# Noncommutative $H^{p}$ spaces associated with type 1 subdiagonal algebras

Ruihan Zhang  School of Mathematics and Statistics, Shaanxi Normal University,
Xian, 710119, People’s Republic of China<EMAIL_ADDRESS>and  Guoxing Ji∗
School of Mathematics and Statistics, Shaanxi Normal University, Xian, 710119,
People’s Republic of China<EMAIL_ADDRESS>
###### Abstract.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra in a $\sigma$-finite von
Neumann algebra $\mathcal{M}$ with respect to a faithful normal conditional
expectation $\Phi$. We consider a Riesz type factorization theorem in
noncommutative $H^{p}$ spaces associated with $\mathfrak{A}$. It is shown that
if $1\leq r,p,q<\infty$ such that $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$, then
for any $h\in H^{r}$, there exist $h_{p}\in H^{p}$ and $h_{q}\in H^{q}$ such
that $h=h_{p}h_{q}$. Beurling type invariant subspace theorem for
noncommutative $L^{p}(1<p<\infty)$ space is obtained. Furthermore, we show
that a $\sigma$-weakly closed subalgebra containing $\mathfrak{A}$ of
$\mathcal{M}$ is also a type 1 subdiagonal algebra. As an application, We
prove that the relative invariant subspace lattice
$Lat_{\mathcal{M}}\mathfrak{A}$ of $\mathfrak{A}$ in $\mathcal{M}$ is
commutative.

###### Key words and phrases:

von Neumann algebra; type 1 subdiagonal algebra; noncommutative $H^{p}$ space;
invariant subspace

###### 1991 Mathematics Subject Classification:

46L52; 47L75; 46K50; 46J15

∗ Corresponding author
This research was supported by the National Natural Science Foundation of
China(No. 11771261).

##  1\. Introduction

Riesz factorization theorem and Beurling invariant subspace theorem in the
classical Hardy spaces are well-known. There were many very interesting
extensions to abstract function algebras. Arveson in [2] introduced the notion
of subdiagonal algebras as the noncommutative analogue of the classical Hardy
space $H^{\infty}(\mathbb{T})$. It is worth noting that there were several
successful examples showing that some famous mature theorems about the
classical Hardy spaces have been extended to the noncommutative Hardy spaces
based on subdiagonal algebras(cf. [3, 4, 5, 6, 13, 14, 15, 17, 18]). Marsalli
and West [24] gave a Riesz factorization theorem for finite noncommutative
$H^{p}$ spaces. One important extension is due to Blecher and Labuschagne on
Beurling type invariant subspace theorems for a finite subdiagonal algebra in
[7]. Recently, Labuschagne in [23] extended their results to noncommutative
$H^{2}$ for maximal subdiagonal algebras in a $\sigma$-finite von Neumann
algebra based on noncommutative $H^{p}$ spaces in [13]. We also discussed
Riesz type factorization theorem in noncommutative $H^{1}$ and Beurling type
invariant subspace theorem for $L^{1}(\mathcal{M})$ associated with a type 1
subdiagonal algebra in the sense that every right invariant subspace of a
maximal subdiagonal algebra in noncommutative $H^{2}$ is of Beurling type in
[15, 16]. Do Riesz factorization theorem and Beurling type invariant subspace
theorem when $1<p<\infty$ hold for type 1 subdiagonal algebras? We consider
these problems in this paper. We firstly recall some notions.

Let $\mathcal{M}$ be a $\sigma$-finite von Neumann algebra acting on a complex
Hilbert $\mathcal{H}$. $Z(\mathcal{M})=\mathcal{M}\cap\mathcal{M}^{\prime}$ is
the center of $\mathcal{M}$, where $\mathcal{M}^{\prime}$ is the commutant of
$\mathcal{M}$. If $Z(\mathcal{M})=\mathbb{C}I$, the multiples of identity $I$,
then $\mathcal{M}$ is said to be a factor. We denote by $\mathcal{M}_{*}$ and
$\mathcal{M}_{p}$ the space of all $\sigma$-weakly continuous linear
functionals of $\mathcal{M}$ and the set of all projections in $\mathcal{M}$
respectively. Let $\Phi$ be a faithful normal conditional expectation from
$\mathcal{M}$ onto a von Neumann subalgebra $\mathfrak{D}$. Arveson [2] gave
the following definition. A subalgebra $\mathfrak{A}$ of $\mathcal{M}$,
containing $\mathfrak{D}$, is called a subdiagonal algebra of $\mathcal{M}$
with respect to $\Phi$ if

(i) $\mathfrak{A}\cap\mathfrak{A}^{*}=\mathfrak{D}$,

(ii) $\Phi$ is multiplicative on $\mathfrak{A}$, and

(iii) $\mathfrak{A}+\mathfrak{A}^{*}$ is $\sigma$-weakly dense in
$\mathcal{M}$.

The algebra $\mathfrak{D}$ is called the diagonal of $\mathfrak{A}$. we may
assume that subdiagonal algebras are $\sigma$-weakly closed without loss
generality(cf.[2]).

We say that $\mathfrak{A}$ is a maximal subdiagonal algebra in $\mathcal{M}$
with respect to $\Phi$ in case that $\mathfrak{A}$ is not properly contained
in any other subalgebra of $\mathcal{M}$ which is subdiagonal with respect to
$\Phi$. Put $\mathfrak{A}_{0}=\\{X\in\mathfrak{A}:\Phi(X)=0\\}$ and
$\mathfrak{A}_{m}=\\{X\in\mathcal{M}:\Phi(AXB)=\Phi(BXA)=0,~{}\forall
A\in\mathfrak{A},~{}B\in\mathfrak{A}_{0}\\}$. By [2, Theorem 2.2.1], we recall
that $\mathfrak{A}_{m}$ is a maximal subdiagonal algebra of $\mathcal{M}$ with
respect to $\Phi$ containing $\mathfrak{A}$. $\mathfrak{A}$ is said to be
finite if there is a faithful normal finite trace $\tau$ on $\mathcal{M}$ such
that $\tau\circ\Phi=\tau$. Finite subdiagonal algebras are maximal
subdiagonal(cf.[8]).

We next recall Haagerup’s noncommutative $L^{p}$ spaces associated with a
general von Neumann algebra $\mathcal{M}$(cf.[9, 26]). Let $\varphi$ be a
faithful normal state on $\mathcal{M}$ and let
$\\{\sigma_{t}^{\varphi}:t\in\mathbb{R}\\}$ be the modular automorphism group
of $\mathcal{M}$ associated with $\varphi$ by Tomita-Takesaki theory. We
consider the crossed product
$\mathcal{N}=\mathcal{M}\rtimes_{\sigma^{\varphi}}\mathbb{R}$ of $\mathcal{M}$
by $\mathbb{R}$ with respect to $\sigma^{\varphi}$. We denote by $\theta$ the
dual action of $\mathbb{R}$ on $\mathcal{N}$. Then
$\\{\theta_{s}:s\in\mathbb{R}\\}$ is an automorphisms group of $\mathcal{N}$.

Note that $\mathcal{M}=\\{X\in\mathcal{N}:\theta_{s}(X)=X,\forall
s\in\mathbb{R}\\}$. $\mathcal{N}$ is a semifinite von Neumann algebra and
there is the normal faithful semifinite trace $\tau$ on $\mathcal{N}$
satisfying

$\tau\circ\theta_{s}=e^{-s}\tau,\ \ \ \forall s\in\mathbb{R}.$

According to Haagerup [9, 26], the noncommutative $L^{p}$ space
$L^{p}(\mathcal{M})$ for each $0<p\leq\infty$ is defined as the set of all
$\tau$-measurable operators $x$ affiliated with $\mathcal{N}$ satisfying

$\theta_{s}(x)=e^{-\frac{s}{p}}x,\ \forall s\in\mathbb{R}.$

There is a linear bijection between the predual $\mathcal{M}_{*}$ of
$\mathcal{M}$ and $L^{1}(\mathcal{M})$: $f\to h_{f}$. If we define
$\mbox{tr}(h_{f})=f(I),f\in$$\mathcal{M}_{*}$, then

$\mbox{tr}(|h_{f}|)=\mbox{tr}(h_{|f|})=|f|(I)=\|f\|$

for all $f\in\mathcal{M}_{*}$ and

$|\mbox{tr}(x)|\leq\mbox{tr}(|x|)$

for all $x\in L^{1}(\mathcal{M})$. As in [9], we define the operator $L_{A}$
and $R_{A}$ on $L^{p}(\mathcal{M})$($1\leq p<\infty)$ by $L_{A}x=Ax$ and
$R_{A}x=xA$ for all $A\in\mathcal{M}$ and $x\in L^{p}(\mathcal{M})$. Note that
$L^{2}(\mathcal{M})$ is a Hilbert space with the inner product
$(a,b)=\mbox{tr}(b^{*}a)$, $\forall a,b\in L^{2}(\mathcal{M})$ and $A\to
L_{A}$( resp. $A\to R_{A}$) is a faithful representation (resp. anti-
representation) of $\mathcal{M}$ on $L^{2}(\mathcal{M})$. We may identify
$\mathcal{M}$ with $L(\mathcal{M})=\\{L_{A}:A\in\mathcal{M}\\}$.

We recall that $\\{\sigma_{t}^{\varphi}:t\in\mathbb{R}\\}$ is the modular
automorphism group of $\mathcal{M}$ associated with $\varphi$. Let $h_{0}$ be
the noncommutative Radon-Nikodym derivative of the dual weight of $\varphi$ on
$\mathcal{N}$ with respect to $\tau$. Then $h_{0}$ is the image($h_{\varphi}$)
of $\varphi$ in $L^{1}(\mathcal{M})$ and we have that the following
representation of $\sigma_{t}^{\varphi}$(cf.[21, 22]):

(1.1) $\sigma_{t}^{\varphi}(X)=h_{0}^{it}Xh_{0}^{-it},\ \forall
t\in\mathbb{R},\ \forall X\in\mathcal{M}.$

It is known that the noncommutative $H^{p}$ space $H^{p}(\mathcal{M})$ and
$H_{0}^{p}(\mathcal{M})$ in $L^{p}(\mathcal{M})$ for any $1\leq p<\infty$ is
$H^{p}=H^{p}(\mathcal{M})=[h_{0}^{\frac{\theta}{p}}\mathfrak{A}h_{0}^{\frac{1-\theta}{p}}]_{p}$
and
$H_{0}^{p}=H_{0}^{p}(\mathcal{M})=[h_{0}^{\frac{\theta}{p}}\mathfrak{A}_{0}h_{0}^{\frac{1-\theta}{p}}]_{p}$
for any $\theta\in[0,1]$ from [13, Definition 2.6] and [14, Proposition 2.1].
If $p=\infty$, then we identify $H^{\infty}$ as $\mathfrak{A}$ and
$H_{0}^{\infty}$ as $\mathfrak{A}_{0}$.

