X. Cao and J. Fang]School of Mathematical Sciences, Hebei Normal University,
Shijiazhuang, Hebei 050016, China Z. Yao]Postdoctoral Research Station of
Mathematics, Hebei Normal University, Shijiazhuang, Hebei 050016, China

# On finite sums of projections and Dixmier’s averaging theorem for type ${\rm
II}_{1}$ factors

Xinyan Cao, Junsheng Fang and Zhaolin Yao [[<EMAIL_ADDRESS><EMAIL_ADDRESS><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 an $X\in\mathcal{M}$, $X=X^{*}$, then there is a 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
$E_{j}XE_{j}=\tau(X)E_{j}$ for all $j=1,...,N$. Equivalently, there is a
unitary operator $U\in\mathcal{M}$ with $U^{N}=I$ and
$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}XU^{j}=\tau(X)I.$ As the first
application, we prove that a positive operator $A\in\mathcal{M}$ can be
written as a finite sum of projections in $\mathcal{M}$ if and only if
$\tau(A)\geq\tau(R_{A})$, where $R_{A}$ is the range projection of $A$. This
result answers affirmatively Question 6.7 of [9]. As the second application,
we show that if $X\in\mathcal{M}$, $X=X^{*}$ and $\tau(X)=0$, then there
exists a nilpotent element $Z\in\mathcal{M}$ such that $X$ is the real part of
$Z$. This result answers affirmatively Question 1.1 of [4]. As the third
application, we show that for $X_{1},...,X_{n}\in\mathcal{M}$, there exist
unitary operators $U_{1},...,U_{k}\in\mathcal{M}$ such that
$\frac{1}{k}\sum_{i=1}^{k}U_{i}^{*}X_{j}U_{i}=\tau(X_{j})I,\quad\forall 1\leq
j\leq n.$ This result is a stronger version of Dixmier’s averaging theorem for
type ${\rm II}_{1}$ factors.

###### Key words and phrases:

finite sums of projections, Dixmier’s averaging theorem, type ${\rm II}$
factors

###### 2010 Mathematics Subject Classification:

46L10, 46L36

The authors are supported by National Natural Science Foundation of China
(Grant No.12071109) and a start up funding from Hebei Normal Univeristy.

##  1\. Introduction

Which positive bounded linear operators on a separable Hilbert space can be
written as strong sums of projections? If the underlying space is finite
dimensional, then a characterization of such operators was obtained by
Fillmore [6]: a finite-dimensional operator is the sum of projections if and
only if it is positive, it has an integer trace and the trace is greater than
or equal to its rank.

The same question can be asked relative to a von Neumann algebra
$\mathcal{M}$. This question was first asked by Kaftal, Ng and Zhang in [10].
In [10], they developed beautiful techniques and addressed the question
completely for positive operators in type ${\rm I}$ and type ${\rm III}$
factors. They proved the following results. Let $\mathcal{M}$ be a type ${\rm
I}$ factor. Given $A\in\mathcal{M}^{+}$. Denote by
$A_{+}=(A-I)\chi_{A}(1,\|A\|]$ the excess part of $A$ and by
$A_{-}=(I-A)\chi_{A}(0,1)$ the defect part of $A$. Then $A$ is a strong sum of
projections if and only if either ${\rm Tr}(A_{+})=\infty$ or ${\rm
Tr}(A_{-})\leq{\rm Tr}(A_{+})<\infty$ and ${\rm Tr}(A_{+})-{\rm
Tr}(A_{-})\in\mathbb{N}\cup\\{0\\}$. Let $\mathcal{M}$ be a type ${\rm III}$
factor. Then $A\in\mathcal{M}$ is a strong sum of projections if and only if
either $\|A\|>1$ or $A$ is a projection. For type ${\rm II}$ factors, they
proved that if $A$ is diagonalizable, then $A$ is a strong sum of projections
if and only if $\tau(A_{+})\geq\tau(A_{-})$. The condition
$\tau(A_{+})\geq\tau(A_{-})$ is necessary even when $A$ is not diagonalizable.
Recall that a positive operator $A$ is said to be diagonalizable if
$A=\sum\gamma_{j}E_{j}$ for some $\gamma_{j}>0$ and mutually orthogonal
projections $\\{E_{j}\\}$ in $\mathcal{M}$. In [2], we proved that if
$\mathcal{M}$ is a type ${\rm II}$ factor, $A\in\mathcal{M}^{+}$ and
$\tau(A_{+})\geq\tau(A_{-})$, then $A$ can be written as the sum of a finite
or infinite collection of projections.

A natural question is the following: which positive operators in a factor von
Neumann algebra can be written as finite sums of projections? This question is
much more interesting and difficult. Note that the question is even open for
positive operators in $\mathcal{B}(\mathcal{H})$. One of the main goals of
this paper is to study this question in type ${\rm II}_{1}$ factors.

As reported in a survey article [[23], Theorem 4.12] by Wu quoting unpublished
joint work of Choi and Wu in 1988, they proved that positive operators with
essential norm greater than 1 are finite sums of projections. This result is
presented together with other related results in their recent paper [[3],
Theorem 2.2]. The special case of $\alpha I$ with $\alpha>1$ follows also from
a delicate analysis which characterizes the numbers $\alpha$ for which $\alpha
I$ is a sum of at least $n$ projections (see in [16] and several other
papers.)

[13] presented a characterization of strict sums of projections in the
multiplier algebras of $\sigma$-unital purely infinite simple $C^{*}$-algebras
and [11] provided sufficient conditions for operators to be finite sums of
projections in the multiplier algebras of $\sigma$-unital purely infinite
simple $C^{*}$-algebras.

In [9], H. Halpern, V. Kaftal, P. Ng and S. Zhang provided some sufficient
conditions and also a necessary condition for a positive operator can be
written as a finite sum of projections. For a type ${\rm II}_{1}$ factor, they
showed a sufficient condition for a positive diagonalizable operator $A$ to be
a finite sum of projections is $\tau(A_{+})>\tau(A_{-})$ or equivalently
$\tau(A)>\tau(R_{A})$.

In this paper, we prove the following result.

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

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and let
$X=X^{*}\in\mathcal{M}$. Then

  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 $U\in\mathcal{M}$ with

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

  4. (4)

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

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

This result is inspired by [3]. To prove this theorem, we develop some new
techniques in this paper. On the other hand, several results, e.g.,
Proposition 2.3, Corollary 2.5, Lemma 2.8 and Lemma 6.1, in [9] play important
roles in the proof.

We mention the following results as applications of the main theorem.

######  Theorem 1.2.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and
$A\in\mathcal{M}^{+}$. Then $A$ is a finite sum of projections if and only if
$\tau(A)\geq\tau(R_{A})$.

The above theorem answers affirmatively Question 6.7 of [9]. The following
theorem answers affirmatively Question 1.1 in [4].

######  Theorem 1.3.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and
$X\in\mathcal{M}$, $X=X^{*}$ and $\tau(X)=0$. Then there exists a nilpotent
element $Z\in\mathcal{M}$ such that $X=ReZ$, i.e., $X$ is the real part of a
nilpotent element $Z$.

The following theorem is a stronger version of Dixmier’s averaging theorem for
type ${\rm II}_{1}$ factors.

######  Theorem 1.4.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and let
$X_{1},...,X_{n}\in\mathcal{M}$. Then there exists unitary operators
$U_{1},...,U_{k}\in\mathcal{M}$ such that

$\frac{1}{k}\sum_{i=1}^{k}U_{i}^{*}X_{j}U_{i}=\tau(X_{j})I,\quad\forall 1\leq
j\leq n.$

The organization of this paper is as follows. Section 2 below is devoted to
some definitions and lemmas which will be very useful. In section 3, we prove
the main theorem stated as above. As applications, we prove Theorem 1.2 and
Theorem 1.3. In section 4, we give some properties of majorization and provide
the third application of the main theorem.

We refer to [3, 7, 11, 12, 13, 14, 17, 18, 19] for interesting results on the
related topics. We refer to [15, 20, 21, 22] for the basic knowledge of von
Neumann algebras.

_Notation_. For every $x\in\mathbb{R}$, we will denote by $\lceil x\rceil$ the
smallest integer $n\geq x$.

Throughout this paper, $(\mathcal{M},\tau)$ is a type ${\rm II}_{1}$ factor
with the faithful normal tracial state $\tau$. $E,F,E^{\prime},F^{\prime},p,q$
will denote projections in $\mathcal{M}$. $\mathcal{M}^{+}$ is the set of
positive operators in $\mathcal{M}$. $\mathbb{N}$ denotes the set of non-
negative integers and $\mathbb{N}^{*}$ is the set of strictly positive
integers.

For a selfadjoint element $A\in\mathcal{M}$, $\chi_{A}$ will denote the
(projection-valued) spectral measure of $A$ and $R_{A}$ the range projection
of $A$.

##  2\. Definitions and Preliminary lemmas

######  Definition 2.1 (Definition 2.2 in [9]).

An operator algebra $\mathcal{M}$ has constants $N_{0}$ and $V_{0}$ if every
selfadjoint operator $a\in\mathcal{M}$ can be decomposed into a real linear
combination $a=\sum_{j=1}^{N_{0}}\alpha_{j}p_{j}$ of $N_{0}$ projections
$p_{j}\in\mathcal{M}$ such that

$\sum_{j=1}^{N_{0}}|\alpha_{j}|\leq V_{0}\|a\|.$

The existence and the estimates of the constants $N_{0}$ and $V_{0}$ for $\rm
II_{1}$ factors are given in Theorem 1-3 in [7]. If $\mathcal{M}$ is of type
$\rm II_{1}$ factor, then $N_{0}=12$ and $V_{0}=14$.

######  Definition 2.2.

A linear combination is called positive combinations of projections if a
positive operator is finite linear combination of projections with positive
coefficients.

######  Proposition 2.3 (Proposition 2.3 in [9]).

Let $\mathcal{M}$ be a type $\rm II_{1}$ factor. Assume that
$a\in\mathcal{M}^{+}$ is invertible. Then $a$ is a positive combination of
$N_{0}+\lceil V_{0}(\|a\|\|a^{-1}\|-1)\rceil+1$ projections in $\mathcal{M}$.

######  Definition 2.4.

A positive operator $a$ such that $a\geq\nu R_{a}$ for some $\nu>0$ is
invertible in the reduced algebra $R_{a}\mathcal{M}R_{a}$. We call such an
element _locally invertible_.

######  Lemma 2.5 (Lemma 2.4 in [9]).

Let $a\in\mathcal{M}^{+}$. Then $a$ is a positive combination of projections
in $\mathcal{M}$ if and only if $a$ is a positive combination of projections
in $R_{a}\mathcal{M}R_{a}$.

######  Corollary 2.6 (Corollary 2.5 in [9]).

Let $a\in\mathcal{M}$ and assume that it is locally invertible, i.e.,
$a\geq\nu R_{a}$ for some $\nu>0$. Then $a$ is a positive combination of
$N_{0}+\lceil V_{0}(\frac{\|a\|}{\nu}-1)\rceil+1$ projections in
$\mathcal{M}$.

