text:
\section*{This is an unnumbered first-level section head}

\section{PRELIMINARIES}
It has been of special interest to study the linear maps between
two
 nonself\-adjoint operator algebras over the last several decades.
 These maps include isomorphisms, isometries and rank preserving
 maps etc. In \cite{MR33:4703}, Ringrose proved that an isomorphism between two
 nest algebras must be spatially implemented by making effective
 use of rank one operators. In \cite{MR87k:47103}, Gilfeather and Moore showed
 that an isomorphism acting between two CSL algebras with
 completely \mbox{distributive} lattices need not be spatially or
 quasi-spatially implemented. But they also showed that if an
 isomorphism preserves the rank of all finite rank operators, then
 it must be quasi-spatially implemented. In \cite{MR17:09095}, Panaia has shown that
 if $\mathfrak{A}_i=\operatorname{Alg}{\mathfrak{L}}_i$ where the $\mathfrak{L}_i$ are finite distributive
 subspaces lattices, then every rank preserving algebraic isomorphism of $\mathfrak{A}_1$ onto
 $\mathfrak{A}_2$ is quasi-spatial. The characterization of
 the rank preserving maps on nest algebras was done by the Chinese
 mathematicians, Shu-Yun Wei and Sheng-Zhao Hou \cite{MR2000m:47099}. Part of their results showed that
some rank preserving maps $\Phi$ on a nest algebra can be
described as $\Phi(T)=ATB^*$ for some $A,B \in B(\mathcal{H})$.
Use of rank one operators was an integral part of their proof. In
this paper, we shall describe the rank preserving maps on
completely distributive commutative subspace lattice algebras
(CDCSL algebras).

In this paper we show that a rank preserving map $\Phi$ on a
completely distributive algebra, CDC algebra, with certain
condition can be described as $\Phi(A)=UAV^*$ for some densely
defined linear transformations $U$ and $V$.

Let $\mathcal{H}$ be a complex separable Hilbert space. A
\emph{subspace lattice} $\mathfrak{L}$ is a strongly closed
lattice of orthogonal projections on $\mathcal{H}$, containing $0$
and $I$. If $\mathfrak{L}$ is a subspace lattice,
$\operatorname{Alg}{\mathfrak{L}}$ denotes the algebra of all bounded operators
on $\mathcal{H}$ that leave invariant every projection in
$\mathfrak{L}$. $\operatorname{Alg}{\mathfrak{L}}$ is a weakly closed subalgebra
of $B(\mathcal{H})$, the algebra of all bounded operators on
$\mathcal{H}$. Dually, if $\mathfrak{A}$ is a subalgebra of
$B(\mathcal{H})$, then $\operatorname{Lat}{\mathfrak{A}}$ is the lattice of all
projections invariant for each operator in $\mathfrak{A}$. An
algebra $\mathfrak{A}$ is reflexive if
$\mathfrak{A}=\operatorname{Alg}\operatorname{Lat}{\mathfrak{A}}$ and a lattice $\mathfrak{L}$
is reflexive if $\mathfrak{L}=\operatorname{Lat}\operatorname{Alg}{\mathfrak{L}}$. A lattice
is a \emph{commutative subspace lattice}, or CSL, if each pair of
projections in $\mathfrak{L}$ commute; $\operatorname{Alg}{\mathfrak{L}}$ is
then called a CSL algebra. All lattices in this paper will be
commutative.

In \cite{MR51:1420}, Arveson showed that every commutative
subspace lattice is reflexive. A totally ordered (and hence
commutative) subspace lattice is a nest and the associated algebra
is a nest algebra. We use the convention that a subspace is
identified with the orthogonal projection onto the subspace. Thus
$E \subseteq F$ is the same as $E \leq F$.

We have two operations between subspaces: meet($\wedge$) and
join($\vee$).

Given any family $\{E_{\alpha}\}_{\alpha \in I}$ of subspaces of a
Hilbert space $\mathcal{H}$, $\wedge_{\alpha\in I}E_{\alpha}$
denotes the greatest subspace contained in each $E_{\alpha}$ and
$\vee_{\alpha\in I}E_{\alpha}$ denotes the smallest subspace
containing each $E_{\alpha}$.

We now restrict our attention to a special type of lattice:
completely distributive lattice. One of the advantages of
completely distributive lattices over other lattices is the
abundance of the rank one operators in the associated algebra;
these operators will be building blocks in the subsequent
discussion. A \emph{completely distributive lattice} is a complete
lattice which satisfies distributive laws expressed for families
of arbitrary cardinality. To be more precise, let $\mathfrak{L}$
be a complete lattice and $I$ be an arbitrary index set. For each
$\alpha \in I$, let $J_{\alpha}$ also be an arbitrary index set,
and for each $\beta \in J_{\alpha}$, let $E_{\beta \alpha}$ denote
an element of $\mathfrak{L}$. $\Pi$ will denote the Cartesian
product of all the $J_{\alpha}$, i.e. the collection of all choice
functions $\varphi:I \rightarrow\cup_{\alpha \in I}J_{\alpha}$
satisfying $\varphi(\alpha) \in J_{\alpha}$, for all $\alpha$.
$\mathfrak{L}$ is completely distributive if it satisfies the
following two identities.

\begin{enumerate}
\item $\wedge_{\alpha \in I}(\vee_{\beta \in
J_{\alpha}}E_{\beta\alpha})=\vee_{\varphi \in \Pi}(\wedge_{\alpha
\in I}E_{\varphi(\alpha)\alpha})$

\item $\vee_{\alpha \in I}(\wedge_{\beta \in
J_{\alpha}}E_{\beta\alpha})=\wedge_{\varphi \in \Pi}(\vee_{\alpha
\in I}E_{\varphi(\alpha)\alpha})$
\end{enumerate}

If $\mathfrak{L}$ is completely distributive and commutative, we
will call $\operatorname{Alg}{\mathfrak{L}}$ a CDC algebra. For further
discussion about the characteristics of CDC algebra, we need to
introduce the notion of rank one operators. We will let $x\otimes
y^{\ast}$ denote the rank one operator defined on $\mathcal{H}$ by
$(x\otimes y^{\ast})(f)=\langle f,y \rangle x$.

 There are plenty of rank one operators and finite rank
operators in a nest algebra. Moreover, these operators have played
a central role in the theory of nest algebra. However, there are
many examples of commutative subspace lattices whose corresponding
algebras do not contain any rank one operators \cite{MR84h:47048}.
But as we discussed earlier, the complete distributivity of a
lattice guarantees a reasonable supply of rank one operators.
Furthermore, one of the surprising characteristics of the CDC
algebra is stated in the following lemma. This lemma is due to
Laurie and Longstaff \cite{MR85b:47052}.

\begin{lemma}\label{wk}
Let $\mathfrak{L}$ be a commutative completely distributive
subspace lattice on a separable Hilbert space $\mathcal{H}$. Let
$R_{\mathfrak{L}}$ denote the linear span of the rank one
operators in $\operatorname{Alg} \mathfrak{L}$. Then $R_{\mathfrak{L}}$ is dense
in $\operatorname{Alg}\mathfrak{L}$ in any of the weak, strong, ultraweak or
ultrastrong topologies.
\end{lemma}

It is an interesting question to ask when $x \otimes y^{\ast}$
belongs to $\operatorname{Alg}\mathfrak{L}$. The following lemma, due to
Longstaff \cite{MR53:1294}, is the answer to this question, and
the use of this lemma is essential.

\begin{lemma}\label{lsl1}
The operator $x \otimes f^{\ast} $ belongs to $\operatorname{Alg} \mathfrak{L}$
if and only if there is a projection $E \in \mathfrak{L}$ such
that $x \in E $ and $f \in E_{-}^{\perp}$.
\end{lemma}

Let $\mathfrak{L}$ be a subspace lattice. For $M \in
\mathfrak{L}$, define $M_{-}$ by $M_{-} = \vee \{N|M \neq N, N \in
\mathfrak{L}\}$. We end this section with a couple of lemmas that
will be used repeatedly.

\begin{lemma}\label{cdl1}
Let $\mathfrak{L}$ be a commutative completely distributive
lattice. Then
\[\vee \{N | N \in \mathfrak{L}, N_{-} \neq I \}=I \]
 and
\[\vee \{N_{-}^{\perp}| N \in \mathfrak{L}, N \neq 0  \}=I \]
\end{lemma}

\begin{lemma}\label{ul}
Let $\mathfrak{L}$ be a subspace lattice and $\{E_k\}_{k=1}^n$ be
a finite subset of $\mathfrak{L}$. Then $(\vee_{k=1}^n
E_k)_-=\vee_{k=1}^n (E_k)_- $ and $(\wedge_{k=1}^n E_k)_- \leq
\wedge_{k=1}^n (E_k)_- $
\end{lemma}
\begin{proof}
It is sufficient to prove that the statements are true in case
$n=2$. Let $E$ and $F$ denote two elements in $\mathfrak{L}$.
Since $E \leq E \vee F$ and $F \leq E \vee F$, we have $E_- \leq
(E \vee F)_-$ and $F_- \leq (E \vee F)_-$. These imply that \[ E_-
\vee F_- \leq (E \vee F)_-. \]

 On the other hand, if $K \in \{G
\in \mathfrak{L} \, | \, G \ngeq E \vee F \}$, then $K \in \{G \in
\mathfrak{L} \, | \, G \ngeq E\}$ or $K \in \{G \in \mathfrak{L}
\, | \, G \ngeq F\}$, so $K \leq E_-$ or $K \leq F_-$. These imply
that $K \leq E_- \vee F_-$. Hence $ (E \vee F)_- \leq E_- \vee
F_-.$ This proves $(1)$.

 Let $E$ and $F$ be two elements in
$\mathfrak{L}$. Since $E \wedge F \leq E$ and $E \wedge F \leq F$,
$(E \wedge F)_- \leq E_-$ and $(E \wedge F)_- \leq F_-$. Therefore
$(E \wedge F)_- \leq E_- \wedge F_-$. The results follow.
\end{proof}

Now let's discuss two important types of linear maps on $\operatorname{Alg}
\mathfrak{L}$.

\begin{definition}
Let $\mathfrak{L}$ be a commutative subspace lattice.
\begin{enumerate}
\item A linear map $\Phi:\operatorname{Alg}\mathfrak{L}\rightarrow
\operatorname{Alg}\mathfrak{L}$ is called an \emph{isomorphism} if $\Phi$ is a
bijection and a multiplication preserving map. \item A linear map
$\Phi:\operatorname{Alg} \mathfrak{L}\rightarrow \operatorname{Alg}\mathfrak{L}$ is called a
\emph{rank preserving map} if $rank(\Phi(A))=rank(A)$ for each
finite rank operator $A$ in $\operatorname{Alg}\mathfrak{L}$.
\end{enumerate}
\end{definition}

The automatic norm continuity of an isomorphism is proved by
Gilfeather and Moore in \cite{MR87k:47103}. In order to discuss
more results about isomorphisms, we need the following
terminology. An isomorphism $\Phi$ is said to be \emph{spatially
implemented} if there is a bounded invertible operator $T$ on
$\mathcal{H}$ so that $\Phi(A)=TAT^{-1}$ for all $A \in
\operatorname{Alg}(\mathfrak{L})$. An isomorphism is said to be
\emph{quasi}-\emph{spatially implemented} if there is a one-to-one
operator with dense domain $\mathcal{D}$ so that $\Phi(A)Tf=TAf$
for all $A \in \operatorname{Alg}\mathcal{L}$ and for all $f \in \mathcal{D}$.

The following theorem characterizes isomorphisms of CSL algebras.
The theorem is due to Gilfeather and Moore \cite{MR87k:47103}.

\begin{theorem}
Let $\mathfrak{L}_1$ and $\mathfrak{L}_2$ be commutative subspace
lattices on Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$,
respectively, and let $\mathfrak{L}_1$ be completely distributive.
Let $\rho:\operatorname{Alg}\mathfrak{L_1}\rightarrow \operatorname{Alg} \mathfrak{L_2}$ be an
algebraic isomorphism. The followings are equivalent.
\begin{enumerate}
\item $\rho$ is quasi-spatially implemented by a closed, injective
linear transformation $T:H_1 \rightarrow H_2$ whose range and
domain are dense. \item $\rho$ is a rank-preserving map.
\end{enumerate}
\end{theorem}

The weak continuity of a rank preserving map $\Phi$ will be used
in the proof of the main Theorem. However, we set the assumption
that \emph{each rank preserving map $\Phi$ is weakly continuous}
from this early stage to avoid any misunderstanding.