We introduce the notion of column $L^{p}$-sum in the noncommutative
$L^{p}$-space $L^{p}(\mathcal{M})(1\leq p\leq\infty)$ for a $\sigma$-finite
von Neumann algebra $\mathcal{M}$ given in [19] by Junge and Sherman. For any
subset $S\subseteq L^{p}(\mathcal{M})$, $[S]_{p}$ denotes the closed linear
span of $S$ in $L^{p}(\mathcal{M})$. If $X$ is a closed subspace of
$L^{p}(\mathcal{M})$, and $\\{X_{i}:i\in\Lambda\\}$ is a family of closed
subspaces of $X$ such that $X=[\bigvee\\{X_{i}:i\in\Lambda\\}]_{p}$ with the
property that $X_{j}^{*}X_{i}=\\{0\\}$ if $i\not=j$, then we say that X is the
internal column $L^{p}$-sum $\oplus_{i\in\Lambda}^{col}X_{i}$. If $p=\infty$,
we assume that $X$ and $\\{X_{i}:i\in\Lambda\\}$ are $\sigma$-weakly closed,
and the closed linear span is taken with the $\sigma$-weak topology. For
symmetry, if $X_{j}X_{i}^{*}=\\{0\\}$ if $i\not=j$ and
$X=[\vee\\{X_{i}:i\in\Lambda\\}]_{p}$, then we say that $X$ is the internal
row $L^{p}$-sum $\oplus_{i\in\Lambda}^{row}X_{i}$.

Following [7, 23], we define the right wandering subspace of a right invariant
subspace $\mathfrak{M}\subseteq L^{2}(\mathcal{M})$, that is,
$\mathfrak{M}\mathfrak{A}\subseteq\mathfrak{M}$, to be the space
$W=\mathfrak{M}\ominus[\mathfrak{M}\mathfrak{A}_{0}]_{2}$. We say that
$\mathfrak{M}$ is of type 1 if $W$ generates $\mathfrak{M}$ as an
$\mathfrak{A}$-module (that is, $\mathfrak{M}=[W\mathfrak{A}]_{2}$). We say
that $\mathfrak{M}$ is of type 2 if $W=\\{0\\}$. Note that every right
invariant subspace $\mathfrak{M}$ is an $L^{2}$-column sum
$\mathfrak{M}=\mathfrak{N}_{1}\oplus^{col}\mathfrak{N}_{2}$, where
$\mathfrak{N}_{i}$ is of type $i$ for $i=1,2$ from [7, Theorem 2.1] and [23,
Theorem 2.3]. In particular, if $\mathfrak{M}$ is of type 1, then
$\mathfrak{M}$ is of the Beurling type, that is, there exists a family of
partial isometries $\\{U_{n}:n\geq 1\\}$ in $\mathcal{M}$ satisfying
$U_{n}^{*}U_{m}=0$ if $n\not=m$ and $U_{n}^{*}U_{n}\in\mathfrak{D}$ such that
$\mathfrak{M}=\oplus_{n}^{col}U_{n}H^{2}$. We refer to [7, 23] for more
details. Symmetrically, we may consider left invariant subspaces for
$\mathfrak{A}$ and a type 1 left invariant subspace $\mathfrak{M}$ may be
represented as $\mathfrak{M}=\oplus_{n\geq 1}^{row}H^{2}V_{n}$ by a family of
partial isometries $\\{V_{n}:n\geq 1\\}$ satisfying $V_{n}V_{m}^{*}=0$ if
$n\not=m$ and $V_{n}V_{n}^{*}\in\mathfrak{D}$ if $n=m$ and this fact will be
used frequently. Motivated by the column(resp. row)-sum as above, we call that
a family of partial isometries $\mathcal{U}=\\{U_{n}:n\geq 1\\}$ in
$\mathcal{M}$ is column(resp. row) orthogonal if If $U_{n}^{*}U_{m}=0$(resp.
$U_{n}U_{m}^{*}=0$) for $n\not=m$ and $U_{n}^{*}U_{n}\in\mathfrak{D}$(resp.
$U_{n}U_{n}^{*}\in\mathfrak{D}$) for all $n$(cf.[16]).

In this paper, we continuously consider type 1 subdiagonal algebras introduced
in [15]. Let $\mathfrak{A}$ be a type 1 subdiagonal algebra with respect to
$\Phi$ with diagonal $\mathfrak{D}$, that is, every invariant right invariant
subspace $\mathfrak{M}\subseteq H^{2}$ of $\mathfrak{A}$ is of type 1. Note
that there is a family of column orthogonal partial isometries
$\mathcal{U}=\\{U_{n}:n\geq 1\\}$ such that
$H_{0}^{2}=\oplus_{n}^{col}U_{n}H^{2}$ and those partial isometries together
with the diagonal $\mathfrak{D}$ consist of generators of
$\mathfrak{A}$(cf.[15]). We will fixed those partial isometries throughout
this paper. We give a Riesz type factorization theorem for non-commutative
$H^{P}(1\leq p<\infty)$ which says that every element in $H^{r}$ is a product
of two elements in $H^{p}$ and $H^{q}$ respectively with
$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$ for any $1\leq r,p,q<\infty$ in Section
2. This gives an answer of a problem in [15]. Moreover, we consider a Beurling
type invariant subspace theorem in $L^{p}(\mathcal{M})$ when $1<p<\infty$ in
Section 3. In Section 4, We determine the structure of those $\sigma$-weakly
closed subalgebras containing the type 1 subdiagonal algebra $\mathfrak{A}$ in
$\mathcal{M}$. It is shown that those subalgebras are of type 1. As an
application, we show that the relative invariant subspace lattice
$Lat_{\mathcal{M}}\mathfrak{A}=\\{E\in\mathcal{M}_{p}:(I-E)AE=0,\ \forall
A\in\mathfrak{A}\\}$ of $\mathfrak{A}$ in the von Neumann algebra
$\mathcal{M}$ is commutative in Section 5.

##  2\. Riesz factorization in noncommutative $H^{p}$ spaces for type 1
subdiagonal algebras

In [15], we consider the Riesz factorization in noncommutative $H^{1}$ for
type 1 subdiagonal algebra $\mathfrak{A}$. It is shown that for any $h\in
H^{1}$, there are two elements $h_{i}\in H^{2}(i=1,2)$ such that
$h=h_{1}h_{2}$. Let $1\leq r,p,q<\infty$ such that
$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$ and $h\in H^{r}$. Does $h=h_{p}h_{q}$
for some $h_{p}\in H^{p}$ and $h_{q}\in H^{q}$? We consider this problem in
this section. For any subset $\mathcal{S}\subseteq L^{p}(\mathcal{M})$ and
$\mathcal{K}\subseteq L^{q}(\mathcal{M})$, denote by
$\mathcal{S}\mathcal{K}=[\\{sk:s\in\mathcal{S},k\in\mathcal{K}\\}]_{r}$, the
closed subspace generated by the subset
$\\{sk:s\in\mathcal{S},k\in\mathcal{K}\\}$ in $L^{r}(\mathcal{M})$. We firstly
recall that an element $h\in H^{p}$ is right(resp. left) outer in
$[h\mathfrak{A}]_{p}=H^{p}$(resp. $[\mathfrak{A}h]_{p}=H^{p}$).

######  Lemma 2.1.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra. Then for any nonzero $f\in
L^{2}(\mathcal{M})$, there exist a left outer element $g\in H^{2}$ and a
contraction $B\in\mathcal{M}$ such that $f=gB$.

###### Proof.

We consider the type 1 subdiagonal algebra $\mathfrak{A}^{*}$ and
$(H^{2})^{*}$,. Then for any $f^{*}\in L^{2}(\mathcal{M})$, there exist a
right outer element $g^{*}\in(H^{2})^{*}$ and a contraction $B\in\mathcal{M}$
such that $f^{*}=B^{*}g^{*}$ by [15, Theorem 3.5]. Then $f=gB$. Note that
$[g^{*}\mathfrak{A}^{*}]_{2}=(H^{2})^{*}$. Thus $[\mathfrak{A}g]_{2}=H^{2}$,
that is, $g\in H^{2}$ is left outer. ∎

Let $\mathfrak{M}\subseteq L^{p}(\mathcal{M})$ be a closed subspace. If
$\mathfrak{M}\mathfrak{A}\subseteq\mathfrak{M}$(resp.
$\mathfrak{A}\mathfrak{M}\subseteq\mathfrak{M}$), then we say that
$\mathfrak{M}$ is a right(resp. left) invariant subspace.

######  Lemma 2.2.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and $2<p<\infty$. If
$\mathfrak{M}\subseteq H^{p}$ is a right invariant subspace such that
$\mathfrak{M}H^{s}=H^{2}$, then $\mathfrak{M}=H^{p}$, where
$\frac{1}{p}+\frac{1}{s}=\frac{1}{2}$.