######  Lemma 2.7 (Lemma 2.8 in [9]).

Let $e$ and $f$ be orthogonal projections in $\mathcal{M}$ with $e\prec f$.
Let $b=be=eb$, $d=df=fd$ be positive elements in $\mathcal{M}$ such that
$d\geq\nu f$ for some $\nu>0$. Then $a:=b+d$ is a positive combination of
projections in $\mathcal{M}$.

######  Lemma 2.8 (Lemma 6.1 in [9]).

Let $e$ and $f$ be mutually orthogonal projections in $\mathcal{M}$,
$0\leq\lambda\leq 1$, and $\mu>0$, and $a:=(1+\mu)e+(1-\lambda)f$. If
$\mu\tau(e)>\lambda\tau(f)$, then $a$ is a finite sum of projections in
$\mathcal{M}$.

######  Lemma 2.9 (Lemma 3.2 in [2]).

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and let
$a\in\mathcal{M}^{+}$ such that $\tau(a)=1$. Then $a$ is a strong sum of
projections.

##  3\. Main theorem

The following lemma is motivated by Proposition 1.5 of [3] and its proof is
similar to the proof of Proposition 1.5 of [3]. We provide a complete proof
for the reader’s convenience.

######  Lemma 3.1.

Let $X\in\mathcal{M}$. Then the following conditions are equivalent.

  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 unitary operator $U\in\mathcal{M}$ with

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

###### Proof.

(1)$\Rightarrow$(2). Set $A:=\frac{X}{\tau(X)}$, then $\tau(A)=1$. By
hypothesis, there are mutually orthogonal nonzero projections
$E_{1},E_{2},...,E_{N}$ such that $\sum_{k=1}^{N}E_{k}=I$ and
$\sum_{k=1}^{N}E_{k}AE_{k}=I$. Set $U:=\sum_{k=1}^{N}e^{\frac{2\pi
ki}{N}}E_{k}$. Then

$U^{*}=\sum_{k=1}^{N}e^{-\frac{2\pi ki}{N}}E_{k}=U^{-1}$

and

$\displaystyle\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}$
$\displaystyle=\frac{1}{N}\sum_{j=0}^{N-1}\sum_{k,l=1}^{N}e^{\frac{2\pi
i(l-k)j}{N}}E_{k}AE_{l}$
$\displaystyle=\frac{1}{N}\sum_{k,l=1}^{N}\left(\sum_{j=0}^{N-1}e^{\frac{2\pi
i(l-k)j}{N}}\right)E_{k}AE_{l}$ $\displaystyle=\sum_{k=1}^{N}E_{k}AE_{k}=I$

since $\sum_{j=0}^{N-1}e^{\frac{2\pi i(l-k)j}{N}}=0$ for $k\neq l$. These show
that $U$ is the required operator.

(2)$\Rightarrow$(1). Assume that $U$ is the unitary operator with
$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}=I$. Then

$A-{U^{*}}^{N}AU^{N}=\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}-\sum_{j=1}^{N}{U^{*}}^{j}AU^{j}=NI-U^{*}\left(\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}\right)U=0.$

This implies that $AU^{N}=U^{N}A$. Let $f(e^{i\theta})=e^{\frac{i\theta}{N}}$
for $\theta\in[0,2\pi)$ and let $W=f(U^{N})^{-1}U$. Then $W$ is a unitary
operator. Since $U$ commutes with $U^{N}$,
$W^{N}=(f(U^{N})^{-1}U)^{N}=f(U^{N})^{-N}U^{N}=U^{-N}U^{N}=I$. Then the
spectrum of $W$ consists of eigenvalues from the set $\\{e^{\frac{2k\pi
i}{N}}:1\leq k\leq N\\}$. Choose $E_{1},E_{2},...,E_{N}\in\mathcal{M}$ to be
spectral projections of $W$ such that $\sum_{k=1}^{N}E_{k}=I$ and
$W=\sum_{k=1}^{N}e^{\frac{2k\pi i}{N}}E_{k}$. As $A$ commutes with $U^{N}$, we
have

$\frac{1}{N}\sum_{j=0}^{N-1}{W^{*}}^{j}AW^{j}=\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}f(U^{N})^{j}Af(U^{N})^{-j}U^{j}=\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}=I.$

Since $WE_{k}=e^{\frac{2k\pi i}{N}}E_{k}$ for each $1\leq k\leq N$, we have

$\displaystyle I$
$\displaystyle=\frac{1}{N}\sum_{k=1}^{N}E_{k}\left(\sum_{j=0}^{N-1}{W^{*}}^{j}AW^{j}\right)E_{k}=\frac{1}{N}\sum_{k=1}^{N}\left(\sum_{j=0}^{N-1}(e^{-\frac{2jk\pi
i}{N}}E_{k})A(e^{\frac{2jk\pi i}{N}}E_{k})\right)$
$\displaystyle=\frac{1}{N}\sum_{k=1}^{N}\left(\sum_{j=0}^{N-1}E_{k}AE_{k}\right)=\sum_{k=1}^{N}E_{k}AE_{k}.$

This proves (1). ∎

######  Lemma 3.2.

Let $A\in\mathcal{M}^{+}$, $\tau(A)=1$ 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 decomposition of the identity into $N$ mutually orthogonal nonzero
projections $E_{j}$, $I=\sum_{j=1}^{N}E_{j}$, for which
$\sum_{j=1}^{N}E_{j}AE_{j}=I$;

  3. (3)

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

$E_{\mathcal{B}^{\prime}\cap\mathcal{M}}(A)=I,$

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

  4. (4)

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

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

###### Proof.

By Lemma 3.1, we have (2)$\Leftrightarrow$(4).

(2)$\Rightarrow$(1). $\sum_{j=1}^{N}E_{j}AE_{j}=I$ is equivalent to
$E_{j}AE_{j}=E_{j}$ for each $1\leq j\leq N$. Let $P_{j}=A^{1/2}E_{j}A^{1/2}$
for $1\leq j\leq N$. Then $P_{j}$ is a projection for $1\leq j\leq N$ and
$A=\sum_{j=1}^{N}P_{j}$.

(1)$\Rightarrow$(2). Let $A=\sum_{j=1}^{N}P_{j}$, where $\\{P_{j}\\}$ are
nonzero projections. Then $1=\tau(A)=\sum_{j=1}^{N}\tau(P_{j})$. We decompose
the identity $I=\sum_{j=1}^{N}F_{j}$ into $N$ mutually orthogonal projections
$F_{j}\sim P_{j}$. Now choose partial isometries $W_{j}\in\mathcal{M}$ with
$P_{j}=W_{j}^{*}W_{j}$ and $F_{j}=W_{j}W_{j}^{*}$. Denote
$B:=\sum_{j=1}^{N}W_{j}$, then

$B^{*}B=\left(\sum_{j=1}^{N}W_{j}\right)^{*}\left(\sum_{j=1}^{N}W_{j}\right)=\sum_{i,j=1}^{N}W_{i}^{*}W_{j}=\sum_{j=1}^{N}W_{j}^{*}W_{j}=\sum_{j=1}^{N}P_{j}=A.$

Let $B=VA^{1/2}$. Then $BB^{*}=VAV^{*}$. Since $\mathcal{M}$ is a ${\rm
II}_{1}$ factor, $V$ can be chosen to be a unitary operator. Moreover,
$F_{j}B=W_{j}$ for every $j$, thus

$\sum_{j=1}^{N}F_{j}VAV^{*}F_{j}=\sum_{j=1}^{N}F_{j}BB^{*}F_{j}=\sum_{j=1}^{N}W_{j}W_{j}^{*}=\sum_{j=1}^{N}F_{j}=I.$

Let $E_{j}=V^{*}F_{j}V$. Then we have (2).

(2)$\Rightarrow$(3). Take $\mathcal{B}$ to be the algebra generated by
$\\{E_{j}\\}_{j=1}^{N}$. Since
$E_{\mathcal{B}^{\prime}\cap\mathcal{M}}(A)=\sum_{j=1}^{N}E_{j}AE_{j}$. Then,
by hypothesis, (3) follows.

(3)$\Rightarrow$(2). Let $E_{1},E_{2},...,E_{N}\in\mathcal{B}$ be the minimal
projections of $\mathcal{B}$ such that $\sum_{j=1}^{N}E_{j}=I$. Then
$E_{\mathcal{B}^{\prime}\cap\mathcal{M}}(A)=\sum_{j=1}^{N}E_{j}AE_{j}=I$.

∎

######  Remark 3.3.

Note that in the above Lemma, (1)$\Leftrightarrow$(2) is true for $N=+\infty$.

######  Lemma 3.4.

Let $E\in\mathcal{M}$ be a projection. Then there is a unitary operator
$U\in\mathcal{M}$ and $N\in\mathbb{N}^{*}$ such that

$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}EU^{j}=\tau(E)I.$

###### Proof.

The case $E=0,I$ being trivial, we assume that $0<E<I$. Then $0<\tau(E)<1$ and
$\frac{1}{\tau(E)}>1$. Set $A:=\frac{1}{\tau(E)}E$. By Lemma 2.8, $A$ is a sum
of $N\in\mathbb{N}^{*}$ nonzero projections in $\mathcal{M}$. Since
$\tau(A)=1$, then by Lemma 3.2, there is a unitary operator $U\in\mathcal{M}$
such that

$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}EU^{j}=\tau(E)I.$

∎

######  Lemma 3.5.

Let $E$ and $F$ be mutually orthogonal projections in $(\mathcal{M},\tau)$,
$\mu\geq 0$, $0\leq\lambda\leq 1$, and $A:=(1+\mu)E+(1-\lambda)F$. If
$\mu\tau(E)\geq\lambda\tau(F)$, then $A$ is a finite sum of projections in
$\mathcal{M}$.

###### Proof.

By considering $(E+F)\mathcal{M}(E+F)$, we may assume that $E+F=I$. By Lemma
2.8, if $\mu\tau(E)>\lambda\tau(F)$, then $A$ is a finite sum of projections
in $\mathcal{M}$. So we may assume that $\mu\tau(E)=\lambda\tau(F)$. Notice
that $A=(1+\mu)E+(1-\lambda)F=(1+\mu)E+(1-\lambda)(I-E)$ and
$\tau(A)=1+\mu\tau(E)-\lambda\tau(I-E)=1$. By Lemma 3.4, there is a unitary
operator $U$ and $N\in\mathbb{N}^{*}$ such that
$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}EU^{j}=\tau(E)I$. Thus

$\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}AU^{j}=\frac{1}{N}\sum_{j=0}^{N-1}{U^{*}}^{j}(I-\lambda
I+(\mu+\lambda)E)U^{j}=I-\lambda I+(\mu+\lambda)\tau(E)I=I.$

So by Lemma 3.2, $A$ is a sum of $N$ nonzero projections in $\mathcal{M}$. ∎

The following lemma is inspired by [8].