\section{RANK PRESERVING LINEAR MAPS}
We begin with a lemma which will play a key role in our result.
For notational convenience, we will let $a \sim b$ represent the
linear dependency of two nonzero vectors $a,b$ in $\mathcal{H}$.
It can be easily verified that $a \sim b$ is an equivalence
relation.

Suppose $\mathfrak{L}$ is a commutative subspace lattice and
$\Phi$ is a rank preserving map from $\operatorname{Alg} \mathfrak{L}$ to $\operatorname{Alg}
\mathfrak{L}$. Let $N$ be an element in $\mathfrak{L}$ with $\dim
N \geq 2, \, \dim N_-^{\perp} \geq 2$. Let $x,y$ be any two
linearly independent vectors in $N$ and $f,g$ be any two linearly
independent vectors in $N_-^{\perp}$. Then we can form four
different rank one operators in $\operatorname{Alg}\mathfrak{L}$ as follows: $x
\otimes f^{\ast},y \otimes f^{\ast},x \otimes g^{\ast},y \otimes
g^{\ast}$.

Since $\Phi$ is a rank preserving map on $\operatorname{Alg}\mathfrak{L}$, we
can consider another four rank one operators in $\operatorname{Alg}
\mathfrak{L}$ which are images of each of the four rank one
operators and write them as follows.
\begin{align*}
  \Phi(x \otimes f^{\ast})&= u \otimes v^{\ast} &&\cdots \cdots (A)    \\
  \Phi(y \otimes f^{\ast})&= p \otimes q^{\ast} &&\cdots \cdots (B)    \\
  \Phi(x \otimes g^{\ast})&= w \otimes z^{\ast} &&\cdots \cdots (C)    \\
  \Phi(y \otimes g^{\ast})&= r \otimes s^{\ast} &&\cdots \cdots (D)
\end{align*}

\begin{lemma} (Four Vectors Lemma)\label{fvl}
 Let $\mathfrak{L}$ be a
commutative subspace lattice and $\Phi$ be a rank preserving map
on $\operatorname{Alg}\mathfrak{L}$. Let $N$ be a subspace in $\mathfrak{L}$
with $ \dim N \geq 2, \, \dim N_-^{\perp} \geq 2$. If we consider
the eight rank one operators described above, then either $(1)$ or
$(2)$ holds for all $x,y \in N$ and $f,g \in N_-^{\perp}$.
 \begin{enumerate}
  \item $u \sim w, v \sim q, p \sim r \, and \, z \sim s$
  \item $v \sim z, u \sim p, w \sim r \, and \, q \sim r$
\end{enumerate}
\end{lemma}

\begin{proof}
Let's consider the following six equalities which are obtained by
adding two of $ (A), (B),(C) \, and \, (D) $.

$(A)+(C):\Phi(x \otimes(f+g)^*)=u\otimes v^*+w\otimes z^*$

$(A)+(B):\Phi((x + y)\otimes f^*)=u\otimes v^*+p\otimes q^*$

$(B)+(D):\Phi(y \otimes(f+g)^*)=p\otimes q^*+r\otimes s^*$

$(C)+(D):\Phi((x+y) \otimes g^*)=w\otimes z^*+r\otimes s^*$

$(A)+(D):\Phi(x \otimes f^*+y \otimes g^*)=u\otimes v^*+r\otimes
s^*$

$(B)+(C):\Phi(x \otimes g^*+y\otimes f^*)=p\otimes q^*+w\otimes
z^*$

Since $x\otimes (f+g)^*$ is a rank one operator and $\Phi$ is a
rank preserving map, $u\otimes v^*+w\otimes z^*$ is also a rank
one operator. This fact implies either $u\sim w$ or $v\sim z$.
Similarly we can argue that either $u\sim p$ or $v\sim q$ is true
by the equality $(A)+(B)$, either $p\sim r$ or $q\sim s$ is true
by the equality $(B)+(D)$ and $w\sim r$ or $z\sim s$ is true by
the fourth equality $(C)+(D)$. Apart from the first four
equalities, the map $\Phi$ maps a rank two operator into a rank
two operator in the equality $(A)+(D)$ and$(B)+(C)$, which gives
the following results
\[u \nsim r, v\nsim s, p\nsim w \, \textrm{and} \, q\nsim z .\]
\underline{Case 1:} Now let's suppose $u\sim w$ is true. Then it
is easily deduced that $v\sim q$. Otherwise $u\sim p$ is true
implying $w \sim p$, which contradicts the relation $w \nsim p$.
Similarly, we can show that $p\sim r, z\sim s$ are true, so the
result follows.
\\
\underline{Case 2:} Suppose $v\sim z$. The proof of this case is
almost identical to the proof of Case 1.
\end{proof}

\begin{remark}
If we assume that \emph{Case 1} holds, then $u \nsim p$.
\end{remark}

If we consider the case $\dim N=1$ and $\dim N_{-}^{\perp}\geq 2$,
we have the following two equalities.
\begin{align*}
 \Phi(x \otimes f^{\ast})&= u \otimes v^{\ast} &&\cdots \cdots (E)
    \\
 \Phi(x \otimes g^{\ast})&= p \otimes q^{\ast} &&\cdots \cdots (F)
\end{align*}
where $x \in N$ and $f,g$ are two linearly independent vectors in
$N_{-}^{\perp}$.
 The following lemma is analogous to Lemma
\ref{fvl} for this case.

\begin{lemma}\label{tvl}
 Let $\mathfrak{L}$ be a
commutative subspace lattice and $\Phi$ be a rank preserving map
on $\operatorname{Alg}\mathfrak{L}$. Let $N$ be a subspace in $\operatorname{Alg}\mathfrak{L}$
with $\dim N=1$ and $\dim N_{-}^{\perp}\geq 2$. If we consider the
two equalities described above, then either $(1)$ or $(2)$ holds
for all $x \in N$ and $f,g \in N_-^{\perp}$.
 \begin{enumerate}
  \item $ u \sim p , \, v \nsim q $
  \item $ u \nsim p , \, v \sim q $
  \end{enumerate}
\end{lemma}

\begin{proof}
By adding the two equalities, the following equality is obtained.

 $(E)+(F): \Phi(x \otimes (f+g)^*)=u \otimes v^* + p \otimes
 q^*$

 From the fact that $\Phi$ maps a rank one operator into a rank one operator,
 it can be easily deduced that either $ u \sim p$ or $ v \sim q$
 is true. Suppose $u \sim p$ and $v \sim q$. Then there exist
 $\lambda, \gamma \in \mathbb{C}$ such that $\lambda u \otimes v^* + \gamma p
 \otimes q^*=0$. Therefore $ \Phi(x \otimes
 (\overline{\lambda} f + \overline{\gamma} g)^*) = \lambda u \otimes v^* + \gamma
 p\otimes q^*=0$ which is a contradiction that $\Phi$ maps a rank one operator into
 a rank zero operator, so the result follows.
 \end{proof}

The next lemma states a different version of Lemma \ref{tvl} for
the case $\dim N \geq 2$ and
 $\dim N_{-}^{\perp}=1$. The proof of this lemma is almost same
 as the proof of Lemma \ref{tvl}.

\begin{lemma} \label{tvl2}
  Let $\mathfrak{L}$ be a
 commutative subspace lattice and $\Phi$ be a rank preserving map
 on $\operatorname{Alg}\mathfrak{L}$. Let $N$ be a subspace in $\operatorname{Alg}\mathfrak{L}$
 with $\dim N \geq 2$ and $\dim N_{-}^{\perp}=1$. If we consider the following two equalities,
  \begin{align*}
   \Phi(x \otimes f^{\ast})&= u \otimes v^{\ast} &&\cdots \cdots (G)
      \\
   \Phi(y \otimes f^{\ast})&= p \otimes q^{\ast} &&\cdots \cdots (H)
  \end{align*}
  where $ x,y $ are two linearly independent vectors in $N$ and
  $f \in N_-^{\perp}$, then one of the following holds.
  \begin{enumerate}
    \item $ u \sim p , \, v \nsim q $
    \item $  u \nsim q , \, v \sim q $
   \end{enumerate}
\end{lemma}

With these three lemmas in hand, we now can show the following
important result.
\begin{lemma}\label{ml}
Let $\mathfrak{L}$ be a commutative subspace lattice and $\Phi$ be
a rank preserving map on $\operatorname{Alg}\mathfrak{L}$. Let $N$ be a nonzero
element in $\mathfrak{L}$ with $N_{-}\neq I$. Then at least one of
the following holds.
\begin{enumerate}
\item There exist a linear map $U$ from $N$ to $\mathcal{H}$ and a
linear map $V$ from $N_{-}^{\perp}$ to $\mathcal{H}$ such that
$\Phi(x \otimes f^*)=U(x) \otimes V(f)^*$ for all $x \in N$ and $f
\in N_-^{\perp}.$
 \item There exist a conjugate linear map $U$ from
$N_{-}^{\perp}$ to $\mathcal{H}$ and a conjugate linear map $V$
from $N$ to $\mathcal{H}$ such that $\Phi(x \otimes f^*)=U(f)
\otimes V(x)^*$ for all $x \in N$ and $f \in N_-^{\perp}.$
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose $(1)$ holds for all $x,y \in N$ and $f,g \in N_-^{\perp}$
in Lemma \ref{fvl}, Lemma \ref{tvl} and Lemma \ref{tvl2}.

\underline{Case 1:} Suppose that $\dim N=1$ and $\dim
N_{-}^{\perp}=1$. Fix $x_1 \in N, f_1 \in
 N_-^{\perp}$. Set $\Phi(x_1 \otimes f_1^*)=u_1 \otimes
 v_1^*$. For any $x \in N, f \in N_-^{\perp}$, there exist
 $\lambda, \gamma \in \mathbb{C}$ such that $x=\lambda x_1, f=\gamma
 f_1$. If we define $U:N \rightarrow \mathcal{H} $ and $V:
 N_-^{\perp} \rightarrow \mathcal{H}$ by
 $U(x)=\lambda u_1$ and $V(f)=\gamma v_1$, then these maps are
 clearly well defined linear maps. However, if we define $U:N \rightarrow \mathcal{H} $
 and  $V:N_-^{\perp} \rightarrow \mathcal{H}$ by $U(x)=\lambda v_1$ and $V(f)=\gamma
 u_1$, these maps are also well defined linear maps. Therefore
 both conclusions $(1)$ and $(2)$ hold in this special case.

\underline{Case 2:} Suppose that $\dim N=1$ and $\dim
N_{-}^{\perp}\geq 2$. Fix $x_1 \in N$. It follows from Lemma
\ref{tvl} (1) that it is possible to to define a map $V$ on
$N_-^{\perp}$ by $\Phi(x_1 \otimes f^*)=u_1 \otimes V(f) $. For
any $f, g \in N_{-}^{\perp}, t\in \mathbb{C}$, since $\Phi$ is
linear,
 \begin{align*}
 \Phi(x_1\otimes (f+tg)^*)&=\Phi(x_1\otimes
 f^*)+\Phi(x_1\otimes(tg)^*)\\
 &=u_1\otimes V(f)^* +\overline{t}u_1\otimes V(g)^*.
 \end{align*}

From the definition of $V$, $\Phi(x_1\otimes (f+tg)^*)=u_1\otimes
V(f+tg)^*$. By comparing the two equalities, we see the map
$V:N_-^{\perp} \rightarrow \mathcal{H}$
 is a linear map on $N_{-}^{\perp}$.

For any $x \in N$, there exists $\lambda \in \mathbb{C}$ such that
$x=\lambda x_1$. Now we can define a linear map $U:N \rightarrow
 \mathcal{H}$ by $U(x)=\lambda u_1$. Therefore the conclusion $(1)$ holds with this assumption.
Likewise, the conclusion $(2)$ follows if we assume $(2)$ in Lemma
\ref{tvl} is true.

\underline{Case 3:} Suppose that $\dim N \geq 2$ and $\dim
N_{-}^{\perp}= 1$. The proof of this case is almost identical to
the proof of case 2.