###### Proof.

Let $\frac{1}{p}+\frac{1}{q}=1$ and $\frac{1}{s}+\frac{1}{t}=1$. Then
$\frac{1}{2}+\frac{1}{s}=\frac{1}{q}$ and
$\frac{1}{p}+\frac{1}{2}=\frac{1}{t}$. Put $\mathfrak{M}^{\bot}=\\{g\in
L^{q}(\mathcal{M}):\mbox{tr}(gm)=0,\forall m\in\mathfrak{M}\\}$. It is
sufficient to show that $\mathfrak{M}^{\bot}=(H^{p})^{\bot}=H_{0}^{q}$. It is
trivial that $\mathfrak{M}^{\bot}\supseteq H_{0}^{q}.$ Take any
$g\in\mathfrak{M}^{\bot}$. Let $g=V|g|$ be the polar decomposition. Then
$g=V|g|^{\frac{q}{2}}|g|^{\frac{q}{s}}=g_{1}g_{2}$, where
$g_{1}=V|g|^{\frac{q}{2}}\in L^{2}(\mathcal{M})$ and
$g_{2}=|g|^{\frac{q}{s}}\in L^{s}(\mathcal{M})$. Then $g_{1}=h_{1}B$ for a
left outer element $h_{1}\in H^{2}$ and a contraction $B\in\mathcal{M}$ by
Lemma 2.1. Put $h_{2}=Bg_{2}\in L^{s}(\mathcal{M})$. Now
$\mbox{tr}(mg)=\mbox{tr}(mg_{1}g_{2})=\mbox{tr}(mh_{1}h_{2})=0$ for all
$m\in\mathfrak{M}$. On the other hand,
$[\mathfrak{M}h_{1}]_{t}=[\mathfrak{M}\mathfrak{A}h_{1}]_{t}=[\mathfrak{M}H_{2}]_{t}=[\mathfrak{M}h_{0}^{\frac{1}{2}}\mathfrak{A}]_{t}=[\mathfrak{M}h_{0}^{\frac{1}{s}}h_{0}^{\frac{1}{p}}\mathfrak{A}]_{t}=[[\mathfrak{M}h_{0}^{\frac{1}{s}}\mathfrak{A}]_{2}H^{p}]_{t}=[H^{2}H^{p}]_{t}=H^{t}$.
Thus $\mbox{tr}(xh_{2})=0$ for all $x\in H^{t}$. It follows that
$h_{2}\in(H^{t})^{\bot}=H_{0}^{s}$. Therefore $g=h_{1}h_{2}\in H_{0}^{q}$.
Consequently, $\mathfrak{M}=H^{p}$. ∎

######  Theorem 2.3.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and $\eta\in
L^{p}(\mathcal{M})(1\leq p<\infty)$ a nonzero vector. Then for any $\delta>0$,
there exist a contraction $B\in\mathcal{M}$ and a right outer element $h\in
H^{p}$ with $\|h\|_{p}<\|\eta\|_{p}+\delta$ such that $\eta=Bh$. If $\eta\in
H^{p}$, then we may have $B\in\mathfrak{A}$.

###### Proof.

If $p=2$, then the result is proved in [15, Theorem 3.5]. Suppose that
$2<p<\infty$ and we take some $t>2$ such that
$\frac{1}{2}=\frac{1}{p}+\frac{1}{t}$. Then $\eta h_{0}^{\frac{1}{t}}\in
L^{2}(\mathcal{M})$.

Put
$x=(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}h_{0}^{\frac{1}{t}}$.
Then we have that $x\in L^{2}(\mathcal{M})$ such that
$\varepsilon^{2}h_{0}\leq x^{*}x$ and $h_{0}^{\frac{1}{t}}\eta^{*}\eta
h_{0}^{\frac{1}{t}}\leq x^{*}x$. By the proof of [15, Theorem 3.5], there are
injective contractions $A,C\in\mathcal{M}$ with dense ranges, a unitary
operator $U\in\mathcal{M}$ and a right outer element $\xi\in H^{2}$, such that
$\varepsilon^{-1}Ax=h_{0}^{\frac{1}{2}}$, $\eta h_{0}^{\frac{1}{t}}=Cx$ and
$x=U\xi$. Note that
$\xi=U^{*}x=U^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}h_{0}^{\frac{1}{t}}$
is a right outer element in $H^{2}$. Then
$H^{2}=[\xi\mathfrak{A}]_{2}=[U^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}h_{0}^{\frac{1}{t}}\mathfrak{A}]_{2}=[U^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}H^{r}]_{2}$.
Put $h=U^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}$.
We note that $h\in H^{p}$. In fact,
$H^{2}=[hh_{0}^{\frac{1}{t}}\mathfrak{A}]_{2}=[h\mathfrak{A}]_{p}H^{t}$ and
$H_{0}^{q}=H^{t}H_{0}^{2}$. Thus $\mbox{tr}(hxy)=0$ for all $x\in H^{t}$ and
$y\in H_{0}^{2}$, which implies that $h\in(H_{0}^{q})^{\bot}=H^{p}$. Therefore
$\mathfrak{M}=[h\mathfrak{A}]_{p}\subseteq H^{p}$ and
$\mathfrak{M}H^{t}=[\mathfrak{M}h_{0}^{\frac{1}{t}}]_{2}=[h\mathfrak{A}h_{0}^{\frac{1}{t}}]_{2}=[\xi\mathfrak{A}]_{2}=H^{2}$.
By Lemma 2.2, $\mathfrak{M}=H^{p}$ and $h$ is right outer. Note that $\eta
h_{0}^{\frac{1}{t}}=CU^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}h_{0}^{\frac{1}{t}}=Chh_{0}^{\frac{1}{t}}$.
We then have $\eta=Ch$. Note that
$\|h\|_{p}=\|U^{*}(\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{1}{2}}\|_{p}=\left(\mbox{tr}((\eta^{*}\eta+\varepsilon^{2}h_{0}^{\frac{2}{p}})^{\frac{p}{2}})\right)^{\frac{1}{p}}\to\|\eta\|_{p}(\varepsilon\to
0)$. We may choose $\varepsilon>0$ so that $\|h\|_{p}<\|\eta\|_{p}+\delta$.

If $1\leq p<2$, then $\frac{1}{p}=\frac{1}{2}+\frac{1}{s}$ for some $s\geq 2$.
Let $\eta=V|\eta|=V|\eta|^{\frac{p}{2}}|\eta|^{\frac{p}{s}}=\eta_{2}\eta_{s}$,
where $\eta_{2}=V|\eta|^{\frac{p}{2}}\in L^{2}(\mathcal{M})$,
$\eta_{s}=|\eta|^{\frac{p}{s}}\in L^{s}(\mathcal{M})$ with
$\|\eta_{2}\|_{2}=\|\eta\|_{p}^{\frac{p}{2}}$ and
$\|\eta_{s}\|_{s}=\|\eta\|_{p}^{\frac{p}{s}}$. We choose two positive numbers
$\delta_{2},\delta_{s}$ such that
$(\|\eta_{2}\|_{2}+\delta_{2})(\|\eta_{s}\|_{s}+\delta_{s})<\|\eta_{2}\|_{2}\|\eta_{s}\|_{s}+\delta=\|\eta\|_{p}+\delta$.
By [15, Theorem 3.5] and the proof as above, $\eta_{s}=Ch_{s}$ for a right
outer element $h_{s}\in H^{s}$ with
$\|h_{s}\|_{s}<\|\eta_{s}\|_{s}+\delta_{s}$ and a contraction
$C\in\mathcal{M}$. Now $\eta_{2}C\in L^{2}(\mathcal{M})$. We again have
$\eta_{2}C=Bh_{2}$ for a right outer element $h_{2}\in H^{2}$ with
$\|h_{2}\|_{2}<\|\eta_{2}\|_{2}+\delta_{2}$ and a contraction
$B\in\mathcal{M}$. It is elementary that $h=h_{2}h_{s}\in H^{p}$ is right
outer such that
$\|h\|_{p}\leq\|h_{2}\|_{2}\|h_{s}\|_{s}<\|\eta_{2}\|_{2}\|\eta_{s}\|_{s}+\delta=\|\eta\|_{p}+\delta$.

If $\eta\in H^{p}$, then $BhA=\eta A\in H^{p}$ for all $A\in\mathfrak{A}$.
Thus $BH^{p}\subseteq H^{p}$ and $B\in\mathfrak{A}$ from [13, Theorem 2.7]. ∎

We now have the following Riesz type factorization theorem in noncommutative
$H^{p}$ spaces for $1\leq p<\infty$ whose proof is similar to [15, Theorem
3.6]. To completeness, we give the detail.

######  Theorem 2.4.

Let $1\leq r,p,q<\infty$ such that $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$ and
let $\mathfrak{A}$ be a type 1 subdiagonal algebra. Then for any $h\in H^{r}$
and $\varepsilon>0$, there exist $h_{p}\in H^{p}$, $h_{q}\in H^{q}$ with
$\|h_{t}\|_{t}<\|h\|_{r}^{\frac{r}{t}}+\varepsilon$ for $t=p,q$ such that
$h=h_{p}h_{q}$. If $h\in H_{0}^{r}$, we may choose one of $h_{t}$ in
$H_{0}^{t}$ for $t=p,q$.

###### Proof.

Take $1<s\leq\infty$ such that $\frac{1}{r}+\frac{1}{s}=1$. Let
$h=V|h|=V|h|^{\frac{r}{p}}|h|^{\frac{r}{q}}$ be the polar decomposition. Then
$|h|^{\frac{r}{t}}\in L^{t}(\mathcal{M})$ with
$\||h|^{\frac{r}{t}}\|_{t}=\|h\|_{r}^{\frac{r}{t}}$ for $t=p,q$. By Theorem
2.3, there exist a contraction $B\in\mathcal{M}$ and a right outer element
$h_{q}\in H^{q}$ with
$\|h_{q}\|_{q}<\||h|^{\frac{r}{q}}\|_{q}+\varepsilon=\|h\|_{r}^{\frac{r}{q}}+\varepsilon$
such that $|h|^{\frac{r}{q}}=Bh_{q}$. Put $h_{p}=V|h|^{\frac{r}{p}}B$. We show
that $h_{p}\in H^{p}$. In fact, $tr(hg)=0$ for all $g\in H^{s}_{0}$ since
$h\in H^{r}$. Then $tr(hg)=tr(h_{p}h_{q}g)=0$ for all $g\in H^{s}_{0}$. Note
that $h_{q}$ is right outer. Then
$[h_{q}H^{s}_{0}]_{p^{\prime}}=[h_{q}\mathfrak{A}H^{s}_{0}]_{p^{\prime}}=H^{q}H^{s}_{0}=H^{p^{\prime}}_{0}$,
where $p^{\prime}>0$ such that $\frac{1}{p}+\frac{1}{p^{\prime}}=1$. Therefore
$h_{p}\in H^{p}$ with
$\|h_{p}\|_{p}\leq\||h|^{\frac{r}{p}}\|_{p}<\|h\|_{r}^{\frac{r}{p}}+\varepsilon$.
Moreover, if $h\in H_{0}^{p}$, then $tr(hg)=0$ for all $g\in H^{s}$. It easily
follows that $h_{p}\in H_{0}^{p}$ in this case.

