# On Diximier’s averaging theorem for operators in type ${\rm II}_{1}$ factors

Shilin Wen  School of Mathematics and Information Sciences, China West Normal
University, Nanchong, Sichuan, 637000, China<EMAIL_ADDRESS>,
Junsheng Fang  School of Mathematical Sciences, Hebei Normal University,
Shijiazhuang, Hebei, 050024, China<EMAIL_ADDRESS>and  Zhaolin Yao
School of Mathematical Sciences, Hebei Normal University, Shijiazhuang, Hebei,
050024, China<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{M}$ be a type ${\rm II_{1}}$ factor and let $\tau$ be the
faithful normal tracial state on $\mathcal{M}$. In this paper, we prove that
given finite elements $X_{1},\cdots X_{n}\in\mathcal{M}$, there is a finite
decomposition of the identity into $N\in\mathbb{N}$ mutually orthogonal
nonzero projections $E_{j}\in\mathcal{M}$, $I=\sum_{j=1}^{N}E_{j}$, such that
$E_{j}X_{i}E_{j}=\tau(X_{i})E_{j}$ for all $j=1,\cdots,N$ and $i=1,\cdots,n$.
Equivalently, there is a unitary operator $U\in\mathcal{M}$ such that
$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}X_{i}U^{j}=\tau(X_{i})I$ for
$i=1,\cdots,n$. This result is a stronger version of Dixmier’s averaging
theorem for type ${\rm II}_{1}$ factors. As the first application, we show
that all elements of trace zero in a type ${\rm II}_{1}$ factor are single
commutators and any self-adjoint elements of trace zero are single self-
commutators. This result answers affirmatively Question 1.1 in [10]. As the
second application, we prove that any self-adjoint element in a type ${\rm
II}_{1}$ factor can be written a linear combination of 4 projections. This
result answers affirmatively Question 6(2) in [15]. As the third application,
we show that if $(\mathcal{M},\tau)$ is a finite factor, $X\in\mathcal{M}$,
then there exists a normal operator $N\in\mathcal{M}$ and a nilpotent operator
$K$ such that $X=N+K$. This result answers affirmatively Question 1.1 in [9].

Shilin Wen was partly supported by NSFC(Grant No.12001437) and the Fundamental
Research Funds of the China West Normal University (21E027)

Junsheng Fang was partly supported by NSFC(Grant No.12071109) and a Start-up
funding of Hebei Normal University.

Zhaolin Yao was partly supported by Science Foundation of Hebei Normal
University(Grant No.13113115)

Keywords: Type ${\rm II}_{1}$ factor, Dixmier’s averaging theorem, Commutator

MSC2010: 47C15

##  1\. Introduction

Let $\mathcal{M}$ be a type ${\rm II_{1}}$ factor and let $\tau$ be the
faithful normal tracial state on $\mathcal{M}$. In [7], the following result
is proved. If $X=X^{*}\in\mathcal{M}$, then there is a decomposition of the
identity into $N\in\mathbb{N}$ mutually orthogonal nonzero projections
$E_{j}$, $I=\sum_{j=1}^{N}E_{j}$, for which $E_{j}XE_{j}=\tau(X)E_{j}$ for all
$j=1,...,N$, equivalently, there is a unitary operator $U\in\mathcal{M}$ with
$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}XU^{j}=\tau(X)I$. A natural question
arises: can we remove the selfadjoint assumption on $X$. In [29], the authors
proved that for an arbitrary operator $X$ in the ultrapower algebra of the
hyperfinite type ${\rm II}_{1}$ factor $\mathcal{R}^{w}$, there is a
decomposition of the identity into $N\in\mathbb{N}$ mutually orthogonal
nonzero projections $E_{j}\in\mathcal{R}^{w}$, $I=\sum_{j=1}^{N}E_{j}$, for
which $E_{j}XE_{j}=\tau(X)E_{j}$ for all $j=1,...,N$. The main result of this
paper generalizes the above results to arbitrary operators in any type ${\rm
II}_{1}$ factors. Precisely, we prove the following result.

######  Theorem 1.1 (Main Theorem).

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor,
$X_{1},\cdots,X_{n}\in\mathcal{M}$. Then there is a family of finitely many
mutually orthogonal non-zero projections $\\{E_{i}\\}_{i=1}^{N}$ in
$\mathcal{M}$ such that $\sum\limits_{i=1}^{N}E_{i}=I$ and
$E_{i}X_{j}E_{i}=\tau(X_{j})E_{i}$ for $1\leq i\leq N$ and $1\leq j\leq n$.
That is, $X_{j}$ can be written as follows

$X_{j}=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr
E_{1}&\tau(X_{j})&(X_{j})_{12}&\cdots&(X_{j})_{1N}\cr
E_{2}&(X_{j})_{21}&\tau(X_{j})&\cdots&(X_{j})_{2N}\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&(X_{j})_{N1}&(X_{j})_{N2}&\cdots&\tau(X_{j})\cr}.$

Equivalently, there is a unitary operator $U\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{i=0}^{N-1}(U^{i})^{\ast}X_{j}U^{i}=\tau(X_{j})I$

for $1\leq j\leq n$.

The main theorem has several interesting applications.

