficult to see that this projection is equal to $(P_{12}[x_2] \vee e_3^M)\wedge (P_{13}[x_3] \vee e_2^M)$.
It follows that $\Phi(P_{x_2, x_3}) = \Phi'(P_{x_2, x_3})$.

Consider an arbitrary nonzero projection $p = (p_{i, j})_{1\leq i, j\leq 3}\in \P(\M_3(\hat{M}))$.
By Zorn's lemma, to show that $\Phi(p) = \Phi'(p)$, it suffices to find a nonzero subprojection $(p\geq)\,\, q\in \P(\M_3(\hat{M}))$ such that $\Phi(q) = \Phi'(q)$.
Note that $p_{ii} = \sum_{1\leq k\leq 3} p_{ik} p_{ik}^*$, hence we see that $p_{ii}\neq 0$ for some $i\in \{1, 2, 3\}$.

If $p_{11}\neq 0$, put $e:= \chi_{(\lVert p_{11}\rVert/2, \lVert p_{11}\rVert]}(p_{11})\in \P(\hat{M})\setminus \{0\}$ and $x_1:= p_{11}^{-1}e\in \hat{M}$.
It follows that the projection $q\in \P(\M_3(\hat{M}))$ onto the subspace
\[
\left\{\begin{pmatrix}
p_{11}\xi \\
p_{21}\xi \\
p_{31}\xi
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \xi\in eK\right\} = \left\{\begin{pmatrix}
\eta \\
p_{21}x_1\eta \\
p_{31}x_1\eta
\end{pmatrix}\in K\oplus K\oplus K \,\middle|\, \eta\in eK\right\}
\]
is a nonzero subprojection of $p$.
Since $q = P_{p_{21}x_1, p_{31}x_1} \wedge ((P_{12}[e^{\perp}]\wedge e_1^M)\vee e_2^M \vee e_3^M)$, we obtain $\Phi(q) = \Phi'(q)$.

If $p_{11}= 0$ and $p_{22}\neq 0$, we have $p\leq e_2^M\vee e_3^M$.
Then a similar discussion applies.
If $p_{11}=p_{22} = 0$, then $p_{33}\in \P(\hat{M})$.
Use the equation
$(P_{13}[1]\vee P_{13}[p_{33}^{\perp}])\wedge e_3^M= p$, which can be verified easily, to obtain the desired conclusion.
\end{proof}

Therefore, the proof of Theorem A is complete in the case $M$ has order $3$.
The same discussion with a slight modification is valid in any case $M$ has order $n$ with $3\leq n<\infty$.
We know that a projection lattice isomorphism preserves central projections because a projection $p$ in a von Neumann algebra $M$ is central if and only if $\{q\in \P(M)\mid p\vee q=1, p\wedge q=0\} =\{p^{\perp}\}$.
Since every von Neumann algebra without type I$_1$ and I$_2$ direct summands decomposes into the direct sum of algebras of order $3\leq n<\infty$, now it easy to complete the proof of Theorem A in the general case.\medskip

In what follows, let us give a proof of Theorem \ref{dye} by Dye (in the case the von Neumann algebras are without commutative direct summands) as an application of Theorem A.
The proof below is partly based on Feldman's argument in \cite[Proof of Theorem 3]{Fe}.

Let $M$ and $N$ be von Neumann algebra without type I$_1$ and I$_2$ direct summands and suppose that $\Phi\colon \P(M)\to\P(N)$ is a lattice isomorphism.
Suppose further that we have $pq = 0$ if and only if $\Phi(p)\Phi(q)=0$ for any pair $p, q\in \P(M)$.
By Theorem A, there exists a unique ring isomorphism $\Psi\colon LS(M)\to LS(N)$ such that $\Phi(l(x))=l(\Psi(x))$ for any $x\in LS(M)$.

Then we have $\Psi(p)=\Phi(p)\in \P(N)$ for every $p\in\P(M)$.
Indeed, since $p^2=p$ and $pp^{\perp}=0$, we have $\Psi(p)^2=\Psi(p)$ and $\Psi(p)\Psi(p^{\perp}) = 0$.
Thus we have $r(\Psi(p))l(\Psi(p^{\perp})) = 0$.
Our assumption implies $l(\Psi(p^{\perp})) =\Phi(p^{\perp})=\Phi(p)^{\perp}= l(\Psi(p))^{\perp}$, thus we obtain $r(\Psi(p))\leq l(\Psi(p))$.
By the equation $(l(\Psi(p))-\Psi(p))\Psi(p)=0$, we obtain $0=(l(\Psi(p))-\Psi(p))l(\Psi(p)) = l(\Psi(p))-\Psi(p)$.
Hence $\Psi(p)=l(\Psi(p)) =\Phi(p)\in \P(N)$.