On the other hand, by taking the pair $(h^{*},(H^{r})^{*})$ if
$h\in(H_{0}^{r})$, we may write $h^{*}$ in the form $h^{*}=g_{q}g_{p}$ with
$g_{q}\in(H_{0}^{q})^{*}$ and $g_{p}\in(H^{p})^{*}$. Then we have
$h=h_{p}h_{q}$ with $h_{p}\in H^{p}$ and $h_{q}\in H_{0}^{q}$ by setting
$h_{q}=g_{q}^{*}$ and $h_{p}=g_{p}^{*}$. ∎

##  3\. Beurling type invariant subspace theorem for type 1 subdiagonal
algebras

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra. For any positive integer
$n\geq 1$, let $\mathfrak{A}_{0}^{n}$ be the $\sigma$-weakly closed ideal of
$\mathfrak{A}$ generated by $\\{a_{1}a_{2}\cdots
a_{n}:a_{j}\in\mathfrak{A}_{0}\\}$. It is elementary that a right invariant
subspace $\mathfrak{M}\subseteq L^{2}(\mathcal{M})$ is of type 1(resp. type 2)
if and only if $\cap_{n\geq
1}[\mathfrak{M}\mathfrak{A}_{0}^{n}]_{2}=\\{0\\}$(resp.
$\mathfrak{M}=[\mathfrak{M}\mathfrak{A}_{0}]_{2}$). Based on this fact, we
introduce similar notions in $L^{p}$ spaces.

######  Definition 3.1.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and $\mathfrak{M}$ a
right(resp. left) invariant subspace in $L^{p}(\mathcal{M})$. Then we say that
$\mathfrak{M}$ is of type 1 if $\cap_{n\geq
1}[\mathfrak{M}\mathfrak{A}_{0}^{n}]_{p}=\\{0\\}$(resp. $\cap_{n\geq
1}[\mathfrak{A}_{0}^{n}\mathfrak{M}]_{p}=\\{0\\}$) and of type 2 if
$\mathfrak{M}=[\mathfrak{M}\mathfrak{A}_{0}]_{p}$(resp.
$\mathfrak{M}=[\mathfrak{A}_{0}\mathfrak{M}]_{p}$).

We recall that there exists a column orthogonal family of partial isometries
$\\{U_{n}:n\geq 1\\}$ in $\mathcal{M}$ such that

(3.1) $H_{0}^{2}=\oplus_{n\geq 1}^{col}U_{n}H^{2}$

when $\mathfrak{A}$ is of type 1.

We note that a Beurling type invariant subspace theorem is given in [16] when
$p=1$ or $p=\infty$. We now consider this theorem in noncommutative
$L^{p}(\mathcal{M})$ for $1<p<\infty$.

######  Lemma 3.2.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra. If $\mathfrak{M}\subseteq
L^{p}(\mathcal{M})$ ($1<p<\infty$) is a right invariant subspace of type 2,
then there exists a projection $E\in\mathcal{M}$ such that
$\mathfrak{M}=EL^{p}(\mathcal{M})$.

###### Proof.

By [15, Theorem 2.7],
$\mathfrak{A}_{0}=\vee\\{\mathfrak{D}U_{i_{1}}U_{i_{2}}\cdots
U_{i_{n}}:i_{j}\geq 1\\}$, where $\\{U_{i}:i\geq 1\\}$ is a family of column
orthogonal partial isometries as in $(3.1)$. Note that $\mathfrak{M}$ is
$\mathfrak{D}$ right invariant and
$[\mathfrak{M}\mathfrak{A}_{0}]_{p}=\mathfrak{M}$. Since
$\vee\\{\mathfrak{M}U_{n}:n\geq
1\\}\supseteq\vee\\{\mathfrak{M}U_{i_{1}}U_{i_{2}}\cdots U_{i_{n}}:i_{k}\geq
1,n\geq 1\\}$, $\mathfrak{M}=\vee\\{\mathfrak{M}U_{n}:n\geq 1\\}$. For any
$m,n\geq 1$ and $x\in\mathfrak{M}$, we have
$R_{U_{m}^{*}}R_{U_{n}}x=xU_{n}U_{m}^{*}\in\mathfrak{M}$ since
$U_{n}U_{m}^{*}\in\mathfrak{D}$ from [15, Proposition 2.3]. Then
$\mathfrak{M}$ is right $\mathcal{M}$ invariant. That is, $\mathfrak{M}$ is an
$\mathcal{M}$ right module. Then by [19, Theorem 3.6], there is a projection
$E\in\mathcal{M}$ such that $\mathfrak{M}=EL^{p}(\mathcal{M})$. In fact, we
also have that the right submodule $\mathfrak{M}$ is column summand in
$L^{p}(\mathcal{M})$ from the proof of this theorem(cf.[19, $p_{19}$]). It
follows that the projection onto $\mathfrak{M}$ commutes with the right
multiplications by $\mathcal{M}$. Thus there is a projection $E\in\mathcal{M}$
such that $\mathfrak{M}=EL^{p}(\mathcal{M})$ by [19, Corollary 2.6]. ∎

For a family of column orthogonal partial isometries
$\mathcal{W}=\\{W_{n}:n\geq 1\\}$, we define a right invariant subspace
$\mathfrak{M}_{\mathcal{W}}^{p}=\oplus_{n\geq 1}^{col}W_{n}H^{p}$ in
$L^{p}(\mathcal{M})$ for any $1\leq p\leq\infty$ as in [16]. If $1\leq
r<\infty$ and $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$ for some $p,q\geq 1$, then
$\mathfrak{M}_{\mathcal{W}}^{r}=[(\oplus_{n\geq
1}W_{n}H^{p})H^{q}]_{r}=\mathfrak{M}_{\mathcal{W}}^{p}H^{q}$. The following
theorem is motivated from [16, Theorem 2.2].

######  Theorem 3.3.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and let
$\mathfrak{M}\subseteq L^{p}(\mathcal{M})$$(1<p<\infty)$ be a nonzero closed
right invariant subspace of $\mathfrak{A}$.

$(i)$ There are right invariant subspaces $\mathfrak{M}_{i}$ of type $i$
$(i=1,2)$ such that
$\mathfrak{M}=\mathfrak{M}_{1}\oplus^{col}\mathfrak{M}_{2}$.

$(ii)$ If $\mathfrak{M}\subseteq L^{p}(\mathcal{M})$ is a right invariant
subspace of type 1, then there exists a family of column orthogonal partial
isometries $\mathcal{W}$ such that
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{p}$.

###### Proof.

$(i)$ Put $\mathfrak{M}_{2}=\cap_{n\geq
1}[\mathfrak{M}\mathfrak{A}_{0}^{n}]_{p}$. Then
$\mathfrak{M}_{2}\subseteq\mathfrak{M}$ is a right invariant subspace of type
2. By Lemma 3.1, there is a projection $E\in\mathcal{M}$, such that
$\mathfrak{M}_{2}=EL^{p}(\mathcal{M})\subseteq\mathfrak{M}$. Put
$\mathfrak{M}_{1}=(I-E)\mathfrak{M}\subseteq\mathfrak{M}$. This mean that
$\mathfrak{M}_{1}$ is closed and right invariant such that
$\mathfrak{M}=\mathfrak{M}_{1}\oplus^{col}\mathfrak{M}_{2}$. Note that
$\cap_{n\geq 1}[\mathfrak{M}_{1}\mathfrak{A}_{0}^{n}]_{p}\subseteq\cap_{n\geq
1}[\mathfrak{M}\mathfrak{A}_{0}^{n}]_{p}\subseteq\mathfrak{M}_{1}\cap\mathfrak{M}_{2}=\\{0\\}$.
Then $\mathfrak{M}_{1}$ is of type 1.

$(ii)$ Firstly, we assume that $1<p<2$ and take $r>2$ with
$\frac{1}{p}=\frac{1}{2}+\frac{1}{r}$. Put

$\displaystyle\mathcal{P}=$ $\displaystyle\\{\mathcal{W}=\\{W_{n}\\}_{n\geq
1}\subseteq\mathcal{M}:\mbox{column orthogonal }$ $\displaystyle\mbox{ partial
isometries such that
}\mathfrak{M}_{\mathcal{W}}^{2}H^{r}\subseteq\mathfrak{M}\\}.$

Note that $\mathcal{P}$ is non-empty. In fact, we show that for any nonzero
$h\in\mathfrak{M}$, $[h\mathfrak{A}]_{p}=(VH^{2})H^{r}$ for a partial isometry
$V\in\mathcal{M}$ with $V^{*}V\in\mathfrak{D}$. Write
$h=(u|h|^{\frac{p}{2}})|h|^{\frac{p}{r}}$. Then $|h|^{\frac{p}{r}}\in
L^{r}(\mathcal{M})$ by Theorem 2.3, there exist a contraction
$B\in\mathcal{M}$ and a right outer element $h_{r}\in H^{r}$ such that
$|h|^{\frac{p}{r}}=Bh_{r}$. Put $h_{2}=u|h|^{\frac{p}{2}}B$. We have
$h=h_{2}h_{r}$. This implies that

$\mathfrak{M}\supseteq[h\mathfrak{A}]_{p}=[h_{2}h_{r}\mathfrak{A}]_{p}=[h_{2}H^{r}]_{p}=[h_{2}\mathfrak{A}h_{0}^{\frac{1}{r}}\mathfrak{A}]_{p}=[h_{2}\mathfrak{A}]_{2}H^{r}$

since $[h_{r}\mathfrak{A}]_{r}=H^{r}$. Note that
$[h_{2}\mathfrak{A}]_{2}\subseteq L^{2}(\mathcal{M})$ is a right invariant
subspace. By [15, Lemma 3.3],
$[h_{2}\mathfrak{A}]_{2}=VH^{2}\oplus^{col}N_{2}$ for a partial isometry
$V\in\mathcal{M}$ such that $V^{*}V\in\mathfrak{D}$ and a right invariant
subspace $N_{2}$ of type 2. Note that

$[N_{2}H^{r}]_{p}=[[N_{2}\mathfrak{A}_{0}]_{2}H^{r}]_{p}=[[N_{2}H^{r}]_{p}\mathfrak{A}_{0}]_{p}.$

It follows that $[N_{2}H^{r}]_{p}=\cap_{n\geq
1}[(N_{2}H^{r})\mathfrak{A}_{0}^{n}]_{p}\subseteq\mathfrak{M}=\\{0\\}$. Thus
$N_{2}=\\{0\\}$ and $[h_{2}\mathfrak{A}]_{2}=VH^{2}$ with
$V^{*}V\in\mathfrak{D}$.