Application on the commutators in type ${\rm II}_{1}$ factors.

An operator $A$ in a von Neumann algebra $\mathcal{M}$ is a commutator in
$\mathcal{M}$ if there exist operators $B$ and $C$ in $\mathcal{M}$ such that
$A=BC-CB$. A self-commutator is a self-adjoint operator in $\mathcal{M}$ which
is of the form $X^{*}X-XX^{*}$ for some operator $X\in\mathcal{M}$. The
problem of specifying which operators in $\mathcal{M}$ are commutators in
$\mathcal{M}$ has been attacked by several authors and has been solved in
certain special cases. K. Shoda in [28] showed that if $\mathcal{M}$ is a type
${\rm I}_{n}$ factor, then a matrix has trace zero if and only if it is a
commutator in $\mathcal{M}$. A. Brown and C. Pearcy showed that if
$\mathcal{M}$ is a factor of type ${\rm I}_{\infty}$, then non-commutators in
$\mathcal{M}$ are exactly the operators that are congruent to a non-zero
scalar modulo the ideal of compact operators [5]. A. Brown and C. Pearcy in
[4] showed that if $\mathcal{M}$ is a factor of type ${\rm III}$ acting on a
separable Hilbert space, then the commutators in $\mathcal{M}$ consist exactly
of the non-scalar operators together with the operator zero. In [6], A. Brown,
C. Pearcy and D. Topping showed that every operator in the strong radical of
$\mathcal{M}$ is a commutator in an arbitrary properly infinite von Neumann
algebra $\mathcal{M}$.

However, The theory is far from complete and one of the most intractable of
unsolved problems is that of determining the commutators in a finite von
Neumann algebra. A commutator in a finite von Neumann algebra must have
central trace zero, and is not unreasonable to hope that the commutators in
such an algebra are exactly the operators with central trace zero.

The case of type ${\rm II}_{1}$ factors remains open. Some partial results are
known. Fack and de la Harpe in [12] showed that every element of trace zero is
a sum of ten commutators (and with control of the norms of the elements). S.
Goldstein and A. Paszkiewicz improved the number(see [15]). They proved that
any self-adjoint operator of central trace zero in a von Neumann algebra of
type ${\rm II}_{1}$ is a sum of four commutators. The best previous estimates
belonged to Marcoux in [20] which showed that every element of trace zero in a
type ${\rm II}_{1}$ factor is a sum of two commutators, and every self-adjoint
element of trace zero is a sum of four or fewer self-commutators. Pearcy and
Topping, in [26], showed that in the type ${\rm II}_{1}$ factors of Wright
every self-adjoint element of trace zero is a single commutator, and hence,
every trace zero element is a sum of two commutators. Recently, K. Dykema and
A. Skripka showed that all nilpotent elements are single commutators and every
normal element with trace zero and purely atomic distribution is a single
commutator(see [10]).

Using our main theorem, we show that all elements of trace zero in a ${\rm
II}_{1}$ factor are single commutators and any self-adjoint elements of trace
zero are single self-commutators. This result solves the problem of
commutators in type ${\rm II}_{1}$ factors [10].

Application on the linear combination of projections in type ${\rm II}_{1}$
factors.

The linear combination of projections in von Neumann algebra factors is an
active topic. For example, any self-adjoint operator acting in a finite
dimensional Hilbert space can be written as a linear combination of a finite
number of projections. A natural question is whether the number of projections
does not depend on the dimension of Hilbert space. If the Hilbert space is
infinite, is that possible? If the elements of algebra can be written as
linear combinations of projections, a question naturally arises: what is the
minimum number of projections?

The first positive results were obtained by Fillmore (see [13]). He proved
that any bounded operator acting on a separable infinite-dimensional Hilbert
space can be written as a linear combination of 257 projections. Further,
Fillmore was able to get down the number of projections to 9 in [14]. Pearcy
and Topping in [25] reduced it to 8, then A. Paszkiewicz to 6 (see [24]).
Using the techniques in [24] , Matsumoto proved in 1984 that 5 is enough. Also
in 1984, Nakamura [22] showed that any self-adjoint operator can be written as
a linear combination of 4 projections.

In [15],[16] and [17], S. Goldstein and A. Paszkiewicz showed that each self-
adjoint operator in type ${\rm II}_{1}$ factors is a linear combination of 12
projections and there exists a self-adjoint operator in any type ${\rm
II}_{1}$ algebra that can not be written as a linear combination of 3
projections; Also for type ${\rm II}_{\infty}$ factors, any self-adjoint
operator can be written as a linear combination of 4 projections and there
exists a self-adjoint operator that can not be written as a linear combination
of 3 projections; For countable decomposable type ${\rm III}$ factors, each
self-adjoint element is a linear combination of 3 projections.

As the second application, we prove that any self-adjoint element in type
${\rm II}_{1}$ factor can be written a linear combination of 4 projections.
This result answers affirmatively Question 6(2) in [15].

Application on the generalization of Schur Theorem.