######  Lemma 3.6.

Let $E,F$ be two nonzero projections in $(\mathcal{M},\tau)$. Then there are
projections $E_{1},E_{2},F_{1},F_{2}$ in $\mathcal{M}$ satisfying

  1. (1)

$E_{1}+E_{2}=E$, $F_{1}+F_{2}=F$;

  2. (2)

$\tau(E_{1})=\tau(E_{2})=\frac{1}{2}\tau(E)$,
$\tau(F_{1})=\tau(F_{2})=\frac{1}{2}\tau(F)$;

  3. (3)

$E_{1}\bot F_{1}$, $E_{2}\bot F_{2}$.

###### Proof.

By the relative position of two projections in a type $\rm{II}_{1}$ factor
(see page 306-308 of [22]), we can write

$E=\begin{pmatrix}P&\\\ &Q\\\ &&0\\\ &&&\begin{pmatrix}I&0\\\
0&0\end{pmatrix}\end{pmatrix},\quad F=\begin{pmatrix}P&\\\ &0\\\ &&R\\\
&&&\begin{pmatrix}h^{2}&hk\\\ hk&k^{2}\end{pmatrix}\end{pmatrix},$

where $P,Q,R$ are projections, $h\geq 0$, $k\geq 0$ and $h^{2}+k^{2}=I$. $h,k$
are contained in a diffuse abelian von Neumann subalgebra $\mathcal{A}$. Hence
there is a projection $e\in\mathcal{A}$ with $\tau_{I}(e)=\frac{1}{2}$. Let
$P_{1},P_{2},Q_{1},Q_{2},R_{1},R_{2}$ be projections satisfying

  1. (1)

$P_{1}+P_{2}=P$, $Q_{1}+Q_{2}=Q$, $R_{1}+R_{2}=R$;

  2. (2)

$\tau(P_{1})=\tau(P_{2})=\frac{1}{2}\tau(P)$,
$\tau(Q_{1})=\tau(Q_{2})=\frac{1}{2}\tau(Q)$,
$\tau(R_{1})=\tau(R_{2})=\frac{1}{2}\tau(R)$.

Set

$E_{1}=\begin{pmatrix}P_{1}&\\\ &Q_{1}\\\ &&0\\\ &&&\begin{pmatrix}e&0\\\
0&0\end{pmatrix}\end{pmatrix},\quad E_{2}=\begin{pmatrix}P_{2}&\\\ &Q_{2}\\\
&&0\\\ &&&\begin{pmatrix}I-e&0\\\ 0&0\end{pmatrix}\end{pmatrix},$

$F_{1}=\begin{pmatrix}P_{2}&\\\ &0\\\ &&R_{1}\\\
&&&\begin{pmatrix}(I-e)h^{2}&(I-e)hk\\\
(I-e)hk&(I-e)k^{2}\end{pmatrix}\end{pmatrix},\quad
F_{2}=\begin{pmatrix}P_{1}&\\\ &0\\\ &&R_{2}\\\
&&&\begin{pmatrix}eh^{2}&ehk\\\ ehk&ek^{2}\end{pmatrix}\end{pmatrix}.$

Then $E_{1},E_{2},F_{1},F_{2}$ satisfy the lemma. ∎

######  Lemma 3.7.

Let $E,F$ be two nonzero projections in $(\mathcal{M},\tau)$ and let
$X=(1+\mu)E+(1-\lambda)F$, where $\mu\geq 0$ and $0\leq\lambda\leq 1$. If
$\mu\tau(E)\geq\lambda\tau(F)$, then $X$ is a finite sum of projections in
$\mathcal{M}$.

###### Proof.

To avoid triviality, assume that $\mu>0$ and $0<\lambda<1$. By Lemma 3.6,
there are projections $E_{1},E_{2},F_{1},F_{2}\in\mathcal{M}$ satisfying

  1. (1)

$E_{1}+E_{2}=E$, $F_{1}+F_{2}=F$;

  2. (2)

$\tau(E_{1})=\tau(E_{2})=\frac{1}{2}\tau(E)$,
$\tau(F_{1})=\tau(F_{2})=\frac{1}{2}\tau(F)$;

  3. (3)

$E_{1}\bot F_{1}$, $E_{2}\bot F_{2}$.

Set $X_{1}:=(1+\mu)E_{1}+(1-\lambda)F_{1}$ and
$X_{2}:=(1+\mu)E_{2}+(1-\lambda)F_{2}$. By the hypothesis,

$\mu\tau(E_{1})=\frac{1}{2}\mu\tau(E)\geq\frac{1}{2}\lambda\tau(F)=\lambda\tau(F_{1});$
$\mu\tau(E_{2})=\frac{1}{2}\mu\tau(E)\geq\frac{1}{2}\lambda\tau(F)=\lambda\tau(F_{2}).$

So by Lemma 3.5, $X_{1}$ and $X_{2}$ are finite sums of projections in
$\mathcal{M}$, respectively, and $X=X_{1}+X_{2}$ is a finite sum of
projections in $\mathcal{M}$. ∎

######  Lemma 3.8.

Let $E_{i},F_{j}$, $1\leq i\leq N,1\leq j\leq M$ be nonzero projections in
$(\mathcal{M},\tau)$ and let
$X=\sum_{i=1}^{N}(1+\mu_{i})E_{i}+\sum_{j=1}^{M}(1-\lambda_{j})F_{j}$, where
$\mu_{i}\geq 0$ and $0\leq\lambda_{j}\leq 1$. If
$\sum_{i=1}^{N}\mu_{i}\tau(E_{i})\geq\sum_{j=1}^{M}\lambda_{j}\tau(F_{j})$,
then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

(1) If $\mu_{1}\tau(E_{1})=\lambda_{1}\tau(F_{1})$, then by Lemma 3.7,
$(1+\mu_{1})E_{1}+(1-\lambda_{1})F_{1}$ is a finite sum of projections and
$\sum_{i=2}^{N}\mu_{i}\tau(E_{i})\geq\sum_{j=2}^{M}\lambda_{j}\tau(F_{j})$.

(2) If $\mu_{1}\tau(E_{1})>\lambda_{1}\tau(F_{1})$, then there exists a
nonzero projection $E_{1}^{\prime}\leq E_{1}$ such that
$\mu_{1}\tau(E_{1}^{\prime})=\lambda_{1}\tau(F_{1})$. Hence
$(1+\mu_{1})E_{1}^{\prime}+(1-\lambda_{1})F_{1}$ is a finite sum of
projections and
$\sum_{i=2}^{N}\mu_{i}\tau(E_{i})+\mu_{1}\tau(E_{1}-E_{1}^{\prime})\geq\sum_{j=2}^{M}\lambda_{j}\tau(F_{j})$.

(3) If $\mu_{1}\tau(E_{1})<\lambda_{1}\tau(F_{1})$, then there exists a
nonzero projection $F_{1}^{\prime}\leq F_{1}$ such that
$\mu_{1}\tau(E_{1})=\lambda_{1}\tau(F_{1}^{\prime})$. Hence
$(1+\mu_{1})E_{1}+(1-\lambda_{1})F_{1}^{\prime}$ is a finite sum of
projections and
$\sum_{i=2}^{N}\mu_{i}\tau(E_{i})\geq\sum_{j=2}^{M}\lambda_{j}\tau(F_{j})+\lambda_{1}\tau(F_{1}-F_{1}^{\prime})$.

By induction, we have that $X$ is a finite sum of projections in
$\mathcal{M}$. ∎

######  Lemma 3.9.

Let $E_{i}$, $1\leq i\leq K,$ be nonzero projections in $(\mathcal{M},\tau)$
and let $X=\sum_{i=1}^{K}\alpha_{i}E_{i}$, where $\alpha_{i}\geq 0$, such that
$\tau(X)=1$. Then $KX$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

Since

$K=\tau(KX)=\sum_{i=1}^{K}(K\alpha_{i})\tau(E_{i})=\sum_{i=1}^{K}\tau(E_{i})+\sum_{i=1}^{K}(K\alpha_{i}-1)\tau(E_{i})\leq
K+\sum_{i=1}^{K}(K\alpha_{i}-1)\tau(E_{i}),$

we have $\sum_{i=1}^{K}(K\alpha_{i}-1)\tau(E_{i})\geq 0$. Without loss of
generality, assume that

$KX=\sum_{i=1}^{N}(1+\mu_{i})E_{i}+\sum_{j=1}^{M}(1-\lambda_{j})F_{j},$

where $\mu_{i}\geq 0,0\leq\lambda_{j}\leq 1$ and $N+M=K$. Then
$\sum_{i=1}^{N}\mu_{i}\tau(E_{i})\geq\sum_{j=1}^{M}\lambda_{j}\tau(F_{j})$. By
Lemma 3.8, $KX$ is a finite sum of projections in $\mathcal{M}$.

∎

######  Remark 3.10.

Let $x\in\mathcal{M}^{+}$. Set $x_{1}:=x\chi_{x}[1,+\infty)$,
$x_{2}:=x\chi_{x}[0,1)$, $p:=\chi_{x}[1,+\infty)$ and $q:=\chi_{x}[0,1)$, then
$x=x_{1}+x_{2}$. To avoid triviality, we may assume that $x_{1},x_{2}\neq 0$.
Notice that

$x_{1}\geq p,\ \ \ \ x_{2}\leq q.$

Thus $\tau_{p}(x_{1}):=1+\mu\geq 1$ and $\tau_{q}(x_{2}):=1-\lambda\leq 1$. So
by Lemma 2.9, $\frac{x_{1}}{1+\mu}=\sum_{i=1}^{\infty}p_{i}$ for some
projections $\\{p_{i}\\}_{i=1}^{\infty}$ in $p\mathcal{M}p$ and
$\frac{x_{2}}{1-\lambda}=\sum_{j=1}^{\infty}q_{j}$ for some projections
$\\{q_{j}\\}_{j=1}^{\infty}$ in $q\mathcal{M}q$. Therefore,

$x=x_{1}+x_{2}=(1+\mu)\sum_{i=1}^{\infty}p_{i}+(1-\lambda)\sum_{j=1}^{\infty}q_{j}.$

The following three lemmas show that $\\{p_{i}\\}_{i=1}^{\infty}$ and
$\\{q_{j}\\}_{j=1}^{\infty}$ have very nice properties.

######  Lemma 3.11.

Let $q_{j}$ be as in Remark 3.10. Then $\forall N\in\mathbb{N}$,
$(1-\lambda)\sum_{j=N+1}^{\infty}q_{j}\leq\bigvee_{j=N+1}^{\infty}q_{j}$.

###### Proof.

For each $N\in\mathbb{N}$,
$(1-\lambda)\sum_{j=N+1}^{\infty}q_{j}\leq(1-\lambda)\sum_{j=1}^{\infty}q_{j}=x_{2}$.
Then

$\displaystyle(1-\lambda)\sum_{j=N+1}^{\infty}q_{j}$
$\displaystyle=\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)\left((1-\lambda)\sum_{j=N+1}^{\infty}q_{j}\right)\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)$
$\displaystyle\leq\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)x_{2}\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)$
$\displaystyle\leq\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)q\left(\bigvee_{j=N+1}^{\infty}q_{j}\right)$
$\displaystyle=\bigvee_{j=N+1}^{\infty}q_{j}.$

∎

######  Lemma 3.12.

Let $p_{i}$ be as in Remark 3.10. Then $\forall N\in\mathbb{N}$,
$(1+\mu)\sum_{i=N+1}^{\infty}p_{i}\geq\bigvee_{i=N+1}^{\infty}p_{i}$.