Note that $h_{r}$ is right outer. This means that
$[h\mathfrak{A}]_{p}=[h_{2}h_{r}\mathfrak{A}]_{p}=[h_{2}H^{r}]_{p}=[h_{2}\mathfrak{A}H^{r}]_{p}=[h_{2}\mathfrak{A}]_{2}H^{r}=(VH^{2})H^{r}$
and $\\{V\\}\in\mathcal{P}$.

We define a partial order in $\mathcal{P}$ by $\mathcal{W}\leq\mathcal{V}$ if
$\mathfrak{M}_{\mathcal{W}}^{2}\subseteq\mathfrak{M}_{\mathcal{V}}^{2}$ for
any $\mathcal{W},\mathcal{V}\in\mathcal{P}$. Let
$\\{\mathcal{W}_{\lambda}:\lambda\in\Lambda\\}\subseteq\mathcal{P}$ be a
totally ordered family in $\mathcal{P}$. Put
$\mathfrak{N}=\vee\\{\mathfrak{M}_{\mathcal{W}_{\lambda}}^{2}:\lambda\in\Lambda\\}$.
Note that $\mathfrak{N}\subseteq L^{2}(\mathcal{M})$ is a right invariant
subspace in $L^{2}(\mathcal{M})$. Then
$\mathfrak{N}=\mathfrak{M}_{\mathcal{W}}^{2}\oplus^{c}\mathfrak{N}_{2}$ for a
family of column orthogonal partial isometries $\mathcal{W}$ and a right
invariant subspace $\mathfrak{N}_{2}$ of type 2 in $L^{2}(\mathcal{M})$. Then
$[\mathfrak{N}_{2}H^{r}]_{p}$ is also a type 2 right invariant subspace in
$L^{p}(\mathcal{M})$. This implies that
$[\mathfrak{N}_{2}H^{r}]_{p}=\cap_{n\geq
1}[(\mathfrak{N}_{2}H^{r})\mathfrak{A}_{0}^{n}]_{p}\subseteq\cap_{n\geq
1}[\mathfrak{M}\mathfrak{A}_{0}^{n}]_{p}=\\{0\\}$. Hence
$\mathfrak{N}_{2}=\\{0\\}$. Since
$\mathfrak{M}_{\mathcal{W}_{\lambda}}^{2}H^{r}\subseteq\mathfrak{M}$,
$\mathfrak{N}H^{r}\subseteq\mathfrak{M}$. Thus $\mathcal{W}\in\mathcal{P}$
with $\mathcal{W}_{\lambda}\leq\mathcal{W}$ for any $\lambda\in\Lambda$. That
is, $\mathcal{W}$ is an upper bound of
$\\{\mathcal{W}_{\lambda}:\lambda\in\Lambda\\}$. By Zorn’s lemma, there exists
a maximal element $\mathcal{W}\in\mathcal{P}$. We show that
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{p}$. Otherwise, assume that there is
an $h\in\mathfrak{M}$ such that $h\notin\mathfrak{M}_{\mathcal{W}}^{p}$. Then
$[h\mathfrak{A}]_{p}=(VH^{2})H^{r}$ for a partial isometry $V\in\mathcal{M}$
with $V^{*}V\in\mathfrak{D}$ as above. Since
$h\notin\mathfrak{M}_{\mathcal{W}}^{p}$,
$VH^{2}\nsubseteq\mathfrak{M}_{\mathcal{W}}^{2}$. Now
$\tilde{\mathfrak{N}}=\vee\\{\mathfrak{M}_{\mathcal{W}}^{2},VH^{2}\\}$ is also
a right invariant subspace in $L^{2}(\mathcal{M})$ with
$\tilde{\mathfrak{N}}H^{r}\subseteq\mathfrak{M}$. In this case, we have
$\tilde{\mathfrak{N}}=\mathfrak{M}_{\mathcal{V}}^{2}$ for a family of column
orthogonal partial isometries $\mathcal{V}\in\mathcal{P}$. It is trivial that
$\mathfrak{M}_{\mathcal{W}}^{2}\subsetneqq\mathfrak{M}_{\mathcal{V}}^{2}$ by a
similar treatment. This is a contradiction. Thus
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{2}H^{r}=\mathfrak{M}_{\mathcal{W}}^{p}$.

We have that for a general right invariant subspace $\mathfrak{M}\subseteq
L^{p}(\mathcal{M})$, there is a right invariant subspace
$\mathfrak{N}\subseteq L^{2}(\mathcal{M})$ such that
$\mathfrak{M}=\mathfrak{N}H^{r}$. In fact, by $(i)$,
$\mathfrak{M}=\mathfrak{M}_{1}\oplus^{col}\mathfrak{M}_{2}=\mathfrak{M}_{\mathcal{W}}^{2}H^{r}\oplus^{col}EL^{2}(\mathcal{M})H^{r}=(\mathfrak{M}_{\mathcal{W}}^{2}\oplus^{col}EL^{2}(\mathcal{M}))H^{r}=\mathfrak{N}H^{r}$,
where
$\mathfrak{N}=\mathfrak{M}_{\mathcal{W}}^{2}\oplus^{col}EL^{2}(\mathcal{M})$.

Let $2<p<\infty$ and let $p$ and $q$ be conjugate exponents, that is,
$\frac{1}{p}+\frac{1}{q}=1$. Then $1<q<2$ and
$\frac{1}{2}=\frac{1}{p}+\frac{1}{r}$ as well as
$\frac{1}{q}=\frac{1}{2}+\frac{1}{r}$ for some $1<r<\infty$. Note that
$\mathfrak{M}H^{r}\subseteq L^{2}(\mathcal{M})$ is right invariant of type 1.
Thus there is a family of column orthogonal partial isometries
$\mathcal{W}=\\{W_{n}:n\geq 1\\}$ such that
$\mathfrak{M}H^{r}=\mathfrak{M}_{\mathcal{W}}^{2}=\mathfrak{M}_{\mathcal{W}}^{p}H^{r}$.

We next show that $\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{p}$. For any
$x\in\mathfrak{M}$, it is elementary that
$xh_{0}^{\frac{1}{r}}=\oplus^{col}_{n\geq
1}W_{n}W_{n}^{*}xh_{0}^{\frac{1}{r}}$ with $W_{n}^{*}xh_{0}^{\frac{1}{r}}\in
H^{2}$ since $W_{n}^{*}xh_{0}^{\frac{1}{r}}\in H^{p}$. Then $W_{n}^{*}x\in
H^{p}$ and $x=\oplus^{col}_{n\geq
1}W_{n}W_{n}^{*}x\in\mathfrak{M}_{\mathcal{W}}^{p}$. Therefore
$\mathfrak{M}\subseteq\mathfrak{M}_{\mathcal{W}}^{p}$.

For any closed subspace $K\subseteq L^{p}(\mathcal{M})$, we put
$K^{\bot}=\\{y\in L^{q}(\mathcal{M}):tr(xy)=0,\forall x\in K\\}$. It is known
that $K^{\bot}$ is left invariant when $K$ is right invariant. By symmetry, we
easily have $\mathfrak{M}^{\bot}=H^{r}\mathfrak{N}=\oplus_{n\geq
1}^{row}H^{q}V_{n}\oplus^{row}L^{q}(\mathcal{M})F$ for a left invariant
subspace $\mathfrak{N}=\oplus_{n\geq
1}^{row}H^{2}V_{n}\oplus^{row}L^{2}(\mathcal{M})F\subseteq L^{2}(\mathcal{M})$
since $1<q<2$ by preceding proof.

We claim that $(\mathfrak{M}H^{r})^{\bot}=\mathfrak{N}$. It is clear that
$(\mathfrak{M}H^{r})^{\bot}\supseteq\mathfrak{N}$. Take any
$y\in(\mathfrak{M}H^{r})^{\bot}$. Then $\mbox{tr}(xhy)=0$ for all
$x\in\mathfrak{M}$ and $h\in H^{r}$. It follows that
$H^{r}y\subseteq\mathfrak{M}^{\bot}=\oplus_{n\geq
1}^{row}H^{q}V_{n}\oplus^{row}L^{q}(\mathcal{M})F$. By a similar treatment as
above, we have $h_{0}^{\frac{1}{r}}y=\oplus_{n\geq
1}^{row}h_{0}^{\frac{1}{r}}yV_{n}^{*}V_{n}\oplus^{row}h_{0}^{\frac{1}{r}}yF$
and thus $y=\oplus_{n\geq
1}^{row}yV_{n}^{*}V_{n}\oplus^{row}yF\in\mathfrak{N}$.

Thus
$\mathfrak{M}^{\bot}=H^{r}\mathfrak{N}=H^{r}(\mathfrak{M}H^{r})^{\bot}=H^{r}(\mathfrak{M}_{\mathcal{W}}^{p}H^{r})^{\bot}=(\mathfrak{M}_{\mathcal{W}}^{p})^{\bot}$
and $\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{p}$. ∎

Let $\mathfrak{M}\subseteq L^{p}(\mathcal{M})$ be a right invariant subspace.
Then
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{p}\oplus^{col}EL^{p}(\mathcal{M})$
for a family of column orthogonal partial isometries $\mathcal{W}$ and a
projection $E$ in $\mathcal{M}$. It follows that
$\mathfrak{M}_{\infty}=\mathfrak{M}_{\mathcal{W}}^{\infty}\oplus E\mathcal{M}$
is a $\sigma$-weakly closed right invariant subspace in $\mathcal{M}$.

######  Corollary 3.4.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra. Then there is a lattice
isomorphism from the lattice of all right invariant subspaces of
$\mathfrak{A}$ in $\mathcal{M}$ onto that in $L^{p}(\mathcal{M})$ for $1\leq
p<\infty$.