We state the famous theorem of Schur. For every matrix $T\in
M_{n}(\mathbb{C})$, there exists a unitary matrix $U\in M_{n}(\mathbb{C})$
such that $U^{-1}TU$ is an upper triangular matrix. The diagonal entries of
$U^{-1}TU$ are the eigenvalues of $T$, repeated up to multiplicity, and $U$
can be chosen so that they appear in any order. Hence each ordering of the
spectrum of $T$ gives a decomposition $T=N+Q$, where $N$ is normal and $Q$ is
nilpotent.

Dykema, Sukochev and Zanin in [11] used a Peano curve covering the support of
the Brown measure of an operator $T$ in a diffuse, finite von Neumann algebra
to give an ordering to the support of the Brown measure, and create a
decomposition $T=N+Q$, where $N$ is normal and $Q$ is s.o.t.-quasinilpotent.
This constructions generalize the Schur upper triangular form of an $n\times
n$ matrix. The normal part $N$ is constructed as the conditional expectation
of $T$ onto an abelian algebra generated by an increasing net of Haagerup-
Schultz projections of $T$. It is natural to ask the question: under what
circumstances is the s.o.t.-quasinilpotent operator $Q$ actually
quasinilpotent.

As the third application, we show that if $(\mathcal{M},\tau)$ be a finite
factor, $X\in\mathcal{M}$, then there exists a normal operator
$N\in\mathcal{M}$ and a nilpotent operator $K$ such that $X=N+K$. This result
answers affirmatively Question 1.1 in [9].

The organization of this paper is as follows. Section 2 below is devoted to
some results which will be very useful in section 3 and section 4. In Section
3, we prove the main theorem. In Section 4, we give three applications of the
main theorem.

##  2\. Preliminaries

######  Theorem 2.1.

(Theorem 1.1 in [7]) Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor,
$X\in\mathcal{M}$, $X=X^{*}$. Then we have the following equivalent results.

  1. (1)

There is a decomposition of the identity into $N\in\mathbb{N}$ mutually
orthogonal nonzero projections $E_{j}$, $I=\sum_{j=1}^{N}E_{j}$, for which
$E_{j}XE_{j}=\tau(X)E_{j}$ for all $j=1,...,N$;

  2. (2)

There is a finite dimensional abelian von Neumann algebra
$\mathcal{B}\subseteq\mathcal{M}$ such that

$E_{\mathcal{B}^{\prime}\cap\mathcal{M}}(X)=\tau(X)I;$

where $E_{\mathcal{B}^{\prime}\cap\mathcal{M}}$ is the conditional expectation
from $\mathcal{M}$ onto $\mathcal{B}^{\prime}\cap\mathcal{M}$;

  3. (3)

There is a unitary operator $U\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}XU^{j}=\tau(X)I;$

  4. (4)

There is a unitary operator $W\in\mathcal{M}$ with $W^{N}=I$ and

$\frac{1}{N}\sum_{j=0}^{N-1}{W^{*}}^{j}XW^{j}=\tau(X)I.$

######  Lemma 2.2.

(Proposition 1.5 in [8] or Lemma 3.2 in [7]) Let $(\mathcal{M},\tau)$ be a
type ${\rm II}_{1}$ factor, $A\in\mathcal{M}^{+}$, $\tau(A)=1$ and
$N\in\mathbb{N}$. Then the following conditions are equivalent.

  1. (1)

$A$ is the sum of $N$ nonzero projections;

  2. (2)

There is a unitary operator $U\in\mathcal{M}$ with

$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}=I.$

##  3\. Main theorem

######  Theorem 3.1.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor,
$X_{1},\cdots,X_{n}\in\mathcal{M}$. Then there is a family of finitely many
mutually orthogonal non-zero projections $\\{E_{i}\\}_{i=1}^{N}$ in
$\mathcal{M}$ such that $\sum\limits_{i=1}^{N}E_{i}=I$ and
$E_{i}X_{j}E_{i}=\tau(X_{j})E_{i}$ for $1\leq i\leq N$ and $1\leq j\leq n$.
That is, $X_{j}$ can be expressed as follows

$X_{j}=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr
E_{1}&\tau(X_{j})&(X_{j})_{12}&\cdots&(X_{j})_{1N}\cr
E_{2}&(X_{j})_{21}&\tau(X_{j})&\cdots&(X_{j})_{2N}\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&(X_{j})_{N1}&(X_{j})_{N2}&\cdots&\tau(X_{j})\cr}.$

Equivalently, there is a unitary operator $U\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{i=0}^{N-1}(U^{i})^{\ast}X_{j}U^{i}=\tau(X_{j})I$

for $1\leq j\leq n$.