\underline{Case 4:}
 Let's fix $x_1\in N$. By Lemma \ref{fvl} (1), we can define a
 map $V$ from $N_{-}^{\perp}$ to $\mathcal{H}$ by $\Phi(x_1 \otimes f^*)=u_1 \otimes
 V(f)^*$. For any $f, g \in N_{-}^{\perp}, t\in \mathbb{C}$, since
 $\Phi$ is linear,
 \begin{align*}
 \Phi(x_1\otimes (f+tg)^*)&=\Phi(x_1\otimes
 f^*)+\Phi(x_1\otimes(tg)^*) \\
 &=u_1\otimes V(f)^* +\overline{t}u_1\otimes V(g)^*.
 \end{align*}

 From the definition of $V$, $\Phi(x_1\otimes (f+tg)^*)=u_1\otimes
 V(f+tg)^*$.
 By comparing the two equalities, we see the map $V$ is a linear
 map on $N_{-}^{\perp}$. Likewise, we can define a linear map $U$
 from $N$ to $\mathcal{H}$ by $\Phi(x\otimes {f_1}^*)=U(x)\otimes
 {v_1}^*$ for some fixed $f_1, v_1 \in N_{-}^{\perp}$. By
 considering
 \begin{align*}
 \Phi(x_1\otimes {f_1}^*)&=U(x_1)\otimes{v_1}^*\\
 &=u_1\otimes V(f_1)^*\\
 &=u_1\otimes{v_1}^* ,
 \end{align*}
 we can make an observation that $U(x_1)=u_1$ and $V(f_1)=v_1$.
 Now for any $x\in N, f\in N_{-}^{\perp}$, we may write
 $\Phi(x\otimes f^*)$ as follows.
 \[ \Phi(x\otimes f^*)=\alpha(x,f)U(x)\otimes V(f)^*\]
 where $\alpha(x,f)$ is a complex valued function.

 We claim that $\alpha(x,f)=1$ for all $x$ and $f$. Note that
 $\Phi(x_1\otimes f^*)=\alpha(x_1,f)u_1\otimes
 V(f)^*=u_1\otimes V(f)^*$, so $\alpha(x_1,f)=1$ for any $f\in
 N_{-}^{\perp}$. Similarly, we observe that $\alpha(x,f_1)=1$ for any
 $x\in N$. Consider
  \begin{align*}
 \Phi((x+sx')\otimes f^*)&=\alpha(x+sx',f)(U(x+sx')\otimes
 V(f)^*)\\
 &=\alpha(x+sx',f)[(U(x)+sU(x'))\otimes V(f)^*]\\
 &=\alpha(x+sx',f)U(x)\otimes V(f)^*+ s\alpha(x+sx',f)U(x')\otimes V(f)^*
  \end{align*}
  where $x$ and $x'$ are linearly independent vectors in $N$.

On the other hand,
  \begin{align*}
  \phi((x+sx')\otimes f^*)&=\Phi(x\otimes f^*)+s\Phi(x'\otimes f^*)\\
  &=\alpha(x,f)U(x)\otimes V(f)^* +s\alpha(x',f)U(x')\otimes
  V(f)^*.
  \end{align*}
 We get the following equality by comparing the two equalities
 above:
\[\alpha(x+sx',f)U(x)+s\alpha(x+sx',f)U(x')=\alpha(x,f)U(x)+s\alpha(x',f)U(x')\]
for all $s\in \mathbb{C}$. Since $U(x)$ and $U(x')$ are linearly
independent, $\alpha(x,f)=\alpha(x+sx',f)=\alpha(x',f)$. Hence,
$\alpha$ is independent of $x$. Likewise we can show that $\alpha$
is independent of $f$. Therefore $\alpha \equiv 1.$ The result
follows.
\end{proof}

What Lemma~\ref{ml} says is that whether a given subspace $N$
satisfies $(1)$ or $(2)$ in the lemma is a characteristic of the
subspace which is associated with the rank preserving map $\Phi$.

\begin{definition}
 Let $\mathfrak{L}$ be a
commutative subspace lattice and $\Phi$ be a rank preserving map
on $\operatorname{Alg}\mathfrak{L}$. Let $N$ be a nonzero subspace in $\operatorname{Alg}
\mathfrak{L}$. The subspace $N$ is called \emph{consistent with
respect to} $\Phi$ if $\Phi$ and $N$ satisfy $(1)$ in Lemma
\ref{ml}. The subspace $N$ is called \emph{twisted with respect
to} $\Phi$ if $\Phi$ and $N$ satisfy $(2)$ in Lemma \ref{ml}.
\end{definition}

In a nest algebra, it is impossible that both $N$ and $N^{\perp}$
are in a nest $\mathcal{N}$ (a totally ordered lattice) if $N \neq
0$ and $N \neq I.$ But a commutative subspace lattice
$\mathfrak{L}$ can have $N$ and $N^{\perp}$ both in it with the
assumption $N \neq 0$ and $N \neq I.$ If there exists a subspace
$N \in \mathfrak{L}$ with $N_- \neq I$ such that $\dim N=1$ and
$N^{\perp} \in \mathfrak{L}$, then $N_-=\vee\{E \in
\mathfrak{L}\,|\, E \ngeq N\} \geq N^{\perp}$ since $N^{\perp} \in
\mathfrak{L}$ and $N^{\perp} \ngeq N$. Hence $N_-^{\perp} \leq N$.
Since $N_- \neq I$, $N_-^{\perp}=N$. \underline{Case 1} in the
proof of Lemma~\ref{ml} shows that such $N$ is both consistent and
twisted with respect to $\Phi$.

\begin{definition}
A nonzero subspace $N$ in $\mathfrak{L}$ with $\dim N=1$ and
$N^{\perp} \in \mathfrak{L}$ is called \emph{isolated}.
\end{definition}

\begin{lemma} \label{isl}
Let $\Phi$ be a rank preserving map on $\operatorname{Alg}\mathfrak{L}$ and $N$
be a subspace in $\mathfrak{L}$. Then $N$ is isolated if and only
if $N$ is both consistent and twisted with respect to $\Phi$.
\end{lemma}
\begin{proof}
It suffices to prove sufficiency. Suppose $N$ is both consistent
and twisted with respect to $\Phi$. Then there exist linear maps
$U_1,V_2$ defined on $N$, and $U_2,V_1$ defined on $N_-^{\perp}$
so that \[\Phi(x \otimes f^*)=U_1x \otimes (V_1f)^*=U_2f \otimes
(V_2x)^*\] \[\Phi(y \otimes f^*)=U_1y \otimes (V_1f)^*=U_2f
\otimes (V_2y)^*\] for arbitrary $x,y \in N$ and $f \in
N_-^{\perp}$. It can be observed that $U_1x \sim U_2f$ and $U_2f
\sim U_1y$, so $U_1x \sim U_1y$. Since the map $U_1$ can not have
nonzero kernel, $x \sim y$. Using the same idea as this with $x
\in N$ and $f,g \in N_-^{\perp}$, we can argue that $f \sim g$.
These facts imply that $\dim N=\dim N_-^{\perp}=1$. Observe that
if $x \in N$, $x \otimes x^*$ is the projection on $N$ since $\dim
N=1$, so $x \otimes x^* \in \operatorname{Alg}\mathfrak{L}$. Therefore, $x \in
N_-^{\perp}$. Then $N=N_-^{\perp}$. Thus $N$ is isolated.
\end{proof}

\begin{lemma}\label{sl}
Let $M$ and $N$ be two subspaces in $\mathfrak{L}$ with $N \leq
M$. Let $\Phi$ be a rank preserving map on $\operatorname{Alg}\mathfrak{L}$.
Then the followings are true.
\begin{enumerate}
\item If $M$ is consistent with respect to $\Phi$, then $N$ is
also consistent with respect to $\Phi$. \item If $M$ is twisted
with respect to $\Phi$, then $N$ is also twisted with respect to
$\Phi$.
\end{enumerate}
 \end{lemma}

\begin{proof}
First, let $M$ be a consistent element in $\mathfrak{L}$ with
respect to $\Phi$. Without loss of generality, we assume $\dim N
\geq 2$. Suppose that $N$ is twisted with respect to $\Phi$. Let
$U_M,V_M$ and $U_N,V_N$ denote the two maps in lemma \ref{ml}
defined on $M$ and $N$ ,respectively. By the assumption, we can
choose two linearly independent vectors $x,y \in N$ and a vector
$f \in M_-^{\perp}$. Note that $N_-^{\perp} \geq M_-^{\perp}$
since $N \leq M$. Since $M$ is consistent with respect to $\Phi$
and $N$ is twisted with respect to $\Phi$, we have
\[\Phi(x \otimes f^*)= U_Mx \otimes (V_Mf)^*=U_Nf \otimes (V_Nx)^*
\]
   \[ \Phi(y \otimes f^*)= U_My \otimes (V_Mf)^*=U_Nf \otimes (V_Ny)^*. \]
Then there exist two complex numbers $\lambda$ and $\gamma$ such
that $U_Mx=\lambda U_Nf$ and $U_My=\gamma U_Nf$. But this
contradicts the fact that $U_Mx$ and $U_My$ are linearly
independent by Lemma \ref{fvl}, so the result $(1)$ follows. The
proof of $(2)$ is essentially the same as the proof of $(1)$.
\end{proof}

\begin{lemma}\label{vil}
Let $\Phi$ be a rank preserving map on $\operatorname{Alg} \mathfrak{L}$. Let
$M$ and $N$ be two non-zero, non-isolated subspaces in
$\mathfrak{L}$ with $M_- \neq I$. If $M$ is consistent with
respect to $\Phi$, and $N$ is twisted with respect to $\Phi$, then
\[M \wedge N=0\] and
\[M_-^{\perp} \wedge N_-^{\perp}=0.\]
\end{lemma}

\begin{proof}
Suppose $M \wedge N \neq 0$. Then $M \wedge N$ is a non-zero
subspace of $M$ and $N$. By Lemma \ref{isl}, $M \wedge N$ is
isolated, so $M \wedge N=(M \wedge N)_-^{\perp}=\langle e \rangle$
for some $e \in M \wedge N$. By Lemma \ref{ul}, we have $(M \wedge
N)_- \leq M_- \wedge N_-$. It follows that $(M \wedge N)_-^{\perp}
\geq M_-^{\perp} \vee N_-^{\perp}$. Thus
$M_-^{\perp}=N_-^{\perp}=(M \wedge N)_-^{\perp}=\langle e
\rangle$. On the other hand, we have $M_- \geq N$ since $N \ngeq
M$, so $\langle e\rangle=M_-^{\perp} \leq N^{\perp}$. Therefore $N
\wedge N^{\perp} \geq \langle e\rangle \neq 0$ which gives a
contradiction. Thus \[M \wedge N=0.\]

 Observe that $M \vee N$ is
neither consistent nor twisted with respect to $\Phi$ by Lemma
\ref{sl}. This implies $(M \vee N)_-=I$ by Lemma \ref{ml}.
Therefore \[M_-^{\perp} \wedge N_-^{\perp}=(M_- \vee
N_-)^{\perp}=(M \vee N)_-^{\perp}=0.\]
\end{proof}

\begin{theorem}\label{dct}
Let $\mathfrak{L}$ be a completely distributive commutative
subspace lattice and $\Phi$ be a rank preserving map on $\operatorname{Alg}
\mathfrak{L}$. Let \[M_i=\vee\{E \in \mathfrak{L}\,|\,E \,
\textrm{is isolated}\},\]
    \[M_c=\vee\{E \in \mathfrak{L}\,|\, E \, \textrm{is consistent with
respect to} \, \Phi \} \ominus M_i\] and
    \[M_t=\vee\{E \in \mathfrak{L}\,|\, E \, \textrm{is twisted with respect
    to}\, \Phi \ \} \ominus M_i.\]

Then \[\operatorname{Alg}\mathfrak{L}=\operatorname{Alg}(M_i\mathfrak{L}M_i) \oplus
\operatorname{Alg}(M_c\mathfrak{L}M_c) \oplus \operatorname{Alg}(M_t\mathfrak{L}M_t) .\]

\end{theorem}

\begin{proof}
By Lemma \ref{vil}, it is clear that $M_i \oplus M_c \oplus
M_t=I$. Since $M_i^{\perp} \in \mathfrak{L}$, $M_c, M_t \in
\mathfrak{L}.$ Note that $M_c^{\perp}=M_i \vee M_t.$ Hence
$M_c^{\perp} \in \mathfrak{L}$. Likewise, $M_t^{\perp} \in
\mathfrak{L}$. The result follows.
\end{proof}