###### Proof.

Let $\mathfrak{M}\subseteq\mathcal{M}$ be a $\sigma$-weakly closed right
invariant subspace. Then
$\mathfrak{M}_{p}=[\mathfrak{M}h_{0}^{\frac{1}{p}}]_{p}=\mathfrak{M}H^{p}\subseteq
L^{p}(\mathcal{M})$ is a right invariant subspace. We show this correspondence
is a bijection. Assume that $\mathfrak{M}H^{p}=\mathfrak{N}H^{p}$ for a
$\sigma$-weakly closed right invariant subspace
$\mathfrak{N}\subseteq\mathcal{M}$. Then we have two column orthogonal
families of partial isometries $\mathcal{W}$ and $\mathcal{V}$ and two
projections $E,F\in\mathcal{M}$ such that
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{\infty}\oplus^{col}E\mathcal{M}$ and
$\mathfrak{N}=\mathfrak{M}_{\mathcal{V}}^{\infty}\oplus^{col}F\mathcal{M}$ by
[16, Theorem 2.2]. For any $A\in\mathfrak{M}$,
$Ah_{0}^{\frac{1}{p}}\in\mathfrak{M}H^{p}=\mathfrak{N}H^{p}$. Then
$Ah_{0}^{\frac{1}{p}}=\oplus_{n\geq 1}V_{n}V_{n}^{*}Ah_{0}^{\frac{1}{p}}\oplus
FAh_{0}^{\frac{1}{p}}$ with $V^{*}_{n}A\in\mathfrak{A}$ and
$FAh_{0}^{\frac{1}{p}}\in FL^{p}(\mathcal{M})$. It follows that $A=\oplus
V_{n}V_{n}^{*}A+FA\in\mathfrak{N}$. That is,
$\mathfrak{M}\subseteq\mathfrak{N}$. By symmetry, we have
$\mathfrak{M}=\mathfrak{N}$. On the other hand, for any right invariant
subspace $\mathfrak{M}_{p}\subseteq L^{p}(\mathcal{M})$, there is a family of
column orthogonal partial isometries $\mathcal{W}$ and a projection
$E\in\mathcal{M}$ such that
$\mathfrak{M}_{p}=\mathfrak{M}_{\mathcal{W}}^{p}\oplus EL^{p}(\mathcal{M})$
from Theorem 2.3. Put
$\mathfrak{M}=\mathfrak{M}_{\mathcal{W}}^{\infty}\oplus^{col}E\mathcal{M}$.
Then $\mathfrak{M}_{p}=\mathfrak{M}H^{p}$. Thus the map:
$\mathfrak{M}\to\mathfrak{M}H^{p}$ is a bijection from the lattice of all
$\sigma-$weakly closed right invariant subspaces in $\mathcal{M}$ onto that of
all right closed invariant subspaces in $L^{p}(\mathcal{M})$. ∎

##  4\. Subalgebras containing a type 1 subdiagonal algebra

We gave a necessary and sufficient condition for a type 1 subdiagonal algebra
$\mathfrak{A}$ to be a maximal subalgebra of $\mathcal{M}$ in [16]. In fact,
we may determine all $\sigma-$weakly closed subalgebras containing
$\mathfrak{A}$ in $\mathcal{M}$. Let $\mathcal{B}$ be a $\sigma-$weakly closed
subalgebra of $\mathcal{M}$ containing $\mathfrak{A}$. Then
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$ is a right invariant subspace in
$L^{2}(\mathcal{M})$. As in [16], let $p$ and $q$ be the projections from
$L^{2}(\mathcal{M})$ onto $L^{2}(\mathfrak{D})$ and the wandering subspace
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}\ominus[\mathfrak{B}h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}]_{2}$
of $[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$ respectively. It is known that both
$p$ and $q$ are in the commutant $(R(\mathfrak{D}))^{\prime}$ of
$R(\mathfrak{D})$. For any two projections $E$ and $F$ in a von Neumann
algebra $\mathcal{M}$, $E\preccurlyeq F$ means that there is a partial
isometry $V\in\mathcal{M}$ such that $V^{*}V=E$ and $VV^{*}\leq F$. If
$E\preccurlyeq F$ and $F\preccurlyeq E$, then $E\sim F$. For an operator $A$,
$R(A)$ denotes the range of $A$.

######  Lemma 4.1.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and $\mathfrak{B}$ as
above. Then $q\leq p$.

###### Proof.

Note that $\mathfrak{B}$ is two-side $\mathfrak{A}$ invariant subspace. Then
there exists a family of column orthogonal partial isometries $\mathcal{W}$
and a projection $E$ in $\mathcal{M}$ such that $\mathfrak{B}=\oplus_{n\geq
1}^{col}W_{n}\mathfrak{A}\oplus^{col}E\mathcal{M}$ as well as
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}=\oplus_{n\geq
1}^{col}W_{n}H^{2}\oplus^{col}EL^{2}(\mathcal{M})$ by Corollary 3.1. This
means that $EL^{2}(\mathcal{M})=\cap_{n\geq
1}[\mathfrak{B}h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}^{n}]_{2}$ . Since
$[\mathfrak{B}h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}^{n}]_{2}$ is left
$\mathfrak{B}$ invariant for any $n$, $EL^{2}(\mathcal{M})$ is also left
$\mathfrak{B}$ invariant. This implies that $EBE=EB$ for any
$B\in\mathfrak{B}$. In particular, $ED=DE$ for all
$D\in\mathfrak{B}\cap\mathfrak{B}^{*}$. We now have
$W_{n}\in\mathfrak{B}\cap\mathfrak{B}^{*}$ for any $n\geq 1$. In fact, it is
trivial that $W_{n}\in\mathfrak{B}$ for any $n\geq 1$. On the other hand,
$W_{n}^{*}\mathfrak{B}=W_{n}^{*}W_{n}\mathfrak{A}\subseteq\mathfrak{A}\subseteq\mathfrak{B}$
since $W_{n}^{*}W_{n}\in\mathfrak{D}$. Then $W_{n}^{*}\in\mathfrak{B}$ for any
$n\geq 1$. Since $EW_{n}=0$ for any $n$, $EW_{n}=W_{n}E=0$ and $\oplus_{n\geq
1}W_{n}W_{n}^{*}=I-E$. Put $\mathfrak{N}=\oplus_{n\geq
1}^{col}W_{n}H^{2}\subseteq[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$. Then
$\mathfrak{N}$ is the type 1 part of the right invariant subspace
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$.

Take any subprojections $p_{1}\leq p$ and $q_{1}\leq q$ such that $p_{1}\sim
q_{1}$ in $(R(\mathfrak{D}))^{\prime}$. We begin by proving that $p_{1}\leq
q$. Put $\mathfrak{M}_{1}=[R(p_{1})\mathfrak{A}]$ and
$\mathfrak{N}_{1}=[R(q_{1})\mathfrak{A}]$. Let
$w_{1}\in(R(\mathfrak{D}))^{\prime}$ be a partial isometry such that
$w_{1}^{*}w_{1}=q_{1}$ and $w_{1}w_{1}^{*}=p_{1}\leq p$ as in [16, Proposition
2.5]. Then $\mathfrak{N}_{1}=W^{*}H^{2}$ for a partial isometry
$W\in\mathcal{M}$ such that
$H^{2}=W\mathfrak{N}_{1}\oplus^{col}(I-WW^{*})H^{2}$ by [16, Proposition 2.5]
again. It follows that $\mathfrak{N}_{1}=W^{*}H^{2}=W^{*}W\mathfrak{N}_{1}$.
Note that
$\mathfrak{N}=[\left([\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}\ominus[\mathfrak{B}h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}]_{2}\right)\mathfrak{A}]_{2}=[R(q)\mathfrak{A}]_{2}$.
We have
$\mathfrak{N}=[R(q_{1})\mathfrak{A}]_{2}\oplus^{col}[(R(q-q_{1}))\mathfrak{A}]_{2}=\mathfrak{N}_{1}\oplus^{col}[(R(q-q_{1}))\mathfrak{A}]_{2}$.
We similarly obtain that
$H^{2}=[R(p_{1})\mathfrak{A}]_{2}\oplus^{col}[(R(p-p_{1}))\mathfrak{A}]_{2}=\mathfrak{M}_{1}\oplus^{col}[(R(p-p_{1}))\mathfrak{A}]_{2}$.
It follows that $W^{*}H^{2}=W^{*}\mathfrak{M}_{1}=\mathfrak{N}_{1}$. Then we
have $\mathfrak{M}_{1}=W\mathfrak{N}_{1}$ since
$W^{*}\mathfrak{M}_{1}=W^{*}H^{2}=W^{*}W\mathfrak{N}_{1}$. We also get
$WW^{*}H^{2}=WW^{*}\mathfrak{M}_{1}=\mathfrak{M}_{1}\subseteq H^{2}$ and so
that $WW^{*}\in\mathfrak{D}$. Note that
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}=[R(q)\mathfrak{A}]_{2}\oplus^{col}EL^{2}(\mathcal{M})$.
We have $[R(q)\mathcal{M}]_{2}=(I-E)L^{2}(\mathcal{M})$ since $\oplus_{n\geq
1}^{col}W_{n}W_{n}^{*}=I-E$. It is elementary that $W$ is an isometry on
$[R(q_{1})\mathcal{M}]_{2}$ while $Wx=0$ for any
$x\in[R(q)\mathcal{M}]_{2}^{\bot}$. It follows that $W(EL^{2}(\mathcal{M}))=0$
and we in turn have $WE=0$. Thus
$W[\mathfrak{B}h_{0}^{\frac{1}{1}}]_{2}=W[R(q)\mathfrak{A}]_{2}=W[R(q_{1})\mathfrak{A}]_{2}=W\mathfrak{N}_{1}=\mathfrak{M}_{1}\subseteq
H^{2}\subseteq[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$. We now have
$W\in\mathfrak{B}$ by [16, Proposition 2.3].