Consider the ring automorphism $x\mapsto \Psi^{-1}(\Psi(x^*)^*)$ of $LS(M)$.
This fixes every projection, hence Lemma \ref{id} implies that $x= \Psi^{-1}(\Psi(x^*)^*)$, or equivalently, $\Psi(x)^*=\Psi(x^*)$ for each $x\in LS(M)$.
It follows that $\Psi$ maps the self-adjoint part of $LS(M)$ onto that of $LS(N)$. Since $\Psi$ preserves squares, $\Psi$ restricted to self-adjoint parts preserves order in both directions.
Since $\Psi(1)=1$, $\Psi$ restricts to a real $^*$-isomorphism from $M$ onto $N$ and extends $\Phi$, which is the desired conclusion.

\section{Ring isomorphisms of locally measurable operator algebras}\label{extra}
By the preceding section, lattice isomorphisms between projection lattices are in one-to-one correspondence with ring isomorphisms between the algebras of locally measurable operators.
Hence the following question is well motivated.
\begin{q}
Let $M$, $N$ be von Neumann algebras.
What is the general form of ring isomorphisms from $LS(M)$ onto $LS(N)$?
\end{q}

\begin{lemma}
Let $M, N$ be general von Neumann algebras.
Let
\[
\begin{split}
M&=\left(\bigoplus_{n\geq 1}M_{\mathrm{I}_n}\right) \oplus M_{\mathrm{I}_{\infty}} \oplus M_{\mathrm{II}_1} \oplus M_{\mathrm{II}_{\infty}} \oplus M_{\mathrm{III}},\\
N&=\left(\bigoplus_{n\geq 1}N_{\mathrm{I}_n}\right) \oplus N_{\mathrm{I}_{\infty}} \oplus N_{\mathrm{II}_1} \oplus N_{\mathrm{II}_{\infty}} \oplus N_{\mathrm{III}}
\end{split}
\]
be the type decompositions, where $M_j$, $N_j$ are von Neumann algebras of type $j$.
Suppose that $\Psi\colon LS(M)\to LS(N)$ is a ring isomorphism.
Then there exist ring isomorphisms $\psi_j\colon LS(M_j)\to LS(N_j)$ such that $\Psi(x) =\psi_j(x)$ for any $x\in LS(M_j)\,\,(\subset LS(M))$.
\end{lemma}
\begin{proof}
It is easy to see that $\Psi$ maps the collection of central projections in $M$ onto that in $N$.
Hence it suffices to show: If $M$, $N$ are of type $j, k\in \{\mathrm{I}_n\mid n\geq 1\}\cup \{\mathrm{I}_{\infty}, \mathrm{II}_1, \mathrm{II}_{\infty}, \mathrm{III}\}$, respectively, then $j=k$.
We consider the lattice isomorphism $\Phi\colon \P(M)\to\P(N)$ as in Proposition \ref{converse}.
It is easy to see that a projection $p\in \P(M)$ is abelian (namely, $pMp$ is an abelian von Neumann algebra) if and only if $\Phi(p)$ is abelian.
Moreover, a projection $p\in \P(M)$ is finite if and only if $\Phi(p)\in \P(N)$ is finite.
Indeed, if $p$ is not finite, then there exist mutually orthogonal nonzero subprojections $p_1, p_2, p_3$ of $p$ such that $p_1\sim p_2\sim p_3\sim p_1+p_2$.
The same argument as in the proof of Lemma \ref{sim} implies $\Phi(p_1)\sim\Phi(p_3)\sim \Phi(p_1+p_2)$, which shows that $\Phi(p)$ is not finite.
Similarly, if $\Phi(p)$ is not finite, then $p$ is not finite.
The rest of the proof is a standard argument of von Neumann algebra theory, and we omit the details.
See e.g.\ \cite[Chapter 6]{KR}.
\end{proof}

Therefore, Question reduces to the case both $M$ and $N$ are of type $j$, $j\in \{\mathrm{I}_n\mid n\geq 1\}\cup \{\mathrm{I}_{\infty}, \mathrm{II}_1, \mathrm{II}_{\infty}, \mathrm{III}\}$.