To prove Theorem 3.1, we may assume that $X_{i}=X_{i}^{*}$ and $\tau(X_{i})=0$
for $1\leq i\leq n$. In the following we assume that $n=2$. The general case
can be proved by mathematical induction. We assume that $H=H^{*}$, $K=K^{*}$
in $\mathcal{M}$ such that $\tau(H)=\tau(K)=0$. By Theorem 2.1, there is a
family of mutually orthogonal nonzero projections $\\{E_{i}\\}_{i=1}^{N}$ in
$\mathcal{M}$ such that $\sum\limits_{i=1}^{N}E_{i}=I$ and $E_{i}HE_{i}=0$ for
$1\leq i\leq N$. So we have

$H=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr E_{1}&0&\ast&\cdots&\ast\cr
E_{2}&\ast&0&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&\ast&\ast&\cdots&0\cr},\quad
K=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr E_{1}&K_{1}&\ast&\cdots&\ast\cr
E_{2}&\ast&K_{2}&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&\ast&\ast&\cdots&K_{N}\cr}.$

By Theorem 2.1 again, there is family of mutually orthogonal nonzero
projections $\\{F_{i}\\}_{i=1}^{L}$ in $\mathcal{M}$ such that
$\sum\limits_{i=1}^{L}F_{i}=I$ and $F_{i}HF_{i}=0$,
$F_{i}KF_{i}=\alpha_{i}F_{i}$ for $1\leq i\leq L$. Now we have

$H=\bordermatrix{&F_{1}&F_{2}&\cdots&F_{L}\cr F_{1}&0&\ast&\cdots&\ast\cr
F_{2}&\ast&0&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
F_{L}&\ast&\ast&\cdots&0\cr},\quad
K=\bordermatrix{&F_{1}&F_{2}&\cdots&F_{L}\cr
F_{1}&\alpha_{1}&\ast&\cdots&\ast\cr
F_{2}&\ast&\alpha_{2}&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
F_{L}&\ast&\ast&\cdots&\alpha_{L}\cr}.$

By mathematical induction, we only need to show that for

$H=\bordermatrix{&F_{1}&F_{2}&\cdots&F_{L}\cr F_{1}&0&\ast&\cdots&\ast\cr
F_{2}&\ast&0&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
F_{L}&\ast&\ast&\cdots&0\cr},\quad
K=\bordermatrix{&F_{1}&F_{2}&\cdots&F_{L}\cr F_{1}&\alpha&\ast&\cdots&\ast\cr
F_{2}&\ast&\alpha&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
F_{L}&\ast&\ast&\cdots&\beta\cr}$

we can rewrite $H$ and $K$ as

$H=\bordermatrix{&P_{1}&P_{2}&\cdots&P_{s}\cr P_{1}&0&\ast&\cdots&\ast\cr
P_{2}&\ast&0&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
P_{s}&\ast&\ast&\cdots&0\cr},\quad
K=\bordermatrix{&P_{1}&P_{2}&\cdots&P_{s}\cr P_{1}&\tau(K)&\ast&\cdots&\ast\cr
P_{2}&\ast&\tau(K)&\cdots&\ast\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
P_{s}&\ast&\ast&\cdots&\tau(K)\cr}$

for a finite decomposition $I=P_{1}+\cdots+P_{s}$. We may assume that
$\tau(K)=0$, i.e.,
$\alpha(\tau(F_{1})+\cdots+\tau(F_{L-1}))+\beta\tau(F_{L})=0$. By cutting
$F_{L}$ into suitable small pieces, i.e., $F_{L}=F_{L,1}+\cdots+F_{L,L-1}$, we
may assume that $\alpha\tau(F_{i})+\beta\tau(F_{L,i})=0$ for $1\leq i\leq
L-1$. Therefore, we only need to consider the case:

$H=\begin{pmatrix}0&\quad\ast\\\ \ast&\quad 0\end{pmatrix},\quad
K=\begin{pmatrix}\alpha&\quad\ast\\\ \ast&\quad\beta\end{pmatrix},$

where $\tau(K)=0$.

We consider the special case $\alpha=1$ and $\beta=-1$ first. Then we can
write

$K=\begin{pmatrix}1&\quad C\\\ C^{*}&\quad-1\end{pmatrix},\quad
H=\begin{pmatrix}0&\quad D\\\ \quad D^{*}&0\end{pmatrix}.$

By the polar decomposition theorem, $D=U|D|$, where $U$ is a unitary operator.
Thus

$\begin{pmatrix}0&D\\\ D^{*}&0\end{pmatrix}=\begin{pmatrix}U&0\\\
0&I\end{pmatrix}\begin{pmatrix}0&|D|\\\
|D|&0\end{pmatrix}\begin{pmatrix}U^{*}&0\\\ 0&I\end{pmatrix}$
$\begin{pmatrix}U^{*}&0\\\ 0&I\end{pmatrix}\begin{pmatrix}1&C\\\
C^{*}&-1\end{pmatrix}\begin{pmatrix}U&0\\\
0&I\end{pmatrix}=\begin{pmatrix}1&U^{*}C\\\ C^{*}U&-1\end{pmatrix}.$

So we may assume that $D\geq 0$. Then there is a maximal abelian von Neumann
algebra $\mathcal{A}$ of $\mathcal{M}$ containing $D$. Choose projections
$P,Q\in\mathcal{A}$ with $P+Q=I$ and $\tau(P)=\tau(Q)=\frac{1}{2}$. Therefore,
we can write

$H=\begin{pmatrix}0&0&PD&0\\\ 0&0&0&QD\\\ PD&0&0&0\\\
0&QD&0&0\end{pmatrix}\text{ and }K=\begin{pmatrix}P&0&*&*\\\ 0&Q&*&*\\\
*&*&-P&0\\\ *&*&0&-Q\end{pmatrix}.$

Note that we can rewrite $H$ and $K$ in the following way (interchange the
second row and the fourth row and interchange the second column and fourth
column of $H$ and $K$, respectively).