On the other hand,
$W^{*}H^{2}=W^{*}\mathfrak{M}_{1}=\mathfrak{N}_{1}\subseteq\mathfrak{N}=\oplus_{n\geq
1}^{col}W_{n}H^{2}$. Then $W^{*}h_{0}^{\frac{1}{2}}=\oplus_{n\geq
1}^{col}W_{n}\xi_{n}$ for some $\xi_{n}\in H^{2}$. It follows that
$W_{n}^{*}W^{*}h_{0}^{\frac{1}{2}}=W_{n}^{*}W_{n}\xi_{n}\in H^{2}$, which
implies that $W_{n}^{*}W^{*}\in\mathfrak{A}\subseteq\mathfrak{B}$. Moreover,
$W_{n}W_{n}^{*}W^{*}h_{0}^{\frac{1}{2}}=W_{n}\xi_{n}$. So we get
$W^{*}h_{0}^{\frac{1}{2}}=\oplus_{n\geq 1}^{col}W_{n}\xi_{n}=\oplus_{n\geq
1}^{col}W_{n}W_{n}^{*}W^{*}h_{0}^{\frac{1}{2}}$. Then $W^{*}=\oplus_{n\geq
1}^{col}W_{n}W_{n}^{*}W^{*}$ since $h_{0}^{\frac{1}{2}}$ is separating. We now
have $W^{*}\in\mathfrak{B}$. It follows that
$\mathfrak{M}_{1}=WW^{*}\mathfrak{M}_{1}=WW^{*}H^{2}\subseteq
WW^{*}[\mathfrak{B}h_{0}^{\frac{1}{2}}]\subseteq
W[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}=W\mathfrak{N}_{1}=\mathfrak{M}_{1}$.
Thus
$WW^{*}H^{2}=WW^{*}[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}=\mathfrak{M}_{1}$
with the wandering subspace $R(p_{1})$. Take any $\xi\in R(p_{1})$. We have

$\langle\xi,Bh_{0}^{\frac{1}{2}}A\rangle=\langle
WW^{*}\xi,Bh_{0}^{\frac{1}{2}}A\rangle=\langle\xi,WW^{*}Bh_{0}^{\frac{1}{2}}A\rangle=0$

for any $B\in\mathfrak{B}$ and $A\in\mathfrak{A}_{0}$. Moreover, we have
$\xi\in R(p_{1})\subseteq R(p)=L^{2}(\mathfrak{D})\subseteq
H^{2}\subseteq[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$. It follows that
$R(p_{1})\subseteq[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}\ominus[\mathfrak{B}h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}]_{2}=R(q)$.
Therefore $p_{1}\leq q$.

We take a maximal family of mutually orthogonal subprojections $\\{p_{n}:n\geq
1\\}$ of $p$ such that $p_{n}\sim q_{n}$ for some subprojections $q_{n}\leq q$
for all $n\geq 1$ in $(R(\mathfrak{D}))^{\prime}$ by Zorn’s lemma. Then we
must have $p_{n}\leq q$ for all $n\geq 1$ as above. Put
$p_{0}=p-(\oplus_{n\geq 1}p_{n})$ and $q_{0}=q-(\oplus_{n\geq 1}p_{n})$, then
$p_{0}$ and $q_{0}$ are in $(\mathcal{R}(\mathfrak{D}))^{\prime}$ and either
$p_{0}$ or $q_{0}$ is $0$ by [20, Theorem 6.2.7].

If $q_{0}=0$, then we have $q\leq p$. If $p_{0}=0$, then we obtain that $p\leq
q$. Next we show that $p=q$. Note that
$\mathfrak{N}=[R(q)\mathfrak{A}]_{2}=[R(p)\mathfrak{A}]_{2}\oplus^{col}[(R(q-p))\mathfrak{A}]_{2}=H^{2}\oplus^{col}[(R(q-p))\mathfrak{A}]_{2}$.
It trivial that $L^{2}(\mathcal{M})=[R(p)\mathcal{M}]_{2}$ is orthogonal with
$[(R(q-p))\mathcal{M}]_{2}$. Then we obtain that $[(R(q-p))\mathcal{M}]_{2}=0$
and $p=q$. Consequently, we have $q\leq p$. ∎

######  Theorem 4.2.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra and $\mathfrak{B}$ a
$\sigma$-weakly closed subalgebra containing $\mathfrak{A}$ of $\mathcal{M}$.
Then the following assertions hold:

$(1)$ There exists a projection $E\in
Lat_{\mathcal{M}}\mathfrak{A}\cap\mathfrak{D}$ such that
$\mathfrak{B}=E\mathcal{M}+(I-E)\mathfrak{A}$.

$(2)$ $\mathfrak{B}$ is a type 1 subdiagonal algebra with respect to the
unique faithful normal expectation $\Psi$ from $\mathcal{M}$ onto the diagonal
$\mathfrak{B}\cap\mathfrak{B}^{*}$ of $\mathfrak{B}$ such that
$\varphi\circ\Psi=\varphi$.

###### Proof.

$(1)$ As in the proof of Lemma 4.1, there exists a family of column orthogonal
partial isometries $\mathcal{W}$ and a projection $E$ in $\mathcal{M}$ such
that $\mathfrak{B}=\oplus_{n\geq
1}^{col}W_{n}\mathfrak{A}\oplus^{col}E\mathcal{M}$ and
$[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}=\oplus_{n\geq
1}^{col}W_{n}H^{2}\oplus^{col}EL^{2}(\mathcal{M})$. Put
$\mathfrak{N}=\oplus_{n\geq
1}^{col}W_{n}H^{2}\subseteq[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$ be the type
1 part of $[\mathfrak{B}h_{0}^{\frac{1}{2}}]_{2}$. Let $p$ and $q$ as in Lemma
4.1. We have $q\leq p$. This means that
$\mathfrak{N}=[R(q)\mathfrak{A}]_{2}\subseteq[R(p)\mathfrak{A}]_{2}=H^{2}$.
Thus we have $W_{n}\in\mathfrak{A}$ since $\mathfrak{N}=\oplus_{n\geq
1}^{col}W_{n}H^{2}\subseteq H^{2}$. Since
$W_{n}^{*}\mathfrak{B}=W_{n}^{*}W_{n}\mathfrak{A}\subseteq\mathfrak{A}$, we
get $W_{n}^{*}\in\mathfrak{A}$. Therefore $W_{n},W_{n}^{*}\in\mathfrak{D}$ and
$E\in\mathfrak{D}\subseteq\mathfrak{A}$ from the fact that $\oplus_{n\geq
1}^{col}W_{n}W_{n}^{*}=I-E\subseteq\mathfrak{D}$. Note that $E\in
Lat_{\mathcal{M}}\mathfrak{B}\subseteq Lat_{\mathcal{M}}\mathfrak{A}$,
$E\in\mathfrak{D}$ and $\mathfrak{B}=\oplus_{n\geq
1}^{col}W_{n}\mathfrak{A}\oplus^{col}E\mathcal{M}$. We have
$(I-E)\mathfrak{B}=(I-E)(\oplus_{n\geq
1}^{col}W_{n}\mathfrak{A})=\oplus_{n\geq
1}^{col}W_{n}\mathfrak{A}\subseteq(I-E)\mathfrak{A}\subseteq(I-E)\mathfrak{B}$
and so $\oplus_{n\geq 1}^{col}W_{n}\mathfrak{A}=(I-E)\mathfrak{A}$. Thus,
$\mathfrak{B}=E\mathcal{M}+(I-E)\mathfrak{A}$.

$(2)$ By (1),
$\tilde{\mathfrak{D}}=\mathfrak{B}\cap\mathfrak{B}^{*}=E\mathcal{M}E+(I-E)\mathfrak{D}$.
We next claim that $\mathfrak{B}$ is $\sigma_{t}^{\varphi}$-invariant. In
fact, $E$ is in the center of $\mathfrak{D}$. Thus we have
$\sigma_{t}^{\varphi}(E)=E$ for all $t\in\mathbb{R}$. It follows that
$\sigma_{t}^{\varphi}(\mathfrak{B})=\mathfrak{B}$ and
$\sigma_{t}^{\varphi}(\tilde{\mathfrak{D}})=\tilde{\mathfrak{D}}$ for all
$t\in\mathbb{R}$. By [25, Chapter IX, Theorem 4.2], there exists unique
faithful normal conditional expectation $\Psi$ from $\mathcal{M}$ onto
$\tilde{\mathfrak{D}}$ such that $\varphi\circ\Psi=\varphi$. We show that
$\mathfrak{B}$ is a maximal subdiagonal algebra of $\mathcal{M}$ with respect
to $\Psi$. In fact,
$\mathfrak{B}_{0}=\\{B\in\mathfrak{B}:\Psi(B)=0\\}=E\mathcal{M}(I-E)+(I-E)\mathfrak{A}_{0}.$
An elementary calculation shows that $\mathfrak{B}_{0}$ is an ideal of
$\mathfrak{B}$. We thus have that $\Psi$ is multiplicative on $\mathfrak{B}$.
It is maximal subdiagonal by [27, Theorem 1.1].