We first consider Question in the case $M, N$ are von Neumann algebras of type I$_n$.
Suppose that $LS(M)$ is ring isomorphic to $LS(N)$.
Since the central projection lattices of $M$ and $N$ are lattice isomorphic, we see that the center of $M$ is $^*$-isomorphic to that of $N$.
Hence there exists a commutative von Neumann algebra $A$ such that $M\cong N\cong \M_n (A)$.
Therefore, it suffices to think about ring automorphisms of $LS(\M_n(A))$, which can be identified with the collection of all $n \times n$ matrices with entries in $LS(A)$.
Note that $A$ can be identified with the algebra $L^{\infty}(\mu)$ of all complex-valued essentially bounded measurable functions (modulo almost-everywhere equivalence) for some measure $\mu$.
Then $LS(A)$ corresponds to $L^0(\mu)$, which denotes the collection of all complex-valued measurable functions.
Remark that any ring automorphism $\psi$ of $LS(A)$ determines a ring automorphism $\psi'$ of $LS(\M_n(A))$ by the formula $\psi'((x_{ij})) = (\psi(x_{ij}))_{ij}$.
The following proposition slightly generalizes (but can be shown by exactly the same argument as in) \cite[Theorem 3.3]{AAKD} by Albeverio, Ayupov, Kudaybergenov and  Djumamuratov.
\begin{proposition}\label{I}
Let $n\geq 1$ be an integer and $A$ be a commutative von Neumann algebra.
Suppose that $\Psi$ is a ring automorphism of $LS(\M_n(A))$.
Then there exist a ring automorphism $\psi\colon LS(A)\to LS(A)$ and an invertible element $y$ in $LS(\M_n(A))$ such that $\Psi(x)=y\psi'(x)y^{-1}$, $x\in LS(\M_n(A))$.
\end{proposition}
\begin{proof}
Note that $\Psi$ restricts to a ring automorphism $\psi$ of the center of $LS(\M_n(A))$, which is canonically isomorphic to $LS(A)$.
Then $\Psi\circ {\psi'}^{-1}$ fixes every element in the center of $LS(\M_n(A))$.
We may apply \cite[Theorem 3.1]{AAKD} to obtain the desired conclusion.
\end{proof}

There exist highly nontrivial examples of ring automorphisms of $LS(A) = L^0(\mu)$ for a commutative von Neumann algebra $A$.
For example, consider the case $A=\C=LS(A)$.
There are many ring automorphisms of $\C$ that are far from real-linear.
Consider the case $\mu$ is an atomless measure.
It is known \cite[{(1)$\Leftrightarrow$(6) of Theorem 3.4}]{Ku} (see also \cite[Remark 6.3]{Ku}) that there exists a (complex-linear) algebra automorphism $\psi$ of $L^0(\mu)$ such that $\psi(p)=p$ for any $p\in \P(A)$ but $\psi\neq\operatorname{id}_{L^0(\mu)}$.
It seems that these examples are beyond the scope of the theory of operator algebras.

In contrast, we may give a purely operator algebraic solution to Question for type I$_{\infty}$ or III in the following manner.
This improves \cite[Theorem 3.8]{AAKD}, in which algebra isomorphisms of the case of type I$_\infty$ were considered.
\begin{thmb}
Let $M, N$ be von Neumann algebras of type I$_{\infty}$ or III.
If $\Psi\colon LS(M)\to LS(N)$ is a ring isomorphism, then there exist a real $^*$-isomorphism $\psi\colon M\to N$ (which extends to a real $^*$-isomorphism from $LS(M)$ onto $LS(N)$) and an invertible element $y\in LS(N)$ such that  $\Psi(x)=y\psi(x)y^{-1}$, $x\in LS(M)$.
\end{thmb}
\begin{proof}
Beware of the fact that $\Psi$ restricts to a lattice isomorphism between the central projection lattices of $M$ and $N$.
We first prove:
\begin{itemize}
\item[\underline{Claim}]
There exists an operator $a\in LS(\Z(N))_+$ such that $\vertiii{\Psi(x)}\leq a$ for any $x\in M\,\,(\subset LS(M))$ with $\lVert x\rVert\leq 1$.
\end{itemize}
Assume that this claim does not hold.
We will obtain a contradiction in Step 4.\smallskip

\noindent
\underline{Step 1}
We prove that there exists a central projection $e$ in $M$ such that for any $n\geq 1$ there exists some $x\in M$ with $\lVert x\rVert \leq 1$ and $\vertiii{\Psi(x)}\geq n\Psi(e)$.