Remember that the operators $U$ and $V$ were constructed with
 the subspace $N$ fixed. We now refer to
them as $U_N$ and $V_N$ and try to fit together the $U_N$'s and
$V_N$'s into operators $U$ and $V$ defined on the whole space
$\mathcal{H}$ such that
\[\Phi(x \otimes f^*)=Ux \otimes (Vf)^* \]
whenever $x \otimes f^* \in \operatorname{Alg}\mathfrak{L}$. Note that if $M_-
 \neq I$, if $M<N$, and if $x \in M$ and $f \in N_-^{\perp}$, then $x \in N$
and $\Phi(x \otimes f^*)=(U_N x) \otimes (V_N f)^*$. On the other
hand, $f \in N_-^{\perp} <M_-^{\perp}$, so $\Phi(x \otimes
f^*)=(U_M x) \otimes (V_M f)^*$. Thus there exists a complex
number $\lambda$ such that $U_M x=\lambda U_N x$ and $V_N
f=\overline{\lambda}V_M f$. Since $x$ and $f$ may vary
independently, $\lambda$ does not depend on $x$ and $f$ but only
on $M$ and $N$. We call it $\lambda_{MN}$ and have
\[ U_M=\lambda_{MN} U_N|M \,\,\, \textrm{and} \,\,\,
V_N=\overline{\lambda_{MN}}V_M|N_-^{\perp} .\]

Let $\mathfrak{F}$ be the collection of all subspaces $N$ in
$\mathfrak{L}$ such that $N \neq 0$ and $N_- \neq I$.

\begin{remark}
In the remainder of this paper we assume that every subspace in
$\mathfrak{F}$ is consistent with respect to $\Phi$.
\end{remark}
 Suppose that $M$ and $N$ lie
in $\mathfrak{F}$. We will say that $M$ and $N$ are
\emph{comparable} if either $M \leq N$ or $N \leq M$. Suppose $M$
and $N$ are comparable. If $M \leq N$, $\lambda_{MN}$ has already
been defined. If $N \leq M$, define
\[ \lambda_{MN}= \frac{1}{\lambda_{NM}}\,\, .\]
Thus $\lambda_{MN}$ is defined whenever $M$ and $N$ are
comparable, and it is easy to check that
$\lambda_{LN}=\lambda_{LM}\lambda_{MN}$ whenever each pair from
$\{L,M,N\}$ is comparable.

\begin{definition}
Let $M$ and $N$ be two subspaces in $\mathfrak{F}$. We define a
\emph{chain} from $M$ to $N$ to be a finite sequence of subspaces
$\{M_0,M_1,...,M_n\}$, each $M_k$ in $\mathfrak{F}$, such that
$M_0=M, M_n=N$, and such that $M_k$ is comparable to $M_{k+1}$ for
each $k=0,1,...,n-1$.

 If $M=N$, the chain $\{M_0,M_1,...,M_n\}$ is
called a \emph{cycle of length} n.
\end{definition}

Suppose $M,N \in \mathfrak{F}.$ If there is a chain
$\{M_0,M_1,...,M_n\}$ from $M$ to $N$, we want to define
$\lambda_{MN}$ to be $\lambda_{MM_1}\lambda_{M_1M_2} \cdots
\lambda_{M_{n-2}M_{n-1}}\lambda_{M_{n-1}N}$. Since there may be
more than one chain from $M$ to $N$, we need to show that such a
product is well defined. The following lemma will prove this. An
elaborate proof of the lemma can be found in \cite{MR87k:47103}.

\begin{lemma} \label{lprt}
Let $\{M_0,M_1,...,M_n\}$ be a cycle in $\mathfrak{F}$. Then
\[ \lambda_{M_0M_1}\lambda_{M_1M_2} \cdots
\lambda_{M_{n-2}M_{n-1}}\lambda_{M_{n-1}M_n}=1  .\]

\end{lemma}

By making use of the Lemma \ref{lprt}, we can now show that
\[\lambda_{MN}=\lambda_{MM_1}\lambda_{M_1M_2} \cdots
\lambda_{M_{n-2}M_{n-1}}\lambda_{M_{n-1}N}\] is well defined if
there is a chain $\{M_0,M_1,...,M_n\}$ from $M$ to $N$.

Suppose that there are two chains
$\{M_0,M_1,M_2,...,M_{n-2},M_{n-1},M_n\}$ and \\
$\{M_0,M_1',M_2',...,M_{n-2}',M_{n-1}',M_n\}$ from $M$ to $N$. We
now form a cycle \[\{M,M_1,...,M_{n-1},N,M_{n-1}',...,M_1',M\}.\]
By applying Lemma \ref{lprt} to this cycle, it follows that
\[\lambda_{MM_1} \cdots \lambda_{M_{n-1}N}\lambda_{NM_{n-1}'}
\cdots \lambda_{M_1'M}=1.\] From this equation we get,
\[\lambda_{MM_1} \cdots \lambda_{M_{n-1}N}=
\frac{1}{\lambda_{NM_{n-1}'}\cdots
\lambda_{M_1'M}}=\lambda_{MM_1'}\cdots \lambda_{M_{n-1}'N}. \]
This shows that $\lambda_{MN}$ is well defined.\\

Recall that $\mathfrak{F}$ denotes the collection of all $N$ in
$\mathfrak{L}$ such that $N \neq 0$ and $N_- \neq I$. Fix $N \in
\mathfrak{L}$ and let
\[\mathfrak{G}_N^n=\{M \in \mathfrak{F}\,|\,M \, \textrm{can be
connected to} \, N \, \textrm{by a chain of length} \, k \leq
n\}.\] Let $\mathfrak{G}_N=\cup_{n}\mathfrak{G}_N^n$.

\begin{lemma}\label{du}
Let $N \in \mathfrak{F}$. Then there exist linear transformations
$U$ with dense domain in $\vee \{M \, | \, M \in \mathfrak{G}_N\}$
and $V$ with dense domain in $\vee \{M_-^{\perp} \, | \, M \in
\mathfrak{G}_N \}$ such that $\Phi(x \otimes f^*)=Ux \otimes
(Vf)^*$ whenever there is $M \in \mathfrak{G}_N$ for which $x \in
M$ and $f \in M_-^{\perp}$.
\end{lemma}

\begin{proof}
For $M \in \mathfrak{G}_N$, we have associated operators $U_M$ and
$V_M$ such that \[\Phi(x \otimes f^*)=U_Mx \otimes (V_Mf)^*,\]
whenever $x \in M$ and $f \in M_-^{\perp}$. Let
$\widetilde{U}_M=\lambda_{NM}U_M$ and
$\widetilde{V}_M=\overline{\lambda}_{MN} V_M$. Since
$\lambda_{MN}\lambda_{NM}=1$, we have $\Phi(x \otimes f^*)=
\widetilde{U}_Mx \otimes (\widetilde{V}_Mf)^*$ for $x \in M$ and
$f \in M_-^{\perp}$. Note that for $L,M \in \mathfrak{G}_N$ with
$L \leq M$, if $\{N,N_1,...,N_{n-1},M \}$ is a chain from $N$ to
$M$, then $\{N,N_1,...,N_{n-1},M,L\}$ is a chain from $N$ to $L$.
Therefore $\lambda_{NL}=\lambda_{NM}\lambda_{ML}$. Thus, if $x \in
L$, we have
$\widetilde{U}_{L}x=\lambda_{NL}U_{L}x=\lambda_{NM}\lambda_{ML}U_{L}x$.
On the other hand, by definition of $\lambda_{LM}$ we have
$U_Lx=\lambda_{LM}U_Mx$, so
$\widetilde{U}_{L}x=\lambda_{NM}\lambda_{ML}\lambda_{LM}U_{M}x=\lambda_{NM}U_{M}x
=\widetilde{U}_{M}x$. Thus, if $L \leq M$, $\widetilde{U}_L$ and
$\widetilde{U}_M$ agree on $L$. Let $\mathcal{M} =
\{x_1+\cdots+x_n \, | \, \textrm{for some positive integer} \, n
\, , \, x_i \in M \, \textrm{for some} \, M \in \mathfrak{G}_N
\}$. Define a linear transformation $U$ on $\mathcal{M}$ by
$U|M=\widetilde{U}_M$. By the coherence of the $\widetilde{U}_M$,
the map $U$ is well defined. Similarly, we can define
$V|M_-^{\perp}=\widetilde{V}_M$. If $x \in M, \, f \in
M_-^{\perp}$ for $M \in \mathfrak{G}_N$, then we have $\Phi(x
\otimes f^*)=Ux \otimes (Vf)^*$.
\end{proof}

\begin{lemma}\label{orth}
For each $N \in \mathfrak{F}$, let $G_N=\vee\{M \, | \, M \in
\mathfrak{G}_N\}$ and $F_N=\vee\{M_-^{\perp} \, | \, M \in
\mathfrak{G}_N \}$. Let $N,N' \in \mathfrak{F}$.
\begin{enumerate}
\item If $N' \in \mathfrak{G}_N$, then $G_N=G_{N'}$ and
$F_N=F_{N'}$
 \item If $N' \notin \mathfrak{G}_N$, then $G_N \perp
G_{N'}$ and $F_N \perp F_{N'}$
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose that $N' \in \mathfrak{G}_N$. Note that if there exists a
chain connecting $N$ to $N'$ and one connecting $N'$ to $M$, then
there is a chain connecting $N$ to $M$, whence
$\mathfrak{G}_N=\mathfrak{G}_{N'}$. Thus $G_N=G_{N'}$ and
$F_N=F_{N'}$. Suppose that $N' \notin \mathfrak{G}_N$; then, for
any $M' \in \mathfrak{G}_{N'}$, $M' \notin \mathfrak{G}_N$. For
such an $M'$, consider the projection $NM'$. If $NM' \neq 0$, then
$NM' \in \mathfrak{G}_N \cap \mathfrak{G}_{M'}=\mathfrak{G}_{M'}
\cap \mathfrak{G}_N$. Since this is impossible, it must be that
$NM'=0$. Likewise, if $M \in \mathfrak{G}_N$, $MM'=0$ and hence
$G_NG_{N'}=0$.
\end{proof}

\begin{theorem} \label{mpn}
Let $\mathfrak{L}$ be a completely distributive commutative
subspace lattice and let $\Phi$ be a weakly continuous rank
preserving map defined on $\operatorname{Alg}\mathfrak{L}$. If we assume that
$\mathfrak{F}=\mathfrak{G}_N$ for some $N \in \mathfrak{F}$ (i.e.
$\operatorname{Alg} \mathfrak{L}$ is irreducible.), then there exist two densely
defined linear transformations $U,V$ such that
\[\Phi(A)=UAV^*\] for all $A \in \operatorname{Alg}\mathfrak{L}$.
\end{theorem}

\begin{proof}
By Lemma \ref{du}, there exist densely defined maps $U$ and $V$
such that $\Phi(x \otimes f^*)=Ux \otimes (Vf)^*$ whenever $x
\otimes f^* \in \operatorname{Alg}\mathfrak{L}$. Since
$\mathfrak{F}=\mathfrak{G}_N$ for some $N \in \mathfrak{F}$, the
domain of $U$ is the nonclosed linear span of $\mathfrak{F}=\{E
\in \mathfrak{L}\,|\, E_-\neq I \}$ and the domain of $V$ is the
nonclosed linear span of $\{M_-^{\perp} \,|\, M \in \mathfrak{F}
\}$. By Lemma \ref{cdl1}, these domains are dense in
$\mathcal{H}$. Recall that $R_{\mathfrak{L}}$ denote the linear
span of the rank one operators in $\operatorname{Alg}\mathfrak{L}$. Therefore,
for any operator $A \in R_{\mathfrak{L}}$, $\Phi(A)=UAV^*$ by the
linearity of $\Phi$. By Lemma \ref{wk}, in CDC algebra it is
guaranteed that there exists a net of operators in
$R_{\mathfrak{L}}$ weakly converging to any operator in
$\operatorname{Alg}\mathfrak{L}$. Let $A$ be an operator in $\operatorname{Alg}\mathfrak{L}$.
Then there exists a net $\{A_{\alpha}\}_{\alpha \in J}$ such that
$A_{\alpha}$ converges weakly to $A$. For each $h \in
\operatorname{dom}V^*$ and $k \in \operatorname{dom}U^*$,
\begin{align*}
\langle UA_{\alpha}V^*h,k \rangle &= \langle A_{\alpha}V^*h, U^*k
\rangle  \\    &\rightarrow \langle AV^*h,U^*k \rangle \,\,
\textrm{since} \, A_{\alpha} \, \textrm{weakly converges to} \, A
\\  &= \langle UAV^*h,k \rangle .
\end{align*}
Thus \[\Phi(A_{\alpha})=UA_{\alpha}V^*
\overset{w}{\longrightarrow} UAV^*.
\]
On the other hand, $\Phi(A_{\alpha})$ converges weakly to
$\Phi(A)$ since $\Phi$ is weakly continuous. Thus $\Phi(A)=UAV^*$
for $A \in \operatorname{Alg}\mathfrak{L}$.