Finally we show that $\mathfrak{B}$ is of type 1. Note that $E$ is in the
center of $\mathfrak{D}$, we have $\Phi(ET(I-E))=0$ for all $T\in\mathcal{M}$.
Moreover,
$E(\mathfrak{A}+\mathfrak{A}^{*})(I-E)=E\mathfrak{A}(I-E)\subseteq\mathfrak{A}$.
We have $E\mathcal{M}(I-E)\subseteq\mathfrak{A}_{0}$. It follows that
$\mathfrak{B}_{0}\subseteq\mathfrak{A}_{0}$. Consequently, $\cap_{n\geq
1}[h_{0}^{\frac{1}{2}}\mathfrak{B}_{0}^{n}]_{2}\subseteq\cap_{n\geq
1}[h_{0}^{\frac{1}{2}}\mathfrak{A}_{0}^{n}]_{2}=\\{0\\}$. Therefore,
$\mathfrak{B}$ is a type 1 subdiagonal algebra from [15, Proposition 2.2]. ∎

##  5\. The relative invariant subspace lattice of a type 1 subdiagonal
algebra

We recall that the invariant subspace lattice of a subalgebra
$\mathcal{A}\subseteq\mathcal{\mathcal{missing}}B(\mathcal{H})$ is the set
$Lat\mathcal{A}=\\{E\in\mathcal{B}(\mathcal{H})_{p}:(I-E)AE=0,\forall
A\in\mathcal{A}\\}$. If $\mathcal{A}\subseteq\mathcal{M}$, then
$Lat_{\mathcal{M}}\mathcal{A}=\\{E\in\mathcal{M}_{p}:(I-E)AE=0,\forall
A\in\mathcal{A}\\}=Lat\mathcal{A}\cap\mathcal{M}_{p}$ is the relative
invariant subspace lattice of $\mathcal{A}$ in $\mathcal{M}$. A subspace
lattice is a nest if it is linearly ordered by inclusion. Gilfeather and
Larson in [10] studied the relative invariant subspace lattices of certain
subalgebras in a von Neumann algebra and asked: If $\mathcal{A}$ is a
subalgebra in a factor von Neumann algebra $\mathcal{M}$ such that
$\mathcal{A}+\mathcal{A}^{*}$ is dense in $\mathcal{M}$ in some topology for
which multiplication is separately continuous, is
$Lat_{\mathcal{M}}\mathcal{A}$ a nest? Although this question was negatively
answered for the case in which $\mathcal{M}=\mathcal{B}(\mathcal{H})$ in [1],
it is still interesting to consider those subalgebras with additional
properties. For example, how is the relative invariant subspace lattice of a
maximal subdiagonal algebra? In [12], we shown that the relative invariant
subspace lattice of a finite subdiagonal algebra is commutative. As an
application of Theorem 4.2, we prove that the relative invariant subspace
lattice $Lat_{\mathcal{M}}\mathfrak{A}$ of a type 1 subdiagonal algebra
$\mathfrak{A}$ in $\mathcal{M}$ is commutative.

We note that the centralizer $\mathcal{M}^{\varphi}$ of von Neumann algebra
$\mathcal{M}$ associated with $\varphi$ is defined to be the set
$\mathcal{M}^{\varphi}=\\{A\in\mathcal{M}:\varphi(AB)=\varphi(BA),\forall
B\in\mathcal{M}\\}$(cf.[20, 25]). In fact, $\mathcal{M}^{\varphi}$ is just the
fixed point algebra of $\mathcal{M}$ with respect to
$\\{\sigma_{t}^{\varphi}\\}_{t\in\mathbb{R}}$:
$\mathcal{M}^{\varphi}=\\{A\in\mathcal{M}:\sigma_{t}^{\varphi}(A)=A,\forall
t\in\mathbb{R}\\}$. We remark that $\mathcal{M}^{\varphi}$ is a finite von
Neumann algebra.

######  Theorem 5.1.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra of $\mathcal{M}$. Then the
relative invariant subspace lattice $Lat_{\mathcal{M}}\mathfrak{A}$ of
$\mathfrak{A}$ in $\mathcal{M}$ is commutative.

###### Proof.

Take any $Q\in Lat_{\mathcal{M}}\mathfrak{A}$ and let $\mathfrak{B}$ be the
$\sigma$-weakly closed subalgebra generated by $Q$ and $\mathfrak{A}$. Then
$\mathfrak{B}$ is a $\sigma_{t}^{\varphi}$-invariant type 1 subdiagonal
algebra by Theorem 4.2. Now $\mathfrak{B}$ can be expressed as
$\mathfrak{B}=\vee\\{Q\mathfrak{A}Q\\}+\vee\\{Q\mathfrak{A}(I-Q)\\}+\vee\\{(I-Q)\mathfrak{A}(I-Q)\\}$.
Since $Q\in Lat_{\mathcal{M}}\mathfrak{A}$, $Q\mathfrak{A}^{*}(I-Q)=0$. It
follows that
$Q\mathcal{M}(I-Q)=\vee\\{Q(\mathfrak{A}+\mathfrak{A}^{*})(I-Q)\\}=\vee\\{Q\mathfrak{A}(I-Q)\\}$.
Thus we have $Q\mathcal{M}(I-Q)\subseteq\mathfrak{B}_{0}$ and
$\Psi(QA(I-Q))=0$ for all $A\in\mathcal{M}$. This implies that
$\varphi(QA(I-Q))=\varphi\circ\Psi(QA(I-Q))=0$ for all $A\in\mathcal{M}$. That
is $Q\mathcal{M}(I-Q)\subseteq\ker\varphi$. Since $\varphi$ is a state, we
also have that $(I-Q)\mathcal{M}Q\subseteq\ker\varphi$. For every
$T\in\mathcal{M}$, $\varphi(TQ)=\varphi(QTQ)+\varphi((I-Q)TQ)$ and
$\varphi(QT)=\varphi(QTQ)+\varphi(QT(I-Q))$. Then we have
$\varphi(TQ)=\varphi(QT)$. It implies that $Q\in\mathcal{M}^{\varphi}$.
Therefore, $Lat_{\mathcal{M}}\mathfrak{A}\subseteq\mathcal{M}^{\varphi}$ and
we in turn obtain that $Lat_{\mathcal{M}}\mathfrak{A}\subseteq
Lat_{\mathcal{M}^{\varphi}}(\mathfrak{A}\cap\mathcal{M}^{\varphi})$. Note that
$\mathcal{M}^{\varphi}$ is a finite von Neumann algebra and
$\mathfrak{A}\cap\mathcal{M}^{\varphi}$ is a finite subdiagonal algebra by
[17, Corollary 2.5]. Thus
$Lat_{\mathcal{M}^{\varphi}}(\mathfrak{A}\cap\mathcal{M}^{\varphi})$ is
commutative by [12, Theorem 2.1]. It is trivial that
$Lat_{\mathcal{M}}\mathfrak{A}$ is also commutative. ∎

######  Corollary 5.2.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra of $\mathcal{M}$. Then for
any $Q\in Lat_{\mathcal{M}}\mathfrak{A}$,
$Q\mathcal{M}(I-Q)\subseteq\mathfrak{A}_{0}$.

###### Proof.

Note that $Q\in\mathcal{M}^{\varphi}$ and
$Q\mathcal{M}(I-Q))\subseteq\ker\varphi$. Thus
$\mbox{tr}(Th_{0}Q)=\mbox{tr}(QTh_{0})=\varphi(QT)=\varphi(TQ)=\mbox{tr}(TQh_{0})$
for all $T\in\mathcal{M}$. That is, $\mbox{tr}(T(h_{0}Q-Qh_{0}))=0$ for all
$T\in\mathcal{M}$. Thus $Qh_{0}=h_{0}Q$. Furthermore,
$\mbox{tr}(Ah_{0}QT(I-Q))=\mbox{tr}(AQh_{0}T(I-Q))=\mbox{tr}((I-Q)AQh_{0}T)=0$
for all $A\in\mathfrak{A}$. It follows that
$QT(I-Q)\in(H^{1})^{\bot}=\mathfrak{A}_{0}$ for all $T\in\mathcal{M}$. ∎

Similar to [12, Theorem 2.4], we next show that
$Lat_{\mathcal{M}}\mathfrak{A}$ is a complete lattice generated by a nest in
$Lat_{\mathcal{M}}\mathfrak{A}$ together with the lattice of projections in
the center of $\mathcal{M}$.

######  Theorem 5.3.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra of $\mathcal{M}$. Then the
relative invariant subspace lattice $Lat_{\mathcal{M}}\mathfrak{A}$ of
$\mathfrak{A}$ in $\mathcal{M}$ is the complete lattice generated by a nest in
$Lat_{\mathcal{M}}\mathfrak{A}$ together with the lattice of projections in
the center of $\mathcal{M}$.

###### Proof.

It is known that $Lat_{\mathcal{M}}\mathfrak{A}$ is commutative by Theorem
5.1. Take any $E,F\in Lat_{\mathcal{M}}\mathfrak{A}$. Then
$E\mathcal{M}(I-E)\subseteq\mathfrak{A}_{0}$ by Corollary 5.2. Thus we get
$(I-F)E\mathcal{M}(I-E)F\subseteq(I-F)\mathfrak{A}F=\\{0\\}$ and we in turn
have $C_{E(I-F)}C_{(I-E)F}=0$, where $C_{P}$ denotes the central carrier of a
projection $P$.

Note that $Lat_{\mathcal{M}}\mathfrak{A}$ contains the lattice of projections
in the center of $\mathcal{M}$. We next show that
$Lat_{\mathcal{M}}\mathfrak{A}$ satisfies the condition of [11, Theorem 4.2].
In fact, for $E,F\in Lat_{\mathcal{M}}\mathfrak{A}$, we have
$C_{E(I-F)}C_{(I-E)F}=0$, then $E(I-F)\leq C_{E(I-F)}$ while $(I-E)F\leq
I-C_{E(I-F)}$. It follows that the reflexive operator algebra

$AlgLat_{\mathcal{M}}\mathfrak{A}=\\{T\in\mathcal{B}(\mathcal{H}):(I-P)TP=0,\forall
P\in Lat_{\mathcal{M}}\mathfrak{A}\\}$

with commutative subspaces lattice $Lat_{\mathcal{M}}\mathfrak{A}$ is a nest
subalgebra of a von Neumann algebra $\mathfrak{R}$ with a nest $\mathcal{N}$
in $Lat_{\mathcal{M}}\mathfrak{A}$ by [11, Theorem 4.2]. Thus
$AlgLat_{\mathcal{M}}\mathfrak{A}=Alg\mathcal{N}\cap\mathfrak{R}$. Note that
$\mathfrak{R}\supseteq AlgLat_{\mathcal{M}}\mathfrak{A}\supseteq
AlgLat_{\mathcal{M}}\mathfrak{A}\cap\mathcal{M}\supseteq\mathfrak{A}$. Then
$\mathfrak{R}\supseteq[\mathfrak{A}+\mathfrak{A}^{*}]_{\infty}=\mathcal{M}$.
Therefore,
$AlgLat_{\mathcal{M}}\mathfrak{A}\cap\mathcal{M}=Alg\mathcal{N}\cap\mathfrak{R}\cap\mathcal{M}=Alg\mathcal{N}\cap\mathcal{M}$.
Similar to [12, Theorem 2.4], we have the desired result. ∎

######  Corollary 5.4.

Let $\mathfrak{A}$ be a type 1 subdiagonal algebra of $\mathcal{M}$. If
$\mathcal{M}$ is a factor, then the relative invariant subspace lattice
$Lat_{\mathcal{M}}\mathfrak{A}$ of $\mathfrak{A}$ is a nest.

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