###### Proof.

Set $z_{1}:=\frac{x_{1}}{1+\mu}$. Then $\tau(z_{1})=1$. By Lemma 2.9, Remark
3.3, and $(2)\Rightarrow(1)$ of the proof of Lemma 3.2,
$p_{i}=z_{1}^{\frac{1}{2}}f_{i}z_{1}^{\frac{1}{2}}$ for some mutually
orthogonal projections $f_{i}$. Set projection
$f:=\sum_{i=N+1}^{\infty}f_{i}\leq p$, then we have that

$(1+\mu)\sum_{i=N+1}^{\infty}p_{i}=(1+\mu)\sum_{i=N+1}^{\infty}z_{1}^{\frac{1}{2}}f_{i}z_{1}^{\frac{1}{2}}=\sum_{i=N+1}^{\infty}x_{1}^{\frac{1}{2}}f_{i}x_{1}^{\frac{1}{2}}=x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}}.$

Since $fx_{1}f\geq fpf\geq f$,
$\sigma(fx_{1}f)\subseteq\\{0\\}\cup[1,+\infty)$. As
$x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}}$ and $fx_{1}f$ are unitary
equivalent, then
$\sigma(x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}})\subseteq\\{0\\}\cup[1,+\infty)$.
Since $x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}}$ restricted to
$R_{(x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}})}H$ is injective,
$\sigma\left(x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}}|_{R_{(x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}})}H}\right)\subseteq[1,+\infty)$.
Since
$R(x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}})=R(\sum_{i=N+1}^{\infty}p_{i})=\bigvee_{i=N+1}^{\infty}p_{i}$,
$(1+\mu)\sum_{i=N+1}^{\infty}p_{i}=x_{1}^{\frac{1}{2}}fx_{1}^{\frac{1}{2}}\geq\bigvee_{i=N+1}^{\infty}p_{i}$.

∎

######  Lemma 3.13.

Let $p_{i}$ and $q_{j}$ be as in Remark 3.10. Then we have the following
properties.

  1. (1)

$p=\chi_{x}[1,+\infty)=\bigvee_{i=1}^{\infty}p_{i}$;

  2. (2)

$q=\chi_{x}[0,1)=\bigvee_{j=1}^{\infty}q_{j}$;

  3. (3)

$\forall N\in\mathbb{N}$,
$\sum_{i=N+1}^{\infty}\tau(p_{i})=\tau\left(\bigvee_{i=N+1}^{\infty}p_{i}\right);$

  4. (4)

$\forall N\in\mathbb{N}$,
$\sum_{j=N+1}^{\infty}\tau(q_{j})=\tau\left(\bigvee_{j=N+1}^{\infty}q_{j}\right).$

###### Proof.

We only give the proofs of (1) and (3). The proofs of (2) and (4) are similar.

(1). Since $(\bigvee_{i=1}^{\infty}p_{i})x_{1}=x_{1}$, then
$\chi_{x}[1,+\infty)\leq\bigvee_{i=1}^{\infty}p_{i}$. As
$p_{i}\leq\chi_{x}[1,+\infty)$ for all $i$, thus
$\bigvee_{i=1}^{\infty}p_{i}\leq\chi_{x}[1,+\infty)$. So
$\chi_{x}[1,+\infty)=\bigvee_{i=1}^{\infty}p_{i}$.

(3). By (1), we have that

$\sum_{i=1}^{\infty}\tau_{p}(p_{i})=\tau_{p}\left(\frac{x_{1}}{1+\mu}\right)=1=\tau_{p}(\chi_{x}[1,+\infty))=\tau_{p}(\bigvee_{i=1}^{\infty}p_{i}).$

If
$\tau\left(\bigvee_{i=N+1}^{\infty}p_{i}\right)<\sum_{i=N+1}^{\infty}\tau(p_{i})$
for some $N$, then
$\tau_{p}\left(\bigvee_{i=N+1}^{\infty}p_{i}\right)<\sum_{i=N+1}^{\infty}\tau_{p}(p_{i})$.
Therefore,

$1=\tau_{p}\left(\bigvee_{i=1}^{\infty}p_{i}\right)\leq\tau_{p}(p_{1})+\cdots+\tau_{p}(p_{N})+\tau_{p}\left(\bigvee_{i=N+1}^{\infty}p_{i}\right)<\sum_{i=1}^{\infty}\tau_{p}(p_{i})=1,$

which is a contradiction. Then we have (3).

∎

######  Remark 3.14.

Let $x\in\mathcal{M}^{+}$ with $\tau(x)\geq 1$. Set
$x_{1}:=x\chi_{x}[1,+\infty)$, $x_{2}:=x\chi_{x}[0,1)$,
$p:=\chi_{x}[1,+\infty)$ and $q:=\chi_{x}[0,1)$, then $x=x_{1}+x_{2}$.

Case 1: $\tau(x)=1$. If $\tau_{p}(x_{1})=1$, then $x_{1}=p$ since $x_{1}\geq
p$. In this case, $\tau_{q}(x_{2})=1$ and $x_{2}=q$. Thus $x=p+q$ is a finite
sum of projections. So in the following we will assume that $\mu>0$ and
therefore $0<\lambda\leq 1$. Then we have that
$\tau_{p}(\frac{x_{1}}{1+\mu})=1$ and $\tau_{q}(\frac{x_{2}}{1-\lambda})=1$.
Since $\tau(x)=1$, $\tau\left(\sum_{i=1}^{\infty}p_{i}\right)=\tau(p)$ and
$\tau\left(\sum_{j=1}^{\infty}q_{j}\right)=\tau(q)$, we have

$1=\tau(x)=\tau(x_{1}+x_{2})=(1+\mu)\sum_{i=1}^{\infty}\tau(p_{i})+(1-\lambda)\sum_{j=1}^{\infty}\tau(q_{j})=1+\mu\sum_{i=1}^{\infty}\tau(p_{i})-\lambda\sum_{j=1}^{\infty}\tau(q_{j}).$

Thus

$\mu\sum_{i=1}^{\infty}\tau(p_{i})=\lambda\sum_{j=1}^{\infty}\tau(q_{j}).$

Case 2: $\tau(x)>1$. Since $\tau\left(\sum_{i=1}^{\infty}p_{i}\right)=\tau(p)$
and $\tau\left(\sum_{j=1}^{\infty}q_{j}\right)=\tau(q)$, similarly, we have

$\mu\sum_{i=1}^{\infty}\tau(p_{i})>\lambda\sum_{j=1}^{\infty}\tau(q_{j}).$

The idea of the proof of the following lemma comes from the proof of Lemma 2.7
in [9].

######  Lemma 3.15.

Let $X=(1+\mu)E+(1-\lambda)\sum_{i=1}^{\infty}F_{i}$, where $\mu>0$,
$0<\lambda\leq 1$, and $E$, $\\{F_{i}\\}_{i=1}^{\infty}$ are nonzero
projections in $(\mathcal{M},\tau)$. If $E$, $\\{F_{i}\\}_{i=1}^{\infty}$
satisfy the following conditions

  1. (1)

$E\bot(\bigvee_{i=1}^{\infty}F_{i});$

  2. (2)

$\sum_{i=1}^{\infty}\tau(F_{i})\leq\tau(E);$

  3. (3)

$(1-\lambda)\sum_{i=1}^{\infty}F_{i}\leq\bigvee_{i=1}^{\infty}F_{i},$

then $X=\sum_{i=1}^{K}\alpha_{i}E_{i}$, where $\alpha_{i}\geq 0$ and $E_{i}$
are some projections in $\mathcal{M}$. Furthermore, $K\leq
15+\lceil\frac{28}{\mu}\rceil$.

###### Proof.

Set $d:=(1+\mu)E$ and $b:=(1-\lambda)\sum_{i=1}^{\infty}F_{i}$, hence $X=b+d$.
Notice that by hypothesis (3),

$\|b\|\leq\|\bigvee_{i=1}^{\infty}F_{i}\|\leq 1.$

Since $\sum_{i=1}^{\infty}\tau(F_{i})\leq\tau(E)$, we can choose a partial
isometry $V\in\mathcal{M}$ for which $V^{*}V=\bigvee_{i=1}^{\infty}F_{i}$ and
$E^{\prime}:=VV^{*}\leq E.$ Define the projections

$q_{-}:=\begin{pmatrix}\frac{b}{\|b\|}&-\left(\frac{b}{\|b\|}-\frac{b^{2}}{\|b\|^{2}}\right)^{\frac{1}{2}}V^{*}\\\
-V\left(\frac{b}{\|b\|}-\frac{b^{2}}{\|b\|^{2}}\right)^{\frac{1}{2}}&E^{\prime}-\frac{1}{\|b\|}VbV^{*}\end{pmatrix}$

and

$q_{+}:=\begin{pmatrix}\frac{b}{\|b\|}&\left(\frac{b}{\|b\|}-\frac{b^{2}}{\|b\|^{2}}\right)^{\frac{1}{2}}V^{*}\\\
V\left(\frac{b}{\|b\|}-\frac{b^{2}}{\|b\|^{2}}\right)^{\frac{1}{2}}&E^{\prime}-\frac{1}{\|b\|}VbV^{*}\end{pmatrix}.$

Then $q_{-}+q_{+}=\frac{2}{\|b\|}b+2E^{\prime}-\frac{2}{\|b\|}VbV^{*}$, and
hence

$X=b+d=\frac{\|b\|}{2}q_{-}+\frac{\|b\|}{2}q_{+}+d-\|b\|E^{\prime}+VbV^{*}.$

Set $A:=d-\|b\|E^{\prime}+VbV^{*}$, then $\|A\|\leq\|d\|+\|VbV^{*}\|\leq
1+\mu+1=2+\mu$. Since

$A=d-\|b\|E^{\prime}+VbV^{*}\geq
d-\|b\|E^{\prime}=(1+\mu)E-\|b\|E^{\prime}\geq(1+\mu-\|b\|)E\geq\mu E,$

it follows by Proposition 2.3 and Lemma 2.5 that the locally invertible
element $A$ can be expressed as a positive linear combination of projections
in $\mathcal{M}$ no more than

$N_{0}+\lceil V_{0}\left(\|A\|\|A^{-1}\|-1\right)\rceil+1\leq 12+\lceil
14\left(\frac{2+\mu}{\mu}-1\right)\rceil+1=13+\lceil\frac{28}{\mu}\rceil.$

So $X$ is a positive combination of $K\leq 15+\lceil\frac{28}{\mu}\rceil$
projections in $\mathcal{M}$. ∎