\end{proof}

Now we are finally in a position to state and prove the main
result of this paper.
\begin{theorem}
Let $\mathfrak{L}$ be a completely distributive commutative
subspace lattice and let $\Phi$ be a weakly continuous rank
preserving map defined on $\operatorname{Alg}\mathfrak{L}$. If we assume that
every subspace is consistent with respect to $\Phi$, then there
exist two densely defined linear transformations $U,V$ such that
\[\Phi(A)=UAV^*\] for all $A \in \operatorname{Alg}\mathfrak{L}$.
\end{theorem}
\begin{proof}
Let $N \in \mathfrak{F}$ and consider the collection
$\mathfrak{G}_N$. If $\mathfrak{G}_N = \mathfrak{F}$, then this is
nothing but the case of Proposition \ref{mpn}. Suppose there
exists $M \in \mathfrak{F}$ with $M \notin \mathfrak{G}_N$ such
that $G_N G_M=G_M G_N=0$. In this way we can form a sequence
$\{G_{N_i}\}$ of mutually orthogonal projections in
$\mathfrak{L}$. We shall suppress the $N$ and write simply $G_i$.
The separability of $\mathcal{H}$ guarantees that there are no
more than countably many $G_i$'s. We have $\vee G_i=I$ since $\vee
\{N \,:\,N \in \mathfrak{F}\}=I$, whence $G_j^{\perp}=\vee_{i \neq
j}G_i$ is also in $\mathfrak{L}$. Therefore, the algebra
$\operatorname{Alg}\mathfrak{L}$ can be written as the direct sum $\sum_i \oplus
\operatorname{Alg}(G_i\mathfrak{L}G_i)$. For any $A \in \operatorname{Alg}\mathfrak{L}$, we
can write $A=\sum_i \oplus A_i$ where $A_i \in
\operatorname{Alg}(G_i\mathfrak{L}G_i).$ By Proposition~\ref{mpn}, there exist
two densely defined linear maps $U_i$ and $V_i$ such that
$\Phi(A_i)=U_iA_iV_i^*$ for each $i$. If we define $U=\sum_i
\oplus U_i$ and $V=\sum_i \oplus V_i$, then $U$ and $V$ are
densely defined on $\mathcal{H}$. Then the result follows.
\end{proof}

\begin{corollary}
Let $\mathcal{H}$ be a finite dimensional Hilbert space and let
$\mathfrak{L}$ be a commutative subspace lattice consisting of
subspaces in $\mathcal{H}.$ Let $\Phi$ be a rank preserving map on
$\operatorname{Alg} \mathfrak{L}$. If we assume that every element in
$\mathfrak{L}$ is consistent with respect to $\Phi$, then there
exists a map $\Psi$ defined by $\Psi(A)=A\Phi^{-1}(I)$ such that
$\Psi \circ \Phi$ is an isomorphism on $\operatorname{Alg} \mathfrak{L}.$
\end{corollary}

\begin{proof}
By Main Theorem, there are two linear maps $U$ and $V$ such that
$\Phi(A)=UAV^*$. Since $\mathcal{H}$ is finite dimensional, it is
true that $\Phi(I)=UV^*$ is invertible. Therefore
$\Psi(A)=A\Phi^{-1}(I)=A(UV^*)^{-1}$ is a well defined map. Then
\begin{align*}
(\Psi \circ \Phi)(A)&=\Phi(A)(V^*)^{-1}U^{-1}
\end{align*}
\begin{align*}
                    &=UAV^*(V^*)^{-1}U^{-1}  \\
                    &=UAU^{-1}
\end{align*}
Thus $\Psi \circ \Phi$ is an isomorphism implemented by $U.$
\end{proof}

\section{EXAMPLES}
In this section we present some examples of rank preserving maps.
Most of the examples we discuss in this section will be either
$\mathcal{A}_{2n}$ or $\mathcal{A}_{\infty}$. The precise
definition of these algebras is given in \cite{MR85g:47062}. But
for our discussion it is enough to say that the algebras
$\mathcal{A}_{2n}$ are tridiagonal matrices, of size $2n \times
2n$, of the form
\[
\begin{bmatrix}
* & * &   &   &   &   &   &   &   & * \\
  & * &   &   &   &   &   &   &   & \\
  & * & * & * &   &   &   &   &   & \\
  &   &   & * &   &   &   &   &   & \\
  &   &   & * & * & * &   &   &   & \\
  &   &   &   &   & \ddots & &   &   & \\
  &   &   &   &   &   &   & * &   & \\
  &   &   &   &   &   &   & * &   & \\
  &   &   &   &   &   &   & * & * & * \\
  &   &   &   &   &   &   &   &   & *
\end{bmatrix}
\]
where all nonstarred entries are $0$. It can be observed that the
associated lattice consists of certain diagonal projections, hence
it is commutative and completely distributive. Thus the algebra
$\mathcal{A}_{2n}$ is reflexive. The algebra
$\mathcal{A}_{\infty}$ consists of infinite matrices of the form
\[
\begin{bmatrix}
* & * &   &   & \\
  & * &   &   & \\
  & * & * & * & \\
  &   &   & * & \\
  &   &   & * & \\
  &   &   &   & \ddots
\end{bmatrix}
.\]

Once again the associated lattice is commutative and completely
distributive, so the algebra $\mathcal{A}_{\infty}$ is reflexive.
\\
\begin{example}
Consider $\mathcal{A}_4$. Define a map $\Phi_1:\mathcal{A}_4
\rightarrow \mathcal{A}_4$ by
\[
 \left[
\begin{matrix}
a & b & 0 & h  \\
0 & c & 0 & 0  \\
0 & d & e & f  \\
0 & 0 & 0 & g
\end{matrix}
\right] \mapsto \left[
\begin{matrix}
e & f & 0 & d  \\
0 & g & 0 & 0  \\
0 & h & a & b  \\
0 & 0 & 0 & c
\end{matrix}
\right] .\] Then it is easy to check that $\Phi_1$ is rank
preserving. Moreover, the map $\Phi_1$ is implemented by
\[
U =
\begin{bmatrix}
0 & 0 & 1 & 0  \\
0 & 0 & 0 & 1  \\
1 & 0 & 0 & 0  \\
0 & 1 & 0 & 0
\end{bmatrix}
.\] In other words, the map $\Phi_1$ is an isomorphism such that
\[\Phi_1(A)=UAU^{-1}.\] If we define a map $\Phi_2:\mathcal{A}_4
\rightarrow \mathcal{A}_4$ by
\[
 \left[
\begin{matrix}
a & b & 0 & h  \\
0 & c & 0 & 0  \\
0 & d & e & f  \\
0 & 0 & 0 & g
\end{matrix}
\right] \mapsto \left[
\begin{matrix}
g & f & 0 & h  \\
0 & e & 0 & 0  \\
0 & d & c & b  \\
0 & 0 & 0 & a
\end{matrix}
\right] ,\] then $\Phi_2$ is also a rank preserving map. But the
map $\Phi_2$ is not an isomorphism. Instead, we can write the map
$\Phi_2$ as
\[\Phi_2(A)=VA^{T}V^{-1}\] where
\[
V =
\begin{bmatrix}
0 & 0 & 0 & 1  \\
0 & 0 & 1 & 0  \\
0 & 1 & 0 & 0  \\
1 & 0 & 0 & 0
\end{bmatrix}
.\]

\end{example}

\begin{example}
Let $\mathcal{H}$ be a separable Hilbert space and
$\mathcal{B}=\{e_k\}_{k=1}^{\infty}$ be an orthonormal basis for
$\mathcal{H}$. Let
$\mathcal{N}=\{0,[e_1],[e_1,e_2],[e_1,e_2,e_3],....\}$. Then
$\mathcal{N}$ is a nest and the corresponding nest algebra is
\[
\operatorname{Alg}\mathcal{N} =
\begin{bmatrix}
* & * & * & * & * & \cdots \\
0 & * & * & * & * & \cdots \\
0 & 0 & * & * & * & \cdots \\
0 & 0 & 0 & * & * & \cdots \\
0 & 0 & 0 & 0 & * & \cdots \\
\vdots  & \vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
.\]

Let $S$ denote the unilateral shift operator such that
$Se_k=e_{k+1}$. If we define $\Phi:\operatorname{Alg}\mathcal{N} \longrightarrow
\operatorname{Alg}\mathcal{N}$ by $\Phi(A)=SAS^*$, then
\[
\Phi :
\begin{bmatrix}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & \cdots \\
0 & a_{22} & a_{23} & a_{24} & a_{25} & \cdots \\
0 & 0 & a_{33} & a_{34} & a_{35} & \cdots \\
0 & 0 & 0 & a_{44} & a_{45} & \cdots \\
0 & 0 & 0 & 0 & a_{55} & \cdots \\
\vdots  & \vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
\mapsto
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & \cdots \\
0 & a_{11} & a_{12} & a_{13} & a_{14} & \cdots \\
0 & 0 & a_{22} & a_{23} & a_{24} & \cdots \\
0 & 0 & 0 & a_{33} & a_{34} & \cdots \\
0 & 0 & 0 & 0 & a_{44} & \cdots \\
\vdots  & \vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}
.\] From this fact we can observe that $\Phi$ is a rank preserving
map on $\operatorname{Alg}\mathcal{N}$ but it is not an onto map.
\end{example}

\begin{example}
Consider the algebra $\mathcal{A}_{\infty}$ which is described at
the beginning of Section 3. Let $\mathfrak{L}$ be the lattice
associated with $\mathcal{A}_{\infty}$. We can easily argue that
for any $M,N \in \mathfrak{L}$, $\mathfrak{G}_M=\mathfrak{G}_N$.
Therefore $\mathcal{A}_{\infty}$ is irreducible. Let $\Phi$ be a
map on $\mathcal{A}_{\infty}$ defined by
\[
\Phi:
\begin{bmatrix}
a & b & 0 & 0 & 0 & 0 & \cdots\\
0 & c & 0 & 0 & 0 & 0 & \\
0 & d & e & f & 0 & 0 & \\
0 & 0 & 0 & g & 0 & 0 & \\
0 & 0 & 0 & h & i & j & \\
 \vdots & & & &   &   & \ddots
\end{bmatrix}
\longrightarrow
\begin{bmatrix}
a & \frac{1}{2} b & 0 & 0 & 0 & 0 & \cdots\\
0 & c & 0 & 0 & 0 & 0 & \\
0 & \frac{3}{2} d & e & \frac{3}{4} f & 0 & 0 & \\
0 & 0 & 0 & g & 0 & 0 & \\
0 & 0 & 0 & \frac{5}{4} h & i & \frac{5}{6} j & \\
 \vdots & & & &   &   & \ddots
\end{bmatrix}
.\] Then it is obvious that the map $\Phi:\mathcal{A}_{\infty}
\rightarrow \mathcal{A}_{\infty}$ preserves rank and it can also
be observed that the map $\Phi$ is implemented by an unbounded
operator
\[ U=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & \cdots\\
0 & 2 & 0 & 0 & 0 & \\
0 & 0 & 3 & 0 & 0 & \\
0 & 0 & 0 & 4 & 0 & \\
0 & 0 & 0 & 0 & 5 & \\
 \vdots & & & &   & \ddots
\end{bmatrix}
.\] In other words, \[\Phi(A)=UAU^{-1}\] for all $A \in \operatorname{Alg}
\mathfrak{L}.$