$H=\begin{pmatrix}0&0&PD&0\\\ 0&0&0&QD\\\ PD&0&0&0\\\
0&QD&0&0\end{pmatrix}\text{ and }K=\begin{pmatrix}P&*&*&0\\\ *&-Q&0&*\\\
*&0&-P&*\\\ 0&*&*&Q\end{pmatrix}.$

Since $\tau(P)=\tau(Q)$, by Theorem 2.1, there are finitely many projections
$\tilde{E_{i}}$ such that

$\tilde{E_{i}}\begin{pmatrix}P&*&\\\ *&-Q&\end{pmatrix}\tilde{E_{i}}=0$

and finitely many projections $\tilde{F_{j}}$ such that

$\tilde{F_{j}}\begin{pmatrix}-P&*&\\\ *&Q&\end{pmatrix}\tilde{F_{j}}=0$

Now the $\\{\tilde{E_{i}},\tilde{F_{j}}\\}$ satisfy
$\sum\tilde{E_{i}}+\sum\tilde{F_{j}}=I$,
$\tilde{E_{i}}H\tilde{E_{i}}=\tilde{E_{i}}K\tilde{E_{i}}=0$ and
$\tilde{F_{j}}H\tilde{F_{j}}=\tilde{F_{j}}K\tilde{F_{j}}=0$.

Now, we consider the general case:

$H=\begin{pmatrix}0&\quad A\\\ A^{*}&\quad 0\end{pmatrix}\begin{matrix}P\\\
Q\end{matrix},\quad K=\begin{pmatrix}\alpha&\quad\ast\\\
\ast&\quad\beta\end{pmatrix}\begin{matrix}P\\\ Q\end{matrix},$

where $\tau(K)=\alpha\tau(P)+\beta\tau(Q)=0$.

######  Lemma 3.2.

Let $A=PAQ$ as above. Then there exist projections $P_{1},P_{2}$ and
$Q_{1},Q_{2}$ such that $P_{1}+P_{2}=P$,
$\tau(P_{1})=\tau(P_{2})=\frac{1}{2}\tau(P)$, $Q_{1}+Q_{2}=Q$,
$\tau(Q_{1})=\tau(Q_{2})=\frac{1}{2}\tau(Q)$ and $P_{1}AQ_{2}=P_{2}AQ_{1}=0$.

###### Proof.

Let $A=V|A|$ be the polar decomposition of $A$. Let $\tilde{Q}=R(A^{*})\leq Q$
and let $\tilde{P}=R(A)\leq P$. Then $V:\tilde{Q}\rightarrow\tilde{P}$ is a
unitary operator. Note that $\tilde{Q}=R(A^{*})=R(A^{*}A)=R(|A|)$. So we have
$\tilde{Q}|A|=|A|=|A|\tilde{Q}$. Let $\mathcal{A}$ be a maximal abelian von
Neumann algebra containing $W^{*}(|A|,\tilde{Q})$. Then there exist two
projections $\tilde{Q}_{1},\tilde{Q}_{2}$ in $\mathcal{A}$ such that
$\tilde{Q}_{1}+\tilde{Q}_{2}=\tilde{Q}$ and
$\tau(\tilde{Q}_{1})=\tau(\tilde{Q}_{2})=\frac{1}{2}\tau(\tilde{Q})$. Note
that $\tilde{Q}_{1}|A|=|A|\tilde{Q}_{1}$ and
$\tilde{Q}_{2}|A|=|A|\tilde{Q}_{2}$. Let $\tilde{P}_{1}=V\tilde{Q}_{1}V^{*}$,
$\tilde{P}_{2}=V\tilde{Q}_{2}V^{*}$. Then
$\tilde{P}_{1}+\tilde{P}_{2}=\tilde{P}$ and
$\tau(\tilde{P}_{1})=\tau(\tilde{P}_{2})=\frac{1}{2}\tau(\tilde{P})$. Note
that

$\tilde{P}_{2}A\tilde{Q}_{1}=V\tilde{Q}_{2}V^{*}V|A|\tilde{Q}_{1}=V\tilde{Q}_{2}|A|\tilde{Q}_{1}=V\tilde{Q}_{2}\tilde{Q}_{1}|A|=0,$
$\tilde{P}_{1}A\tilde{Q}_{2}=V\tilde{Q}_{1}V^{*}V|A|\tilde{Q}_{2}=V\tilde{Q}_{1}|A|\tilde{Q}_{2}=V\tilde{Q}_{1}\tilde{Q}_{2}|A|=0.$