######  Lemma 3.16.

Let $X=(1+\mu)E+(1-\lambda)\sum_{i=1}^{\infty}F_{i}$, where $\mu>0$,
$0<\lambda\leq 1$, and $E$, $\\{F_{i}\\}_{i=1}^{\infty}$ are nonzero
projections in $(\mathcal{M},\tau)$. If $E$, $\\{F_{i}\\}_{i=1}^{\infty}$
satisfy the following conditions

  1. (1)

$E\bot(\bigvee_{i=1}^{\infty}F_{i});$

  2. (2)

$\lambda\sum_{i=1}^{\infty}\tau(F_{i})<\mu\tau(E);$

  3. (3)

$\forall
N\in\mathbb{N},(1-\lambda)\sum_{i=N+1}^{\infty}F_{i}\leq\bigvee_{i=N+1}^{\infty}F_{i},$

then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

Set $\varepsilon:=\mu\tau(E)-\lambda\sum_{i=1}^{\infty}\tau(F_{i})$, then
$\varepsilon>0$ by (2). Since $\mathcal{M}$ is of type $\rm{II}_{1}$ factor,
we can decompose $E=E_{1}+E_{2}$ into the sum of two mutually orthogonal
projections such that

$0<\tau(E_{2})<\frac{\varepsilon}{3(2+\mu)K},$

where $K=15+\lceil\frac{28}{\mu}\rceil$. As
$\sum_{i=1}^{\infty}\tau(F_{i})<+\infty$, then for $\tau(E_{2})$, there exists
an $N\in\mathbb{N}$ such that

$\tau(E_{2})\geq\sum_{j=N+1}^{\infty}\tau(F_{j})\ \ \text{and}\ \
\mu\tau(E_{2})>\lambda\sum_{j=N+1}^{\infty}\tau(F_{j}).$

Set $Z:=(1+\mu)E_{2}+(1-\lambda)\sum_{j=N+1}^{\infty}F_{j}$ and
$Q:=E_{2}\vee(\bigvee_{j=N+1}^{\infty}F_{j})$. Then by above inequalities,

$s:=\tau_{Q}(Z)=\frac{\tau(E_{2})+\sum_{j=N+1}^{\infty}\tau(F_{j})+\mu\tau(E_{2})-\lambda\sum_{j=N+1}^{\infty}\tau(F_{j})}{\tau(Q)}\geq
1.$

Thus by Lemma 3.15, $Z$ can be expressed as a positive linear combination of
no more than $K$ projections and so is $\frac{Z}{s}$. Set
$\frac{Z}{s}=\sum_{j=1}^{K}\alpha_{j}q_{j}$ for some projections
$q_{j}\in\mathcal{M}$ and positive scalars $\alpha_{j}$. Notice that
$\tau_{Q}(\frac{Z}{s})=1$. Then by Lemma 3.9,
$\frac{KZ}{s}=\sum_{i=1}^{n}p_{i}$ for some projections $p_{i}\in\mathcal{M}$.
It implies $KZ=\sum_{i=1}^{n}sp_{i}$. If $s>1$, then by Lemma 2.8, $sp_{i}$ is
a finite sum of projections for every $i$; If $s=1$, then $KZ$ is already a
finite sum of projections. So without loss of generality, we can assume that
$KZ=\sum_{i=1}^{m}f_{i}$ for some projections $f_{i}\in\mathcal{M}$. Thus

$\displaystyle X$
$\displaystyle=(1+\mu)E_{1}+(1-\lambda)\sum_{i=1}^{N}F_{i}+Z$
$\displaystyle=(1+\mu)E_{1}+(1-\lambda)\sum_{i=1}^{N}F_{i}+\frac{1}{K}\sum_{i=1}^{m}f_{i}$
$\displaystyle=(1+\mu)E_{1}+(1-\lambda)\sum_{i=1}^{N}F_{i}+\left(1-(1-\frac{1}{K})\right)\sum_{i=1}^{m}f_{i}.$

Next we will show that

$\mu\tau(E_{1})-\lambda\sum_{i=1}^{N}\tau(F_{i})-\left(1-\frac{1}{K}\right)\sum_{i=1}^{m}\tau(f_{i})>0.$

For if the above inequality is true, then by Lemma 3.8, $X$ is a finite sum of
projections in $\mathcal{M}$.

As

$\displaystyle\sum_{i=1}^{m}\tau(f_{i})$
$\displaystyle=\tau(KZ)=K\left((1+\mu)\tau(E_{2})+(1-\lambda)\sum_{j=N+1}^{\infty}\tau(F_{j})\right)$
$\displaystyle\leq
K\left((1+\mu)\tau(E_{2})+\sum_{j=N+1}^{\infty}\tau(F_{j})\right)$
$\displaystyle\leq
K\left((1+\mu)\tau(E_{2})+\tau(E_{2})\right)<\frac{\varepsilon}{3},$

thus

$\displaystyle\mu\tau(E_{1})-\lambda\sum_{i=1}^{N}\tau(F_{i})-\left(1-\frac{1}{K}\right)\sum_{i=1}^{m}\tau(f_{i})$
$\displaystyle>\mu\tau(E)-\mu\tau(E_{2})-\lambda\sum_{i=1}^{\infty}\tau(F_{i})-\sum_{i=1}^{m}\tau(f_{i})$
$\displaystyle>\mu\tau(E)-\lambda\sum_{i=1}^{\infty}\tau(F_{i})-\mu\tau(E_{2})-\frac{\varepsilon}{3}$
$\displaystyle>\varepsilon-\frac{\mu\varepsilon}{3(2+\mu)K}-\frac{\varepsilon}{3}>\frac{\varepsilon}{3}>0.$

This completes the proof.

∎

The proof of the following lemma is similar to the proof of Lemma 3.16,
however, there exist some differences. So we give a complete proof.

######  Lemma 3.17.

Let $X=(1-\lambda)F+(1+\mu)\sum_{i=1}^{\infty}E_{i}$, where $\mu>0$,
$0<\lambda\leq 1$, and $F$, $\\{E_{i}\\}_{i=1}^{\infty}$ are nonzero
projections in $(\mathcal{M},\tau)$. If $F$, $\\{E_{i}\\}_{i=1}^{\infty}$
satisfy the following conditions

  1. (1)

$F\bot(\bigvee_{i=1}^{\infty}E_{i});$

  2. (2)

$\lambda\tau(F)<\mu\sum_{i=1}^{\infty}\tau(E_{i});$

  3. (3)

$\forall
N\in\mathbb{N},\bigvee_{i=N+1}^{\infty}E_{i}\leq(1+\mu)\sum_{i=N+1}^{\infty}E_{i},$

then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

By (3), we have $(1+\mu)\sum_{i=N+1}^{\infty}E_{i}$ is locally invertible for
every $N$(i.e., invertible in the algebra
$(\bigvee_{i=N+1}^{\infty}E_{i})\mathcal{M}(\bigvee_{i=N+1}^{\infty}E_{i})$),
thus $(1+\mu)\sum_{i=N+1}^{\infty}E_{i}$ is positive combination of no more
than $K_{N+1}:=N_{0}+\lceil
V_{0}(\|(1+\mu)\sum_{i=N+1}^{\infty}E_{i}\|-1)\rceil+1$ projections, where
$N_{0}=12,V_{0}=14$, by Corollary 2.6. Set $K:=N_{0}+\lceil
V_{0}(\|(1+\mu)\sum_{i=1}^{\infty}E_{i}\|-1)\rceil+1$. As
$(1+\mu)\sum_{i=N+1}^{\infty}E_{i}\leq(1+\mu)\sum_{i=1}^{\infty}E_{i}$, we
have $K_{N+1}\leq K$ for every $N$. So $(1+\mu)\sum_{i=N+1}^{\infty}E_{i}$ can
be expressed as a positive combination of no more than $K$ projections for
every $N$.

Set $\varepsilon:=\mu\sum_{i=1}^{\infty}\tau(E_{i})-\lambda\tau(F)$, then
$\varepsilon>0$ by (2). As $\sum_{i=1}^{\infty}\tau(E_{i})<+\infty$, then
there exists an $N\in\mathbb{N}$ such that
$\sum_{i=N+1}^{\infty}\tau(E_{i})<\frac{\varepsilon}{3(1+\mu)K}$. Set
$E:=\bigvee_{i=N+1}^{\infty}E_{i}$ and $Z:=(1+\mu)\sum_{i=N+1}^{\infty}E_{i}$.
Then by (3), we have
$s:=\tau_{E}(Z)=(1+\mu)\sum_{i=N+1}^{\infty}\tau_{E}(E_{i})\geq\tau_{E}(E)=1.$
Therefore $\tau_{E}(\frac{Z}{s})=1$. Then by Lemma 3.9,
$\frac{KZ}{s}=\sum_{i=1}^{n}p_{i}$ for some projections $p_{i}\in\mathcal{M}$.
It implies $KZ=\sum_{i=1}^{n}sp_{i}$. If $s>1$, then by Lemma 2.8, $sp_{i}$ is
a finite sum of projections for every $i$. If $s=1$, then $KZ$ is already a
finite sum of projections. So without loss of generality, we can assume that
$KZ=\sum_{i=1}^{m}f_{i}$ for some projections $f_{i}\in\mathcal{M}$. Thus

$\displaystyle X$ $\displaystyle=(1-\lambda)F+(1+\mu)\sum_{i=1}^{N}E_{i}+Z$
$\displaystyle=(1-\lambda)F+(1+\mu)\sum_{i=1}^{N}E_{i}+\frac{1}{K}\sum_{i=1}^{m}f_{i}$
$\displaystyle=(1-\lambda)F+(1+\mu)\sum_{i=1}^{N}E_{i}+\left(1-(1-\frac{1}{K})\right)\sum_{i=1}^{m}f_{i}.$

Next we will show that

$\mu\sum_{i=1}^{N}\tau(E_{i})-\lambda\tau(F)-\left(1-\frac{1}{K}\right)\sum_{i=1}^{m}\tau(f_{i})>0.$

For if the above inequality is true, then by Lemma 3.8, $X$ is a finite sum of
projections in $\mathcal{M}$.

Since

$\sum_{i=1}^{m}\tau(f_{i})=\tau(KZ)=K\left((1+\mu)\sum_{i=N+1}^{\infty}\tau(E_{i})\right)<\frac{\varepsilon}{3},$

we have

$\displaystyle\mu\sum_{i=1}^{N}\tau(E_{i})-\lambda\tau(F)-\left(1-\frac{1}{K}\right)\sum_{i=1}^{m}\tau(f_{i})$
$\displaystyle>\mu\sum_{i=1}^{\infty}\tau(E_{i})-\lambda\tau(F)-\mu\sum_{i=N+1}^{\infty}\tau(E_{i})-\frac{\varepsilon}{3}$
$\displaystyle>\varepsilon-\frac{\mu\varepsilon}{3(1+\mu)K}-\frac{\varepsilon}{3}>\frac{\varepsilon}{3}>0.$

This completes the proof. ∎

######  Lemma 3.18.

Let $X\in\mathcal{M}^{+}$. If $\tau(X)>1$, then $X$ is a finite sum of
projections in $\mathcal{M}$.