\end{example}

\bibliographystyle{amsalpha}

redpajama_set_name: RedPajamaArXiv


text:
PCMag Australia Reviews Consumer Electronics Digital Cameras
Fujifilm X-Pro2 Review
The Fujifilm X-Pro2 puts an optical viewfinder in a slim mirrorless design, making it one of the most fun—and capable—cameras on the market.
byJim Fisher Oct 28, 2016
Quick focus and 8.2fps burst shooting.
Large buffer.
Unique hybrid optical/electronic viewfinder.
Focus selection joystick.
Film emulation modes.
All-weather build.
Dual SD slots.
Optical viewfinder not well-suited for telephoto lenses.
Fixed LCD doesn't support touch input.
Video limited to 1080p.
Omits built-in flash.
Expensive.
In the five years since the introduction of Fujifilm's X-Pro1 ($399.00 at Amazon) , the mirrorless camera space has improved by leaps and bounds. The long-awaited X-Pro2 ($1,699, body only) takes everything that photographers love about its predecessor and improve upon it. The sensor resolution has been upped to 24MP, the EVF is crisper, and the camera's focus and shooting rate are significantly faster. If you love shooting with an optical viewfinder, and don't want to shell out the big bucks for a Leica M ($2,969.00 at Amazon) rangefinder, you'll find the X-Pro2 a joy to use. It's a very strong performer, though it faces very stiff competition in the premium mirrorless camera space, including our Editors' Choice, the Sony Alpha 6300 ($589.00 at Amazon) .
The X-Pro2 ($799.95 at Amazon) is styled after a classic rangefinder camera, complete with a viewfinder placed in the top corner, rather than at the center as you find with mirrorless cameras styled after SLRs. It's finished in black, and measures 3.3 by 5.5 by 1.8 inches (HWD) and weighs 15.7 ounces without a lens. There's no built-in flash, but you can mount an external strobe or a PocketWizard ($89.00 at Amazon) in the hot shoe.
The camera handles quite well. An integrated grip makes it comfortable to use handheld, and its size makes it equally comfortable to use with a wrist or shoulder strap. The body has a natural rear thumb rest, but many photographers will add an accessory, like the folding thumb rest from Lensmate. I used the X-Pro2 both with and without the Lensmate add-on and found that, for the most part, I preferred the feel of the camera in my hands with it installed. But it can get in the way of controls, specifically the top control dials. It folds in half to give you better access, but if you frequently adjust shutter speed or EV compensation, you may find the Lensmate to be more of a pain than a plus.
When set to EVF mode, bringing the camera to your eye shows you just what the lens is seeing, complete with a depth of field and focus preview. The 2.4-million-dot EVF refreshes at 85 frames per second, delivering a smooth, clear image to your eye. A diopter adjustment is included, so you can tune the image to best match your eyesight, giving some eyeglass wearers the ability to use the camera sans glasses.
Switching to the optical finder gives you a clear, perfectly focused view of the world, complete with a frame line that adjusts based on the attached lens, and automatically moves based on the set focus distance, compensating for parallax. The finder has two level of magnification, a wide-angle setting (0.36x magnification) for 18, 23, 27, and 32mm lenses, and a tighter 0.6x magnification option for 35, 56, 60, and 90mm primes. It's a little more versatile than a rangefinder with a fixed viewfinder (like the digital Leica M series and its 0.72x finder), as the magnification changes based on the focal length of the attached lens.
The optical finder isn't a great choice for zoom lenses, however. Frame lines do change dynamically as you zoom, but by the time you move from the wide-angle setting to the telephoto extreme of a lens like the Fujinon 18-135mm ($899.95 at B&H Photo Video) , you're dealing with a very small frame line. But attaching a big zoom to the X-Pro2 defeats the camera's purpose; like the rangefinders it is styled after, this is a body that pairs best with compact primes.
And Fujifilm has some very good primes in its lineup. The most appealing for the X-Pro2 are the XF 23mm F2 WR ($449.95) and XF 35mm F2 R WR ($399.00 at Amazon) , both of which are quite small and feature weather-sealed designs that complement the sealed X-Pro2. When paired with a WR lens, the X-Pro2 is an all-weather camera. And, if you're using a prime lens, pressing the button at the center of the viewfinder toggle switch will preview the field of view of various other focal lengths, which can help you decide whether or not to switch lenses for a shot.
The toggle switch is on the front plate, between the lens mount and handgrip. Pulling it away from the lens switches between the optical and electronic views. When the X-Pro2 is set to manual focus mode and the optical viewfinder is enabled, pushing the lever toward the lens adds a small electronic window to the bottom corner of the finder. It shows a magnified view of the selected focus area, which allows you to adjust manual focus with precision, without sacrificing the optical view. It's a feature inherited from the X100T ($889.95 at Amazon) , and while it's helpful if you frequently use manual focus, it's something I barely used, as the autofocus system in the X-Pro2 is outstanding.
The focus mode switch is also located on the face plate—it has S (Single), C (Continuous), and M (Manual) settings. There's also a front control dial, located in the upper right corner. Top controls are all located to the right of the centered hot shoe. A combination shutter speed/ISO dial occupies much of the real estate; a center button must be depressed to turn it out of the A (automatic) position, but the wheel turns freely otherwise, clicking into place at full-stop speeds from 1-second through 1/8,000-second. To change the ISO you need to grip and lift the wheel by its edges; turning the lifted wheel sets the ISO, which is visible in a small cutout window. It too has an automatic setting, as well as Low (ISO 100) and H (ISO 25600) marks, with standard settings (ISO 200-12800) adjustable in third-stop increments.
The power switch surrounds the shutter release, which is threaded to accept a standard release cable or a soft release button. Also on top are a dedicated EV dial (-3 to +3EV in third-stop setttings), along with a C setting for additional adjustment range (-5 to +5EV) via the front control dial, and the programmable Fn button. By default, Fn starts and stops video recording.
Rear buttons start at the top, with the View Mode sitting to the right of the viewfinder. It switches between the rear display, the finder, or an eye sensor that automatically switches between them. Next to it are dedicated buttons to adjust the metering pattern and activate exposure lock (AE-L), and the rear control dial. Playback, Delete, and Display/Back buttons sit to the right of the rear LCD. They're joined by a four-way directional controller with a center Menu/OK button. The up direction is dedicated to the Drive mode, but the down, left, and right controls are programmable.
A small joystick sits on the rear plate. It's used to directly change the active focus point. It's a very useful addition that was absent on the X-Pro1, as older Fuji cameras required you to press a direction button and then use the rear directional pad to manually select a focus point, a cumbersome process.
The final two buttons sit at the right rear edge. AF-L locks focus for a shot, so you can recompose an image without changing the point of focus. It's joined by Q, which brings up an on-screen menu that provides quick access to many shooting settings. You can adjust the focus area, white balance, noise reduction level, file format, film emulation, and image toning parameters (dynamic range, highlight and shadow clipping, color saturation, and sharpness). And you can save any group of those settings as a custom profile. The Q menu also allows you to change LCD brightness, set the self-timer, control external flash settings, and toggle face and eye detection settings, but those parameters aren't saved in custom profiles.
The rear display is a 3-inch, 1,620k-dot LCD. It's not touch sensitive, nor is it mounted on a hinge, but it is very sharp and bright. The X-Pro2 isn't really designed for video use—the Fujifilm X-T2, with its 4K capabilities and tilting rear display is better suited for that—and as much as I like tap-to-focus on mirrorless cameras, the focus selection joystick is so convenient to use that I didn't find myself missing it here.
Fujifilm offers a number of in-camera film emulation modes for JPG shooters, as well as in-camera Raw processing that can apply any of the emulations to Raw shots after you've shot them. I adore the Classic Chrome look—which is Fujifilm's version of Kodachrome—for its muted tones for color, and the Acros black-and-white mode. Monochrome shooting is aided by the ability to apply the effect of a yellow, red, or green filter, without having to invest in physical lens filters.
You can also apply a varying amount of grain to images—Grain Effect settings include Off, Weak, and Strong. I prefer a grainy, rough look to images and left it on Strong for most of my shooting with the X-Pro2, but you can set it to match your taste. In addition to Classic Chrome and Acros, there are Provia (Standard), Velvia (Vivid), Astia (Soft), Pro Negative High Contrast, Pro Negative Standard Contrast, Monochrome, and Sepia settings. I'm normally a Raw shooter, but when I pick up a Fujifilm model I invariably find myself being very happy with the out-of-camera JPGs, as the film emulation modes deliver the look I want in a photo.
The X-Pro2 has Wi-Fi, which is ubiquitous to the point where its omission would be surprising. It's easy enough to transfer images and videos to your smartphone for social sharing via the Fujifilm Cam Remote app, a free download for Android and iOS devices.
The app also supports remote control. If the camera is set to manual mode, full manual control is available via the app. But if the body is locked into shutter priority or aperture priority mode when you start the remote, the remote app will be locked in kind. You do have the ability to tap to focus in any mode, and you can always change the film simulation mode, set the white balance, toggle the self-timer, and start or stop video recording via the app.
The X-Pro2 supports dual SD cards, with both slots taking advantage of the speed delivered by UHS-II media in slot 1. Slot 2 is rated for UHS-I. It features a bevy or ports and connections, including a standard hot shoe, a PC Sync flash socket, micro USB, micro HDMI, and a 3.5mm port for a wired microphone or remote control. In-camera battery charging is not supported. A dedicated external battery charger is included; it features a removable cord rather than an integrated plug.
Performance and Image Quality
The X-Pro2 is a speed demon. It starts, focuses, and fires in about 1.4 seconds, locks focus in about 0.1-second, and can track subjects and capture images at 8.2 frames per second. The high burst rate is matched by a strong shooting buffer. You can manage 27 Raw+JPG, 29 Raw, or 255 JPGs before the shooting rate slows.
See How We Test Digital Cameras
The autofocus system is a huge step up from the X-Pro1. If you take the time to apply a free firmware update, it uses the same algorithm as the X-T2. While I think the XT-2's body is a better match for a telezoom like the excellent Fujinon 100-400mm ($1,899.00 at Amazon) , the lens can track subjects with the same adroitness on the X-Pro2 as with the X-T2. The X-Pro2 doesn't offer the same level of customization as the X-T2, which allows you to discretely set the tracking, speed tracking, and zone area switching sensitivity, and to select from another of case types to best match the action you're shooting, but the X-Pro2 does incorporate face and eye detection, a plus for street photographers.
With default focus settings enabled, I managed a superb focus hit rate when shooting action in the field along with the Fujinon 18-135mm lens. The camera wasn't quite as amazing when I switched to the testing lab, as out of the box it missed several shots in our standard continuous focus test, which shoots at targets moving toward and away from the camera. With a simple change, setting the AF-C Release/Focus priority to Focus rather than Release, the X-Pro2 nailed every shot on this test, while maintaining its 8.2fps shooting rate. This may not be the first camera you think of when it comes to tracking moving action with a telezoom, but it's adept at the task nonetheless.
The X-Pro2 is the first Fujifilm X model to use a new 24MP image sensor. It maintains the X-Trans color filter array we've seen in past models. It's a more complex design than the Bayer sensors used by most digital cameras. The 6-by-6 color filter array used by the X-Trans chip is more complex, creating more natural looking grain at higher ISOs. It also eliminates color moiré in images, so there's no need to add a slight blur to images via an optical low pass filter (OLPF).
I'm a fan of the X-Trans look, although it can be troublesome for certain Raw conversion software to handle. Our standard Raw converter is Adobe Lightroom, and while Adobe has improved its X-Trans development by leaps and bounds when compared with early efforts, it still lags behind some other options. You'll notice that Raw files, converted in Lightroom, don't look quite as crisp as JPG counterparts in the crops included in our slideshow. The best results I've gotten from X-Trans Raw files is from Iridient Developer, a Mac-only Raw converter. It's worth trying (a free trial version is available) if you use an X-Trans camera and work on a Mac.
Of course, there are advantages to shooting in Raw—I was able to reign in highlights and open up shadows in the shot below from the Raw file. The JPG blew out the background completely and left the subjects in dim light in more shadow than I'd like to see. The X-Pro2 delivers incredible dynamic range when working in Raw, which allows you to save shots with extremely mixed lighting like this one.
Fujifilm sells the X-Pro2 as a body only. I shot our ISO test scene using the Fujinon 90mm ($1,249.00 at Amazon) prime at f/5.6. Imatest shows that noise is kept under 1.5 percent through ISO 12800 when shooting JPGs at default settings. At the top ISO 25600 setting the camera shows about 1.9 percent noise in images. Both are excellent results, but a pure number doesn't tell the entire story, as in-camera noise reduction affects how much detail is retained when you push the camera to its highest sensitivities.
Images are crisp with no evidence of loss of quality through ISO 800. Details are slightly smudged at ISO 1600 and 3200, and more noticeably so at ISO 6400 and up. Detail isn't as strong as the Sony Alpha 6300 at higher sensitivities; the Alpha's output is crisper through ISO 6400, it's a toss-up at ISO 12800, and I'd say the X-Pro2 wins the day at ISO 25600.
In terms of Raw image detail, the X-Pro2 keeps up with the Alpha 6300 through ISO 3200. But the Alpha shows more detail and contrast at ISO 6400 and 12800, with the gap closing and ISO 25600 results ending in a tie. The Sony has an extra gear, an ISO 51200 setting the X-Pro2 doesn't support. The Alpha 6300 is the best APS-C camera we've seen at high ISO, so the fact that the X-Pro2 doesn't quite measure up to the high mark set by the Alpha 6300 shouldn't be considered a deal breaker—I'd still qualify its image quality as excellent.
Where it lags behind more noticeably is in video capture. Footage is recorded in 720p or 1080p resolution at 24, 25, 30, 50, or 60fps. Video quality is a big step forward when compared with other Fuji cameras that use the older 16MP X-Trans sensor. Details are crisp, autofocus is quick, and there's no evidence of moiré. But the X-Pro2 is a 1080p camera in a 4K world—and competing models, including Fuji's own X-T2, as well as the Sony Alpha 6300 and Sony Alpha 6500, support video capture at 4K quality.
Despite the resolution limitation, the X-Pro2 does have some solid pluses as a video camera. You can apply any film look to video footage, and there is a microphone input. The internal microphone picks up voices close to the camera with clarity, but also picks up background noise, a typical result for an in-camera mic. When paired with a stabilized lens handheld footage is smooth and steady, but the X-Pro2 doesn't offer in-body stabilization, a feature that's included with two of the competitors on the horizon, the Alpha 6500 and the Olympus OM-D E-M1 Mark II.
I absolutely adore the Fujifilm X-Pro2. It's compact, sealed for use in inclement weather, and backed by a strong lens system with many top-quality, affordable options. It also delivers the best image quality we've seen to date in an X series camera, offers a control layout that makes it very easy to take control exposure, and includes a hybrid optical/electronic viewfinder that sets it way apart from rest of the mirrorless camera pack. Modern electronic viewfinders, including the one in the X-Pro2, are fantastic. But there's something special about bringing a camera to your eye and seeing a crisp, bright view of the world in the eyepiece. The X-Pro2's optical finder makes it special.
But despite my admiration for the camera, I'm not naming the X-Pro2 our Editors' Choice. The Sony Alpha 6300 betters it in many ways, including a faster 11.1fps shooting rate, 4K video recording, and an attractive $1,000 price point, making it a more well-rounded alternative. That's not to say the Sony is perfect—the company's APS-C lens system isn't as full of strong performers when compared with Fuji, although the Alpha 6300 can mount full-frame Sony FE lenses, which goes a long way to bridge that gap.
Of cameras currently on the market, the Fujifilm X-T2 is the strongest threat to the Alpha 6000. It's $100 less than the X-Pro2, built just as tough, and should deliver the same image quality, more customizable autofocus, and 4K video. (I've yet to review the X-T2, which, but it's the next camera we'll be looking at from Fujifilm.) Also looming on the horizon is the Sony Alpha 6500, which should at least be the equal of the Alpha 6300 in terms of image quality and autofocus speed, and adds in-body image stabilization, a larger shooting buffer, and a touch screen, all for $1,400. And there's the Olympus OM-D E-M1 Mark II, which hasn't received official US pricing yet, but promises to shoot Raw images at 18fps with tracking and at an incredible 60fps with locked focus and exposure.
The X-Pro2 is the clear choice for any photographer who wants to get as close to the rangefinder experience as possible without shelling out the big bucks for a real mechanically coupled Leica rangefinder. The optical finder is a real treat, especially when working with a compact prime lens in the wide to standard angle range. And if you're anything like me, the various film emulation settings will keep you reaching for the X-Pro2 again and again.
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About Jim Fisher
Images, and the devices that capture them, are my focus. I've covered cameras at PCMag for the past 10 years, which has given me a front row seat for the DSLR to mirrorless transition, the smartphone camera revolution, and the mainstream adoption of drones for aerial imaging. You can find me on Instagram @jamespfisher.
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redpajama_set_name: RedPajamaCommonCrawl