Let $Q_{1}^{\prime},Q_{2}^{\prime}$ be projections such that
$Q_{1}^{\prime}+Q_{2}^{\prime}=Q-\tilde{Q}$ and
$\tau(Q_{1}^{\prime})=\tau(Q_{2}^{\prime})=\frac{1}{2}\tau(Q-\tilde{Q})$. Let
$P_{1}^{\prime},P_{2}^{\prime}$ be projections such that
$P_{1}^{\prime}+P_{2}^{\prime}=P-\tilde{P}$ and
$\tau(P_{1}^{\prime})=\tau(P_{2}^{\prime})=\frac{1}{2}\tau(P-\tilde{P})$. Let
$Q_{1}=Q_{1}^{\prime}+\tilde{Q}_{1}$, $Q_{2}=Q_{2}^{\prime}+\tilde{Q}_{2}$,
$P_{1}=P_{1}^{\prime}+\tilde{P}_{1}$, $P_{2}=P_{2}^{\prime}+\tilde{P}_{2}$.
Then $P_{1},P_{2},Q_{1},Q_{2}$ satisfy the lemma. ∎

Now we come back to the proof of the general case. Write

$H=\begin{pmatrix}0&0&\ast&0\\\ 0&0&0&\ast\\\ \ast&0&0&0\\\
0&\ast&0&0\end{pmatrix}\begin{matrix}P_{1}\\\ P_{2}\\\ Q_{1}\\\
Q_{2}\end{matrix}=\begin{pmatrix}0&0&\ast&\ast\\\ 0&0&\ast&\ast\\\
\ast&\ast&0&0\\\ \ast&\ast&0&0\end{pmatrix}\begin{matrix}P_{1}\\\ Q_{2}\\\
P_{2}\\\ Q_{1}\end{matrix},$ $K=\begin{pmatrix}\alpha&0&\ast&\ast\\\
0&\alpha&\ast&\ast\\\ \ast&\ast&\beta&0\\\
\ast&\ast&0&\beta\end{pmatrix}\begin{matrix}P_{1}\\\ P_{2}\\\ Q_{1}\\\
Q_{2}\end{matrix}=\begin{pmatrix}\alpha&\ast&\ast&\ast\\\
\ast&\beta&\ast&\ast\\\ \ast&\ast&\alpha&\ast\\\
\ast&\ast&\ast&\beta\end{pmatrix}\begin{matrix}P_{1}\\\ Q_{2}\\\ P_{2}\\\
Q_{1}\end{matrix}.$

Since
$\alpha\tau(P_{1})+\beta\tau(Q_{2})=\alpha\tau(P_{2})+\beta\tau(Q_{1})=0$, by
Theorem 2.1, there are finitely many projections $E_{i}$ such that
$E_{1}+\cdots+E_{N}=I$ and $E_{i}HE_{i}=E_{i}KE_{i}=0$ for $1\leq i\leq N$.

##  4\. Applications

The following theorem answers affirmatively Question 1.1 in [10].

######  Theorem 4.1.

If $(\mathcal{M},\tau)$ is a type ${\rm II}_{1}$ factor and $A\in\mathcal{M}$
with $\tau(A)=0$, then A is a single commutator. If $A$ is a self-adjoint
operator with $\tau(A)=0$, then $A$ is a self-commutator.

###### Proof.

By Theorem 3.1, there is a family of finitely many mutually orthogonal non-
zero projections $\\{E_{i}\\}_{i=1}^{N}$ in $\mathcal{M}$ such that
$\sum\limits_{i=1}^{N}E_{i}=I$ and $E_{i}AE_{i}=0$ for $1\leq i\leq N$. That
is, $A$ can be written as follows

$A=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr E_{1}&0&A_{12}&\cdots&A_{1N}\cr
E_{2}&A_{21}&0&\cdots&A_{2N}\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&A_{N1}&A_{N2}&\cdots&0\cr}.$

Then $A=BC-CB$, where

$B=\begin{pmatrix}0&B_{12}&\cdots&B_{1N}\\\ B_{21}&0&\cdots&B_{2N}\\\
\vdots&\vdots&\ddots&\vdots\\\ B_{N1}&B_{N2}&\cdots&0\end{pmatrix},\quad
C=\begin{pmatrix}1&0&\cdots&0\\\ 0&e^{\frac{2\pi i}{N}}&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&\cdots&e^{\frac{(N-1)2\pi
i}{N}}\end{pmatrix}$

and where the $B_{kj}$ are chosen so that $(e^{\frac{(j-1)2\pi
i}{N}}-e^{\frac{(k-1)2\pi i}{N}})B_{kj}=A_{kj},1\leq k,j\leq N$ and $k\neq j$.

Now, suppose that $A$ is a self-adjoint element with $\tau(A)=0$. By Theorem
3.1 again

$A=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr E_{1}&0&A_{12}&\cdots&A_{1N}\cr
E_{2}&A_{12}^{*}&0&\cdots&A_{2N}\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&A_{1N}^{*}&A_{2N}^{*}&\cdots&0\cr}.$

Then $A=D-CDC^{*}$, where

$D=\begin{pmatrix}0&D_{12}&\cdots&D_{1N}\\\ D_{12}^{*}&0&\cdots&D_{2N}\\\
\vdots&\vdots&\ddots&\vdots\\\
D_{1N}^{*}&D_{2N}^{*}&\cdots&0\end{pmatrix},\quad
C=\begin{pmatrix}1&0&\cdots&0\\\ 0&e^{\frac{2\pi i}{N}}&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots\\\ 0&0&\cdots&e^{\frac{(N-1)2\pi
i}{N}}\end{pmatrix}$

and the $D_{kj}$ are chosen so that $(1-e^{\frac{(k-j)2\pi
i}{N}})D_{kj}=A_{kj},1\leq k,j\leq N$ and $k\neq j$. Since $D$ is a self-
adjoint operator, then for large enough $\lambda\in\mathbb{R}$,
$D+\lambda\geqslant 0$. Set $X:=C(D+\lambda)^{\frac{1}{2}}$, we have
$A=X^{*}X-XX^{*}$. ∎

The following lemma is Proposition 2.10 of [19]. We provide a complete proof
for reader’s convenience.