Assume for a while that the center $\Z(M)$ of $M$ admits a faithful normal state $\tau\colon \Z(M)\to \C$.
For each positive integer $n$, consider the collection
\[
{E}_n := \{ e\in \P(\Z(M))\mid \text{ there exists }x\in M \text{ with } \lVert x\rVert \leq 1 \text{ and } \vertiii{\Psi(x)}\geq n\Psi(e)\}.
\]
Suppose that $e, f$ belong to this collection.
Take $x, y \in M$ such that $\lVert x\rVert,  \lVert y\rVert\leq 1$ and $\vertiii{\Psi(x)}\geq n\Psi(e)$, $\vertiii{\Psi(y)}\geq n\Psi(f)$.
Then the element $x':=xe+ye^{\perp}$ satisfies $\lVert x'\rVert\leq 1$ and
\[
\begin{split}
\vertiii{\Psi(x')} &= \vertiii{\Psi(xe+ye^{\perp})} \\
&= \vertiii{\Psi(x)\Psi(e)+\Psi(y)\Psi(e)^{\perp}}\\
&=\vertiii{\Psi(x)}\Psi(e) + \vertiii{\Psi(y)}\Psi(e)^{\perp}\\
&\geq n\Psi(e) + n\Psi(f)\Psi(e)^{\perp}\\
&= n\Psi(e)\vee\Psi(f) = n\Psi(e\vee f).
\end{split}
\]
Hence we have $e\vee f\in {E}_n$, which implies that ${E}_n$ is upward directed.
Put $c_n:= \sup\{\tau(e)\mid e\in {E}_n\}$.
We may take an increasing sequence $\{e^{(k)}\}\subset {E}_n$ such that $\tau(e^{(k)})\to c_n$ as $k\to\infty$.
For each $k$ take $x^{(k)} \in M$ such that $\lVert x^{(k)}\rVert\leq 1$ and $\vertiii{\Psi(x^{(k)})}\geq n\Psi(e^{(k)})$.
Some calculations show that the element
\[
x'':= x^{(1)}e^{(1)} + \sum_{k\geq 2} x^{(k)}(e^{(k)}-e^{(k-1)})\in M
\]
satisfies $\lVert x''\rVert\leq 1$ and $\vertiii{\Psi(x'')}\geq n\Psi(e^{(k)})$ for every $k$.
This implies that for the projection $e_n := \bigvee{E}_n\in \P(\Z(M))$ there exists  $x_n\in \P(\Z(M))$ such that $\lVert x_n\rVert\leq 1$ and $\vertiii{\Psi(x_n)}\geq n\Psi(e_n)$.

Clearly, $\{e_n\}$ is a decreasing sequence.
Assume that $e_n\to 0$ as $n\to \infty$, then the element $a=\Psi(1+\sum_{n\geq 1}e_n)\in LS(\Z(N))_+$ satisfies the property of Claim, which contradicts our assumption.
Hence we have $e_n\to e\in \P(\Z(M))\setminus\{0\}$ as $n\to \infty$, and $e$ satisfies the desired property.
Since every von Neumann algebra can be decomposed into the direct sum of von Neumann algebras whose centers admit faithful normal states, the same holds for arbitrary $M$ and $N$.\smallskip

Considering the restriction of $\Psi$ to a ring isomorphism from $LS(Me)$ onto $LS(N\Psi(e))$, we may assume that for any $n\geq 1$ there exists some $x\in M$ with $\lVert x\rVert \leq 1$ and $\vertiii{\Psi(x)}\geq n$.\smallskip

\noindent\underline{Step 2}
We prove that for any $a\in LS(\Z(N))_+$ there exists some $x\in M$ with $\lVert x\rVert \leq 1$ and $\vertiii{\Psi(x)}\geq a$.