text:
New delivery service launches in Eau Claire
Story by Bridget Kelley, Sports EditorNovember 13, 2018Leave a Comment
More stories from Bridget Kelley
EatStreet launches, donates to local food pantry
Gyro King on Water Street is one of several businesses adopting EatStreet's delivery service.
Photo by Sam Farley
EatStreet, a Madison-based company, launched their service in several cities across Wisconsin this past week, including Eau Claire.
EatStreet was created in 2010 by Matt Howard, co-founder and CEO of EatStreet, and two of his friends. The trio were students at UW-Madison and were using an ordering service that charged a fee for using it, Howard said. Thus, Badgerbites was born. Badgerbites later became EatStreet as the business expanded.
EatStreet was originally just an ordering service but eventually evolved into a delivery service. Howard said customers are able to see where their food is in the ordering process using the order-tracking feature on their app.
"We really are focused on making the ordering process as easy as possible," Howard said.
EatStreet has served in over 250 cities across the United States and delivers from over 15,000 restaurants, according to their website. Howard said EatStreet provides food from local restaurants in addition to major chains, giving the customer more variety.
EatStreet currently provides delivery from 28 local and chain restaurants in Eau Claire, including The Goat Coffee House and Buffalo Wild Wings. It also offers free delivery in Eau Claire until Dec. 6 as a promotion of their launch.
Howard said Eau Claire was a natural next stop for the company. The company is available primarily in college towns, as their target demographic is college-aged students. Eau Claire does not currently have many delivery services, so it made sense for EatStreet to launch in Eau Claire, Howard said.
Ellie Decker, a first-year nursing student, said that EatStreet is convenient for those who do not have a lot of time to cook or transportation to go get food.
"I think that it'd be great, especially for the university community, because a lot of students can benefit from that," Decker said. "We're busy and, especially with winter, it's nice to have things delivered."
EatStreet opened four new markets in Wisconsin and toured the state in a custom green double-decker bus called the Double-Decker Hunger Wrecker, according to the company's Twitter. On their "Haulin' Ass for Hunger Tour," the Hunger Wrecker made stops in Eau Claire, Green Bay, Appleton and La Crosse.
Howard said it has always been important to him and the company to give back to their community. Besides promoting their launch in these cities, EatStreet donated to food pantries in each of the cities they visited. The company donated $20,000, including a $2000 donation to Feed My People in Eau Claire. The company also spent time volunteering in local food pantries.
"We wanted to help the local communities where we are launching," Howard said.
Kelley can be reached at [email protected]
Tags: delivery, EatStreet, eau claire, Matt Howard
Bridget Kelley, Sports Editor
Bridget Kelley is a second-year journalism student and Sports Editor for The Spectator. Bridget enjoys babies, coffee and oxford commas. Kelley has not yet found gluten-free donuts, so keep suggestions coming.
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text:
About Baltic Sea Properties
Baltic Sea Properties is a Norwegian real estate company investing in the Baltics.
Baltic Sea Properties is a Norwegian company focusing on commercial real estate in the Baltics. The company's operations is organised trough fully-owned subsidiaries which are all registered in Lithuania.
The company owns and manages about 130 000 square metres of commercial property in the Baltics.
The portfolio consists of approx. 130.000 sqm commercial real estate, including:
The company also manages other investors' portfolios in the retail and office sector of Lithuania.
The Team - Board and Management
LARS CHRISTIAN BERGER
Lcb@balticsea.no
Lars Christian Berger has more than 10 years of experience from finance and real estate management. He has worked as an asset manager for several real estate funds and managed apx. 300.000 sqm commercial property and development projects.
Lars has been the CEO of Baltic Sea Properties since November 2017, and holds a M.A. in Business & Finance from Heriot Watt University, Edinburgh.
SIGITAS JAUTAKIS
MD - Local Operations
sj@balticsea.no
Sigitas Jautakis has more than 15 years of experience from the Baltic real estate market and has worked with transactions, tenant advisory, and property management. He was previously the CEO of Newsec Asset & Property Management in the Baltics and has been the MD of BSP's Lithuanian operations since 2015.
Sigitas holds a M.Sc. in Production Economics and Leadership from the University of Vilnius.
MILDA MALINAUSKIENE
mm@balticsea.no
Milda Malinauskiene has over 13 years of experience working in the real estate sector and over 18 years in finance as a senior accountant, auditor and finance director in Lithuania, Ireland and UK. She has worked as a project co-ordinator and a chief accountant for a commercial real estate portfolio owned by Canadian Investment Fund, as an investment manager, which involved asset/property management, real estate development and finances for a fund management company.
Milda joined Baltic Sea Properties as an investment manager in May 2019.
She graduated in finance and accounting from Economic College, and took advanced studies in Griffith College, Dublin as an ACCA candidate.
JULIJA NIKITARAVICIENE
Head of Administration and Legal
jn@balticsea.no
Julija Nikitaraviciene is an attorney at law and has 15 years of experience in the fields of real estate transactions, corporate, M&A and financing. She is also an associate partner at Ellex Valiunas.
Julija is responsible for administration and daily operations of BSP group companies, including legal matters.
Julija holds a Master of Laws (LL.M) degree from the Vilnius University.
INGA MOCKUVIENE
im@balticsea.no
Inga Mockuviene has over 16 years of experience working in finances, 10 of them in real estate sector.
Inga has been responsible for accounting and finances of Baltic Sea Properties in Lithuania since 2014. She joined the company as a Chief Accountant in June 2019.
Inga is educated in Biology and Ecology from Vilnius University.
RITA MATONIENE
rm@balticsea.no
Rita Matoniene has over 12 years of experience in accounting, 8 of them in real estate sector. Rita joined Baltic Sea Properties as an accountant in May 2019.
Rita studied accounting and finances. She is educated in Business Management from Vilnius Gediminas Technical University.
GEDIMINAS KLIMAS
gk@balticsea.no
Gedminas Klimas has more than 15 years of experience within technical management of buildings and construction projects. He previously worked for Newsec for 8 years as the leader of the department for technical maintenance, and had through this the responsibility for all property classes.
Gediminas is currently responsible for the technical management of BSP's portfolio and management clients, including construction management for all developments and maintenance of projects in Lithuania.
He is educated a Civil Engineer from the Vilnius Technological University.
JAMES ANDREW CLARKE
Executive Chairman of the Board
james@bntp.lt
James Clarke has experience from leading roles in Lithuanian real estate companies (i.ea. Baltic Equity Group and Baltic Real-Estate Developments) and owns several logistics, shopping centres and business parks.
He holds an Executive MBA from Baltic Management Institute and B.Sc. in Construction Management from Waterford Institute of Technology. He is also educated within corporate governance for board members from Baltic Institute of Corporate Governance.
James has been a board member since October 2017.
Bjørn Bjøro
bere-ber@online.no
+47 993 22 623
Bjørn Bjøro has for more than 25 years been member of the congregation of Centralkirken's financial committee, managing more than NOK 125 million in equity.
He has also been a board member of Baltinor Property for many years, investing in industrial real estate in Lithuania.
He is board member of Eiendomskapital Buskerud AS, chairman of the board in Eiendomskapital Norge V AS and has had the project management responsibilities for upgrading these properties.
Bjørn has been a board member since May 2018.
JOHN AFSETH
ja@balticsea.no
John Afseth has long experience as chairman and board member from multiple companies admitted to trade on Merkur Market, Oslo Axess and Oslo Stock Exchange (e.g. Gentian Diagnostics, Aqua BioTechnology ASA and Simtronics ASA)
John is also board member of Emerging Europe Commercial Properties AS, another Norwegian real estate company in Lithuania.
He has been a board member of BSP since 2010, and holds a Ph.D. in microbiologics and pharmacology from the University of Oslo.
JOHN DAVID MOSVOLD
mosvold@holstein.no
John D. Mosvold has extensive experience with real estate and shipping projects both in Norway and the Baltics, both as chairman of the board, board member and investor.
He has been active in the Baltics since the latter half of the 90's, and brings with him important knowledge both professionally and geographically.
John is educated from BI Norwegian Business School, and has been a board member of BSP since 2015.
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0252 OSLO,
post@balticsea.no
c/o Hagen Berger og Aas AS,
Postboks 2076 Vika,
Baltic Sea Properties
Baltic Sea Properties er et norsk eiendomsselskap med eiendom i Baltikum.
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A Temporada de Fórmula 3000 de 1997 foi a décima-terceira da história da categoria. O campeão foi o brasileiro Ricardo Zonta, da Super Nova Racing e o vice foi o colombiano Juan Pablo Montoya, da RSM Marko. A diferença entre os dois pilotos foi de apenas 1,5 ponto (39 para Zonta, 37,5 para Montoya)

Equipes e pilotos
Todos as equipes competiram com chassi Lola T96/50 e motores Zytek-Judd.

Classificação

Notas
 Ricardo Zonta foi desclassificado do GP de Silverstone por irregularidades no câmbio de seu carro.
 Realizada sob chuva, a etapa de Nürburgring foi encerrada antes do tempo depois de um violento acidente com o inglês Dino Morelli. Com isso, metade dos pontos foi atribuída.
 Juan Pablo Montoya foi desclassificado do GP da Bélgica por andar com o carro danificado.
 Jamie Davies e Grégoire de Galzain foram excluídos do resultado final da etapa de Mugello por não terem comparecido ao briefing. Tom Kristensen também foi punido com desclassificação, desta vez por uma irregularidade em seu carro.
 Rui Águas foi desclassificado do GP de Hockenheim por direção perigosa.
 Cyrille Sauvage, que terminou o GP de Jerez em segundo lugar, foi desclassificado porque um membro de sua equipe, a Draco Engineering, entrou no parque fechado. Oliver Tichy herdou a posição e Gareth Rees, que terminara em quarto, completou o pódio.
 Kurt Mollekens foi excluído do resultado final do GP de Silverstone por não ter comparecido ao briefing.