######  Lemma 4.2.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and
$A^{*}=A\in\mathcal{M}$ with $\tau(A)=0$. If there is a unitary
$U\in\mathcal{M}$ for which $A+U^{*}AU=0$, then $A$ can be expressed as
$A=aP+bQ$ with projections $P,Q\in\mathcal{M}$ and $a,b\in\mathbb{R}$.

###### Proof.

Set $B:=\frac{A}{\|A\|}$, then $\tau(B+I)=1$ and $B+I\geq 0$. Since
$A+U^{*}AU=0$, $\frac{(B+I)+U^{*}(B+I)U}{2}=I$. By Lemma 2.2, there exist
projections $P,Q\in\mathcal{M}$ such that $B+I=P+Q$. Thus we have
$A=\|A\|P-\|A\|(I-Q)$. ∎

The idea for the proof of the following theorem comes from [27]. This result
answers affirmatively Question 6(2) in [15].

######  Theorem 4.3.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and
$A^{*}=A\in\mathcal{M}$, then $A$ is a linear combination of 4 projections.

###### Proof.

When $\tau(A)\neq 0$, by considering $\frac{A}{\tau(A)}$, we may assume that
$\tau(A)=1$. The proof for the case when $\tau(A)=0$ is similar. By
considering $R_{A}\mathcal{M}R_{A}$, we may assume that $R_{A}=I$. Let
$\mathcal{A}$ be a maximal abelian von Neumann algebra containing $A$. Since
$\mathcal{M}$ is a type ${\rm II}_{1}$ factor, $\mathcal{A}$ is diffuse. Thus,
there exist projections $E_{1},E_{2}\in\mathcal{A}$ such that $E_{1}+E_{2}=I$
and $E_{1}\sim E_{2}$. Note that $\mathcal{M}\cong M_{2}(\mathbb{C})\otimes
E_{1}\mathcal{M}E_{1}$. Denote $\mathcal{N}:=E_{1}\mathcal{M}E_{1}$.
Therefore, $A-I$ can be written as

$A-I=\begin{pmatrix}A_{1}&0\\\ 0&A_{2}\end{pmatrix}\in
M_{2}(\mathbb{C})\otimes\mathcal{N},$

where $A_{1},A_{2}$ are self-adjoint operators in $\mathcal{N}$. Write

$A=\begin{pmatrix}A_{1}+A_{2}&0\\\ 0&0\end{pmatrix}+\begin{pmatrix}-A_{2}&0\\\
0&A_{2}\end{pmatrix}+\begin{pmatrix}I&0\\\ 0&I\end{pmatrix}$

Since $\tau(A-I)=0$,
$\tau_{E_{1}}(A_{1}+A_{2})=\frac{\tau(A_{1}+A_{2})}{\tau(E_{1})}=0$. By
Theorem 4.1, there exists $X\in\mathcal{N}$ such that
$A_{1}+A_{2}=X^{*}X-XX^{*}$. We can assume that $X$ is invertible and
$X^{*}X>3I$, because of the invariance property of the commutator
$X^{*}X-XX^{*}=(X^{*}+tI)(X+tI)-(X+tI)(X^{*}+tI),t\in\mathbb{C}$. Then

$A=\begin{pmatrix}X^{*}X-XX^{*}&0\\\
0&0\end{pmatrix}+\begin{pmatrix}-A_{2}&0\\\
0&A_{2}\end{pmatrix}+\begin{pmatrix}I&0\\\ 0&I\end{pmatrix}$
$=\begin{pmatrix}X^{*}X&0\\\
0&-XX^{*}\end{pmatrix}+\begin{pmatrix}-XX^{*}&0\\\
0&XX^{*}\end{pmatrix}+\begin{pmatrix}-A_{2}&0\\\
0&A_{2}\end{pmatrix}+\begin{pmatrix}I&0\\\ 0&I\end{pmatrix}$
$=\begin{pmatrix}X^{*}X&0\\\
0&2I-XX^{*}\end{pmatrix}+\begin{pmatrix}-XX^{*}-A_{2}+I&0\\\
0&XX^{*}+A_{2}-I\end{pmatrix}.$

The last term of the above equality can be written as $\lambda P_{1}-\lambda
P_{2}$ by Lemma 4.2, where $P_{1},P_{2}\in\mathcal{M}$ are projections and
$\lambda\in\mathbb{R}$. So to prove this theorem, we only need to prove the
first term of the above equality is a linear combination of 2 projections.
This conclusion can be found in [23]. For the convenience of the reader, we
will rewrite the proof.