Let $a\in LS(\Z(N))_+$.
We may take a sequence of mutually orthogonal central projections $\{ f_n \}$ such that
$a\leq \sum_{n\geq 1}nf_n$.
For each $n$, take $x_n \in M$ such that $\lVert x_n\rVert\leq 1$ and $\vertiii{\Psi(x_n)}\geq nf_n$.
Some calculations show that the element $x:=  \sum_{n\geq 1} x_n\Psi^{-1}(f_n)$
satisfies $\lVert x\rVert\leq 1$ and $\vertiii{\Psi(x)}\geq \sum_{n\geq 1} nf_n \geq a$.\smallskip

\noindent\underline{Step 3}
We prove: For any $p\in \P(M)$ with $p\sim p^{\perp}$ and any $a\in LS(\Z(N))_+$, there exists an element $x\in M$ with $pxp=x$, $\lVert x\rVert\leq 1$ and $\vertiii{\Psi(x)}\geq a$.

Take a partial isometry $v\in M$ such that $vv^*=p$ and $v^*v=p^{\perp}$.
Since $\Psi$ is a ring isomorphism, for any $x\in M$, we have
\[
\begin{split}
\Psi(x) &= \Psi(pxp+pxp^{\perp}+p^{\perp}xp+p^{\perp}xp^{\perp})\\
&= \Psi(pxp)+\Psi(pxv^*)\Psi(v)+\Psi(v^*)\Psi(vxp)+\Psi(v^*)\Psi(vxv^*)\Psi(v).
\end{split}
\]
For a given $a\in LS(\Z(N))_+$, put
\[
b:= 4a+4a \vertiii{\Psi(v)}+ 4a \vertiii{\Psi(v^*)}+ 4a\vertiii{\Psi(v)}
\vertiii{\Psi(v^*)}\in LS(\Z(N))_+.
\]
The preceding step implies there exists $x\in M$ with $\lVert x\rVert\leq 1$ and
\[
\begin{split}
b&\leq \vertiii{\Psi(x)}\\
&\leq \vertiii{\Psi(pxp)}+\vertiii{\Psi(pxv^*)}\vertiii{\Psi(v)}+\vertiii{\Psi(v^*)}\vertiii{\Psi(vxp)}+\vertiii{\Psi(v^*)}\vertiii{\Psi(vxv^*)}\vertiii{\Psi(v)}.
\end{split}
\]
Hence there exists a quadruple $f_1, f_2, f_3, f_4$ of central projections in $N$ such that $f_1+f_2+f_3+f_4=1$, $\vertiii{\Psi(pxp)}f_1\geq bf_1/4$, $\vertiii{\Psi(pxv^*)}\vertiii{\Psi(v)}f_2\geq bf_2/4$, $\vertiii{\Psi(v^*)}\vertiii{\Psi(vxp)}f_3\geq bf_3/4$ and $\vertiii{\Psi(v^*)}\vertiii{\Psi(vxv^*)}\vertiii{\Psi(v)}f_4\geq bf_4/4$.
Put
\[
x' := pxp\Psi^{-1}(f_1) + pxv^*\Psi^{-1}(f_2) + vxp\Psi^{-1}(f_3) + vxv^*\Psi^{-1}(f_4).
\]
Then we have $px'p=x'$, $\lVert x'\rVert \leq 1$ and
\[
\begin{split}
\vertiii{\Psi(x')}&= \vertiii{\Psi(pxp)f_1 + \Psi(pxv^*)f_2 + \Psi(vxp)f_3 + \Psi(vxv^*)f_4}\\
&= \vertiii{\Psi(pxp)}f_1 + \vertiii{\Psi(pxv^*)}f_2 + \vertiii{\Psi(vxp)}f_3 +\vertiii{\Psi(vxv^*)}f_4\\
&\geq \frac{1}{4}b (f_1 + \vertiii{\Psi(v)}^{-1}f_2 + \vertiii{\Psi(v^*)}^{-1}f_3 + \vertiii{\Psi(v)}^{-1}\vertiii{\Psi(v^*)}^{-1}f_4)\geq a.
\end{split}
\]
(Note that $\vertiii{\Psi(v)}$, $\vertiii{\Psi(v^*)}$ are invertible in $LS(\Z(N))$.)
\smallskip