###### Proof.

By Remark 3.10, Remark 3.14, Lemma 3.11, Lemma 3.12 and Lemma 3.13,
$X=(1+\mu)\sum_{i=1}^{\infty}E_{i}+\sum_{j=1}^{\infty}(1-\lambda)F_{j}$, where
$\mu>0$, $0<\lambda\leq 1$, and $E_{i},F_{j}$ are nonzero projections in
$(\mathcal{M},\tau)$ such that

  1. (1)

$(\bigvee_{i=1}^{\infty}E_{i})\bot(\bigvee_{j=1}^{\infty}F_{j});$

  2. (2)

$\mu\sum_{i=1}^{\infty}\tau(E_{i})>\lambda\sum_{j=1}^{\infty}\tau(F_{j});$

  3. (3)

$\forall
N\in\mathbb{N},\bigvee_{i=N+1}^{\infty}E_{i}\leq(1+\mu)\sum_{i=N+1}^{\infty}E_{i};$

  4. (4)

$\forall
N\in\mathbb{N},(1-\lambda)\sum_{j=N+1}^{\infty}F_{j}\leq\bigvee_{j=N+1}^{\infty}F_{j}.$

Set

$\varepsilon:=\mu\sum_{i=1}^{\infty}\tau(E_{i})-\lambda\sum_{j=1}^{\infty}\tau(F_{j}),$

thus $\varepsilon>0$ by (2). Since $\sum_{j=1}^{\infty}\tau(F_{j})<+\infty$,
there is a smallest integer $n_{1}$ for which
$0<\mu\tau(E_{1})-\lambda\sum_{j=n_{1}}^{\infty}\tau(F_{j})$. If
$\mu\tau(E_{1})-\lambda\sum_{j=n_{1}}^{\infty}\tau(F_{j})\geq\frac{\varepsilon}{3}$,
then we can decompose
$F_{n_{1}-1}=F_{n_{1}-1}^{\prime}+F_{n_{1}-1}^{\prime\prime}$ into the sum of
two mutually orthogonal projections with
$0<\mu\tau(E_{1})-\lambda\sum_{j=n_{1}}^{\infty}\tau(F_{j})-\lambda\tau(F_{n_{1}-1}^{\prime})<\frac{\varepsilon}{3}$.
Then without loss of generality, we may assume that there exists
$N,M\in\mathbb{N}$ large enough such that

$0<\mu\tau(E_{1})-\lambda\sum_{j=N+1}^{\infty}\tau(F_{j})<\frac{\varepsilon}{3}\
\ \text{and}\ \
0<\mu\sum_{i=M+1}^{\infty}\tau(E_{i})-\lambda\tau(F_{1})<\frac{\varepsilon}{3}.$

Set

$X_{1}:=(1+\mu)E_{1}+(1-\lambda)\sum_{j=N+1}^{\infty}F_{j};$
$X_{2}:=(1-\lambda)F_{1}+(1+\mu)\sum_{i=M+1}^{\infty}E_{i};$
$X_{3}:=(1+\mu)\sum_{i=2}^{M}E_{i}+(1-\lambda)\sum_{j=2}^{N}F_{j}.$

Then $X=X_{1}+X_{2}+X_{3}$. By Lemma 3.16 and Lemma 3.17, $X_{1}$ and $X_{2}$
are finite sums of projections. Now, we consider $X_{3}$. Since

$\displaystyle\mu\sum_{i=2}^{M}\tau(E_{i})-\lambda\sum_{j=2}^{N}\tau(F_{j})$
$\displaystyle=\left(\mu\sum_{i=1}^{\infty}\tau(E_{i})-\lambda\sum_{j=1}^{\infty}\tau(F_{j})\right)-\left(\mu\tau(E_{1})-\lambda\sum_{j=N+1}^{\infty}\tau(F_{j})\right)-\left(\mu\sum_{i=M+1}^{\infty}\tau(E_{i})-\lambda\tau(F_{1})\right)$
$\displaystyle>\varepsilon-\frac{\varepsilon}{3}-\frac{\varepsilon}{3}>0.$

Hence, by Lemma 3.8, $X_{3}$ is a finite sum of projections. so $X$ is a
finite sum of projections. ∎

######  Lemma 3.19.

Let $X=(1+\mu)E+(1-\lambda)\sum_{i=1}^{\infty}F_{i}$, where $\mu>0$,
$0<\lambda\leq 1$, and $E$, $\\{F_{i}\\}_{i=1}^{\infty}$ are nonzero
projections in $(\mathcal{M},\tau)$. If $E$, $\\{F_{i}\\}_{i=1}^{\infty}$
satisfy the following conditions

  1. (1)

$E\bot(\bigvee_{i=1}^{\infty}F_{i});$

  2. (2)

$\lambda\sum_{i=1}^{\infty}\tau(F_{i})=\mu\tau(E);$

  3. (3)

$(1-\lambda)\sum_{i=1}^{\infty}F_{i}\leq\bigvee_{i=1}^{\infty}F_{i};$

  4. (4)

$\sum_{i=1}^{\infty}\tau(F_{i})=\tau(\bigvee_{i=1}^{\infty}F_{i}),$

then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

Set $F:=\bigvee_{i=1}^{\infty}F_{i}$. By considering $(E+F)\mathcal{M}(E+F)$,
we may assume that $E+F=I$. Thus $\tau(X)=1$ by (2) and (4). Set
$Z:=\frac{X-I}{\mu}$, then $\tau(Z)=0$ and $\tau(I-Z)=1$. By (3),

$\displaystyle I-Z$
$\displaystyle=\left(1+\frac{1}{\mu}\right)F-\left(\frac{1-\lambda}{\mu}\right)\sum_{i=1}^{\infty}F_{i}$
$\displaystyle\geq\frac{1}{\mu}\left((1+\mu)(1-\lambda)\sum_{i=1}^{\infty}F_{i}-(1-\lambda)\sum_{i=1}^{\infty}F_{i}\right)$
$\displaystyle=(1-\lambda)\sum_{i=1}^{\infty}F_{i}\geq 0$

and

$\displaystyle\tau_{F}(I-Z)$
$\displaystyle=\left(1+\frac{1}{\mu}\right)\tau_{F}(F)-\left(\frac{1-\lambda}{\mu}\right)\sum_{i=1}^{\infty}\tau_{F}(F_{i})$
$\displaystyle=1+\frac{1}{\mu}+\frac{\lambda}{\mu}-\frac{1}{\mu}=1+\frac{\lambda}{\mu}>1.$

Then by Lemma 3.18, $I-Z$ is a finite sum of $N$ projections in $\mathcal{M}$.
So by Lemma 3.2, there are nonzero mutually orthogonal projections
$P_{1},...,P_{N}$ such that $\sum_{i=1}^{N}P_{i}=I$ and
$P_{i}(I-Z)P_{i}=P_{i}$ for all $i=1,2,...,N$, which implies that
$P_{i}ZP_{i}=0$ for all $i=1,2,...,N$. Since $X=I+\mu Z$, then we also have
$P_{i}XP_{i}=P_{i}(I+\mu Z)P_{i}=P_{i}$ for all $i=1,2,...,N$. Notice that

$P_{i}=P_{i}X^{\frac{1}{2}}X^{\frac{1}{2}}P_{i}=(X^{\frac{1}{2}}P_{i})^{*}(X^{\frac{1}{2}}P_{i}),$

for all $i=1,2,...,N$. Thus $X^{\frac{1}{2}}P_{i}$ is a partial isometry. So
we conclude that $X^{\frac{1}{2}}P_{i}X^{\frac{1}{2}}=Q_{i}$ is a projection
for all $i=1,...,N$. As $\sum_{i=1}^{N}P_{i}=I$, we have
$X=Q_{1}+\cdots+Q_{N}$.

∎

######  Lemma 3.20.

Let $X=(1-\lambda)F+(1+\mu)\sum_{i=1}^{\infty}E_{i}$, where $\mu>0$,
$0<\lambda\leq 1$, and $F$, $\\{E_{i}\\}_{i=1}^{\infty}$ are nonzero
projections in $(\mathcal{M},\tau)$. If $F$, $\\{E_{i}\\}_{i=1}^{\infty}$
satisfy the following conditions

  1. (1)

$F\bot(\bigvee_{i=1}^{\infty}E_{i});$

  2. (2)

$\lambda\tau(F)=\mu\sum_{i=1}^{\infty}\tau(E_{i});$

  3. (3)

$\bigvee_{i=1}^{\infty}E_{i}\leq(1+\mu)\sum_{i=1}^{\infty}E_{i};$

  4. (4)

$\sum_{i=1}^{\infty}\tau(E_{i})=\tau(\bigvee_{i=1}^{\infty}E_{i}),$

then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

Set $E:=\bigvee_{i=1}^{\infty}E_{i}$. By considering $(E+F)\mathcal{M}(E+F)$,
we may assume that $E+F=I$. Thus $\tau(X)=1$ by (2) and (4). Set
$Z:=\frac{X-I}{\lambda}$, then $\tau(Z)=0$ and $\tau(I+Z)=1$. Notice that by
(3)

$\displaystyle I+Z$
$\displaystyle=\left(1-\frac{1}{\lambda}\right)E+\left(\frac{1+\mu}{\lambda}\right)\sum_{i=1}^{\infty}E_{i}$
$\displaystyle\geq\left(1-\frac{1}{\lambda}\right)E+\frac{1}{\lambda}\bigvee_{i=1}^{\infty}E_{i}=E\geq
0$

and

$\displaystyle\tau_{E}(I+Z)$
$\displaystyle=\left(1-\frac{1}{\lambda}\right)\tau_{E}(E)+\left(\frac{1+\mu}{\lambda}\right)\sum_{i=1}^{\infty}\tau_{E}(E_{i})$
$\displaystyle=\left(1-\frac{1}{\lambda}\right)+\left(\frac{1+\mu}{\lambda}\right)\tau_{E}\left(\bigvee_{i=1}^{\infty}E_{i}\right)$
$\displaystyle=1-\frac{1}{\lambda}+\frac{1+\mu}{\lambda}=1+\frac{\mu}{\lambda}>1.$

Then by Lemma 3.18, $I+Z$ is a finite sum of $N$ projections in $\mathcal{M}$.
So by Lemma 3.2, there are nonzero mutually orthogonal projections
$P_{1},...,P_{N}$ such that $\sum_{i=1}^{N}P_{i}=I$ and
$P_{i}(I+Z)P_{i}=P_{i}$ for all $i=1,2,...,N$, which implies that
$P_{i}ZP_{i}=0$ for all $i=1,2,...,N$. Since $X=I+\lambda Z$, then we also
have $P_{i}XP_{i}=P_{i}(I+\lambda Z)P_{i}=P_{i}$ for all $i=1,2,...,N$. Notice
that

$P_{i}=P_{i}X^{\frac{1}{2}}X^{\frac{1}{2}}P_{i}=(X^{\frac{1}{2}}P_{i})^{*}(X^{\frac{1}{2}}P_{i}),$

for all $i=1,2,...,N$. Then $X^{\frac{1}{2}}P_{i}$ is a partial isometry. So
we conclude that $X^{\frac{1}{2}}P_{i}X^{\frac{1}{2}}=Q_{i}$ is a projection
for all $i=1,...,N$. Since $\sum_{i=1}^{N}P_{i}=I$, we have
$X=Q_{1}+\cdots+Q_{N}$. ∎

######  Lemma 3.21.

Let $X=(1+\mu)\sum_{i=1}^{\infty}E_{i}+(1-\lambda)\sum_{j=1}^{\infty}F_{j}$,
where $\mu>0$, $0<\lambda\leq 1$, and $E_{i},F_{j}$ are nonzero projections in
$(\mathcal{M},\tau)$. If $E_{i},F_{j}$ satisfy the following conditions

  1. (1)

$(\bigvee_{i=1}^{\infty}E_{i})\bot(\bigvee_{j=1}^{\infty}F_{j});$

  2. (2)

$\mu\sum_{i=1}^{\infty}\tau(E_{i})=\lambda\sum_{j=1}^{\infty}\tau(F_{j});$

  3. (3)

$\forall
N\in\mathbb{N},\bigvee_{i=N+1}^{\infty}E_{i}\leq(1+\mu)\sum_{i=N+1}^{\infty}E_{i};$

  4. (4)

$\forall
N\in\mathbb{N},(1-\lambda)\sum_{j=N+1}^{\infty}F_{j}\leq\bigvee_{j=N+1}^{\infty}F_{j},$

then $X$ is a finite sum of projections in $\mathcal{M}$.