Set $B:=\begin{pmatrix}X^{*}X&0\\\ 0&2I-XX^{*}\end{pmatrix}$. Since we may
assume $X$ is invertible in $\mathcal{N}$, by the polar decomposition theorem,
there is a unitary operator $U\in\mathcal{N}$ and an invertible positive
operator $|X|$ such that $X=U|X|$. Then we can verify

$B-I=\begin{pmatrix}X^{*}X-I&0\\\
0&I-XX^{*}\end{pmatrix}=\begin{pmatrix}I&0\\\
0&U\end{pmatrix}\begin{pmatrix}X^{*}X-I&0\\\
0&I-X^{*}X\end{pmatrix}\begin{pmatrix}I&0\\\ 0&U^{*}\end{pmatrix}$

Since $X^{*}X>3I$, $X^{*}X-I$ is an invertible positive operator. Take any
real $d$ such that $X^{*}X-I<dI$, and define $S_{1}$ and $S_{2}$ by

$S_{1}=\frac{1}{1+d}(X^{*}X-1+d(X^{*}X-1)^{-1})\quad\text{and}\quad
S_{2}=\frac{1}{d-1}(-X^{*}X+1+d(X^{*}X-1)^{-1}).$

Define $P_{3}$ and $P_{4}$ by

$P_{3}=\frac{1}{2}\begin{pmatrix}I+S_{1}&(I-S_{1}^{2})^{\frac{1}{2}}\\\
(I-S_{1}^{2})^{\frac{1}{2}}&I-S_{1}\end{pmatrix}\quad\text{and}\quad
P_{4}=\frac{1}{2}\begin{pmatrix}I+S_{2}&(I-S_{2}^{2})^{\frac{1}{2}}\\\
(I-S_{2}^{2})^{\frac{1}{2}}&I-S_{2}\end{pmatrix}.$

It can be verified that $P_{3},P_{4}$ are projections in
$\mathcal{M}_{2}(\mathbb{C})\otimes\mathcal{N}$. Define $a=1+d$ and $b=1-d$.
By calculation we can obtain that

$\begin{pmatrix}X^{*}X-I&0\\\ 0&I-X^{*}X\end{pmatrix}=aP_{3}+bP_{4}-I.$

Therefore, $B$ is a linear combination of 2 projections. The proof is
completed. ∎

The following result answers affirmatively Question 1.1 in [9].

######  Theorem 4.4.

Let $(\mathcal{M},\tau)$ be a finite factor, $X\in\mathcal{M}$. Then there
exists a normal operator $N\in\mathcal{M}$ and a nilpotent operator $K$ such
that $X=N+K$.

###### Proof.

We only need to consider that $(\mathcal{M},\tau)$ is a type ${\rm II}_{1}$
factor. By Theorem 3.1, there is a family of finitely many mutually orthogonal
non-zero projections $\\{E_{i}\\}_{i=1}^{N}$ in $\mathcal{M}$ such that
$\sum\limits_{i=1}^{N}E_{i}=I$ and $E_{i}XE_{i}=\tau(X)E_{i}$ for $1\leq i\leq
N$. That is, $X$ can be written as follows

$X=\bordermatrix{&E_{1}&E_{2}&\cdots&E_{N}\cr
E_{1}&\tau(X)&X_{12}&\cdots&X_{1N}\cr
E_{2}&X_{21}&\tau(X)&\cdots&X_{2N}\cr\vdots&\vdots&\vdots&\ddots&\vdots\cr
E_{N}&X_{N1}&X_{N2}&\cdots&\tau(X)\cr}.$

Set $\lambda:=\tau(X)$, then $\lambda=\exp(i\theta)|\lambda|$ for some
$\theta$.

Let

$S=\begin{pmatrix}|\lambda|&\exp(-i\theta)X_{12}&\cdots&\exp(-i\theta)X_{1N}\\\
\exp(i\theta)X_{12}^{\ast}&|\lambda|&\cdots&\exp(-i\theta)X_{2N}\\\
\vdots&\vdots&\ddots&\vdots\\\
\exp(i\theta)X_{1N}^{\ast}&\exp(i\theta)X_{2N}^{\ast}&\cdots&|\lambda|\end{pmatrix},$

then $S$ is self-adjoint.

Define

$K=X-\exp(i\theta)\begin{pmatrix}|\lambda|&\exp(-i\theta)X_{12}&\cdots&\exp(-i\theta)X_{1N}\\\
\exp(i\theta)X_{12}^{\ast}&|\lambda|&\cdots&\exp(-i\theta)X_{2N}\\\
\vdots&\vdots&\ddots&\vdots\\\
\exp(i\theta)X_{1N}^{\ast}&\exp(i\theta)X_{2N}^{\ast}&\cdots&|\lambda|\end{pmatrix}$
$=\begin{pmatrix}0&0&\cdots&0\\\
X_{21}-\exp(2i\theta)X_{12}^{\ast}&0&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots\\\
X_{N1}-\exp(2i\theta)X_{1N}^{\ast}&X_{N2}-\exp(2i\theta)X_{2N}^{\ast}&\cdots&0\end{pmatrix}.$

Then $K$ is a strictly lower-triangular operator and therefore nilpotent.
Write $N=\exp(i\theta)S$. Then $N$ is normal and $X=N+K$.

∎

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