\noindent\underline{Step 4}
Since $M$ is properly infinite, we may take a sequence $\{p_n\}_{n\geq 1}$ of mutually orthogonal projections in $M$ such that $p_n\sim p_n^{\perp}$, $n\geq 1$.
By Step 3, for each $n\geq 1$, we may take an element $x_n\in M$ with $p_nx_np_n=x_n$, $\lVert x_n\rVert\leq1$ and $\vertiii{\Psi(x_n)}\geq n\vertiii{\Psi(p_n)}$.
Put $x:=\sum_{n\geq 1} x_n\in M$ (which is well-defined since $p_n$, $n\geq 1$, are mutually orthogonal).
For every $n\geq 1$, we have
\[
\vertiii{\Psi(x)}\vertiii{\Psi(p_n)}\geq\vertiii{\Psi(x)\Psi(p_n)}=\vertiii{\Psi(xp_n)} = \vertiii{\Psi(x_n)}\geq n\vertiii{\Psi(p_n)}.
\]
Since $\vertiii{\Psi(p_n)}$ is invertible in $LS(\Z(N))$, we obtain $\vertiii{\Psi(x)}\geq n$ for all $n\in \N$, a contradiction.
This completes the proof of Claim.\smallskip

\noindent\underline{Step 5}
It follows that there exists an element $a\in LS(\Z(N))_+$ such that $\vertiii{\Psi(x)}\leq a$ for any $x\in M$ with $\lVert x\rVert\leq 1$.
By the same discussion applied to $\Psi^{-1}$, we also obtain an element $a'\in LS(\Z(M))_+$ such that $\vertiii{\Psi^{-1}(y)}\leq a'$  for any $y\in N$ with $\lVert y \rVert \leq 1$.
We may take a sequence $e_n$ of central projections in $M$ such that $e_n\nearrow 1$ and $\Psi$ restricts to a norm-bicontinuous ring isomorphism $\Psi_n$ from $M e_n$ onto $N\Psi(e_n)$, $n\geq 1$.
By Lemma \ref{isoms} we may verify the statement for each $\Psi_n$, which suffices to complete the proof.
\end{proof}

\begin{corollary}
Let $M, N$ be von Neumann algebras of type I$_{\infty}$ or III.
Suppose that $\Phi\colon \P(M)\to \P(N)$ is a lattice isomorphism.
Then there exist a real $^*$-isomorphism $\psi\colon M\to N$ and an invertible element $y\in LS(N)$ such that $\Phi(p)=l(y\psi(p))$, $p\in \P(M)$.
\end{corollary}

\section{Questions}\label{question}
The author skeptically conjectures that the same as Theorem B holds for type II von Neumann algebras:
\begin{conjecture}
Let $M$ and $N$ be von Neumann algebras of type II.
Suppose that $\Psi\colon LS(M)\to LS(N)$ is a ring isomorphism.
Then there exist an invertible operator $y\in LS(N)$ and a real $^*$-isomorphism $\psi\colon M\to N$ such that $\Psi(x) = y\psi(x)y^{-1}$ for any $x\in LS(M)$.
\end{conjecture}

Not much is known about the structure of the algebra $LS(M)$ for a type II (in particular, II$_1$) von Neumann algebra $M$.
The author does not know whether or not such a $\Psi$ is automatically real-linear even in the case $M$ and $N$ are (say, approximately finite dimensional) II$_1$ factors.
Note that $LS(M)$ cannot have a Banach algebra structure because of the fact that an element of $LS(M)$ can have an empty or dense spectral set.
Hence it seems to be difficult to make use of \emph{automatic continuity} results on algebra isomorphisms as in \cite{Da}.
However, the author suspects that at least the following weaker statement holds:
\begin{conjecture}
Let $M$ and $N$ be von Neumann algebras of type II.
If $\P(M)$ and $\P(N)$ are lattice isomorphic, or equivalently, if $LS(M)$ and $LS(N)$ are ring isomorphic, then $M$ and $N$ are real $^*$-isomorphic (or equivalently, $M$ and $N$ are Jordan $^*$-isomorphic).
\end{conjecture}

In another direction, we compare Theorem A with von Neumann's theory of complemented modular lattices and regular rings.
Von Neumann axiomatized projection lattices of type II$_1$ von Neumann algebras, and completed the amazing theory on the correspondence between the vast classes of complemented modular lattices and regular rings.
Let us briefly recall this theory in \cite[Part II]{N}.
\begin{definition}
A lattice $L$ with greatest element $1$ and least element $0$ is \emph{complemented} if for each $a\in L$ there exists $b\in L$ such that $a\vee b=1$, $a\wedge b=0$.
A lattice $L$ is \emph{modular} if the equation $(a\vee b)\wedge c=a\vee (b\wedge c)$ holds for any $a, b, c\in L$ with $a\leq c$.