###### Proof.

Without loss of generality, we may assume that there exists $N,M\in\mathbb{N}$
large enough such that

$\mu\tau(E_{1})=\lambda\sum_{j=N+1}^{\infty}\tau(F_{j})\ \ \text{and}\ \
\mu\sum_{i=M+1}^{\infty}\tau(E_{i})=\lambda\tau(F_{1}).$

Set

$X_{1}:=(1+\mu)E_{1}+(1-\lambda)\sum_{j=N+1}^{\infty}F_{j};$
$X_{2}:=(1-\lambda)F_{1}+(1+\mu)\sum_{i=M+1}^{\infty}E_{i};$
$X_{3}:=\sum_{i=2}^{M}(1+\mu)E_{i}+(1-\lambda)\sum_{j=2}^{N}F_{j}.$

Then $X=X_{1}+X_{2}+X_{3}$. By Lemma 3.19 and Lemma 3.20, $X_{1}$ and $X_{2}$
are finite sums of projections. Now, we consider $X_{3}$. Since

$\displaystyle\mu\sum_{i=2}^{M}\tau(E_{i})-\lambda\sum_{j=2}^{N}\tau(F_{j})$
$\displaystyle=\left(\mu\sum_{i=1}^{\infty}\tau(E_{i})-\lambda\sum_{j=1}^{\infty}\tau(F_{j})\right)-\left(\mu\tau(E_{1})-\lambda\sum_{j=N+1}^{\infty}\tau(F_{j})\right)-\left(\mu\sum_{i=M+1}^{\infty}\tau(E_{i})-\lambda\tau(F_{1})\right)$
$\displaystyle=0.$

Hence, by Lemma 3.8, $X_{3}$ is a finite sum of projections, so is $X$. ∎

######  Lemma 3.22.

Let $X\in\mathcal{M}^{+}$. If $\tau(X)=1$, then $X$ is a finite sum of
projections in $\mathcal{M}$.

###### Proof.

By Remark 3.10, we can assume that

$X=(1+\mu)\sum_{i=1}^{\infty}p_{i}+(1-\lambda)\sum_{j=1}^{\infty}q_{j},$

where $\mu>0,0<\lambda\leq 1$ and $p_{i},q_{j}$ satisfy the properties of
Lemma 3.11, Lemma 3.12 and Lemma 3.13. Furthermore, we have that

$1=\tau(X)=1+\mu\sum_{i=1}^{\infty}\tau(p_{i})-\lambda\sum_{j=1}^{\infty}\tau(q_{j}).$

It implies that
$\mu\sum_{i=1}^{\infty}\tau(p_{i})=\lambda\sum_{j=1}^{\infty}\tau(q_{j})$.
Then by Lemma 3.21, it follows that $X$ is a finite sum of projections in
$\mathcal{M}$.

∎

######  Theorem 3.23.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor, $X\in\mathcal{M}$,
$X=X^{*}$. Then we have the following 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 $U\in\mathcal{M}$ with

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

  4. (4)

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

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

###### Proof.

We first consider the case $X\geq 0$. Assume that $\tau(X)=s$. It implies that
$\tau(\frac{X}{s})=1$. Then by Lemma 3.22, $\frac{X}{s}$ is a finite sum of
$N$ projections. Thus by Lemma 3.1, there is a unitary operator $U$ with

$\frac{1}{N}\sum_{i=0}^{N-1}{U^{*}}^{i}\left(\frac{X}{s}\right)U^{i}=I.$

Therefore,

$\frac{1}{N}\sum_{i=0}^{N-1}{U^{*}}^{i}XU^{i}=sI.$

Now, let $X$ be a self-adjoint operator. Since $-\|X\|I\leq X$, we have
$\|X\|I+X\geq 0$. By above argument, there exists a unitary operator $U$ with

$\frac{1}{N}\sum_{i=0}^{N-1}{U^{*}}^{i}(\|X\|I+X)U^{i}=\tau(\|X\|I+X)I=\|X\|I+sI.$

Thus

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

Therefore, by Lemma 3.1 and Lemma 3.2, we have the theorem. ∎

The following theorem answers affirmatively Question 6.7 in [9].

######  Theorem 3.24.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and
$A\in\mathcal{M}^{+}$. Then $A$ is a finite sum of projections if and only if
$\tau(A)\geq\tau(R_{A})$.

###### Proof.

$``\Rightarrow"$. Let $A=P_{1}+\cdots+P_{N}$ with $P_{i}$ nonzero projections
and $N\in\mathbb{N}$. By Kaplansky’s parallelogram law [[15], Theorem, 6.1.7],
we have

$\tau(R_{A})=\tau\left(\bigvee_{i=1}^{N}P_{i}\right)\leq\sum_{i=1}^{N}\tau(P_{i})=\tau(A).$

$``\Leftarrow"$. By considering $R_{A}\mathcal{M}R_{A}$, we may assume that
$R_{A}=I$. Then $\tau(A)\geq 1$. By Lemma 3.18 and Lemma 3.22, it implies that
$A$ is a finite sum of projections in $\mathcal{M}$.

∎

The following theorem answers affirmatively the Question 1.1 in [4].

######  Theorem 3.25.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor, $X\in\mathcal{M}$,
$X=X^{*}$ and $\tau(X)=0$. Then there exists a nilpotent element
$Z\in\mathcal{M}$ such that $X=ReZ$, i.e., $X$ is the real part of a nilpotent
element $Z$.

###### Proof.

By Theorem 3.23, there are $N$ mutually orthogonal nonzero projections $E_{j}$
such that $I=\sum_{j=1}^{N}E_{j}$ and $E_{j}XE_{j}=0$ for all $j=1,...,N$.
Since $X$ is self-adjoint, we can write

$X=\begin{pmatrix}0&X_{12}&X_{13}&\cdots&X_{1N}\\\
X_{12}^{*}&0&X_{23}&\cdots&X_{2N}\\\ X_{13}^{*}&X_{23}^{*}&0&\cdots&X_{3N}\\\
\vdots&\vdots&\ddots&\ddots&\vdots\\\
X_{1N}^{*}&X_{2N}^{*}&X_{3N}^{*}&\cdots&0\end{pmatrix}$

with respect to the decomposition $I=\sum_{j=1}^{N}E_{j}$. Set

$Z:=\begin{pmatrix}0&X_{12}&X_{13}&\cdots&X_{1N}\\\
0&0&X_{23}&\cdots&X_{2N}\\\ 0&0&0&\cdots&X_{3N}\\\
\vdots&\vdots&\ddots&\ddots&\vdots\\\ 0&0&0&\cdots&0\end{pmatrix}.$

We have that $Z$ is a nilpotent operator with $X=Z+Z^{*}$. This shows that $X$
is the real part of $Z$. ∎

##  4\. Dixmier’s averaging theorem for type ${\rm II}_{1}$ factors

######  Definition 4.1.

$A$ is said to be majorized by $B$, in notion $A\prec B$, if there are unitary
operators $U_{1},...,U_{N}\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{i=1}^{N}U_{i}^{*}BU_{i}=A.$

######  Proposition 4.2.

If $A\prec B$ and $B\prec C$, then $A\prec C$.

###### Proof.

If

$\frac{1}{N}\sum_{i=1}^{N}U_{i}^{*}BU_{i}=A\quad\text{and}\quad\frac{1}{M}\sum_{j=1}^{M}V_{j}^{*}CV_{j}=B,$

then

$\frac{1}{NM}\sum_{i=1}^{N}\sum_{j=1}^{M}U_{i}^{*}V_{j}^{*}CV_{j}U_{i}=A.$

∎

######  Lemma 4.3.

Set $B:=\begin{pmatrix}E_{1}BE_{1}&\cdots&E_{1}BE_{n}\\\ \vdots&&\vdots\\\
E_{n}BE_{1}&\cdots&E_{n}BE_{n}\end{pmatrix}$ with respect to decomposition
$I=E_{1}+\cdots+E_{n}$. Then $A\prec B$, where
$A=\begin{pmatrix}E_{1}BE_{1}&\cdots&0\\\ \vdots&\ddots&\vdots\\\
0&\cdots&E_{n}BE_{n}\end{pmatrix}$.

###### Proof.

For the $n\times n$ diagonal matrix. Let the first position is 1 and the other
position is 1 or -1 with respect to decomposition $I=E_{1}+\cdots+E_{n}$. Then
there are $2^{n-1}$ unitary operators denoted by $U_{1},...,U_{2^{n-1}}$. It
can prove that

$\frac{1}{2^{n-1}}\sum_{i=1}^{2^{n-1}}U_{i}^{*}BU_{i}=A.$

∎

######  Lemma 4.4.

If $(E_{a,b})_{a,b\in B}$ is a self-adjoint system of $n\times n$ matrix units
for a von Neumann algebra $\mathcal{M}$, it is known that $\mathcal{M}\cong
M_{n}(\mathds{C})\otimes\mathcal{N}$ for some von Neumann algebra
$\mathcal{N}$. In this algebra,
$A:=\begin{pmatrix}\frac{A_{1}+A_{2}+\cdots+A_{n}}{n}&\cdots&0\\\
\vdots&\ddots&\vdots\\\
0&\cdots&\frac{A_{1}+A_{2}+\cdots+A_{n}}{n}\end{pmatrix}\prec\begin{pmatrix}A_{1}&\cdots&0\\\
\vdots&\ddots&\vdots\\\ 0&\cdots&A_{n}\end{pmatrix}:=B,$ where
$A_{i}\in\mathcal{N}^{+}$ for $i=1,...,n$.