1997
1997 no automobilismo

redpajama_set_name: RedPajamaWikipedia


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RSF decries Ukrainian security raids on TV channel, reporters
Photo credit: AFP/radiosvoboda.org
Reporters Without Borders (RSF) condemns last week's searches of the homes of three Ukrainian journalists and the Kiev-based commercial TV channel 1+1 in response to the leak of an audio recording of the prime minister that led him to offer his resignation – an offer eventually rejected by the president. Journalists' sources must be protected, Reporters sans frontières said.
The Security Service of Ukraine (SBU) has been targeting 1+1's investigative journalism programme "Sekretni Materialy" (Secret Documents) ever since it broadcast the recording, which shook the Ukrainian political landscape.
On 5 February, the SBU searched 1+1's headquarters, the homes of Denis Danko and Yevhen Kuksin, who are members of the TV channel's journalistic investigations unit, and the home of Oleksandr Nizovtsev, a journalist who used to work for 1+1. The SBU claimed to be looking for illegally obtained material although the three journalists insist they obtained the leaked recording legally.
"These searches violated the confidentiality of journalists' sources because the seized computer material contained potentially sensitive data," said Jeanne Cavelier, the head of RSF's Eastern Europe and Central Asia desk. "Police and judicial access to journalistic sources must be limited to extremely exceptional circumstances that are strictly determined by the law. We are alarmed by the repeated attempts to violate the confidentiality of sources in Ukraine, including the harassment of journalists working for the Radio Free Europe programme "Skhemy" in October 2019."
Prime Minister Oleksi Honcharuk offered to resign on 15 January following publication of the audio recording made without his knowledge, in which he is heard questioning President Volodymyr Zelensky's competence in economic matters during a private meeting with the head of the national bank.
The president refused to accept his resignation and ordered the police to find out who made the recording. During a meeting attended by the head of 1+1 on 6 February, he said he had proof that the journalists had played a direct role.
1+1 , which is owned by the oligarch Ihor Kolomoisky, published an open letter to President Zelensky on 6 February accusing the authorities of violating the freedom of expression that is guaranteed by article 34 of Ukraine's constitution, and of an "attempt to put political pressure on an independent media outlet."
The parliamentary committees on freedom of expression and on humanitarian and information issues held special meetings the same day and condemned the "obstruction of the work of journalists." Two Ukrainian NGOs, the Institute of Mass Information (IMI) and DetektorMedia, issued a statement reminding the security services and journalists of the need to respect the law.
1+1 has also said it plans to file a complaint against the SBU for several violations of legal procedure during the searches. The searches should have been limited to the offices of the journalists named in the judicial investigation but were also carried out in other areas of the TV channel.
Two computers were seized in violation of article 168 of the criminal procedure code, which allows the police to copy data files but not seize equipment. In the search of Danko's home, the SBU seized a tablet computer, three USB sticks and wedding photos, and no warrant was provided during the raid on Kuksin's home. The SBU responded to the criticism by flatly denying any procedural violations and claiming that that these three journalists were not the only targets of the investigation.
Ukraine is ranked 102nd out of 180 countries in RSF's 2019 World Press Freedom Index.


There are many organisations who you can contact for help, advice and support. Here is a selection of the organisations who may be able to help you.
Non urgent medical advice. In an emergency phone 999 for an ambulance.
Advice and support from the University and Students' Union for lesbian, gay, bisexual, trans, queer and intersex students.
The local police service. In an emergency phone 999.
Information on learning English and studying in the UK.
Information on managing money, borrowing money, and how to deal with debt problems.


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Home > News & Events > Retail Energy Supply Association Responds to COVID-19 Pandemic
Retail Energy Supply Association Responds to COVID-19 Pandemic
California, Connecticut, Delaware, Federal, Georgia, Illinois, Maine, Maryland, Massachusetts, Michigan, New Hampshire, New Jersey, New York, Ohio, Pennsylvania, Rhode Island, Texas, Virginia, Washington DC, Wisconsin
(Harrisburg, PA) April 20, 2020 – Retail Energy Supply Association (RESA), the nation's leading trade association representing competitive retail energy suppliers, is responding to the pandemic by actively monitoring and sharing critical, real-time information on COVID-19 regulatory directives by state.
RESA recognizes the need for relevant, timely data because the rapidly changing COVID-19 environment presents a unique set of challenges ranging from changing customer load profiles, to a surge in residential usage versus commercial usage, to unexpected delays in starting new capital projects, to managing increased supply demands and more.
"This pandemic has upended all our lives and the way we conduct business," said president of RESA, Matt White. "The information RESA is monitoring, aggregating and disseminating daily ranges from the newest COVID-19 regulatory directives, to shelter in place orders, to price-gouging statutes by state."
The expectation of demand is especially high during COVID-19 because of emergency and medical facilities, coupled with stay-at-home residents' usage. The significant variations in supply and demand requires retail energy suppliers to not only be responsive, but aware of irregular use patterns in this unique environment.
"As we watch the COVID-19 situation unfold, we will monitor the impact this public health crisis is causing to competitive energy retailers," said RESA executive director, Tracy McCormick. "Not only is the information we are sharing valuable to our members, but also to regulators, legislators, and our partners. We will continue to propose and recommend solutions that best serve our industry."
RESA is committed to protecting the safety of its members, co-workers, staff, partners and their families. The association strongly supports following federal, state and local orders to ensure a safe work environment. RESA will continue to hold its meetings remotely until further notice.
The Retail Energy Supply Association is a broad and diverse group of retail energy suppliers who share the common vision that competitive retail energy markets deliver a more efficient, customer-oriented outcome than a regulated utility structure. RESA is devoted to working with all stakeholders to promote vibrant and sustainable competitive retail energy markets for residential, commercial and industrial consumers. For more information visit: www.resausa.org. Follow RESA on LinkedIn, Facebook and Twitter.
For all media inquires, please contact:
Amy Power
Stacey Gaswirth
Together with our members, we play a critical role in educating key stakeholders and are active in support of or opposition to bills that affect competitive electricity markets, consumers, and RESA initiated legislation.
Read more about RESA


El Paso Community College (EPCC) Summer Repertory will open the season with Boots. Performances will be June 23, 24, 25, 30, July 1 and 2 at 8:00 p.m. and Sunday matinees on June 26 and July 3 @ 2:30 pm.
Boots is a new comedy written by Stephanie Karr and Mark Watts and directed by Matthew Robert Smith. If you like Country-Western bars you'll wanna two-step on in to Boots, a west Texas Bar where you can mingle with the reg'lars.
A friendly bunch, more or less, with characters the likes of which you're no doubt familiar. And, who's the dude always sittin' in the dark corner?
The EPCC Summer Repertory performs at the EPCC Transmountain Campus Forum Theater (9570 Gateway North) General Admission is $15, Non-EPCC students and military are $10 and EPCC students/staff and senior citizens are $7. Tickets are available at the door. The box office opens two hours before each performance.


For over 29 years we have delivered physical therapy, chiropractic, and massage therapy to patients on the North Shore. Beginning in Lake Forest and expanding to Highland Park and Lake Bluff, we have helped over 10,000 patients with conditions ranging from common to severely complicated. We work closely with your surgeon or primary care physician to ensure a safe, successful, and lasting rehabilitation. Under the direction of our founder David A. Johnson, DC, PT, DPT, North Shore Spinal & Sports Rehabilitation (NSSSR)has created an experienced team of physical therapist, chiropractor, and massage therapist professionals in both our Highland Park and Lake Bluff locations.
Our Director of Clinics, David A. Johnson, DC, PT, DPT, and his staff will support you in your recovery. Whether you are dealing with back pain, neck pain, headache or a sports injury, we are here to help you every step of the way. So if you're looking for exceptional pain relief and rehabilitation in the North Shore Chicago area, please call us.
Choose the multidisciplinary care team who continues to set the standard for helping the toughest cases achieve lasting results.
To schedule an appointment, please call our office in Lake Bluff at (847) 295-0920 or Highland Park at (847) 432-4077 . We also serve the communities of Deerfield and Glencoe, Illinois and nearby locations. For your convenience, you can use our online Request an Appointment form.


HEATED/COOLED SEATS, REMOTE START, BACKUP CAMERA, SIRIUSXM RADIO, BLUETOOTH, 6-Speed Automatic Electronic with Overdrive, 4WD, Jet Black w/Leather Appointed Seat Trim, 4-Wheel Disc Brakes, Auto-Dimming Inside Rear-View Mirror, Auxiliary External Transmission Oil Cooler, Deep-Tinted Glass, Driver Alert Package, Electronic Autotrac Transfer Case, EZ Lift & Lower Tailgate, Front Frame-Mounted Black Recovery Hooks, Front Halogen Fog Lamps, Heavy-Duty Rear Locking Differential, Leather Wrapped Heated Steering Wheel, Preferred Equipment Group 4SA, Rear 60/40 Folding Bench Seat (Folds Up), Rear Vision Camera w/Dynamic Guide Lines, Rear Wheelhouse Liners, Remote Keyless Entry, SLT Preferred Package, Theft Deterrent System (Unauthorized Entry), Trailering Equipment, Ultrasonic Front & Rear Park Assist. CARFAX One-Owner. Clean CARFAX.
Truck stuff. Haulin` cargo, wheelin` through dirt `n mud, towin` that 30 ft. Camper. Truck stuff, the stuff you need to do, the stuff that you can do with this fantastic GMC Sierra 1500. Need to pull your buddy`s Prius out of the ditch? Better have a truck. Need to bring a load of cold, hard steel to the work site? You better not try that in the minivan, it`s best used for haulin` soccer balls and spilled cheese puffs. You`re doing truck stuff, and for truck stuff, you need this GMC Sierra 1500. Find out why our dealerships have won DealerRater.com DEALER OF THE YEAR a whopping 10 TIMES! Call, email, or live chat with one of our friendly sales professionals now to schedule your test drive!
Hello, I would like more information about the 2014 GMC Sierra 1500, stock# CT05245A.
I would like test drive the 2014 GMC Sierra 1500, stock# CT05245A.
Please text me about this 2014 GMC Sierra 1500 (1GTV2VEC7EZ100941).


In March of 2009, Josh Freese (a session drummer who's played with every band you've ever heard of) combined preorder-funding, tiered pricing, and a huge dollop of punk rock ridiculousness to fund his second album, Since 1972. For $7, you could download the album. For $50, you would get a signed CD, a T-shirt, and a personal thank-you phone call from Freese.
For $1,000, Freese would wash your car, take you out drinking, and then you'd give each other haircuts in the parking lot of the Long Beach courthouse. For $20,000, Freese, Tool's Maynard James Keenan, and Devo's Mark Mothersbaugh would take you mini-golfing, and then drop you off on the side of the freeway.
The publicity stunt was a huge success, getting write-ups everywhere from Wired to NPR, and "crowdfunding" took off. This was a huge boon to Kickstarter, a crowdfunding service that launched a month after the release of Since 1972, and quickly established itself as the market leader.
Crowdfunding services – websites that act as both a social network to connect projects with backers and as a marketplace or escrow house for project funding – have become a popular business model in the last two years, and several more have sprung up alongside Kickstarter, each with their own perks, quirks, and twists on the basic model.
Crowdfunding is a natural fit for an independent game developer who needs to connect with an audience and secure funding at the same time; rather than having to "prove" your game to a publisher, you're "proving" it directly to your customers, and you don't have to go out on a limb funding it yourself with no idea of whether or not you'll sell enough copies to recoup your investment.
While Kickstarter is the most well-known crowdfunding service, it may not be the best fit for every video game project, so we're taking a look at five different crowdfunding services for video game projects to help you decide what will work best for your needs.
Kickstarter, the leader of the pack, has successfully funded over 6,500 projects since launching in April of 2009, 67 of which have been video games. IndieGoGo actually launched over a year before Kickstarter, but limited its funding to film projects until 2010.
RocketHub and ulule are two young up-and-comers who launched last year and are taking the basic crowdfunding model in interesting new directions; RocketHub is building gamification and an incubator of sorts on top of the basic model, while ulule is coupling its crowdfunding service with a message board and keeping costs to a bare minimum to create a very friendly and inviting space. Finally, 8-Bit Funding is the underdog of the bunch, launched at the beginning of this year as a crowdfunding service exclusively for video games.