Let $L$ be a complemented modular lattice.
Two elements $a, b\in L$ are said to be \emph{perspective} if there exists $c\in L$ such that $a\vee c=1=b\vee c$ and $a\wedge c=0=b\wedge c$.
Let $n$ be a positive integer.
We say $L$ has \emph{order $n$} if there exist pairwise perspective elements $a_1, a_2, \ldots, a_n\in L$ with $a_1\vee a_2\vee\cdots\vee a_n=1$ and $\bigl(\bigvee_{i\in I_1}a_i\bigr) \wedge \bigl(\bigvee_{j\in I_2}a_j\bigr)=0$ for any disjoint $I_1, I_2\subset \{1, 2, \ldots, n\}$.
\end{definition}

\begin{definition}
A (\emph{von Neumann}) \emph{regular ring} is a ring $R$ with unit such that for each $x\in R$ there exists $y\in R$ such that $xyx=x$.

Let $R$ be a regular ring.
A right ideal $\mathfrak{a}$ of $R$ is \emph{principal} if it is generated by one element of $R$.
\end{definition}
Let $M$ be a von Neumann algebra.
Then $\P(M)$ is a complemented lattice.
It is not difficult to show that the following three conditions are equivalent.
\begin{itemize}
\item The von Neumann algebra $M$ is finite.
\item The lattice $\P(M)$ is modular.
\item The ring $LS(M)$ is regular.
\end{itemize}

\begin{theorem}[von Neumann]
If $R$ is a regular ring, then the collection $L$ of all principal right ideals of $R$, ordered by inclusion, forms a complemented modular lattice.
\end{theorem}
We call $L$ in the statement of the preceding theorem the \emph{right ideal lattice of $R$}.
\begin{theorem}[von Neumann]
Let $R_1, R_2$ be regular rings with right ideal lattices $L_1, L_2$, respectively. Suppose that $L_1$ has order $n\geq 3$. If $\Phi\colon L_1\to L_2$ is a lattice isomorphism, then there exists a unique ring isomorphism $\Psi\colon R_1\to R_2$ such that $\Phi(\mathfrak{a})=\Psi(\mathfrak{a})$, $\mathfrak{a}\in L_1$.
\end{theorem}
\begin{theorem}[von Neumann]
Let $L$ be a complemented modular lattice with order $n\geq 4$.
Then there exists a regular ring $R$ such that the right ideal lattice of $R$ is lattice isomorphic to $L$.
\end{theorem}
Let $M$ be a finite von Neumann algebra.
Let $\mathfrak{a}\subset LS(M)$ be a principal right ideal generated by $a\in LS(M)$.
It is an easy exercise to show that $\mathfrak{a}=\{x\in LS(M)\mid l(x)\leq l(a)\}$.
Hence we obtain an identification of the right ideal lattice of $LS(M)$ with the projection lattice $\P(M)$.
In particular, Theorem \ref{vN} is a corollary of von Neumann's results above.
See also the article \cite{Go} by Goodearl, which deals with the history of the study of regular rings in connection with functional analysis.

Von Neumann's theory, applied to the setting of von Neumann algebras, is valid only for finite von Neumann algebras.
In this paper, we proved that there exists a complete correspondence between lattice isomorphisms and ring isomorphisms in the general setting of von Neumann algebras.
Hence the author believes that one might be able to generalize von Neumann's theory to a broader class of lattices that covers projection lattices of any von Neumann algebras (of fixed order $n\geq 3$ or $4$).
This is left as a research program in the future.\medskip

\textbf{Acknowledgements} \quad
The author appreciates Yasuyuki Kawahigashi, who is the advisor of the author, for invaluable support.
The author is also indebted to Peter \v{S}emrl and Yuhei Suzuki, who encouraged the author to begin working on lattice isomorphisms.
This work was supported by Leading Graduate Course for Frontiers of Mathematical Sciences and Physics (FMSP) and JSPS Research Fellowship for Young Scientists (KAKENHI Grant Number 19J14689), MEXT, Japan.
Part of this work was completed during the author's visit to Gilles Pisier in IMJ-PRG, Paris.
This visit was supported by JSPS Overseas Challenge Program for Young Researchers.


TITLE: Minimum viscosity of liquids
QUESTION [23 upvotes]: In a lecture by Purcell he mentions that he notices that there aren't any liquids with viscosities much less than that of water, even though they go up seemingly unbounded. In an endnote (endnote 1 in that copy), he mentioned that Weisskopf found a reason, but I haven't been able to find what that reason actually was.
Note that this is not about the AdS-CFT lower bound.