###### Proof.

Set $U_{1}=I$, $U_{i}:=\bordermatrix{&1&2&\cdots&i&i+1&\cdots&n\cr
1&0&0&\cdots&1&0&\cdots&0\cr
2&0&0&\cdots&0&1&\cdots&0\cr\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\ddots&\vdots\cr
n+2-i&1&0&\cdots&0&0&\cdots&0\cr
n+3-i&0&1&\cdots&0&0&\cdots&0\cr\vdots&\vdots&\vdots&\cdots&\vdots&\vdots&\ddots&0\cr
n&0&0&\cdots&0&0&\cdots&0},$ where $i=2,...,n-1$, and
$U_{n}=\begin{pmatrix}0&\cdots&1\\\ \vdots&\ddots&\vdots\\\
1&\cdots&0\end{pmatrix}.$ Then it can prove that

$\frac{1}{n}\sum_{i=1}^{n}U_{i}^{*}BU_{i}=A.$

∎

######  Proposition 4.5.

Let $A,B_{1},B_{2}\in\mathcal{M}$ and $A\prec B_{1}$. If $B_{1}$ is unitary
equivalent with $B_{2}$, then $A\prec B_{2}$.

###### Proof.

Since $A\prec B_{1}$, then by definition, there are unitary operators
$U_{1},...,U_{N}\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{i=1}^{N}U_{i}^{*}B_{1}U_{i}=A.$

As $B_{1}$ is unitary equivalent with $B_{2}$, then $B_{1}=V^{*}B_{2}V$ for
some unitary operator $V$. Thus

$\frac{1}{N}\sum_{i=1}^{N}U_{i}^{*}V^{*}B_{2}VU_{i}=A.$

∎

The following lemma is a generalization of Corollary 7.3 of [1].

######  Lemma 4.6.

If $(E_{a,b})_{a,b\in B}$ is a self-adjoint system of $n\times n$ matrix units
for a von Neumann algebra $\mathcal{M}$, it is known that $\mathcal{M}\cong
M_{n}(\mathds{C})\otimes\mathcal{N}$ for some von Neumann algebra
$\mathcal{N}$. In the algebra $\mathcal{M}\cong
M_{n}(\mathds{C})\otimes\mathcal{N}$,

$\begin{pmatrix}A_{1}&0&\cdots&0\\\ 0&A_{2}&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&A_{n}\end{pmatrix}\prec\begin{pmatrix}A_{1}+A_{2}+\cdots+A_{n}&0&\cdots&0\\\
0&0&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\ 0&0&\cdots&0\end{pmatrix},$

where $A_{i}\in\mathcal{N}^{+}$ for $i=1,2,...,n$

###### Proof.

First, we consider $2\times 2$ matrix. Write $\begin{pmatrix}A+B&0\\\
0&0\end{pmatrix}=\begin{pmatrix}A^{\frac{1}{2}}&B^{\frac{1}{2}}\\\
0&0\end{pmatrix}\begin{pmatrix}A^{\frac{1}{2}}&0\\\
B^{\frac{1}{2}}&0\end{pmatrix},$ which is unitary equivalent with
$\begin{pmatrix}A^{\frac{1}{2}}&0\\\
B^{\frac{1}{2}}&0\end{pmatrix}\begin{pmatrix}A^{\frac{1}{2}}&B^{\frac{1}{2}}\\\
0&0\end{pmatrix}.$ By the proof of Lemma 4.3, $\begin{pmatrix}A&0\\\
0&B\end{pmatrix}\prec\begin{pmatrix}A&A^{\frac{1}{2}}B^{\frac{1}{2}}\\\
B^{\frac{1}{2}}A^{\frac{1}{2}}&B\end{pmatrix}=\begin{pmatrix}A^{\frac{1}{2}}&0\\\
B^{\frac{1}{2}}&0\end{pmatrix}\begin{pmatrix}A^{\frac{1}{2}}&B^{\frac{1}{2}}\\\
0&0\end{pmatrix}.$ Then by Proposition 4.5, we have $\begin{pmatrix}A&0\\\
0&B\end{pmatrix}\prec\begin{pmatrix}A+B&0\\\ 0&0\end{pmatrix}.$ Next we will
prove that $\begin{pmatrix}A&0&0\\\ 0&B&0\\\
0&0&C\end{pmatrix}\prec\begin{pmatrix}A+B+C&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix},$ where $A,B,C\in\mathcal{N}^{+}$. Similarly to the above
argument.

$\begin{pmatrix}A+B+C&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix}=\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&C^{\frac{1}{2}}\\\
0&0&0\\\ 0&0&0\end{pmatrix}\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&0\\\ 0&0&0\\\
C^{\frac{1}{2}}&0&0\end{pmatrix},$

which is unitary equivalent with

$\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&0\\\ 0&0&0\\\
C^{\frac{1}{2}}&0&0\end{pmatrix}\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&C^{\frac{1}{2}}\\\
0&0&0\\\ 0&0&0\end{pmatrix}.$

Thus

$\begin{pmatrix}A+B&0&0\\\ 0&0&0\\\
0&0&C\end{pmatrix}\prec\begin{pmatrix}A+B&0&(A+B)^{\frac{1}{2}}C^{\frac{1}{2}}\\\
0&0&0\\\
C^{\frac{1}{2}}(A+B)^{\frac{1}{2}}&0&C\end{pmatrix}=\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&0\\\
0&0&0\\\
C^{\frac{1}{2}}&0&0\end{pmatrix}\begin{pmatrix}(A+B)^{\frac{1}{2}}&0&C^{\frac{1}{2}}\\\
0&0&0\\\ 0&0&0\end{pmatrix}.$

Then we have

$\begin{pmatrix}A+B&0&0\\\ 0&0&0\\\
0&0&C\end{pmatrix}\prec\begin{pmatrix}A+B+C&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix}.$

Since by above argument,

$\begin{pmatrix}A&0&0\\\ 0&B&0\\\
0&0&C\end{pmatrix}\prec\begin{pmatrix}A+B&0&0\\\ 0&0&0\\\ 0&0&C\end{pmatrix}.$

So we prove that

$\begin{pmatrix}A&0&0\\\ 0&B&0\\\
0&0&C\end{pmatrix}\prec\begin{pmatrix}A+B+C&0&0\\\ 0&0&0\\\
0&0&0\end{pmatrix}.$

By the same argument we have that

$\begin{pmatrix}A_{1}&0&\cdots&0\\\ 0&A_{2}&\cdots&0\\\
\vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&A_{n}\end{pmatrix}\prec\begin{pmatrix}A_{1}+A_{2}+\cdots+A_{n}&0&\cdots&0\\\
0&0&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\ 0&0&\cdots&0\end{pmatrix}.$

∎

######  Proposition 4.7.

Let $A\in\mathcal{M}$ be an injective positive map with dense range. Then the
following conditions are equivalent.

  1. (1)

$\begin{pmatrix}B&0\\\ 0&0\end{pmatrix}\prec\begin{pmatrix}A&0\\\
0&0\end{pmatrix};$

  2. (2)

$B\prec A.$

###### Proof.

$(1)\Rightarrow(2)$. By hypothesis, there are unitary operators
$U_{1},...,U_{N}\in\mathcal{M}$ such that

$\frac{1}{N}\sum_{i=1}^{N}U_{i}^{*}\begin{pmatrix}A&0\\\
0&0\end{pmatrix}U_{i}=\begin{pmatrix}B&0\\\ 0&0\end{pmatrix}.$

Set $U_{i}:=\begin{pmatrix}U_{11}^{i}&U_{12}^{i}\\\
U_{21}^{i}&U_{22}^{i}\end{pmatrix}.$ Then we have
$\frac{1}{N}\sum_{i=1}^{N}(U_{21}^{i})^{*}AU_{21}^{i}=0.$ Since
$(U_{21}^{i})^{*}AU_{21}^{i}\geq 0$ for every $i=1,...,N$, then
$(U_{21}^{i})^{*}AU_{21}^{i}=0$ for every $i=1,...,N$. Write
$(U_{21}^{i})^{*}AU_{21}^{i}=(U_{21}^{i})^{*}A^{\frac{1}{2}}A^{\frac{1}{2}}U_{21}^{i}$,
thus $A^{\frac{1}{2}}U_{21}^{i}=0$. As $A$ is injective, it implies that
$U_{21}^{i}=0$ for every $i=1,...,N$. Since $U_{i}$ is a unitary operator, it
implies that $(U_{11}^{i})^{*}U_{11}^{i}=I$ and
$U_{11}^{i}(U_{11}^{i})^{*}+U_{12}^{i}(U_{12}^{i})^{*}=I$. As $\mathcal{M}$ is
finite factor, then $(U_{11}^{i})^{*}U_{11}^{i}=I$ implies that
$U_{11}^{i}(U_{11}^{i})^{*}=I$. So we have $U_{12}^{i}=0$ for every
$i=1,...,N$. Therefore we conclude that

$\frac{1}{N}(U_{11}^{i})^{*}AU_{11}^{i}=B.$

Then we have (2). ∎

######  Corollary 4.8.

Let $(\mathcal{M},\tau)$ be a type $\rm{II}_{1}$ factor and $A\in\mathcal{M}$
be an injective positive map with dense range. Then the following conditions
are equivalent.

  1. (1)

$\begin{pmatrix}\tau(A)I&0\\\ 0&0\end{pmatrix}\prec\begin{pmatrix}A&0\\\
0&0\end{pmatrix};$

  2. (2)

$\tau(A)I\prec A.$

The following theorem is a stronger version of Dixmier’s averaging theorem for
type ${\rm II}_{1}$ factors.

######  Theorem 4.9.

Let $(\mathcal{M},\tau)$ be a type ${\rm II}_{1}$ factor and let
$X_{1},...,X_{n}\in\mathcal{M}$. Then there exist unitary operators
$U_{1},...,U_{k}\in\mathcal{M}$ such that

$\frac{1}{k}\sum_{i=1}^{k}U_{i}^{*}X_{j}U_{i}=\tau(X_{j})I,\quad\forall 1\leq
j\leq n.$

###### Proof.

Clearly, we may assume that $X_{i}=X_{i}^{*}$ for $1\leq i\leq n$. For
$X_{1}$, by Theorem 3.23, there exist unitary operators $V_{1},...,V_{r}$ such
that

$\frac{1}{r}\sum_{i=1}^{r}V_{i}^{*}X_{1}V_{i}=\tau(X_{1})I.$

Let

$A=\frac{1}{r}\sum_{i=1}^{r}V_{i}^{*}X_{2}V_{i}.$

Then $A=A^{*}$ and $\tau(A)=\tau(X_{2})$. By Theorem 3.23, there exist unitary
operators $W_{1},...,W_{s}$ such that

$\frac{1}{s}\sum_{i=1}^{s}W_{i}^{*}AW_{i}=\tau(X_{2})I.$

Now it is easy to see that $W_{i}V_{j}$ satisfies

$\frac{1}{rs}\sum_{i=1}^{s}\sum_{j=1}^{r}W_{i}^{*}V_{j}^{*}X_{k}V_{j}W_{i}=\tau(X_{k})I,\quad
k=1,2.$

Now, by induction we have the theorem. ∎

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