REPLY [4 votes]: If "not much less than water" means "not an order of magnitude lower than water at room temperature" this is probably correct. However there are substances like http://en.wikipedia.org/wiki/Pentane with a viscosity 4 times less than that of water or http://en.wikipedia.org/wiki/Acetone with a viscosity 3 times less than water (at 20 degrees celcius). At lower temperatures (77K) you find liquid nitrogen with an viscosity less than 5 times that of water at room temperature. On the other side at 1K helium becomes a superfluid with effectively zero viscosity http://en.wikipedia.org/wiki/Superfluid_helium-4) In that sense viscosity seems to behave similar to electric resistance, being lower with lower temperature. So without reading Weisskopf I guess it's simply room temperature which provides an lower limit to the viscosity of a fluid.


TITLE: How do you go from insights to formal solutions?
QUESTION [7 upvotes]: My question pertains to the problem of translating ideas and insights to the solution (of some problem) into formal notations, thus expressing the answer formally.
In other words, having a bunch of ideas (pertaining to the solution) and transforming them into a well composed, formal solution is a process which I often find difficult.
It might be because (at times) I am not aware of enough literature in the problem domain and so have the problem of not knowing how to express my ideas. However, I would like to know if this is something that can be improved upon and if so, how?

REPLY [8 votes]: One purpose of writing down a solution to a problem in the first place is that it needs to be utterly convincing to your readers.  If you think you have an argument, try writing it down fully, first in whatever language / notation you wish.  Beware of any instances where you tell yourself things like "It is clear that..." or "The following just has to be true."  Then, ask yourself: does it have any holes in it? Am I hand-waving/cheating anywhere?  Can I fully explain every detail if asked?
If you find that your work cannot withstand such questions, then you still have some more to figure out.  For example, I find that I am not completely sure of my proofs until I try writing them down, and sometimes in the process of writing, I discover problems that I wouldn't have imagined.  If I leave writing until last minute, I often find myself in danger.  On the other hand, if you can address every issue, then you can go back and rewrite your work up formally, using the proper notation and language so that your peers will be willing to read your work.
Unfortunately, there are no shortcuts here either.  It takes reading/presenting a lot of papers until you are familiar with the proper style.  One advantage of getting comfortable with writing in proper style is that after a while, you'll be writing proofs "properly" from the first attempt and won't have to do the rewriting step.


text:
\section{Normal to Cycloid passes through Bottom of Generating Circle}
Tags: Cycloids, Normal to Cycloid passes through Bottom of Generating Circle

\begin{theorem}
Let $C$ be a [[Definition:Cycloid|cycloid]] generated by the equations:
: $x = a \left({\theta - \sin \theta}\right)$
: $y = a \left({1 - \cos \theta}\right)$
Then the [[Definition:Normal to Curve|normal]] to $C$ at a point $P$ on $C$ passes through the bottom of the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$.
\end{theorem}

\begin{proof}
From [[Normal to Cycloid]], the equation for the [[Definition:Normal to Curve|normal]] to $C$ at a point $P = \left({x, y}\right)$ is given by:
{{ProofWanted|We don't even have the definition of a normal to a curve posted up yet, so there's a long way to go here.}}
\end{proof}

\begin{proof}
From [[Tangent to Cycloid passes through Top of Generating Circle]], the [[Definition:Tangent to Curve|tangent]] to $C$ at a point $P = \left({x, y}\right)$ passes through the top of the [[Definition:Generating Circle of Cycloid|generating circle]].
By definition, the [[Definition:Normal to Curve|normal]] to $C$ at $P$ is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$.
From [[Thales' Theorem]], the [[Definition:Normal to Curve|normal]], the [[Definition:Tangent to Curve|tangent]] and the [[Definition:Diameter of Circle|diameter]] of the [[Definition:Generating Circle of Cycloid|generating circle]] form a [[Definition:Right Triangle|right triangle]].
Thus the [[Definition:Normal to Curve|normal]] to $C$ at $P$ meets the [[Definition:Generating Circle of Cycloid|generating circle]] at the opposite end of the [[Definition:Diameter of Circle|diameter]] to the [[Definition:Tangent to Curve|tangent]].
Hence the result.
{{qed}}
\end{proof}

config: wiki